chapter
stringlengths 1.97k
1.53M
| path
stringlengths 47
241
|
|---|---|
Learning Objectives
• To understand periodic trends in atomic radii.
• To predict relative ionic sizes within an isoelectronic series.
Although some people fall into the trap of visualizing atoms and ions as small, hard spheres similar to miniature table-tennis balls or marbles, the quantum mechanical model tells us that their shapes and boundaries are much less definite than those images suggest. As a result, atoms and ions cannot be said to have exact sizes; however, some atoms are larger or smaller than others, and this influences their chemistry. In this section, we discuss how atomic and ion “sizes” are defined and obtained.
Atomic Radii
Recall that the probability of finding an electron in the various available orbitals falls off slowly as the distance from the nucleus increases. This point is illustrated in Figure $1$ which shows a plot of total electron density for all occupied orbitals for three noble gases as a function of their distance from the nucleus. Electron density diminishes gradually with increasing distance, which makes it impossible to draw a sharp line marking the boundary of an atom.
Figure $1$ also shows that there are distinct peaks in the total electron density at particular distances and that these peaks occur at different distances from the nucleus for each element. Each peak in a given plot corresponds to the electron density in a given principal shell. Because helium has only one filled shell (n = 1), it shows only a single peak. In contrast, neon, with filled n = 1 and 2 principal shells, has two peaks. Argon, with filled n = 1, 2, and 3 principal shells, has three peaks. The peak for the filled n = 1 shell occurs at successively shorter distances for neon (Z = 10) and argon (Z = 18) because, with a greater number of protons, their nuclei are more positively charged than that of helium. Because the 1s2 shell is closest to the nucleus, its electrons are very poorly shielded by electrons in filled shells with larger values of n. Consequently, the two electrons in the n = 1 shell experience nearly the full nuclear charge, resulting in a strong electrostatic interaction between the electrons and the nucleus. The energy of the n = 1 shell also decreases tremendously (the filled 1s orbital becomes more stable) as the nuclear charge increases. For similar reasons, the filled n = 2 shell in argon is located closer to the nucleus and has a lower energy than the n = 2 shell in neon.
Figure $1$ illustrates the difficulty of measuring the dimensions of an individual atom. Because distances between the nuclei in pairs of covalently bonded atoms can be measured quite precisely, however, chemists use these distances as a basis for describing the approximate sizes of atoms. For example, the internuclear distance in the diatomic Cl2 molecule is known to be 198 pm. We assign half of this distance to each chlorine atom, giving chlorine a covalent atomic radius ($r_{cov}$), which is half the distance between the nuclei of two like atoms joined by a covalent bond in the same molecule, of 99 pm or 0.99 Å (Figure $\PageIndex{2a}$). Atomic radii are often measured in angstroms (Å), a non-SI unit: 1 Å = 1 × 10−10 m = 100 pm.
In a similar approach, we can use the lengths of carbon–carbon single bonds in organic compounds, which are remarkably uniform at 154 pm, to assign a value of 77 pm as the covalent atomic radius for carbon. If these values do indeed reflect the actual sizes of the atoms, then we should be able to predict the lengths of covalent bonds formed between different elements by adding them. For example, we would predict a carbon–chlorine distance of 77 pm + 99 pm = 176 pm for a C–Cl bond, which is very close to the average value observed in many organochlorine compounds. A similar approach for measuring the size of ions is discussed later in this section.
Covalent atomic radii can be determined for most of the nonmetals, but how do chemists obtain atomic radii for elements that do not form covalent bonds? For these elements, a variety of other methods have been developed. With a metal, for example, the metallic atomic radius ($r_{met}$) is defined as half the distance between the nuclei of two adjacent metal atoms in the solid (Figure $\PageIndex{2b}$). For elements such as the noble gases, most of which form no stable compounds, we can use what is called the van der Waals atomic radius ($r_{vdW}$), which is half the internuclear distance between two nonbonded atoms in the solid (Figure $\PageIndex{2c}$). This is somewhat difficult for helium which does not form a solid at any temperature. An atom such as chlorine has both a covalent radius (the distance between the two atoms in a $\ce{Cl2}$ molecule) and a van der Waals radius (the distance between two Cl atoms in different molecules in, for example, $\ce{Cl2(s)}$ at low temperatures). These radii are generally not the same (Figure $\PageIndex{2d}$).
Because it is impossible to measure the sizes of both metallic and nonmetallic elements using any one method, chemists have developed a self-consistent way of calculating atomic radii using the quantum mechanical functions. Although the radii values obtained by such calculations are not identical to any of the experimentally measured sets of values, they do provide a way to compare the intrinsic sizes of all the elements and clearly show that atomic size varies in a periodic fashion (Figure $3$).
In the periodic table, atomic radii decrease from left to right across a row and increase from top to bottom down a column. Because of these two trends, the largest atoms are found in the lower left corner of the periodic table, and the smallest are found in the upper right corner (Figure $4$).
Trends in atomic size result from differences in the effective nuclear charges ($Z_{eff}$) experienced by electrons in the outermost orbitals of the elements. For all elements except H, the effective nuclear charge is always less than the actual nuclear charge because of shielding effects. The greater the effective nuclear charge, the more strongly the outermost electrons are attracted to the nucleus and the smaller the atomic radius.
Atomic radii decrease from left to right across a row and increase from top to bottom down a column.
The atoms in the second row of the periodic table (Li through Ne) illustrate the effect of electron shielding. All have a filled 1s2 inner shell, but as we go from left to right across the row, the nuclear charge increases from +3 to +10. Although electrons are being added to the 2s and 2p orbitals, electrons in the same principal shell are not very effective at shielding one another from the nuclear charge. Thus the single 2s electron in lithium experiences an effective nuclear charge of approximately +1 because the electrons in the filled 1s2 shell effectively neutralize two of the three positive charges in the nucleus. (More detailed calculations give a value of Zeff = +1.26 for Li.) In contrast, the two 2s electrons in beryllium do not shield each other very well, although the filled 1s2 shell effectively neutralizes two of the four positive charges in the nucleus. This means that the effective nuclear charge experienced by the 2s electrons in beryllium is between +1 and +2 (the calculated value is +1.66). Consequently, beryllium is significantly smaller than lithium. Similarly, as we proceed across the row, the increasing nuclear charge is not effectively neutralized by the electrons being added to the 2s and 2p orbitals. The result is a steady increase in the effective nuclear charge and a steady decrease in atomic size (Figure $5$).
The increase in atomic size going down a column is also due to electron shielding, but the situation is more complex because the principal quantum number n is not constant. As we saw in Chapter 2, the size of the orbitals increases as n increases, provided the nuclear charge remains the same. In group 1, for example, the size of the atoms increases substantially going down the column. It may at first seem reasonable to attribute this effect to the successive addition of electrons to ns orbitals with increasing values of n. However, it is important to remember that the radius of an orbital depends dramatically on the nuclear charge. As we go down the column of the group 1 elements, the principal quantum number n increases from 2 to 6, but the nuclear charge increases from +3 to +55!
As a consequence the radii of the lower electron orbitals in Cesium are much smaller than those in lithium and the electrons in those orbitals experience a much larger force of attraction to the nucleus. That force depends on the effective nuclear charge experienced by the the inner electrons. If the outermost electrons in cesium experienced the full nuclear charge of +55, a cesium atom would be very small indeed. In fact, the effective nuclear charge felt by the outermost electrons in cesium is much less than expected (6 rather than 55). This means that cesium, with a 6s1 valence electron configuration, is much larger than lithium, with a 2s1 valence electron configuration. The effective nuclear charge changes relatively little for electrons in the outermost, or valence shell, from lithium to cesium because electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge. Even though cesium has a nuclear charge of +55, it has 54 electrons in its filled 1s22s22p63s23p64s23d104p65s24d105p6 shells, abbreviated as [Xe]5s24d105p6, which effectively neutralize most of the 55 positive charges in the nucleus. The same dynamic is responsible for the steady increase in size observed as we go down the other columns of the periodic table. Irregularities can usually be explained by variations in effective nuclear charge.
Not all Electrons shield equally
Electrons in the same principal shell are not very effective at shielding one another from the nuclear charge, whereas electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge.
Example $1$
On the basis of their positions in the periodic table, arrange these elements in order of increasing atomic radius: aluminum, carbon, and silicon.
Given: three elements
Asked for: arrange in order of increasing atomic radius
Strategy:
1. Identify the location of the elements in the periodic table. Determine the relative sizes of elements located in the same column from their principal quantum number n. Then determine the order of elements in the same row from their effective nuclear charges. If the elements are not in the same column or row, use pairwise comparisons.
2. List the elements in order of increasing atomic radius.
Solution:
A These elements are not all in the same column or row, so we must use pairwise comparisons. Carbon and silicon are both in group 14 with carbon lying above, so carbon is smaller than silicon (C < Si). Aluminum and silicon are both in the third row with aluminum lying to the left, so silicon is smaller than aluminum (Si < Al) because its effective nuclear charge is greater.
B Combining the two inequalities gives the overall order: C < Si < Al.
Exercise $1$
On the basis of their positions in the periodic table, arrange these elements in order of increasing size: oxygen, phosphorus, potassium, and sulfur.
Answer
O < S < P < K
Atomic Radius: Atomic Radius, YouTube(opens in new window) [youtu.be]
Ionic Radii and Isoelectronic Series
An ion is formed when either one or more electrons are removed from a neutral atom to form a positive ion (cation) or when additional electrons attach themselves to neutral atoms to form a negative one (anion). The designations cation or anion come from the early experiments with electricity which found that positively charged particles were attracted to the negative pole of a battery, the cathode, while negatively charged ones were attracted to the positive pole, the anode.
Ionic compounds consist of regular repeating arrays of alternating positively charged cations and negatively charges anions. Although it is not possible to measure an ionic radius directly for the same reason it is not possible to directly measure an atom’s radius, it is possible to measure the distance between the nuclei of a cation and an adjacent anion in an ionic compound to determine the ionic radius (the radius of a cation or anion) of one or both. As illustrated in Figure $6$, the internuclear distance corresponds to the sum of the radii of the cation and anion. A variety of methods have been developed to divide the experimentally measured distance proportionally between the smaller cation and larger anion. These methods produce sets of ionic radii that are internally consistent from one ionic compound to another, although each method gives slightly different values. For example, the radius of the Na+ ion is essentially the same in NaCl and Na2S, as long as the same method is used to measure it. Thus despite minor differences due to methodology, certain trends can be observed.
A comparison of ionic radii with atomic radii (Figure $7$) shows that a cation, having lost an electron, is always smaller than its parent neutral atom, and an anion, having gained an electron, is always larger than the parent neutral atom. When one or more electrons is removed from a neutral atom, two things happen: (1) repulsions between electrons in the same principal shell decrease because fewer electrons are present, and (2) the effective nuclear charge felt by the remaining electrons increases because there are fewer electrons to shield one another from the nucleus. Consequently, the size of the region of space occupied by electrons decreases and the ion shrinks (compare Li at 167 pm with Li+ at 76 pm). If different numbers of electrons can be removed to produce ions with different charges, the ion with the greatest positive charge is the smallest (compare Fe2+ at 78 pm with Fe3+ at 64.5 pm). Conversely, adding one or more electrons to a neutral atom causes electron–electron repulsions to increase and the effective nuclear charge to decrease, so the size of the probability region increases and the ion expands (compare F at 42 pm with F at 133 pm).
Cations are always smaller than the neutral atom and anions are always larger.
Because most elements form either a cation or an anion but not both, there are few opportunities to compare the sizes of a cation and an anion derived from the same neutral atom. A few compounds of sodium, however, contain the Na ion, allowing comparison of its size with that of the far more familiar Na+ ion, which is found in many compounds. The radius of sodium in each of its three known oxidation states is given in Table $1$. All three species have a nuclear charge of +11, but they contain 10 (Na+), 11 (Na0), and 12 (Na) electrons. The Na+ ion is significantly smaller than the neutral Na atom because the 3s1 electron has been removed to give a closed shell with n = 2. The Na ion is larger than the parent Na atom because the additional electron produces a 3s2 valence electron configuration, while the nuclear charge remains the same.
Table $1$: Experimentally Measured Values for the Radius of Sodium in Its Three Known Oxidation States
Na+ Na0 Na
Electron Configuration 1s22s22p6 1s22s22p63s1 1s22s22p63s2
Radius (pm) 102 154* 202
*The metallic radius measured for Na(s). †Source: M. J. Wagner and J. L. Dye, “Alkalides, Electrides, and Expanded Metals,” Annual Review of Materials Science 23 (1993): 225–253.
Ionic radii follow the same vertical trend as atomic radii; that is, for ions with the same charge, the ionic radius increases going down a column. The reason is the same as for atomic radii: shielding by filled inner shells produces little change in the effective nuclear charge felt by the outermost electrons. Again, principal shells with larger values of n lie at successively greater distances from the nucleus.
Because elements in different columns tend to form ions with different charges, it is not possible to compare ions of the same charge across a row of the periodic table. Instead, elements that are next to each other tend to form ions with the same number of electrons but with different overall charges because of their different atomic numbers. Such a set of species is known as an isoelectronic series. For example, the isoelectronic series of species with the neon closed-shell configuration (1s22s22p6) is shown in Table $3$.
The sizes of the ions in this series decrease smoothly from N3− to Al3+. All six of the ions contain 10 electrons in the 1s, 2s, and 2p orbitals, but the nuclear charge varies from +7 (N) to +13 (Al). As the positive charge of the nucleus increases while the number of electrons remains the same, there is a greater electrostatic attraction between the electrons and the nucleus, which causes a decrease in radius. Consequently, the ion with the greatest nuclear charge (Al3+) is the smallest, and the ion with the smallest nuclear charge (N3−) is the largest. The neon atom in this isoelectronic series is not listed in Table $3$, because neon forms no covalent or ionic compounds and hence its radius is difficult to measure.
Ion Radius (pm) Atomic Number
Table $3$: Radius of Ions with the Neon Closed-Shell Electron Configuration. Source: R. D. Shannon, “Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides,” Acta Crystallographica 32, no. 5 (1976): 751–767.
N3− 146 7
O2− 140 8
F 133 9
Na+ 98 11
Mg2+ 79 12
Al3+ 57 13
Example $2$
Based on their positions in the periodic table, arrange these ions in order of increasing radius: Cl, K+, S2−, and Se2.
Given: four ions
Asked for: order by increasing radius
Strategy:
1. Determine which ions form an isoelectronic series. Of those ions, predict their relative sizes based on their nuclear charges. For ions that do not form an isoelectronic series, locate their positions in the periodic table.
2. Determine the relative sizes of the ions based on their principal quantum numbers n and their locations within a row.
Solution:
A We see that S and Cl are at the right of the third row, while K and Se are at the far left and right ends of the fourth row, respectively. K+, Cl, and S2− form an isoelectronic series with the [Ar] closed-shell electron configuration; that is, all three ions contain 18 electrons but have different nuclear charges. Because K+ has the greatest nuclear charge (Z = 19), its radius is smallest, and S2− with Z = 16 has the largest radius. Because selenium is directly below sulfur, we expect the Se2 ion to be even larger than S2−.
B The order must therefore be K+ < Cl < S2− < Se2.
Exercise $2$
Based on their positions in the periodic table, arrange these ions in order of increasing size: Br, Ca2+, Rb+, and Sr2+.
Answer
Ca2+ < Sr2+ < Rb+ < Br
Summary
Ionic radii share the same vertical trend as atomic radii, but the horizontal trends differ due to differences in ionic charges. A variety of methods have been established to measure the size of a single atom or ion. The covalent atomic radius (rcov) is half the internuclear distance in a molecule with two identical atoms bonded to each other, whereas the metallic atomic radius (rmet) is defined as half the distance between the nuclei of two adjacent atoms in a metallic element. The van der Waals radius (rvdW) of an element is half the internuclear distance between two nonbonded atoms in a solid. Atomic radii decrease from left to right across a row because of the increase in effective nuclear charge due to poor electron screening by other electrons in the same principal shell. Moreover, atomic radii increase from top to bottom down a column because the effective nuclear charge remains relatively constant as the principal quantum number increases. The ionic radii of cations and anions are always smaller or larger, respectively, than the parent atom due to changes in electron–electron repulsions, and the trends in ionic radius parallel those in atomic size. A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an isoelectronic series, shows a clear correlation between increasing nuclear charge and decreasing size.
Contributors and Attributions
Learning Objectives
• To master the concept of electron affinity as a measure of the energy required to add an electron to an atom or ion.
• To recognize the inverse relationship of ionization energies and electron affinities
The electron affinity ($EA$) of an element $E$ is defined as the energy change that occurs when an electron is added to a gaseous atom or ion:
$E_{(g)}+e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA \label{7.5.1}$
Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). This sign convention is consistent with a negative value corresponded to the energy change for an exothermic process, which is one in which heat is released (Figure $1$).
The chlorine atom has the most negative electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element:
$\ce{ Cl(g) + e^- \rightarrow Cl^- (g)} \;\;\; EA=-346\; kJ/mol \label{7.5.2}$
In contrast, beryllium does not form a stable anion, so its effective electron affinity is
$\ce{ Be(g) + e^- \rightarrow Be^- (g)} \;\;\; EA \ge 0 \label{7.5.3}$
Nitrogen is unique in that it has an electron affinity of approximately zero. Adding an electron neither releases nor requires a significant amount of energy:
$\ce{ N(g) + e^- \rightarrow N^- (g)} \;\;\; EA \approx 0 \label{7.5.4}$
Generally, electron affinities become more negative across a row of the periodic table.
In general, electron affinities of the main-group elements become less negative as we proceed down a column. This is because as n increases, the extra electrons enter orbitals that are increasingly far from the nucleus.
Atoms with the largest radii, which have the lowest ionization energies (affinity for their own valence electrons), also have the lowest affinity for an added electron. There are, however, two major exceptions to this trend:
1. The electron affinities of elements B through F in the second row of the periodic table are less negative than those of the elements immediately below them in the third row. Apparently, the increased electron–electron repulsions experienced by electrons confined to the relatively small 2p orbitals overcome the increased electron–nucleus attraction at short nuclear distances. Fluorine, therefore, has a lower affinity for an added electron than does chlorine. Consequently, the elements of the third row (n = 3) have the most negative electron affinities. Farther down a column, the attraction for an added electron decreases because the electron is entering an orbital more distant from the nucleus. Electron–electron repulsions also decrease because the valence electrons occupy a greater volume of space. These effects tend to cancel one another, so the changes in electron affinity within a family are much smaller than the changes in ionization energy.
2. The electron affinities of the alkaline earth metals become more negative from Be to Ba. The energy separation between the filled ns2 and the empty np subshells decreases with increasing n, so that formation of an anion from the heavier elements becomes energetically more favorable.
The equations for second and higher electron affinities are analogous to those for second and higher ionization energies:
$E_{(g)} + e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA_1 \label{7.5.5}$
$E^-_{(g)} + e^- \rightarrow E^{2-}_{(g)} \;\;\; \text{energy change=}EA_2 \label{7.5.6}$
As we have seen, the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons. For example, the first electron affinity of oxygen is −141 kJ/mol, but the second electron affinity is +744 kJ/mol:
$O_{(g)} + e^- \rightarrow O^-_{(g)} \;\;\; EA_1=-141 \;kJ/mol \label{7.5.7}$
$O^-_{(g)} + e^- \rightarrow O^{2-}_{(g)} \;\;\; EA_2=+744 \;kJ/mol \label{7.5.8}$
Thus the formation of a gaseous oxide ($O^{2−}$) ion is energetically quite unfavorable (estimated by adding both steps):
$O_{(g)} + 2e^- \rightarrow O^{2-}_{(g)} \;\;\; EA=+603 \;kJ/mol \label{7.5.9}$
Similarly, the formation of all common dianions (such as $S^{2−}$) or trianions (such as $P^{3−}$) is energetically unfavorable in the gas phase.
While first electron affinities can be negative, positive, or zero, second electron affinities are always positive.
Electron Affinity: Electron Affinity, YouTube(opens in new window) [youtu.be] (opens in new window)
If energy is required to form both positively charged cations and monatomic polyanions, why do ionic compounds such as $MgO$, $Na_2S$, and $Na_3P$ form at all? The key factor in the formation of stable ionic compounds is the favorable electrostatic interactions between the cations and the anions in the crystalline salt.
Example $1$: Contrasting Electron Affinities of Sb, Se, and Te
Based on their positions in the periodic table, which of Sb, Se, or Te would you predict to have the most negative electron affinity?
Given: three elements
Asked for: element with most negative electron affinity
Strategy:
1. Locate the elements in the periodic table. Use the trends in electron affinities going down a column for elements in the same group. Similarly, use the trends in electron affinities from left to right for elements in the same row.
2. Place the elements in order, listing the element with the most negative electron affinity first.
Solution:
A We know that electron affinities become less negative going down a column (except for the anomalously low electron affinities of the elements of the second row), so we can predict that the electron affinity of Se is more negative than that of Te. We also know that electron affinities become more negative from left to right across a row, and that the group 15 elements tend to have values that are less negative than expected. Because Sb is located to the left of Te and belongs to group 15, we predict that the electron affinity of Te is more negative than that of Sb. The overall order is Se < Te < Sb, so Se has the most negative electron affinity among the three elements.
Exercise $1$: Contrasting Electron Affinities of Rb, Sr, and Xe
Based on their positions in the periodic table, which of Rb, Sr, or Xe would you predict to most likely form a gaseous anion?
Answer
Rb
Summary
The electron affinity (EA) of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/08%3A_Electron_Configuration_and_Chemical_Periodicity/8.03%3A_Trends_in_Three_Atomic_Properties.txt |
Learning Objectives
• To understand the basic properties separating Metals from Nonmetals and Metalloids
An element is the simplest form of matter that cannot be split into simpler substances or built from simpler substances by any ordinary chemical or physical method. There are 118 elements known to us, out of which 92 are naturally occurring, while the rest have been prepared artificially. Elements are further classified into metals, non-metals, and metalloids based on their properties, which are correlated with their placement in the periodic table.
Metallic Elements Nonmetallic elements
Table $1$: Characteristic properties of metallic and non-metallic elements:
Distinguishing luster (shine) Non-lustrous, various colors
Malleable and ductile (flexible) as solids Brittle, hard or soft
Conduct heat and electricity Poor conductors
Metallic oxides are basic, ionic Nonmetallic oxides are acidic, covalent
Form cations in aqueous solution Form anions, oxyanions in aqueous solution
Metals
With the exception of hydrogen, all elements that form positive ions by losing electrons during chemical reactions are called metals. Thus metals are electropositive elements with relatively low ionization energies. They are characterized by bright luster, hardness, ability to resonate sound and are excellent conductors of heat and electricity. Metals are solids under normal conditions except for Mercury.
Physical Properties of Metals
Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include:
• State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).
• Luster: Metals have the quality of reflecting light from their surface and can be polished e.g., gold, silver and copper.
• Malleability: Metals have the ability to withstand hammering and can be made into thin sheets known as foils. For example, a sugar cube sized chunk of gold can be pounded into a thin sheet that will cover a football field.
• Ductility: Metals can be drawn into wires. For example, 100 g of silver can be drawn into a thin wire about 200 meters long.
• Hardness: All metals are hard except sodium and potassium, which are soft and can be cut with a knife.
• Valency: Metals typically have 1 to 3 electrons in the outermost shell of their atoms.
• Conduction: Metals are good conductors because they have free electrons. Silver and copper are the two best conductors of heat and electricity. Lead is the poorest conductor of heat. Bismuth, mercury and iron are also poor conductors
• Density: Metals have high density and are very heavy. Iridium and osmium have the highest densities whereas lithium has the lowest density.
• Melting and Boiling Points: Metals have high melting and boiling points. Tungsten has the highest melting and boiling points whereas mercury has the lowest. Sodium and potassium also have low melting points.
Chemical Properties of Metals
Metals are electropositive elements that generally form basic or amphoteric oxides with oxygen. Other chemical properties include:
• Electropositive Character: Metals tend to have low ionization energies, and typically lose electrons (i.e. are oxidized) when they undergo chemical reactions They normally do not accept electrons. For example:
• Alkali metals are always 1+ (lose the electron in s subshell)
• Alkaline earth metals are always 2+ (lose both electrons in s subshell)
• Transition metal ions do not follow an obvious pattern, 2+ is common (lose both electrons in s subshell), and 1+ and 3+ are also observed
$\ce{Na^0 \rightarrow Na^+ + e^{-}} \label{1.1}$
$\ce{Mg^0 \rightarrow Mg^{2+} + 2e^{-}} \label{1.2}$
$\ce{Al^0 \rightarrow Al^{3+} + 3e^{-}} \label{1.3}$
Compounds of metals with non-metals tend to be ionic in nature. Most metal oxides are basic oxides and dissolve in water to form metal hydroxides:
$\ce{Na2O(s) + H2O(l) \rightarrow 2NaOH(aq)}\label{1.4}$
$\ce{CaO(s) + H2O(l) \rightarrow Ca(OH)2(aq)} \label{1.5}$
Metal oxides exhibit their basic chemical nature by reacting with acids to form metal salts and water:
$\ce{MgO(s) + HCl(aq) \rightarrow MgCl2(aq) + H2O(l)} \label{1.6}$
$\ce{NiO(s) + H2SO4(aq) \rightarrow NiSO4(aq) + H2O(l)} \label{1.7}$
Example $1$
What is the chemical formula for aluminum oxide?
Solution
Al has a 3+ charge, the oxide ion is $O^{2-}$, thus $Al_2O_3$.
Example $2$
Would you expect it to be solid, liquid or gas at room temperature?
Solutions
Oxides of metals are characteristically solid at room temperature
Example $3$
Write the balanced chemical equation for the reaction of aluminum oxide with nitric acid:
Metal oxide + acid -> salt + water
$\ce{Al2O3(s) + 6HNO3(aq) \rightarrow 2Al(NO3)3(aq) + 3H2O(l)} \nonumber$
Nonmetals
Elements that tend to gain electrons to form anions during chemical reactions are called non-metals. These are electronegative elements with high ionization energies. They are non-lustrous, brittle and poor conductors of heat and electricity (except graphite). Non-metals can be gases, liquids or solids.
Physical Properties of Nonmetals
• Physical State: Most of the non-metals exist in two of the three states of matter at room temperature: gases (oxygen) and solids (carbon). Only bromine exists as a liquid at room temperature.
• Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets.
• Conduction: They are poor conductors of heat and electricity.
• Luster: These have no metallic luster and do not reflect light.
• Melting and Boiling Points: The melting points of non-metals are generally lower than metals, but are highly variable.
• Seven non-metals exist under standard conditions as diatomic molecules: $\ce{H2(g)}$, $\ce{N2(g)}$, $\ce{O2(g)}$, $\ce{F2(g)}$, $\ce{Cl2(g)}$, $\ce{Br2(l)}$, $\ce{I2(s)}$.
Chemical Properties of Nonmetals
Non-metals have a tendency to gain or share electrons with other atoms. They are electronegative in character. Nonmetals, when reacting with metals, tend to gain electrons (typically attaining noble gas electron configuration) and become anions:
$\ce{3Br2(l) + 2Al(s) \rightarrow 2AlBr3(s)} \nonumber$
Compounds composed entirely of nonmetals are covalent substances. They generally form acidic or neutral oxides with oxygen that that dissolve in water to form acids:
$\ce{CO2(g) + H2O(l)} \rightarrow \underset{\text{carbonic acid}}{\ce{H2CO3(aq)}} \nonumber$
As you may know, carbonated water is slightly acidic (carbonic acid).
Nonmetal oxides can combine with bases to form salts.
$\ce{CO2(g) + 2NaOH(aq) \rightarrow Na2CO3(aq) + H2O(l)} \nonumber$
Metalloids
Metalloids have properties intermediate between the metals and nonmetals. Metalloids are useful in the semiconductor industry. Metalloids are all solid at room temperature. They can form alloys with other metals. Some metalloids, such as silicon and germanium, can act as electrical conductors under the right conditions, thus they are called semiconductors. Silicon for example appears lustrous, but is not malleable nor ductile (it is brittle - a characteristic of some nonmetals). It is a much poorer conductor of heat and electricity than the metals. The physical properties of metalloids tend to be metallic, but their chemical properties tend to be non-metallic. The oxidation number of an element in this group can range from +5 to -2, depending on the group in which it is located.
Table $2$: Elements categorized into metals, non-metals and metalloids.
Metals Non-metals Metalloids
Gold Oxygen Silicon
Silver Carbon Boron
Copper Hydrogen Arsenic
Iron Nitrogen Antimony
Mercury Sulfur Germanium
Zinc Phosphorus
Metallic character is strongest for the elements in the leftmost part of the periodic table, and tends to decrease as we move to the right in any period (nonmetallic character increases with increasing electronegativity and ionization energy values). Within any group of elements (columns), the metallic character increases from top to bottom (the electronegativity and ionization energy values generally decrease as we move down a group). This general trend is not necessarily observed with the transition metals.
Contributors and Attributions
The elements within the same group of the periodic table tend to exhibit similar physical and chemical properties. Four major factors affect reactivity of metals: nuclear charge, atomic radius, shielding effect and sublevel arrangement (of electrons). Metal reactivity relates to ability to lose electrons (oxidize), form basic hydroxides, form ionic compounds with non-metals. In general, the bigger the atom, the greater the ability to lose electrons. The greater the shielding, the greater the ability to lose electrons. Therefore, metallic character increases going down the table, and decreases going across -- so the most active metal is towards the left and down.
Group 1: The Alkali Metals
The word "alkali" is derived from an Arabic word meaning "ashes". Many sodium and potassium compounds were isolated from wood ashes ($\ce{Na2CO3}$ and $\ce{K2CO3}$ are still occasionally referred to as "soda ash" and "potash"). In the alkali group, as we go down the group we have elements Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs) and Francium (Fr). Several physical properties of these elements are compared in Table $1$. These elements have all only one electron in their outermost shells. All the elements show metallic properties and have valence +1, hence they give up electron easily.
Table $1$: General Properties of Group I Metals
Element Electronic Configuration Melting Point (°C) Density (g/cm3) Atomic Radius Ionization Energy (kJ/mol)
Lithium $[He]2s^1$ 181 0.53 1.52 520
Sodium $[Ne]3s^1$ 98 0.97 1.86 496
Potassium $[Ar]4s^1$ 63 0.86 2.27 419
Rubidium $[Kr]5s^1$ 39 1.53 2.47 403
Cesium $[Xe]6s^1$ 28 1.88 2.65 376
As we move down the group (from Li to Fr), the following trends are observed (Table $1$):
• All have a single electron in an 's' valence orbital
• The melting point decreases
• The density increases
• The atomic radius increases
• The ionization energy decreases (first ionization energy)
The alkali metals have the lowest $I_1$ values of the elements
This represents the relative ease with which the lone electron in the outer 's' orbital can be removed.
The alkali metals are very reactive, readily losing 1 electron to form an ion with a 1+ charge:
$M \rightarrow M^+ + e- \nonumber$
Due to this reactivity, the alkali metals are found in nature only as compounds. The alkali metals combine directly with most nonmetals:
• React with hydrogen to form solid hydrides
$2M_{(s)} + H_{2(g)} \rightarrow 2MH(s) \nonumber$
(Note: hydrogen is present in the metal hydride as the hydride H- ion)
• React with sulfur to form solid sulfides
$2M_{(s)} + S_{(s)} \rightarrow M_2S_{(s)} \nonumber$
React with chlorine to form solid chlorides
$2M_{(s)} + Cl_{2(g)} \rightarrow 2MCl_{(s)} \nonumber$
Alkali metals react with water to produce hydrogen gas and alkali metal hydroxides; this is a very exothermic reaction (Figure $1$).
$2M_{(s)} + 2H_2O_{(l)} \rightarrow 2MOH_{(aq)} + H_{2(g)} \nonumber$
The reaction between alkali metals and oxygen is more complex:
• A common reaction is to form metal oxides which contain the O2- ion
$4Li_{(s)} + O_{2 (g)} \rightarrow \underbrace{2Li_2O_{(s)}}_{\text{lithium oxide}} \nonumber$
Other alkali metals can form metal peroxides (contains O22- ion)
$2Na(s) + O_{2 (g)} \rightarrow \underbrace{Na_2O_{2(s)}}_{\text{sodium peroxide}} \nonumber$
K, Rb and Cs can also form superoxides (O2- ion)
$K(s) + O_{2 (g)} \rightarrow \underbrace{KO_{2(s)}}_{\text{potassium superoxide}} \nonumber$
Colors via Absorption
The color of a chemical is produced when a valence electron in an atom is excited from one energy level to another by visible radiation. In this case, the particular frequency of light that excites the electron is absorbed. Thus, the remaining light that you see is white light devoid of one or more wavelengths (thus appearing colored). Alkali metals, having lost their outermost electrons, have no electrons that can be excited by visible radiation. Alkali metal salts and their aqueous solution are colorless unless they contain a colored anion.
Colors via Emission
When alkali metals are placed in a flame the ions are reduced (gain an electron) in the lower part of the flame. The electron is excited (jumps to a higher orbital) by the high temperature of the flame. When the excited electron falls back down to a lower orbital a photon is released. The transition of the valence electron of sodium from the 3p down to the 3s subshell results in release of a photon with a wavelength of 589 nm (yellow)
Flame colors:
• Lithium: crimson red
• Sodium: yellow
• Potassium: lilac
Group 2: The Alkaline Earth Metals
Compared with the alkali metals, the alkaline earth metals are typically harder, more dense, melt at a higher temperature. The first ionization energies ($I_1$) of the alkaline earth metals are not as low as the alkali metals. The alkaline earth metals are therefore less reactive than the alkali metals (Be and Mg are the least reactive of the alkaline earth metals). Several physical properties of these elements are compared in Table $2$.
Table $2$: General Properties of Group 2 Metals
Element Electronic Configuration Melting Point (°C) Density (g/cm3) Atomic Radius Ionization Energy (kJ/mol)
Beryllium $[He]2s^2$ 1278 1.85 1.52 899
Magnesium $[Ne]3s^2$ 649 1.74 1.60 738
Calcium $[Ar]4s^2$ 839 1.54 1.97 590
Strontium $[Kr]5s^2$ 769 2.54 2.15 549
Barium $[Xe]6s^2$ 725 3.51 2.17 503
Calcium, and elements below it, react readily with water at room temperature:
$Ca_{(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(aq)} + H_{2(g)} \nonumber$
The tendency of the alkaline earths to lose their two valence electrons is demonstrated in the reactivity of Mg towards chlorine gas and oxygen:
$Mg_{(s)} + Cl_{2(g)} \rightarrow MgCl_{2(s)} \nonumber$
$2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)} \nonumber$
The 2+ ions of the alkaline earth metals have a noble gas like electron configuration and are thus form colorless or white compounds (unless the anion is itself colored). Flame colors:
• Calcium: brick red
• Strontium: crimson red
• Barium: green
Contributors and Attributions
Learning Objectives
• To gain a descriptive understanding of the chemical properties of Hydrogen, the group 16, 17 and 18 elements.
Non-metallic character is the ability to be reduced (be an oxidizing agent), form acidic hydroxides, form covalent compounds with non-metals. These characteristics increase with a larger nuclear charge and smaller radius, with no increase in shielding. The most active non-metal would be the one farthest up and to the right -- not including the noble gases (non-reactive.)
Hydrogen
Hydrogen has a 1s1 electron configuration and is placed above the alkali metal group. Hydrogen is a non-metal, which occurs as a gas (H2) under normal conditions.
• Its ionization energy is considerably higher (due to lack of shielding, and thus higher $Z_{eff}$) than the rest of the Group 1 metals and is more like the nonmetals
• Hydrogen generally reacts with other nonmetals to form molecular compounds (typically highly exothermic)
• Hydrogen reacts with active metals to form metal hydrides which contain the H- hydride ion:
$2Na_{(s)} + H_{2(g)} \rightarrow 2NaH_{(s)} \label{7.8.1}$
• Hydrogen can also lose an electron to yield the aqueous $H^+_{(aq)}$ hydronium ion.
Group 16: The Oxygen Family
As we proceed down group 16 the elements become more metallic in nature:
• Oxygen is a gas, the rest are solids
• Oxygen, sulfur and selenium are nonmetals
• Tellurium is a metalloid with some metal properties
• Polonium is a metal
Oxygen can be found in two molecular forms, O2 and O3 (ozone). These two forms of oxygen are called allotropes (different forms of the same element in the same state)
$3O_{2(g)} \rightarrow 2O_{3(g)}\;\;\; \Delta H = 284.6\; kJ / mol \label{7.8.2}$
the reaction is endothermic, thus ozone is less stable that O2
Oxygen has a great tendency to attract electrons from other elements (i.e. to "oxidize" them)
• Oxygen in combination with metals is almost always present as the O2- ion (which has noble gas electronic configuration and is particularly stable)
• Two other oxygen anions are observed: peroxide (O22-) and superoxide (O2-)
Sulfur
Sulfur also exists in several allotropic forms, the most common stable allotrope is the yellow solid S8 (an 8 member ring of sulfur atoms). Like oxygen, sulfur has a tendency to gain electrons from other elements, and to form sulfides (which contain the S2- ion). This is particular true for the active metals:
$16Na_{(s)} + S_{8(s)} \rightarrow 8Na_2S_{(s)}\label{7.8.3}$
Note: most sulfur in nature is present as a metal-sulfur compound. Sulfur chemistry is more complex than that of oxygen.
Group 17: The Halogens
"Halogen" is derived from the Greek meaning "salt formers"
• Astatine is radioactive and rare, and some of its properties are unknown
• All the halogens are nonmetals
• Each element consists of diatomic molecules under standard conditions
Colors of diatomic halogens: (not flame colors)
• Fluorine: pale yellow
• Chlorine: yellow green
• Bromine: reddish brown
• Iodine: violet vapor
The halogens have some of the most negative electron affinities (i.e. large exothermic process in gaining an electron from another element)
$X_2 + 2e^- \rightarrow 2X^-\label{7.8.4}$
• Fluorine and chlorine are the most reactive halogens (highest electron affinities). Fluorine will remove electrons from almost any substance (including several of the noble gases from Group 18).
Note
The chemistry of the halogens is dominated by their tendency to gain electrons from other elements (forming a halide ion)
In 1992, 22.3 billion pounds of chlorine was produced. Both chlorine and sodium can be produced by electrolysis of molten sodium chloride (table salt). The electricity is used to strip electrons from chloride ions and transfer them to sodium ions to produce chlorine gas and solid sodium metal
Chlorine reacts slowly with water to produce hydrochloric acid and hypochlorous acid:
$Cl_{2(g)} + H_2O_{(l)} \rightarrow HCl_{(aq)} + HOCl_{(aq)}\label{7.8}.5$
Hypochlorous acid is a disinfectant, thus chlorine is a useful addition to swimming pool water
The halogens react with most metals to form ionic halides:
$Cl_{2(g)} + 2Na_{(s)} \rightarrow 2NaCl_{(s)}\label{7.8.6}$
Group 18: The Noble Gases
• Nonmetals
• Gases at room temperature
• monoatomic
• completely filled 's' and 'p' subshells
• large first ionization energy, but this decreases somewhat as we move down the group
Rn is highly radioactive and some of its properties are unknown
They are exceptionally unreactive. It was reasoned that if any of these were reactive, they would most likely be Rn, Xe or Kr where the first ionization energies were lower.
Note
In order to react, they would have to be combined with an element which had a high tendency to remove electrons from other atoms. Such as fluorine.
Compounds of noble gases to date:
$XeF_2$ $XeF_4$ $XeF_6$
only one compound with Kr has been made
$KrF_2$
No compounds observed with He, Ne, or Ar; they are truly inert gases.
Contributors and Attributions
Learning Objectives
• To gain a descriptive understanding of the chemical properties of Hydrogen, the group 16, 17 and 18 elements.
Non-metallic character is the ability to be reduced (be an oxidizing agent), form acidic hydroxides, form covalent compounds with non-metals. These characteristics increase with a larger nuclear charge and smaller radius, with no increase in shielding. The most active non-metal would be the one farthest up and to the right -- not including the noble gases (non-reactive.)
Hydrogen
Hydrogen has a 1s1 electron configuration and is placed above the alkali metal group. Hydrogen is a non-metal, which occurs as a gas (H2) under normal conditions.
• Its ionization energy is considerably higher (due to lack of shielding, and thus higher $Z_{eff}$) than the rest of the Group 1 metals and is more like the nonmetals
• Hydrogen generally reacts with other nonmetals to form molecular compounds (typically highly exothermic)
• Hydrogen reacts with active metals to form metal hydrides which contain the H- hydride ion:
$2Na_{(s)} + H_{2(g)} \rightarrow 2NaH_{(s)} \label{7.8.1}$
• Hydrogen can also lose an electron to yield the aqueous $H^+_{(aq)}$ hydronium ion.
Group 16: The Oxygen Family
As we proceed down group 16 the elements become more metallic in nature:
• Oxygen is a gas, the rest are solids
• Oxygen, sulfur and selenium are nonmetals
• Tellurium is a metalloid with some metal properties
• Polonium is a metal
Oxygen can be found in two molecular forms, O2 and O3 (ozone). These two forms of oxygen are called allotropes (different forms of the same element in the same state)
$3O_{2(g)} \rightarrow 2O_{3(g)}\;\;\; \Delta H = 284.6\; kJ / mol \label{7.8.2}$
the reaction is endothermic, thus ozone is less stable that O2
Oxygen has a great tendency to attract electrons from other elements (i.e. to "oxidize" them)
• Oxygen in combination with metals is almost always present as the O2- ion (which has noble gas electronic configuration and is particularly stable)
• Two other oxygen anions are observed: peroxide (O22-) and superoxide (O2-)
Sulfur
Sulfur also exists in several allotropic forms, the most common stable allotrope is the yellow solid S8 (an 8 member ring of sulfur atoms). Like oxygen, sulfur has a tendency to gain electrons from other elements, and to form sulfides (which contain the S2- ion). This is particular true for the active metals:
$16Na_{(s)} + S_{8(s)} \rightarrow 8Na_2S_{(s)}\label{7.8.3}$
Note: most sulfur in nature is present as a metal-sulfur compound. Sulfur chemistry is more complex than that of oxygen.
Group 17: The Halogens
"Halogen" is derived from the Greek meaning "salt formers"
• Astatine is radioactive and rare, and some of its properties are unknown
• All the halogens are nonmetals
• Each element consists of diatomic molecules under standard conditions
Colors of diatomic halogens: (not flame colors)
• Fluorine: pale yellow
• Chlorine: yellow green
• Bromine: reddish brown
• Iodine: violet vapor
The halogens have some of the most negative electron affinities (i.e. large exothermic process in gaining an electron from another element)
$X_2 + 2e^- \rightarrow 2X^-\label{7.8.4}$
• Fluorine and chlorine are the most reactive halogens (highest electron affinities). Fluorine will remove electrons from almost any substance (including several of the noble gases from Group 18).
Note
The chemistry of the halogens is dominated by their tendency to gain electrons from other elements (forming a halide ion)
In 1992, 22.3 billion pounds of chlorine was produced. Both chlorine and sodium can be produced by electrolysis of molten sodium chloride (table salt). The electricity is used to strip electrons from chloride ions and transfer them to sodium ions to produce chlorine gas and solid sodium metal
Chlorine reacts slowly with water to produce hydrochloric acid and hypochlorous acid:
$Cl_{2(g)} + H_2O_{(l)} \rightarrow HCl_{(aq)} + HOCl_{(aq)}\label{7.8}.5$
Hypochlorous acid is a disinfectant, thus chlorine is a useful addition to swimming pool water
The halogens react with most metals to form ionic halides:
$Cl_{2(g)} + 2Na_{(s)} \rightarrow 2NaCl_{(s)}\label{7.8.6}$
Group 18: The Noble Gases
• Nonmetals
• Gases at room temperature
• monoatomic
• completely filled 's' and 'p' subshells
• large first ionization energy, but this decreases somewhat as we move down the group
Rn is highly radioactive and some of its properties are unknown
They are exceptionally unreactive. It was reasoned that if any of these were reactive, they would most likely be Rn, Xe or Kr where the first ionization energies were lower.
Note
In order to react, they would have to be combined with an element which had a high tendency to remove electrons from other atoms. Such as fluorine.
Compounds of noble gases to date:
$XeF_2$ $XeF_4$ $XeF_6$
only one compound with Kr has been made
$KrF_2$
No compounds observed with He, Ne, or Ar; they are truly inert gases. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/08%3A_Electron_Configuration_and_Chemical_Periodicity/8.04%3A_Atomic_Properties_and_Chemical_Reactivity.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
7.1: Development of the Periodic Table
Conceptual Problems
1. Johannes Dobereiner is credited with developing the concept of chemical triads. Which of the group 15 elements would you expect to compose a triad? Would you expect B, Al, and Ga to act as a triad? Justify your answers.
2. Despite the fact that Dobereiner, Newlands, Meyer, and Mendeleev all contributed to the development of the modern periodic table, Mendeleev is credited with its origin. Why was Mendeleev’s periodic table accepted so rapidly?
3. How did Moseley’s contribution to the development of the periodic table explain the location of the noble gases?
4. The eka- naming scheme devised by Mendeleev was used to describe undiscovered elements.
1. Use this naming method to predict the atomic number of eka-mercury, eka-astatine, eka-thallium, and eka-hafnium.
2. Using the eka-prefix, identify the elements with these atomic numbers: 79, 40, 51, 117, and 121.
Numerical Problem
1. Based on the data given, complete the table.
Species Molar Mass (g/mol) Density (g/cm3) Molar Volume (cm3/mol)
A 40.078 25.85
B 39.09 0.856
C 32.065 16.35
D 1.823 16.98
E 26.98 9.992
F 22.98 0.968
Plot molar volume versus molar mass for these substances. According to Meyer, which would be considered metals and which would be considered nonmetals?
Numerical Answer
1. Species Molar Mass (g/mol) Density (g/cm3) Molar Volume (cm3/mol)
A 40.078 1.550 25.85
B 39.09 0.856 45.67
C 32.065 1.961 16.35
D 30.95 1.823 16.98
E 26.98 2.700 9.992
F 22.98 0.968 23.7
2. Meyer found that the alkali metals had the highest molar volumes, and that molar volumes decreased steadily with increasing atomic mass, then leveled off, and finally rose again. The elements located on the rising portion of a plot of molar volume versus molar mass were typically nonmetals. If we look at the plot of the data in the table, we can immediately identify those elements with the largest molar volumes (A, B, F) as metals located on the left side of the periodic table. The element with the smallest molar volume (E) is aluminum. The plot shows that the subsequent elements (C, D) have molar volumes that are larger than that of E, but smaller than those of A and B. Thus, C and D are most likely to be nonmetals (which is the case: C = sulfur, D = phosphorus).
7.2: Effective Nuclear Charge
Conceptual Problems
1. What happens to the energy of a given orbital as the nuclear charge Z of a species increases? In a multielectron atom and for a given nuclear charge, the Zeff experienced by an electron depends on its value of l. Why?
2. The electron density of a particular atom is divided into two general regions. Name these two regions and describe what each represents.
3. As the principal quantum number increases, the energy difference between successive energy levels decreases. Why? What would happen to the electron configurations of the transition metals if this decrease did not occur?
4. Describe the relationship between electron shielding and Zeff on the outermost electrons of an atom. Predict how chemical reactivity is affected by a decreased effective nuclear charge.
5. If a given atom or ion has a single electron in each of the following subshells, which electron is easier to remove?
• 2s, 3s
• 3p, 4d
• 2p, 1s
• 3d, 4s
7.3: Sizes of Atoms and Ions
Conceptual Problems
1. The electrons of the 1s shell have a stronger electrostatic attraction to the nucleus than electrons in the 2s shell. Give two reasons for this.
2. Predict whether Na or Cl has the more stable 1s2 shell and explain your rationale.
3. Arrange K, F, Ba, Pb, B, and I in order of decreasing atomic radius.
4. Arrange Ag, Pt, Mg, C, Cu, and Si in order of increasing atomic radius.
5. Using the periodic table, arrange Li, Ga, Ba, Cl, and Ni in order of increasing atomic radius.
6. Element M is a metal that forms compounds of the type MX2, MX3, and MX4, where X is a halogen. What is the expected trend in the ionic radius of M in these compounds? Arrange these compounds in order of decreasing ionic radius of M.
7. The atomic radii of Na and Cl are 190 and 79 pm, respectively, but the distance between sodium and chlorine in NaCl is 282 pm. Explain this discrepancy.
8. Are shielding effects on the atomic radius more pronounced across a row or down a group? Why?
9. What two factors influence the size of an ion relative to the size of its parent atom? Would you expect the ionic radius of S2− to be the same in both MgS and Na2S? Why or why not?
10. Arrange Br, Al3+, Sr2+, F, O2−, and I in order of increasing ionic radius.
11. Arrange P3−, N3−, Cl, In3+, and S2− in order of decreasing ionic radius.
12. How is an isoelectronic series different from a series of ions with the same charge? Do the cations in magnesium, strontium, and potassium sulfate form an isoelectronic series? Why or why not?
13. What isoelectronic series arises from fluorine, nitrogen, magnesium, and carbon? Arrange the ions in this series by
1. increasing nuclear charge.
2. increasing size.
14. What would be the charge and electron configuration of an ion formed from calcium that is isoelectronic with
1. a chloride ion?
2. Ar+?
Conceptual Answers
1. The 1s shell is closer to the nucleus and therefore experiences a greater electrostatic attraction. In addition, the electrons in the 2s subshell are shielded by the filled 1s2 shell, which further decreases the electrostatic attraction to the nucleus.
2.
3. Ba > K > Pb > I > B > F
4.
5.
6.
7. The sum of the calculated atomic radii of sodium and chlorine atoms is 253 pm. The sodium cation is significantly smaller than a neutral sodium atom (102 versus 154 pm), due to the loss of the single electron in the 3s orbital. Conversely, the chloride ion is much larger than a neutral chlorine atom (181 versus 99 pm), because the added electron results in greatly increased electron–electron repulsions within the filled n = 3 principal shell. Thus, transferring an electron from sodium to chlorine decreases the radius of sodium by about 50%, but causes the radius of chlorine to almost double. The net effect is that the distance between a sodium ion and a chloride ion in NaCl is greater than the sum of the atomic radii of the neutral atoms.
Numerical Problems
1. Plot the ionic charge versus ionic radius using the following data for Mo: Mo3+, 69 pm; Mo4+, 65 pm; and Mo5+, 61 pm. Then use this plot to predict the ionic radius of Mo6+. Is the observed trend consistent with the general trends discussed in the chapter? Why or why not?
2. Internuclear distances for selected ionic compounds are given in the following table.
1. If the ionic radius of Li+ is 76 pm, what is the ionic radius of each of the anions?
LiF LiCl LiBr LiI
Distance (pm) 209 257 272 296
• What is the ionic radius of Na+?
NaF NaCl NaBr NaI
Distance (pm) 235 282 298 322
• Arrange the gaseous species Mg2+, P3−, Br, S2−, F, and N3− in order of increasing radius and justify your decisions.
7.4: Ionization Energy
Conceptual Problems
1. Identify each statement as either true or false and explain your reasoning.
1. Ionization energies increase with atomic radius.
2. Ionization energies decrease down a group.
3. Ionization energies increase with an increase in the magnitude of the electron affinity.
4. Ionization energies decrease diagonally across the periodic table from He to Cs.
5. Ionization energies depend on electron configuration.
6. Ionization energies decrease across a row.
2. Based on electronic configurations, explain why the first ionization energies of the group 16 elements are lower than those of the group 15 elements, which is contrary to the general trend.
3. The first through third ionization energies do not vary greatly across the lanthanides. Why? How does the effective nuclear charge experienced by the ns electron change when going from left to right (with increasing atomic number) in this series?
4. Most of the first row transition metals can form at least two stable cations, for example iron(II) and iron(III). In contrast, scandium and zinc each form only a single cation, the Sc3+ and Zn2+ ions, respectively. Use the electron configuration of these elements to provide an explanation.
5. Of the elements Nd, Al, and Ar, which will readily form(s) +3 ions? Why?
6. Orbital energies can reverse when an element is ionized. Of the ions B3+, Ga3+, Pr3+, Cr3+, and As3+, in which would you expect this reversal to occur? Explain your reasoning.
7. The periodic trends in electron affinities are not as regular as periodic trends in ionization energies, even though the processes are essentially the converse of one another. Why are there so many more exceptions to the trends in electron affinities compared to ionization energies?
8. Elements lying on a lower right to upper left diagonal line cannot be arranged in order of increasing electronegativity according to where they occur in the periodic table. Why?
9. Why do ionic compounds form, if energy is required to form gaseous cations?
10. Why is Pauling’s definition of electronegativity considered to be somewhat limited?
11. Based on their positions in the periodic table, arrange Sb, O, P, Mo, K, and H in order of increasing electronegativity.
12. Based on their positions in the periodic table, arrange V, F, B, In, Na, and S in order of decreasing electronegativity.
Conceptual Answers
5. Both Al and Nd will form a cation with a +3 charge. Aluminum is in Group 13, and loss of all three valence electrons will produce the Al3+ ion with a noble gas configuration. Neodymium is a lanthanide, and all of the lanthanides tend to form +3 ions because the ionization potentials do not vary greatly across the row, and a +3 charge can be achieved with many oxidants.
11. K < Mo ≈ Sb < P ≈ H < O
Numerical Problems
1. The following table gives values of the first and third ionization energies for selected elements:
Number of Electrons Element I1 (E → E+ + e, kJ/mol) Element I3 (E2+ → E3+ + e, kJ/mol)
11 Na 495.9 Al 2744.8
12 Mg 737.8 Si 3231.6
13 Al 577.6 P 2914.1
14 Si 786.6 S 3357
15 P 1011.9 Cl 3822
16 S 999.6 Ar 3931
17 Cl 1251.2 K 4419.6
18 Ar 1520.6 Ca 4912.4
Plot the ionization energies versus the number of electrons. Explain why the slopes of the I1 and I3 plots are different, even though the species in each row of the table have the same electron configurations.
2. Would you expect the third ionization energy of iron, corresponding to the removal of an electron from a gaseous Fe2+ ion, to be larger or smaller than the fourth ionization energy, corresponding to the removal of an electron from a gaseous Fe3+ ion? Why? How would these ionization energies compare to the first ionization energy of Ca?
3. Which would you expect to have the highest first ionization energy: Mg, Al, or Si? Which would you expect to have the highest third ionization energy. Why?
4. Use the values of the first ionization energies given in Figure 7.11 to construct plots of first ionization energy versus atomic number for (a) boron through oxygen in the second period; and (b) oxygen through tellurium in group 16. Which plot shows more variation? Explain the reason for the variation in first ionization energies for this group of elements.
5. Arrange Ga, In, and Zn in order of increasing first ionization energies. Would the order be the same for second and third ionization energies? Explain your reasoning.
6. Arrange each set of elements in order of increasing magnitude of electron affinity.
1. Pb, Bi, and Te
2. Na, K, and Rb
3. P, C, and Ge
7. Arrange each set of elements in order of decreasing magnitude of electron affinity.
1. As, Bi, and N
2. O, F, and Ar
3. Cs, Ba, and Rb
8. Of the species F, O, Al3+, and Li+, which has the highest electron affinity? Explain your reasoning.
9. Of the species O, N2−, Hg2+, and H+, which has the highest electron affinity? Which has the lowest electron affinity? Justify your answers.
10. The Mulliken electronegativity of element A is 542 kJ/mol. If the electron affinity of A is −72 kJ/mol, what is the first ionization energy of element A? Use the data in the following table as a guideline to decide if A is a metal, a nonmetal, or a semimetal. If 1 g of A contains 4.85 × 1021 molecules, what is the identity of element A?
Na Al Si S Cl
EA (kJ/mol) −59.6 −41.8 −134.1 −200.4 −348.6
I (kJ/mol) 495.8 577.5 786.5 999.6 1251.2
11. Based on their valence electron configurations, classify the following elements as either electrical insulators, electrical conductors, or substances with intermediate conductivity: S, Ba, Fe, Al, Te, Be, O, C, P, Sc, W, Na, B, and Rb.
12. Using the data in Problem 10, what conclusions can you draw with regard to the relationship between electronegativity and electrical properties? Estimate the approximate electronegativity of a pure element that is very dense, lustrous, and malleable.
13. Of the elements Al, Mg, O2, Ti, I2, and H2, which, if any, would you expect to be a good reductant? Explain your reasoning.
14. Of the elements Zn, B, Li, Se, Co, and Br2, which if any, would you expect to be a good oxidant? Explain your reasoning.
15. Determine whether each species is a good oxidant, a good reductant, or neither.
1. Ba
2. Mo
3. Al
4. Ni
5. O2
6. Xe
16. Determine whether each species is a good oxidant, a good reductant, or neither.
1. Ir
2. Cs
3. Be
4. B
5. N
6. Po
7. Ne
17. Of the species I2, O, Zn, Sn2+, and K+, choose which you would expect to be a good oxidant. Then justify your answer.
18. Based on the valence electron configuration of the noble gases, would you expect them to have positive or negative electron affinities? What does this imply about their most likely oxidation states? their reactivity?
Numerical Answers
1. The general features of both plots are roughly the same, with a small peak at 12 electrons and an essentially level region from 15–16 electrons. The slope of the I3 plot is about twice as large as the slope of the I1 plot, however, because the I3 values correspond to removing an electron from an ion with a +2 charge rather than a neutral atom. The greater charge increases the effect of the steady rise in effective nuclear charge across the row.
2.
3. Electron configurations: Mg, 1s22s22p63s2; Al, 1s22s22p63s23p1; Si, 1s22s22p63s23p2; First ionization energies increase across the row due to a steady increase in effective nuclear charge; thus, Si has the highest first ionization energy. The third ionization energy corresponds to removal of a 3s electron for Al and Si, but for Mg it involves removing a 2p electron from a filled inner shell; consequently, the third ionization energy of Mg is the highest.
4.
5.
6.
1. Bi > As > N
2. F > O >> Ar
3. Rb > Cs > Ba
7.
8. Hg2+ > H+ > O > N2−; Hg2+ has the highest positive charge plus a relatively low energy vacant set of orbitals (the 6p subshell) to accommodate an added electron, giving it the greatest electron affinity; N2− has a greater negative charge than O, so electron–electron repulsions will cause its electron affinity to be even lower (more negative) than that of O.
9.
10. insulators: S, O, C (diamond), P; conductors: Ba, Fe, Al, C (graphite), Be, Sc, W, Na, Rb; Te and B are semimetals and semiconductors.
11.
12. Mg, Al, Ti, and H2
13.
1. reductant
2. neither
3. reductant
4. reductant
5. oxidant
6. neither
14.
15. I2 is the best oxidant, with a moderately strong tendency to accept an electron to form the I ion, with a closed shell electron configuration. O would probably also be an oxidant, with a tendency to add an electron to form salts containing the oxide ion, O2−. Zn and Sn2+ are all reductants, while K+ has no tendency to act as an oxidant or a reductant.
7.5: Electron Affinities
see above question to tease out
10: The Shapes of Molecules
Thumbnail: Structural formula of sulfur tetrafluoride, showing dimensions. (Public Domain; Benjah-bmm27). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/08%3A_Electron_Configuration_and_Chemical_Periodicity/8.E%3A_Electron_Configuration_and_Chemical_Periodicity_%28Exercises%29.txt |
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
00: Tools of Quantitative Chemistry
1
1.01: Matter: Elements and Atoms Compounds and Molecules
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
01: Introduction to Chemistry
John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. This theory explains several concepts that are relevant in the observable world: the composition of a pure gold necklace, what makes the pure gold necklace different than a pure silver necklace, and what occurs when pure gold is mixed with pure copper. Before discussing the atomic theory, this article explains the theories that Dalton used as a basis for his theory: the law of conservation of mass and the law of constant composition.
Law of Conservation of Mass: (1766-1844)
The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction; in other words, mass is conserved. The law of conservation of mass was formulated by Antoine Lavoisier (1743-1794) as a result of his combustion experiment, in which he observed that the mass of his original substance—a glass vessel, tin, and air—was equal to the mass of the produced substance—the glass vessel, “tin calx”, and the remaining air.
Historically, this was a difficult concept for scientists to grasp. If this law was true, then how could a large piece of wood be reduced to a small pile of ashes? The wood clearly has a greater mass than the ashes. From this observation scientists concluded that mass had been lost. However, the illustration below shows that the burning of word does follow the law of conservation of mass. Scientists did not take into account the gases that play a critical role in this reaction.
Law of Constant Composition
Joseph Proust (1754-1826) formulated the law of constant composition (also called the law of definite proportions). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. The illustration below depicts this law; 31 grams of H2O and 8 grams of H2O are made up of the same percent of hydrogen and oxygen.
Dalton's Atomic Theory
1. Each chemical element is composed of extremely small particles that are indivisible and cannot be seen by the naked eye, called atoms. Atoms can neither be created nor destroyed. Pictured below is a helium atom. The purple and red dots represent the neutrons and protons in the nucleus. The black area around the nucleus represent the electron cloud. The following sections discuss this further.
2. All atoms of an element are alike in mass and other properties, but the atoms of one element differ from all other elements. For example, gold and silver have different atomic masses and different properties.
Gold Silver
Atomic Mass: 196.97 Atomic Mass: 107.87
Figure 4 (Gold): Courtesy of Chris Ralph that released this image into the public domain. Figure 5 (silver): Courtesy of resourcescommittee.house.gov/.../photogallery/
3. For each compound, different elements combine in a simple numerical ratio. The illustration below describes this rule. The second equation for the reaction is incorrect because half of an atom does not exist.
Atomic theory can be used to answers the questions presented above. A pure gold necklace is made up of atoms. A pure gold necklace and a pure silver necklace are different because they have different atoms. Pure gold mixed with pure copper forms rose gold. The gold and copper atoms combine in a simple numerical ratio.
Dalton's theory has not proven to be correct under all circumstances. The first rule was proven incorrect when scientists divided atoms in a process called nuclear fission. The second rule was proven incorrect by the discovery that not all atoms of the same element have the same mass; there are different isotopes. However, these failures do not justify discarding the atomic theory. It correctly explains the law of conservation of mass: if atoms of an element are indestructible, then the same atom must be present after a chemical reaction as before and, and the mass must constant. Dalton’s atomic theory also explains the law of constant composition: if all the atoms of an element are alike in mass and if atoms unite in fixed numerical ratios, the percent composition of a compound must have a unique value without regards to the sample analyzed. The atomic theory led to the creation of the law of multiple proportions.
Law of Multiple Proportions
The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers. The illustration of the third rule of the atomic theory correctly depicts this law.
Discovering Electrons
The first cathode-ray tube (CRT) was invented by Michael Faraday (1791-1867). Cathode rays are a type of radiation emitted by the negative terminal, the cathode, and were discovered by passing electricity through nearly-evacuated glass tubes. The radiation crosses the evacuated tube to the positive terminal, the anode. Cathode rays produced by the CRT are invisible and can only be detected by light emitted by the materials that they strike, called phosphors, painted at the end of the CRT to reveal the path of the cathode rays. These phosphors showed that cathode rays travel in straight lines and have properties independent of the cathode material (whether it is gold, silver, etc.). Another significant property of cathode rays is that they are deflected by magnetic and electric fields in a manner that is identical to negatively charged material. Due to these observations, J.J. Thompson (1856-1940) concluded that cathode rays are negatively charged particles that are located in all atoms. It was George Stoney who first gave the term electrons to the cathode rays. The below figures depict the way that the cathode ray is effected by magnetics. The cathode ray is always attracted by the positive magnet and deflected by the negative magnets.
The Plum Pudding Model
After Thompson discovered the electron, he proposed the plum pudding model of an atom, which states that the electrons float in positively-charged material. This model was named after the plum-pudding dessert.
Discovery of the Proton
In 1909, Ernest Rutherford (1871-1937) performed a series of experiments studying the inner structure of atoms using alpha particles. Rutherford knew that alpha particles are significantly more massive than electrons and positively charged. Using the plum-pudding model for reference, Rutherford predicted that particles in an alpha beam would largely pass through matter unaffected, with a small number of particles slightly deflected. The particles would only be deflected if they happened to come into contact with electrons. According to the plum pudding model, this occurrence would be very unlikely. In order to test his hypothesis, Rutherford shot a beam of alpha particles at a thin piece of gold foil. Around the gold foil Rutherford placed sheets of zinc sulfide. These sheets produced a flash of light when struck by an alpha particle. However, this experiment produced results that contradicted Rutherford's hypothesis. Rutherford observed that the majority of the alpha particles went through the foil; however, some particles were slightly deflected, a small number were greatly deflected, and another small number were thrown back in nearly the direction from which they had come. Figure 10 shows Rutherford's prediction based off of the plum-pudding model (pink) and the observed large deflections of the alpha particles (gold).
To account for these observations, Rutherford devised a model called the nuclear atom. In this model, the positive charge is held in an extremely small area called the nucleus, located in the middle of the atom. Outside of the nucleus the atom is largely composed of empty space. This model states that there were positive particles within the nucleus, but failed to define what these particles are. Rutherford discovered these particles in 1919, when he conducted an experiment that scattered alpha particles against nitrogen atoms. When the alpha particles and nitrogen atoms collided protons were released.
The Discovery of the Neutron
In 1933, James Chadwick (1891-1974) discovered a new type of radiation that consisted of neutral particles. It was discovered that these neutral atoms come from the nucleus of the atom. This last discovery completed the atomic model.
Example Problems
1. Basic concept check: When 32.0 grams (g) of methane are burned in 128.0 g of oxygen, 88.0 g of carbon dioxide and 72.0 g of water are produced. Which law is this an example of? (a) Law of Definite Proportions (b) Law of Conservation of Mass or (c) Law of Multiple Proportions.
The answer is (b) Law of Conservation of Mass. The number of grams of reactants (32.0 g of methane and 128.0 g of oxygen = 160.0 g total) is equal to the number of grams of product (88.0 g of carbon dioxide and 72.0 g of water = 160.0 g total).
2. Law of Conservation of Mass: 8.00 grams (g) of methane are burned in 32.00 g of oxygen. The reaction produces 22.00 g of carbon dioxide and an unmeasured mass of water. What mass of water is produced?
The answer is 18.00 g of water. Because the only products are water and carbon dioxide, their total mass must equal the total mass of the reactants, methane and oxygen. 8.00 g of methane + 32.00 g of oxygen = 40.00 total g of reactants. Because the total mass of the reactants equals the total mass of the products, the total mass of the products is also 40.00 g. Thus, 40.00 total g of products = 22.00 g carbon dioxide + unknown mass water. 40.00 total g of products - 22.00 g carbon dioxide = 18.00 g water.
3. Law of Definite Proportions: Two experiments using sodium and chlorine are performed. In the first experiment, 4.36 grams (g) sodium are reacted with 32.24 g of chlorine, using up all the sodium. 11.08 g of sodium chloride was produced in the first experiment. In the second experiment, 4.20 g of chlorine reacted with 20.00 g of sodium, using up all the chlorine. 6.92 g of of sodium chloride was produced in the second experiment. Show that these results are consistent with the law of constant composition.
To solve, determine the percent of sodium in each sample of sodium chloride. There is 4.36 g sodium for every 11.08 g of sodium chloride in the first experiment. The amount of sodium in the sodium chloride for the second experiment must be found. This is found by subtracted the known amount of reacted chlorine (4.20 g) from the amount of sodium chloride (6.92 g). 6.92 g sodium chloride - 4.20 g chlorine = 2.72 g sodium.
Thus, the percent of sodium in each sample is represented below:
% Na = (4.36 g Na)/(11.08 g NaCl) x 100% = 39.4% Na % Na = (2.72 g Na)/(6.92 g NaCl) x 100% = 39.3%
The slight difference in compositions is due to significant figures: each percent has an uncertainty of .01% in either direction. The two samples of sodium chloride have the same composition.
4. Law of Conservation of Mass: 36.0 grams (g) of wood are burned in oxygen. The products of this reaction weigh 74.4 g. (a) What mass of oxygen is needed in this reaction? (b) What mass of oxygen is needed to burn 8.00 lb of wood? 1 lb = 453.59237 g.
1. The answer is 38.4 g of oxygen. The total mass of the products is 74.4 g. Thus, the total mass of the reactants must equal 74.4 g as well. Thus, 74.4 g products - 36.0 g wood reactant = 38.4 g oxygen reactant.
2. The answer is 8.53 lb of oxygen. From, (a) that it takes 38.4 g of oxygen to burn 18.0 g of wood. First, convert both of these values to pounds (alternatively, the 8.00 lb can be converted to grams).
36.0 g wood x (1lb)/(453.59237 g) = .0793664144 lb wood
38.4 g oxygen x (1 lb)/(453.59237 g) = .0846575087 lb oxygen
Now two ratios equal to each other can be set up to determine the unknown mass of oxygen.
(0.0793664144 lb wood)/(.0846575087 lb oxygen) = (8.00 lb wood)/(unknown mass oxygen)
Solving reveals that it requires 8.53 lb of oxygen to burn 8.00 lb of wood.
5. Law of Definite Proportions: A sample of methane contains only carbon and hydrogen, with 3.00 grams (g) of carbon for every 1.00 g of hydrogen. How much hydrogen should be present in a different, 50.0 g same of methane?
The answer is 12.5 g of hydrogen. If there are 3.00 g of carbon present for every 1.00 g of hydrogen, we can assume the smallest whole number combination of these elements in that ratio to be 4.00 g of methane:
50.0 g methane x (1.00 g hydrogen)/(4.00 g methane) = 12.5 g of hydrogen.
• Description of Atomic Theory - http://en.Wikipedia.org/wiki/Atomic_theory
• Description of Atomic Theory - http://www.visionlearning.com/librar...wer.php?mid=50
• Cathode Ray Demonstration - www.youtube.com/watch?v=XU8nMKkzbT8
• Theoretical Demonstration of Conservation of Mass - www.youtube.com/watch?v=dExpJAECSL8
• Actual Demonstration of Conservation of Mass - www.youtube.com/watch?v=J5hM1...eature=related
• The alpha scattering experiment-http://www.youtube.com/watch?v=5pZj0u_XMbc | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/00%3A_Tools_of_Quantitative_Chemistry/1B.3%3A__Scientific_Notation_and_Significant_Figures_%286%29.txt |
The difference between a physical reaction and a chemical reaction is composition. In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition. Although we call them physical "reactions," no reaction is actually occurring. In order for a reaction to take place, there must be a change in the elemental composition of the substance in question. Thus, we shall simply refer to physical "reactions" as physical changes from now on.
Introduction
Physical changes are limited to changes that result in a difference in display without changing the composition. Some common changes (but not limited to) are:
• Texture
• Color
• Temperature
• Shape
• Change of State (Boiling Point and Melting Point are significant factors in determining this change.)
Physical properties include many other aspects of a substance. The following are (but not limited to) physical properties.
• Luster
• Malleability
• Ability to be drawn into a thin wire
• Density
• Viscosity
• Solubility
• Mass
• Volume
Any change in these physical properties is referred to as a physical change. For further information, please refer to Properties of Matter.
Chemical changes, on the other hand, are quite different. A chemical change occurs when the substance's composition is changed. When bonds are broken and new ones are formed a chemical change occurs. The following are indicators of chemical changes:
• Change in Temperature
• Change in Color
• Noticeable Odor (after reaction has begun)
• Formation of a Precipitate
• Formation of Bubbles
Note: When two or more reactants are mixed and a change in temperature, color, etc. is noticed, a chemical reaction is probably occurring. These are not definite indicators; a chemical reaction may not be occurring. A change in color is not always a chemical change. If one were to change the color of a substance in a non-chemical reaction scenario, such as painting a car, the change is physical and not chemical. This is because the composition of the car has not changed. Proceed with caution.
Common Physical Changes
Texture
The texture of a substance can differ with a physical change. For example, if a piece of wood was sanded, waxed, and polished, it would have a very different texture than it initially had as a rough piece of wood.
(left) Rough plank boardwalk, Quebec City, Canada (right) Finished mountain ash floor. (CC BY-SA 4.0; WikiPedant and CC BY-SA 2.5; MarkAnthonyBoyle, respectively).
As you can see, the texture of the finished wood is much smoother than the initial grainy wood.
Color
The changing of color of a substance is not necessarily an indicator of a chemical change. For example, changing the color of a metal does not change its physical properties. However, in a chemical reaction, a color change is usually an indicator that a reaction is occurring. Painting the metal car does not changing the composition of the metallic substance.
Robotic arm applying paint on car parts. Image use with permission (CC BY-SA 4.0l RoboGuru).
Temperature
Although we cannot see temperature change, unless if a change of state is occurring, it is a physical change.
Hot metalwork. (CC BY-SA-NC 2.0; flagstaffotos.com.au)
One cannot see the pan physically changing shape, color, texture, or any of the other physical properties. However, if one were to touch the pan, it would be incredibly hot and could cause a burn. Sitting idle in a cupboard, this pan would be cold. One cannot assess this change only through visual exposure; the use of a thermometer or other instrument is necessary.
Shape
The shape of an object can be changed and the object will still remain true to its chemical composition. For example, if one were to fold money, as shown by the figure below, the money is still chemically the same.
Origami Money
Change of State
The change of state is likewise a physical change. In this scenario, one can observe a number of physical properties changing, such as viscosity and shape. As ice turns into water, it does not retain a solid shape and now becomes a viscous fluid. The physical "reaction" for the change of ice into liquid water is:
$\ce{H_2O_{(s)} \rightarrow H_2O_{(l)}}\nonumber$
The following are the changes of state:
Solid → Liquid Melting
Liquid → Gas Vaporization
Liquid → Solid Freezing
Gas → Liquid Condensation
Solid → Gas Sublimation
• If heat is added to a substance, such as in melting, vaporization, and sublimation, the process is endothermic. In this instance, heat is increasing the speed of the molecules causing them move faster.
• If heat is removed from a substance, such as in freezing and condensation, then process is exothermic. In this instance, heat is decreasing the speed of the molecules causing them move slower.
Physical Properties
Luster
The luster of an element is defined as the way it reacts to light. Luster is a quality of a metal. Almost all of the metals, transition metals, and metalloids are lustrous. The non-metals and gases are not lustrous. For example, oxygen and bromine are not lustrous. Shown below is are lustrous paper clips.
Lustrous Paperclips
Malleability
Malleability is also a quality of metals. Metals are said to be malleable. This means that the metals can deform under an amount of stress. For example, if you can hit a metal with a mallet and it deforms, it is malleable. Also, a paperclip can be shaped with bare hands.
Bent Paperclip
The image shows the malleability of a certain metal as stress is applied to it.
Ability to be drawn into a thin wire
In materials science, this property is called ductility. For example, raw copper can be obtained and it can be purified and wrapped into a cord. Once again, this property is characteristic of mainly metals, nonmetals do not possess this quality.
Copper Wire
Density
The density of an object is its mass divided by its volume ($d=m/v$). A substance will have a higher density if it has more mass in a fixed amount of volume. For example, take a ball of metal, roughly the size of a baseball, compressed from raw metal. Compare this to a baseball made of paper. The baseball made of metal has a much greater weight to it in the same amount of volume. Therefore the baseball made out of metal has a much higher density. The density of an object will also determine whether it will sink or float in a particular chemical. Water for example has a density of 1 g/cm3. Any substance with a density lower than that will float, while any substance with a density above that will sink.
Viscosity
Viscosity is defined to be the resistance to deformation of a particular chemical substance when a force is applied to it. In the example below, one can see two cubes falling into two different test tubes. The upper substance shows a violent reaction to the dropping of the cube. The lower substance simply engulfs it slowly without much reaction. The upper substance has a lower viscosity relative to the lower substance, which has very high viscosity. One may even think of viscosity in terms of thickness. The substance with more thickness has higher viscosity than a substance that is deemed "thin." Water has a lower viscosity than honey or magma, which have relatively high viscosities.
Viscosity of Fluids
Common Chemical Changes
The follow are all indicators of chemical reactions. For further information on chemical reactions, please refer to Chemical Reactions.
Change in Temperature
A change in temperature is characteristic of a chemical change. During an experiment, one could dip a thermometer into a beaker or Erlenmeyer Flask to verify a temperature change. If temperature increases, as it does in most reactions, a chemical change is likely to be occurring. This is different from the physical temperature change. During a physical temperature change, one substance, such as water is being heated. However, in this case, one compound is mixed in with another, and these reactants produce a product. When the reactants are mixed, the temperature change caused by the reaction is an indicator of a chemical change.
As an example of a exothermic reaction, if $Fe_2O_3$ is mixed with Al and ignighted (often with burning Mg), then the thermite reaciton is initiated
$Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + \text{Heat}\nonumber$
This reaction generates heat as a product and is (very) exothermic.
However, physical changes can be exothermic or endothermic. The melting of an ice cube, which is endothermic, is a change in a physical property and not composition. Thus, it is a physical change.
Change in Color
A change in color is also another characteristic of a chemical reaction taking place. For example, if one were to observe the rusting of metal over time, one would realized that the metal has changed color and turned orange. This change in color is evidence of a chemical reaction. However, one must be careful; sometimes a change in color is simply the mixing of two colors, but no real change in the composition of the substances in question.
The reaction above is that of the rusting of iron.
$\ce{4Fe + 3O_2 + 6H_2O \rightarrow 4Fe(OH)_3}\nonumber$
Noticeable Odor
When two or more compounds or elements are mixed and a scent or odor is present, a chemical reaction has taken place. For example, when an egg begins to smell, (a rotten egg) a chemical reaction has taken place. This is the result of a chemical decomposition.
Spoiled Egg
Formation of a Precipitate
The formation of a precipitate may be one of the most common signs of a chemical reaction taking place. A precipitate is defined to be a solid that forms inside of a solution or another solid. Precipitates should not be confused with suspensions, which are solutions that are homogeneous fluids with particles floating about in them. For instance, when a soluble carbonate reacts with Barium, a Barium Carbonate precipitate can be observed.
Test Tube
Reaction:
$\ce{Ba^{2+}(aq) + CO^{2-}3(aq) \rightarrow BaCO3(s)}\nonumber$
For further information, please refer to Classification of Matter.
Formation of Bubbles
The formation of bubbles, or rather a gas, is another indicator of a chemical reaction taking place. When bubbles form, a temperature change could also be taking place. Temperature change and formation of bubbles often occur together. For example, in the following image, one can see a gas spewing. This is the formation of a gas.
Gas Formation
However, most reactions are much more subtle. For instance, if the following reaction occurs, one may notice Carbon Dioxide bubbles forming. If there is enough Hydrochloric Acid, bubbles are visible. If there is not, one can't readily notice the change:
$\ce{Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2}\nonumber$
Exercise $1$
Which of the following is a chemical reaction?
1. Freezing liquid Mercury
2. Adding yellow to blue to make green
3. Cutting a piece of paper into two pieces
4. Dropping a sliced orange into a vat of Sodium Hydroxide
5. Filling a balloon with natural air
Answer
D
Exercise $2$
Which of the following is a physical reaction?
1. Shattering Glass with a baseball
2. Corroding Metal
3. Fireworks Exploding
4. Lighting a match
5. Baking a cake
Answer
A
Exercise $3$
Which of the following is a chemical reaction?
1. Painting a wall blue
2. A bicycle rusting
3. Ice cream melting
4. Scratching a key across a desk
5. Making a sand castle
Answer
B
Exercise $4$
Which of the following is a physical reaction?
1. Frying an egg
2. Digesting carrots
3. A Macbook falling out of a window
4. Creating ATP in the human body
5. Dropping a fizzy tablet into a glass of water
Answer
C
Exercise $5$
Write C for Chemical Reaction or P for Physical Reaction.
1. Burning Leaves
2. Cutting Diamonds
3. Crushing a pencil
4. The salivary amylase enzyme that breaks down food in the mouth
5. Salt mixing in with water
Answer
a) C
b) P
c) P
d) C
e) Neither. This is one of the gray areas of chemical change and physical change. Although the salt has dissociated into Sodium and Chloride ions, it is still salt in water. Salt, initially is actually just a conglomerate of sodium and chloride ions and by dissociating them, just the arrangement of the ions has changed. Please click here for more information
All images are courtesy of http://www.sxc.hu, which provides royalty free images that are free to be copied without restrictions. The viscosity image is also free to be duplicated as per permission of author on Wikipedia.com. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/01%3A_Introduction_to_Chemistry/1.02%3A_Physical_and_Chemical_Properties.txt |
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
02: Atoms Molecules and Ions
Learning Objectives
• to know the meaning of isotopes and atomic masses.
Atomic and Molecular Weights
The subscripts in chemical formulas, and the coefficients in chemical equations represent exact quantities. $\ce{H_2O}$, for example, indicates that a water molecule comprises exactly two atoms of hydrogen and one atom of oxygen. The following equation:
$\ce{C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H2O(l)} \label{Eq1}$
not only tells us that propane reacts with oxygen to produce carbon dioxide and water, but that 1 molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. Since counting individual atoms or molecules is a little difficult, quantitative aspects of chemistry rely on knowing the masses of the compounds involved.
Atoms of different elements have different masses. Early work on the separation of water into its constituent elements (hydrogen and oxygen) indicated that 100 grams of water contained 11.1 grams of hydrogen and 88.9 grams of oxygen:
$\text{100 grams Water} \rightarrow \text{11.1 grams Hydrogen} + \text{88.9 grams Oxygen} \label{Eq2}$
Later, scientists discovered that water was composed of two atoms of hydrogen for each atom of oxygen. Therefore, in the above analysis, in the 11.1 grams of hydrogen there were twice as many atoms as in the 88.9 grams of oxygen. Therefore, an oxygen atom must weigh about 16 times as much as a hydrogen atom:
$\dfrac{\dfrac{88.9\;g\;Oxygen}{1\; atom}}{\dfrac{111\;g\;Hydrogen}{2\;atoms}} = 16 \label{Eq3}$
Hydrogen, the lightest element, was assigned a relative mass of '1', and the other elements were assigned 'atomic masses' relative to this value for hydrogen. Thus, oxygen was assigned an atomic mass of 16. We now know that a hydrogen atom has a mass of 1.6735 x 10-24 grams, and that the oxygen atom has a mass of 2.6561 X 10-23 grams. As we saw earlier, it is convenient to use a reference unit when dealing with such small numbers: the atomic mass unit. The atomic mass unit (amu) was not standardized against hydrogen, but rather, against the 12C isotope of carbon (amu = 12).
Thus, the mass of the hydrogen atom (1H) is 1.0080 amu, and the mass of an oxygen atom (16O) is 15.995 amu. Once the masses of atoms were determined, the amu could be assigned an actual value:
1 amu = 1.66054 x 10-24 grams conversely: 1 gram = 6.02214 x 1023 amu
Mass Numbers and Atomic Mass of Elements: Mass Numbers and Atomic Mass of Elements, YouTube(opens in new window) [youtu.be]
Average Atomic Mass
Although the masses of the electron, the proton, and the neutron are known to a high degree of precision (Table 2.3.1), the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1H (hydrogen) and 2H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions.
Because atoms are much too small to measure individually and do not have charges, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses (Figure $1$). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest.
The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu or u), defined as one-twelfth of the mass of one atom of 12C. Because the masses of all other atoms are calculated relative to the 12C standard, 12C is the only atom listed in Table 2.3.2 whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10−24 g.
Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2H to the mass of 12C, so the absolute mass of 2H is
$\rm{\text{mass of }^2H \over \text{mass of }^{12}C} \times \text{mass of }^{12}C = 0.167842 \times 12 \;amu = 2.104104\; amu \label{Eq4}$
The masses of the other elements are determined in a similar way.
The periodic table lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes in Table 2.3.2 reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes. Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12C (mass = 12 amu by definition) and 1.11% 13C (mass = 13.003355 amu). The percent abundance of 14C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as follows:
$\rm(0.9889 \times 12 \;amu) + (0.0111 \times 13.003355 \;amu) = 12.01 \;amu \label{Eq5}$
Carbon is predominantly 12C, so its average atomic mass should be close to 12 amu, which is in agreement with this calculation.
The value of 12.01 is shown under the symbol for C in the periodic table, although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes.
Example $1$: Bromine
Naturally occurring bromine consists of the two isotopes listed in the following table:
Solutions to Example 2.4.1
Isotope Exact Mass (amu) Percent Abundance (%)
79Br 78.9183 50.69
81Br 80.9163 49.31
Calculate the atomic mass of bromine.
Given: exact mass and percent abundance
Asked for: atomic mass
Strategy:
1. Convert the percent abundances to decimal form to obtain the mass fraction of each isotope.
2. Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass.
3. Add together the weighted masses to obtain the atomic mass of the element.
4. Check to make sure that your answer makes sense.
Solution:
A The atomic mass is the weighted average of the masses of the isotopes. In general, we can write
atomic mass of element = [(mass of isotope 1 in amu) (mass fraction of isotope 1)] + [(mass of isotope 2) (mass fraction of isotope 2)] + …
Bromine has only two isotopes. Converting the percent abundances to mass fractions gives
$\ce{^{79}Br}: {50.69 \over 100} = 0.5069 \nonumber$
$\ce{^{81}Br}: {49.31 \over 100} = 0.4931 \nonumber$
B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass:
$\ce{^{79}Br}: 79.9183 \;amu \times 0.5069 = 40.00\; amu$
$\ce{^{81}Br}: 80.9163 \;amu \times 0.4931 = 39.90 \;amu$
C The sum of the weighted masses is the atomic mass of bromine is
40.00 amu + 39.90 amu = 79.90 amu
D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%.
Exercise $1$
Magnesium has the three isotopes listed in the following table:
Solutions to Example 2.4.1
Isotope Exact Mass (amu) Percent Abundance (%)
24Mg 23.98504 78.70
25Mg 24.98584 10.13
26Mg 25.98259 11.17
Use these data to calculate the atomic mass of magnesium.
Answer
24.31 amu
Finding the Averaged Atomic Weight of an Element: Finding the Averaged Atomic Weight of an Element(opens in new window) [youtu.be]
Summary
The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.01%3A_Atoms%3A_Their_Composition_and_Structure.txt |
Atoms that have the same atomic number (number of protons), but different mass numbers (number of protons and neutrons) are called isotopes. There are naturally occurring isotopes and isotopes that are artificially produced. Isotopes are separated through mass spectrometry; MS traces show the relative abundance of isotopes vs. mass number (mass : charge ratio).
Introduction
As mentioned before, isotopes are atoms that have the same atomic number, but different mass numbers. Isotopes are denoted the same way as nuclides, but they are often symbolized only with the mass numbers because isotopes of the same element have the the same atomic number. Carbon, for example, has two naturally occurring isotopes, \(^{12}_6C\) and \(^{13}_6C\). Because both of these isotopes have 6 protons, they are often written as \(^{12}C\) and \(^{13}C\). \(^{12}C\) has 6 neutrons, and \(^{13}C\) has 7 neutrons.
Of all the elements on the periodic table, only 21 are pure elements. Pure, or monotopic, elements are those elements with only one naturally occurring nuclide. The following lists the 21 pure elements:
1. \(^{27}_{13}Al\)
2. \(^{75}_{33}As\)
3. \(^{9}_4Be\)
4. \(^{209}_{83}Bi\)
5. \(^{133}_{55}Cs\)
6. \(^{59}_{27}Co\)
7. \(^{19}_9F\)
8. \(^{197}_{79}Au\)
9. \(^{165}_{67}Ho\)
10. \(^{127}_{53}I\)
11. \(^{55}_{25}Mn\)
12. \(^{93}_{41}Nb\)
13. \(^{31}_{15}P\)
14. \(^{141}_{59}Pr\)
15. \(^{103}_{45}Rh\)
16. \(^{45}_{21}Sc\)
17. \(^{23}_{11}Na\)
18. \(^{159}_{65}Tb\)
19. \(^{232}_{90}Th\)
20. \(^{169}_{69}Tm\)
21. \(^{89}_{39}Y\)
Isotopes of the other elements either occur naturally or are artificially produced.
Natural and Artificial Isotopes
Most elements have naturally occurring isotopes. Percent natural abundances indicate which isotopes of any given element are predominant (occur in greater abundance) and which only occur in trace amounts. Mercury, for example, has seven naturally occurring isotopes: \(^{196}Hg\), \(^{198}Hg\), \(^{199}Hg\), \(^{200}Hg\), \(^{201}Hg\), \(^{202}Hg\), \(^{204}Hg\); these have the percent natural abundances of 0.146%, 10.02%, 16.84%, 23.13%, 13.22%, 29.80%, and 6.85%, respectively. It is clear that \(^{202}Hg\) occurs with greatest abundance, and \(^{200}Hg\) is the next most abundant, but the other isotopes only occur in small traces.
Note: The sum of the percent natural abundances of all the isotopes of any given element must total 100%.
There are 20 elements with only artificially produced isotopes. The majority of these are heavier elements; the lightest elements with artificial isotopes are \(^{43}Tc\) and \(^{61}Pm\). The other elements that only have artificial isotopes are those with atomic numbers of 84-88 and 89-103, otherwise known as the actinoids, but excluding \(^{90}Th\) and \(^{92}U\).
Some naturally occurring and artificially produced isotopes are radioactive. The nucleus of a radioactive isotope is unstable; radioactive isotopes spontaneously decay, emitting alpha, beta, and gamma rays until they reach a stability, usually in the state of a different element. Bismuth (\(^{209}_{83}Bi\)) has the highest atomic and mass number of all the stable nuclides. All nuclides with atomic number and mass number greater than 83 and 209, respectively, are radioactive. However, there are some lighter nuclides that are radioactive. For example, hydrogen has two naturally occurring stable isotopes, \(^{1}H\) and \(^{2}H\) (deuterium), and a third naturally occurring radioactive isotope, \(^{3}H\) (tritium).
Radioisotope Dating
The presence of certain radioisotopes in an object can be used to determine its age. Carbon dating is based on the fact that living plants absorb stable \(^{12}C\), \(^{13}C\) and radioactive \(^{14}C\) from the atmosphere, and animals absorb them from the plants. An organism no longer absorbs carbon after it dies, its age can be determined by measuring the ratio of \(^{13}C\) to \(^{14}C\) in the sample and extrapolating based on its decay rate.
Art forgeries are often detected by similar means. \(^{137}Cs\) and \(^{90}Sr\) do not occur naturally and are only present in the atmosphere today because of nuclear weapons. Any object created before July 1945, then, would have neither of these elements, so finding them through mass spectrometry or other means would indicate that it was created later.
Isotopic Masses, Percent Natural Abundance, and Weighted-Average Atomic Mass
Because most elements occur as isotopes and different isotopes have different masses, the atomic mass of an element is the average of the isotopic masses, weighted according to their naturally occurring abundances; this is the mass of each element recorded on the periodic table, also known as the relative atomic mass (Ar). Treating isotopic masses in weighted averages gives greater importance to the isotope with greatest percent natural abundance. Below is a general equation to calculate the atomic mass of an element based on percent natural abundance and isotopic masses:
* fractional abundance is the percent abundance divided by 100%
Bromine has two naturally occurring isotopes: bromine-79 has a mass of 78.9183 u and an abundance of 50.69%, and bromine-81 has a mass of 80.92 u and an abundance of 49.31%. The equation above can be used to solve for the relative atomic mass of bromine:
atomic mass of Br = (0.5069 x 78.9183 u) + (0.4931 x 80.92 u) = 79.91 u
This is the relative atomic number of bromine that is listed on the periodic table.
Comparing their isotopic masses of any given element to the relative atomic mass of the element reveals that the Ar is very close to the isotope that occurs most frequently. Thus, the isotope whose isotopic mass is closest to the atomic mass of the element is the isotope that occurs in the greatest abundance.
Mass Spectrometry
Mass spectrometry is a technique that can be used to distinguish between isotopes of a given element. A mass spectrometer separates each isotope by mass number. Each isotope is characterized by a peak (of given intensity) according to its relative abundance. The most intense peak corresponds to the isotope that occurs in the largest relative natural abundance, and vice versa. Refer to Mass Spectrometry: Isotope Effects.
Example:
The mass spectrum of strontium has four different peaks, varying in intensity. The four peaks indicate that there are four isotopes of strontium. The four isotopes of strontium have isotopic mass numbers of 84, 86, 87, and 88, and relative abundances of 0.56%, 9.86%, 7.00%, and 82.58%, respectively. The intensity of the peak corresponds to the abundance. \(^{84}Sr\) has the smallest peak, which corresponds to its relative abundance of 0.56%, whereas \(^{88}Sr\) has the largest peak, which corresponds to its relative abundance of 82.58%. This indicates that \(^{88}Sr\) is the isotope that occurs in highest amounts.
Problems
1) Find the number of protons and neutrons in the following isotopes: a) \(^{20}Ne\), \(^{21}Ne\), \(^{22}Ne\); b) \(^{84}Sr\), \(^{86}Sr\), \(^{87}Sr\), \(^{88}Sr\); c) \(^{102}Pd\), \(^{104}Pd\), \(^{105}Pd\), \(^{106}Pd\), \(^{108}Pd\), \(^{110}Pd\)
(Answer: a) \(^{20}Ne\) has 10 p and 10 n; \(^{21}Ne\) has 10 p and 11 n; \(^{22}Ne\) has 10 p and 12 n; b) \(^{84}Sr\) has 38 p and 46 n, \(^{86}Sr\) has 38 p and 48 n, \(^{87}Sr\) has 38 p and 49 n, \(^{88}Sr\) has 38 p and 50 n; c) \(^{102}Pd\) has 46 p and 56 n, \(^{104}Pd\) has 46 p and 58 n, \(^{105}Pd\) has 46 p and 59 n, \(^{106}Pd\) has 46 p and 60 n, \(^{108}Pd\) has 46 p and 62 n, \(^{110}Pd\) has 46 p and 64 n)
2) An isotope of lead has an isotopic mass of 208. Give the appropriate notation for the isotope.
(Answer: \(^{208}Pb\))
3) Thallium has two naturally occurring istotopes, \(^{203}Tl\) and \(^{205}Tl\). Tl-205 has an abundance of 70.48%. What is the percent natural abundance of Tl-203?
(Answer: 29.52%)
4) Chlorine has two naturally occuring isotopes, \(^{35}Cl\) and \(^{37}Cl\), with masses of 34.97 u and 36.97 u, respectively. Which of these two isotopes occur in greater abundance?
(Answer: \(^{35}Cl\))
5. Potassium has three naturally occurring isotopes: \(^{39}K\), \(^{40}K\), and \(^{41}K\), with masses (and abundances) of 38.96 u (93.3%), 39.96 u (0.012%), and 40.96 u (6.7%), respectively. Calculate the relative atomic mass of K.
(Answer: 39.10 u)
Review Questions
1. What do the terms 'monotopic' and 'radioactive' mean?
2. On the mass spectrum of any given atom, what does the number and intensity of the peak(s) tell you?
3. What is the relationship between isotopic masses, percent natural abundance, and weighted-average atomic mass? | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.02%3A_Isotopes.txt |
Mass is a basic physical property of matter. The mass of an atom or a molecule is referred to as the atomic mass. The atomic mass is used to find the average mass of elements and molecules and to solve stoichiometry problems.
Introduction
In chemistry, there are many different concepts of mass. It is often assumed that atomic mass is the mass of an atom indicated in unified atomic mass units (u). However, the book Quantities, Units and Symbols in Physical Chemistry published by the IUPAC clearly states:
"Neither the name of the physical quantity, nor the symbol used to denote it, should imply a particular choice of unit."
The name "atomic mass" is used for historical reasons, and originates from the fact that chemistry was the first science to investigate the same physical objects on macroscopic and microscopic levels. In addition, the situation is rendered more complicated by the isotopic distribution. On the macroscopic level, most mass measurements of pure substances refer to a mixture of isotopes. This means that from a physical stand point, these mixtures are not pure. For example, the macroscopic mass of oxygen (O2) does not correspond to the microscopic mass of O2. The former usually implies a certain isotopic distribution, whereas the latter usually refers to the most common isotope (16O2). Note that the former is now often referred to as the "molecular weight" or "atomic weight".
Mass Concepts in Chemistry
name in chemistry physical meaning symbol units
atomic mass mass on microscopic scale m, ma Da, u, kg, g
molecular mass mass of a molecule m Da, u, kg, g
isotopic mass mass of a specific isotope Da, u, kg, g
mass of entity mass of a chemical formula m, mf Da, u, kg, g
average mass average mass of a isotopic distribution m Da, u, kg, g
molar mass average mass per mol M = m/n kg/mol or g/mol
atomic weight average mass of an element Ar = m / mu unitless
molecular weight average mass of a molecule Mr = m / mu unitless
relative atomic mass ratio of mass m and and the atomic mass constant mu Ar = m / mu unitless
atomic mass constant mu = m(12C)/12 mu = 1 Da = 1 u Da, u, kg, g
relative molecular mass ratio of mass m of a molecule and and the atomic mass constant mu Mr = m / mu unitless
relative molar mass ? ? ?
mass number nucleon number A nucleons, or unitless
integer mass nucleon number * Da m Da, u
nominal mass integer mass of molecule consisting of most abundant isotopes m Da, u
exact mass mass of molecule calculated from the mass of its isotopes (in contrast of measured ba a mass spectrometer) Da, u, kg, g
accurate mass mass (not normal mass) Da, u, kg, g
These concepts are further explained below.
Average Mass
Isotopes are atoms with the same atomic number, but different mass numbers. A different mass size is due to the difference in the number of neutrons that an atom contains. Although mass numbers are whole numbers, the actual masses of individual atoms are never whole numbers (except for carbon-12, by definition). This explains how lithium can have an atomic mass of 6.941 Da. The atomic masses on the periodic table take these isotopes into account, weighing them based on their abundance in nature; more weight is given to the isotopes that occur most frequently in nature. Average mass of the element E is defined as:
$m(E) = \sum_{n=1} m(I_n) \times p(I_n)$
where ∑ represents a n-times summation over all isotopes $I_n$ of element E, and p(I) represents the relative abundance of the isotope I.
Example 1
Find the average atomic mass of boron using the Table 1 below:
Mass and abundance of Boron isotopes
n isotope In mass m (Da) isotopic abundance p
1 10B 10.013 0.199
2 11B 11.009 0.801
Solution
The average mass of Boron is:
$m(B) = (10.013\ Da)(0.199) + (11.009\ Da)(0.801) = 1.99\ Da + 8.82\ Da = 10.81\ Da$
Relative Mass
Traditionally it was common practice in chemistry to avoid using any units when indicating atomic masses (e.g. masses on microscopic scale). Even today, it is common to hear a chemist say, "12C has exactly mass 12". However, because mass is not a dimensionless quantity, it is clear that a mass indication needs a unit. Chemists have tried to rationalize the omission of a unit; the result is the concept of relative mass, which strictly speaking is not even a mass but a ratio of two masses. Rather than using a unit, these chemists claim to indicate the ratio of the mass they want to indicate and the atomic mass constant mu which is defined analogous to the unit they want to avoid. Hence the relative atomic mass of the mass m is defined as:
$A_r = \dfrac{m}{m_u}$
The quantity is now dimensionless. As this unit is confusing and against the standards of modern metrology, the use of relative mass is discouraged.
Molecular Weight, Atomic Weight, Weight vs. Mass
Until recently, the concept of mass was not clearly distinguished from the concept of weight. In colloquial language this is still the case. Many people indicate their "weight" when they actually mean their mass. Mass is a fundamental property of objects, whereas weight is a force. Weight is the force F exerted on a mass m by a gravitational field. The exact definition of the weight is controversial. The weight of a person is different on ground than on a plane. Strictly speaking, weight even changes with location on earth.
When discussing atoms and molecules, the mass of a molecule is often referred to as the "molecular weight". There is no univerally-accepted definition of this term; however, mosts chemists agree that it means an average mass, and many consider it dimensionless. This would make "molecular weight" a synonym to "average relative mass".
Integer Mass
Because the proton and the neutron have similar mass, and the electron has a very small mass compared to the former, most molecules have a mass that is close to an integer value when measured in daltons. Therefore it is quite common to only indicate the integer mass of molecules. Integer mass is only meaningful when using dalton (or u) units.
Accurate Mass
Many mass spectrometers can determine the mass of a molecule with accuracy exceeding that of the integer mass. This measurement is therefore called the accurate mass of the molecule. Isotopes (and hence molecules) have atomic masses that are not integer masses due to a mass defect caused by binding energy in the nucleus.
Units
The atomic mass is usually measured in the units unified atomic mass unit (u), or dalton (Da). Both units are derived from the carbon-12 isotope, as 12 u is the exact atomic mass of that isotope. So 1 u is 1/12 of the mass of a carbon-12 isotope:
1 u = 1 Da = m(12C)/12
The first scientists to measure atomic mass were John Dalton (between 1803 and 1805) and Jons Jacoband Berzelius (between 1808 and 1826). Early atomic mass theory was proposed by the English chemist William Prout in a series of published papers in 1815 and 1816. Known was Prout's Law, Prout suggested that the known elements had atomic weights that were whole number multiples of the atomic mass of hydrogen. Berzelius demonstrated that this is not always the case by showing that chlorine (Cl) has a mass of 35.45, which is not a whole number multiple of hydrogen's mass.
Some chemists use the atomic mass unit (amu). The amu was defined differently by physicists and by chemists:
• Physics: 1 amu = m(16O)/16
• Chemistry: 1 amu = m(O)/16
Chemists used oxygen in the naturally occurring isotopic distribution as the reference. Because the isotopic distribution in nature can change, this definition is a moving target. Therefore, both communities agreed to the compromise of using m(12C)/12 as the new unit, naming it the "unified atomic mass unit" (u). Hence, the amu is no longer in use; those who still use it do so with the definition of the u in mind. For this reason, the dalton (Da) is increasingly recommended as the accurate mass unit.
Neither u nor Da are SI units, but both are recognized by the SI.
Molar Mass
The molar mass is the mass of one mole of substance, whether the substance is an element or a compound. A mole of substance is equal to Avogadro's number (6.023×1023) of that substance. The molar mass has units of g/mol or kg/mol. When using the unit g/mol, the numerical value of the molar mass of a molecule is the same as its average mass in daltons:
• Average mass of C: 12.011 Da
• Molar mass of C: 12.011 g/mol
This allows for a smooth transition from the microscopic world, where mass is measured in daltons, to the macroscopic world, where mass is measured in kilograms.
Example 2
What is the molar mass of phenol, C6H5OH?
Average mass m = 6 × 12.011 Da + 6 × 1.008 Da + 1 × 15.999 Da = 94.113 Da
Molar mass = 94.113 g/mol = 0.094113 kg/mo
Measuring Masses in the Atomic Scale
Masses of atoms and molecules are measured by mass spectrometry. Mass spectrometry is a technique that measures the mass-to-charge ratio (m/q) of ions. It requires that all molecules and atoms to be measured be ionized. The ions are then separated in a mass analyzer according to their mass-to-charge ratio. The charge of the measured ion can then be determined, because it is a multiple of the elementary charge. The the ion's mass can be deduced. The average masses indicated in the periodic table are then calculated using the isotopic abundances, as explained above.
The masses of all isotopes have been measured with very high accuracy. Therefore, it is much simpler and more accurate to calculate the mass of a molecule of interest as a sum of its isotopes than measuring it with a commercial mass spectrometer.
Note that the same is not true on the nucleon scale. The mass of an isotope cannot be calculated accurately as the sum of its particles (given in the table below); this would ignore the mass defect caused by the binding energy of the nucleons, which is significant.
Table 2: Mass of three sub-atomic particles
Particle SI (kg) Atomic (Da) Mass Number A
Proton 1.6726×10-27 1.0073 1
Neutron 1.6749×10-27 1.0087 1
Electron 9.1094×10-31 0.00054858 0
As shown in Table 2, the mass of an electron is relatively small; it contributes less than 1/1000 to the overall mass of the atom.
Where to Find Atomic Mass
The atomic mass found on the Periodic Table (below the element's name) is the average atomic mass. For example, for Lithium:
The red arrow indicates the atomic mass of lithium. As shown in Table 2 above and mathematically explained below, the masses of a protons and neutrons are about 1u. This, however, does not explain why lithium has an atomic mass of 6.941 Da where 6 Da is expected. This is true for all elements on the periodic table. The atomic mass for lithium is actually the average atomic mass of its isotopes. This is discussed further in the next section.
One particularly useful way of writing an isotope is as follows:
Applications
Applications Include:
1. Average Molecular Mass
2. Stoichiometry
Note: One particularly important relationship is illustrated by the fact that an atomic mass unit is equal to 1.66 × 10-24 g. This is the reciprocal of Avogadro's constant, and it is no coincidence:
$\dfrac{\rm Atomic~Mass~(g)}{1 {\rm g}} \times \dfrac{1 {\rm mol}}{6.022 \times 10^{23}} = \dfrac{\rm Mass~(g)}{1 {\rm atom}}$
Because a mol can also be expressed as gram × atoms,
$1\ u = \frac{M_u\ (molar\ mass\ unit)}{N_A\ (Avogadro's\ Number)} = 1\ \frac{g}{mol\ N_A}$
1u = Mu(molar mass unit)/NA(Avogadro's Number)=1g/mol/NA
NA known as Avogadro's number (Avogadro's constant) is equal to 6.023×1023 atoms.
Atomic mass is particularly important when dealing with stoichiometry.
Practice Problems
1. What is the molecular mass of radium bicarbonate, Ra(HCO3)2?
2. List the following, from least to greatest, in terms of their number of neutrons, and then atomic mass: 14N, 42Cl, 25Na, 10Be
3. A new element, Zenium, has 3 isotopes, 59Ze, 61Ze, and 67Ze, with abundances of 62%, 27%, and 11% respectively. What is the atomic mass of Zenium?
4. An isotope with a mass number of 55 has 5 more neutrons than protons. What element is it?
5. How much mass does 3.71 moles of Fluorine have?
6. How many grams are there in 4.3 × 1022 molecules of POCl3?
7. How many moles are there in 23 grams of sodium carbonate?
Solutions
a) Molecular mass of Ra(HCO3)2
= 226 + 2(1.01 u + 12.01 u + (16.00 u)(3)) = 348 u or g/mol
b) Number of neutrons: 10Be, 14N, 25Na, 42Cl
Atomic Mass: 10Be, 14N, 25Na, 42Cl
Note: It is the same increasing order for both number of neutrons and atomic mass because more neutrons means more mass.
c) Atomic mass of Zenium:
(59 u)(0.62) + (61 u)(0.27) + (67 u)(0.11)
= 37 u + 16 u + 7.4 u
= 60.4 u or g/mol
d) Mn
e) (3.71 moles F2)(19 × 2 g/mol F2)
= (3.71 mol F2)(38 g/mol F2)
= 140 g F2
f) (4.3 × 1022 molecules POCI3)(1 mol/6.022 × 1023 molecules POCI3)(30.97 + 16.00 + 35.45 x 3 g/mol POCI3)
= (4.3 × 1022 molecules POCI3)(mol/6.022 × 1023 molecules POCI3)(153.32 g/mol POCI3)
= 11 g POCI3
g) (23 g Na2CO3)(1 mol/22.99 × 2 + 12.01 + 16.00 × 3 g Na2CO3)
= (23 g Na2CO3)(1 mol/105.99 g Na2CO3)
= (0.22 mol Na2CO3)
Contributors and Attributions
• Gunitika Dandona (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.03%3A_Isotope_Abundance_and_Atomic_Weight.txt |
Molecular geometry is the 3-dimensional shape that a molecule occupies in space. It is determined by the central atom and the surrounding atoms and electron pairs. The shape of most molecules can be predicted using the Valence Shell Electron Pair Repulsion (VSEPR) method. This method states a few rules to help one determine the shape of a substance without using high technology methods such as X-ray crystallography, NMR Spectroscopy, or electron microscopy. Some of the most common shapes that can be taken are linear, trigonal planar, tetrahedral, pyramidal, and angular (or bent).
2.07: Ionic Compounds
There are many types of chemical bonds and forces that bind molecules together. The two most basic types of bonds are characterized as either ionic or covalent. In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same electronegativity share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.
Introduction
Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion. Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal.
Ionic bonding is observed because metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in their valence shells tend to readily accept electrons to achieve noble gas configuration. In ionic bonding, more than 1 electron can be donated or received to satisfy the octet rule. The charges on the anion and cation correspond to the number of electrons donated or received. In ionic bonds, the net charge of the compound must be zero.
This sodium molecule donates the lone electron in its valence orbital in order to achieve octet configuration. This creates a positively charged cation due to the loss of electron.
This chlorine atom receives one electron to achieve its octet configuration, which creates a negatively charged anion.
The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of the electrostatic attraction between the particles. At the ideal interatomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ion.
Example \(1\): Chloride Salts
In this example, the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.
In this example, the magnesium atom is donating both of its valence electrons to chlorine atoms. Each chlorine atom can only accept 1 electron before it can achieve its noble gas configuration; therefore, 2 atoms of chlorine are required to accept the 2 electrons donated by the magnesium. Notice that the net charge of the compound is 0.
Covalent Bonding
Covalent bonding is the sharing of electrons between atoms. This type of bonding occurs between two atoms of the same element or of elements close to each other in the periodic table. This bonding occurs primarily between nonmetals; however, it can also be observed between nonmetals and metals.
If atoms have similar electronegativities (the same affinity for electrons), covalent bonds are most likely to occur. Because both atoms have the same affinity for electrons and neither has a tendency to donate them, they share electrons in order to achieve octet configuration and become more stable. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small for ionic bonding to occur. For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet. To form ionic bonds, Carbon molecules must either gain or lose 4 electrons. This is highly unfavorable; therefore, carbon molecules share their 4 valence electrons through single, double, and triple bonds so that each atom can achieve noble gas configurations. Covalent bonds include interactions of the sigma and pi orbitals; therefore, covalent bonds lead to formation of single, double, triple, and quadruple bonds.
Example \(2\): \(PCl_3\)
In this example, a phosphorous atom is sharing its three unpaired electrons with three chlorine atoms. In the end product, all four of these molecules have 8 valence electrons and satisfy the octet rule.
Bonding in Organic Chemistry
Ionic and covalent bonds are the two extremes of bonding. Polar covalent is the intermediate type of bonding between the two extremes. Some ionic bonds contain covalent characteristics and some covalent bonds are partially ionic. For example, most carbon-based compounds are covalently bonded but can also be partially ionic. Polarity is a measure of the separation of charge in a compound. A compound's polarity is dependent on the symmetry of the compound and on differences in electronegativity between atoms. Polarity occurs when the electron pushing elements, found on the left side of the periodic table, exchanges electrons with the electron pulling elements, on the right side of the table. This creates a spectrum of polarity, with ionic (polar) at one extreme, covalent (nonpolar) at another, and polar covalent in the middle.
Both of these bonds are important in organic chemistry. Ionic bonds are important because they allow the synthesis of specific organic compounds. Scientists can manipulate ionic properties and these interactions in order to form desired products. Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding. Covalent bonding allows molecules to share electrons with other molecules, creating long chains of compounds and allowing more complexity in life.
Problems
1. Are these compounds ionic or covalent?
2. In the following reactions, indicate whether the reactants and products are ionic or covalently bonded.
a)
b) Clarification: What is the nature of the bond between sodium and amide? What kind of bond forms between the anion carbon chain and sodium?
c)
Solutions
• 1) From left to right: Covalent, Ionic, Ionic, Covalent, Covalent, Covalent, Ionic.
• 2a) All products and reactants are ionic.
• 2b) From left to right: Covalent, Ionic, Ionic, Covalent, Ionic, Covalent, Covalent, Ionic.
• 2c) All products and reactants are covalent. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.05%3A_Molecular_Formulas_and_Models.txt |
Generally, there are two types of inorganic compounds that can be formed: ionic compounds and molecular compounds. Nomenclature is the process of naming chemical compounds with different names so that they can be easily identified as separate chemicals. Inorganic compounds are compounds that do not deal with the formation of carbohydrates, or simply all other compounds that do not fit into the description of an organic compound. For example, organic compounds include molecules with carbon rings and/or chains with hydrogen atoms (see picture below). Inorganic compounds, the topic of this section, are every other molecule that does not include these distinctive carbon and hydrogen structures.
Compounds between Metals and Nonmetals (Cation and Anion)
Compounds made of a metal and nonmetal are commonly known as Ionic Compounds, where the compound name has an ending of –ide. Cations have positive charges while anions have negative charges. The net charge of any ionic compound must be zero which also means it must be electrically neutral. For example, one Na+ is paired with one Cl-; one Ca2+ is paired with two Br-. There are two rules that must be followed through:
• The cation (metal) is always named first with its name unchanged
• The anion (nonmetal) is written after the cation, modified to end in –ide
Table 1: Cations and Anions:
+1 Charge +2 Charge -1 Charge -2 Charge -3 Charge -4 Charge
Group 1A elements Group 2A elements Group 7A elements Group 6A elements Group 5A elements Group 4A elements
Hydrogen: H+ Beryllium: Be2+ Hydride: H- Oxide: O2- Nitride: N3- Carbide: C4-
Lithium: Li+ Magnesium: Mg2+ Fluoride: F- Sulfide: S2- Phosphide: P3-
Soduim: Na+ Calcium: Ca2+ Chloride: Cl-
Potassium: K+ Strontium: Sr2+ Bromide: Br-
Rubidium: Rb+ Barium: Ba2+ Iodide: I-
Cesium: Cs+
Example 1
Na+ + Cl- = NaCl; Ca2+ + 2Br- = CaBr2
Sodium + Chlorine = Sodium Chloride; Calcium + Bromine = Calcium Bromide
The transition metals may form more than one ion, thus it is needed to be specified which particular ion we are talking about. This is indicated by assigning a Roman numeral after the metal. The Roman numeral denotes the charge and the oxidation state of the transition metal ion. For example, iron can form two common ions, Fe2+ and Fe3+. To distinguish the difference, Fe2+ would be named iron (II) and Fe3+ would be named iron (III).
Table of Transition Metal and Metal Cations:
+1 Charge +2 Charge +3 Charge +4 Charge
Copper(I): Cu+ Copper(II): Cu2+ Aluminum: Al3+ Lead(IV): Pb4+
Silver: Ag+ Iron(II): Fe2+ Iron(III): Fe3+ Tin(IV): Sn4+
Cobalt(II): Co2+ Cobalt(III): Co3+
Tin(II): Sn2+
Lead(II): Pb2+
Nickel: Ni2+
Zinc: Zn2+
Example 2
Ions: Fe2++ 2Cl- Fe3++ 3Cl-
Compound: FeCl2 FeCl3
Nomenclature Iron (II) Chloride Iron (III) Chloride
However, some of the transition metals' charges have specific Latin names. Just like the other nomenclature rules, the ion of the transition metal that has the lower charge has the Latin name ending with -ous and the one with the the higher charge has a Latin name ending with -ic. The most common ones are shown in the table below:
Transition Metal Ion with Roman Numeral Latin name
Copper (I): Cu+ Cuprous
Copper (II): Cu2+ Cupric
Iron (II): Fe2+ Ferrous
Iron (III): Fe3+ Ferric
Lead (II): Pb2+ Plumbous
Lead (IV): Pb4+ Plumbic
Mercury (I): Hg22+ Mercurous
Mercury (II): Hg2+ Mercuric
Tin (II): Sn2+ Stannous
Tin (IV): Sn4+ Stannic
Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion. Instead of using Roman numerals, the different ions can also be presented in plain words. The metal is changed to end in –ous or –ic.
• -ous ending is used for the lower oxidation state
• -ic ending is used for the higher oxidation state
Example 3
Compound Cu2O CuO FeCl2 FeCl3
Charge Charge of copper is +1 Charge of copper is +2 Charge of iron is +2 Charge of iron is +3
Nomenclature Cuprous Oxide Cupric Oxide Ferrous Chloride Ferric Chloride
However, this -ous/-ic system is inadequate in some cases, so the Roman numeral system is preferred. This system is used commonly in naming acids, where H2SO4 is commonly known as Sulfuric Acid, and H2SO3 is known as Sulfurous Acid.
Compounds between Nonmetals and Nonmetals
Compounds that consist of a nonmetal bonded to a nonmetal are commonly known as Molecular Compounds, where the element with the positive oxidation state is written first. In many cases, nonmetals form more than one binary compound, so prefixes are used to distinguish them.
# of Atoms 1 2 3 4 5 6 7 8 9 10
Prefixes Mono- Di- Tri- Tetra- Penta- Hexa- Hepta- Octa- Nona- Deca-
Example 4
CO = carbon monoxide BCl3 = borontrichloride
CO2 = carbon dioxide N2O5 =dinitrogen pentoxide
The prefix mono- is not used for the first element. If there is not a prefix before the first element, it is assumed that there is only one atom of that element.
Binary Acids
Although HF can be named hydrogen fluoride, it is given a different name for emphasis that it is an acid. An acid is a substance that dissociates into hydrogen ions (H+) and anions in water. A quick way to identify acids is to see if there is an H (denoting hydrogen) in front of the molecular formula of the compound. To name acids, the prefix hydro- is placed in front of the nonmetal modified to end with –ic. The state of acids is aqueous (aq) because acids are found in water.
Some common binary acids include:
HF (g) = hydrogen fluoride -> HF (aq) = hydrofluoric acid
HBr (g) = hydrogen bromide -> HBr (aq) = hydrobromic acid
HCl (g) = hydrogen chloride -> HCl (aq) = hydrochloric acid
H2S (g) = hydrogen sulfide -> H2S (aq) = hydrosulfuricacid
It is important to include (aq) after the acids because the same compounds can be written in gas phase with hydrogen named first followed by the anion ending with –ide.
Example 5
hypo____ite ____ite ____ate per____ate
ClO- ClO2- ClO3- ClO4-
hypochlorite chlorite chlorate perchlorate
---------------->
As indicated by the arrow, moving to the right, the following trends occur:
Increasing number of oxygen atoms
Increasing oxidation state of the nonmetal
(Usage of this example can be seen from the set of compounds containing Cl and O)
This occurs because the number of oxygen atoms are increasing from hypochlorite to perchlorate, yet the overall charge of the polyatomic ion is still -1. To correctly specify how many oxygen atoms are in the ion, prefixes and suffixes are again used.
Polyatomic Ions
In polyatomic ions, polyatomic (meaning two or more atoms) are joined together by covalent bonds. Although there may be a element with positive charge like H+, it is not joined with another element with an ionic bond. This occurs because if the atoms formed an ionic bond, then it would have already become a compound, thus not needing to gain or loose any electrons. Polyatomic anions are more common than polyatomic cations as shown in the chart below. Polyatomic anions have negative charges while polyatomic cations have positive charges. To indicate different polyatomic ions made up of the same elements, the name of the ion is modified according to the example below:
Table: Common Polyatomic ions
Name: Cation/Anion Formula
Ammonium ion NH4+
Hydronium ion
H3O+
Acetate ion
C2H3O2-
Arsenate ion
AsO43-
Carbonate ion
CO32-
Hypochlorite ion
ClO-
Chlorite ion
ClO2-
Chlorate ion
ClO3-
Perchlorate ion
ClO4-
Chromate ion
CrO42-
Dichromate ion
Cr2O72-
Cyanide ion
CN-
Hydroxide ion
OH-
Nitrite ion
NO2-
Nitrate ion
NO3-
Oxalate ion
C2O42-
Permanganate ion
MnO4-
Phosphate ion
PO43-
Sulfite ion
SO32-
Sulfate ion
SO42-
Thiocyanate ion
SCN-
Thiosulfate ion
S2O32-
To combine the topic of acids and polyatomic ions, there is nomenclature of aqueous acids. Such acids include sulfuric acid (H2SO4) or carbonic acid (H2CO3). To name them, follow these quick, simple rules:
1. If the ion ends in -ate and is added with an acid, the acid name will have an -ic ending. Examples: nitrate ion (NO3-) + H+ (denoting formation of acid) = nitric acid (HNO3)
2. If the ion ends in -ite and is added with an acid, then the acid name will have an -ous ending. Example: nitite ion (NO2-) + H+ (denoting formation of acid) = nitrous acid (HNO2)
Problems
1. What is the correct formula for Calcium Carbonate?
a. Ca+ + CO2-
b. CaCO2-
c. CaCO3
d. 2CaCO3
2. What is the correct name for FeO?
a. Iron oxide
b. Iron dioxide
c. Iron(III) oxide
d. Iron(II) oxide
3. What is the correct name for Al(NO3)3?
a. Aluminum nitrate
b. Aluminum(III) nitrate
c. Aluminum nitrite
d. Aluminum nitrogen trioxide
4. What is the correct formula of phosphorus trichloride?
a. P2Cl2
b. PCl3
c. PCl4
d. P4Cl2
5. What is the correct formula of lithium perchlorate?
a. Li2ClO4
b. LiClO2
c. LiClO
d. None of these
6. Write the correct name for these compounds.
a. BeC2O4:
b. NH4MnO4:
c. CoS2O3:
7. What is W(HSO4)5?
8. How do you write diphosphorus trioxide?
9. What is H3P?
10. By adding oxygens to the molecule in number 9, we now have H3PO4? What is the name of this molecule?
Answer
1.C; Calcium + Carbonate --> Ca2+ + CO32- --> CaCO3
2.D; FeO --> Fe + O2- --> Iron must have a charge of +2 to make a neutral compound --> Fe2+ + O2- --> Iron(II) Oxide
3.A; Al(NO3)3 --> Al3+ + (NO3-)3 --> Aluminum nitrate
4.B; Phosphorus trichloride --> P + 3Cl --> PCl3
5.D, LiClO4; Lithium perchlorate --> Li+ + ClO4- --> LiClO4
6. a. Beryllium Oxalate; BeC2O4 --> Be2+ + C2O42- --> Beryllium Oxalate
b. Ammonium Permanganate; NH4MnO4 --> NH4+ + MnO4- --> Ammonium Permanganate
c. Cobalt (II) Thiosulfate; CoS2O3 --> Co + S2O32- --> Cobalt must have +2 charge to make a neutral compund --> Co2+ + S2O32- --> Cobalt(II) Thiosulfate
7. Tungsten (V) hydrogen sulfate
8. P2O3
9. Hydrophosphoric Acid
10. Phosphoric Acid
Contributors and Attributions
• Pui Yan Ho (UCD), Alex Moskaluk (UCD), Emily Nguyen (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.08%3A_Naming_Ionic_Compounds.txt |
Generally, there are two types of inorganic compounds that can be formed: ionic compounds and molecular compounds. Nomenclature is the process of naming chemical compounds with different names so that they can be easily identified as separate chemicals. Inorganic compounds are compounds that do not deal with the formation of carbohydrates, or simply all other compounds that do not fit into the description of an organic compound. For example, organic compounds include molecules with carbon rings and/or chains with hydrogen atoms (see picture below). Inorganic compounds, the topic of this section, are every other molecule that does not include these distinctive carbon and hydrogen structures.
Compounds between Metals and Nonmetals (Cation and Anion)
Compounds made of a metal and nonmetal are commonly known as Ionic Compounds, where the compound name has an ending of –ide. Cations have positive charges while anions have negative charges. The net charge of any ionic compound must be zero which also means it must be electrically neutral. For example, one Na+ is paired with one Cl-; one Ca2+ is paired with two Br-. There are two rules that must be followed through:
• The cation (metal) is always named first with its name unchanged
• The anion (nonmetal) is written after the cation, modified to end in –ide
Table 1: Cations and Anions:
+1 Charge +2 Charge -1 Charge -2 Charge -3 Charge -4 Charge
Group 1A elements Group 2A elements Group 7A elements Group 6A elements Group 5A elements Group 4A elements
Hydrogen: H+ Beryllium: Be2+ Hydride: H- Oxide: O2- Nitride: N3- Carbide: C4-
Lithium: Li+ Magnesium: Mg2+ Fluoride: F- Sulfide: S2- Phosphide: P3-
Soduim: Na+ Calcium: Ca2+ Chloride: Cl-
Potassium: K+ Strontium: Sr2+ Bromide: Br-
Rubidium: Rb+ Barium: Ba2+ Iodide: I-
Cesium: Cs+
Example 1
Na+ + Cl- = NaCl; Ca2+ + 2Br- = CaBr2
Sodium + Chlorine = Sodium Chloride; Calcium + Bromine = Calcium Bromide
The transition metals may form more than one ion, thus it is needed to be specified which particular ion we are talking about. This is indicated by assigning a Roman numeral after the metal. The Roman numeral denotes the charge and the oxidation state of the transition metal ion. For example, iron can form two common ions, Fe2+ and Fe3+. To distinguish the difference, Fe2+ would be named iron (II) and Fe3+ would be named iron (III).
Table of Transition Metal and Metal Cations:
+1 Charge +2 Charge +3 Charge +4 Charge
Copper(I): Cu+ Copper(II): Cu2+ Aluminum: Al3+ Lead(IV): Pb4+
Silver: Ag+ Iron(II): Fe2+ Iron(III): Fe3+ Tin(IV): Sn4+
Cobalt(II): Co2+ Cobalt(III): Co3+
Tin(II): Sn2+
Lead(II): Pb2+
Nickel: Ni2+
Zinc: Zn2+
Example 2
Ions: Fe2++ 2Cl- Fe3++ 3Cl-
Compound: FeCl2 FeCl3
Nomenclature Iron (II) Chloride Iron (III) Chloride
However, some of the transition metals' charges have specific Latin names. Just like the other nomenclature rules, the ion of the transition metal that has the lower charge has the Latin name ending with -ous and the one with the the higher charge has a Latin name ending with -ic. The most common ones are shown in the table below:
Transition Metal Ion with Roman Numeral Latin name
Copper (I): Cu+ Cuprous
Copper (II): Cu2+ Cupric
Iron (II): Fe2+ Ferrous
Iron (III): Fe3+ Ferric
Lead (II): Pb2+ Plumbous
Lead (IV): Pb4+ Plumbic
Mercury (I): Hg22+ Mercurous
Mercury (II): Hg2+ Mercuric
Tin (II): Sn2+ Stannous
Tin (IV): Sn4+ Stannic
Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion. Instead of using Roman numerals, the different ions can also be presented in plain words. The metal is changed to end in –ous or –ic.
• -ous ending is used for the lower oxidation state
• -ic ending is used for the higher oxidation state
Example 3
Compound Cu2O CuO FeCl2 FeCl3
Charge Charge of copper is +1 Charge of copper is +2 Charge of iron is +2 Charge of iron is +3
Nomenclature Cuprous Oxide Cupric Oxide Ferrous Chloride Ferric Chloride
However, this -ous/-ic system is inadequate in some cases, so the Roman numeral system is preferred. This system is used commonly in naming acids, where H2SO4 is commonly known as Sulfuric Acid, and H2SO3 is known as Sulfurous Acid.
Compounds between Nonmetals and Nonmetals
Compounds that consist of a nonmetal bonded to a nonmetal are commonly known as Molecular Compounds, where the element with the positive oxidation state is written first. In many cases, nonmetals form more than one binary compound, so prefixes are used to distinguish them.
# of Atoms 1 2 3 4 5 6 7 8 9 10
Prefixes Mono- Di- Tri- Tetra- Penta- Hexa- Hepta- Octa- Nona- Deca-
Example 4
CO = carbon monoxide BCl3 = borontrichloride
CO2 = carbon dioxide N2O5 =dinitrogen pentoxide
The prefix mono- is not used for the first element. If there is not a prefix before the first element, it is assumed that there is only one atom of that element.
Binary Acids
Although HF can be named hydrogen fluoride, it is given a different name for emphasis that it is an acid. An acid is a substance that dissociates into hydrogen ions (H+) and anions in water. A quick way to identify acids is to see if there is an H (denoting hydrogen) in front of the molecular formula of the compound. To name acids, the prefix hydro- is placed in front of the nonmetal modified to end with –ic. The state of acids is aqueous (aq) because acids are found in water.
Some common binary acids include:
HF (g) = hydrogen fluoride -> HF (aq) = hydrofluoric acid
HBr (g) = hydrogen bromide -> HBr (aq) = hydrobromic acid
HCl (g) = hydrogen chloride -> HCl (aq) = hydrochloric acid
H2S (g) = hydrogen sulfide -> H2S (aq) = hydrosulfuricacid
It is important to include (aq) after the acids because the same compounds can be written in gas phase with hydrogen named first followed by the anion ending with –ide.
Example 5
hypo____ite ____ite ____ate per____ate
ClO- ClO2- ClO3- ClO4-
hypochlorite chlorite chlorate perchlorate
---------------->
As indicated by the arrow, moving to the right, the following trends occur:
Increasing number of oxygen atoms
Increasing oxidation state of the nonmetal
(Usage of this example can be seen from the set of compounds containing Cl and O)
This occurs because the number of oxygen atoms are increasing from hypochlorite to perchlorate, yet the overall charge of the polyatomic ion is still -1. To correctly specify how many oxygen atoms are in the ion, prefixes and suffixes are again used.
Polyatomic Ions
In polyatomic ions, polyatomic (meaning two or more atoms) are joined together by covalent bonds. Although there may be a element with positive charge like H+, it is not joined with another element with an ionic bond. This occurs because if the atoms formed an ionic bond, then it would have already become a compound, thus not needing to gain or loose any electrons. Polyatomic anions are more common than polyatomic cations as shown in the chart below. Polyatomic anions have negative charges while polyatomic cations have positive charges. To indicate different polyatomic ions made up of the same elements, the name of the ion is modified according to the example below:
Table: Common Polyatomic ions
Name: Cation/Anion Formula
Ammonium ion NH4+
Hydronium ion
H3O+
Acetate ion
C2H3O2-
Arsenate ion
AsO43-
Carbonate ion
CO32-
Hypochlorite ion
ClO-
Chlorite ion
ClO2-
Chlorate ion
ClO3-
Perchlorate ion
ClO4-
Chromate ion
CrO42-
Dichromate ion
Cr2O72-
Cyanide ion
CN-
Hydroxide ion
OH-
Nitrite ion
NO2-
Nitrate ion
NO3-
Oxalate ion
C2O42-
Permanganate ion
MnO4-
Phosphate ion
PO43-
Sulfite ion
SO32-
Sulfate ion
SO42-
Thiocyanate ion
SCN-
Thiosulfate ion
S2O32-
To combine the topic of acids and polyatomic ions, there is nomenclature of aqueous acids. Such acids include sulfuric acid (H2SO4) or carbonic acid (H2CO3). To name them, follow these quick, simple rules:
1. If the ion ends in -ate and is added with an acid, the acid name will have an -ic ending. Examples: nitrate ion (NO3-) + H+ (denoting formation of acid) = nitric acid (HNO3)
2. If the ion ends in -ite and is added with an acid, then the acid name will have an -ous ending. Example: nitite ion (NO2-) + H+ (denoting formation of acid) = nitrous acid (HNO2)
Problems
1. What is the correct formula for Calcium Carbonate?
a. Ca+ + CO2-
b. CaCO2-
c. CaCO3
d. 2CaCO3
2. What is the correct name for FeO?
a. Iron oxide
b. Iron dioxide
c. Iron(III) oxide
d. Iron(II) oxide
3. What is the correct name for Al(NO3)3?
a. Aluminum nitrate
b. Aluminum(III) nitrate
c. Aluminum nitrite
d. Aluminum nitrogen trioxide
4. What is the correct formula of phosphorus trichloride?
a. P2Cl2
b. PCl3
c. PCl4
d. P4Cl2
5. What is the correct formula of lithium perchlorate?
a. Li2ClO4
b. LiClO2
c. LiClO
d. None of these
6. Write the correct name for these compounds.
a. BeC2O4:
b. NH4MnO4:
c. CoS2O3:
7. What is W(HSO4)5?
8. How do you write diphosphorus trioxide?
9. What is H3P?
10. By adding oxygens to the molecule in number 9, we now have H3PO4? What is the name of this molecule?
Answer
1.C; Calcium + Carbonate --> Ca2+ + CO32- --> CaCO3
2.D; FeO --> Fe + O2- --> Iron must have a charge of +2 to make a neutral compound --> Fe2+ + O2- --> Iron(II) Oxide
3.A; Al(NO3)3 --> Al3+ + (NO3-)3 --> Aluminum nitrate
4.B; Phosphorus trichloride --> P + 3Cl --> PCl3
5.D, LiClO4; Lithium perchlorate --> Li+ + ClO4- --> LiClO4
6. a. Beryllium Oxalate; BeC2O4 --> Be2+ + C2O42- --> Beryllium Oxalate
b. Ammonium Permanganate; NH4MnO4 --> NH4+ + MnO4- --> Ammonium Permanganate
c. Cobalt (II) Thiosulfate; CoS2O3 --> Co + S2O32- --> Cobalt must have +2 charge to make a neutral compund --> Co2+ + S2O32- --> Cobalt(II) Thiosulfate
7. Tungsten (V) hydrogen sulfate
8. P2O3
9. Hydrophosphoric Acid
10. Phosphoric Acid
Contributors and Attributions
• Pui Yan Ho (UCD), Alex Moskaluk (UCD), Emily Nguyen (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.10%3A_Naming_Binary_Nonmetal_Compounds.txt |
The number of moles in a system can be determined using the atomic mass of an element, which can be found on the periodic table. This mass is usually an average of the abundant forms of that element found on earth. An element's mass is listed as the average of all its isotopes on earth.
Avogadro's Constant
One mole of oxygen atoms contains $6.02214179 \times 10^{23}$ oxygen atoms. Also, one mole of nitrogen atoms contains $6.02214179 \times 10^{23}$ nitrogen atoms. The number $6.02214179 \times 10^{23}$ is called Avogadro's number ($N_A$) or Avogadro's constant, after the 19th century scientist Amedeo Avogadro.
Each carbon-12 atom weighs about $1.99265 \times 10^{-23}\; g$; therefore,
$(1.99265 \times 10^{-23}\; g) \times (6.02214179 \times 10^{23}\; atoms) = 12\; g\; \text{ of carbon-12} \nonumber$
Applications of the Mole
The mass of a mole of substance is called the molar mass of that substance. The molar mass is used to convert grams of a substance to moles and is used often in chemistry. The molar mass of an element is found on the periodic table, and it is the element's atomic weight in grams/mole (g/mol). If the mass of a substance is known, the number of moles in the substance can be calculated. Converting the mass, in grams, of a substance to moles requires a conversion factor of (one mole of substance/molar mass of substance).
The mole concept is also applicable to the composition of chemical compounds. For instance, consider methane, CH4. This molecule and its molecular formula indicate that per mole of methane there is 1 mole of carbon and 4 moles of hydrogen. In this case, the mole is used as a common unit that can be applied to a ratio as shown below:
$2 \text{ mol H } + 1 \text{ mol O }= 1 \text{ mol } \ce{H2O} \nonumber$
In this this chemical reactions, the moles of H and O describe the number of atoms of each element that react to form 1 mol of $\ce{H_2O}$.
To think about what a mole means, one should relate it to quantities such as dozen or pair. Just as a pair can mean two shoes, two books, two pencils, two people, or two of anything else, a mole means 6.02214179×1023 of anything. Using the following relation:
$\text{1 mole} = 6.02214179 \times 10^{23}$
is analogous to saying:
$\text{1 Dozen} = \text{12 eggs}$
It is quite difficult to visualize a mole of something because Avogadro's constant is extremely large. For instance, consider the size of one single grain of wheat. If all the people who have existed in Earth's history did nothing but count individual wheat grains for their entire lives, the total number of wheat grains counted would still be much less than Avogadro's constant; the number of wheat grains produced throughout history does not even approach Avogadro's Number.
Example $1$: Converting Mass to Moles
How many moles of potassium ($\ce{K}$) atoms are in 3.04 grams of pure potassium metal?
Solution
In this example, multiply the mass of $\ce{K}$ by the conversion factor (inverse molar mass of potassium):
$\dfrac{1\; mol\; K}{39.10\; grams \;K} \nonumber$
39.10 grams is the molar mass of one mole of $\ce{K}$; cancel out grams, leaving the moles of $\ce{K}$:
$3.04\; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0778\; mol\; K \nonumber$
Similarly, if the moles of a substance are known, the number grams in the substance can be determined. Converting moles of a substance to grams requires a conversion factor of molar mass of substance/one mole of substance. One simply needs to follow the same method but in the opposite direction.
Example $2$: Converting Moles to mass
How many grams are 10.78 moles of Calcium ($\ce{Ca}$)?
Solution
Multiply moles of Ca by the conversion factor (molar mass of calcium) 40.08 g Ca/ 1 mol Ca, which then allows the cancelation of moles, leaving grams of Ca.
$10.78 \cancel{\;mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\; \cancel{mol\; Ca}}\right) = 432.1\; g\; Ca \nonumber$
The total number of atoms in a substance can also be determined by using the relationship between grams, moles, and atoms. If given the mass of a substance and asked to find the number of atoms in the substance, one must first convert the mass of the substance, in grams, to moles, as in Example $1$. Then the number of moles of the substance must be converted to atoms. Converting moles of a substance to atoms requires a conversion factor of Avogadro's constant (6.02214179×1023) / one mole of substance. Verifying that the units cancel properly is a good way to make sure the correct method is used.
Example $3$: Atoms to Mass
How many atoms are in a 3.5 g sample of sodium (Na)?
Solution
$3.5\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.152\; mol\; Na \nonumber$
$0.152\; \cancel{mol\; Na} \left(\dfrac{6.02214179\times 10^{23}\; atoms\; Na}{1\;\cancel{ mol\; Na}}\right) = 9.15 \times 10^{22}\; atoms\; of\; Na \nonumber$
In this example, multiply the grams of Na by the conversion factor 1 mol Na/ 22.98 g Na, with 22.98g being the molar mass of one mole of Na, which then allows cancelation of grams, leaving moles of Na. Then, multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.
Using Avogadro's constant, it is also easy to calculate the number of atoms or molecules present in a substance (Table $1$). By multiplying the number of moles by Avogadro's constant, the mol units cancel out, leaving the number of atoms. The following table provides a reference for the ways in which these various quantities can be manipulated:
Table $1$: Conversion Factors
Known Information Multiply By Result
Mass of substance (g) 1/ Molar mass (mol/g) Moles of substance
Moles of substance (mol) Avogadro's constant (atoms/mol) Atoms (or molecules)
Mass of substance (g) 1/Molar mass (mol/g) × Avogadro's constant (atoms/mol)) Atoms (or molecules)
Example $4$: Mass to Moles
How many moles are in 3.00 grams of potassium (K)?
Solution
$3.00 \; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0767\; mol\; K \nonumber$
In this example, multiply the mass of K by the conversion factor:
$\dfrac{1\; mol\; K}{39.10\; grams\; K} \nonumber$
39.10 grams is the molar mass of one mole of K. Grams can be canceled, leaving the moles of K.
Example $5$: Moles to Mass
How many grams is in 10.00 moles of calcium (Ca)?
Solution
This is the calculation in Example $2$ performed in reverse. Multiply moles of Ca by the conversion factor 40.08 g Ca/ 1 mol Ca, with 40.08 g being the molar mass of one mole of Ca. The moles cancel, leaving grams of Ca:
$10.00\; \cancel{mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\;\cancel{ mol\; Ca}}\right) = 400.8\; grams \;of \;Ca \nonumber$
The number of atoms can also be calculated using Avogadro's Constant (6.02214179×1023) / one mole of substance.
Example $6$: Mass to Atoms
How many atoms are in a 3.0 g sample of sodium (Na)?
Solution
Convert grams to moles
$3.0\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.130\; mol\; Na \nonumber$
Convert moles to atoms
$0.130548\; \cancel{ mol\; Na} \left(\dfrac{6.02214179 \times 10^{23}\; atoms \;Na}{1\; \cancel{ mol\; Na}}\right) = 7.8 \times 10^{22} \; atoms\; of\; \; Na \nonumber$
Summary
The mole, abbreviated mol, is an SI unit which measures the number of particles in a specific substance. One mole is equal to $6.02214179 \times 10^{23}$ atoms, or other elementary units such as molecules.
Problems
1. Using a periodic table, give the molar mass of the following:
1. H
2. Se
3. Ne
4. Cs
5. Fe
2. Convert to moles and find the total number of atoms.
1. 5.06 grams of oxygen
2. 2.14 grams of K
3. 0.134 kg of Li
3. Convert the following to grams
1. 4.5 mols of C
2. 7.1 mols of Al
3. 2.2 mols of Mg
4. How many moles are in the product of the reaction
1. 6 mol H + 3 mol O → ? mol H2O
2. 1 mol Cl + 1 mol Cl → ? mol Cl2
3. 5 mol Na + 4 mol Cl → ? mol NaCl
Answers
1. Question 2
1. 1.008 g/mol
2. 78.96 g/mol
3. 20.18 g/mol
4. 132.91g/mol
5. 55.85 g/mol
2. Question 2
2. 5.06g O (1mol/16.00g)= 0.316 mol of O
0.316 mols (6.022x1023 atoms/ 1mol) = 1.904x1023 atoms of O
3. 2.14g K (1mol/39.10g)= 0.055 mol of K
0.055 mols (6.022x1023 atoms/ 1mol) = 3.312x1022 atoms of K
4. 0.134kg Li (1000g/1kg)= 134g Li (1mol/6.941g)= 19.3 mols Li
19.3 (6.022x1023 atoms/ 1mol) = 1.16x1025 atoms of Li
1. Question 3
1. 4.5 mols of C (12.011g/1mol) = 54.05 g of C
2. 7.1 mols of Al (26.98g/1mol) = 191.56 g of Al
3. 2.2 mols of Mg (24.31g/1mol) = 53.48 g of MG
2. Question 4
1. 8. 6 mol H + 3 mol O → 3 mol H2O
2. 9. 1 mol Cl + 1 mol Cl → 1 mol Cl2
3. 10. 5 mol Na + 4 mol Cl → 4 mol NaCl + 1 mol Na (excess) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.11%3A_Atoms_and_the_Mole.txt |
The number of moles in a system can be determined using the atomic mass of an element, which can be found on the periodic table. This mass is usually an average of the abundant forms of that element found on earth. An element's mass is listed as the average of all its isotopes on earth.
Avogadro's Constant
One mole of oxygen atoms contains $6.02214179 \times 10^{23}$ oxygen atoms. Also, one mole of nitrogen atoms contains $6.02214179 \times 10^{23}$ nitrogen atoms. The number $6.02214179 \times 10^{23}$ is called Avogadro's number ($N_A$) or Avogadro's constant, after the 19th century scientist Amedeo Avogadro.
Each carbon-12 atom weighs about $1.99265 \times 10^{-23}\; g$; therefore,
$(1.99265 \times 10^{-23}\; g) \times (6.02214179 \times 10^{23}\; atoms) = 12\; g\; \text{ of carbon-12} \nonumber$
Applications of the Mole
The mass of a mole of substance is called the molar mass of that substance. The molar mass is used to convert grams of a substance to moles and is used often in chemistry. The molar mass of an element is found on the periodic table, and it is the element's atomic weight in grams/mole (g/mol). If the mass of a substance is known, the number of moles in the substance can be calculated. Converting the mass, in grams, of a substance to moles requires a conversion factor of (one mole of substance/molar mass of substance).
The mole concept is also applicable to the composition of chemical compounds. For instance, consider methane, CH4. This molecule and its molecular formula indicate that per mole of methane there is 1 mole of carbon and 4 moles of hydrogen. In this case, the mole is used as a common unit that can be applied to a ratio as shown below:
$2 \text{ mol H } + 1 \text{ mol O }= 1 \text{ mol } \ce{H2O} \nonumber$
In this this chemical reactions, the moles of H and O describe the number of atoms of each element that react to form 1 mol of $\ce{H_2O}$.
To think about what a mole means, one should relate it to quantities such as dozen or pair. Just as a pair can mean two shoes, two books, two pencils, two people, or two of anything else, a mole means 6.02214179×1023 of anything. Using the following relation:
$\text{1 mole} = 6.02214179 \times 10^{23}$
is analogous to saying:
$\text{1 Dozen} = \text{12 eggs}$
It is quite difficult to visualize a mole of something because Avogadro's constant is extremely large. For instance, consider the size of one single grain of wheat. If all the people who have existed in Earth's history did nothing but count individual wheat grains for their entire lives, the total number of wheat grains counted would still be much less than Avogadro's constant; the number of wheat grains produced throughout history does not even approach Avogadro's Number.
Example $1$: Converting Mass to Moles
How many moles of potassium ($\ce{K}$) atoms are in 3.04 grams of pure potassium metal?
Solution
In this example, multiply the mass of $\ce{K}$ by the conversion factor (inverse molar mass of potassium):
$\dfrac{1\; mol\; K}{39.10\; grams \;K} \nonumber$
39.10 grams is the molar mass of one mole of $\ce{K}$; cancel out grams, leaving the moles of $\ce{K}$:
$3.04\; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0778\; mol\; K \nonumber$
Similarly, if the moles of a substance are known, the number grams in the substance can be determined. Converting moles of a substance to grams requires a conversion factor of molar mass of substance/one mole of substance. One simply needs to follow the same method but in the opposite direction.
Example $2$: Converting Moles to mass
How many grams are 10.78 moles of Calcium ($\ce{Ca}$)?
Solution
Multiply moles of Ca by the conversion factor (molar mass of calcium) 40.08 g Ca/ 1 mol Ca, which then allows the cancelation of moles, leaving grams of Ca.
$10.78 \cancel{\;mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\; \cancel{mol\; Ca}}\right) = 432.1\; g\; Ca \nonumber$
The total number of atoms in a substance can also be determined by using the relationship between grams, moles, and atoms. If given the mass of a substance and asked to find the number of atoms in the substance, one must first convert the mass of the substance, in grams, to moles, as in Example $1$. Then the number of moles of the substance must be converted to atoms. Converting moles of a substance to atoms requires a conversion factor of Avogadro's constant (6.02214179×1023) / one mole of substance. Verifying that the units cancel properly is a good way to make sure the correct method is used.
Example $3$: Atoms to Mass
How many atoms are in a 3.5 g sample of sodium (Na)?
Solution
$3.5\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.152\; mol\; Na \nonumber$
$0.152\; \cancel{mol\; Na} \left(\dfrac{6.02214179\times 10^{23}\; atoms\; Na}{1\;\cancel{ mol\; Na}}\right) = 9.15 \times 10^{22}\; atoms\; of\; Na \nonumber$
In this example, multiply the grams of Na by the conversion factor 1 mol Na/ 22.98 g Na, with 22.98g being the molar mass of one mole of Na, which then allows cancelation of grams, leaving moles of Na. Then, multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.
Using Avogadro's constant, it is also easy to calculate the number of atoms or molecules present in a substance (Table $1$). By multiplying the number of moles by Avogadro's constant, the mol units cancel out, leaving the number of atoms. The following table provides a reference for the ways in which these various quantities can be manipulated:
Table $1$: Conversion Factors
Known Information Multiply By Result
Mass of substance (g) 1/ Molar mass (mol/g) Moles of substance
Moles of substance (mol) Avogadro's constant (atoms/mol) Atoms (or molecules)
Mass of substance (g) 1/Molar mass (mol/g) × Avogadro's constant (atoms/mol)) Atoms (or molecules)
Example $4$: Mass to Moles
How many moles are in 3.00 grams of potassium (K)?
Solution
$3.00 \; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0767\; mol\; K \nonumber$
In this example, multiply the mass of K by the conversion factor:
$\dfrac{1\; mol\; K}{39.10\; grams\; K} \nonumber$
39.10 grams is the molar mass of one mole of K. Grams can be canceled, leaving the moles of K.
Example $5$: Moles to Mass
How many grams is in 10.00 moles of calcium (Ca)?
Solution
This is the calculation in Example $2$ performed in reverse. Multiply moles of Ca by the conversion factor 40.08 g Ca/ 1 mol Ca, with 40.08 g being the molar mass of one mole of Ca. The moles cancel, leaving grams of Ca:
$10.00\; \cancel{mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\;\cancel{ mol\; Ca}}\right) = 400.8\; grams \;of \;Ca \nonumber$
The number of atoms can also be calculated using Avogadro's Constant (6.02214179×1023) / one mole of substance.
Example $6$: Mass to Atoms
How many atoms are in a 3.0 g sample of sodium (Na)?
Solution
Convert grams to moles
$3.0\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.130\; mol\; Na \nonumber$
Convert moles to atoms
$0.130548\; \cancel{ mol\; Na} \left(\dfrac{6.02214179 \times 10^{23}\; atoms \;Na}{1\; \cancel{ mol\; Na}}\right) = 7.8 \times 10^{22} \; atoms\; of\; \; Na \nonumber$
Summary
The mole, abbreviated mol, is an SI unit which measures the number of particles in a specific substance. One mole is equal to $6.02214179 \times 10^{23}$ atoms, or other elementary units such as molecules.
Problems
1. Using a periodic table, give the molar mass of the following:
1. H
2. Se
3. Ne
4. Cs
5. Fe
2. Convert to moles and find the total number of atoms.
1. 5.06 grams of oxygen
2. 2.14 grams of K
3. 0.134 kg of Li
3. Convert the following to grams
1. 4.5 mols of C
2. 7.1 mols of Al
3. 2.2 mols of Mg
4. How many moles are in the product of the reaction
1. 6 mol H + 3 mol O → ? mol H2O
2. 1 mol Cl + 1 mol Cl → ? mol Cl2
3. 5 mol Na + 4 mol Cl → ? mol NaCl
Answers
1. Question 2
1. 1.008 g/mol
2. 78.96 g/mol
3. 20.18 g/mol
4. 132.91g/mol
5. 55.85 g/mol
2. Question 2
2. 5.06g O (1mol/16.00g)= 0.316 mol of O
0.316 mols (6.022x1023 atoms/ 1mol) = 1.904x1023 atoms of O
3. 2.14g K (1mol/39.10g)= 0.055 mol of K
0.055 mols (6.022x1023 atoms/ 1mol) = 3.312x1022 atoms of K
4. 0.134kg Li (1000g/1kg)= 134g Li (1mol/6.941g)= 19.3 mols Li
19.3 (6.022x1023 atoms/ 1mol) = 1.16x1025 atoms of Li
1. Question 3
1. 4.5 mols of C (12.011g/1mol) = 54.05 g of C
2. 7.1 mols of Al (26.98g/1mol) = 191.56 g of Al
3. 2.2 mols of Mg (24.31g/1mol) = 53.48 g of MG
2. Question 4
1. 8. 6 mol H + 3 mol O → 3 mol H2O
2. 9. 1 mol Cl + 1 mol Cl → 1 mol Cl2
3. 10. 5 mol Na + 4 mol Cl → 4 mol NaCl + 1 mol Na (excess) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.12%3A_Molecules_Compounds_and_the_Mole.txt |
Which one weighs more, a kilogram of feathers or a kilogram of bricks? Though many people will say that a kilogram of bricks is heavier, they actually weigh the same! However, many people are caught up by the concept of density, which causes them to answer the question incorrectly. A kilogram of feathers clearly takes up more space, but this is because it is less "dense." But what is density, and how can we determine it?
Introduction
Density ($\rho$) is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary. The equation is as follows:
$Density = \dfrac{Mass}{Volume} \nonumber$
or just
$\rho = \dfrac{m}{v} \label{dens}$
Based on Equation $\ref{dens}$, it's clear that density can, and does, vary from element to element and substance to substance due to differences in the relation of mass and volume. Let us break it down one step further. What are mass and volume? We cannot understand density until we know its parts: mass and volume. The following two sections will teach you all the information you need to know about volume and mass to properly solve and manipulate the density equation.
Mass
Mass concerns the quantity of matter in an object. The SI unit for mass is the kilogram (kg), although grams (g) are commonly used in the laboratory to measure smaller quantities. Often, people mistake weight for mass. Weight concerns the force exerted on an object as a function of mass and gravity. This can be written as
$\text{Weight} = \text{mass} \times \text{gravity} \nonumber$
$Weight = {m}{g}$
Hence, weight changes due to variations in gravity and acceleration. For example, the mass of a 1 kg cube will continue to be 1 kg whether it is on the top of a mountain, the bottom of the sea, or on the moon, but its weight will differ. Another important difference between mass and weight is how they are measured. Weight is measured with a scale, while mass must be measured with a balance. Just as people confuse mass and weight, they also confuse scales and balances. A balance counteracts the effects of gravity while a scale incorporates it. There are two types of balances found in the laboratory: electronic and manual. With a manual balance, you find the unknown mass of an object by adjusting or comparing known masses until equilibrium is reached.
Volume
Volume describes the quantity of three dimensional space than an object occupies. The SI unit for volume is meters cubed (m3), but milliliters (mL), centimeters cubed (cm3), and liters (L) are more common in the laboratory. There are many equations to find volume. Here are just a few of the easy ones:
Volume = (length)3 or (length)(width)(height) or (base area)(height)
Density: A Further Investigation
We know all of density's components, so let's take a closer look at density itself. The unit most widely used to express density is g/cm3 or g/mL, though the SI unit for density is technically kg/m3. Grams per centimeter cubed is equivalent to grams per milliliter (g/cm3 = g/mL). To solve for density, simply follow the equation d = m/v. For example, if you had a metal cube with mass 7.0 g and volume 5.0 cm3, the density would be
$\rho = \dfrac{7\,g}{5\,cm^3}= 1.4\, g/cm^3 \nonumber$
Sometimes, you have to convert units to get the correct units for density, such as mg to g or in3 to cm3.
Density can be used to help identify an unknown element. Of course, you have to know the density of an element with respect to other elements. Below is a table listing the density of a few elements from the Periodic Table at standard conditions for temperature and pressure, or STP corresponding to a temperature of 273 K (0° Celsius) and 1 atmosphere of pressure.
Element Name and Symbol Density (g/cm3) Atomic Number
Table $1$: Density of elements
Hydrogen (H) 0.000089 $(8.9 \times 10^{-5})$ at 0 °C and 1 Atm. pressumre 1
Helium (He) 0.000164 $(1.64 \times 10^{-4})$ at 0 °C and 1 Atm. pressure 2
Aluminum (Al) 2.7 13
Zinc (Zn) 7.13 30
Tin (Sn) 7.31 50
Iron (Fe) 7.87 26
Nickel (Ni) 8.9 28
Cobalt (Co) 8.9 27
Copper (Cu) 8.96 29
Silver (Ag) 10.5 47
Lead (Pb) 11.35 82
Mercury (Hg) 11.55 80
Gold (Au) 19.32 79
Platinum (Pt) 21.45 78
Osmium (Os) 22.6 76
As can be seen from the table, the most dense element is Osmium (Os) with a density of 22.6 g/cm3. The least dense element is Hydrogen (H) with a density of 0.09 g/cm3.
Density and Temperature
Density generally decreases with increasing temperature and likewise increases with decreasing temperatures. This is because volume differs according to temperature. Volume increases with increasing temperature. If you are curious as to why the density of a pure substance could vary with temperature, check out the ChemWiki page on Van Der Waal interactions. Below is a table showing the density of pure water with differing temperatures.
Temperature (C) Density (g/cm3)
Table $2$: Density of water as a function of temperature
100 0.9584
80 0.9718
60 0.9832
40 0.9922
30 0.9957
25 0.997
22 0.9978
20 0.9982
15 0.9991
10 0.9997
4 1.000
0 (liquid) .9998
0 (solid) 0.9150
As can be seen from Table $2$, the density of water decreases with increasing temperature. Liquid water also shows an exception to this rule from 0 degrees Celsius to 4 degrees Celsius, where it increases in density instead of decreasing as expected. Looking at the table, you can also see that ice is less dense than water. This is unusual as solids are generally denser than their liquid counterparts. Ice is less dense than water due to hydrogen bonding. In the water molecule, the hydrogen bonds are strong and compact. As the water freezes into the hexagonal crystals of ice, these hydrogen bonds are forced farther apart and the volume increases. With this volume increase comes a decrease in density. This explains why ice floats to the top of a cup of water: the ice is less dense.
Even though the rule of density and temperature has its exceptions, it is still useful. For example, it explains how hot air balloons work.
Density and Pressure
Density increases with increasing pressure because volume decreases as pressure increases. And since density=mass/volume , the lower the volume, the higher the density. This is why all density values in the Periodic Table are recorded at STP, as mentioned in the section "Density and the Periodic Table." The decrease in volume as related to pressure is explained in Boyle's Law: $P_1V_1 = P_2V_2$ where P = pressure and V = volume. This idea is explained in the figure below. More about Boyle's Law, as well as the other gas laws, can be found here.
Archimedes' Principle
The Greek scientist Archimedes made a significant discovery in 212 B.C. The story goes that Archimedes was asked to find out for the King if his goldsmith was cheating him by replacing his gold for the crown with silver, a cheaper metal. Archimedes did not know how to find the volume of an irregularly shaped object such as the crown, even though he knew he could distinguish between elements by their density. While meditating on this puzzle in a bath, Archimedes recognized that when he entered the bath, the water rose. He then realized that he could use a similar process to determine the density of the crown! He then supposedly ran through the streets naked shouting "Eureka," which means "I found it!" in Latin.
Archimedes then tested the king's crown by taking a genuine gold crown of equal mass and comparing the densities of the two. The king's crown displaced more water than the gold crown of the same mass, meaning that the king's crown had a greater volume and thus had a smaller density than the real gold crown. The king's "gold" crown, therefore, was not made of pure gold. Of course, this tale is disputed today because Archimedes was not precise in all his measurements, which would make it hard to determine accurately the differences between the two crowns.
Archimedes' Principle states that if an object has a greater density than the liquid that it is placed into, it will sink and displace a volume of liquid equal to its own. If it has a smaller density, it will float and displace a mass of liquid equal to its own. If the density is equal, it will not sink or float. This principle also explains why balloons filled with helium float. Balloons, as we learned in the section concerning density and temperature, float because they are less dense than the surrounding air. Helium is less dense than the atmospheric air, so it rises. Archimedes' Principle can also be used to explain why boats float. Boats, including all the air space, within their hulls, are far less dense than water. Boats made of steel can float because they displace their mass in water without submerging all the way.
Table $3$ below gives the densities of a few liquids to put things into perspective.
Liquid
Density in kg/m3
Density in g/cm3
Table $3$: Density of select liquids
2-Methoxyethanol
964.60
0.9646
Acetic Acid
1049.10
1.049
Acetone
789.86
0.7898
Alcohol, ethyl
785.06
0.7851
Alcohol, methyl
786.51
0.7865
Ammonia
823.35
0.8234
Benzene
873.81
0.8738
Water, pure
1000.00
1.000
Percent Composition
Percent composition is very simple. Percent composition tells you by mass what percent of each element is present in a compound. A chemical compound is the combination of two or more elements. If you are studying a chemical compound, you may want to find the percent composition of a certain element within that chemical compound. The equation for percent composition is (mass of element/molecular mass) x 100.
Steps to calculating the percent composition of the elements in an compound
1. Find the molar mass of all the elements in the compound in grams per mole.
2. Find the molecular mass of the entire compound.
3. Divide the component's molar mass by the entire molecular mass.
4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.
Tips for solving:
1. The percent composition of all elements in a compounds must add up to 100%. In a binary compound, you can find the % of the first element, then do 100%-(% first element) to get (% second element)
2. If using a calculator, you can store the overall molar mass to a variable such as "A". This will speed up calculations, and reduce errors.
Example $1$: Phosphorus Pentachloride
What is the percent composition of phosphorus and chlorine in $PCl_5$?
Solution
Find the molar mass of all the elements in the compound in grams per mole.
• $P$: $1 \times 30.975 \,g/mol = 30.75\, g/mol$
• $Cl$: $5 \times 35.453 \, g/mol = 177.265\, g/mol$
Find the molecular mass of the entire compound.
• $PCl_5$: $1 \times 30.975 \,g/mol + 5 \times 35.453 \, g/mol = 208.239 \, g/mol$
Divide the component's molar mass by the entire molecular mass.
• $P$: $\dfrac{30.75 \, g/mol}{208.239\, g/mol} \times 100\% = 14.87\%$
• $Cl$: $\dfrac{177.265 \, g/mol}{208.239\, g/mol} \times 100\% = 85.13 \%$
Therefore, in $PCl_5$ is 14.87% phosphorus and 85.13% chlorine by mass.
Example $2$: HCl
What is the percent composition of each element in hydrochloric acid (HCl).
Solution
First find the molar mass of hydrogen.
$H = 1.00794 \,g \nonumber$
Now find the molecular mass of the HCl molecule:
$1.00794\,g + 35.4527\,g = 36.46064\,g \nonumber$
Follow steps 3 and 4:
$\left(\dfrac{1.00794\,g}{36.46064\,g}\right) \times 100\% = 2.76\% \nonumber$
Now just subtract to find the percent by mass of chlorine in the compound:
$100\%-2.76\% = 97.24\% \nonumber$
Therefore, $HCl$ is 2.76% hydrogen and 97.24% chlorine by mass.
Percent Composition in Everyday Life
Percent composition plays an important role in everyday life. It is more than just the amount of chlorine in your swimming pool because it concerns everything from the money in your pocket to your health and how you live. The next two sections describe percent composition as it relates to you.
Nutrition Labels
The nutrition label found on the container of every bit of processed food sold by the local grocery store employs the idea of percent composition. On all nutrition labels, a known serving size is broken down in five categories: Total Fat, Cholesterol, Sodium, Total Carbohydrate, and Protein. These categories are broken down into further subcategories, including Saturated Fat and Dietary Fiber. The mass for each category, except Protein, is then converted to percent of Daily Value. Only two subcategories, Saturated Fat and Dietary Fiber are converted to percent of Daily Value. The Daily Value is based on a the mass of each category recommended per day per person for a 2000 calorie diet. The mass of protein is not converted to percent because their is no recommended daily value for protein. Following is a picture outlining these ideas.
For example, if you wanted to know the percent by mass of the daily value for sodium you are eating when you eat one serving of the food with this nutrition label, then go to the category marked sodium. Look across the same row and read the percent written. If you eat one serving of this food, then you will have consumed about 9% of your daily recommended value for sodium. To find the percent mass of fat in the whole food, you could divide 3.5 grams by 15 grams, and see that this snack is 23.33% fat.
Penny: The Lucky Copper Coin
The penny should be called "the lucky copper coated coin." The penny has not been made of solid copper since part of 1857. After 1857, the US government started adding other cheaper metals to the mix. The penny, being only one cent, is literally not worth its weight in copper. People could melt copper pennies and sell the copper for more than the pennies were worth. After 1857, nickel was mixed with the more expensive copper. After 1864, the penny was made of bronze. Bronze is 95% copper and 5% zinc and tin. For one year, 1943, the penny had no copper in it due to the expense of the World War II. It was just zinc coated steel. After 1943 until 1982, the penny went through periods where it was brass or bronze.
Today, the penny in America is 2.5% copper with 97.5% zinc. The copper coats the outside of the penny while the inner portion is zinc. For comparison's sake, the penny in Canada is 94% steel, 1.5% nickel, and 4.5% copper.
The percent composition of a penny may actually affect health, particularly the health of small children and pets. Since the newer pennies are made mainly of zinc instead of copper, they are a danger to a child's health if ingested. Zinc is very susceptible to acid. If the thin copper coating is scratched and the hydrochloric acid present in the stomach comes into contact with the zinc core it could cause ulcers, anemia, kidney and liver damage, or even death in severe cases. Three important factors in penny ingestion are time, pH of the stomach, and amount of pennies ingested. Of course, the more pennies swallowed, the more danger of an overdose of zinc. The more acidic the environment, the more zinc will be released in less time. This zinc is then absorbed and sent to the liver where it begins to cause damage. In this kind of situation, time is of the essence. The faster the penny is removed, the less zinc is absorbed. If the penny or pennies are not removed, organ failure and death can occur.
Below is a picture of a scratched penny before and after it had been submerged in lemon juice. Lemon juice has a similar pH of 1.5-2.5 when compared to the normal human stomach after food has been consumed. Time elapsed: 36 hours.
As you can see, the copper is vastly unharmed by the lemon juice. That's why pennies made before 1982 with mainly copper (except the 1943 penny) are relatively safe to swallow. Chances are they would pass through the digestive system naturally before any damage could be done. Yet, it is clear that the zinc was partially dissolved even though it was in the lemon juice for only a limited amount of time. Therefore, the percent composition of post 1982 pennies is hazardous to your health and the health of your pets if ingested.
Summary
Density and percent composition are important concepts in chemistry. Each have basic components as well as broad applications. Components of density are: mass and volume, both of which can be more confusing than at first glance. An application of the concept of density is determining the volume of an irregular shape using a known mass and density. Determining Percent Composition requires knowing the mass of entire object or molecule and the mass of its components. In the laboratory, density can be used to identify an element, while percent composition is used to determine the amount, by mass, of each element present in a chemical compound. In daily life, density explains everything from why boats float to why air bubbles will try to escape from soda. It even affects your health because bone density is very important. Similarly, percent composition is commonly used to make animal feed and compounds such as the baking soda found in your kitchen.
Density Problems
These problems are meant to be easy in the beginning and then gradually become more challenging. Unless otherwise stated, answers should be in g/mL or the equivalent g/cm3.
1. If you have a 2.130 mL sample of acetic acid with mass .002234 kg, what is the density?
2. Calculate the density of a .03020 L sample of ethyl alcohol with a mass of 23.71002 g.
3. Find the density of a sample that has a volume of 36.5 L and a mass of 10.0 kg.
4. Find the volume in mL of an object that has a density of 10.2 g/L and a mass of 30.0 kg.
5. Calculate the mass in grams of an object with a volume of 23.5 mL and density of 10.0 g/L.
6. Calculate the density of a rectangular prism made of metal. The dimensions of the prism are: 5cm by 4cm by 5cm. The metal has a mass of 50 grams.
7. Find the denstiy of an unknown liquid in a beaker. The beaker's mass is 165g when there is no liquid present. With the unknown liquid, the total mass is 309g. The volume of the unknown is 125mL.
8. Determine the density in g/L of an unknown with the following information. A 55 gallon tub weighs 137.5lb when empty and 500.0 lb when filled with the unknown.
9. A ring has a mass of 5.00g and a volume of 0.476 mL. Is it pure silver?
10. What is the density of the solid in the image if the mass is 40 g? Make your answer have 3 significant figures.
11) Below is a model of a pyramid found at an archeological dig made of an unknown substance. It is too large to find the volume by submerging it in water. Also, the scientists refuse to remove a piece to test because this pyramid is a part of history. Its height is 150.0m. The length of its base is 75.0m and the width is 50.0m. The mass of this pyramid is 5.50x105kg. What is the density?
Density Problem Solutions
1. 1.049 g/mL
2. 0.7851 g/mL
3. 0.274 g/mL
4. 2.94 x 106 mL
5. 0.3.27 kg
6. 0.5 g/cm3
7. 1.15 g/mL
8. 790 g/L
9. Yes
10. 0.195 g/cm3
11. 29.3 g/cm3
Percent Composition Problems
These problems will follow the same pattern of difficulty as those of density.
1. Calculate the percent by mass of each element in Cesium Fluoride (CsF).
2. Calculate the percent by mass of each element present in carbon tetrachloride (CCl4)
3. A solution of salt and water is 33.0% salt by mass and has a density of 1.50 g/mL. What mass of the salt in grams is in 5.00L of this solution?
4. A solution of water and HCl contains 25% HCl by mass. The density of the solution is 1.05 g/mL. If you need 1.7g of HCl for a reaction, what volume of this solution will you use?
5. A solution containing 42% NaOH by mass has a density of 1.30 g/mL. What mass, in kilograms, of NaOH is in 6.00 L of this solution?
Percent Composition Problem Solutions
1. CsF is 87.5% Cs and 12.5% F by mass
2. CCl4is 92.2% Cl and 7.8% C by mass
3. 2480g
4. 6.5mL
5. 2.38 kg
References
1. AUTOR , ARQUIMEDES , and Thomas Little . The Works of Archimedes . Courier Dover Publications, 2002.
2. Chande, D. and T. Fisher (2003). "Have a Penny? Need a Penny? Eliminating the One-Cent Coin from Circulation." Canadian Public Policy/Analyse de Politiques 29(4): 511-517.
3. Jefferson, T. (1999). "A Thought for Your Pennies." JAMA 281(2): 122.
4. Petrucci , Ralph , William Harwood , and Geoffrey Herring . Principles and Modern Application. ninth . New Jersey : Peason Eduation , 2007.
5. Rauch, F., H. Plotkin, et al. (2003). "Bone Mass, Size, and Density in Children and Adolescents with Osteogenesis Imperfecta: Effect of Intravenous Pamidronate Therapy." Journal of Bone and Mineral Research 18: 610-614.
6. Richardson, J., S. Gwaltney-Brant, et al. (2002). "Zinc Toxicosis from Penny Ingestion in Dogs." Vet Med 97(2): 96-99.
7. Tate, J. "Archimedes’ Discoveries: A Closer Look." | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.13%3A_Percent_Composition.txt |
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balance reactions.
Balancing
In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The reactants are displayed on the left side of the equation and the products are shown on the right, with the separation of either a single or double arrow that signifies the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, but we will not go into detail about it in this module. To balance an equation, it is necessary that there are the same number of atoms on the left side of the equation as the right. One can do this by raising the coefficients.
Reactants to Products
A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as well as their states, and the proportion for how much of each particle reacts or is formed relative to one another, through the stoichiometric coefficient. The following equation demonstrates the typical format of a chemical equation:
$\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber$
In the above equation, the elements present in the reaction are represented by their chemical symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and products, though the elements they are paired up with often change in a reaction. In this reaction, sodium ($Na$), hydrogen ($H$), and chloride ($Cl$) are the elements present in both reactants, so based on the law of conservation of mass, they are also present on the product side of the equations. Displaying each element is important when using the chemical equation to convert between elements.
Stoichiometric Coefficients
In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation:
$\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber$
we can determine that 2 moles of $HCl$ will react with 2 moles of $Na_{(s)}$ to form 2 moles of $NaCl_{(aq)}$ and 1 mole of $H_{2(g)}$. If we know how many moles of $Na$ reacted, we can use the ratio of 2 moles of $NaCl$ to 2 moles of Na to determine how many moles of $NaCl$ were produced or we can use the ratio of 1 mole of $H_2$ to 2 moles of $Na$ to convert to $NaCl$. This is known as the coefficient factor. The balanced equation makes it possible to convert information about the change in one reactant or product to quantitative data about another reactant or product. Understanding this is essential to solving stoichiometric problems.
Example 1
Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.
$\ce{Pb(OH)4 + H2SO4 \rightarrow Pb(SO4)2 +H2O} \nonumber$
Solution
Start by counting the number of atoms of each element.
UNBALANCED
Element
Reactant (# of atoms)
Product (# of atoms)
Pb
1
1
O
8
9
H
6
2
S
1
2
The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of $H_2SO_4$ to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of $H_2O$ where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced.
$\ce{ Pb(OH)4 + 2 H2SO4 \rightarrow Pb(SO4)2 + 4H2O} \nonumber$
BALANCED
Element
Reactant (# of atoms)
Product (# of atoms)
Pb
1
1
O
12
12
H
8
8
S
2
2
Balancing reactions involves finding least common multiples between numbers of elements present on both sides of the equation. In general, when applying coefficients, add coefficients to the molecules or unpaired elements last.
A balanced equation ultimately has to satisfy two conditions.
1. The numbers of each element on the left and right side of the equation must be equal.
2. The charge on both sides of the equation must be equal. It is especially important to pay attention to charge when balancing redox reactions.
Stoichiometry and Balanced Equations
In stoichiometry, balanced equations make it possible to compare different elements through the stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show how stoichiometric factors are useful.
Example 2
There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How many party invitations can be sent?
Solution
The equation for this can be written as
$\ce{I + 2S \rightarrow IS2}\nonumber$
where
• $I$ represents invitations,
• $S$ represents stamps, and
• $IS_2$ represents the sent party invitations consisting of one invitation and two stamps.
Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation.
Invitations Stamps Party Invitations Sent
In this example are all the reactants (stamps and invitations) used up? No, and this is normally the case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the one that runs out first, prevents the reaction from continuing and determines the maximum amount of product that can be formed.
Example 3
What is the limiting reagent in this example?
Solution
Stamps, because there was only enough to send out invitations, whereas there were enough invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can be solved using stoichiometric factors.
12 I x (1IS2/1I) = 12 IS2 possible
20 S x (1IS2/2S) = 10 IS2 possible
When there is no limiting reagent because the ratio of all the reactants caused them to run out at the same time, it is known as stoichiometric proportions.
Types of Reactions
There are 6 basic types of reactions.
• Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical and O2
• Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a complex product.
• Decomposition: Decomposition is when complex reactants are broken down into simpler products.
• Single Displacement: Single displacement is when an element from on reactant switches with an element of the other to form two new reactants.
• Double Displacement: Double displacement is when two elements from on reactants switched with two elements of the other to form two new reactants.
• Acid-Base: Acid- base reactions are when two reactants form salts and water.
Molar Mass
Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass
Example 4
What is the molar mass of H2O?
Solution
$\text{Molar mass} = 2 \times (1.00794\; g/mol) + 1 \times (15.9994\; g/mol) = 18.01528\; g/mol \nonumber$
Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of products or vice versa.
Example 5: Combustion of Propane
Propane ($\ce{C_3H_8}$) burns in this reaction:
$\ce{C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2} \nonumber$
If 200 g of propane is burned, how many g of $H_2O$ is produced?
Solution
Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products you must convert from grams of $C_3H_8$ to moles of $C_3H_8$ then from moles of $C_3H_8$ to moles of $H_2O$. Then convert from moles of $H_2O$ to grams of $H_2O$.
• Step 1: 200 g $C_3H_8$ is equal to 4.54 mol $C_3H_8$.
• Step 2: Since there is a ratio of 4:1 $H_2O$ to $C_3H_8$, for every 4.54 mol $C_3H_8$ there are 18.18 mol $H_2O$.
• Step 3: Convert 18.18 mol $H_2O$ to g $H_2O$. 18.18 mol $H_2O$ is equal to 327.27 g $H_2O$.
Variation in Stoichiometric Equations
Almost every quantitative relationship can be converted into a ratio that can be useful in data analysis.
Density
Density ($\rho$) is calculated as mass/volume. This ratio can be useful in determining the volume of a solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse relationship would be used.
Volume x (Mass/Volume) = Mass
Mass x (Volume/Mass) = Volume
Percent Mass
Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule.
Example 6
A substance is 5% carbon by mass. If the total mass of the substance is 10 grams, what is the mass of carbon in the sample? How many moles of carbon are there?
Solution
10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon
0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon
Molarity
Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions.
Example 7
How much 5 M stock solution is needed to prepare 100 mL of 2 M solution?
Solution
100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution.
These ratios of molarity, density, and mass percent are useful in complex examples ahead.
Determining Empirical Formulas
An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present.
Example 8: Combustion of Organic Molecules
1.000 gram of an organic molecule burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO2 and 0.599 g of H2O. What is the empirical formula of the organic molecule?
Solution
This is a combustion reaction. The problem requires that you know that organic molecules consist of some combination of carbon, hydrogen, and oxygen elements. With that in mind, write the chemical equation out, replacing unknown numbers with variables. Do not worry about coefficients here.
$\ce{C_xH_yO_z(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)} \nonumber$
Since all the moles of C and H in CO2 and H2O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample.
0.0333mol CO2 (1mol C/ 1mol CO2) = 0.0333mol C in unknown
0.599g H2O (1mol H2O/ 18.01528g H2O)(2mol H/ 1mol H2O) = 0.0665 mol H in unknown
Calculate the final moles of oxygen by taking the sum of the moles of oxygen in CO2 and H2O. This will give you the number of moles from both the unknown organic molecule and the O2 so you must subtract the moles of oxygen transferred from the O2.
Moles of oxygen in CO2:
0.0333mol CO2 (2mol O/1mol CO2) = 0.0666 mol O
Moles of oxygen in H2O:
0.599g H2O (1mol H2O/18.01528 g H2O)(1mol O/1mol H2O) = 0.0332 mol O
Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O2 reactant.
0.333mol CO2(44.0098g CO2/ 1mol CO2) = 1.466g CO2
1.466g CO2 + 0.599g H2O - 1.000g unknown organic = 1.065g O2
Moles of oxygen in O2
1.065g O2(1mol O2/ 31.9988g O2)(2mol O/1mol O2) = 0.0666mol O
Moles of oxygen in unknown
(0.0666mol O + 0.0332 mol O) - 0.0666mol O = 0.0332 mol O
Construct a mole ratio for C, H, and O in the unknown and divide by the smallest number.
(1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol O
From this ratio, the empirical formula is calculated to be CH2O.
Determining Molecular Formulas
To determine a molecular formula, first determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, divide the molecular mass by the molar mass of the empirical formula (calculated by finding the sum the total atomic masses of all the elements in the empirical formula). Multiply the subscripts of the molecular formula by this answer to get the molecular formula.
Example 9
In the example above, it was determined that the unknown molecule had an empirical formula of CH2O.
1. Find the molar mass of the empircal formula CH2O.
12.011g C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH2O
2. Determine the molecular mass experimentally. For our compound, it is 120.056 g/mol.
3. Divide the experimentally determined molecular mass by the mass of the empirical formula.
(120.056 g/mol) / (30.026 g/mol) = 3.9984
4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a slight error in the experimentally determined molecular mass. If the answer is not close to a whole number, there was either an error in the calculation of the empirical formula or a large error in the determination of the molecular mass.
5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular formula.
CH2O * 4 = ?
C: 1 * 4 = 4
H: 2 * 4 = 8
O 1 * 4 = 4
CH2O * 4 = C4H8O4
6. Check your result by calculating the molar mass of the molecular formula and comparing it to the experimentally determined mass.
molar mass of C4H8O4= 120.104 g/mol
experimentally determined mass = 120.056 g/mol
% error = | theoretical - experimental | / theoretical * 100%
% error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%
% error = 0.040 %
Example 10: Complex Stoichiometry Problem
An amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. The alloy's density is 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm3. He accidentally breaks off a 1.203 cm3 piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. Assuming the acid reacts with all the iron(II) and not with the copper, how many grams of H2(g) are released into the atmosphere because of the amateur's carelessness? (Note that the situation is fiction.)
Solution
Step 1: Write a balanced equation after determining the products and reactants. In this situation, since we assume copper does not react, the reactants are only H+(aq) and Fe(s). The given product is H2(g) and based on knowledge of redox reactions, the other product must be Fe2+(aq).
$\ce{Fe(s) + 2H^{+}(aq) \rightarrow H2(g) + Fe^{2+}(aq)} \nonumber$
Step 2: Write down all the given information
Alloy density = (3.15g alloy/ 1L alloy)
x grams of alloy = 45% copper = (45g Cu(s)/100g alloy)
x grams of alloy = 55% iron(II) = (55g Fe(s)/100g alloy)
1 liter alloy = 1000cm3 alloy
alloy sample = 1.203cm3 alloy
Step 3: Answer the question of what is being asked. The question asks how much H2(g) was produced. You are expected to solve for the amount of product formed.
Step 4: Start with the compound you know the most about and use given ratios to convert it to the desired compound.
Convert the given amount of alloy reactant to solve for the moles of Fe(s) reacted.
1.203cm3 alloy(1liter alloy/1000cm3 alloy)(3.15g alloy/1liter alloy)(55g Fe(s)/100g alloy)(1mol Fe(s)/55.8g Fe(s))=3.74 x 10-5 mol Fe(s)
Make sure all the units cancel out to give you moles of $\ce{Fe(s)}$. The above conversion involves using multiple stoichiometric relationships from density, percent mass, and molar mass.
The balanced equation must now be used to convert moles of Fe(s) to moles of H2(g). Remember that the balanced equation's coefficients state the stoichiometric factor or mole ratio of reactants and products.
3.74 x 10-5 mol Fe (s) (1mol H2(g)/1mol Fe(s)) = 3.74 x 10-5 mol H2(g)
Step 5: Check units
The question asks for how many grams of H2(g) were released so the moles of H2(g) must still be converted to grams using the molar mass of H2(g). Since there are two H in each H2, its molar mass is twice that of a single H atom.
molar mass = 2(1.00794g/mol) = 2.01588g/mol
3.74 x 10-5 mol H2(g) (2.01588g H2(g)/1mol H2 (g)) = 7.53 x 10-5 g H2(g) released
Problems
Stoichiometry and balanced equations make it possible to use one piece of information to calculate another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and see if you can use what you learned to solve the following problems.
1) Why are the following equations not considered balanced?
1. $H_2O_{(l)} \rightarrow H_{2(g)} + O_{2(g)}$
2. $Zn_{(s)} + Au^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ag_{(s)}$
2) Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions. Write the balanced chemical equation for this reaction.
3) Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M solution?
4) If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the limiting reagent and how many moles of water are produced? The unbalanced equation is provided below.
$\ce{CH4(g) + O2(g) \rightarrow CO2(g) + H2O(l)} \nonumber$
5) A 0.777g sample of an organic compound is burned completely. It produces 1.42g CO2 and 0.388g H2O. Knowing that all the carbon and hydrogen atoms in CO2 and H2O came from the 0.777g sample, what is the empirical formula of the organic compound?
Contributors and Attributions
• Joseph Nijmeh (UCD), Mark Tye (DVC) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.14%3A_Empirical_and_Molecular_Formulas.txt |
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balance reactions.
Balancing
In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The reactants are displayed on the left side of the equation and the products are shown on the right, with the separation of either a single or double arrow that signifies the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, but we will not go into detail about it in this module. To balance an equation, it is necessary that there are the same number of atoms on the left side of the equation as the right. One can do this by raising the coefficients.
Reactants to Products
A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as well as their states, and the proportion for how much of each particle reacts or is formed relative to one another, through the stoichiometric coefficient. The following equation demonstrates the typical format of a chemical equation:
$\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber$
In the above equation, the elements present in the reaction are represented by their chemical symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and products, though the elements they are paired up with often change in a reaction. In this reaction, sodium ($Na$), hydrogen ($H$), and chloride ($Cl$) are the elements present in both reactants, so based on the law of conservation of mass, they are also present on the product side of the equations. Displaying each element is important when using the chemical equation to convert between elements.
Stoichiometric Coefficients
In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation:
$\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber$
we can determine that 2 moles of $HCl$ will react with 2 moles of $Na_{(s)}$ to form 2 moles of $NaCl_{(aq)}$ and 1 mole of $H_{2(g)}$. If we know how many moles of $Na$ reacted, we can use the ratio of 2 moles of $NaCl$ to 2 moles of Na to determine how many moles of $NaCl$ were produced or we can use the ratio of 1 mole of $H_2$ to 2 moles of $Na$ to convert to $NaCl$. This is known as the coefficient factor. The balanced equation makes it possible to convert information about the change in one reactant or product to quantitative data about another reactant or product. Understanding this is essential to solving stoichiometric problems.
Example 1
Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.
$\ce{Pb(OH)4 + H2SO4 \rightarrow Pb(SO4)2 +H2O} \nonumber$
Solution
Start by counting the number of atoms of each element.
UNBALANCED
Element
Reactant (# of atoms)
Product (# of atoms)
Pb
1
1
O
8
9
H
6
2
S
1
2
The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of $H_2SO_4$ to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of $H_2O$ where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced.
$\ce{ Pb(OH)4 + 2 H2SO4 \rightarrow Pb(SO4)2 + 4H2O} \nonumber$
BALANCED
Element
Reactant (# of atoms)
Product (# of atoms)
Pb
1
1
O
12
12
H
8
8
S
2
2
Balancing reactions involves finding least common multiples between numbers of elements present on both sides of the equation. In general, when applying coefficients, add coefficients to the molecules or unpaired elements last.
A balanced equation ultimately has to satisfy two conditions.
1. The numbers of each element on the left and right side of the equation must be equal.
2. The charge on both sides of the equation must be equal. It is especially important to pay attention to charge when balancing redox reactions.
Stoichiometry and Balanced Equations
In stoichiometry, balanced equations make it possible to compare different elements through the stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show how stoichiometric factors are useful.
Example 2
There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How many party invitations can be sent?
Solution
The equation for this can be written as
$\ce{I + 2S \rightarrow IS2}\nonumber$
where
• $I$ represents invitations,
• $S$ represents stamps, and
• $IS_2$ represents the sent party invitations consisting of one invitation and two stamps.
Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation.
Invitations Stamps Party Invitations Sent
In this example are all the reactants (stamps and invitations) used up? No, and this is normally the case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the one that runs out first, prevents the reaction from continuing and determines the maximum amount of product that can be formed.
Example 3
What is the limiting reagent in this example?
Solution
Stamps, because there was only enough to send out invitations, whereas there were enough invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can be solved using stoichiometric factors.
12 I x (1IS2/1I) = 12 IS2 possible
20 S x (1IS2/2S) = 10 IS2 possible
When there is no limiting reagent because the ratio of all the reactants caused them to run out at the same time, it is known as stoichiometric proportions.
Types of Reactions
There are 6 basic types of reactions.
• Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical and O2
• Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a complex product.
• Decomposition: Decomposition is when complex reactants are broken down into simpler products.
• Single Displacement: Single displacement is when an element from on reactant switches with an element of the other to form two new reactants.
• Double Displacement: Double displacement is when two elements from on reactants switched with two elements of the other to form two new reactants.
• Acid-Base: Acid- base reactions are when two reactants form salts and water.
Molar Mass
Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass
Example 4
What is the molar mass of H2O?
Solution
$\text{Molar mass} = 2 \times (1.00794\; g/mol) + 1 \times (15.9994\; g/mol) = 18.01528\; g/mol \nonumber$
Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of products or vice versa.
Example 5: Combustion of Propane
Propane ($\ce{C_3H_8}$) burns in this reaction:
$\ce{C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2} \nonumber$
If 200 g of propane is burned, how many g of $H_2O$ is produced?
Solution
Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products you must convert from grams of $C_3H_8$ to moles of $C_3H_8$ then from moles of $C_3H_8$ to moles of $H_2O$. Then convert from moles of $H_2O$ to grams of $H_2O$.
• Step 1: 200 g $C_3H_8$ is equal to 4.54 mol $C_3H_8$.
• Step 2: Since there is a ratio of 4:1 $H_2O$ to $C_3H_8$, for every 4.54 mol $C_3H_8$ there are 18.18 mol $H_2O$.
• Step 3: Convert 18.18 mol $H_2O$ to g $H_2O$. 18.18 mol $H_2O$ is equal to 327.27 g $H_2O$.
Variation in Stoichiometric Equations
Almost every quantitative relationship can be converted into a ratio that can be useful in data analysis.
Density
Density ($\rho$) is calculated as mass/volume. This ratio can be useful in determining the volume of a solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse relationship would be used.
Volume x (Mass/Volume) = Mass
Mass x (Volume/Mass) = Volume
Percent Mass
Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule.
Example 6
A substance is 5% carbon by mass. If the total mass of the substance is 10 grams, what is the mass of carbon in the sample? How many moles of carbon are there?
Solution
10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon
0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon
Molarity
Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions.
Example 7
How much 5 M stock solution is needed to prepare 100 mL of 2 M solution?
Solution
100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution.
These ratios of molarity, density, and mass percent are useful in complex examples ahead.
Determining Empirical Formulas
An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present.
Example 8: Combustion of Organic Molecules
1.000 gram of an organic molecule burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO2 and 0.599 g of H2O. What is the empirical formula of the organic molecule?
Solution
This is a combustion reaction. The problem requires that you know that organic molecules consist of some combination of carbon, hydrogen, and oxygen elements. With that in mind, write the chemical equation out, replacing unknown numbers with variables. Do not worry about coefficients here.
$\ce{C_xH_yO_z(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)} \nonumber$
Since all the moles of C and H in CO2 and H2O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample.
0.0333mol CO2 (1mol C/ 1mol CO2) = 0.0333mol C in unknown
0.599g H2O (1mol H2O/ 18.01528g H2O)(2mol H/ 1mol H2O) = 0.0665 mol H in unknown
Calculate the final moles of oxygen by taking the sum of the moles of oxygen in CO2 and H2O. This will give you the number of moles from both the unknown organic molecule and the O2 so you must subtract the moles of oxygen transferred from the O2.
Moles of oxygen in CO2:
0.0333mol CO2 (2mol O/1mol CO2) = 0.0666 mol O
Moles of oxygen in H2O:
0.599g H2O (1mol H2O/18.01528 g H2O)(1mol O/1mol H2O) = 0.0332 mol O
Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O2 reactant.
0.333mol CO2(44.0098g CO2/ 1mol CO2) = 1.466g CO2
1.466g CO2 + 0.599g H2O - 1.000g unknown organic = 1.065g O2
Moles of oxygen in O2
1.065g O2(1mol O2/ 31.9988g O2)(2mol O/1mol O2) = 0.0666mol O
Moles of oxygen in unknown
(0.0666mol O + 0.0332 mol O) - 0.0666mol O = 0.0332 mol O
Construct a mole ratio for C, H, and O in the unknown and divide by the smallest number.
(1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol O
From this ratio, the empirical formula is calculated to be CH2O.
Determining Molecular Formulas
To determine a molecular formula, first determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, divide the molecular mass by the molar mass of the empirical formula (calculated by finding the sum the total atomic masses of all the elements in the empirical formula). Multiply the subscripts of the molecular formula by this answer to get the molecular formula.
Example 9
In the example above, it was determined that the unknown molecule had an empirical formula of CH2O.
1. Find the molar mass of the empircal formula CH2O.
12.011g C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH2O
2. Determine the molecular mass experimentally. For our compound, it is 120.056 g/mol.
3. Divide the experimentally determined molecular mass by the mass of the empirical formula.
(120.056 g/mol) / (30.026 g/mol) = 3.9984
4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a slight error in the experimentally determined molecular mass. If the answer is not close to a whole number, there was either an error in the calculation of the empirical formula or a large error in the determination of the molecular mass.
5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular formula.
CH2O * 4 = ?
C: 1 * 4 = 4
H: 2 * 4 = 8
O 1 * 4 = 4
CH2O * 4 = C4H8O4
6. Check your result by calculating the molar mass of the molecular formula and comparing it to the experimentally determined mass.
molar mass of C4H8O4= 120.104 g/mol
experimentally determined mass = 120.056 g/mol
% error = | theoretical - experimental | / theoretical * 100%
% error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%
% error = 0.040 %
Example 10: Complex Stoichiometry Problem
An amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. The alloy's density is 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm3. He accidentally breaks off a 1.203 cm3 piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. Assuming the acid reacts with all the iron(II) and not with the copper, how many grams of H2(g) are released into the atmosphere because of the amateur's carelessness? (Note that the situation is fiction.)
Solution
Step 1: Write a balanced equation after determining the products and reactants. In this situation, since we assume copper does not react, the reactants are only H+(aq) and Fe(s). The given product is H2(g) and based on knowledge of redox reactions, the other product must be Fe2+(aq).
$\ce{Fe(s) + 2H^{+}(aq) \rightarrow H2(g) + Fe^{2+}(aq)} \nonumber$
Step 2: Write down all the given information
Alloy density = (3.15g alloy/ 1L alloy)
x grams of alloy = 45% copper = (45g Cu(s)/100g alloy)
x grams of alloy = 55% iron(II) = (55g Fe(s)/100g alloy)
1 liter alloy = 1000cm3 alloy
alloy sample = 1.203cm3 alloy
Step 3: Answer the question of what is being asked. The question asks how much H2(g) was produced. You are expected to solve for the amount of product formed.
Step 4: Start with the compound you know the most about and use given ratios to convert it to the desired compound.
Convert the given amount of alloy reactant to solve for the moles of Fe(s) reacted.
1.203cm3 alloy(1liter alloy/1000cm3 alloy)(3.15g alloy/1liter alloy)(55g Fe(s)/100g alloy)(1mol Fe(s)/55.8g Fe(s))=3.74 x 10-5 mol Fe(s)
Make sure all the units cancel out to give you moles of $\ce{Fe(s)}$. The above conversion involves using multiple stoichiometric relationships from density, percent mass, and molar mass.
The balanced equation must now be used to convert moles of Fe(s) to moles of H2(g). Remember that the balanced equation's coefficients state the stoichiometric factor or mole ratio of reactants and products.
3.74 x 10-5 mol Fe (s) (1mol H2(g)/1mol Fe(s)) = 3.74 x 10-5 mol H2(g)
Step 5: Check units
The question asks for how many grams of H2(g) were released so the moles of H2(g) must still be converted to grams using the molar mass of H2(g). Since there are two H in each H2, its molar mass is twice that of a single H atom.
molar mass = 2(1.00794g/mol) = 2.01588g/mol
3.74 x 10-5 mol H2(g) (2.01588g H2(g)/1mol H2 (g)) = 7.53 x 10-5 g H2(g) released
Problems
Stoichiometry and balanced equations make it possible to use one piece of information to calculate another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and see if you can use what you learned to solve the following problems.
1) Why are the following equations not considered balanced?
1. $H_2O_{(l)} \rightarrow H_{2(g)} + O_{2(g)}$
2. $Zn_{(s)} + Au^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ag_{(s)}$
2) Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions. Write the balanced chemical equation for this reaction.
3) Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M solution?
4) If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the limiting reagent and how many moles of water are produced? The unbalanced equation is provided below.
$\ce{CH4(g) + O2(g) \rightarrow CO2(g) + H2O(l)} \nonumber$
5) A 0.777g sample of an organic compound is burned completely. It produces 1.42g CO2 and 0.388g H2O. Knowing that all the carbon and hydrogen atoms in CO2 and H2O came from the 0.777g sample, what is the empirical formula of the organic compound?
Contributors and Attributions
• Joseph Nijmeh (UCD), Mark Tye (DVC) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.15%3A_Determining_Formulas_from_Mass_Data.txt |
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
04: Stoichiometry: Quantitative Information About Chemical Reactions
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
05: Principles of Chemical Reactivity: Energy and Chemical Reactions
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
06: The Chemistry of Fuels and Energy Resources
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
07: The Structure of Atoms and Periodic Trends
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
08: Milestones in the Development of Chemistry and the Modern View of Atoms and Molecules
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
9.1: Valence Bond Theory
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
09: Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals
Valence Bond (VB) Theory looks at the interaction between atoms to explain chemical bonds. It is one of the two common theories that helps describe the bonding between atoms. The other theory is Molecular Orbital Theory. Take note that these are theories and should be treated as such; they are not always perfect.
Introduction
Valence Bond Theory has its roots in Gilbert Newton Lewis’s paper The Atom and The Molecule. Possibly unaware that Lewis’s model existed, Walter Heitler and Fritz London came up with the idea that resonance and wavefunctions contributed to chemical bonds, in which they used dihydrogen as an example. Their theory was equivalent to Lewis’s theory, with the difference of quantum mechanics being developed. Nonetheless, Heitler and London's theory proved to be successful, providing Linus Pauling and John C. Slater with an opportunity to assemble a general chemical theory containing all of these ideas. Valence Bond Theory was the result, which included the ideas of resonance, covalent-ionic superposition, atomic orbital overlap, and hybridization to describe chemical bonds.
Atomic Orbital Overlap
Bonds are formed between atoms because the atomic orbitals overlap and that the electrons in those orbitals are localized within that overlap. They have a higher probability of being found within that bond; we will return to this statement when we talk about wavefunctions. Dihydrogen (H2) is a simple diatomic gas that has been used to illustrate this idea; however, let's look at Cl2 as a simple example. Cl has seven valence electrons. From its Lewis structure, one can see that Cl has a radical. That sole radical indicates that Cl can bond once. As a general rule, the number of unpaired electrons denotes how many bonds that atom can make. Since there is only one unpaired electron in each Cl, those electrons interact to bond. In this case, the 3p orbitals overlap. Lone pairs can be seen as the orbitals not interacting with each other. The idea of atomic orbitals overlapping works well for simple molecules, such as diatomic gases, but more complex molecules cannot be explained simply by the overlap of atomic orbitals, especially if they defy the octet rule when drawing out Lewis structures and if they bond beyond their predicted amount.
One small concept to add in here is that an orbital from one atoms can be overlap with the other orbital of the second atom. This will eventually give one of the two results. First, the two orbitals have correct symmetry to interact or "mix" (4). Second, the two orbitals do not have correct symmetry to interact or "mix" (4). In the case of two orbitals do not interact, there will be no bonding interaction, which means one of the atomic orbitals will not contribute into the bonding. Another word, it will not affected because of the presence of other atomic orbitals. For further explaination, when the two wavefunctions are knowns as constructive interaction, which means the orbitals do not interact with one another. And if the two orbitals are interacting, they are knowns as destructive interaction where their wavefunctions have opposite site (4).
In the case of two orbitals do interact with each others, it will become molecular orbitals, and it is knowns as constructive interaction, resulting bonding orbitals. On the other hand, if destructive interaction, it will create an antibonding interaction. The two molecular orbitals are the result from adding and subtracting the two wavefunctions (4).
Atomic Orbital Hybridization
Consider phosphorus pentaflouride (PF5). P, the central atom, has five valence electrons, with three lone electrons. Thus, P should only be able to form three bonds. However, in order for PF5 to exist with P as the central atoms, P must be able to bond five times. This event is described by orbital hybridization. In order to create degenerate hybrid orbitals that allow atoms to bond well beyond their normal amount. Orbitals involved in hybridization are the s, p, and d atomic orbitals. There are certain principles that must be followed:
1. Orbitals are not magically lost or gained. The number of orbitals mixed must match the number of hybrid orbitals acquired.
2. Always start from an s orbital. Build your way up to p orbitals, and then d orbitals as necessary. These are orbitals from the atom in question. In this case, it is the P atom.
Hybrid orbitals consist of sp, sp2, sp3, sp2d, sp3d, and sp3d2 (3). Returning back to PF5, let's see how hybrid orbitals can describe its bonds. Remember that P can only bond three times, but we need it to bond five times. In order to bond five times, five orbitals must be singularly filled in. Bonds occur when orbitals with only one electron are spin paired with the electron from another atom. The five orbitals can be acquired by sp3d orbital hybridization. Starting off with a total of five orbitals will result in five hybrid orbitals. Note that since P is in period 3, it has d orbitals in the same energy level. Thus, 3d orbitals can be used to hybridize, even though electrons do not occupy it. The process is outlined in the original figure below. Remember to follow the aufbau principle, Hund's rule, and the Pauli exclusion principle when assigning electrons to their orbitals. Note that valence electrons are only shown. Other orbitals and their corresponding electrons are not hybridized and are not involved in bonding. Although they are not shown, they still exist.
Now, P has five hybrid orbitals. Remember that P has five valence electrons. Electrons too are not lost or gained, so those electrons transfer over to the hybrid orbitals. Fill them in accordingly, following Hund's rule and the Pauli exclusion principle. Note that the hybrid orbitals are now degenerate, so the aufbau principle doesn't apply for this case. From sp3d hybridization, P has five unpaired electrons and can now bond with five F.
Generally, if two bonds are needed, then use an sp; for three, use sp2; the number of bonds needed equals the number of orbitals that need to be hybridized. Please take into account that lone pairs apply as well. If there is a lone pair when you complete the Lewis structure, then the number of required hybrid orbitals equals the number of lone pairs plus the number of bonds; this is also called the coordinate number. PF5, although not shown, does not have lone pairs. The type of hybrid orbitals also corresponds to the molecular shape. In the case of PF5, its shape as defined by VSEPR is trigonal bypyramidal. If hybrid orbitals are involved, the shape can also be determined by the type of hybridization. As another example, sp is linear, and sp2 is trigonal planar. sp2d is square planar. Think over carefully why the type of hybridization also determines the shape. An example problem will aid you in your thinking in the problems section.
How the wavefunction Applies to VB Theory
The wavefunction describes the state of an electron. From the name of the function, one can derive that the electron can behave like a wave. This wave-like behavior of the electrons defines the shapes of the orbitals. Thus, it makes sense that wavefunctions are related to the Valence Bond Theory. If orbitals overlap to create bonds, and orbital shapes and the state of an electron is described by the wavefunction, then it makes sense that the overlap of orbitals (the bonds) can be described by wavefunctions as well. Thus, covalent and ionic bonds can be described by wavefunctions. Covalent and ionic representations of a bond represent the same bond, but they differ in how the electrons are placed. This is called resonance. Different intermolecular interactions give rise to different wavefunctions. Recall from Hund's Rule and the Pauli exclusion principle that electrons must be spin paired when the right conditions are met. Due to this, there are two separate ways to represent a covalent bond in terms of electron spin, which is related to the wavefunction. The figure shows the two different cases for two electrons in a bond. Thus, it is possible to write two different wavefunctions that describe each case, which can be superimposed to describe the overall covalent bond (2). The superposition of the covalent bond and ionic bond wavefunctions will result in an overall wavefunction that describes the state of the molecule. Due to this module being an overview of Valence Bond Theory, the full details will not be covered.
The wavefunction squared will result in a probability density. The wavefunction alone has no physical significance; however, when the wavefunction is squared, the square wavefunction can determine where the electron is most likely located. Recall that when atomic orbitals overlap, the electrons are localized and more likely to be found within that overlap. In terms of electrostatic interactions, this bond will result in some form of equilibrium between all of the electrostatic forces (between the electrons and the electrons with each nucleus) (1). If the overlap is too far in, there is a net repulsion force; the electrons will also be forced apart. If the atomic orbital overlap is too small, the net attraction force is very small; the electron will have a smaller chance of remaining in the overlap since they will be more attracted to their own nucleus. If the overlap is just right, then the electrons are attracted to both nuclei and more likely to stay in the overlapped area. Thus, wavefunctions and electrostatic forces can both explain why electrons are localized within their bonds.
Reliability of Valence Bond Theory and Uses
As one can see, Valence Bond Theory can help describe how bonds are formed. However, there are some notable failures when it comes to Valence Bond Theory. One such failure is dioxygen. Valence Bond Theory fails to predict dioxygen's paramagnitism; it predicts that oxygen is diamagnetic. A species is paramagnetic if electrons are not spin paired and diamagnetic if the electrons are spin paired. Since Valence Bond theory begins with the basis that atomic orbitals overlap to create bonds and through that reasoning, one can see that electrons are spin paired when bonds overlap, dioxygen is indeed predicted to be diamagnetic if Valence Bond Theory is used. In reality, that is not the case. Also, sp2d and sp3 both have a coordinate number of four. Thus, Valence Bond Theory cannot predict whether the molecule is a square planar or the other shape (3). One must correctly draw the Lewis structure and use VSEPR to determine the shape.
Problems
1. Draw the Lewis structures HF, CO2, and C2H2. Which orbitals are involved with bonding for HF? Now, use hybridization to explain the bonds of CO2 and C2H2. Can you account for the single, double, and triple bonds via orbitals overlapping and hybridization? It is not required to draw the hybridization process, but it might help you in your thinking.
2. Using VSEPR, one can derive that IBr5 has an octahedral parent shape. Now, use Valence Bond Theory to account for all of the bonds and verify that the parent shape is indeed an octahedral. Why can you determine the parent shape of a molecule based on Valence Bond Theory? Show your work.
3. Determine the family shapes correlated with the sp3, sp3d, and sp3d2 hybridization. Does this mean that all molecules with the respectable hybridization will have the same shape? Explain.
4. Predict which of these atoms/molecules is diamagnetic using Valence Bond Theory: H, H2, and NO.
5. Briefly summarize Valence Bond Theory. Is it always reliable?
Solutions
HF: Hydrogen has one electron in the 1s orbital. Flourine has two electrons in the 2s orbital, and five in the 2p orbitals. These can be derived from the electron configurations. We are only interested in the valence electrons, so the lower shell can be ignored. Remember to fill in the orbitals accordingly. The orbitals availiable to bond are Hydrogen's 1s orbital and one of Flourine's 2p orbitals. Thus, the bond must take place between these two orbitals. This is a sigma bond. A bond of this type, or one of similar orientation, such as two s orbitals, are the bonds associated with hybrid orbitals when hybridization takes place.
CO2: C has four valence electrons; two are in 2s and two are in 2p. O has six valence electrons; two in 2s and four in 2p. Notice that there is a double bond. Now, hybridize to explain the bonds. When determining the coordinate number, double bonds and triple bonds count as one bond. Thus, the coordinate number of C is 2 and the corresponding hybridization is sp. The diagram below shows the sp hybridization process for C. Notice that two 2p orbital remains unhybridized. We fill in the hybrid orbitals with two of C's electrons so that C has space available to bond with the two O's. The hybrid orbitals describes two sigma bonds between C and the O's; one 2p orbital from each O is used in this process. Two electrons remain for C, which are allocated to the unhybridized orbitals. Now, C can form the other two bonds for a total of four which is what is expected due to the presence of two double bonds. The unhybridized orbitals overlap O's 2p orbitals and forms what is called a pi bond. There is one pi bond and one sigma bond between C and each O. The lone pairs of the O's are the filled 2s orbital and the final and filled 2p orbital. Thus, a double bond is explained by one sigma bond associated with the hybrid orbitals and a pi bond associated with unhybridized p orbitals.
C2H2: C has four valence electrons (two in 2s and two in 2p) and H has one (one in 1s). There is a total of ten valence electrons. From the Lewis structure, one can see that there is a triple bond between the two C's. For each C, one can explain the bonds through sp hybridization (a triple bond and one single bond). This process is similar to CO2. However, in this case, C's available unhybridized 2p orbitals bond together with the unhybridized p orbitals of the other C. Now, there are two pi bonds and one sigma bond between the C's; there is one sigma bond between each C and H bond. Thus, a triple bond is explained by one sigma bond associated with the hybrid orbitals and two pi bonds associated with unhybridized p orbitals.
2. I is the central atom. Usually, the least electronegative atom is the central atom. Drawing the Lewis structure reveals that I can only form one bond. However, with the use of hybridization, I can form more bonds. Showing that is the goal. First, find out how many valence electrons IBr5 has. This is simply derived by adding all of the valence electrons from each atom; all have 7 valence electrons, so (6)(7)=42. Then, use Lewis structure and allocate all of the electrons into the correct places. You will find that the Lewis structure looks like the picture below.
Notice that there is a lone pair. Remember that the number of hybrid orbitals needed is the summation of lone pairs and bonds. This is a total of six; thus a sp3d2 hybridization is needed. This process is shown below. Take note that the lone pair is the only spin paired electrons in the hybrid orbitals. Since the coordinate number is six, then the family shape of this molecule is an octahedral shape. Even the lone pairs are accounted for when hybridization takes place. It is because we used the coordinate number (determined by using Lewis dot structure first) to determiine the type of hybridization that hybridization does indeed give the correct family shape. There are of course exceptions, such as sp2d and sp3. We cannot predict the shape based on hybridization alone for these.
3. The shapes are tetrahedral, trigonal bypyramidal, and octahedral respectively. Not all molecules will have the shape associated with the hybridization type because lone pairs must be taken into account as well. For example, look at problem number two. The coordinate number is six. That tells you that the parent shape is an octahedral. However, the shape of IBr5 will only be that shape if the lone pair was another atom. The parent shape will not always match the actual shape of the molecule. It gives you a place to start and is only true if lone pairs do not exist. "Removing" one bond away from the octahedral shape will give the result of the square-base pyramidal shape.
4. One can solve this problem simply by counting the valence electrons. H has only one electron; it is not paired with any other electron and thus must be paramagnetic. H2 has two valence electrons; thus, those electrons are spin paired and H2 is diamagnetic. NO has eleven valence electrons; it is paramagnetic. In general, if an atom/molecule has an odd number of electrons, then that atom/molecule is paramagnetic. It is diamagnetic if it has an even number of electrons; however, like the case of O2, this does not always work.
5. Valence Bond Theory looks at the interaction between orbitals to describe bonds. It can also be used to derive the shape of the molecule in question, as well as determining whether or not an atom/molecule is diamagnetic or paramagnetic; however Valence Bond Theory is not always reliable. It fails in some cases. One must always remember that this is a theory. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/03%3A_Chemical_Reactions.txt |
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
10: Carbon: More Than Just Another Element
This page was auto-generated because a user created a sub-page to this page.
10.4: Alcohols Ethers and Amines
Alcohols are some of the most important molecules in organic chemistry. They can be prepared from and converted into many different types of compounds. Alcohols contain the hydroxy functional group (-OH), bonded to a carbon atom of an alkyl or substituted alkyl group. The functional group of an alcohol is the hydroxyl group, –OH. Unlike the alkyl halides, this group has two reactive covalent bonds, the C–O bond and the O–H bond. The electronegativity of oxygen is substantially greater than that of carbon and hydrogen. Consequently, the covalent bonds of this functional group are polarized so that oxygen is electron rich and both carbon and hydrogen are electrophilic, as shown in the figure below.
Indeed, the dipolar nature of the O–H bond is such that alcohols are much stronger acids than alkanes (by roughly 1030 times), and nearly that much stronger than ethers. The most reactive site in an alcohol molecule is the hydroxyl group, despite the fact that the O–H bond strength is significantly greater than that of the C–C, C–H and C–O bonds, demonstrating again the difference between thermodynamic and chemical stability.
10.5: Compounds with a Carbonyl Group
A carbonyl group is a chemically organic functional group composed of a carbon atom double-bonded to an oxygen atom --> [C=O] The simplest carbonyl groups are aldehydes and ketones usually attached to another carbon compound. These structures can be found in many aromatic compounds contributing to smell and taste.
Introduction
Before going into anything in depth be sure to understand that the C=O entity itself is known as the "Carbonyl group" while the members of this group are called "carbonyl compounds" --> X-C=O. The carbon and oxygen are usually sp2 hybridized and planar.
Carbonyl Group Double Bonds
The double bonds in alkenes and double bonds in carbonyl groups are VERY different in terms of reactivity. The C=C is less reactive due to C=O electronegativity attributed to the oxygen and its two lone pairs of electrons. One pair of the oxygen lone pairs are located in 2s while the other pair are in 2p orbital where its axis is directed perpendicular to the direction of the pi orbitals. The Carbonyl groups properties are directly tied to its electronic structure as well as geometric positioning. For example, the electronegativity of oxygen also polarizes the pi bond allowing the single bonded substituent connected to become electron withdrawing.
*Note: Both the pi bonds are in phase (top and botom blue ovals)
The double bond lengths of a carbonyl group is about 1.2 angstroms and the strength is about 176-179 kcal/mol). It is possible to correlate the length of a carbonyl bond with its polarity; the longer the bond meaing the lower the polarity. For example, the bond length in C=O is larger in acetaldehyde than in formaldehyde (this of course takes into account the inductive effect of CH3 in the compound).
Polarization
As discussed before, we understand that oxygen has two lone pairs of electrons hanging around. These electrons make the oxygen more electronegative than carbon. The carbon is then partially postive (electrophillic) and the oxygen partially negative (nucleophillic). The polarizability is denoted by a lowercase delta and a positive or negative superscript depending. For example, carbon would have d+ and oxygen delta^(-). The polarization of carbonyl groups also effects the boiling point of aldehydes and ketones to be higher than those of hydrocarbons in the same amount. The larger the carbonyl compound the less soluble it is in water. If the compound exceeds six carbons it then becomes insoluble.
*For more information about carbonyl solubility, look in the "outside links" section
*Amides are the most stable of the carbonyl couplings due to the high-resonance stabilization between nitrogen-carbon and carbon-oxygen.
Nucleophile Addition to a Carbonyl Group
C=O is prone to additions and nucleophillic attack because or carbon's positive charge and oxygen's negative charge. The resonance of the carbon partial positive charge allows the negative charge on the nucleophile to attack the Carbonyl group and become a part of the structure and a positive charge (usually a proton hydrogen) attacks the oxygen. Just a reminder, the nucleophile is a good acid therefore "likes protons" so it will attack the side with a positive charge.
*Remember: due to the electronegative nature of oxygen the carbon is partially positive and oxygen is partially negative
1 2 3
1. The Nucleophile (Nu) attacks the positively charged carbon and pushes one of the double bond electrons onto oxygen to give it a negative charge.
2. The Nucleophile is now a part of the carbonl structure with a negatively cahrged oxygen and a Na+ "floating" around.
3. The negatively charged oxygen attacks the proton (H+) to give the resulting product above.
Problems
1. What is the hybridization of the carbon in the C=O? the oxygen?
2. Illustrate the correct partial positive/negative or polarization of a formaldehyde.
3. Is carboxylic acid soluble in water? enone? acetaldehyde?
Answers
1. sp2;sp2
2. partial positive on the carbon and partial negative on the oxygen
3. yes; yes; yes
Contributors
• Sharleen Agvateesiri | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/10%3A_Carbon%3A_More_Than_Just_Another_Element/10.3%3A_Aromatic_Compounds.txt |
Functional groups in organic compounds
Functional groups are structural units within organic compounds that are defined by specific bonding arrangements between specific atoms. The structure of capsaicin, the compound discussed in the beginning of this chapter, incorporates several functional groups, labeled in the figure below and explained throughout this section.
As we progress in our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups behave in organic reactions.
The 'default' in organic chemistry (essentially, the lack of any functional groups) is given the term alkane, characterized by single bonds between carbon and carbon, or between carbon and hydrogen. Methane, CH4, is the natural gas you may burn in your furnace. Octane, C8H18, is a component of gasoline.
Alkanes
Alkenes (sometimes called olefins) have carbon-carbon double bonds, and alkynes have carbon-carbon triple bonds. Ethene, the simplest alkene example, is a gas that serves as a cellular signal in fruits to stimulate ripening. (If you want bananas to ripen quickly, put them in a paper bag along with an apple - the apple emits ethene gas, setting off the ripening process in the bananas). Ethyne, commonly called acetylene, is used as a fuel in welding blow torches.
Alkenes and alkynes
In chapter 2, we will study the nature of the bonding on alkenes and alkynes, and learn that that the bonding in alkenes is trigonal planar in in alkynes is linear. Furthermore, many alkenes can take two geometric forms: cis or trans. The cis and trans forms of a given alkene are different molecules with different physical properties because, as we will learn in chapter 2, there is a very high energy barrier to rotation about a double bond. In the example below, the difference between cis and trans alkenes is readily apparent.
We will have more to say about the subject of cis and trans alkenes in chapter 3, and we will learn much more about the reactivity of alkenes in chapter 14.
Alkanes, alkenes, and alkynes are all classified as hydrocarbons, because they are composed solely of carbon and hydrogen atoms. Alkanes are said to be saturated hydrocarbons, because the carbons are bonded to the maximum possible number of hydrogens - in other words, they are saturated with hydrogen atoms. The double and triple-bonded carbons in alkenes and alkynes have fewer hydrogen atoms bonded to them - they are thus referred to as unsaturated hydrocarbons. As we will see in chapter 15, hydrogen can be added to double and triple bonds, in a type of reaction called 'hydrogenation'.
The aromatic group is exemplified by benzene (which used to be a commonly used solvent on the organic lab, but which was shown to be carcinogenic), and naphthalene, a compound with a distinctive 'mothball' smell. Aromatic groups are planar (flat) ring structures, and are widespread in nature. We will learn more about the structure and reactions of aromatic groups in chapters 2 and 14.
Aromatics
When the carbon of an alkane is bonded to one or more halogens, the group is referred to as a alkyl halide or haloalkane. Chloroform is a useful solvent in the laboratory, and was one of the earlier anesthetic drugs used in surgery. Chlorodifluoromethane was used as a refrigerant and in aerosol sprays until the late twentieth century, but its use was discontinued after it was found to have harmful effects on the ozone layer. Bromoethane is a simple alkyl halide often used in organic synthesis. Alkyl halides groups are quite rare in biomolecules.
Haloalkanes
In the alcohol functional group, a carbon is single-bonded to an OH group (the OH group, by itself, is referred to as a hydroxyl). Except for methanol, all alcohols can be classified as primary, secondary, or tertiary. In a primary alcohol, the carbon bonded to the OH group is also bonded to only one other carbon. In a secondary alcohol and tertiary alcohol, the carbon is bonded to two or three other carbons, respectively. When the hydroxyl group is directly attached to an aromatic ring, the resulting group is called a phenol. The sulfur analog of an alcohol is called a thiol (from the Greek thio, for sulfur).
Alcohols, phenols, and thiols
Note that the definition of a phenol states that the hydroxyl oxygen must be directly attached to one of the carbons of the aromatic ring. The compound below, therefore, is not a phenol - it is a primary alcohol.
The distinction is important, because as we will see later, there is a significant difference in the reactivity of alcohols and phenols.
The deprotonated forms of alcohols, phenols, and thiols are called alkoxides, phenolates, and thiolates, respectively. A protonated alcohol is an oxonium ion.
In an ether functional group, a central oxygen is bonded to two carbons. Below is the structure of diethyl ether, a common laboratory solvent and also one of the first compounds to be used as an anesthetic during operations. The sulfur analog of an ether is called a thioether or sulfide.
Ethers and sulfides
Amines are characterized by nitrogen atoms with single bonds to hydrogen and carbon. Just as there are primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary amines. Ammonia is a special case with no carbon atoms.
One of the most important properties of amines is that they are basic, and are readily protonated to form ammonium cations. In the case where a nitrogen has four bonds to carbon (which is somewhat unusual in biomolecules), it is called a quaternary ammonium ion.
Amines
Note: Do not be confused by how the terms 'primary', 'secondary', and 'tertiary' are applied to alcohols and amines - the definitions are different. In alcohols, what matters is how many other carbons the alcohol carbon is bonded to, while in amines, what matters is how many carbons the nitrogen is bonded to.
Phosphate and its derivative functional groups are ubiquitous in biomolecules. Phosphate linked to a single organic group is called a phosphate ester; when it has two links to organic groups it is called a phosphate diester. A linkage between two phosphates creates a phosphate anhydride.
Organic phosphates
Chapter 9 of this book is devoted to the structure and reactivity of the phosphate group.
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
A group with a carbon-nitrogen double bond is called an imine, or sometimes a Schiff base (in this book we will use the term 'imine'). The chemistry of aldehydes, ketones, and imines will be covered in chapter 10.
Aldehydes, ketones, and imines
When a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to an oxygen, nitrogen, or sulfur, the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a set of related functional groups. The eponymous member of this family is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl group. The conjugate base of a carboxylic acid is a carboxylate. Other derivatives are carboxylic esters (usually just called 'esters'), thioesters, amides, acyl phosphates, acid chlorides, and acid anhydrides. With the exception of acid chlorides and acid anhydrides, the carboxylic acid derivatives are very common in biological molecules and/or metabolic pathways, and their structure and reactivity will be discussed in detail in chapter 11.
Carboxylic acid derivatives
Finally, a nitrile group is characterized by a carbon triple-bonded to a nitrogen.
Nitriles
A single compound often contains several functional groups, particularly in biological organic chemistry. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
The hormone testosterone, the amino acid phenylalanine, and the glycolysis metabolite dihydroxyacetone phosphate all contain multiple functional groups, as labeled below.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological organic chemistry. Table 9 in the tables section at the back of this book provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text.
Exercise 1.12: Identify the functional groups (other than alkanes) in the following organic compounds. State whether alcohols and amines are primary, secondary, or tertiary.
Solutions to exercises
Exercise 1.13: Draw one example each of compounds fitting the descriptions below, using line structures. Be sure to designate the location of all non-zero formal charges. All atoms should have complete octets (phosphorus may exceed the octet rule). There are many possible correct answers for these, so be sure to check your structures with your instructor or tutor.
a) a compound with molecular formula C6H11NO that includes alkene, secondary amine, and primary alcohol functional groups
b) an ion with molecular formula C3H5O6P 2- that includes aldehyde, secondary alcohol, and phosphate functional groups.
c) A compound with molecular formula C6H9NO that has an amide functional group, and does not have an alkene group.
Naming organic compounds
A system has been devised by the International Union of Pure and Applied Chemistry (IUPAC, usually pronounced eye-you-pack) for naming organic compounds. While the IUPAC system is convenient for naming relatively small, simple organic compounds, it is not generally used in the naming of biomolecules, which tend to be quite large and complex. It is, however, a good idea (even for biologists) to become familiar with the basic structure of the IUPAC system, and be able to draw simple structures based on their IUPAC names.
Naming an organic compound usually begins with identify what is referred to as the 'parent chain', which is the longest straight chain of carbon atoms. We’ll start with the simplest straight chain alkane structures. CH4 is called methane, and C2H6 ethane. The table below continues with the names of longer straight-chain alkanes: be sure to commit these to memory, as they are the basis for the rest of the IUPAC nomenclature system (and are widely used in naming biomolecules as well).
Names for straight-chain alkanes:
1 carbon: methane
2 carbons: ethane
3 carbons: propane
4 carbons: butane
5 carbons: pentane
6 carbons: hexane
7 carbons: heptane
8 carbons: octane
9 carbons: nonane
10 carbons: decane
Substituents branching from the main parent chain are located by a carbon number, with the lowest possible numbers being used (for example, notice in the example below that the compound on the left is named 1-chlorobutane, not 4-chlorobutane). When the substituents are small alkyl groups, the terms methyl, ethyl, and propyl are used.
Other common names for hydrocarbon substituent groups isopropyl, tert-butyl and phenyl.
Notice in the example below, an ‘ethyl group’ (in blue) is not treated as a substituent, rather it is included as part of the parent chain, and the methyl group is treated as a substituent. The IUPAC name for straight-chain hydrocarbons is always based on the longest possible parent chain, which in this case is four carbons, not three.
Cyclic alkanes are called cyclopropane, cyclobutane, cyclopentane, cyclohexane, and so on:
In the case of multiple substituents, the prefixes di, tri, and tetra are used.
Functional groups have characteristic suffixes. Alcohols, for example, have ‘ol’ appended to the parent chain name, along with a number designating the location of the hydroxyl group. Ketones are designated by ‘one’.
Alkenes are designated with an 'ene' ending, and when necessary the location and geometry of the double bond are indicated. Compounds with multiple double bonds are called dienes, trienes, etc.
Some groups can only be present on a terminal carbon, and thus a locating number is not necessary: aldehydes end in ‘al’, carboxylic acids in ‘oic acid’, and carboxylates in ‘oate’.
Ethers and sulfides are designated by naming the two groups on either side of the oxygen or sulfur.
If an amide has an unsubstituted –NH2 group, the suffix is simply ‘amide’. In the case of a substituted amide, the group attached to the amide nitrogen is named first, along with the letter ‘N’ to clarify where this group is located. Note that the structures below are both based on a three-carbon (propan) parent chain.
For esters, the suffix is 'oate'. The group attached to the oxygen is named first.
All of the examples we have seen so far have been simple in the sense that only one functional group was present on each molecule. There are of course many more rules in the IUPAC system, and as you can imagine, the IUPAC naming of larger molecules with multiple functional groups, ring structures, and substituents can get very unwieldy very quickly. The illicit drug cocaine, for example, has the IUPAC name 'methyl (1R,2R,3S,5S)-3-(benzoyloxy)-8-methyl-8-azabicyclo[3.2.1] octane-2-carboxylate' (this name includes designations for stereochemistry, which is a structural issue that we will not tackle until chapter 3).
You can see why the IUPAC system is not used very much in biological organic chemistry - the molecules are just too big and complex. A further complication is that, even outside of a biological context, many simple organic molecules are known almost universally by their ‘common’, rather than IUPAC names. The compounds acetic acid, chloroform, and acetone are only a few examples.
In biochemistry, nonsystematic names (like 'cocaine', 'capsaicin', 'pyruvate' or 'ascorbic acid') are usually used, and when systematic nomenclature is employed it is often specific to the class of molecule in question: different systems have evolved, for example, for fats and for carbohydrates. We will not focus very intensively in this text on IUPAC nomenclature or any other nomenclature system, but if you undertake a more advanced study in organic or biological chemistry you may be expected to learn one or more naming systems in some detail.
Exercise 1.14: Give IUPAC names for acetic acid, chloroform, and acetone.
Exercise 1.15: Draw line structures of the following compounds, based on what you have learned about the IUPAC nomenclature system:
a) methylcyclohexane
b) 5-methyl-1-hexanol
c) 2-methyl-2-butene
d) 5-chloropentanal
e) 2,2-dimethylcyclohexanone
f) 4-penteneoic acid
g) N-ethyl-N-cyclopentylbutanamide
Solutions to exercises
Drawing abbreviated organic structures
Often when drawing organic structures, chemists find it convenient to use the letter 'R' to designate part of a molecule outside of the region of interest. If we just want to refer in general to a functional group without drawing a specific molecule, for example, we can use 'R groups' to focus attention on the group of interest:
The 'R' group is a convenient way to abbreviate the structures of large biological molecules, especially when we are interested in something that is occurring specifically at one location on the molecule. For example, in chapter 15 when we look at biochemical oxidation-reduction reactions involving the flavin molecule, we will abbreviate a large part of the flavin structure which does not change at all in the reactions of interest:
As an alternative, we can use a 'break' symbol to indicate that we are looking at a small piece or section of a larger molecule. This is used commonly in the context of drawing groups on large polymers such as proteins or DNA.
Finally, 'R' groups can be used to concisely illustrate a series of related compounds, such as the family of penicillin-based antibiotics.
Using abbreviations appropriately is a very important skill to develop when studying organic chemistry in a biological context, because although many biomolecules are very large and complex (and take forever to draw!), usually we are focusing on just one small part of the molecule where a change is taking place.
As a rule, you should never abbreviate any atom involved in a bond-breaking or bond-forming event that is being illustrated: only abbreviate that part of the molecule which is not involved in the reaction of interest.
For example, carbon #2 in the reactant/product below most definitely is involved in bonding changes, and therefore should not be included in the 'R' group.
If you are unsure whether to draw out part of a structure or abbreviate it, the safest thing to do is to draw it out.
Exercise 1.16:
a) If you intend to draw out the chemical details of a reaction in which the methyl ester functional group of cocaine (see earlier figure) was converted to a carboxylate plus methanol, what would be an appropriate abbreviation to use for the cocaine structure (assuming that you only wanted to discuss the chemistry specifically occurring at the ester group)?
b) Below is the (somewhat complicated) reaction catalyzed by an enzyme known as 'Rubisco', by which plants 'fix' carbon dioxide. Carbon dioxide and the oxygen of water are colored red and blue respectively to help you see where those atoms are incorporated into the products. Propose an appropriate abbreviation for the starting compound (ribulose 1,5-bisphosphate), using two different 'R' groups, R1 and R2.
Solutions to exercises
Kahn Academy video tutorial on functional groups
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/10%3A_Carbon%3A_More_Than_Just_Another_Element/10.6%3A_Functional_Groups.txt |
Polymers are long chain, giant organic molecules are assembled from many smaller molecules called monomers. Polymers consist of many repeating monomer units in long chains, sometimes with branching or cross-linking between the chains. A polymer is analogous to a necklace made from many small beads (monomers). A chemical reaction forming polymers from monomers is called polymerization, of which there are many types. A common name for many synthetic polymer materials is plastic, which comes from the Greek word "plastikos", suitable for molding or shaping.
In the following illustrated example, many monomers called styrene are polymerized into a long chain polymer called polystyrene. The squiggly lines indicate that the polymer molecule extends further at both the left and right ends. In fact, polymer molecules are often hundreds or thousands of monomer units long.
Introduction
Many objects in daily use from packing, wrapping, and building materials include half of all polymers synthesized. Other uses include textiles, many electronic appliance casings, CD's, automobile parts, and many others are made from polymers. A quarter of the solid waste from homes is plastic materials - some of which may be recycled as shown in the table below.
Some products, such as adhesives, are made to include monomers which can be polymerized by the user in their application.
Types of Polymers
There are many types of polymers including synthetic and natural polymers.
Synthetic polymers
• Plastics
• Elastomers - solids with rubber-like qualities
• Rubber (carbon backbone often from hydrocarbon monomers)
• silicones (backbone of alternating silicon and oxygen atoms).
• Fibers
• Solid materials of intermediate characteristics
• Gels or viscous liquids
Classification of Polymers
• Homopolymers: These consist of chains with identical bonding linkages to each monomer unit. This usually implies that the polymer is made from all identical monomer molecules. These may be represented as : -[A-A-A-A-A-A]- Homopolymers are commonly named by placing the prefix poly in front of the constituent monomer name. For example, polystyrene is the name for the polymer made from the monomer styrene (vinylbenzene).
• Copolymers: These consist of chains with two or more linkages usually implying two or more different types of monomer units. These may be represented as : -[A-B-A-B-A-B]-
Polymers classified by mode of polymerization
• Addition Polymers: The monomer molecules bond to each other without the loss of any other atoms. Addition polymers from alkene monomers or substituted alkene monomers are the biggest groups of polymers in this class. Ring opening polymerization can occur without the loss of any small molecules.
• Condensation Polymers: Usually two different monomer combine with the loss of a small molecule, usually water. Most polyesters and polyamides (nylon) are in this class of polymers. Polyurethane Foam in graphic above.
Polymers classified by Physical Response to Heating
Thermoplastics
Plastics that soften when heated and become firm again when cooled. This is the more popular type of plastic because the heating and cooling may be repeated and the thermoplastic may be reformed.
Thermosets
These are plastics that soften when heated and can be molded, but harden permanently. They will decompose when reheated. An example is Bakelite, which is used in toasters, handles for pots and pans, dishes, electrical outlets and billiard balls.
Recycled Plastics
Recycle Code Abbreviation and Chemical Name of Plastic Types of Uses and Examples
1 PET - polyethylene terephthalate Many types of clear plastic consumer bottles, including clear, 2-liter beverage bottles
2 HDPE - High density polyethylene Milk jugs, detergent bottles, some water bottles, some grocery plastic bags
3 PVC - Polyvinyl chloride Plastic drain pipe, shower curtains, some water bottles
4 LDPE - Low density polyethylene Plastic garbage and other bags, garment bags, snap-on lids such as coffee can lids
5 PP - Polypropylene Many translucent (or opaque) plastic containers; containers for some products such as yogurt, soft butter, or margarine; aerosol can tops; rigid bottle caps; candy wrappers; bottoms of bottles
6 PS - Polystyrene Hard clear plastic cups, foam cups, eating utensils, deli food containers, toy model kits, some packing popcorn
7 Other Polycarbonate is a common type, Biodegradable, Some packing popcorn
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
11: Gases and Their Properties
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/10%3A_Carbon%3A_More_Than_Just_Another_Element/10.7%3A_Polymers.txt |
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
11: Intermolecular Forces and Liquids
Learning Objectives
• To be familiar with the kinetic molecular description of liquids.
The physical properties of a substance depends upon its physical state. Water vapor, liquid water and ice all have the same chemical properties, but their physical properties are considerably different. In general covalent bonds determine: molecular shape, bond energies, chemical properties, while intermolecular forces (non-covalent bonds) influence the physical properties of liquids and solids. The kinetic molecular theory of gases gives a reasonably accurate description of the behavior of gases. A similar model can be applied to liquids, but it must take into account the nonzero volumes of particles and the presence of strong intermolecular attractive forces.
The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. The intermolecular forces are attractive forces that try to draw the particles together (Figure \(2\)). A discussed previously, gasses are very sensitive to temperatures and pressure. However, these also affect liquids and solids too. Heating and cooling can change the kinetic energy of the particles in a substance, and so, we can change the physical state of a substance by heating or cooling it. Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.
Below is an overview of the general properties of the three different phases of matter.
Properties of Gases
• A collection of widely separated molecules
• The kinetic energy of the molecules is greater than any attractive forces between the molecules
• The lack of any significant attractive force between molecules allows a gas to expand to fill its container
• If attractive forces become large enough, then the gases exhibit non-ideal behavior
Properties of Liquids
• The intermolecular attractive forces are strong enough to hold molecules close together
• Liquids are more dense and less compressible than gasses
• Liquids have a definite volume, independent of the size and shape of their container
• The attractive forces are not strong enough, however, to keep neighboring molecules in a fixed position and molecules are free to move past or slide over one another
Thus, liquids can be poured and assume the shape of their containers.
Properties of Solids
• The intermolecular forces between neighboring molecules are strong enough to keep them locked in position
• Solids (like liquids) are not very compressible due to the lack of space between molecules
• If the molecules in a solid adopt a highly ordered packing arrangement, the structures are said to be crystalline
Due to the strong intermolecular forces between neighboring molecules, solids are rigid.
• Cooling a gas may change the state to a liquid
• Cooling a liquid may change the state to a solid
• Increasing the pressure on a gas may change the state to a liquid
• Increasing the pressure on a liquid may change the state to a solid
Physical Properties of Liquids
In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. In contrast, the molecules in liquids are very close together, with essentially no empty space between them. As in gases, however, the molecules in liquids are in constant motion, and their kinetic energy (and hence their speed) depends on their temperature. We begin our discussion by examining some of the characteristic properties of liquids to see how each is consistent with a modified kinetic molecular description.
The properties of liquids can be explained using a modified version of the kinetic molecular theory of gases described previously. This model explains the higher density, greater order, and lower compressibility of liquids versus gases; the thermal expansion of liquids; why they diffuse; and why they adopt the shape (but not the volume) of their containers. A kinetic molecular description of liquids must take into account both the nonzero volumes of particles and the presence of strong intermolecular attractive forces. Solids and liquids have particles that are fairly close to one another, and are thus called "condensed phases" to distinguish them from gases
• Density: The molecules of a liquid are packed relatively close together. Consequently, liquids are much denser than gases. The density of a liquid is typically about the same as the density of the solid state of the substance. Densities of liquids are therefore more commonly measured in units of grams per cubic centimeter (g/cm3) or grams per milliliter (g/mL) than in grams per liter (g/L), the unit commonly used for gases.
• Molecular Order: Liquids exhibit short-range order because strong intermolecular attractive forces cause the molecules to pack together rather tightly. Because of their higher kinetic energy compared to the molecules in a solid, however, the molecules in a liquid move rapidly with respect to one another. Thus unlike the ions in the ionic solids, the molecules in liquids are not arranged in a repeating three-dimensional array. Unlike the molecules in gases, however, the arrangement of the molecules in a liquid is not completely random.
• Compressibility: Liquids have so little empty space between their component molecules that they cannot be readily compressed. Compression would force the atoms on adjacent molecules to occupy the same region of space.
• Thermal Expansion: The intermolecular forces in liquids are strong enough to keep them from expanding significantly when heated (typically only a few percent over a 100°C temperature range). Thus the volumes of liquids are somewhat fixed. Notice from Table S1 (with a shorten version in Table \(1\)) that the density of water, for example, changes by only about 3% over a 90-degree temperature range.
Table \(1\): The Density of Water at Various Temperatures
T (°C) Density (g/cm3)
0 0.99984
30 0.99565
60 0.98320
90 0.96535
• Diffusion: Molecules in liquids diffuse because they are in constant motion. A molecule in a liquid cannot move far before colliding with another molecule, however, so the mean free path in liquids is very short, and the rate of diffusion is much slower than in gases.
• Fluidity: Liquids can flow, adjusting to the shape of their containers, because their molecules are free to move. This freedom of motion and their close spacing allow the molecules in a liquid to move rapidly into the openings left by other molecules, in turn generating more openings, and so forth (Figure \(3\)). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/11%3A_Intermolecular_Forces_and_Liquids/11.1%3A_States_of__Matter_and_Intermolecular_Forces.txt |
Ion-Dipole Interactions
Ion-Dipole Forces are involved in solutions where an ionic compound can be dissolved into a molecular solvent with polar covalent bonds, like that of the solution of table salt (NaCl) into water. For example, the sodium ion/water cluster interaction is approximately 50 KJ/mol.
$Na^+ ↔ (H_2O)_n \tag{1}$
Figure 2a: Ion-Dipole interaction. Note the oxygen end of dipole is closer to the sodium than the hydrogen end, and so the net interaction is attractive.
The name "Ion dipole forces" describes what they are, which simply speaking, are the result of the Coulombic electrostatic interactions between an ion and the charged ends of a dipole. Here the term "Intermolecular Force" is a misnomer, even though it is commonly used, as these are the forces between ions with molecules possessing a dipole moment (unequal sharing of valence electrons). To gain an understanding we can start by looking at the Coulombic potential between two ions (eq. 2).
$V= \dfrac{q_1q_2}{(4\pi\epsilon_0) r} \tag{2}$
(ion-ion potential)
• $r$ is the distance of separation (note that the potential goes to zero when they are separated by infinity)
• $q$ is the charge of the ions
Note, in eq. (2) this potential can be either attractive (opposite charges) or repulsive (like charges), and that a negative potential leads to an attractive bonding type interaction. That is, the potential reaches zero when the distance between them (r) approaches infinity, and so a bond has a negative potential, in that in accordance with the first law of thermodynamics, you add energy to break a bond, and the broken bond has zero energy (review potential well diagrams).
So what is the difference between Ion-Ion and Ion-Dipole interactions?
$V= \dfrac{q_1\mu_2}{(4\pi\epsilon_0) r} \tag{2}$
$V=-\dfrac{\mid q_1\mid\mu_2}{(4\pi\epsilon_0) r^2} \tag{3}$
(ion-dipole potential)
• r is the distance of separation.
• q is the charge of the ion ( only the magnitude of the charge is shown here.)
• $\mu$ is the permanent dipole moment of the polar molecule.
There are several differences between ion-ion (eq.2) and ion-dipole (eq.3) types of interactions. First, is that the dipole has both a positive and negative end, leading to two interactions, one being attractive and the other repulsive. In the case of the sodium cation, the negative oxygen end of the dipole is attractive, and the positive hydrogen end is repulsive. This means the dipole aligns as in figure 11.2a, with the oxygen distance to the sodium being smaller than the hydrogen end of the dipole, and thus the attractive interaction is greater than the repulsive. As μ is positive, we use the absolute value sign around the charge, and place a minus sign in front of the equation, to indicate that the net ion-dipole interaction is a lowering of the system’s energy (attractive).
The second thing to note is that the potential drops off quicker, as Eq. 3 is an inverse square relationship to the radius (1/r2), while eq. 2 is a linear inverse relations ship (1/r). This means the ion-dipole are a shorter range interaction and diminish more rapidly the father the polar molecule is from the ion. This is logical, because the relative distance of the two dipoles from the ion become less significant the farther away they are. For example, if a cation had a radius of 200pm and the distance between the centers of positive and negative charges in a polar molecule molecule were also 200 pm, and they were touching each other, the radius for the negative end would be 100 pm, while the positive would be 300 pm, that is, the attractive interaction is 3 times closer than the repulsive. Now, if we moved them to a separation of 500 pm, the distances become 600 and 800, or only 3/4ths as close. That is, as you move the polar molecule away from the charge the repulsive and attractive interactions not only diminish, but also converge to more similar values,
Two are Repulsive:
• Positive Ion - Positive Dipole Repulsions (between cation and positive end of dipole)
• Negative Ion - Negative Dipole Repulsions (between anion and negative end of dipole)
Two are Attractive:
• Positive Ion - Negative Dipole Attractions (between the positive ion and the negative end of the dipole)
• Negative Ion - Positive Dipole Attractions (between the negative ion and the positive end of the dipole).
In visualizing these types of interactions one also needs to be realize that there are attractions between the ions with each other, and between the polar molecules with each other.
__________________ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/11%3A_Intermolecular_Forces_and_Liquids/11.2%3A_Interactions_between_Ion_and_Molecules_with_a_Permanent_Dipole.txt |
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
12: The Solid State
A unit cell is the most basic and least volume consuming repeating structure of any solid. It is used to visually simplify the crystalline patterns solids arrange themselves in. When the unit cell repeats itself, the network is called a lattice.
Introduction
The work of Auguste Bravais in the early 19th century revealed that there are only fourteen different lattice structures (often referred to as Bravais lattices). These fourteen different structures are derived from seven crystal systems, which indicate the different shapes a unit cell take and four types of lattices, which tells how the atoms are arranged within the unit. The kind of cell one would pick for any solid would be dependent upon how the latices are arranged on top of one another. A method called X-ray Diffraction is used to determine how the crystal is arranged. X-ray Diffraction consists of a X-ray beam being fired at a solid, and from the diffraction of the beams calculated by Bragg's Law the configuration can be determined.
The unit cell has a number of shapes, depending on the angles between the cell edges and the relative lengths of the edges. It is the basic building block of a crystal with a special arrangement of atoms. The unit cell of a crystal can be completely specified by three vectors, a, b, c that form the edges of a parallelepiped. A crystal structure and symmetry is also considered very important because it takes a role in finding a cleavage, an electronic band structure, and an optical property. There are seven crystal systems that atoms can pack together to produce 3D space lattice.
The unit cell is generally chosen so as to contain only complete complement of the asymmetric units. In a geometrical arrangement, the crystal systems are made of three set of ($a$, $b$, $c$), which are coincident with unit cell edges and lengths. The lengths of a, b, c are called the unit cell dimensions, and their directions define the major crystallographic axes. A unit cell can be defined by specifying a, b, c, or alternately by specifying the lengths |a|, |b|, |c| and the angles between the vectors, $\alpha$, $\beta$, and $\gamma$ as shown in Fig. 1.1. Unit cell cannot have lower or higher symmetry than the aggregate of asymmetric units. There are seven crystal systems and particular kind of unit cell chosen will be determined by the type of symmetry elements relating the asymmetric units in the cell.
The unit cell is chosen to contain only one complete complement of the asymmetric units, which is called primitive (P). Unit cells that contain an asymmetric unit greater than one set are called centered or nonprimitive unit cells. The additional asymmetric unit sets are related to the first simple fractions of unit cells edges. For example, (1/2, 1/2, 1/2) for the body centered cell $I$ and (1/2,1/2, 0) for the single-face-centered cell $C$. The units can be completely specified by three vectors (a, b, c) and the lengths of vectors in angstroms are called the unit cell dimensions. Vectors directions are defined the major crystallographic axes. Unit cell can also be defied by specifying the lengths (|a|, |b|, |c|) and the angles between the vectors ($\alpha$, $\beta$, and $\gamma$) as shown in Fig.1.1.
Table 1: includes the allowable unit cell types found in crystals and their distinguishing characteristics.
Crystal System Types of Lattices (number of particles) Description of Cell
Cubic Simple or Primitive (1) Face centered (4) Body centered (2) Quadratic prism that is equilateral and equiangular (i.e. a cube)
Tetragonal Simple or Primitive (1) Body centered (2) Quadratic prism that has two equal edges and one different sized edge, and is equiangular. (i.e. a rectangular prism with a square base)
Orthorhombic Simple or Primitive (1) Face centered (4) Body centered (2) End or Base centered (2) Quadratic prism that has no equal edges, and is equiangular. (i.e. a rectangular prism without any square base faces)
Monoclinic Simple or Primitive (1) End or Base centered (2) Quadratic prism that has no equal edges, two edges at 90 degrees of each other, and one thats not at 90. (i.e. a parallelogram extended to some distant not equal to the width of the base)
Rhombohedral Simple or Primitive (1) Quadratic prism that has equal length edges, but one slanted side (i.e. a slanted cube)
Triclinic Simple or Primitive (1) Quadratic prism with no equal length edges, two unequal angles and one angle at 90 degrees.
Hexagonal Simple or Primitive (1) Hexagonal prism
Note: Edges refers to all parallel edges. And all parallel edges are equal to each other.
This table describes the fourteen different kind of unit cells available. As you can see not every crystal systems can have all the different types of lattices.
Volumes
Calculating the volume for a unit cell is the same as calculating the volume for any prism - base area multiplied by height. The equations for each different crystal system are as follow:
Crystal System Volume =
Cubic abc
Tetragonal abc
Orthorhombic abc
Monoclinic abc sin(?), where ? is the acute non 90 degree angle.
Rhombohedral abc sin(60°)
Triclinic abc ((1- cos²? - cos²? - cos²?) + 2(cos(?) cos(?) cos(?))½
Hexagonal abc sin(60°)
Note: a, b, & c are represent the edges. ?, ?, & ? are the angles.
Most calculations involving unit cells can be solved with the formula: density = Mass/Volume. Then in addition to the obvious three the number of particles per cell can also be calculated by the density/molar mass.
Density - Particles
It can easily be seen that not all the particles are complete in the unit cell form. For the fractional particles in unit cell, its corner particles will always sum to one whole particle, its face particles (for face centered lattices) will sum to three whole particles, and for the base particles will sum to one.
Contributors and Attributions
• Minh Nguyen (UCD), Mandeep Singh (UCD)
12.2: Structures and Formulas of Ionic Solids
Contributors and Attributions
• Nathalie Interiano | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/12%3A_The_Solid_State/12.1%3A_Crystall_Lattices_and_Unit_Cells.txt |
Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. These compounds have an additional stability due to the lattice energy of the solid structure. However, lattice energy cannot be directly measured. The Born-Haber cycle allows us to understand and determine the lattice energies of ionic solids.
Introduction
This module will introduce the idea of lattice energy, as well as one process that allows us to calculate it: the Born-Haber Cycle. In order to use the Born-Haber Cycle, there are several concepts that we must understand first.
Lattice Energy
Lattice Energy is a type of potential energy that may be defined in two ways. In one definition, the lattice energy is the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. This definition causes the value for the lattice energy to always be positive, since this will always be an endothermic reaction. The other definition says that lattice energy is the reverse process, meaning it is the energy released when gaseous ions bind to form an ionic solid. As implied in the definition, this process will always be exothermic, and thus the value for lattice energy will be negative. Its values are usually expressed with the units kJ/mol.
Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favorable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point.
Born-Haber Cycle
There are several important concept to understand before the Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid; ionization energy, electron affinity, dissociation energy, sublimation energy, heat of formation, and Hess's Law.
• Ionization Energy is the energy required to remove an electron from a neutral atom or an ion. This process always requires an input of energy, and thus will always have a positive value. In general, ionization energy increases across the periodic table from left to right, and decreases from top to bottom. There are some excepts, usually due to the stability of half-filled and completely filled orbitals.
• Electron Affinity is the energy released when an electron is added to a neutral atom or an ion. Usually, energy released would have a negative value, but due to the definition of electron affinity, it is written as a positive value in most tables. Therefore, when used in calculating the lattice energy, we must remember to subtract the electron affinity, not add it. In general, electron affinity increases from left to right across the periodic table and decreases from top to bottom.
• Dissociation energy is the energy required to break apart a compound. The dissociation of a compound is always an endothermic process, meaning it will always require an input of energy. Therefore, the change in energy is always positive. The magnitude of the dissociation energy depends on the electronegativity of the atoms involved.
• Sublimation energy is the energy required to cause a change of phase from solid to gas, bypassing the liquid phase. This is an input of energy, and thus has a positive value. It may also be referred to as the energy of atomization.
• The heat of formation is the change in energy when forming a compound from its elements. This may be positive or negative, depending on the atoms involved and how they interact.
• Hess's Law states that the overall change in energy of a process can be determined by breaking the process down into steps, then adding the changes in energy of each step. The Born-Haber Cycle is essentially Hess's Law applied to an ionic solid.
Using the Born-Haber Cycle
The values used in the Born-Haber Cycle are all predetermined changes in enthalpy for the processes described in the section above. Hess' Law allows us to add or subtract these values, which allows us to determine the lattice energy.
Step 1
Determine the energy of the metal and nonmetal in their elemental forms. (Elements in their natural state have an energy level of zero.) Subtract from this the heat of formation of the ionic solid that would be formed from combining these elements in the appropriate ration. This is the energy of the ionic solid, and will be used at the end of the process to determine the lattice energy.
Step 2
The Born-Haber Cycle requires that the elements involved in the reaction are in their gaseous forms. Add the changes in enthalpy to turn one of the elements into its gaseous state, and then do the same for the other element.
Step 3
Metals exist in nature as single atoms and thus no dissociation energy needs to be added for this element. However, many nonmetals will exist as polyatomic species. For example, Cl exists as Cl2 in its elemental state. The energy required to change Cl2 into 2Cl atoms must be added to the value obtained in Step 2.
Step 4
Both the metal and nonmetal now need to be changed into their ionic forms, as they would exist in the ionic solid. To do this, the ionization energy of the metal will be added to the value from Step 3. Next, the electron affinity of the nonmetal will be subtracted from the previous value. It is subtracted because it is a release of energy associated with the addition of an electron.
*This is a common error due to confusion caused by the definition of electron affinity, so be careful when doing this calculation.
Step 5
Now the metal and nonmetal will be combined to form the ionic solid. This will cause a release of energy, which is called the lattice energy. The value for the lattice energy is the difference between the value from Step 1 and the value from Step 4.
--------------------------------------------------------------------------------------------------------------------------------------------
The diagram below is another representation of the Born-Haber Cycle.
Equation
The Born-Haber Cycle can be reduced to a single equation:
Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy
*Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.
Rearrangement to solve for lattice energy gives the equation:
Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities)
Problems
1. Define lattice energy, ionization energy, and electron affinity.
2. What is Hess' Law?
3. Find the lattice energy of KF(s).
Note: Values can be found in standard tables.
4. Find the lattice energy of MgCl2(s).
5. Which one of the following has the greatest lattice energy?
1. A) MgO
2. B) NaC
3. C) LiCl
4. D) MgCl2
6. Which one of the following has the greatest Lattice Energy?
1. NaCl
2. CaCl2
3. AlCl3
4. KCl
Solutions
1. Lattice energy: The difference in energy between the expected experimental value for the energy of the ionic solid and the actual value observed. More specifically, this is the energy gap between the energy of the separate gaseous ions and the energy of the ionic solid.
Ionization energy: The energy change associated with the removal of an electron from a neutral atom or ion.
Electron affinity: The release of energy associated with the addition of an electron to a neutral atom or ion.
2. Hess' Law states that the overall energy of a reaction may be determined by breaking down the process into several steps, then adding together the changes in energy of each step.
3. Lattice Energy= [-436.68-89-(0.5*158)-418.8-(-328)] kJ/mol= -695.48 kJ/mol
4. Lattice Energy= [-641.8-146-243-(737.7+1450.6)-(2*-349)] kJ/mol= -2521.1 kJ/mol
5. MgO. It has ions with the largest charge.
6. AlCl3. According to the periodic trends, as the radius of the ion increases, lattice energy decreases. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/12%3A_The_Solid_State/12.3%3A_Bonding_in_Ionic_Compounds%3A_Lattice_Energy.txt |
In the early 1900's, Paul Drüde came up with the "sea of electrons" metallic bonding theory by modeling metals as a mixture of atomic cores (atomic cores = positive nuclei + inner shell of electrons) and valence electrons. Metallic bonds occur among metal atoms. Whereas ionic bonds join metals to non-metals, metallic bonding joins a bulk of metal atoms. A sheet of aluminum foil and a copper wire are both places where you can see metallic bonding in action.
Metals tend to have high melting points and boiling points suggesting strong bonds between the atoms. Even a soft metal like sodium (melting point 97.8°C) melts at a considerably higher temperature than the element (neon) which precedes it in the Periodic Table. Sodium has the electronic structure 1s22s22p63s1. When sodium atoms come together, the electron in the 3s atomic orbital of one sodium atom shares space with the corresponding electron on a neighboring atom to form a molecular orbital - in much the same sort of way that a covalent bond is formed.
The difference, however, is that each sodium atom is being touched by eight other sodium atoms - and the sharing occurs between the central atom and the 3s orbitals on all of the eight other atoms. Each of these eight is in turn being touched by eight sodium atoms, which in turn are touched by eight atoms - and so on and so on, until you have taken in all the atoms in that lump of sodium. All of the 3s orbitals on all of the atoms overlap to give a vast number of molecular orbitals that extend over the whole piece of metal. There have to be huge numbers of molecular orbitals, of course, because any orbital can only hold two electrons.
The electrons can move freely within these molecular orbitals, and so each electron becomes detached from its parent atom. The electrons are said to be delocalized. The metal is held together by the strong forces of attraction between the positive nuclei and the delocalized electrons (Figure $1$).
This is sometimes described as "an array of positive ions in a sea of electrons". If you are going to use this view, beware! Is a metal made up of atoms or ions? It is made of atoms. Each positive center in the diagram represents all the rest of the atom apart from the outer electron, but that electron has not been lost - it may no longer have an attachment to a particular atom, but it's still there in the structure. Sodium metal is therefore written as $\ce{Na}$, not $\ce{Na^+}$.
Example $1$: Metallic bonding in magnesium
Use the sea of electrons model to explain why Magnesium has a higher melting point (650 °C) than sodium (97.79 °C).
Solution
If you work through the same argument above for sodium with magnesium, you end up with stronger bonds and hence a higher melting point.
Magnesium has the outer electronic structure 3s2. Both of these electrons become delocalized, so the "sea" has twice the electron density as it does in sodium. The remaining "ions" also have twice the charge (if you are going to use this particular view of the metal bond) and so there will be more attraction between "ions" and "sea".
More realistically, each magnesium atom has 12 protons in the nucleus compared with sodium's 11. In both cases, the nucleus is screened from the delocalized electrons by the same number of inner electrons - the 10 electrons in the 1s2 2s2 2p6 orbitals. That means that there will be a net pull from the magnesium nucleus of 2+, but only 1+ from the sodium nucleus.
So not only will there be a greater number of delocalized electrons in magnesium, but there will also be a greater attraction for them from the magnesium nuclei. Magnesium atoms also have a slightly smaller radius than sodium atoms, and so the delocalized electrons are closer to the nuclei. Each magnesium atom also has twelve near neighbors rather than sodium's eight. Both of these factors increase the strength of the bond still further.
Note: Transition metals tend to have particularly high melting points and boiling points. The reason is that they can involve the 3d electrons in the delocalization as well as the 4s. The more electrons you can involve, the stronger the attractions tend to be.
Bulk properties of metals
Metals have several qualities that are unique, such as the ability to conduct electricity and heat, a low ionization energy, and a low electronegativity (so they will give up electrons easily to form cations). Their physical properties include a lustrous (shiny) appearance, and they are malleable and ductile. Metals have a crystal structure but can be easily deformed. In this model, the valence electrons are free, delocalized, mobile, and not associated with any particular atom. This model may account for:
• Conductivity: Since the electrons are free, if electrons from an outside source were pushed into a metal wire at one end (Figure $2$), the electrons would move through the wire and come out at the other end at the same rate (conductivity is the movement of charge).
• Malleability and Ductility: The electron-sea model of metals not only explains their electrical properties but their malleability and ductility as well. The sea of electrons surrounding the protons acts like a cushion, and so when the metal is hammered on, for instance, the overall composition of the structure of the metal is not harmed or changed. The protons may be rearranged but the sea of electrons with adjust to the new formation of protons and keep the metal intact. When one layer of ions in an electron sea moves along one space with respect to the layer below it, the crystal structure does not fracture but is only deformed (Figure $3$).
• Heat capacity: This is explained by the ability of free electrons to move about the solid.
• Luster: The free electrons can absorb photons in the "sea," so metals are opaque-looking. Electrons on the surface can bounce back light at the same frequency that the light hits the surface, therefore the metal appears to be shiny.
However, these observations are only qualitative, and not quantitative, so they cannot be tested. The "Sea of Electrons" theory stands today only as an oversimplified model of how metallic bonding works.
In a molten metal, the metallic bond is still present, although the ordered structure has been broken down. The metallic bond is not fully broken until the metal boils. That means that boiling point is actually a better guide to the strength of the metallic bond than melting point is. On melting, the bond is loosened, not broken. The strength of a metallic bond depends on three things:
1. The number of electrons that become delocalized from the metal
2. The charge of the cation (metal).
3. The size of the cation.
A strong metallic bond will be the result of more delocalized electrons, which causes the effective nuclear charge on electrons on the cation to increase, in effect making the size of the cation smaller. Metallic bonds are strong and require a great deal of energy to break, and therefore metals have high melting and boiling points. A metallic bonding theory must explain how so much bonding can occur with such few electrons (since metals are located on the left side of the periodic table and do not have many electrons in their valence shells). The theory must also account for all of a metal's unique chemical and physical properties.
Expanding the Range of Bonding Possible
Previously, we argued that bonding between atoms can classified as range of possible bonding between ionic bonds (fully charge transfer) and covalent bonds (fully shared electrons). When two atoms of slightly differing electronegativities come together to form a covalent bond, one atom attracts the electrons more than the other; this is called a polar covalent bond. However, simple “ionic” and “covalent” bonding are idealized concepts and most bonds exist on a two-dimensional continuum described by the van Arkel-Ketelaar Triangle (Figure $4$).
Bond triangles or van Arkel–Ketelaar triangles (named after Anton Eduard van Arkel and J. A. A. Ketelaar) are triangles used for showing different compounds in varying degrees of ionic, metallic and covalent bonding. In 1941 van Arkel recognized three extreme materials and associated bonding types. Using 36 main group elements, such as metals, metalloids and non-metals, he placed ionic, metallic and covalent bonds on the corners of an equilateral triangle, as well as suggested intermediate species. The bond triangle shows that chemical bonds are not just particular bonds of a specific type. Rather, bond types are interconnected and different compounds have varying degrees of different bonding character (for example, polar covalent bonds).
Using electronegativity - two compound average electronegativity on x-axis of Figure $4$.
$\sum \chi = \dfrac{\chi_A + \chi_B}{2} \label{sum}$
and electronegativity difference on y-axis,
$\Delta \chi = | \chi_A - \chi_B | \label{diff}$
we can rate the dominant bond between the compounds. On the right side of Figure $4$ (from ionic to covalent) should be compounds with varying difference in electronegativity. The compounds with equal electronegativity, such as $\ce{Cl2}$ (chlorine) are placed in the covalent corner, while the ionic corner has compounds with large electronegativity difference, such as $\ce{NaCl}$ (table salt). The bottom side (from metallic to covalent) contains compounds with varying degree of directionality in the bond. At one extreme is metallic bonds with delocalized bonding and at the other are covalent bonds in which the orbitals overlap in a particular direction. The left side (from ionic to metallic) is meant for delocalized bonds with varying electronegativity difference.
The Three Extremes in bonding
In general:
• Metallic bonds have low $\Delta \chi$ and low average $\sum\chi$.
• Ionic bonds have moderate-to-high $\Delta \chi$ and moderate values of average $\sum \chi$.
• Covalent bonds have moderate to high average $\sum \chi$ and can exist with moderately low $\Delta \chi$.
Example $2$
Use the tables of electronegativities (Table A2) and Figure $4$ to estimate the following values
• difference in electronegativity ($\Delta \chi$)
• average electronegativity in a bond ($\sum \chi$)
• percent ionic character
• likely bond type
for the selected compounds:
1. $\ce{AsH}$ (e.g., in arsine $AsH$)
2. $\ce{SrLi}$
3. $\ce{KF}$.
Solution
a: $\ce{AsH}$
• The electronegativity of $\ce{As}$ is 2.18
• The electronegativity of $\ce{H}$ is 2.22
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{2.18 + 2.22}{2} \[4pt] &= 2.2 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= 2.18 - 2.22 \[4pt] &= 0.04 \end{align*}
• From Figure $4$, the bond is fairly nonpolar and has a low ionic character (10% or less)
• The bonding is in the middle of a covalent bond and a metallic bond
b: $\ce{SrLi}$
• The electronegativity of $\ce{Sr}$ is 0.95
• The electronegativity of $\ce{Li}$ is 0.98
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{0.95 + 0.98}{2} \[4pt] &= 0.965 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= 0.98 - 0.95 \[4pt] &= 0.025 \end{align*}
• From Figure $4$, the bond is fairly nonpolar and has a low ionic character (~3% or less)
• The bonding is likely metallic.
c: $\ce{KF}$
• The electronegativity of $\ce{K}$ is 0.82
• The electronegativity of $\ce{F}$ is 3.98
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{0.82 + 3.98}{2} \[4pt] &= 2.4 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= | 0.82 - 3.98 | \[4pt] &= 3.16 \end{align*}
• From Figure $4$, the bond is fairly polar and has a high ionic character (~75%)
• The bonding is likely ionic.
Exercise $2$
Contrast the bonding of $\ce{NaCl}$ and silicon tetrafluoride.
Answer
$\ce{NaCl}$ is an ionic crystal structure, and an electrolyte when dissolved in water; $\Delta \chi =1.58$, average $\sum \chi =1.79$, while silicon tetrafluoride is covalent (molecular, non-polar gas; $\Delta \chi =2.08$, average $\sum \chi =2.94$.
13: Solutions and Their Behavior
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
14: Chemical Kinetics: The Rates of Chemical Reactions
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
15: Principles of Chemical Reactivity: Equilibria
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
16: Principles of Chemical Reactivity: The Chemistry of Acids and Bases
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
17.E: Exercises
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
17: Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria
1
18: Principles of Chemical Reactivity: Entropy and Free Energy
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
19: Principles of Chemical Reactivity: Electron Transfer Reactions
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
20: Environmental Chemistry- Earth's Environment Energy and Sustainability
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/12%3A_The_Solid_State/12.4%3A_Bonding_in_Metals_and_Semiconductors.txt |
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
21: The Chemistry of the Main Group Elements
Learning Objectives
• To describe the physical and chemical properties of hydrogen and predict its reactivity.
We now turn from an overview of periodic trends to a discussion of the s-block elements, first by focusing on hydrogen, whose chemistry is sufficiently distinct and important to be discussed in a category of its own. Most versions of the periodic table place hydrogen in the upper left corner immediately above lithium, implying that hydrogen, with a 1s1 electron configuration, is a member of group 1. In fact, the chemistry of hydrogen does not greatly resemble that of the metals of Group 1. Indeed, some versions of the periodic table place hydrogen above fluorine in Group 17 because the addition of a single electron to a hydrogen atom completes its valence shell.
Although hydrogen has an ns1 electron configuration, its chemistry does not resemble that of the Group 1 metals.
Isotopes of Hydrogen
Hydrogen, the most abundant element in the universe, is the ultimate source of all other elements by the process of nuclear fusion. Table $1$ "The Isotopes of Hydrogen" compares the three isotopes of hydrogen, all of which contain one proton and one electron per atom. The most common isotope is protium (1H or H), followed by deuterium (2H or D), which has an additional neutron. The rarest isotope of hydrogen is tritium (3H or T), which is produced in the upper atmosphere by a nuclear reaction when cosmic rays strike nitrogen and other atoms; it is then washed into the oceans by rainfall. Tritium is radioactive, decaying to 3He with a half-life of only 12.32 years. Consequently, the atmosphere and oceans contain only a very low, steady-state level of tritium. The term hydrogen and the symbol H normally refer to the naturally occurring mixture of the three isotopes.
Table $1$: The Isotopes of Hydrogen
Protium Deuterium Tritium
symbol $\mathrm{_1^1H}$ $\mathrm{_1^2H}$ $\mathrm{_1^3H}$
neutrons 0 1 2
mass (amu) 1.00783 2.0140 3.01605
abundance (%) 99.9885 0.0115 ~10−17
half-life (years) 12.32
boiling point of X2 (K) 20.28 23.67 25
melting point/boiling point of X2O (°C) 0.0/100.0 3.8/101.4 4.5/?
The different masses of the three isotopes of hydrogen cause them to have different physical properties. Thus H2, D2, and T2 differ in their melting points, boiling points, densities, and heats of fusion and vaporization. In 1931, Harold Urey and coworkers discovered deuterium by slowly evaporating several liters of liquid hydrogen until a volume of about 1 mL remained. When that remaining liquid was vaporized and its emission spectrum examined, they observed new absorption lines in addition to those previously identified as originating from hydrogen. The natural abundance of tritium, in contrast, is so low that it could not be detected by similar experiments; it was first prepared in 1934 by a nuclear reaction.
Harold Urey (1893–1981)
Urey won the Nobel Prize in Chemistry in 1934 for his discovery of deuterium (2H). Urey was born and educated in rural Indiana. After earning a BS in zoology from the University of Montana in 1917, Urey changed career directions. He earned his PhD in chemistry at Berkeley with G. N. Lewis and subsequently worked with Niels Bohr in Copenhagen. During World War II, Urey was the director of war research for the Atom Bomb Project at Columbia University. In later years, his research focused on the evolution of life. In 1953, he and his graduate student, Stanley Miller, showed that organic compounds, including amino acids, could be formed by passing an electric discharge through a mixture of compounds thought to be present in the atmosphere of primitive Earth.
Because the normal boiling point of D2O is 101.4°C (compared to 100.0°C for H2O), evaporation or fractional distillation can be used to increase the concentration of deuterium in a sample of water by the selective removal of the more volatile H2O. Thus bodies of water that have no outlet, such as the Great Salt Lake and the Dead Sea, which maintain their level solely by evaporation, have significantly higher concentrations of deuterated water than does lake or seawater with at least one outlet. A more efficient way to obtain water highly enriched in deuterium is by prolonged electrolysis of an aqueous solution. Because a deuteron (D+) has twice the mass of a proton (H+), it diffuses more slowly toward the electrode surface. Consequently, the gas evolved at the cathode is enriched in H, the species that diffuses more rapidly, favoring the formation of H2 over D2 or HD. Meanwhile, the solution becomes enriched in deuterium. Deuterium-rich water is called heavy water because the density of D2O (1.1044 g/cm3 at 25°C) is greater than that of H2O (0.99978 g/cm3). Heavy water was an important constituent of early nuclear reactors.
Because deuterons diffuse so much more slowly, D2O will not support life and is actually toxic if administered to mammals in large amounts. The rate-limiting step in many important reactions catalyzed by enzymes involves proton transfer. The transfer of D+ is so slow compared with that of H+ because bonds to D break more slowly than those to H, so the delicate balance of reactions in the cell is disrupted. Nonetheless, deuterium and tritium are important research tools for biochemists. By incorporating these isotopes into specific positions in selected molecules, where they act as labels, or tracers, biochemists can follow the path of a molecule through an organism or a cell. Tracers can also be used to provide information about the mechanism of enzymatic reactions.
Bonding in Hydrogen and Hydrogen-Containing Compounds
The 1s1 electron configuration of hydrogen indicates a single valence electron. Because the 1s orbital has a maximum capacity of two electrons, hydrogen can form compounds with other elements in three ways (Figure $1$):
1. Losing its electron to form a proton (H+) with an empty 1s orbital. The proton is a Lewis acid that can accept a pair of electrons from another atom to form an electron-pair bond. In the acid–base reactions, the proton always binds to a lone pair of electrons on an atom in another molecule to form a polar covalent bond. If the lone pair of electrons belongs to an oxygen atom of a water molecule, the result is the hydronium ion (H3O+).
2. Accepting an electron to form a hydride ion (H), which has a filled 1s2 orbital. Hydrogen reacts with relatively electropositive metals, such as the alkali metals (group 1) and alkaline earth metals (group 2), to form ionic hydrides, which contain metal cations and H ions.
3. Sharing its electron with an electron on another atom to form an electron-pair bond. With a half-filled 1s1 orbital, the hydrogen atom can interact with singly occupied orbitals on other atoms to form either a covalent or a polar covalent electron-pair bond, depending on the electronegativity of the other atom.
Hydrogen can also act as a bridge between two atoms. One familiar example is the hydrogen bond, an electrostatic interaction between a hydrogen bonded to an electronegative atom and an atom that has one or more lone pairs of electrons (Figure $2$). An example of this kind of interaction is the hydrogen bonding network found in water (Figure $2$). Hydrogen can also form a three-center bond (or electron-deficient bond), in which a hydride bridges two electropositive atoms. Compounds that contain hydrogen bonded to boron and similar elements often have this type of bonding. The B–H–B units found in boron hydrides cannot be described in terms of localized electron-pair bonds.
Because the H atom in the middle of such a unit can accommodate a maximum of only two electrons in its 1s orbital, the B–H–B unit can be described as containing a hydride that interacts simultaneously with empty sp3 orbitals on two boron atoms (Figure $3$). In these bonds, only two bonding electrons are used to hold three atoms together, making them electron-deficient bonds. You encountered a similar phenomenon in the discussion of π bonding in ozone and the nitrite ion. Recall that in both these cases, we used the presence of two electrons in a π molecular orbital extending over three atoms to explain the fact that the two O–O bond distances in ozone and the two N–O bond distances in nitrite are the same, which otherwise can be explained only by the use of resonance structures.
Hydrogen can lose its electron to form H+, accept an electron to form H, share its electron, hydrogen bond, or form a three-center bond.
Synthesis, Reactions, and Compounds of Hydrogen
The first known preparation of elemental hydrogen was in 1671, when Robert Boyle dissolved iron in dilute acid and obtained a colorless, odorless, gaseous product. Hydrogen was finally identified as an element in 1766, when Henry Cavendish showed that water was the sole product of the reaction of the gas with oxygen. The explosive properties of mixtures of hydrogen with air were not discovered until early in the 18th century; they partially caused the spectacular explosion of the hydrogen-filled dirigible Hindenburg in 1937 (Figure $4$). Due to its extremely low molecular mass, hydrogen gas is difficult to condense to a liquid (boiling point = 20.3 K), and solid hydrogen has one of the lowest melting points known (13.8 K).
The most common way to produce small amounts of highly pure hydrogen gas in the laboratory was discovered by Boyle: reacting an active metal (M), such as iron, magnesium, or zinc, with dilute acid:
$M_{(s)} + 2H^+_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)} \label{21.1}$
Hydrogen gas can also be generated by reacting metals such as aluminum or zinc with a strong base:
$\mathrm{Al(s)}+\mathrm{OH^-(aq)}+\mathrm{3H_2O(l)}\rightarrow\frac{3}{2}\mathrm{H_2(g)}+\mathrm{[Al(OH)_4]^-(aq)} \label{21.2}$
Solid commercial drain cleaners such as Drano use this reaction to generate gas bubbles that help break up clogs in a drainpipe. Hydrogen gas is also produced by reacting ionic hydrides with water. Because ionic hydrides are expensive, however, this reaction is generally used for only specialized purposes, such as producing HD gas by reacting a hydride with D2O:
$MH_{(s)} + D_2O(l) \rightarrow HD_{(g)} + M^+(aq) + OD^−_{(aq)} \label{21.3}$
On an industrial scale, H2 is produced from methane by means of catalytic steam reforming, a method used to convert hydrocarbons to a mixture of CO and H2 known as synthesis gas, or syngas. The process is carried out at elevated temperatures (800°C) in the presence of a nickel catalyst:
$\mathrm{CH_4(g)}+\mathrm{H_2O(g)}\xrightarrow{\mathrm{Ni}}\mathrm{CO(g)}+\mathrm{3H_2(g)} \label{21.4}$
Most of the elements in the periodic table form binary compounds with hydrogen, which are collectively referred to as hydrides. Binary hydrides in turn can be classified in one of three ways, each with its own characteristic properties. Covalent hydrides contain hydrogen bonded to another atom via a covalent bond or a polar covalent bond. Covalent hydrides are usually molecular substances that are relatively volatile and have low melting points. Ionic hydrides contain the hydride ion as the anion with cations derived from electropositive metals. Like most ionic compounds, they are typically nonvolatile solids that contain three-dimensional lattices of cations and anions. Unlike most ionic compounds, however, they often decompose to H2(g) and the parent metal after heating. Metallic hydrides are formed by hydrogen and less electropositive metals such as the transition metals. The properties of metallic hydrides are usually similar to those of the parent metal. Consequently, metallic hydrides are best viewed as metals that contain many hydrogen atoms present as interstitial impurities.
Covalent hydrides are relatively volatile and have low melting points; ionic hydrides are generally nonvolatile solids in a lattice framework.
Summary and Key Takeaway
Hydrogen can lose an electron to form a proton, gain an electron to form a hydride ion, or form a covalent bond or polar covalent electron-pair bond. The three isotopes of hydrogen—protium (1H or H), deuterium (2H or D), and tritium (3H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell. Hydrogen can form compounds that contain a proton (H+), a hydride ion (H), an electron-pair bond to H, a hydrogen bond, or a three-center bond (or electron-deficient bond), in which two electrons are shared between three atoms. Hydrogen gas can be generated by reacting an active metal with dilute acid, reacting Al or Zn with a strong base, or industrially by catalytic steam reforming, which produces synthesis gas, or syngas. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.03%3A__Hydrogen.txt |
Learning Objectives
1. To describe how the alkali metals are isolated.
2. To be familiar with the reactions, compounds, and complexes of the alkali metals.
The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
Humphry Davy (1778–1829)
Davy was born in Penzance, Cornwall, England. He was a bit of a wild man in the laboratory, often smelling and tasting the products of his experiments, which almost certainly shortened his life. He discovered the physiological effects that cause nitrous oxide to be called “laughing gas” (and became addicted to it!), and he almost lost his eyesight in an explosion of nitrogen trichloride (NCl3), which he was the first to prepare. Davy was one of the first to recognize the utility of Alessandro Volta’s “electric piles” (batteries). By connecting several “piles” in series and inserting electrodes into molten salts of the alkali metals and alkaline earth metals, he was able to isolate six previously unknown elements as pure metals: sodium, potassium, calcium, strontium, barium, and magnesium. He also discovered boron and was the first to prepare phosphine (PH3) and hydrogen telluride (H2Te), both of which are highly toxic.
Robert Wilhelm Bunsen (1811–1899)
Bunsen was born and educated in Göttingen, Germany. His early work dealt with organic arsenic compounds, whose highly toxic nature and explosive tendencies almost killed him and did cost him an eye. He designed the Bunsen burner, a reliable gas burner, and used it and emission spectra to discover cesium (named for its blue line) and rubidium (named for its red line).
Preparation of the Alkali Metals
Because the alkali metals are among the most potent reductants known, obtaining them in pure form requires a considerable input of energy. Pure lithium and sodium for example, are typically prepared by the electrolytic reduction of molten chlorides:
$\mathrm{LiCl(l)}\rightarrow\mathrm{Li(l)}+\frac{1}{2}\mathrm{Cl_2(g)} \label{21.15}$
In practice, CaCl2 is mixed with LiCl to lower the melting point of the lithium salt. The electrolysis is carried out in an argon atmosphere rather than the nitrogen atmosphere typically used for substances that are highly reactive with O2 and water because Li reacts with nitrogen gas to form lithium nitride (Li3N). Metallic sodium is produced by the electrolysis of a molten mixture of NaCl and CaCl2. In contrast, potassium is produced commercially from the reduction of KCl by Na, followed by the fractional distillation of K(g). Although rubidium and cesium can also be produced by electrolysis, they are usually obtained by reacting their hydroxide salts with a reductant such as Mg:
$2RbOH_{(s)} + Mg_{(s)} \rightarrow 2Rb_{(l)} + Mg(OH)_{2(s)} \label{21.6}$
Massive deposits of essentially pure NaCl and KCl are found in nature and are the major sources of sodium and potassium. The other alkali metals are found in low concentrations in a wide variety of minerals, but ores that contain high concentrations of these elements are relatively rare. No concentrated sources of rubidium are known, for example, even though it is the 16th most abundant element on Earth. Rubidium is obtained commercially by isolating the 2%–4% of Rb present as an impurity in micas, minerals that are composed of sheets of complex hydrated potassium–aluminum silicates.
Alkali metals are recovered from silicate ores in a multistep process that takes advantage of the pH-dependent solubility of selected salts of each metal ion. The steps in this process are leaching, which uses sulfuric acid to dissolve the desired alkali metal ion and Al3+ from the ore; basic precipitation to remove Al3+ from the mixture as Al(OH)3; selective precipitation of the insoluble alkali metal carbonate; dissolution of the salt again in hydrochloric acid; and isolation of the metal by evaporation and electrolysis. Figure $1$ illustrates the isolation of liquid lithium from a lithium silicate ore by this process.
General Properties of the Alkali Metals
Various properties of the group 1 elements are summarized in Table $1$. In keeping with overall periodic trends, the atomic and ionic radii increase smoothly from Li to Cs, and the first ionization energies decrease as the atoms become larger. As a result of their low first ionization energies, the alkali metals have an overwhelming tendency to form ionic compounds where they have a +1 charge. All the alkali metals have relatively high electron affinities because the addition of an electron produces an anion (M−) with an ns2 electron configuration. The densities of the elements generally increase from Li to Cs, reflecting another common trend: because the atomic masses of the elements increase more rapidly than the atomic volumes as you go down a group, the densest elements are near the bottom of the periodic table. An unusual trend in the group 1 elements is the smooth decrease in the melting and boiling points from Li to Cs. As a result, Cs (melting point = 28.5°C) is one of only three metals (the others are Ga and Hg) that are liquids at body temperature (37°C).
Table $1$: Selected Properties of the Group 1 Elements
Lithium Sodium Potassium Rubidium Cesium Francium
*The values cited are for four-coordinate ions except for Rb+ and Cs+, whose values are given for the six-coordinate ion.
atomic symbol Li Na K Rb Cs Fr
atomic number 3 11 19 37 55 87
atomic mass 6.94 22.99 39.10 85.47 132.91 223
valence electron configuration 2s1 3s1 4s1 5s1 6s1 7s1
melting point/boiling point (°C) 180.5/1342 97.8/883 63.5/759 39.3/688 28.5/671 27/—
density (g/cm3) at 25°C 0.534 0.97 0.89 1.53 1.93
atomic radius (pm) 167 190 243 265 298
first ionization energy (kJ/mol) 520 496 419 403 376 393
most common oxidation state +1 +1 +1 +1 +1 +1
ionic radius (pm)* 76 102 138 152 167
electron affinity (kJ/mol) −60 −53 −48 −47 −46
electronegativity 1.0 0.9 0.8 0.8 0.8 0.7
standard electrode potential (E°, V) −3.04 −2.71 −2.93 −2.98 −3.03
product of reaction with O2 Li2O Na2O2 KO2 RbO2 CsO2
type of oxide basic basic basic basic basic
product of reaction with N2 Li3N none none none none
product of reaction with X2 LiX NaX KX RbX CsX
product of reaction with H2 LiH NaH KH RbH CsH
The standard reduction potentials (E°) of the alkali metals do not follow the trend based on ionization energies. Unexpectedly, lithium is the strongest reductant, and sodium is the weakest (Table $1$). Because Li+ is much smaller than the other alkali metal cations, its hydration energy is the highest. The high hydration energy of Li+ more than compensates for its higher ionization energy, making lithium metal the strongest reductant in aqueous solution. This apparent anomaly is an example of how the physical or the chemical behaviors of the elements in a group are often determined by the subtle interplay of opposing periodic trends.
Reactions and Compounds of the Alkali Metals
All alkali metals are electropositive elements with an ns1 valence electron configuration, forming the monocation (M+) by losing the single valence electron. Because removing a second electron would require breaking into the (n − 1) closed shell, which is energetically prohibitive, the chemistry of the alkali metals is largely that of ionic compounds that contain M+ ions. However, as we discuss later, the lighter group 1 elements also form a series of organometallic compounds that contain polar covalent M–C bonds.
All the alkali metals react vigorously with the halogens (group 17) to form the corresponding ionic halides, where $X$ is a halogen:
$2M_{(s)} + X_{2(s, l, g)} \rightarrow 2M^+X^−_{(s)} \label{21.7}$
Similarly, the alkali metals react with the heavier chalcogens (sulfur, selenium, and tellurium in group 16) to produce metal chalcogenides, where Y is S, Se, or Te:
$2M_{(s)} + Y_{(s)} \rightarrow M_2Y_{(s)} \label{21.8}$
When excess chalcogen is used, however, a variety of products can be obtained that contain chains of chalcogen atoms, such as the sodium polysulfides (Na2Sn, where n = 2–6). For example, Na2S3 contains the S32− ion, which is V shaped with an S–S–S angle of about 103°. The one-electron oxidation product of the trisulfide ion (S3) is responsible for the intense blue color of the gemstones lapis lazuli and blue ultramarine (Figure $2$).
Reacting the alkali metals with oxygen, the lightest element in group 16, is more complex, and the stoichiometry of the product depends on both the metal:oxygen ratio and the size of the metal atom. For instance, when alkali metals burn in air, the observed products are Li2O (white), Na2O2 (pale yellow), KO2 (orange), RbO2 (brown), and CsO2 (orange). Only Li2O has the stoichiometry expected for a substance that contains two M+ cations and one O2− ion. In contrast, Na2O2 contains the O22− (peroxide) anion plus two Na+ cations. The other three salts, with stoichiometry MO2, contain the M+ cation and the O2 (superoxide) ion. Because O2− is the smallest of the three oxygen anions, it forms a stable ionic lattice with the smallest alkali metal cation (Li+). In contrast, the larger alkali metals—potassium, rubidium, and cesium—react with oxygen in air to give the metal superoxides. Because the Na+ cation is intermediate in size, sodium reacts with oxygen to form a compound with an intermediate stoichiometry: sodium peroxide. Under specific reaction conditions, however, it is possible to prepare the oxide, peroxide, and superoxide salts of all five alkali metals, except for lithium superoxide (LiO2).
The chemistry of the alkali metals is largely that of ionic compounds containing the M+ ions.
The alkali metal peroxides and superoxides are potent oxidants that react, often vigorously, with a wide variety of reducing agents, such as charcoal or aluminum metal. For example, Na2O2 is used industrially for bleaching paper, wood pulp, and fabrics such as linen and cotton. In submarines, Na2O2 and KO2 are used to purify and regenerate the air by removing the CO2 produced by respiration and replacing it with O2. Both compounds react with CO2 in a redox reaction in which O22− or O2 is simultaneously oxidized and reduced, producing the metal carbonate and O2:
$2Na_2O_{2(s)} + 2CO_{2(g)} \rightarrow 2Na_2CO_{3(s)} + O_{2(g)} \label{21.9}$
$4KO_{2(s)} + 2CO_{2(g)} \rightarrow 2K_2CO_{3(s)} + 3O_{2(g)} \label{21.10}$
The presence of water vapor, the other product of respiration, makes KO2 even more effective at removing CO2 because potassium bicarbonate, rather than potassium carbonate, is formed:
$4KO_{2(s)} + 4CO_{2(g)} + 2H_2O_{(g)} \rightarrow 4KHCO_{3(s)} + 3O_{2(g)} \label{21.11}$
Notice that 4 mol of CO2 are removed in this reaction, rather than 2 mol in Equation 21.10.
Lithium, the lightest alkali metal, is the only one that reacts with atmospheric nitrogen, forming lithium nitride (Li3N). Lattice energies again explain why the larger alkali metals such as potassium do not form nitrides: packing three large K+ cations around a single relatively small anion is energetically unfavorable. In contrast, all the alkali metals react with the larger group 15 elements phosphorus and arsenic to form metal phosphides and arsenides (where Z is P or As):
$12M_{(s)} + Z_{4(s)} \rightarrow 4M_3Z_{(s)} \label{21.12}$
Because of lattice energies, only lithium forms a stable oxide and nitride.
The alkali metals react with all group 14 elements, but the compositions and properties of the products vary significantly. For example, reaction with the heavier group 14 elements gives materials that contain polyatomic anions and three-dimensional cage structures, such as K4Si4 whose structure is shown here. In contrast, lithium and sodium are oxidized by carbon to produce a compound with the stoichiometry M2C2 (where M is Li or Na):
$2M_{(s)} + 2C_{(s)} \rightarrow M_2C_{2(s)} \label{21.13}$
The same compounds can be obtained by reacting the metal with acetylene (C2H2). In this reaction, the metal is again oxidized, and hydrogen is reduced:
$2M_{(s)} + C_2H_{2(g)} \rightarrow M_2C_{2(s)} + H_{2(g)} \label{21.14}$
The acetylide ion (C22−), formally derived from acetylene by the loss of both hydrogens as protons, is a very strong base. Reacting acetylide salts with water produces acetylene and MOH(aq).
The heavier alkali metals (K, Rb, and Cs) also react with carbon in the form of graphite. Instead of disrupting the hexagonal sheets of carbon atoms, however, the metals insert themselves between the sheets of carbon atoms to give new substances called graphite intercalation compounds (part (a) in Figure $3$). The stoichiometries of these compounds include MC60 and MC48, which are black/gray; MC36 and MC24, which are blue; and MC8, which is bronze (part (b) in Figure $3$). The remarkably high electrical conductivity of these compounds (about 200 times greater than graphite) is attributed to a net transfer of the valence electron of the alkali metal to the graphite layers to produce, for example, K+C8.
All the alkali metals react directly with gaseous hydrogen at elevated temperatures to produce ionic hydrides (M+H):
$2M_{(s)} + H_{2(g)} \rightarrow 2MH_{(s)} \label{21.15a}$
All are also capable of reducing water to produce hydrogen gas:
$\mathrm{M(s)}+\mathrm{H_2O(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{MOH(aq)} \label{21.16}$
Although lithium reacts rather slowly with water, sodium reacts quite vigorously (Figure $4$), and the heavier alkali metals (K, Rb, and Cs) react so vigorously that they invariably explode. This trend, which is not consistent with the relative magnitudes of the reduction potentials of the elements, serves as another example of the complex interplay of different forces and phenomena—in this case, kinetics and thermodynamics. Although the driving force for the reaction is greatest for lithium, the heavier metals have lower melting points. The heat liberated by the reaction causes them to melt, and the larger surface area of the liquid metal in contact with water greatly accelerates the reaction rate.
Alkali metal cations are found in a wide variety of ionic compounds. In general, any alkali metal salt can be prepared by reacting the alkali metal hydroxide with an acid and then evaporating the water:
$2MOH_{(aq)} + H_2SO_{4(aq)} \rightarrow M_2SO_{4(aq)} + 2H_2O_{(l)} \label{21.17}$
$MOH_{(aq)} + HNO_{3(aq)} \rightarrow MNO_{3(aq)} + H_2O_{(l)} \label{21.18}$
Hydroxides of alkali metals also can react with organic compounds that contain an acidic hydrogen to produce a salt. An example is the preparation of sodium acetate (CH3CO2Na) by reacting sodium hydroxide and acetic acid:
$CH_3CO_2H_{(aq)} + NaOH_{(s)} \rightarrow CH_3CO_2Na_{(aq)} + H_2O_{(l)} \label{21.19}$
Soap is a mixture of the sodium and potassium salts of naturally occurring carboxylic acids, such as palmitic acid [CH3(CH2)14CO2H] and stearic acid [CH3(CH2)16CO2H]. Lithium salts, such as lithium stearate [CH3(CH2)14CO2Li], are used as additives in motor oils and greases.
Complexes of the Alkali Metals
Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to react with simple Lewis bases to form metal complexes. Complex formation is most significant for the smallest cation (Li+) and decreases with increasing radius. In aqueous solution, for example, Li+ forms the tetrahedral [Li(H2O)4]+ complex. In contrast, the larger alkali metal cations form octahedral [M(H2O)6]+ complexes. Complex formation is primarily due to the electrostatic interaction of the metal cation with polar water molecules. Because of their high affinity for water, anhydrous salts that contain Li+ and Na+ ions (such as Na2SO4) are often used as drying agents. These compounds absorb trace amounts of water from nonaqueous solutions to form hydrated salts, which are then easily removed from the solution by filtration.
Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to form complexes with simple Lewis bases.
Electrostatic interactions also allow alkali metal ions to form complexes with certain cyclic polyethers and related compounds, such as crown ethers and cryptands. As discussed in Chapter 13, crown ethers are cyclic polyethers that contain four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms, and crown ethers with rings of different sizes prefer to bind metal ions that fit into the cavity. For example, 14-crown-4, with the smallest cavity that can accommodate a metal ion, has the highest affinity for Li+, whereas 18-crown-6 forms the strongest complexes with K+.
Cryptands are more nearly spherical analogues of crown ethers and are even more powerful and selective complexing agents. Cryptands consist of three chains containing oxygen that are connected by two nitrogen atoms (part (b) in Figure 13.7). They can completely surround (encapsulate) a metal ion of the appropriate size, coordinating to the metal by a lone pair of electrons on each O atom and the two N atoms. Like crown ethers, cryptands with different cavity sizes are highly selective for metal ions of particular sizes. Crown ethers and cryptands are often used to dissolve simple inorganic salts such as KMnO4 in nonpolar organic solvents.
Liquid Ammonia Solutions
A remarkable feature of the alkali metals is their ability to dissolve reversibly in liquid ammonia. Just as in their reactions with water, reacting alkali metals with liquid ammonia eventually produces hydrogen gas and the metal salt of the conjugate base of the solvent—in this case, the amide ion (NH2) rather than hydroxide:
$\mathrm{M(s)}+\mathrm{NH_3(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{M^+(am)}+\mathrm{NH_2^-(am)} \label{21.20}$
where the (am) designation refers to an ammonia solution, analogous to (aq) used to indicate aqueous solutions. Without a catalyst, the reaction in Equation 21.20 tends to be rather slow. In many cases, the alkali metal amide salt (MNH2) is not very soluble in liquid ammonia and precipitates, but when dissolved, very concentrated solutions of the alkali metal are produced. One mole of Cs metal, for example, will dissolve in as little as 53 mL (40 g) of liquid ammonia. The pure metal is easily recovered when the ammonia evaporates.
Solutions of alkali metals in liquid ammonia are intensely colored and good conductors of electricity due to the presence of solvated electrons (e, NH3), which are not attached to single atoms. A solvated electron is loosely associated with a cavity in the ammonia solvent that is stabilized by hydrogen bonds. Alkali metal–liquid ammonia solutions of about 3 M or less are deep blue (Figure $5$) and conduct electricity about 10 times better than an aqueous NaCl solution because of the high mobility of the solvated electrons. As the concentration of the metal increases above 3 M, the color changes to metallic bronze or gold, and the conductivity increases to a value comparable with that of the pure liquid metals.
In addition to solvated electrons, solutions of alkali metals in liquid ammonia contain the metal cation (M+), the neutral metal atom (M), metal dimers (M2), and the metal anion (M). The anion is formed by adding an electron to the singly occupied ns valence orbital of the metal atom. Even in the absence of a catalyst, these solutions are not very stable and eventually decompose to the thermodynamically favored products: M+NH2 and hydrogen gas (Equation 21.20). Nonetheless, the solvated electron is a potent reductant that is often used in synthetic chemistry.
Organometallic Compounds of the Group 1 Elements
Compounds that contain a metal covalently bonded to a carbon atom of an organic species are called organometallic compounds. The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components. Because of its small size, lithium, for example, forms an extensive series of covalent organolithium compounds, such as methyllithium (LiCH3), which are by far the most stable and best-known group 1 organometallic compounds. These volatile, low-melting-point solids or liquids can be sublimed or distilled at relatively low temperatures and are soluble in nonpolar solvents. Like organic compounds, the molten solids do not conduct electricity to any significant degree. Organolithium compounds have a tendency to form oligomers with the formula (RLi)n, where R represents the organic component. For example, in both the solid state and solution, methyllithium exists as a tetramer with the structure shown in Figure $6$, where each triangular face of the Li4 tetrahedron is bridged by the carbon atom of a methyl group. Effectively, the carbon atom of each CH3 group is using a single pair of electrons in an sp3 hybrid lobe to bridge three lithium atoms, making this an example of two-electron, four-center bonding. Clearly, such a structure, in which each carbon atom is apparently bonded to six other atoms, cannot be explained using any of the electron-pair bonding schemes. Molecular orbital theory can explain the bonding in methyllithium, but the description is beyond the scope of this text.
• The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components.
• Organosodium and organopotassium compounds are more ionic than organolithium compounds. They contain discrete M+ and R ions and are insoluble or only sparingly soluble in nonpolar solvents.
Uses of the Alkali Metals
Because sodium remains liquid over a wide temperature range (97.8–883°C), it is used as a coolant in specialized high-temperature applications, such as nuclear reactors and the exhaust valves in high-performance sports car engines. Cesium, because of its low ionization energy, is used in photosensors in automatic doors, toilets, burglar alarms, and other electronic devices. In these devices, cesium is ionized by a beam of visible light, thereby producing a small electric current; blocking the light interrupts the electric current and triggers a response.
Compounds of sodium and potassium are produced on a huge scale in industry. Each year, the top 50 industrial compounds include NaOH, used in a wide variety of industrial processes; Na2CO3, used in the manufacture of glass; K2O, used in porcelain glazes; and Na4SiO4, used in detergents.
Several other alkali metal compounds are also important. For example, Li2CO3 is one of the most effective treatments available for manic depression or bipolar disorder. It appears to modulate or dampen the effect on the brain of changes in the level of neurotransmitters, which are biochemical substances responsible for transmitting nerve impulses between neurons. Consequently, patients who take “lithium” do not exhibit the extreme mood swings that characterize this disorder.
$1$
For each application, choose the more appropriate substance based on the properties and reactivities of the alkali metals and their compounds. Explain your choice in each case.
1. For a reaction that requires a strong base in a solution of tetrahydrofuran (THF), would you use LiOH or CsOH?
2. To extinguish a fire caused by burning lithium metal, would you use water, CO2, N2 gas, or sand (SiO2)?
3. Both LiNO3and CsNO3 are highly soluble in acetone (2-propanone). Which of these alkali metal salts would you use to precipitate I from an acetone solution?
Given: application and selected alkali metals
Asked for: appropriate metal for each application
Strategy:
Use the properties and reactivities discussed in this section to determine which alkali metal is most suitable for the indicated application.
Solution
1. Both LiOH and CsOH are ionic compounds that contain the hydroxide anion. Li+, however, is much smaller than Cs+, so the Li+ cation will be more effectively solvated by the oxygen of THF with its lone pairs of electrons. This difference will have two effects: (1) LiOH is likely to be much more soluble than CsOH in the nonpolar solvent, which could be a significant advantage, and (2) the solvated Li+ ions are less likely to form tight ion pairs with the OH ions in the relatively nonpolar solution, making the OH more basic and thus more reactive. Thus LiOH is the better choice.
2. Lithium is a potent reductant that reacts with water to form LiOH and H2 gas, so adding a source of hydrogen such as water to a lithium fire is likely to produce an explosion. Lithium also reacts with oxygen and nitrogen in the air to form Li2O and Li3N, respectively, so we would not expect nitrogen to extinguish a lithium fire. Because CO2 is a gaseous molecule that contains carbon in its highest accessible oxidation state (+4), adding CO2 to a strong reductant such as Li should result in a vigorous redox reaction. Thus water, N2, and CO2 are all unsuitable choices for extinguishing a lithium fire. In contrast, sand is primarily SiO2, which is a network solid that is not readily reduced. Smothering a lithium fire with sand is therefore the best choice.
3. The salt with the smaller cation has the higher lattice energy, and high lattice energies tend to decrease the solubility of a salt. However, the solvation energy of the cation is also important in determining solubility, and small cations tend to have higher solvation energies. Recall that high solvation energies tend to increase the solubility of ionic substances. Thus CsI should be the least soluble of the alkali metal iodides, and LiI the most soluble. Consequently, CsNO3 is the better choice.
$1$
Indicate which of the alternative alkali metals or their compounds given is more appropriate for each application.
1. drying agent for an organic solvent—Li2SO4 or Rb2SO4
2. removing trace amounts of N2 from highly purified Ar gas—Li, K, or Cs
3. reacting with an alkyl halide (formula RX) to prepare an organometallic compound (formula MR)—Li or K
Answer
1. Li2SO4
2. Li
3. Li
$2$
Predict the products of each reaction and then balance each chemical equation.
1. Na(s) + O2(g) →
2. Li2O(s) + H2O(l) →
3. K(s) + CH3OH(l) →
4. Li(s) + CH3Cl(l) →
5. Li3N(s) + KCl(s) →
Given: reactants
Asked for: products and balanced chemical equation
Strategy:
A Determine whether one of the reactants is an oxidant or a reductant or a strong acid or a strong base. If so, a redox reaction or an acid–base reaction is likely to occur. Identify the products of the reaction.
B If a reaction is predicted to occur, balance the chemical equation.
Solution
1. A Sodium is a reductant, and oxygen is an oxidant, so a redox reaction is most likely. We expect an electron to be transferred from Na (thus forming Na+) to O2. We now need to determine whether the reduced product is a superoxide (O2), peroxide (O22−), or oxide (O2−). Under normal reaction conditions, the product of the reaction of an alkali metal with oxygen depends on the identity of the metal. Because of differences in lattice energy, Li produces the oxide (Li2O), the heavier metals (K, Rb, Cs) produce the superoxide (MO2), and Na produces the peroxide (Na2O2).
B The balanced chemical equation is 2Na(s) + O2(g) → Na2O2(s).
1. A Li2O is an ionic salt that contains the oxide ion (O2−), which is the completely deprotonated form of water and thus is expected to be a strong base. The other reactant, water, is both a weak acid and a weak base, so we can predict that an acid–base reaction will occur.
B The balanced chemical equation is Li2O(s) + H2O(l) → 2LiOH(aq).
1. A Potassium is a reductant, whereas methanol is both a weak acid and a weak base (similar to water). A weak acid produces H+, which can act as an oxidant by accepting an electron to form $\frac{1}{2}\mathrm{H_2}$. This reaction, therefore, is an acid dissociation that is driven to completion by a reduction of the protons as they are released.
B The balanced chemical equation is as follows: $\mathrm{K(s)}+\mathrm{CH_3OH(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{CH_3OK(soln)}$.
1. A One of the reactants is an alkali metal, a potent reductant, and the other is an alkyl halide. Any compound that contains a carbon–halogen bond can, in principle, be reduced, releasing a halide ion and forming an organometallic compound. That outcome seems likely in this case because organolithium compounds are among the most stable organometallic compounds known.
B Two moles of lithium are required to balance the equation: 2Li(s) + CH3Cl(l) → LiCl(s) + CH3Li(soln).
1. A Lithium nitride and potassium chloride are largely ionic compounds. The nitride ion (N3−) is a very strong base because it is the fully deprotonated form of ammonia, a weak acid. An acid–base reaction requires an acid as well as a base, however, and KCl is not acidic. What about a redox reaction? Both substances contain ions that have closed-shell valence electron configurations. The nitride ion could act as a reductant by donating electrons to an oxidant and forming N2. KCl is not an oxidant, however, and a redox reaction requires an oxidant as well as a reductant.
B We conclude that the two substances will not react with each other.
$2$
Predict the products of each reaction and balance each chemical equation.
1. K(s) + N2(g) →
2. Li3N(s) + H2O(l) →
3. Na(s) + (CH3)2NH(soln) →
4. C6H5Li(soln) + D2O(l) → C6H5D(l) + LiOD(soln)
5. CH3CH2Cl(soln) + 2Li →
Answer
1. no reaction
2. Li3N(s) + 3H2O(l) → NH3(aq) + 3LiOH(aq)
3. $\mathrm{Na(s)}+\mathrm{(CH_3)_2NH(soln)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{Na[(CH_3)_2N](soln)}$
4. C6H5Li(soln) + D2O(l) → C6H5D(l) + LiOD(soln)
5. CH3CH2Cl(soln) + 2Li → CH3CH2Li(soln) + LiCl(soln)
Summary
The alkali metals are potent reductants whose chemistry is largely that of ionic compounds containing the M+ ion. Alkali metals have only a weak tendency to form complexes with simple Lewis bases. The first alkali metals to be isolated (Na and K) were obtained by passing an electric current through molten potassium and sodium carbonates. The alkali metals are among the most potent reductants known; most can be isolated by electrolysis of their molten salts or, in the case of rubidium and cesium, by reacting their hydroxide salts with a reductant. They can also be recovered from their silicate ores using a multistep process. Lithium, the strongest reductant, and sodium, the weakest, are examples of the physical and chemical effects of opposing periodic trends. The alkali metals react with halogens (group 17) to form ionic halides; the heavier chalcogens (group 16) to produce metal chalcogenides; and oxygen to form compounds, whose stoichiometry depends on the size of the metal atom. The peroxides and superoxides are potent oxidants. The only alkali metal to react with atmospheric nitrogen is lithium. Heavier alkali metals react with graphite to form graphite intercalation compounds, substances in which metal atoms are inserted between the sheets of carbon atoms. With heavier group 14 elements, alkali metals react to give polyatomic anions with three-dimensional cage structures. All alkali metals react with hydrogen at high temperatures to produce the corresponding hydrides, and all reduce water to produce hydrogen gas. Alkali metal salts are prepared by reacting a metal hydroxide with an acid, followed by evaporation of the water. Both Li and Na salts are used as drying agents, compounds that are used to absorb water. Complexing agents such as crown ethers and cryptands can accommodate alkali metal ions of the appropriate size. Alkali metals can also react with liquid ammonia to form solutions that slowly decompose to give hydrogen gas and the metal salt of the amide ion (NH2). These solutions, which contain unstable solvated electrons loosely associated with a cavity in the solvent, are intensely colored, good conductors of electricity, and excellent reductants. Alkali metals can react with organic compounds that contain an acidic proton to produce salts. They can also form organometallic compounds, which have properties that differ from those of their metallic and organic components. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.04%3A__The_Alkali_Metals_Group_1A.txt |
Learning Objectives
• To describe how to isolate the alkaline earth metals.
• To be familiar with the reactions, compounds, and complexes of the alkaline earth metals.
Like the alkali metals, the alkaline earth metals are so reactive that they are never found in elemental form in nature. Because they form +2 ions that have very negative reduction potentials, large amounts of energy are needed to isolate them from their ores. Four of the six group 2 elements—magnesium (Mg), calcium (Ca), strontium (Sr), and barium (Ba)—were first isolated in the early 19th century by Sir Humphry Davy, using a technique similar to the one he used to obtain the first alkali metals. In contrast to the alkali metals, however, compounds of the alkaline earth metals had been recognized as unique for many centuries. In fact, the name alkali comes from the Arabic al-qili, meaning “ashes,” which were known to neutralize acids. Medieval alchemists found that a portion of the ashes would melt on heating, and these substances were later identified as the carbonates of sodium and potassium ($M_2CO_3$). The ashes that did not melt (but did dissolve in acid), originally called alkaline earths, were subsequently identified as the alkaline earth oxides (MO). In 1808, Davy was able to obtain pure samples of Mg, Ca, Sr, and Ba by electrolysis of their chlorides or oxides.
Beryllium (Be), the lightest alkaline earth metal, was first obtained in 1828 by Friedrich Wöhler in Germany and simultaneously by Antoine Bussy in France. The method used by both men was reduction of the chloride by the potent “new” reductant, potassium:
$\mathrm{BeCl_2(s)}+\mathrm{2K(s)}\xrightarrow\Delta\mathrm{Be(s)}+\mathrm{2KCl(s)} \label{Eq1}$
Radium was discovered in 1898 by Pierre and Marie Curie, who processed tons of residue from uranium mines to obtain about 120 mg of almost pure $RaCl_2$. Marie Curie was awarded the Nobel Prize in Chemistry in 1911 for its discovery. Because of its low abundance and high radioactivity however, radium has few uses.
Preparation of the Alkaline Earth Metals
The alkaline earth metals are produced for industrial use by electrolytic reduction of their molten chlorides, as indicated in this equation for calcium:
$CaCl_{2\;(l)} \rightarrow Ca_{(l)} + Cl_{2\;(g)} \label{Eq2}$
The group 2 metal chlorides are obtained from a variety of sources. For example, $BeCl_2$ is produced by reacting $HCl$ with beryllia ($BeO$), which is obtained from the semiprecious stone beryl $[Be_3Al_2(SiO_3)_6]$.
Chemical reductants can also be used to obtain the group 2 elements. For example, magnesium is produced on a large scale by heating a form of limestone called dolomite (CaCO3·MgCO3) with an inexpensive iron/silicon alloy at 1150°C. Initially $CO_2$ is released, leaving behind a mixture of $CaO$ and MgO; Mg2+ is then reduced:
$2CaO·MgO_{(s)} + Fe/Si_{(s)} \rightarrow 2Mg(l) + Ca_2SiO_{4\;(s)} + Fe(s) \label{Eq3}$
An early source of magnesium was an ore called magnesite ($MgCO_3$) from the district of northern Greece called Magnesia. Strontium was obtained from strontianite ($SrCO_3$) found in a lead mine in the town of Strontian in Scotland. The alkaline earth metals are somewhat easier to isolate from their ores, as compared to the alkali metals, because their carbonate and some sulfate and hydroxide salts are insoluble.
General Properties of the Alkaline Earth Metals
Several important properties of the alkaline earth metals are summarized in Table $1$. Although many of these properties are similar to those of the alkali metals (Table $1$), certain key differences are attributable to the differences in the valence electron configurations of the two groups (ns2 for the alkaline earth metals versus ns1 for the alkali metals).
Table $1$: Selected Properties of the Group 2 Elements
Beryllium Magnesium Calcium Strontium Barium Radium
*The values cited are for six-coordinate ions except for Be2+, for which the value for the four-coordinate ion is given.
atomic symbol Be Mg Ca Sr Ba Ra
atomic number 4 12 20 38 56 88
atomic mass 9.01 24.31 40.08 87.62 137.33 226
valence electron configuration 2s2 3s2 4s2 5s2 6s2 7s2
melting point/boiling point (°C) 1287/2471 650/1090 842/1484 777/1382 727/1897 700/—
density (g/cm3) at 25°C 1.85 1.74 1.54 2.64 3.62 ~5
atomic radius (pm) 112 145 194 219 253
first ionization energy (kJ/mol) 900 738 590 549 503
most common oxidation state +2 +2 +2 +2 +2 +2
ionic radius (pm)* 45 72 100 118 135
electron affinity (kJ/mol) ≥ 0 ≥ 0 −2 −5 −14
electronegativity 1.6 1.3 1.0 1.0 0.9 0.9
standard electrode potential (E°, V) −1.85 −2.37 −2.87 −2.90 −2.91 −2.8
product of reaction with O2 BeO MgO CaO SrO BaO2
type of oxide amphoteric weakly basic basic basic basic
product of reaction with N2 none Mg3N2 Ca3N2 Sr3N2 Ba3N2
product of reaction with X2 BeX2 MgX2 CaX2 SrX2 BaX2
product of reaction with H2 none MgH2 CaH2 SrH2 BaH2
As with the alkali metals, the atomic and ionic radii of the alkaline earth metals increase smoothly from Be to Ba, and the ionization energies decrease. As we would expect, the first ionization energy of an alkaline earth metal, with an ns2 valence electron configuration, is always significantly greater than that of the alkali metal immediately preceding it. The group 2 elements do exhibit some anomalies, however. For example, the density of Ca is less than that of Be and Mg, the two lightest members of the group, and Mg has the lowest melting and boiling points. In contrast to the alkali metals, the heaviest alkaline earth metal (Ba) is the strongest reductant, and the lightest (Be) is the weakest. The standard electrode potentials of Ca and Sr are not very different from that of Ba, indicating that the opposing trends in ionization energies and hydration energies are of roughly equal importance.
One major difference between the group 1 and group 2 elements is their electron affinities. With their half-filled ns orbitals, the alkali metals have a significant affinity for an additional electron. In contrast, the alkaline earth metals generally have little or no tendency to accept an additional electron because their ns valence orbitals are already full; an added electron would have to occupy one of the vacant np orbitals, which are much higher in energy.
Reactions and Compounds of the Alkaline Earth Metals
With their low first and second ionization energies, the group 2 elements almost exclusively form ionic compounds that contain M2+ ions. As expected, however, the lightest element (Be), with its higher ionization energy and small size, forms compounds that are largely covalent. Some compounds of Mg2+ also have significant covalent character. Hence organometallic compounds like those discussed for Li in group 1 are also important for Be and Mg in group 2.
The group 2 elements almost exclusively form ionic compounds containing M2+ ions.
All alkaline earth metals react vigorously with the halogens (group 17) to form the corresponding halides (MX2). Except for the beryllium halides, these compounds are all primarily ionic in nature, containing the M2+ cation and two X anions. The beryllium halides, with properties more typical of covalent compounds, have a polymeric halide-bridged structure in the solid state, as shown for BeCl2. These compounds are volatile, producing vapors that contain the linear X–Be–X molecules predicted by the valence-shell electron-pair repulsion (VSEPR) model. As expected for compounds with only four valence electrons around the central atom, the beryllium halides are potent Lewis acids. They react readily with Lewis bases, such as ethers, to form tetrahedral adducts in which the central beryllium is surrounded by an octet of electrons:
$BeCl_{2(s)} + 2(CH_3CH_2)_2O_{(l)} \rightarrow BeCl_2[O(CH_2CH_3)_2]_{2(soln)} \label{Eq4}$
Because of their higher ionization energy and small size, both Be and Mg form organometallic compounds.
The reactions of the alkaline earth metals with oxygen are less complex than those of the alkali metals. All group 2 elements except barium react directly with oxygen to form the simple oxide MO. Barium forms barium peroxide (BaO2) because the larger O22− ion is better able to separate the large Ba2+ ions in the crystal lattice. In practice, only BeO is prepared by direct reaction with oxygen, and this reaction requires finely divided Be and high temperatures because Be is relatively inert. The other alkaline earth oxides are usually prepared by the thermal decomposition of carbonate salts:
$\mathrm{MCO_3(s)}\xrightarrow\Delta\mathrm{MO(s)}+\mathrm{CO_2(g)} \label{Eq5}$
The reactions of the alkaline earth metals with the heavier chalcogens (Y) are similar to those of the alkali metals. When the reactants are present in a 1:1 ratio, the binary chalcogenides (MY) are formed; at lower M:Y ratios, salts containing polychalcogenide ions (Yn2−) are formed.
In the reverse of Equation $\ref{Eq5}$, the oxides of Ca, Sr, and Ba react with CO2 to regenerate the carbonate. Except for BeO, which has significant covalent character and is therefore amphoteric, all the alkaline earth oxides are basic. Thus they react with water to form the hydroxides—M(OH)2:
$MO_{(s)} + H_2O_{(l)} \rightarrow M^{2+}_{(aq)} + 2OH^−_{(aq)} \label{Eq6}$
and they dissolve in aqueous acid. Hydroxides of the lighter alkaline earth metals are insoluble in water, but their solubility increases as the atomic number of the metal increases. Because BeO and MgO are much more inert than the other group 2 oxides, they are used as refractory materials in applications involving high temperatures and mechanical stress. For example, MgO (melting point = 2825°C) is used to coat the heating elements in electric ranges.
The carbonates of the alkaline earth metals also react with aqueous acid to give CO2 and H2O:
$MCO_{3(s)} + 2H^+_{(aq)} \rightarrow M^{2+}_{(aq)} + CO_{2(g)} + H_2O_{(l)} \label{Eq7}$
The reaction in Equation $\ref{Eq7}$ is the basis of antacids that contain MCO3, which is used to neutralize excess stomach acid.
The trend in the reactivities of the alkaline earth metals with nitrogen is the opposite of that observed for the alkali metals. Only the lightest element (Be) does not react readily with N2 to form the nitride (M3N2), although finely divided Be will react at high temperatures. The higher lattice energy due to the highly charged M2+ and N3− ions is apparently sufficient to overcome the chemical inertness of the N2 molecule, with its N≡N bond. Similarly, all the alkaline earth metals react with the heavier group 15 elements to form binary compounds such as phosphides and arsenides with the general formula M3Z2.
Higher lattice energies cause the alkaline earth metals to be more reactive than the alkali metals toward group 15 elements.
When heated, all alkaline earth metals, except for beryllium, react directly with carbon to form ionic carbides with the general formula MC2. The most important alkaline earth carbide is calcium carbide (CaC2), which reacts readily with water to produce acetylene. For many years, this reaction was the primary source of acetylene for welding and lamps on miners’ helmets. In contrast, beryllium reacts with elemental carbon to form Be2C, which formally contains the C4− ion (although the compound is covalent). Consistent with this formulation, reaction of Be2C with water or aqueous acid produces methane:
$Be_2C_{(s)} + 4H_2O_{(l)} \rightarrow 2Be(OH)_{2(s)} + CH_{4(g)} \label{Eq8}$
Beryllium does not react with hydrogen except at high temperatures (1500°C), although BeH2 can be prepared at lower temperatures by an indirect route. All the heavier alkaline earth metals (Mg through Ba) react directly with hydrogen to produce the binary hydrides (MH2). The hydrides of the heavier alkaline earth metals are ionic, but both BeH2 and MgH2 have polymeric structures that reflect significant covalent character. All alkaline earth hydrides are good reducing agents that react rapidly with water or aqueous acid to produce hydrogen gas:
$CaH_{2(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(s)} + 2H_{2(g)} \label{Eq9}$
Like the alkali metals, the heavier alkaline earth metals are sufficiently electropositive to dissolve in liquid ammonia. In this case, however, two solvated electrons are formed per metal atom, and no equilibriums involving metal dimers or metal anions are known. Also, like the alkali metals, the alkaline earth metals form a wide variety of simple ionic salts with oxoanions, such as carbonate, sulfate, and nitrate. The nitrate salts tend to be soluble, but the carbonates and sulfates of the heavier alkaline earth metals are quite insoluble because of the higher lattice energy due to the doubly charged cation and anion. The solubility of the carbonates and the sulfates decreases rapidly down the group because hydration energies decrease with increasing cation size.
The solubility of alkaline earth carbonate and sulfates decrease down the group because the hydration energies decrease.
Complexes of the Alkaline Earth Metals
Because of their higher positive charge (+2) and smaller ionic radii, the alkaline earth metals have a much greater tendency to form complexes with Lewis bases than do the alkali metals. This tendency is most important for the lightest cation (Be2+) and decreases rapidly with the increasing radius of the metal ion.
The alkaline earth metals have a substantially greater tendency to form complexes with Lewis bases than do the alkali metals.
The chemistry of Be2+ is dominated by its behavior as a Lewis acid, forming complexes with Lewis bases that produce an octet of electrons around beryllium. For example, Be2+ salts dissolve in water to form acidic solutions that contain the tetrahedral [Be(H2O)4]2+ ion. Because of its high charge-to-radius ratio, the Be2+ ion polarizes coordinated water molecules, thereby increasing their acidity:
$[Be(H_2O)_4]^{2+}_{(aq)} \rightarrow [Be(H_2O)_3(OH)]^+_{(aq)} + H^+_{(aq)} \label{Eq10}$
Similarly, in the presence of a strong base, beryllium and its salts form the tetrahedral hydroxo complex: [Be(OH)4]2−. Hence beryllium oxide is amphoteric. Beryllium also forms a very stable tetrahedral fluoride complex: [BeF4]2−. Recall that beryllium halides behave like Lewis acids by forming adducts with Lewis bases (Equation $\ref{Eq4}$).
The heavier alkaline earth metals also form complexes, but usually with a coordination number of 6 or higher. Complex formation is most important for the smaller cations (Mg2+ and Ca2+). Thus aqueous solutions of Mg2+ contain the octahedral [Mg(H2O)6]2+ ion. Like the alkali metals, the alkaline earth metals form complexes with neutral cyclic ligands like the crown ethers and cryptands discussed in Section 21.3.
Organometallic Compounds Containing Group 2 Elements
Like the alkali metals, the lightest alkaline earth metals (Be and Mg) form the most covalent-like bonds with carbon, and they form the most stable organometallic compounds. Organometallic compounds of magnesium with the formula RMgX, where R is an alkyl or aryl group and X is a halogen, are universally called Grignard reagents, after Victor Grignard (1871–1935), the French chemist who discovered them. Grignard reagents can be used to synthesize various organic compounds, such as alcohols, aldehydes, ketones, carboxylic acids, esters, thiols, and amines.
Uses of the Alkaline Earth Metals
Elemental magnesium is the only alkaline earth metal that is produced on a large scale (about 5 × 105 tn per year). Its low density (1.74 g/cm3 compared with 7.87 g/cm3 for iron and 2.70 g/cm3 for aluminum) makes it an important component of the lightweight metal alloys used in aircraft frames and aircraft and automobile engine parts (Figure $1$). Most commercial aluminum actually contains about 5% magnesium to improve its corrosion resistance and mechanical properties. Elemental magnesium also serves as an inexpensive and powerful reductant for the production of a number of metals, including titanium, zirconium, uranium, and even beryllium, as shown in the following equation:
$TiCl_{4\;(l)} + 2Mg(s) \rightarrow Ti_{(s)} + 2MgCl_{2\;(s)} \label{11}$
The only other alkaline earth that is widely used as the metal is beryllium, which is extremely toxic. Ingestion of beryllium or exposure to beryllium-containing dust causes a syndrome called berylliosis, characterized by severe inflammation of the respiratory tract or other tissues. A small percentage of beryllium dramatically increases the strength of copper or nickel alloys, which are used in nonmagnetic, nonsparking tools (such as wrenches and screwdrivers), camera springs, and electrical contacts. The low atomic number of beryllium gives it a very low tendency to absorb x-rays and makes it uniquely suited for applications involving radioactivity. Both elemental Be and BeO, which is a high-temperature ceramic, are used in nuclear reactors, and the windows on all x-ray tubes and sources are made of beryllium foil.
Millions of tons of calcium compounds are used every year. As discussed in earlier chapters, CaCl2 is used as “road salt” to lower the freezing point of water on roads in cold temperatures. In addition, CaCO3 is a major component of cement and an ingredient in many commercial antacids. “Quicklime” (CaO), produced by heating CaCO3 (Equation $\ref{Eq5}$), is used in the steel industry to remove oxide impurities, make many kinds of glass, and neutralize acidic soil. Other applications of group 2 compounds described in earlier chapters include the medical use of BaSO4 in “barium milkshakes” for identifying digestive problems by x-rays and the use of various alkaline earth compounds to produce the brilliant colors seen in fireworks.
Example $1$
For each application, choose the most appropriate substance based on the properties and reactivities of the alkaline earth metals and their compounds. Explain your choice in each case. Use any tables you need in making your decision, such as Ksp values (Table 17.1), lattice energies (Table 8.1), and band-gap energies.
1. To neutralize excess stomach acid that causes indigestion, would you use BeCO3, CaCO3, or BaCO3?
2. To remove CO2 from the atmosphere in a space capsule, would you use MgO, CaO, or BaO?
3. As a component of the alloy in an automotive spark plug electrode, would you use Be, Ca, or Ba?
Given: application and selected alkaline earth metals
Asked for: most appropriate substance for each application
Strategy:
Based on the discussion in this section and any relevant information elsewhere in this book, determine which substance is most appropriate for the indicated use.
Solution
1. All the alkaline earth carbonates will neutralize an acidic solution by Equation $\ref{Eq7}$. Because beryllium and its salts are toxic, however, BeCO3 cannot be used as an antacid. Of the remaining choices, CaCO3 is somewhat more soluble than BaCO3 (according to the Ksp values in Table 17.1), suggesting that it will act more rapidly. Moreover, the formula mass of CaCO3 is 100.1 amu, whereas that of BaCO3 is almost twice as large. Therefore, neutralizing a given amount of acid would require twice the mass of BaCO3 compared with CaCO3. Furthermore, reaction of BaCO3 with acid produces a solution containing Ba2+ ions, which are toxic. (Ba2+ is a stimulant that can cause ventricular fibrillation of the heart.) Finally, CaCO3 is produced on a vast scale, so CaCO3 is likely to be significantly less expensive than any barium compound. Consequently, CaCO3 is the best choice for an antacid.
2. This application involves reacting CO2 with an alkaline earth oxide to form the carbonate, which is the reverse of the thermal decomposition reaction in which MCO3 decomposes to CO2 and the metal oxide MO (Equation $\ref{Eq5}$). Owing to their higher lattice energies, the smallest alkaline earth metals should form the most stable oxides. Hence their carbonates should decompose at the lowest temperatures, as is observed (BeCO3 decomposes at 100°C; BaCO3 at 1360°C). If the carbonate with the smallest alkaline earth metal decomposes most readily, we would expect the reverse reaction (formation of a carbonate) to occur most readily with the largest metal cation (Ba2+). Hence BaO is the best choice.
3. The alloy in a spark plug electrode must release electrons and promote their flow across the gap between the electrodes at high temperatures. Of the three metals listed, Ba has the lowest ionization energy and thus releases electrons most readily. Heating a barium-containing alloy to high temperatures will cause some ionization to occur, providing the initial step in forming a spark.
Exercise $1$
Which of the indicated alkaline earth metals or their compounds is most appropriate for each application?
1. drying agent for removing water from the atmosphere—CaCl2, MgSO4, or BaF2
2. removal of scale deposits (largely CaCO3) in water pipes—HCl(aq) or H2SO4(aq)
3. removal of traces of N2 from purified argon gas—Be, Ca, or Ba
Answer
1. MgSO4
2. HCl
3. Ba
Example $2$
Predict the products of each reaction and then balance each chemical equation.
1. CaO(s) + HCl(g) →
2. MgO(s) + excess OH(aq) →
3. $\mathrm{CaH_2(s)}+\mathrm{TiO_2(s)}\xrightarrow\Delta$
Given: reactants
Asked for: products and balanced chemical equation
Strategy:
Follow the procedure given in Example 3 to predict the products of each reaction and then balance each chemical equation.
Solution
1. A Gaseous HCl is an acid, and CaO is a basic oxide that contains the O2− ion. This is therefore an acid–base reaction that produces CaCl2 and H2O.
B The balanced chemical equation is $CaO_{(s)} + 2HCl_{(g)} → CaCl_{2(aq)} + H_2O_{(l)}$
1. A Magnesium oxide is a basic oxide, so it can either react with water to give a basic solution or dissolve in an acidic solution. Hydroxide ion is also a base. Because we have two bases but no acid, an acid–base reaction is impossible. A redox reaction is not likely because MgO is neither a good oxidant nor a good reductant.
B We conclude that no reaction occurs.
1. A Because CaH2 contains the hydride ion (H), it is a good reductant. It is also a strong base because H ions can react with H+ ions to form H2. Titanium oxide (TiO2) is a metal oxide that contains the metal in its highest oxidation state (+4 for a group 4 metal); it can act as an oxidant by accepting electrons. We therefore predict that a redox reaction will occur, in which H is oxidized and Ti4+ is reduced. The most probable reduction product is metallic titanium, but what is the oxidation product? Oxygen must appear in the products, and both CaO and H2O are stable compounds. The +1 oxidation state of hydrogen in H2O is a sign that an oxidation has occurred (2H → 2H+ + 4e).
B The balanced chemical equation is
$\mathrm{CaH_2(s)}+\mathrm{TiO_2(s)}\xrightarrow\Delta\mathrm{Ti(s)}+\mathrm{CaO(s)}+\mathrm{H_2O(l)}$
We could also write the products as Ti(s) + Ca(OH)2(s).
Exercise $2$
Predict the products of each reaction and then balance each chemical equation.
1. $\mathrm{BeCl_2(s)}+\mathrm{Mg(s)}\xrightarrow\Delta$
2. BaCl2(aq) + Na2SO4(aq) →
3. BeO(s) + OH(aq) + H2O(l) →
Answer
1. $\mathrm{BeCl_2(s)}+\mathrm{Mg(s)}\xrightarrow\Delta\mathrm{Be(s)}+ \mathrm{MgCl_2(s)}$
2. BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
3. BeO(s) + 2OH(aq) + H2O(l) → [Be(OH)4]2−(aq)
Summary
Group 2 elements almost exclusively form ionic compounds containing the M2+ ion, they are more reactive toward group 15 elements, and they have a greater tendency to form complexes with Lewis bases than do the alkali metals. Pure samples of most of the alkaline earth metals can be obtained by electrolysis of the chlorides or oxides. Beryllium was first obtained by the reduction of its chloride; radium chloride, which is radioactive, was obtained through a series of reactions and separations. In contrast to the alkali metals, the alkaline earth metals generally have little or no affinity for an added electron. All alkaline earth metals react with the halogens to produce the corresponding halides, with oxygen to form the oxide (except for barium, which forms the peroxide), and with the heavier chalcogens to form chalcogenides or polychalcogenide ions. All oxides except BeO react with CO2 to form carbonates, which in turn react with acid to produce CO2 and H2O. Except for Be, all the alkaline earth metals react with N2 to form nitrides, and all react with carbon and hydrogen to form carbides and hydrides. Alkaline earth metals dissolve in liquid ammonia to give solutions that contain two solvated electrons per metal atom. The alkaline earth metals have a greater tendency than the alkali metals to form complexes with crown ethers, cryptands, and other Lewis bases. The most important alkaline earth organometallic compounds are Grignard reagents (RMgX), which are used to synthesize organic compounds. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.05%3A__The_Alkaline_Earth_Elements_Group_2A.txt |
Learning Objectives
• To understand the trends in properties and the reactivity of the group 13 elements.
Group 13 is the first group to span the dividing line between metals and nonmetals, so its chemistry is more diverse than that of groups 1 and 2, which include only metallic elements. Except for the lightest element (boron), the group 13 elements are all relatively electropositive; that is, they tend to lose electrons in chemical reactions rather than gain them. Although group 13 includes aluminum, the most abundant metal on Earth, none of these elements was known until the early 19th century because they are never found in nature in their free state. Elemental boron and aluminum, which were first prepared by reducing B2O3 and AlCl3, respectively, with potassium, could not be prepared until potassium had been isolated and shown to be a potent reductant. Indium (In) and thallium (Tl) were discovered in the 1860s by using spectroscopic techniques, long before methods were available for isolating them. Indium, named for its indigo (deep blue-violet) emission line, was first observed in the spectrum of zinc ores, while thallium (from the Greek thallos, meaning “a young, green shoot of a plant”) was named for its brilliant green emission line. Gallium (Ga; Mendeleev’s eka-aluminum) was discovered in 1875 by the French chemist Paul Émile Lecoq de Boisbaudran during a systematic search for Mendeleev’s “missing” element in group 13.
Group 13 elements are never found in nature in their free state.
Preparation and General Properties of the Group 13 Elements
As reductants, the group 13 elements are less powerful than the alkali metals and alkaline earth metals. Nevertheless, their compounds with oxygen are thermodynamically stable, and large amounts of energy are needed to isolate even the two most accessible elements—boron and aluminum—from their oxide ores.
Although boron is relatively rare (it is about 10,000 times less abundant than aluminum), concentrated deposits of borax [Na2B4O5(OH)4·8H2O] are found in ancient lake beds (Figure $1$) and were used in ancient times for making glass and glazing pottery. Boron is produced on a large scale by reacting borax with acid to produce boric acid [B(OH)3], which is then dehydrated to the oxide (B2O3). Reduction of the oxide with magnesium or sodium gives amorphous boron that is only about 95% pure:
$\mathrm{Na_2B_4O_5(OH)_4\cdot8H_2O(s)}\xrightarrow{\textrm{acid}}\mathrm{B(OH)_3(s)}\xrightarrow{\Delta}\mathrm{B_2O_3(s)} \label{Eq1}$
$\mathrm{B_2O_3(s)}+\mathrm{3Mg(s)}\xrightarrow{\Delta}\mathrm{2B(s)}+\mathrm{3MgO(s)} \label{Eq2}$
Pure, crystalline boron, however, is extremely difficult to obtain because of its high melting point (2300°C) and the highly corrosive nature of liquid boron. It is usually prepared by reducing pure BCl3 with hydrogen gas at high temperatures or by the thermal decomposition of boron hydrides such as diborane (B2H6):
$\mathrm{BCl_3(g)}+\frac{3}{2}\mathrm{H_2(g)}\rightarrow\mathrm{B(s)}+\mathrm{3HCl(g)} \label{Eq3}$
$B_2H_{6(g)} \rightarrow 2B_{(s)} + 3H_{2(g)} \label{Eq4}$
The reaction shown in Equation $\ref{Eq3}$ is used to prepare boron fibers, which are stiff and light. Hence they are used as structural reinforcing materials in objects as diverse as the US space shuttle and the frames of lightweight bicycles that are used in races such as the Tour de France. Boron is also an important component of many ceramics and heat-resistant borosilicate glasses, such as Pyrex, which is used for ovenware and laboratory glassware.
In contrast to boron, deposits of aluminum ores such as bauxite, a hydrated form of Al2O3, are abundant. With an electrical conductivity about twice that of copper on a weight for weight basis, aluminum is used in more than 90% of the overhead electric power lines in the United States. However, because aluminum–oxygen compounds are stable, obtaining aluminum metal from bauxite is an expensive process. Aluminum is extracted from oxide ores by treatment with a strong base, which produces the soluble hydroxide complex [Al(OH)4]. Neutralization of the resulting solution with gaseous CO2 results in the precipitation of Al(OH)3:
$2[Al(OH)_4]^−_{(aq)} + CO_{2(g)} \rightarrow 2Al(OH)_{3(s)} + CO^{2−}_{3(aq)} + H_2O_{(l)} \label{Eq5}$
Thermal dehydration of Al(OH)3 produces Al2O3, and metallic aluminum is obtained by the electrolytic reduction of Al2O3 using the Hall–Heroult process. Of the group 13 elements, only aluminum is used on a large scale: for example, each Boeing 777 airplane is about 50% aluminum by mass.
The other members of group 13 are rather rare: gallium is approximately 5000 times less abundant than aluminum, and indium and thallium are even scarcer. Consequently, these metals are usually obtained as by-products in the processing of other metals. The extremely low melting point of gallium (29.6°C), however, makes it easy to separate from aluminum. Due to its low melting point and high boiling point, gallium is used as a liquid in thermometers that have a temperature range of almost 2200°C. Indium and thallium, the heavier group 13 elements, are found as trace impurities in sulfide ores of zinc and lead. Indium is used as a crushable seal for high-vacuum cryogenic devices, and its alloys are used as low-melting solders in electronic circuit boards. Thallium, on the other hand, is so toxic that the metal and its compounds have few uses. Both indium and thallium oxides are released in flue dust when sulfide ores are converted to metal oxides and SO2. Until relatively recently, these and other toxic elements were allowed to disperse in the air, creating large “dead zones” downwind of a smelter. The flue dusts are now trapped and serve as a relatively rich source of elements such as In and Tl (as well as Ge, Cd, Te, and As).
Table $1$ summarizes some important properties of the group 13 elements. Notice the large differences between boron and aluminum in size, ionization energy, electronegativity, and standard reduction potential, which is consistent with the observation that boron behaves chemically like a nonmetal and aluminum like a metal. All group 13 elements have ns2np1 valence electron configurations, and all tend to lose their three valence electrons to form compounds in the +3 oxidation state. The heavier elements in the group can also form compounds in the +1 oxidation state formed by the formal loss of the single np valence electron. Because the group 13 elements generally contain only six valence electrons in their neutral compounds, these compounds are all moderately strong Lewis acids.
Table $1$: Selected Properties of the Group 13 Elements
Property Boron Aluminum* Gallium Indium Thallium
*This is the name used in the United States; the rest of the world inserts an extra i and calls it aluminium.
The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate ions in the most common oxidation state, except for Al3+, for which the value for the four-coordinate ion is given. The B3+ ion is not a known species; the radius cited is an estimated four-coordinate value.
§X is Cl, Br, or I. Reaction with F2 gives the trifluorides (MF3) for all group 13 elements.
atomic symbol B Al Ga In Tl
atomic number 5 13 31 49 81
atomic mass (amu) 10.81 26.98 69.72 114.82 204.38
valence electron configuration 2s22p1 3s23p1 4s24p1 5s25p1 6s26p1
melting point/boiling point (°C) 2075/4000 660/2519 29.7/2204 156.6/2072 304/1473
density (g/cm3) at 25°C 2.34 2.70 5.91 7.31 11.8
atomic radius (pm) 87 118 136 156 156
first ionization energy (kJ/mol) 801 578 579 558 589
most common oxidation state +3 +3 +3 +3 +1
ionic radius (pm) −25 54 62 80 162
electron affinity (kJ/mol) −27 −42 −40 −39 −37
electronegativity 2.0 1.6 1.8 1.8 1.8
standard reduction potential (E°, V) −0.87 −1.66 −0.55 −0.34 +0.741 of M3+(aq)
product of reaction with O2 B2O3 Al2O3 Ga2O3 In2O3 Tl2O
type of oxide acidic amphoteric amphoteric amphoteric basic
product of reaction with N2 BN AlN GaN InN none
product of reaction with X2§ BX3 Al2X6 Ga2X6 In2X6 TlX
Neutral compounds of the group 13 elements are electron deficient, so they are generally moderately strong Lewis acids.
In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Zeff) that results from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Zeff increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium.
Anomalies in periodic trends among Ga, In, and Tl can be explained by the increase in the effective nuclear charge due to poor shielding.
Reactions and Compounds of Boron
Elemental boron is a semimetal that is remarkably unreactive; in contrast, the other group 13 elements all exhibit metallic properties and reactivity. We therefore consider the reactions and compounds of boron separately from those of other elements in the group. All group 13 elements have fewer valence electrons than valence orbitals, which generally results in delocalized, metallic bonding. With its high ionization energy, low electron affinity, low electronegativity, and small size, however, boron does not form a metallic lattice with delocalized valence electrons. Instead, boron forms unique and intricate structures that contain multicenter bonds, in which a pair of electrons holds together three or more atoms.
Elemental boron forms multicenter bonds, whereas the other group 13 elements exhibit metallic bonding.
The basic building block of elemental boron is not the individual boron atom, as would be the case in a metal, but rather the B12 icosahedron. Because these icosahedra do not pack together very well, the structure of solid boron contains voids, resulting in its low density (Figure $3$). Elemental boron can be induced to react with many nonmetallic elements to give binary compounds that have a variety of applications. For example, plates of boron carbide (B4C) can stop a 30-caliber, armor-piercing bullet, yet they weigh 10%–30% less than conventional armor. Other important compounds of boron with nonmetals include boron nitride (BN), which is produced by heating boron with excess nitrogen (Equation $\ref{Eq22.6}$); boron oxide (B2O3), which is formed when boron is heated with excess oxygen (Equation $\ref{Eq22.7}$); and the boron trihalides (BX3), which are formed by heating boron with excess halogen (Equation $\ref{Eq22.8}$).
$\mathrm{2B(s)}+\mathrm{N_2(g)}\xrightarrow{\Delta}\mathrm{2BN(s)} \label{Eq22.6}$
$\mathrm{4B(s)} + \mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2B_2O_3(s)}\label{Eq22.7}$
$\mathrm{2B(s)} +\mathrm{3X_2(g)}\xrightarrow{\Delta}\mathrm{2BX_3(g)}\label{Eq22.8}$
As is typical of elements lying near the dividing line between metals and nonmetals, many compounds of boron are amphoteric, dissolving in either acid or base.
Boron nitride is similar in many ways to elemental carbon. With eight electrons, the B–N unit is isoelectronic with the C–C unit, and B and N have the same average size and electronegativity as C. The most stable form of BN is similar to graphite, containing six-membered B3N3 rings arranged in layers. At high temperature and pressure, hexagonal BN converts to a cubic structure similar to diamond, which is one of the hardest substances known. Boron oxide (B2O3) contains layers of trigonal planar BO3 groups (analogous to BX3) in which the oxygen atoms bridge two boron atoms. It dissolves many metal and nonmetal oxides, including SiO2, to give a wide range of commercially important borosilicate glasses. A small amount of CoO gives the deep blue color characteristic of “cobalt blue” glass.
At high temperatures, boron also reacts with virtually all metals to give metal borides that contain regular three-dimensional networks, or clusters, of boron atoms. The structures of two metal borides—ScB12 and CaB6—are shown in Figure $4$. Because metal-rich borides such as ZrB2 and TiB2 are hard and corrosion resistant even at high temperatures, they are used in applications such as turbine blades and rocket nozzles.
Boron hydrides were not discovered until the early 20th century, when the German chemist Alfred Stock undertook a systematic investigation of the binary compounds of boron and hydrogen, although binary hydrides of carbon, nitrogen, oxygen, and fluorine have been known since the 18th century. Between 1912 and 1936, Stock oversaw the preparation of a series of boron–hydrogen compounds with unprecedented structures that could not be explained with simple bonding theories. All these compounds contain multicenter bonds. The simplest example is diborane (B2H6), which contains two bridging hydrogen atoms (part (a) in Figure $5$. An extraordinary variety of polyhedral boron–hydrogen clusters is now known; one example is the B12H122− ion, which has a polyhedral structure similar to the icosahedral B12 unit of elemental boron, with a single hydrogen atom bonded to each boron atom.
A related class of polyhedral clusters, the carboranes, contain both CH and BH units; an example is shown here. Replacing the hydrogen atoms bonded to carbon with organic groups produces substances with novel properties, some of which are currently being investigated for their use as liquid crystals and in cancer chemotherapy.
The enthalpy of combustion of diborane (B2H6) is −2165 kJ/mol, one of the highest values known:
$B_2H_{6(g)} + 3O_{2(g)} \rightarrow B_2O_{3(s)} + 3H_2O(l)\;\;\; ΔH_{comb} = −2165\; kJ/mol \label{Eq 22.9}$
Consequently, the US military explored using boron hydrides as rocket fuels in the 1950s and 1960s. This effort was eventually abandoned because boron hydrides are unstable, costly, and toxic, and, most important, B2O3 proved to be highly abrasive to rocket nozzles. Reactions carried out during this investigation, however, showed that boron hydrides exhibit unusual reactivity.
Because boron and hydrogen have almost identical electronegativities, the reactions of boron hydrides are dictated by minor differences in the distribution of electron density in a given compound. In general, two distinct types of reaction are observed: electron-rich species such as the BH4 ion are reductants, whereas electron-deficient species such as B2H6 act as oxidants.
Example $1$
For each reaction, explain why the given products form.
1. B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(l)
2. BCl3(l) + 3H2O(l) → B(OH)3(aq) + 3HCl(aq)
3. $\mathrm{2BI_3(s)}+\mathrm{3H_2(g)}\xrightarrow{\Delta}\frac{1}{6}\mathrm{B_{12}(s)}+\mathrm{6HI(g)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Molecular oxygen is an oxidant. If the other reactant is a potential reductant, we expect that a redox reaction will occur. Although B2H6 contains boron in its highest oxidation state (+3), it also contains hydrogen in the −1 oxidation state (the hydride ion). Because hydride is a strong reductant, a redox reaction will probably occur. We expect that H will be oxidized to H+ and O2 will be reduced to O2−, but what are the actual products? A reasonable guess is B2O3 and H2O, both stable compounds.
2. Neither BCl3 nor water is a powerful oxidant or reductant, so a redox reaction is unlikely; a hydrolysis reaction is more probable. Nonmetal halides are acidic and react with water to form a solution of the hydrohalic acid and a nonmetal oxide or hydroxide. In this case, the most probable boron-containing product is boric acid [B(OH)3].
3. We normally expect a boron trihalide to behave like a Lewis acid. In this case, however, the other reactant is elemental hydrogen, which usually acts as a reductant. The iodine atoms in BI3 are in the lowest accessible oxidation state (−1), and boron is in the +3 oxidation state. Consequently, we can write a redox reaction in which hydrogen is oxidized and boron is reduced. Because compounds of boron in lower oxidation states are rare, we expect that boron will be reduced to elemental boron. The other product of the reaction must therefore be HI.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. $\mathrm{B_2H_6(g)}+\mathrm{H_2O(l)}\xrightarrow{\Delta}$
2. $\mathrm{BBr_3(l)}+\mathrm{O_2(g)}\rightarrow$
3. $\mathrm{B_2O_3(s)}+\mathrm{Ca(s)}\xrightarrow{\Delta}$
Answer
1. $\mathrm{B_2H_6(g)}+\mathrm{H_2O(l)}\xrightarrow{\Delta}\mathrm{2B(OH)_3(s)}+\mathrm{6H_2(g)}$
2. $\mathrm{BBr_3(l)}+\mathrm{O_2(g)}\rightarrow\textrm{no reaction}$
3. $\mathrm{6B_2O_3(s)}+18\mathrm{Ca(s)}\xrightarrow{\Delta}\mathrm{B_{12}(s)}+\mathrm{18CaO(s)}$
Reactions and Compounds of the Heavier Group 13 Elements
All four of the heavier group 13 elements (Al, Ga, In, and Tl) react readily with the halogens to form compounds with a 1:3 stoichiometry:
$2M_{(s)} + 3X_{2(s,l,g)} \rightarrow 2MX_{3(s)} \text{ or } M_2X_6 \label{Eq10}$
The reaction of Tl with iodine is an exception: although the product has the stoichiometry TlI3, it is not thallium(III) iodide, but rather a thallium(I) compound, the Tl+ salt of the triiodide ion (I3). This compound forms because iodine is not a powerful enough oxidant to oxidize thallium to the +3 oxidation state.
Of the halides, only the fluorides exhibit behavior typical of an ionic compound: they have high melting points (>950°C) and low solubility in nonpolar solvents. In contrast, the trichorides, tribromides, and triiodides of aluminum, gallium, and indium, as well as TlCl3 and TlBr3, are more covalent in character and form halogen-bridged dimers (part (b) in Figure $4$). Although the structure of these dimers is similar to that of diborane (B2H6), the bonding can be described in terms of electron-pair bonds rather than the delocalized electron-deficient bonding found in diborane. Bridging halides are poor electron-pair donors, so the group 13 trihalides are potent Lewis acids that react readily with Lewis bases, such as amines, to form a Lewis acid–base adduct:
$Al_2Cl_{6(soln)} + 2(CH_3)_3N_{(soln)} \rightarrow 2(CH_3)_3N:AlCl_{3(soln)} \label{Eq11}$
In water, the halides of the group 13 metals hydrolyze to produce the metal hydroxide ($M(OH)_3\)): \[MX_{3(s)} + 3H_2O_{(l)} \rightarrow M(OH)_{3(s)} + 3HX_{(aq)} \label{Eq12}$
In a related reaction, Al2(SO4)3 is used to clarify drinking water by the precipitation of hydrated Al(OH)3, which traps particulates. The halides of the heavier metals (In and Tl) are less reactive with water because of their lower charge-to-radius ratio. Instead of forming hydroxides, they dissolve to form the hydrated metal complex ions: [M(H2O)6]3+.
Of the group 13 halides, only the fluorides behave as typical ionic compounds.
Like boron (Equation $\ref{Eq22.7}$), all the heavier group 13 elements react with excess oxygen at elevated temperatures to give the trivalent oxide (M2O3), although Tl2O3 is unstable:
$\mathrm{4M(s)}+\mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2M_2O_3(s)} \label{Eq13}$
Aluminum oxide (Al2O3), also known as alumina, is a hard, high-melting-point, chemically inert insulator used as a ceramic and as an abrasive in sandpaper and toothpaste. Replacing a small number of Al3+ ions in crystalline alumina with Cr3+ ions forms the gemstone ruby, whereas replacing Al3+ with a mixture of Fe2+, Fe3+, and Ti4+ produces blue sapphires. The gallium oxide compound MgGa2O4 gives the brilliant green light familiar to anyone who has ever operated a xerographic copy machine. All the oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric, which is consistent with their location along the diagonal line of the periodic table, also dissolving in concentrated aqueous base to form solutions that contain M(OH)4 ions.
Group 13 trihalides are potent Lewis acids that react with Lewis bases to form a Lewis acid–base adduct.
Aluminum, gallium, and indium also react with the other group 16 elements (chalcogens) to form chalcogenides with the stoichiometry M2Y3. However, because Tl(III) is too strong an oxidant to form a stable compound with electron-rich anions such as S2−, Se2−, and Te2−, thallium forms only the thallium(I) chalcogenides with the stoichiometry Tl2Y. Only aluminum, like boron, reacts directly with N2 (at very high temperatures) to give AlN, which is used in transistors and microwave devices as a nontoxic heat sink because of its thermal stability; GaN and InN can be prepared using other methods. All the metals, again except Tl, also react with the heavier group 15 elements (pnicogens) to form the so-called III–V compounds, such as GaAs. These are semiconductors, whose electronic properties, such as their band gaps, differ from those that can be achieved using either pure or doped group 14 elements. For example, nitrogen- and phosphorus-doped gallium arsenide (GaAs1−x−yPxNy) is used in the displays of calculators and digital watches.
All group 13 oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric.
Unlike boron, the heavier group 13 elements do not react directly with hydrogen. Only the aluminum and gallium hydrides are known, but they must be prepared indirectly; AlH3 is an insoluble, polymeric solid that is rapidly decomposed by water, whereas GaH3 is unstable at room temperature.
Complexes of Group 13 Elements
Boron has a relatively limited tendency to form complexes, but aluminum, gallium, indium, and, to some extent, thallium form many complexes. Some of the simplest are the hydrated metal ions [M(H2O)63+], which are relatively strong Brønsted–Lowry acids that can lose a proton to form the M(H2O)5(OH)2+ ion:
$[M(H_2O)_6]^{3+}_{(aq)} \rightarrow M(H_2O)_5(OH)^{2+}_{(aq)} + H^+_{(aq)} \label{Eq14}$
Group 13 metal ions also form stable complexes with species that contain two or more negatively charged groups, such as the oxalate ion. The stability of such complexes increases as the number of coordinating groups provided by the ligand increases.
Example $2$
For each reaction, explain why the given products form.
1. $\mathrm{2Al(s)} + \mathrm{Fe_2O_3(s)}\xrightarrow{\Delta}\mathrm{2Fe(l)} + \mathrm{Al_2O_3(s)}$
2. $\mathrm{2Ga(s)} + \mathrm{6H_2O(l)}+ \mathrm{2OH^-(aq)}\xrightarrow{\Delta}\mathrm{3H_2(g)} + \mathrm{2Ga(OH)^-_4(aq)}$
3. $\mathrm{In_2Cl_6(s)}\xrightarrow{\mathrm{H_2O(l)}}\mathrm{2In^{3+}(aq)}+\mathrm{6Cl^-(aq)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Aluminum is an active metal and a powerful reductant, and Fe2O3 contains Fe(III), a potential oxidant. Hence a redox reaction is probable, producing metallic Fe and Al2O3. Because Al is a main group element that lies above Fe, which is a transition element, it should be a more active metal than Fe. Thus the reaction should proceed to the right. In fact, this is the thermite reaction, which is so vigorous that it produces molten Fe and can be used for welding.
2. Gallium lies immediately below aluminum in the periodic table and is amphoteric, so it will dissolve in either acid or base to produce hydrogen gas. Because gallium is similar to aluminum in many of its properties, we predict that gallium will dissolve in the strong base.
3. The metallic character of the group 13 elements increases with increasing atomic number. Indium trichloride should therefore behave like a typical metal halide, dissolving in water to form the hydrated cation.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. LiH(s) + Al2Cl6(soln)→
2. Al2O3(s) + OH(aq)→
3. Al(s) + N2(g) $\xrightarrow{\Delta}$
4. Ga2Cl6(soln) + Cl(soln)→
Answer
1. 8LiH(s) + Al2Cl6(soln)→2LiAlH4(soln) + 6LiCl(s)
2. Al2O3(s) + 2OH(aq) + 3H2O(l) → 2Al(OH)4(aq)
3. 2Al(s) + N2(g) $\xrightarrow{\Delta}$ 2AlN(s)
4. Ga2Cl6(soln) + 2Cl(soln) → 2GaCl4(soln)
Summary
Compounds of the group 13 elements with oxygen are thermodynamically stable. Many of the anomalous properties of the group 13 elements can be explained by the increase in Zeff moving down the group. Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior. Many of the inconsistencies observed in the properties of the group 13 elements can be explained by the increase in Zeff that arises from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Instead of forming a metallic lattice with delocalized valence electrons, boron forms unique aggregates that contain multicenter bonds, including metal borides, in which boron is bonded to other boron atoms to form three-dimensional networks or clusters with regular geometric structures. All neutral compounds of the group 13 elements are electron deficient and behave like Lewis acids. The trivalent halides of the heavier elements form halogen-bridged dimers that contain electron-pair bonds, rather than the delocalized electron-deficient bonds characteristic of diborane. Their oxides dissolve in dilute acid, although the oxides of aluminum and gallium are amphoteric. None of the group 13 elements reacts directly with hydrogen, and the stability of the hydrides prepared by other routes decreases as we go down the group. In contrast to boron, the heavier group 13 elements form a large number of complexes in the +3 oxidation state. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.06%3A__Boron_Aluminum_and_the_Group_3A_Elements.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 14 elements.
The elements of group 14 show a greater range of chemical behavior than any other family in the periodic table. Three of the five elements—carbon, tin, and lead—have been known since ancient times. For example, some of the oldest known writings are Egyptian hieroglyphics written on papyrus with ink made from lampblack, a finely divided carbon soot produced by the incomplete combustion of hydrocarbons (Figure $1$). Activated carbon is an even more finely divided form of carbon that is produced from the thermal decomposition of organic materials, such as sawdust. Because it adsorbs many organic and sulfur-containing compounds, activated carbon is used to decolorize foods, such as sugar, and to purify gases and wastewater.
Tin and lead oxides and sulfides are easily reduced to the metal by heating with charcoal, a discovery that must have occurred by accident when prehistoric humans used rocks containing their ores for a cooking fire. However, because tin and copper ores are often found together in nature, their alloy—bronze—was probably discovered before either element, a discovery that led to the Bronze Age. The heaviest element in group 14, lead, is such a soft and malleable metal that the ancient Romans used thin lead foils as writing tablets, as well as lead cookware and lead pipes for plumbing. (Recall that the atomic symbols for tin and lead come from their Latin names: Sn for stannum and Pb for plumbum.)
Although the first glasses were prepared from silica (silicon oxide, SiO2) around 1500 BC, elemental silicon was not prepared until 1824 because of its high affinity for oxygen. Jöns Jakob Berzelius was finally able to obtain amorphous silicon by reducing Na2SiF6 with molten potassium. The crystalline element, which has a shiny blue-gray luster, was not isolated until 30 yr later. The last member of the group 14 elements to be discovered was germanium, which was found in 1886 in a newly discovered silver-colored ore by the German chemist Clemens Winkler, who named the element in honor of his native country.
Preparation and General Properties of the Group 14 Elements
The natural abundance of the group 14 elements varies tremendously. Elemental carbon, for example, ranks only 17th on the list of constituents of Earth’s crust. Pure graphite is obtained by reacting coke, an amorphous form of carbon used as a reductant in the production of steel, with silica to give silicon carbide (SiC). This is then thermally decomposed at very high temperatures (2700°C) to give graphite:
$\mathrm{SiO_2(s)}+\mathrm{3C(s)}\xrightarrow{\Delta}\mathrm{SiC(s)}+\mathrm{2CO(g)} \label{$1$}$
$\mathrm{SiC(s)}\xrightarrow{\Delta}\mathrm{Si(s)}+\mathrm{C(graphite)} \label{$2$}$
One allotrope of carbon, diamond, is metastable under normal conditions, with a ΔG°f of 2.9 kJ/mol versus graphite. At pressures greater than 50,000 atm, however, the diamond structure is favored and is the most stable form of carbon. Because the structure of diamond is more compact than that of graphite, its density is significantly higher (3.51 g/cm3 versus 2.2 g/cm3). Because of its high thermal conductivity, diamond powder is used to transfer heat in electronic devices.
The most common sources of diamonds on Earth are ancient volcanic pipes that contain a rock called kimberlite, a lava that solidified rapidly from deep inside the Earth. Most kimberlite formations, however, are much newer than the diamonds they contain. In fact, the relative amounts of different carbon isotopes in diamond show that diamond is a chemical and geological “fossil” older than our solar system, which means that diamonds on Earth predate the existence of our sun. Thus diamonds were most likely created deep inside Earth from primordial grains of graphite present when Earth was formed (part (a) in Figure $2$). Gem-quality diamonds can now be produced synthetically and have chemical, optical, and physical characteristics identical to those of the highest-grade natural diamonds.
Although oxygen is the most abundant element on Earth, the next most abundant is silicon, the next member of group 14. Pure silicon is obtained by reacting impure silicon with Cl2 to give SiCl4, followed by the fractional distillation of the impure SiCl4 and reduction with H2:
$\mathrm{SiCl_4(l)}+\mathrm{2H_2(g)}\xrightarrow{\Delta}\mathrm{Si(s)}+\mathrm{4HCl(g)} \label{$3$}$
Several million tons of silicon are annually produced with this method. Amorphous silicon containing residual amounts of hydrogen is used in photovoltaic devices that convert light to electricity, and silicon-based solar cells are used to power pocket calculators, boats, and highway signs, where access to electricity by conventional methods is difficult or expensive. Ultrapure silicon and germanium form the basis of the modern electronics industry (part (b) in Figure $2$).
In contrast to silicon, the concentrations of germanium and tin in Earth’s crust are only 1–2 ppm. The concentration of lead, which is the end product of the nuclear decay of many radionuclides, is 13 ppm, making lead by far the most abundant of the heavy group 14 elements. No concentrated ores of germanium are known; like indium, germanium is generally recovered from flue dusts obtained by processing the ores of metals such as zinc. Because germanium is essentially transparent to infrared radiation, it is used in optical devices.
Tin and lead are soft metals that are too weak for structural applications, but because tin is flexible, corrosion resistant, and nontoxic, it is used as a coating in food packaging. A “tin can,” for example, is actually a steel can whose interior is coated with a thin layer (1–2 µm) of metallic tin. Tin is also used in superconducting magnets and low-melting-point alloys such as solder and pewter. Pure lead is obtained by heating galena (PbS) in air and reducing the oxide (PbO) to the metal with carbon, followed by electrolytic deposition to increase the purity:
$\mathrm{PbS(s)}+\frac{3}{2}\mathrm{O_2(g)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\mathrm{SO_2(g)} \label{$4$}$
$\mathrm{PbO(s)}+\mathrm{C(s)}\xrightarrow{\Delta}\mathrm{Pb(l)}+\mathrm{CO(g)} \label{$5$}$
or
$\mathrm{PbO(s)}+\mathrm{CO(g)}\xrightarrow{\Delta}\mathrm{Pb(l)}+\mathrm{CO_2(g)} \label{$6$}$
By far the single largest use of lead is in lead storage batteries. The group 14 elements all have ns2np2 valence electron configurations. All form compounds in which they formally lose either the two np and the two ns valence electrons or just the two np valence electrons, giving a +4 or +2 oxidation state, respectively. Because covalent bonds decrease in strength with increasing atomic size and the ionization energies for the heavier elements of the group are higher than expected due to a higher Zeff, the relative stability of the +2 oxidation state increases smoothly from carbon to lead.
The relative stability of the +2 oxidation state increases, and the tendency to form catenated compounds decreases, from carbon to lead in group 14.
Recall that many carbon compounds contain multiple bonds formed by π overlap of singly occupied 2p orbitals on adjacent atoms. Compounds of silicon, germanium, tin, and lead with the same stoichiometry as those of carbon, however, tend to have different structures and properties. For example, CO2 is a gas that contains discrete O=C=O molecules, whereas the most common form of SiO2 is the high-melting solid known as quartz, the major component of sand. Instead of discrete SiO2 molecules, quartz contains a three-dimensional network of silicon atoms that is similar to the structure of diamond but with an oxygen atom inserted between each pair of silicon atoms. Thus each silicon atom is linked to four other silicon atoms by bridging oxygen atoms.
The tendency to catenate—to form chains of like atoms—decreases rapidly as we go down group 14 because bond energies for both the E–E and E–H bonds decrease with increasing atomic number (where E is any group 14 element). Consequently, inserting a CH2 group into a linear hydrocarbon such as n-hexane is exergonic (ΔG° = −45 kJ/mol), whereas inserting an SiH2 group into the silicon analogue of n-hexane (Si6H14) actually costs energy (ΔG° ≈ +25 kJ/mol). As a result of this trend, the thermal stability of catenated compounds decreases rapidly from carbon to lead.
In Table $1$ "Selected Properties of the Group 14 Elements" we see, once again, that there is a large difference between the lightest element (C) and the others in size, ionization energy, and electronegativity. As in group 13, the second and third elements (Si and Ge) are similar, and there is a reversal in the trends for some properties, such as ionization energy, between the fourth and fifth elements (Sn and Pb). As for group 13, these effects can be explained by the presence of filled (n − 1)d and (n − 2)f subshells, whose electrons are relatively poor at screening the outermost electrons from the higher nuclear charge.
Table $1$: Selected Properties of the Group 14 Elements
Property Carbon Silicon Germanium Tin Lead
*The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate +4 ions in the most common oxidation state, except for C4+ and Si4+, for which values for the four-coordinate ion are estimated.
X is Cl, Br, or I. Reaction with F2 gives the tetrafluorides (EF4) for all group 14 elements, where E represents any group 14 element.
atomic symbol C Si Ge Sn Pb
atomic number 6 14 32 50 82
atomic mass (amu) 12.01 28.09 72.64 118.71 207.2
valence electron configuration* 2s22p2 3s23p2 4s24p2 5s25p2 6s26p2
melting point/boiling point (°C) 4489 (at 10.3 MPa)/3825 1414/3265 939/2833 232/2602 327/1749
density (g/cm3) at 25°C 2.2 (graphite), 3.51 (diamond) 2.33 5.32 7.27(white) 11.30
atomic radius (pm) 77 (diamond) 111 125 145 154
first ionization energy (kJ/mol) 1087 787 762 709 716
most common oxidation state +4 +4 +4 +4 +4
ionic radius (pm) ≈29 ≈40 53 69 77.5
electron affinity (kJ/mol) −122 −134 −119 −107 −35
electronegativity 2.6 1.9 2.0 2.0 1.8
standard reduction potential (E°, V) (for EO2 → E in acidic solution) 0.21 −0.86 −0.18 −0.12 0.79
product of reaction with O2 CO2, CO SiO2 GeO2 SnO2 PbO
type of oxide acidic (CO2) acidic neutral (CO) amphoteric amphoteric amphoteric
product of reaction with N2 none Si3N4 none Sn3N4 none
product of reaction with X2 CX4 SiX4 GeX4 SnX4 PbX2
product of reaction with H2 CH4 none none none none
The group 14 elements follow the same pattern as the group 13 elements in their periodic properties.
Reactions and Compounds of Carbon
Carbon is the building block of all organic compounds, including biomolecules, fuels, pharmaceuticals, and plastics, whereas inorganic compounds of carbon include metal carbonates, which are found in substances as diverse as fertilizers and antacid tablets, halides, oxides, carbides, and carboranes. Like boron in group 13, the chemistry of carbon differs sufficiently from that of its heavier congeners to merit a separate discussion.
The structures of the allotropes of carbon—diamond, graphite, fullerenes, and nanotubes—are distinct, but they all contain simple electron-pair bonds (Figure 7.18). Although it was originally believed that fullerenes were a new form of carbon that could be prepared only in the laboratory, fullerenes have been found in certain types of meteorites. Another possible allotrope of carbon has also been detected in impact fragments of a carbon-rich meteorite; it appears to consist of long chains of carbon atoms linked by alternating single and triple bonds, (–C≡C–C≡C–)n. Carbon nanotubes (“buckytubes”) are being studied as potential building blocks for ultramicroscale detectors and molecular computers and as tethers for space stations. They are currently used in electronic devices, such as the electrically conducting tips of miniature electron guns for flat-panel displays in portable computers.
Although all the carbon tetrahalides (CX4) are known, they are generally not obtained by the direct reaction of carbon with the elemental halogens (X2) but by indirect methods such as the following reaction, where X is Cl or Br:
$CH_{4(g)} + 4X_{2(g)} \rightarrow CX_{4(l,s)} + 4HX_{(g)} \label{$7$}$
The carbon tetrahalides all have the tetrahedral geometry predicted by the valence-shell electron-pair repulsion (VSEPR) model, as shown for CCl4 and CI4. Their stability decreases rapidly as the halogen increases in size because of poor orbital overlap and increased crowding. Because the C–F bond is about 25% stronger than a C–H bond, fluorocarbons are thermally and chemically more stable than the corresponding hydrocarbons, while having a similar hydrophobic character. A polymer of tetrafluoroethylene (F2C=CF2), analogous to polyethylene, is the nonstick Teflon lining found on many cooking pans, and similar compounds are used to make fabrics stain resistant (such as Scotch-Gard) or waterproof but breathable (such as Gore-Tex).
The stability of the carbon tetrahalides decreases with increasing size of the halogen due to increasingly poor orbital overlap and crowding.
Carbon reacts with oxygen to form either CO or CO2, depending on the stoichiometry. Carbon monoxide is a colorless, odorless, and poisonous gas that reacts with the iron in hemoglobin to form an Fe–CO unit, which prevents hemoglobin from binding, transporting, and releasing oxygen in the blood (see Figure 23.26). In the laboratory, carbon monoxide can be prepared on a small scale by dehydrating formic acid with concentrated sulfuric acid:
$\mathrm{HCO_2H(l)}\xrightarrow{\mathrm{H_2SO_4(l)}}\mathrm{CO(g)}+\mathrm{H_3O^+(aq)}+\mathrm{HSO^-_4}\label{$8$}$
Carbon monoxide also reacts with the halogens to form the oxohalides (COX2). Probably the best known of these is phosgene (Cl2C=O), which is highly poisonous and was used as a chemical weapon during World War I:
$\mathrm{CO(g)}+\mathrm{Cl_2(g)}\xrightarrow{\Delta}\textrm{Cl}_2\textrm{C=O(g)}\label{$9$}$
Despite its toxicity, phosgene is an important industrial chemical that is prepared on a large scale, primarily in the manufacture of polyurethanes.
Carbon dioxide can be prepared on a small scale by reacting almost any metal carbonate or bicarbonate salt with a strong acid. As is typical of a nonmetal oxide, CO2 reacts with water to form acidic solutions containing carbonic acid (H2CO3). In contrast to its reactions with oxygen, reacting carbon with sulfur at high temperatures produces only carbon disulfide (CS2):
$\mathrm{C(s)}+\mathrm{2S(g)}\xrightarrow{\Delta}\mathrm{CS_2(g)}\label{$1$0}$
The selenium analog CSe2 is also known. Both have the linear structure predicted by the VSEPR model, and both are vile smelling (and in the case of CSe2, highly toxic), volatile liquids. The sulfur and selenium analogues of carbon monoxide, CS and CSe, are unstable because the C≡Y bonds (Y is S or Se) are much weaker than the C≡O bond due to poorer π orbital overlap.
$\pi$ bonds between carbon and the heavier chalcogenides are weak due to poor orbital overlap.
Binary compounds of carbon with less electronegative elements are called carbides. The chemical and physical properties of carbides depend strongly on the identity of the second element, resulting in three general classes: ionic carbides, interstitial carbides, and covalent carbides. The reaction of carbon at high temperatures with electropositive metals such as those of groups 1 and 2 and aluminum produces ionic carbides, which contain discrete metal cations and carbon anions. The identity of the anions depends on the size of the second element. For example, smaller elements such as beryllium and aluminum give methides such as Be2C and Al4C3, which formally contain the C4− ion derived from methane (CH4) by losing all four H atoms as protons. In contrast, larger metals such as sodium and calcium give carbides with stoichiometries of Na2C2 and CaC2. Because these carbides contain the C4− ion, which is derived from acetylene (HC≡CH) by losing both H atoms as protons, they are more properly called acetylides. Reacting ionic carbides with dilute aqueous acid results in protonation of the anions to give the parent hydrocarbons: CH4 or C2H2. For many years, miners’ lamps used the reaction of calcium carbide with water to produce a steady supply of acetylene, which was ignited to provide a portable lantern.
The reaction of carbon with most transition metals at high temperatures produces interstitial carbides. Due to the less electropositive nature of the transition metals, these carbides contain covalent metal–carbon interactions, which result in different properties: most interstitial carbides are good conductors of electricity, have high melting points, and are among the hardest substances known. Interstitial carbides exhibit a variety of nominal compositions, and they are often nonstoichiometric compounds whose carbon content can vary over a wide range. Among the most important are tungsten carbide (WC), which is used industrially in high-speed cutting tools, and cementite (Fe3C), which is a major component of steel.
Elements with an electronegativity similar to that of carbon form covalent carbides, such as silicon carbide (SiC; Equation $\ref{Eq1}$) and boron carbide (B4C). These substances are extremely hard, have high melting points, and are chemically inert. For example, silicon carbide is highly resistant to chemical attack at temperatures as high as 1600°C. Because it also maintains its strength at high temperatures, silicon carbide is used in heating elements for electric furnaces and in variable-temperature resistors.
Carbides formed from group 1 and 2 elements are ionic. Transition metals form interstitial carbides with covalent metal–carbon interactions, and covalent carbides are chemically inert.
Example $1$
For each reaction, explain why the given product forms.
1. CO(g) + Cl2(g) → Cl2C=O(g)
2. CO(g) + BF3(g) → F3B:C≡O(g)
3. Sr(s) + 2C(s) $\xrightarrow{\Delta}$ SrC2(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Because the carbon in CO is in an intermediate oxidation state (+2), CO can be either a reductant or an oxidant; it is also a Lewis base. The other reactant (Cl2) is an oxidant, so we expect a redox reaction to occur in which the carbon of CO is further oxidized. Because Cl2 is a two-electron oxidant and the carbon atom of CO can be oxidized by two electrons to the +4 oxidation state, the product is phosgene (Cl2C=O).
2. Unlike Cl2, BF3 is not a good oxidant, even though it contains boron in its highest oxidation state (+3). Nor can BF3 behave like a reductant. Like any other species with only six valence electrons, however, it is certainly a Lewis acid. Hence an acid–base reaction is the most likely alternative, especially because we know that CO can use the lone pair of electrons on carbon to act as a Lewis base. The most probable reaction is therefore the formation of a Lewis acid–base adduct.
3. Typically, both reactants behave like reductants. Unless one of them can also behave like an oxidant, no reaction will occur. We know that Sr is an active metal because it lies far to the left in the periodic table and that it is more electropositive than carbon. Carbon is a nonmetal with a significantly higher electronegativity; it is therefore more likely to accept electrons in a redox reaction. We conclude, therefore, that Sr will be oxidized, and C will be reduced. Carbon forms ionic carbides with active metals, so the reaction will produce a species formally containing either C4− or C22−. Those that contain C4− usually involve small, highly charged metal ions, so Sr2+ will produce the acetylide (SrC2) instead.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$
2. C(s) + H2O(l) →
3. NaHCO3(s) + H2SO4(aq) →
Answer
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$ CO2(g)
2. C(s) + H2O(l) → no reaction
3. NaHCO3(s) + H2SO4(aq) → CO2(g) + NaHSO4(aq) + H2O(l)
Reactions and Compounds of the Heavier Group 14 Elements
Although silicon, germanium, tin, and lead in their +4 oxidation states often form binary compounds with the same stoichiometry as carbon, the structures and properties of these compounds are usually significantly different from those of the carbon analogues. Silicon and germanium are both semiconductors with structures analogous to diamond. Tin has two common allotropes: white (β) tin has a metallic lattice and metallic properties, whereas gray (α) tin has a diamond-like structure and is a semiconductor. The metallic β form is stable above 13.2°C, and the nonmetallic α form is stable below 13.2°C. Lead is the only group 14 element that is metallic in both structure and properties under all conditions.
Video $1$: Time lapse tin pest reaction.
Based on its position in the periodic table, we expect silicon to be amphoteric. In fact, it dissolves in strong aqueous base to produce hydrogen gas and solutions of silicates, but the only aqueous acid that it reacts with is hydrofluoric acid, presumably due to the formation of the stable SiF62− ion. Germanium is more metallic in its behavior than silicon. For example, it dissolves in hot oxidizing acids, such as HNO3 and H2SO4, but in the absence of an oxidant, it does not dissolve in aqueous base. Although tin has an even more metallic character than germanium, lead is the only element in the group that behaves purely as a metal. Acids do not readily attack it because the solid acquires a thin protective outer layer of a Pb2+ salt, such as PbSO4.
All group 14 dichlorides are known, and their stability increases dramatically as the atomic number of the central atom increases. Thus CCl2 is dichlorocarbene, a highly reactive, short-lived intermediate that can be made in solution but cannot be isolated in pure form using standard techniques; SiCl2 can be isolated at very low temperatures, but it decomposes rapidly above −150°C, and GeCl2 is relatively stable at temperatures below 20°C. In contrast, SnCl2 is a polymeric solid that is indefinitely stable at room temperature, whereas PbCl2 is an insoluble crystalline solid with a structure similar to that of SnCl2.
The stability of the group 14 dichlorides increases dramatically from carbon to lead.
Although the first four elements of group 14 form tetrahalides (MX4) with all the halogens, only fluorine is able to oxidize lead to the +4 oxidation state, giving PbF4. The tetrahalides of silicon and germanium react rapidly with water to give amphoteric oxides (where M is Si or Ge):
$MX_{4(s,l)} + 2H_2O_{(l)} \rightarrow MO_{2(s)} + 4HX_{(aq)} \label{$1$1}$
In contrast, the tetrahalides of tin and lead react with water to give hydrated metal ions. Because of the stability of its +2 oxidation state, lead reacts with oxygen or sulfur to form PbO or PbS, respectively, whereas heating the other group 14 elements with excess O2 or S8 gives the corresponding dioxides or disulfides, respectively. The dioxides of the group 14 elements become increasingly basic as we go down the group.
The dioxides of the group 14 elements become increasingly basic down the group.
Because the Si–O bond is even stronger than the C–O bond (~452 kJ/mol versus ~358 kJ/mol), silicon has a strong affinity for oxygen. The relative strengths of the C–O and Si–O bonds contradict the generalization that bond strengths decrease as the bonded atoms become larger. This is because we have thus far assumed that a formal single bond between two atoms can always be described in terms of a single pair of shared electrons. In the case of Si–O bonds, however, the presence of relatively low-energy, empty d orbitals on Si and nonbonding electron pairs in the p or spn hybrid orbitals of O results in a partial π bond (Figure $3$). Due to its partial π double bond character, the Si–O bond is significantly stronger and shorter than would otherwise be expected. A similar interaction with oxygen is also an important feature of the chemistry of the elements that follow silicon in the third period (P, S, and Cl). Because the Si–O bond is unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Because silicon–oxygen bonds are unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Compounds with anions that contain only silicon and oxygen are called silicates, whose basic building block is the SiO44− unit:
The number of oxygen atoms shared between silicon atoms and the way in which the units are linked vary considerably in different silicates. Converting one of the oxygen atoms from terminal to bridging generates chains of silicates, while converting two oxygen atoms from terminal to bridging generates double chains. In contrast, converting three or four oxygens to bridging generates a variety of complex layered and three-dimensional structures, respectively.
In a large and important class of materials called aluminosilicates, some of the Si atoms are replaced by Al atoms to give aluminosilicates such as zeolites, whose three-dimensional framework structures have large cavities connected by smaller tunnels (Figure $4$). Because the cations in zeolites are readily exchanged, zeolites are used in laundry detergents as water-softening agents: the more loosely bound Na+ ions inside the zeolite cavities are displaced by the more highly charged Mg2+ and Ca2+ ions present in hard water, which bind more tightly. Zeolites are also used as catalysts and for water purification.
Silicon and germanium react with nitrogen at high temperature to form nitrides (M3N4):
$3Si_{(l)} + 2N_{2(g)} \rightarrow Si_3N_{4(s)} \label{$1$2}$
Silicon nitride has properties that make it suitable for high-temperature engineering applications: it is strong, very hard, and chemically inert, and it retains these properties to temperatures of about 1000°C.
Because of the diagonal relationship between boron and silicon, metal silicides and metal borides exhibit many similarities. Although metal silicides have structures that are as complex as those of the metal borides and carbides, few silicides are structurally similar to the corresponding borides due to the significantly larger size of Si (atomic radius 111 pm versus 87 pm for B). Silicides of active metals, such as Mg2Si, are ionic compounds that contain the Si4− ion. They react with aqueous acid to form silicon hydrides such as SiH4:
$Mg_2Si_{(s)} + 4H^+_{(aq)} \rightarrow 2Mg^{2+}_{(aq)} + SiH_{4(g)} \label{$1$3}$
Unlike carbon, catenated silicon hydrides become thermodynamically less stable as the chain lengthens. Thus straight-chain and branched silanes (analogous to alkanes) are known up to only n = 10; the germanium analogues (germanes) are known up to n = 9. In contrast, the only known hydride of tin is SnH4, and it slowly decomposes to elemental Sn and H2 at room temperature. The simplest lead hydride (PbH4) is so unstable that chemists are not even certain it exists. Because E=E and E≡E bonds become weaker with increasing atomic number (where E is any group 14 element), simple silicon, germanium, and tin analogues of alkenes, alkynes, and aromatic hydrocarbons are either unstable (Si=Si and Ge=Ge) or unknown. Silicon-based life-forms are therefore likely to be found only in science fiction.
The stability of group 14 hydrides decreases down the group, and E=E and E≡E bonds become weaker.
The only important organic derivatives of lead are compounds such as tetraethyllead [(CH3CH2)4Pb]. Because the Pb–C bond is weak, these compounds decompose at relatively low temperatures to produce alkyl radicals (R·), which can be used to control the rate of combustion reactions. For 60 yr, hundreds of thousands of tons of lead were burned annually in automobile engines, producing a mist of lead oxide particles along the highways that constituted a potentially serious public health problem. (Example 6 in Section 22.3 examines this problem.) The use of catalytic converters reduced the amount of carbon monoxide, nitrogen oxides, and hydrocarbons released into the atmosphere through automobile exhausts, but it did nothing to decrease lead emissions. Because lead poisons catalytic converters, however, its use as a gasoline additive has been banned in most of the world.
Compounds that contain Si–C and Si–O bonds are stable and important. High-molecular-mass polymers called silicones contain an (Si–O–)n backbone with organic groups attached to Si (Figure $5$). The properties of silicones are determined by the chain length, the type of organic group, and the extent of cross-linking between the chains. Without cross-linking, silicones are waxes or oils, but cross-linking can produce flexible materials used in sealants, gaskets, car polishes, lubricants, and even elastic materials, such as the plastic substance known as Silly Putty.
Example $2$
For each reaction, explain why the given products form.
1. Pb(s) + Cl2(g) → PbCl2(s)
2. Mg2Si(s) + 4H2O(l) → SiH4(g) + 2Mg(OH)2(s)
3. GeO2(s) + 4OH(aq) → GeO44−(aq) + 2H2O(l)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Lead is a metal, and chlorine is a nonmetal that is a strong oxidant. Thus we can expect a redox reaction to occur in which the metal acts as a reductant. Although lead can form compounds in the +2 and +4 oxidation states, Pb4+ is a potent oxidant (the inert-pair effect). Because lead prefers the +2 oxidation state and chlorine is a weaker oxidant than fluorine, we expect PbCl2 to be the product.
2. This is the reaction of water with a metal silicide, which formally contains the Si4− ion. Water can act as either an acid or a base. Because the other compound is a base, we expect an acid–base reaction to occur in which water acts as an acid. Because Mg2Si contains Si in its lowest possible oxidation state, however, an oxidation–reduction reaction is also a possibility. But water is a relatively weak oxidant, so an acid–base reaction is more likely. The acid (H2O) transfers a proton to the base (Si4−), which can accept four protons to form SiH4. Proton transfer from water produces the OH ion, which will combine with Mg2+ to give magnesium hydroxide.
3. We expect germanium dioxide (GeO2) to be amphoteric because of the position of germanium in the periodic table. It should dissolve in strong aqueous base to give an anionic species analogous to silicate.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. PbO2(s) $\xrightarrow{\Delta}$
2. GeCl4(s) + H2O(l) →
3. Sn(s) + HCl(aq) →
Answer
1. $\mathrm{PbO_2(s)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\frac{1}{2}\mathrm{O_2(g)}$
2. GeCl4(s) + 2H2O(l) → GeO2(s) + 4HCl(aq)
3. Sn(s) + 2HCl(aq) → Sn2+(aq) + H2(g) + 2Cl(aq)
Summary
The group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. The group 14 elements show the greatest range of chemical behavior of any group in the periodic table. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff, the stability of the +2 oxidation state increases from carbon to lead. The tendency to form multiple bonds and to catenate decreases as the atomic number increases. The stability of the carbon tetrahalides decreases as the halogen increases in size because of poor orbital overlap and steric crowding. Carbon forms three kinds of carbides with less electronegative elements: ionic carbides, which contain metal cations and C4− (methide) or C22− (acetylide) anions; interstitial carbides, which are characterized by covalent metal–carbon interactions and are among the hardest substances known; and covalent carbides, which have three-dimensional covalent network structures that make them extremely hard, high melting, and chemically inert. Consistent with periodic trends, metallic behavior increases down the group. Silicon has a tremendous affinity for oxygen because of partial Si–O π bonding. Dioxides of the group 14 elements become increasingly basic down the group and their metallic character increases. Silicates contain anions that consist of only silicon and oxygen. Aluminosilicates are formed by replacing some of the Si atoms in silicates by Al atoms; aluminosilicates with three-dimensional framework structures are called zeolites. Nitrides formed by reacting silicon or germanium with nitrogen are strong, hard, and chemically inert. The hydrides become thermodynamically less stable down the group. Moreover, as atomic size increases, multiple bonds between or to the group 14 elements become weaker. Silicones, which contain an Si–O backbone and Si–C bonds, are high-molecular-mass polymers whose properties depend on their compositions.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.07%3A__Silicon_and_the_Group_4A_Elements.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 15 elements: the pnicogens.
Like the group 14 elements, the lightest member of group 15, nitrogen, is found in nature as the free element, and the heaviest elements have been known for centuries because they are easily isolated from their ores. Antimony (Sb) was probably the first of the pnicogens to be obtained in elemental form and recognized as an element. Its atomic symbol comes from its Roman name: stibium. It is found in stibnite (Sb2S3), a black mineral that has been used as a cosmetic (an early form of mascara) since biblical times, and it is easily reduced to the metal in a charcoal fire (Figure $1$). The Egyptians used antimony to coat copper objects as early as the third millennium BC, and antimony is still used in alloys to improve the tonal quality of bells.
In the form of its yellow sulfide ore, orpiment (As2S3), arsenic (As) has been known to physicians and professional assassins since ancient Greece, although elemental arsenic was not isolated until centuries later. The history of bismuth (Bi), in contrast, is more difficult to follow because early alchemists often confused it with other metals, such as lead, tin, antimony, and even silver (due to its slightly pinkish-white luster). Its name comes from the old German wismut, meaning “white metal.” Bismuth was finally isolated in the 15th century, and it was used to make movable type for printing shortly after the invention of the Gutenberg printing process in 1440. Bismuth is used in printing because it is one of the few substances known whose solid state is less dense than the liquid. Consequently, its alloys expand as they cool, filling a mold completely and producing crisp, clear letters for typesetting.
Phosphorus was discovered in 1669 by the German alchemist Hennig Brandt, who was looking for the “philosophers’ stone,” a mythical substance capable of converting base metals to silver or gold. Believing that human urine was the source of the key ingredient, Brandt obtained several dozen buckets of urine, which he allowed to putrefy. The urine was distilled to dryness at high temperature and then condensed; the last fumes were collected under water, giving a waxy white solid that had unusual properties. For example, it glowed in the dark and burst into flames when removed from the water. (Unfortunately for Brandt, however, it did not turn lead into gold.) The element was given its current name (from the Greek phos, meaning “light,” and phoros, meaning “bringing”) in the 17th century. For more than a century, the only way to obtain phosphorus was the distillation of urine, but in 1769 it was discovered that phosphorus could be obtained more easily from bones. During the 19th century, the demand for phosphorus for matches was so great that battlefields and paupers’ graveyards were systematically scavenged for bones. Early matches were pieces of wood coated with elemental phosphorus that were stored in an evacuated glass tube and ignited when the tube was broken (which could cause unfortunate accidents if the matches were kept in a pocket!).
Unfortunately, elemental phosphorus is volatile and highly toxic. It is absorbed by the teeth and destroys bone in the jaw, leading to a painful and fatal condition called “phossy jaw,” which for many years was accepted as an occupational hazard of working in the match industry.
Although nitrogen is the most abundant element in the atmosphere, it was the last of the pnicogens to be obtained in pure form. In 1772, Daniel Rutherford, working with Joseph Black (who discovered CO2), noticed that a gas remained when CO2 was removed from a combustion reaction. Antoine Lavoisier called the gas azote, meaning “no life,” because it did not support life. When it was discovered that the same element was also present in nitric acid and nitrate salts such as KNO3 (nitre), it was named nitrogen. About 90% of the nitrogen produced today is used to provide an inert atmosphere for processes or reactions that are oxygen sensitive, such as the production of steel, petroleum refining, and the packaging of foods and pharmaceuticals.
Preparation and General Properties of the Group 15 Elements
Because the atmosphere contains several trillion tons of elemental nitrogen with a purity of about 80%, it is a huge source of nitrogen gas. Distillation of liquefied air yields nitrogen gas that is more than 99.99% pure, but small amounts of very pure nitrogen gas can be obtained from the thermal decomposition of sodium azide:
$\mathrm{2NaN_3(s)\xrightarrow{\Delta}2Na(l)+3N_2(g)} \label{Eq1}$
In contrast, Earth’s crust is relatively poor in nitrogen. The only important nitrogen ores are large deposits of KNO3 and NaNO3 in the deserts of Chile and Russia, which were apparently formed when ancient alkaline lakes evaporated. Consequently, virtually all nitrogen compounds produced on an industrial scale use atmospheric nitrogen as the starting material. Phosphorus, which constitutes only about 0.1% of Earth’s crust, is much more abundant in ores than nitrogen. Like aluminum and silicon, phosphorus is always found in combination with oxygen, and large inputs of energy are required to isolate it.
The other three pnicogens are much less abundant: arsenic is found in Earth’s crust at a concentration of about 2 ppm, antimony is an order of magnitude less abundant, and bismuth is almost as rare as gold. All three elements have a high affinity for the chalcogens and are usually found as the sulfide ores (M2S3), often in combination with sulfides of other heavy elements, such as copper, silver, and lead. Hence a major source of antimony and bismuth is flue dust obtained by smelting the sulfide ores of the more abundant metals.
In group 15, as elsewhere in the p block, we see large differences between the lightest element (N) and its congeners in size, ionization energy, electron affinity, and electronegativity (Table $1$). The chemical behavior of the elements can be summarized rather simply: nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. With their ns2np3 valence electron configurations, all form compounds by losing either the three np valence electrons to form the +3 oxidation state or the three np and the two ns valence electrons to give the +5 oxidation state, whose stability decreases smoothly from phosphorus to bismuth. In addition, the relatively large magnitude of the electron affinity of the lighter pnicogens enables them to form compounds in the −3 oxidation state (such as NH3 and PH3), in which three electrons are formally added to the neutral atom to give a filled np subshell. Nitrogen has the unusual ability to form compounds in nine different oxidation states, including −3, +3, and +5. Because neutral covalent compounds of the trivalent pnicogens contain a lone pair of electrons on the central atom, they tend to behave as Lewis bases.
Table $1$: Selected Properties of the Group 15 Elements
Property Nitrogen Phosphorus Arsenic Antimony Bismuth
*The configuration shown does not include filled d and f subshells. For white phosphorus. For gray arsenic. §The values cited are for six-coordinate ions in the indicated oxidation states. The N5+, P5+, and As5+ ions are not known species. ||The chemical form of the elements in these oxidation states varies considerably. For N, the reaction is NO3 + 3H+ + 2e HNO2 + H2O; for P and As, it is $\ce{H3EO4 + 2H^{+} + 2e^{−} → H3EO3 + H2O}$; and for Sb it is $\ce{Sb2O5 + 4e^{-} + 10H^{+} → 2Sb^{3+} + 5H2O}$.
atomic symbol N P As Sb Bi
atomic number 7 15 33 51 83
atomic mass (amu) 14.01 30.97 74.92 121.76 209.98
valence electron configuration* 2s22p3 3s23p3 4s24p3 5s25p3 6s26p3
melting point/boiling point (°C) −210/−196 44.15/281c 817 (at 3.70 MPa)/603 (sublimes) 631/1587 271/1564
density (g/cm3) at 25°C 1.15 (g/L) 1.82 5.75 6.68 9.79
atomic radius (pm) 56 98 114 133 143
first ionization energy (kJ/mol) 1402 1012 945 831 703
common oxidation state(s) −3 to +5 +5, +3, −3 +5, +3 +5, +3 +3
ionic radius (pm)§ 146 (−3), 16 (+3) 212 (−3), 44 (+3) 58 (+3) 76 (+3), 60 (+5) 103 (+3)
electron affinity (kJ/mol) 0 −72 −78 −101 −91
electronegativity 3.0 2.2 2.2 2.1 1.9
standard reduction potential (E°, V) (EV → EIII in acidic solution)|| +0.93 −0.28 +0.56 +0.65
product of reaction with O2 NO2, NO P4O6, P4O10 As4O6 Sb2O5 Bi2O3
type of oxide acidic (NO2), neutral (NO, N2O) acidic acidic amphoteric basic
product of reaction with N2 none none none none
product of reaction with X2 none PX3, PX5 AsF5, AsX3 SbF5, SbCl5, SbBr3, SbI3 BiF5, BiX3
product of reaction with H2 none none none none none
In group 15, the stability of the +5 oxidation state decreases from P to Bi.
Because neutral covalent compounds of the trivalent group 15 elements have a lone pair of electrons on the central atom, they tend to be Lewis bases.
Reactions and Compounds of Nitrogen
Like carbon, nitrogen has four valence orbitals (one 2s and three 2p), so it can participate in at most four electron-pair bonds by using sp3 hybrid orbitals. Unlike carbon, however, nitrogen does not form long chains because of repulsive interactions between lone pairs of electrons on adjacent atoms. These interactions become important at the shorter internuclear distances encountered with the smaller, second-period elements of groups 15, 16, and 17. Stable compounds with N–N bonds are limited to chains of no more than three N atoms, such as the azide ion (N3).
Nitrogen is the only pnicogen that normally forms multiple bonds with itself and other second-period elements, using π overlap of adjacent np orbitals. Thus the stable form of elemental nitrogen is N2, whose N≡N bond is so strong (DN≡N = 942 kJ/mol) compared with the N–N and N=N bonds (DN–N = 167 kJ/mol; DN=N = 418 kJ/mol) that all compounds containing N–N and N=N bonds are thermodynamically unstable with respect to the formation of N2. In fact, the formation of the N≡N bond is so thermodynamically favored that virtually all compounds containing N–N bonds are potentially explosive.
Again in contrast to carbon, nitrogen undergoes only two important chemical reactions at room temperature: it reacts with metallic lithium to form lithium nitride, and it is reduced to ammonia by certain microorganisms. At higher temperatures, however, N2 reacts with more electropositive elements, such as those in group 13, to give binary nitrides, which range from covalent to ionic in character. Like the corresponding compounds of carbon, binary compounds of nitrogen with oxygen, hydrogen, or other nonmetals are usually covalent molecular substances.
Few binary molecular compounds of nitrogen are formed by direct reaction of the elements. At elevated temperatures, N2 reacts with H2 to form ammonia, with O2 to form a mixture of NO and NO2, and with carbon to form cyanogen (N≡C–C≡N); elemental nitrogen does not react with the halogens or the other chalcogens. Nonetheless, all the binary nitrogen halides (NX3) are known. Except for NF3, all are toxic, thermodynamically unstable, and potentially explosive, and all are prepared by reacting the halogen with NH3 rather than N2. Both nitrogen monoxide (NO) and nitrogen dioxide (NO2) are thermodynamically unstable, with positive free energies of formation. Unlike NO, NO2 reacts readily with excess water, forming a 1:1 mixture of nitrous acid (HNO2) and nitric acid (HNO3):
$\ce{2NO2(g) + H2O(l) \rightarrow HNO2(aq) + HNO3(aq)} \label{Eq2}$
Nitrogen also forms N2O (dinitrogen monoxide, or nitrous oxide), a linear molecule that is isoelectronic with CO2 and can be represented as N=N+=O. Like the other two oxides of nitrogen, nitrous oxide is thermodynamically unstable. The structures of the three common oxides of nitrogen are as follows:
Few binary molecular compounds of nitrogen are formed by the direct reaction of the elements.
At elevated temperatures, nitrogen reacts with highly electropositive metals to form ionic nitrides, such as Li3N and Ca3N2. These compounds consist of ionic lattices formed by Mn+ and N3− ions. Just as boron forms interstitial borides and carbon forms interstitial carbides, with less electropositive metals nitrogen forms a range of interstitial nitrides, in which nitrogen occupies holes in a close-packed metallic structure. Like the interstitial carbides and borides, these substances are typically very hard, high-melting materials that have metallic luster and conductivity.
Nitrogen also reacts with semimetals at very high temperatures to produce covalent nitrides, such as Si3N4 and BN, which are solids with extended covalent network structures similar to those of graphite or diamond. Consequently, they are usually high melting and chemically inert materials.
Ammonia (NH3) is one of the few thermodynamically stable binary compounds of nitrogen with a nonmetal. It is not flammable in air, but it burns in an O2 atmosphere:
$\ce{4NH3(g) + 3O2(g) \rightarrow 2N2(g) + 6H2O(g)} \label{Eq3}$
About 10% of the ammonia produced annually is used to make fibers and plastics that contain amide bonds, such as nylons and polyurethanes, while 5% is used in explosives, such as ammonium nitrate, TNT (trinitrotoluene), and nitroglycerine. Large amounts of anhydrous liquid ammonia are used as fertilizer.
Nitrogen forms two other important binary compounds with hydrogen. Hydrazoic acid (HN3), also called hydrogen azide, is a colorless, highly toxic, and explosive substance. Hydrazine (N2H4) is also potentially explosive; it is used as a rocket propellant and to inhibit corrosion in boilers.
B, C, and N all react with transition metals to form interstitial compounds that are hard, high-melting materials.
Example $1$
For each reaction, explain why the given products form when the reactants are heated.
1. Sr(s) + N2O(g) $\xrightarrow{\Delta}$ SrO(s) + N2(g)
2. NH4NO2(s) $\xrightarrow{\Delta}$ N2(g) + 2H2O(g)
3. Pb(NO3)2(s) $\xrightarrow{\Delta}$ PbO2(s) + 2NO2(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. As an alkali metal, strontium is a strong reductant. If the other reactant can act as an oxidant, then a redox reaction will occur. Nitrous oxide contains nitrogen in a low oxidation state (+1), so we would not normally consider it an oxidant. Nitrous oxide is, however, thermodynamically unstable (ΔH°f > 0 and ΔG°f > 0), and it can be reduced to N2, which is a stable species. Consequently, we predict that a redox reaction will occur.
2. When a substance is heated, a decomposition reaction probably will occur, which often involves the release of stable gases. In this case, ammonium nitrite contains nitrogen in two different oxidation states (−3 and +3), so an internal redox reaction is a possibility. Due to its thermodynamic stability, N2 is the probable nitrogen-containing product, whereas we predict that H and O will combine to form H2O.
3. Again, this is probably a thermal decomposition reaction. If one element is in an usually high oxidation state and another in a low oxidation state, a redox reaction will probably occur. Lead nitrate contains the Pb2+ cation and the nitrate anion, which contains nitrogen in its highest possible oxidation state (+5). Hence nitrogen can be reduced, and we know that lead can be oxidized to the +4 oxidation state. Consequently, it is likely that lead(II) nitrate will decompose to lead(IV) oxide and nitrogen dioxide when heated. Even though PbO2 is a powerful oxidant, the release of a gas such as NO2 can often drive an otherwise unfavorable reaction to completion (Le Chatelier’s principle). Note, however, that PbO2 will probably decompose to PbO at high temperatures.
Exercise 23.3.1
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. NO(g) + H2O(l) $\xrightarrow{\Delta}$
2. NH4NO3(s) $\xrightarrow{\Delta}$
3. Sr(s) + N2(g) →
Answer
1. NO(g) + H2O(l) $\xrightarrow{\Delta}$ no reaction
2. NH4NO3(s) $\xrightarrow{\Delta}$ N2O(g) + 2H2O(g)
3. 3Sr(s) + N2(g) → Sr3N2(s)
Reactions and Compounds of the Heavier Pnicogens
Like the heavier elements of group 14, the heavier pnicogens form catenated compounds that contain only single bonds, whose stability decreases rapidly as we go down the group. For example, phosphorus exists as multiple allotropes, the most common of which is white phosphorus, which consists of P4 tetrahedra and behaves like a typical nonmetal. As is typical of a molecular solid, white phosphorus is volatile, has a low melting point (44.1°C), and is soluble in nonpolar solvents. It is highly strained, with bond angles of only 60°, which partially explains why it is so reactive and so easily converted to more stable allotropes. Heating white phosphorus for several days converts it to red phosphorus, a polymer that is air stable, virtually insoluble, denser than white phosphorus, and higher melting, properties that make it much safer to handle. A third allotrope of phosphorus, black phosphorus, is prepared by heating the other allotropes under high pressure; it is even less reactive, denser, and higher melting than red phosphorus. As expected from their structures, white phosphorus is an electrical insulator, and red and black phosphorus are semiconductors. The three heaviest pnicogens—arsenic, antimony, and bismuth—all have a metallic luster, but they are brittle (not ductile) and relatively poor electrical conductors.
As in group 14, the heavier group 15 elements form catenated compounds that contain only single bonds, whose stability decreases as we go down the group.
The reactivity of the heavier pnicogens decreases as we go down the column. Phosphorus is by far the most reactive of the pnicogens, forming binary compounds with every element in the periodic table except antimony, bismuth, and the noble gases. Phosphorus reacts rapidly with O2, whereas arsenic burns in pure O2 if ignited, and antimony and bismuth react with O2 only when heated. None of the pnicogens reacts with nonoxidizing acids such as aqueous HCl, but all dissolve in oxidizing acids such as HNO3. Only bismuth behaves like a metal, dissolving in HNO3 to give the hydrated Bi3+ cation.
The reactivity of the heavier group 15 elements decreases as we go down the column.
The heavier pnicogens can use energetically accessible 3d, 4d, or 5d orbitals to form dsp3 or d2sp3 hybrid orbitals for bonding. Consequently, these elements often have coordination numbers of 5 or higher. Phosphorus and arsenic form halides (e.g., AsCl5) that are generally covalent molecular species and behave like typical nonmetal halides, reacting with water to form the corresponding oxoacids (in this case, H3AsO4). All the pentahalides are potent Lewis acids that can expand their coordination to accommodate the lone pair of a Lewis base:
$\ce{AsF5(soln) + F^{−}(soln) \rightarrow AsF^{−}6(soln)} \label{Eq4}$
In contrast, bismuth halides have extended lattice structures and dissolve in water to produce hydrated ions, consistent with the stronger metallic character of bismuth.
Except for BiF3, which is essentially an ionic compound, the trihalides are volatile covalent molecules with a lone pair of electrons on the central atom. Like the pentahalides, the trihalides react rapidly with water. In the cases of phosphorus and arsenic, the products are the corresponding acids, $\ce{H3PO3}$ and $\ce{H3AsO3}$, where E is P or As:
$\ce{EX3(l) + 3H2O(l) \rightarrow H3EO3(aq) + 3HX(aq)} \label{Eq5}$
Phosphorus halides are also used to produce insecticides, flame retardants, and plasticizers.
Phosphorus has the greatest ability to form π bonds with elements such as O, N, and C.
With energetically accessible d orbitals, phosphorus and, to a lesser extent, arsenic are able to form π bonds with second-period atoms such as N and O. This effect is even more important for phosphorus than for silicon, resulting in very strong P–O bonds and even stronger P=O bonds. The first four elements in group 15 also react with oxygen to produce the corresponding oxide in the +3 oxidation state. Of these oxides, P4O6 and As4O6 have cage structures formed by inserting an oxygen atom into each edge of the P4 or As4 tetrahedron (part (a) in Figure $2$), and they behave like typical nonmetal oxides. For example, P4O6 reacts with water to form phosphorous acid (H3PO3). Consistent with its position between the nonmetal and metallic oxides, Sb4O6 is amphoteric, dissolving in either acid or base. In contrast, Bi2O3 behaves like a basic metallic oxide, dissolving in acid to give solutions that contain the hydrated Bi3+ ion. The two least metallic elements of the heavier pnicogens, phosphorus and arsenic, form very stable oxides with the formula E4O10 in the +5 oxidation state (part (b) in Figure $2$. In contrast, Bi2O5 is so unstable that there is no absolute proof it exists.
The heavier pnicogens form sulfides that range from molecular species with three-dimensional cage structures, such as P4S3 (part (c) in Figure $2$), to layered or ribbon structures, such as Sb2S3 and Bi2S3, which are semiconductors. Reacting the heavier pnicogens with metals produces substances whose properties vary with the metal content. Metal-rich phosphides (such as M4P) are hard, high-melting, electrically conductive solids with a metallic luster, whereas phosphorus-rich phosphides (such as MP15) are lower melting and less thermally stable because they contain catenated Pn units. Many organic or organometallic compounds of the heavier pnicogens containing one to five alkyl or aryl groups are also known. Because of the decreasing strength of the pnicogen–carbon bond, their thermal stability decreases from phosphorus to bismuth.
The thermal stability of organic or organometallic compounds of group 15 decreases down the group due to the decreasing strength of the pnicogen–carbon bond.
Example $2$
For each reaction, explain why the given products form.
1. $\mathrm{Bi(s) +\frac{3}{2}Br(l)\rightarrow BiBr_3(s)}$
2. 2(CH3)3As(l) + O2(g) → 2(CH3)3As=O(s)
3. PBr3(l) + 3H2O(l) → H3PO3(aq) + 3HBr(aq)
4. As(s) + Ga(s) $\xrightarrow{\Delta}$ GaAs(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Bromine is an oxidant, and bismuth is a metal that can be oxidized. Hence a redox reaction is likely to occur. To identify the product, recall that bismuth can form compounds in either the +3 or +5 oxidation state. The heaviest pnicogen, bismuth is rather difficult to oxidize to the +5 oxidation state because of the inert-pair effect. Hence the product will probably be bismuth(III) bromide.
2. Trimethylarsine, with a lone pair of electrons on the arsenic atom, can act as either a Lewis base or a reductant. If arsenic is oxidized by two electrons, then oxygen must be reduced, most probably by two electrons to the −2 oxidation state. Because As(V) forms strong bonds to oxygen due to π bonding, the expected product is (CH3)3As=O.
3. Phosphorus tribromide is a typical nonmetal halide. We expect it to react with water to produce an oxoacid of P(III) and the corresponding hydrohalic acid.Because of the strength of the P=O bond, phosphorous acid (H3PO3) is actually HP(O)(OH)2, which contains a P=O bond and a P–H bond.
4. Gallium is a metal with a strong tendency to act as a reductant and form compounds in the +3 oxidation state. In contrast, arsenic is a semimetal. It can act as a reductant to form compounds in the +3 or +5 oxidation state, or it can act as an oxidant, accepting electrons to form compounds in the −3 oxidation state. If a reaction occurs, then a binary compound will probably form with a 1:1 ratio of the elements. GaAs is an example of a III-V compound, many of which are used in the electronics industry.
Exercise $2$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. PCl5(s) + H2O(l) →
2. Bi2O5(s) $\xrightarrow{\Delta}$
3. Ca3P2(s) + H+(aq) →
4. NaNH2(s) + PH3(soln) →
Answer
1. PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq)
2. Bi2O5(s) $\xrightarrow{\Delta}$ Bi2O3(s) + O2(g)
3. Ca3P2(s) + 6H+(aq) → 2PH3(g) + 3Ca2+(aq)
4. NaNH2(s) + PH3(soln) → NaPH2(s) + NH3(soln)
Summary
The reactivity of the heavier group 15 elements decreases down the group, as does the stability of their catenated compounds. In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. Nitrogen forms compounds in nine different oxidation states. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect. Due to their higher electronegativity, the lighter pnicogens form compounds in the −3 oxidation state. Because of the presence of a lone pair of electrons on the pnicogen, neutral covalent compounds of the trivalent pnicogens are Lewis bases. Nitrogen does not form stable catenated compounds because of repulsions between lone pairs of electrons on adjacent atoms, but it does form multiple bonds with other second-period atoms. Nitrogen reacts with electropositive elements to produce solids that range from covalent to ionic in character. Reaction with electropositive metals produces ionic nitrides, reaction with less electropositive metals produces interstitial nitrides, and reaction with semimetals produces covalent nitrides. The reactivity of the pnicogens decreases with increasing atomic number. Compounds of the heavier pnicogens often have coordination numbers of 5 or higher and use dsp3 or d2sp3 hybrid orbitals for bonding. Because phosphorus and arsenic have energetically accessible d orbitals, these elements form π bonds with second-period atoms such as O and N. Phosphorus reacts with metals to produce phosphides. Metal-rich phosphides are hard, high-melting, electrically conductive solids with metallic luster, whereas phosphorus-rich phosphides, which contain catenated phosphorus units, are lower melting and less thermally stable. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.08_Nitrogen_Phosphorus_and_the_Group_5A_Elements.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 16 elements: the chalcogens.
The chalcogens are the first group in the p block to have no stable metallic elements. All isotopes of polonium (Po), the only metal in group 16, are radioactive, and only one element in the group, tellurium (Te), can even be described as a semimetal. As in groups 14 and 15, the lightest element of group 16, oxygen, is found in nature as the free element.
Of the group 16 elements, only sulfur was known in ancient times; the others were not discovered until the late 18th and 19th centuries. Sulfur is frequently found as yellow crystalline deposits of essentially pure S8 in areas of intense volcanic activity or around hot springs. As early as the 15th century BC, sulfur was used as a fumigant in Homeric Greece because, when burned, it produces SO2 fumes that are toxic to most organisms, including vermin hiding in the walls and under the floors of houses. Hence references to sulfur are common in ancient literature, frequently in the context of religious purification. In fact, the association of sulfur with the divine was so pervasive that the prefixes thio- (meaning “sulfur”) and theo- (meaning “god”) have the same root in ancient Greek. Though used primarily in the production of sulfuric acid, sulfur is also used to manufacture gunpowder and as a cross-linking agent for rubber, which enables rubber to hold its shape but retain its flexibility.
Group 16 is the first group in the p block with no stable metallic elements.
Oxygen was not discovered until 1771, when the Swedish pharmacist Carl Wilhelm Scheele found that heating compounds such as KNO3, Ag2CO3, and HgO produced a colorless, odorless gas that supported combustion better than air. The results were not published immediately, however, so Scheele’s work remained unknown until 1777. Unfortunately, this was nearly two years after a paper by the English chemist Joseph Priestley had been published, describing the isolation of the same gas by using a magnifying glass to focus the sun’s rays on a sample of HgO. Oxygen is used primarily in the steel industry during the conversion of crude iron to steel using the Bessemer process. Another important industrial use of oxygen is in the production of TiO2, which is commonly used as a white pigment in paints, paper, and plastics.
Tellurium was discovered accidentally in 1782 by the Austrian chemist Franz Joseph Müller von Reichenstein, the chief surveyor of mines in Transylvania who was also responsible for the analysis of ore samples. The silvery-white metal had the same density as antimony but very different properties. Because it was difficult to analyze, Müller called it metallum problematicum (meaning “difficult metal”). The name tellurium (from the Latin tellus, meaning “earth”) was coined by another Austrian chemist, Martin Klaproth, who demonstrated in 1798 that Müller’s “difficult metal” was actually a new element. Tellurium is used to color glass and ceramics, in the manufacture of blasting caps, and in thermoelectric devices.
Jöns Jakob Berzelius (1779–1848)
Berzelius was born into a well-educated Swedish family, but both parents died when he was young. He studied medicine at the University of Uppsala, where his experiments with electroshock therapy caused his interests to turn to electrochemistry. Berzelius devised the system of chemical notation that we use today. In addition, he discovered six elements (cerium, thorium, selenium, silicon, titanium, and zirconium).
The heaviest chalcogen, polonium, was isolated after an extraordinary effort by Marie Curie. Although she was never able to obtain macroscopic quantities of the element, which she named for her native country of Poland, she demonstrated that its chemistry required it to be assigned to group 16. Marie Curie was awarded a second Nobel Prize in Chemistry in 1911 for the discovery of radium and polonium.
Preparation and General Properties of the Group 16 Elements
Oxygen is by far the most abundant element in Earth’s crust and in the hydrosphere (about 44% and 86% by mass, respectively). The same process that is used to obtain nitrogen from the atmosphere produces pure oxygen. Oxygen can also be obtained by the electrolysis of water, the decomposition of alkali metal or alkaline earth peroxides or superoxides, or the thermal decomposition of simple inorganic salts, such as potassium chlorate in the presence of a catalytic amount of MnO2:
$\mathrm{2KClO_3(s)\overset{MnO_2(s)}{\underset{\Delta}\rightleftharpoons}2KCl(s)+3O_2(g)} \label{22.4.1}$
Unlike oxygen, sulfur is not very abundant, but it is found as elemental sulfur in rock formations overlying salt domes, which often accompany petroleum deposits (Figure $1$). Sulfur is also recovered from H2S and organosulfur compounds in crude oil and coal and from metal sulfide ores such as pyrite (FeS2).
Because selenium and tellurium are chemically similar to sulfur, they are usually found as minor contaminants in metal sulfide ores and are typically recovered as by-products. Even so, they are as abundant in Earth’s crust as silver, palladium, and gold. One of the best sources of selenium and tellurium is the “slime” deposited during the electrolytic purification of copper. Both of these elements are notorious for the vile odors of many of their compounds. For example, when the body absorbs even trace amounts of tellurium, dimethyltellurium [(CH3)2Te] is produced and slowly released in the breath and perspiration, resulting in an intense garlic-like smell that is commonly called “tellurium breath.”
With their ns2np4 electron configurations, the chalcogens are two electrons short of a filled valence shell. Thus in reactions with metals, they tend to acquire two additional electrons to form compounds in the −2 oxidation state. This tendency is greatest for oxygen, the chalcogen with the highest electronegativity. The heavier, less electronegative chalcogens can lose either four np electrons or four np and two ns electrons to form compounds in the +4 and +6 oxidation state, respectively, as shown in Table Figure $1$. As with the other groups, the lightest member in the group, in this case oxygen, differs greatly from the others in size, ionization energy, electronegativity, and electron affinity, so its chemistry is unique. Also as in the other groups, the second and third members (sulfur and selenium) have similar properties because of shielding effects. Only polonium is metallic, forming either the hydrated Po2+ or Po4+ ion in aqueous solution, depending on conditions.
Table $1$: Selected Properties of the Group 16 Elements
Property Oxygen Sulfur Selenium Tellurium Polonium
*The configuration shown does not include filled d and f subshells.
The values cited for the hexacations are for six-coordinate ions and are only estimated values.
atomic mass (amu) 16.00 32.07 78.96 127.60 209
atomic number 8 16 34 52 84
atomic radius (pm) 48 88 103 123 135
atomic symbol O S Se Te Po
density (g/cm3) at 25°C 1.31 (g/L) 2.07 4.81 6.24 9.20
electron affinity (kJ/mol) −141 −200 −195 −190 −180
electronegativity 3.4 2.6 2.6 2.1 2.0
first ionization energy (kJ/mol) 1314 1000 941 869 812
ionic radius (pm) 140 (−2) 184 (−2), 29 (+6) 198 (−2), 42 (+6) 221 (−2), 56 (+6) 230 (−2), 97 (+4)
melting point/boiling point (°C) −219/−183 115/445 221/685 450/988 254/962
normal oxidation state(s) −2 +6, +4, −2 +6, +4, −2 +6, +4, −2 +2 (+4)
product of reaction with H2 H2O H2S H2Se none none
product of reaction with N2 NO, NO2 none none none none
product of reaction with O2 SO2 SeO2 TeO2 PoO2
product of reaction with X2 O2F2 SF6, S2Cl2, S2Br2 SeF6, SeX4 TeF6, TeX4 PoF4, PoCl2, PoBr2
standard reduction potential (E°, V) (E0 → H2E in acidic solution) +1.23 +0.14 −0.40 −0.79 −1.00
type of oxide acidic acidic amphoteric basic
valence electron configuration* 2s22p4 3s23p4 4s24p4 5s25p4 6s26p4
Reactions and Compounds of Oxygen
As in groups 14 and 15, the lightest group 16 member has the greatest tendency to form multiple bonds. Thus elemental oxygen is found in nature as a diatomic gas that contains a net double bond: O=O. As with nitrogen, electrostatic repulsion between lone pairs of electrons on adjacent atoms prevents oxygen from forming stable catenated compounds. In fact, except for O2, all compounds that contain O–O bonds are potentially explosive. Ozone, peroxides, and superoxides are all potentially dangerous in pure form. Ozone (O3), one of the most powerful oxidants known, is used to purify drinking water because it does not produce the characteristic taste associated with chlorinated water. Hydrogen peroxide (H2O2) is so thermodynamically unstable that it has a tendency to undergo explosive decomposition when impure:
$2H_2O_{2(l)} \rightarrow 2H_2O_{(l)} + O_{2(g)} \;\;\; ΔG^o = −119\; kJ/mol \label{1}$
As in groups 14 and 15, the lightest element in group 16 has the greatest tendency to form multiple bonds.
Despite the strength of the O=O bond ($D_\mathrm{O_2}$ = 494 kJ/mol), $O_2$ is extremely reactive, reacting directly with nearly all other elements except the noble gases. Some properties of O2 and related species, such as the peroxide and superoxide ions, are in Table $2$. With few exceptions, the chemistry of oxygen is restricted to negative oxidation states because of its high electronegativity (χ = 3.4). Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Oxygen is second only to fluorine in its ability to stabilize high oxidation states of metals in both ionic and covalent compounds. For example, AgO is a stable solid that contains silver in the unusual Ag(II) state, whereas OsO4 is a volatile solid that contains Os(VIII). Because oxygen is so electronegative, the O–H bond is highly polar, creating a large bond dipole moment that makes hydrogen bonding much more important for compounds of oxygen than for similar compounds of the other chalcogens.
Table $2$: Some Properties of O2 and Related Diatomic Species
Species Bond Order Number of Unpaired e O–O Distance (pm)*
*Source of data: Lauri Vaska, “Dioxygen-Metal Complexes: Toward a Unified View,” Accounts of Chemical Research 9 (1976): 175.
O2+ 2.5 1 112
O2 2 2 121
O2 1.5 1 133
O22− 1 0 149
Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements that lie on or near the diagonal band of semimetals are generally amphoteric. A few oxides, such as CO and PbO2, are neutral and do not react with water, aqueous acid, or aqueous base. Nonmetal oxides are typically covalent compounds in which the bonds between oxygen and the nonmetal are polarized (Eδ+–Oδ−). Consequently, a lone pair of electrons on a water molecule can attack the partially positively charged E atom to eventually form an oxoacid. An example is reacting sulfur trioxide with water to form sulfuric acid:
$H_2O_{(l)} + SO_{3(g)} \rightarrow H_2SO_{4(aq)} \label{2}$
The oxides of the semimetals and of elements such as Al that lie near the metal/nonmetal dividing line are amphoteric, as we expect:
$Al_2O_{3(s)} + 6H^+_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 3H_2O_{(l)} \label{3}$
$Al_2O_{3(s)} + 2OH^−_{(aq)} + 3H_2O_{(l)} \rightarrow 2Al(OH)^−_{4(aq)} \label{4}$
Oxides of metals tend to be basic, oxides of nonmetals tend to be acidic, and oxides of elements in or near the diagonal band of semimetals are generally amphoteric.
Example $1$
For each reaction, explain why the given products form.
1. Ga2O3(s) + 2OH(aq) + 3H2O(l) → 2Ga(OH)4(aq)
2. 3H2O2(aq) + 2MnO4(aq) + 2H+(aq) → 3O2(g) + 2MnO2(s) + 4H2O(l)
3. KNO3(s) $\xrightarrow{\Delta}$ KNO(s) + O2(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Gallium is a metal. We expect the oxides of metallic elements to be basic and therefore not to react with aqueous base. A close look at the periodic table, however, shows that gallium is close to the diagonal line of semimetals. Moreover, aluminum, the element immediately above gallium in group 13, is amphoteric. Consequently, we predict that gallium will behave like aluminum (Equation $\ref{4}$).
2. Hydrogen peroxide is an oxidant that can accept two electrons per molecule to give two molecules of water. With a strong oxidant, however, H2O2 can also act as a reductant, losing two electrons (and two protons) to produce O2. Because the other reactant is permanganate, which is a potent oxidant, the only possible reaction is a redox reaction in which permanganate is the oxidant and hydrogen peroxide is the reductant. Recall that reducing permanganate often gives MnO2, an insoluble brown solid. Reducing MnO4 to MnO2 is a three-electron reduction, whereas the oxidation of H2O2 to O2 is a two-electron oxidation.
3. This is a thermal decomposition reaction. Because KNO3 contains nitrogen in its highest oxidation state (+5) and oxygen in its lowest oxidation state (−2), a redox reaction is likely. Oxidation of the oxygen in nitrate to atomic oxygen is a two-electron process per oxygen atom. Nitrogen is likely to accept two electrons because oxoanions of nitrogen are known only in the +5 (NO3) and +3 (NO2) oxidation states.
Exercise $2$
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. SiO2(s) + H+(aq) →
2. NO(g) + O2(g) →
3. SO3(g) + H2O(l) →
4. H2O2(aq) + I(aq) →
Answer
1. SiO2(s) + H+(aq) → no reaction
2. 2NO(g) + O2(g) → 2NO2(g)
3. SO3(g) + H2O(l) → H2SO4(aq)
4. H2O2(aq) + 2I(aq) → I2(aq) + 2OH(aq)
Reactions and Compounds of the Heavier Chalcogens
Because most of the heavier chalcogens (group 16) and pnicogens (group 15) are nonmetals, they often form similar compounds. For example, both third-period elements of these groups (phosphorus and sulfur) form catenated compounds and form multiple allotropes. Consistent with periodic trends, the tendency to catenate decreases as we go down the column.
Sulfur and selenium both form a fairly extensive series of catenated species. For example, elemental sulfur forms S8 rings packed together in a complex “crankshaft” arrangement (Figure $2$), and molten sulfur contains long chains of sulfur atoms connected by S–S bonds. Moreover, both sulfur and selenium form polysulfides (Sn2−) and polyselenides (Sen2−), with n ≤ 6. The only stable allotrope of tellurium is a silvery white substance whose properties and structure are similar to those of one of the selenium allotropes. Polonium, in contrast, shows no tendency to form catenated compounds. The striking decrease in structural complexity from sulfur to polonium is consistent with the decrease in the strength of single bonds and the increase in metallic character as we go down the group.
As in group 15, the reactivity of elements in group 16 decreases from lightest to heaviest. For example, selenium and tellurium react with most elements but not as readily as sulfur does. As expected for nonmetals, sulfur, selenium, and tellurium do not react with water, aqueous acid, or aqueous base, but all dissolve in strongly oxidizing acids such as HNO3 to form oxoacids such as H2SO4. In contrast to the other chalcogens, polonium behaves like a metal, dissolving in dilute HCl to form solutions that contain the Po2+ ion.
Just as with the other groups, the tendency to catenate, the strength of single bonds, and reactivity decrease down the group.
Fluorine reacts directly with all chalcogens except oxygen to produce the hexafluorides (YF6), which are extraordinarily stable and unreactive compounds. Four additional stable fluorides of sulfur are known; thus sulfur oxidation states range from +1 to +6 (Figure $2$). In contrast, only four fluorides of selenium (SeF6, SeF4, FSeSeF, and SeSeF2) and only three of tellurium (TeF4, TeF6, and Te2F10) are known.
Direct reaction of the heavier chalcogens with oxygen at elevated temperatures gives the dioxides (YO2), which exhibit a dramatic range of structures and properties. The dioxides become increasingly metallic in character down the group, as expected, and the coordination number of the chalcogen steadily increases. Thus SO2 is a gas that contains V-shaped molecules (as predicted by the valence-shell electron-pair repulsion model), SeO2 is a white solid with an infinite chain structure (each Se is three coordinate), TeO2 is a light yellow solid with a network structure in which each Te atom is four coordinate, and PoO2 is a yellow ionic solid in which each Po4+ ion is eight coordinate.
The dioxides of sulfur, selenium, and tellurium react with water to produce the weak, diprotic oxoacids (H2YO3—sulfurous, selenous, and tellurous acid, respectively). Both sulfuric acid and selenic acid (H2SeO4) are strong acids, but telluric acid [Te(OH)6] is quite different. Because tellurium is larger than either sulfur or selenium, it forms weaker π bonds to oxygen. As a result, the most stable structure for telluric acid is Te(OH)6, with six Te–OH bonds rather than Te=O bonds. Telluric acid therefore behaves like a weak triprotic acid in aqueous solution, successively losing the hydrogen atoms bound to three of the oxygen atoms. As expected for compounds that contain elements in their highest accessible oxidation state (+6 in this case), sulfuric, selenic, and telluric acids are oxidants. Because the stability of the highest oxidation state decreases with increasing atomic number, telluric acid is a stronger oxidant than sulfuric acid.
The stability of the highest oxidation state of the chalcogens decreases down the column.
Sulfur and, to a lesser extent, selenium react with carbon to form an extensive series of compounds that are structurally similar to their oxygen analogues. For example, CS2 and CSe2 are both volatile liquids that contain C=S or C=Se bonds and have the same linear structure as CO2. Because these double bonds are significantly weaker than the C=O bond, however, CS2, CSe2, and related compounds are less stable and more reactive than their oxygen analogues. The chalcogens also react directly with nearly all metals to form compounds with a wide range of stoichiometries and a variety of structures. Metal chalcogenides can contain either the simple chalcogenide ion (Y2−), as in Na2S and FeS, or polychalcogenide ions (Yn2−), as in FeS2 and Na2S5.
The dioxides of the group 16 elements become increasingly basic, and the coordination number of the chalcogen steadily increases down the group.
Ionic chalcogenides like Na2S react with aqueous acid to produce binary hydrides such as hydrogen sulfide (H2S). Because the strength of the Y–H bond decreases with increasing atomic radius, the stability of the binary hydrides decreases rapidly down the group. It is perhaps surprising that hydrogen sulfide, with its familiar rotten-egg smell, is much more toxic than hydrogen cyanide (HCN), the gas used to execute prisoners in the “gas chamber.” Hydrogen sulfide at relatively low concentrations deadens the olfactory receptors in the nose, which allows it to reach toxic levels without detection and makes it especially dangerous.
Example $2$
For each reaction, explain why the given product forms or no reaction occurs.
1. SO2(g) + Cl2(g) → SO2Cl2(l)
2. SF6(g) + H2O(l) → no reaction
3. 2Se(s) + Cl2(g) → Se2Cl2(l)
Given: balanced chemical equations
Asked for: why the given products (or no products) form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form or why no reaction occurs.
Solution
1. One of the reactants (Cl2) is an oxidant. If the other reactant can be oxidized, then a redox reaction is likely. Sulfur dioxide contains sulfur in the +4 oxidation state, which is 2 less than its maximum oxidation state. Sulfur dioxide is also known to be a mild reducing agent in aqueous solution, producing sulfuric acid as the oxidation product. Hence a redox reaction is probable. The simplest reaction is the formation of SO2Cl2 (sulfuryl chloride), which is a tetrahedral species with two S–Cl and two S=O bonds.
1. Sulfur hexafluoride is a nonmetallic halide. Such compounds normally react vigorously with water to produce an oxoacid of the nonmetal and the corresponding hydrohalic acid. In this case, however, we have a highly stable species, presumably because all of sulfur’s available orbitals are bonding orbitals. Thus SF6 is not likely to react with water.
2. Here we have the reaction of a chalcogen with a halogen. The halogen is a good oxidant, so we can anticipate that a redox reaction will occur. Only fluorine is capable of oxidizing the chalcogens to a +6 oxidation state, so we must decide between SeCl4 and Se2Cl2 as the product. The stoichiometry of the reaction determines which of the two is obtained: SeCl4 or Se2Cl2.
Exercise $2$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. Te(s) + Na(s) $\xrightarrow{\Delta}$
2. SF4(g) + H2O(l) →
3. CH3SeSeCH3(soln) + K(s) →
4. Li2Se(s) + H+(aq) →
Answer
1. Te(s) + 2Na(s) → Na2Te(s)
2. SF4(g) + 3H2O(l) → H2SO3(aq) + 4HF(aq)
3. CH3SeSeCH3(soln) + 2K(s) → 2KCH3Se(soln)
4. Li2Se(s) + 2H+(aq) → H2Se(g) + 2Li+(aq)
Summary
The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements. It does not form stable catenated compounds, however, due to repulsions between lone pairs of electrons on adjacent atoms. Because of its high electronegativity, the chemistry of oxygen is generally restricted to compounds in which it has a negative oxidation state, and its bonds to other elements tend to be highly polar. Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements along the dividing line between metals and nonmetals are amphoteric. The reactivity, the strength of multiple bonds to oxygen, and the tendency to form catenated compounds all decrease down the group, whereas the maximum coordination numbers increase. Because Te=O bonds are comparatively weak, the most stable oxoacid of tellurium contains six Te–OH bonds. The stability of the highest oxidation state (+6) decreases down the group. Double bonds between S or Se and second-row atoms are weaker than the analogous C=O bonds because of reduced orbital overlap. The stability of the binary hydrides decreases down the group. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.09%3A__Oxygen_Sulfur_and_the_Group_6A_Elements.txt |
Learning Objectives
• To understand the periodic trends and reactivity of the group 17 elements: the halogens.
Because the halogens are highly reactive, none is found in nature as the free element. Hydrochloric acid, which is a component of aqua regia (a mixture of HCl and HNO3 that dissolves gold), and the mineral fluorspar (CaF2) were well known to alchemists, who used them in their quest for gold. Despite their presence in familiar substances, none of the halogens was even recognized as an element until the 19th century.
Because the halogens are highly reactive, none is found in nature as the free element.
Chlorine was the first halogen to be obtained in pure form. In 1774, Carl Wilhelm Scheele (the codiscoverer of oxygen) produced chlorine by reacting hydrochloric acid with manganese dioxide. Scheele was convinced, however, that the pale green gas he collected over water was a compound of oxygen and hydrochloric acid. In 1811, Scheele’s “compound” was identified as a new element, named from the Greek chloros, meaning “yellowish green” (the same stem as in chlorophyll, the green pigment in plants). That same year, a French industrial chemist, Bernard Courtois, accidentally added too much sulfuric acid to the residue obtained from burned seaweed. A deep purple vapor was released, which had a biting aroma similar to that of Scheele’s “compound.” The purple substance was identified as a new element, named iodine from the Greek iodes, meaning “violet.” Bromine was discovered soon after by a young French chemist, Antoine Jérôme Balard, who isolated a deep red liquid with a strong chlorine-like odor from brine from the salt marshes near Montpellier in southern France. Because many of its properties were intermediate between those of chlorine and iodine, Balard initially thought he had isolated a compound of the two (perhaps ICl). He soon realized, however, that he had discovered a new element, which he named bromine from the Greek bromos, meaning “stench.” Currently, organic chlorine compounds, such as PVC (polyvinylchloride), consume about 70% of the Cl2 produced annually; organobromine compounds are used in much smaller quantities, primarily as fire retardants.
Because of the unique properties of its compounds, fluorine was believed to exist long before it was actually isolated. The mineral fluorspar (now called fluorite [CaF2]) had been used since the 16th century as a “flux,” a low-melting-point substance that could dissolve other minerals and ores. In 1670, a German glass cutter discovered that heating fluorspar with strong acid produced a solution that could etch glass. The solution was later recognized to contain the acid of a new element, which was named fluorine in 1812. Elemental fluorine proved to be very difficult to isolate, however, because both HF and F2 are extraordinarily reactive and toxic. After being poisoned three times while trying to isolate the element, the French chemist Henri Moissan succeeded in 1886 in electrolyzing a sample of KF in anhydrous HF to produce a pale green gas (Figure $1$). For this achievement, among others, Moissan narrowly defeated Mendeleev for the Nobel Prize in Chemistry in 1906. Large amounts of fluorine are now consumed in the production of cryolite (Na3AlF6), a key intermediate in the production of aluminum metal. Fluorine is also found in teeth as fluoroapatite [Ca5(PO4)3F], which is formed by reacting hydroxyapatite [Ca5(PO4)3OH] in tooth enamel with fluoride ions in toothpastes, rinses, and drinking water.
The heaviest halogen is astatine (At), which is continuously produced by natural radioactive decay. All its isotopes are highly radioactive, and the most stable has a half-life of only about 8 h. Consequently, astatine is the least abundant naturally occurring element on Earth, with less than 30 g estimated to be present in Earth’s crust at any one time.
Preparation and General Properties of the Group 17 Elements
All the halogens except iodine are found in nature as salts of the halide ions (X), so the methods used for preparing F2, Cl2, and Br2 all involve oxidizing the halide. Reacting CaF2 with concentrated sulfuric acid produces gaseous hydrogen fluoride:
$CaF_{2(s)} + H_2SO_{4(l)} \rightarrow CaSO_{4(s)} + 2HF_{(g)} \label{1}$
Fluorine is produced by the electrolysis of a 1:1 mixture of HF and K+HF2 at 60–300°C in an apparatus made of Monel, a highly corrosion-resistant nickel–copper alloy:
$KHF_2\cdot HF(l) \xrightarrow{electrolysis}F_2(g) + H_2(g) \label{2}$
Fluorine is one of the most powerful oxidants known, and both F2 and HF are highly corrosive. Consequently, the production, storage, shipping, and handling of these gases pose major technical challenges.
Although chlorine is significantly less abundant than fluorine, elemental chlorine is produced on an enormous scale. Fortunately, large subterranean deposits of rock salt (NaCl) are found around the world (Figure $2$), and seawater consists of about 2% NaCl by mass, providing an almost inexhaustible reserve. Inland salt lakes such as the Dead Sea and the Great Salt Lake are even richer sources, containing about 23% and 8% NaCl by mass, respectively. Chlorine is prepared industrially by the chloralkali process, which uses the following reaction:
$2NaCl_{(aq)} +2H_2O_{(l)} \xrightarrow{electrolysis} 2NaOH(aq) + Cl_{2(g)} + H_{2(g)} \label{3}$
Bromine is much less abundant than fluorine or chlorine, but it is easily recovered from seawater, which contains about 65 mg of Br per liter. Salt lakes and underground brines are even richer sources; for example, the Dead Sea contains 4 g of Br per liter. Iodine is the least abundant of the nonradioactive halogens, and it is a relatively rare element. Because of its low electronegativity, iodine tends to occur in nature in an oxidized form. Hence most commercially important deposits of iodine, such as those in the Chilean desert, are iodate salts such as Ca(IO3)2. The production of iodine from such deposits therefore requires reduction rather than oxidation. The process is typically carried out in two steps: reduction of iodate to iodide with sodium hydrogen sulfite, followed by reaction of iodide with additional iodate:
$2IO^−_{3(aq)} + 6HSO^−_{3(aq)} \rightarrow 2I^−_{(aq)} + 6SO^2−_{4(aq)} + 6H^+_{(aq)} \label{4}$
$5I^−_{(aq)} + IO^−_{3(aq)} + 6H^+_{(aq)} \rightarrow 3I_{2(s)} + 3H_2O_{(l)} \label{5}$
Because the halogens all have ns2np5 electron configurations, their chemistry is dominated by a tendency to accept an additional electron to form the closed-shell ion (X). Only the electron affinity and the bond dissociation energy of fluorine differ significantly from the expected periodic trends shown in Table $1$. Electron–electron repulsion is important in fluorine because of its small atomic volume, making the electron affinity of fluorine less than that of chlorine. Similarly, repulsions between electron pairs on adjacent atoms are responsible for the unexpectedly low F–F bond dissociation energy. (As discussed earlier, this effect is also responsible for the weakness of O–O, N–N, and N–O bonds.)
Electrostatic repulsions between lone pairs of electrons on adjacent atoms cause single bonds between N, O, and F to be weaker than expected.
Table $1$: Selected Properties of the Group 17 Elements
Property Fluorine Chlorine Bromine Iodine Astatine
*The configuration shown does not include filled d and f subshells. The values cited are for the six-coordinate anion (X−).
atomic symbol F Cl Br I At
atomic number 9 17 35 53 85
atomic mass (amu) 19.00 35.45 79.90 126.90 210
valence electron configuration* 2s22p5 3s23p5 4s24p5 5s25p5 6s26p5
melting point/boiling point (°C) −220/−188 −102/−34.0 −7.2/58.8 114/184 302/—
density (g/cm3) at 25°C 1.55 (g/L) 2.90 (g/L) 3.10 4.93
atomic radius (pm) 42 79 94 115 127
first ionization energy (kJ/mol) 1681 1251 1140 1008 926
normal oxidation state(s) −1 −1 (+1, +3, +5, +7) −1 (+1, +3, +5, +7) −1 (+1, +3, +5, +7) −1, +1
ionic radius (pm) 133 181 196 220
electron affinity (kJ/mol) −328 −349 −325 −295 −270
electronegativity 4.0 3.2 3.0 2.7 2.2
standard reduction potential (E°, V) (X2 → X in basic solution) +2.87 +1.36 +1.07 +0.54 +0.30
dissociation energy of X2(g) (kJ/mol) 158.8 243.6 192.8 151.1 ~80
product of reaction with O2 O2F2 none none none none
type of oxide acidic acidic acidic acidic acidic
product of reaction with N2 none none none none none
product of reaction with H2 HF HCl HBr HI HAt
Because it is the most electronegative element in the periodic table, fluorine forms compounds in only the −1 oxidation state. Notice, however, that all the halogens except astatine have electronegativities greater than 2.5, making their chemistry exclusively that of nonmetals. The halogens all have relatively high ionization energies, but the energy required to remove electrons decreases substantially as we go down the column. Hence the heavier halogens also form compounds in positive oxidation states (+1, +3, +5, and +7), derived by the formal loss of ns and np electrons.
Because ionization energies decrease down the group, the heavier halogens form compounds in positive oxidation states (+1, +3, +5, and +7).
Reactions and Compounds of the Halogens
Fluorine is the most reactive element in the periodic table, forming compounds with every other element except helium, neon, and argon. The reactions of fluorine with most other elements range from vigorous to explosive; only O2, N2, and Kr react slowly. There are three reasons for the high reactivity of fluorine:
1. Because fluorine is so electronegative, it is able to remove or at least share the valence electrons of virtually any other element.
2. Because of its small size, fluorine tends to form very strong bonds with other elements, making its compounds thermodynamically stable.
3. The F–F bond is weak due to repulsion between lone pairs of electrons on adjacent atoms, reducing both the thermodynamic and kinetic barriers to reaction.
With highly electropositive elements, fluorine forms ionic compounds that contain the closed-shell F ion. In contrast, with less electropositive elements (or with metals in very high oxidation states), fluorine forms covalent compounds that contain terminal F atoms, such as SF6. Because of its high electronegativity and 2s22p5 valence electron configuration, fluorine normally participates in only one electron-pair bond. Only a very strong Lewis acid, such as AlF3, can share a lone pair of electrons with a fluoride ion, forming AlF63−.
Oxidative strength decreases down group 17.
The halogens (X2) react with metals (M) according to the general equation
$M_{(s,l)} + nX_{2(s,l,g)} \rightarrow MX_{n(s,l)} \label{6}$
For elements that exhibit multiple oxidation states fluorine tends to produce the highest possible oxidation state and iodine the lowest. For example, vanadium reacts with the halogens to give VF5, VCl4, VBr4, and VI3.
Metal halides in the +1 or +2 oxidation state, such as CaF2, are typically ionic halides, which have high melting points and are often soluble in water. As the oxidation state of the metal increases, so does the covalent character of the halide due to polarization of the M–X bond. With its high electronegativity, fluoride is the least polarizable, and iodide, with the lowest electronegativity, is the most polarizable of the halogens. Halides of small trivalent metal ions such as Al3+ tend to be relatively covalent. For example, AlBr3 is a volatile solid that contains bromide-bridged Al2Br6 molecules. In contrast, the halides of larger trivalent metals, such as the lanthanides, are essentially ionic. For example, indium tribromide (InBr3) and lanthanide tribromide (LnBr3) are all high-melting-point solids that are quite soluble in water.
As the oxidation state of the metal increases, the covalent character of the corresponding metal halides also increases due to polarization of the M–X bond.
All halogens react vigorously with hydrogen to give the hydrogen halides (HX). Because the H–F bond in HF is highly polarized (Hδ+–Fδ−), liquid HF has extensive hydrogen bonds, giving it an unusually high boiling point and a high dielectric constant. As a result, liquid HF is a polar solvent that is similar in some ways to water and liquid ammonia; after a reaction, the products can be recovered simply by evaporating the HF solvent. (Hydrogen fluoride must be handled with extreme caution, however, because contact of HF with skin causes extraordinarily painful burns that are slow to heal.) Because fluoride has a high affinity for silicon, aqueous hydrofluoric acid is used to etch glass, dissolving SiO2 to give solutions of the stable SiF62− ion.
Glass etched with hydrogen flouride.© Thinkstock
Except for fluorine, all the halogens react with water in a disproportionation reaction, where X is Cl, Br, or I:
$X_{2(g,l,s)} + H_2O_{(l)} \rightarrow H^+_{(aq)} + X^−_{(aq)} + HOX_{(aq)} \label{7}$
The most stable oxoacids are the perhalic acids, which contain the halogens in their highest oxidation state (+7). The acid strengths of the oxoacids of the halogens increase with increasing oxidation state, whereas their stability and acid strength decrease down the group. Thus perchloric acid (HOClO3, usually written as HClO4) is a more potent acid and stronger oxidant than perbromic acid. Although all the oxoacids are strong oxidants, some, such as HClO4, react rather slowly at low temperatures. Consequently, mixtures of the halogen oxoacids or oxoanions with organic compounds are potentially explosive if they are heated or even agitated mechanically to initiate the reaction. Because of the danger of explosions, oxoacids and oxoanions of the halogens should never be allowed to come into contact with organic compounds.
Both the acid strength and the oxidizing power of the halogen oxoacids decrease down the group.
The halogens react with one another to produce interhalogen compounds, such as ICl3, BrF5, and IF7. In all cases, the heavier halogen, which has the lower electronegativity, is the central atom. The maximum oxidation state and the number of terminal halogens increase smoothly as the ionization energy of the central halogen decreases and the electronegativity of the terminal halogen increases. Thus depending on conditions, iodine reacts with the other halogens to form IFn (n = 1–7), ICl or ICl3, or IBr, whereas bromine reacts with fluorine to form only BrF, BrF3, and BrF5 but not BrF7. The interhalogen compounds are among the most powerful Lewis acids known, with a strong tendency to react with halide ions to give complexes with higher coordination numbers, such as the IF8 ion:
$IF_{7(l)} + KF_{(s)} \rightarrow KIF_{8(s)} \label{8}$
All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7). The interhalogen compounds are also potent oxidants and strong fluorinating agents; contact with organic materials or water can result in an explosion.
All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7), but the importance of the higher oxidation states generally decreases down the group.
Example $1$
For each reaction, explain why the given products form.
1. ClF3(g) + Cl2(g) → 3ClF(g)
2. 2KI(s) + 3H2SO4(aq) → I2(aq) + SO2(g) + 2KHSO4(aq) + 2H2O(l)
3. Pb(s) + 2BrF3(l) → PbF4(s) + 2BrF(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. When the reactants have the same element in two different oxidation states, we expect the product to have that element in an intermediate oxidation state. We have Cl3+ and Cl0 as reactants, so a possible product would have Cl in either the +1 or +2 oxidation state. From our discussion, we know that +1 is much more likely. In this case, Cl2 is behaving like a reductant rather than an oxidant.
2. At first glance, this appears to be a simple acid–base reaction, in which sulfuric acid transfers a proton to I to form HI. Recall, however, that I can be oxidized to I2. Sulfuric acid contains sulfur in its highest oxidation state (+6), so it is a good oxidant. In this case, the redox reaction predominates.
3. This is the reaction of a metallic element with a very strong oxidant. Consequently, a redox reaction will occur. The only question is whether lead will be oxidized to Pb(II) or Pb(IV). Because BrF3 is a powerful oxidant and fluorine is able to stabilize high oxidation states of other elements, it is likely that PbF4 will be the product. The two possible reduction products for BrF3 are BrF and Br2. The actual product will likely depend on the ratio of the reactants used. With excess BrF3, we expect the more oxidized product (BrF). With lower ratios of oxidant to lead, we would probably obtain Br2 as the product.
Exercise $1$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. CaCl2(s) + H3PO4(l) →
2. GeO2(s) + HF(aq) →
3. Fe2O3(s) + HCl(g) $\xrightarrow{\Delta}$
4. NaClO2(aq) + Cl2(g) →
Answer
1. CaCl2(s) + H3PO4(l) → 2HCl(g) + Ca(HPO4)(soln)
2. GeO2(s) + 6HF(aq) → GeF62−(aq) + 2H2O(l) + 2H+(aq)
3. Fe2O3(s) + 6HCl(g) $\xrightarrow{\Delta}$ 2FeCl3(s) + 3H2O(g)
4. 2NaClO2(aq) + Cl2(g) → 2ClO2(g) + 2NaCl(aq)
Summary
The halogens are highly reactive. All halogens have relatively high ionization energies, and the acid strength and oxidizing power of their oxoacids decreases down the group. The halogens are so reactive that none is found in nature as the free element; instead, all but iodine are found as halide salts with the X ion. Their chemistry is exclusively that of nonmetals. Consistent with periodic trends, ionization energies decrease down the group. Fluorine, the most reactive element in the periodic table, has a low F–F bond dissociation energy due to repulsions between lone pairs of electrons on adjacent atoms. Fluorine forms ionic compounds with electropositive elements and covalent compounds with less electropositive elements and metals in high oxidation states. All the halogens react with hydrogen to produce hydrogen halides. Except for F2, all react with water to form oxoacids, including the perhalic acids, which contain the halogens in their highest oxidation state. Halogens also form interhalogen compounds; the heavier halogen, with the lower electronegativity, is the central atom. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.10%3A__The_Halogens_Group_7A.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 18 elements: the noble gases.
The noble gases were all isolated for the first time within a period of only five years at the end of the 19th century. Their very existence was not suspected until the 18th century, when early work on the composition of air suggested that it contained small amounts of gases in addition to oxygen, nitrogen, carbon dioxide, and water vapor. Helium was the first of the noble gases to be identified, when the existence of this previously unknown element on the sun was demonstrated by new spectral lines seen during a solar eclipse in 1868. Actual samples of helium were not obtained until almost 30 years later, however. In the 1890s, the English physicist J. W. Strutt (Lord Rayleigh) carefully measured the density of the gas that remained after he had removed all O2, CO2, and water vapor from air and showed that this residual gas was slightly denser than pure N2 obtained by the thermal decomposition of ammonium nitrite. In 1894, he and the Scottish chemist William Ramsay announced the isolation of a new “substance” (not necessarily a new element) from the residual nitrogen gas. Because they could not force this substance to decompose or react with anything, they named it argon (Ar), from the Greek argos, meaning “lazy.” Because the measured molar mass of argon was 39.9 g/mol, Ramsay speculated that it was a member of a new group of elements located on the right side of the periodic table between the halogens and the alkali metals. He also suggested that these elements should have a preferred valence of 0, intermediate between the +1 of the alkali metals and the −1 of the halogens.
J. W. Strutt (Lord Rayleigh) (1842–1919)
Lord Rayleigh was one of the few members of British higher nobility to be recognized as an outstanding scientist. Throughout his youth, his education was repeatedly interrupted by his frail health, and he was not expected to reach maturity. In 1861 he entered Trinity College, Cambridge, where he excelled at mathematics. A severe attack of rheumatic fever took him abroad, but in 1873 he succeeded to the barony and was compelled to devote his time to the management of his estates. After leaving the entire management to his younger brother, Lord Rayleigh was able to devote his time to science. He was a recipient of honorary science and law degrees from Cambridge University.
Sir William Ramsay (1852–1916)
Born and educated in Glasgow, Scotland, Ramsay was expected to study for the Calvanist ministry. Instead, he became interested in chemistry while reading about the manufacture of gunpowder. Ramsay earned his PhD in organic chemistry at the University of Tübingen in Germany in 1872. When he returned to England, his interests turned first to physical chemistry and then to inorganic chemistry. He is best known for his work on the oxides of nitrogen and for the discovery of the noble gases with Lord Rayleigh.
In 1895, Ramsey was able to obtain a terrestrial sample of helium for the first time. Then, in a single year (1898), he discovered the next three noble gases: krypton (Kr), from the Greek kryptos, meaning “hidden,” was identified by its orange and green emission lines; neon (Ne), from the Greek neos, meaning “new,” had bright red emission lines; and xenon (Xe), from the Greek xenos, meaning “strange,” had deep blue emission lines. The last noble gas was discovered in 1900 by the German chemist Friedrich Dorn, who was investigating radioactivity in the air around the newly discovered radioactive elements radium and polonium. The element was named radon (Rn), and Ramsay succeeded in obtaining enough radon in 1908 to measure its density (and thus its atomic mass). For their discovery of the noble gases, Rayleigh was awarded the Nobel Prize in Physics and Ramsay the Nobel Prize in Chemistry in 1904. Because helium has the lowest boiling point of any substance known (4.2 K), it is used primarily as a cryogenic liquid. Helium and argon are both much less soluble in water (and therefore in blood) than N2, so scuba divers often use gas mixtures that contain these gases, rather than N2, to minimize the likelihood of the “bends,” the painful and potentially fatal formation of bubbles of N2(g) that can occur when a diver returns to the surface too rapidly.
Preparation and General Properties of the Group 18 Elements
Fractional distillation of liquid air is the only source of all the noble gases except helium. Although helium is the second most abundant element in the universe (after hydrogen), the helium originally present in Earth’s atmosphere was lost into space long ago because of its low molecular mass and resulting high mean velocity. Natural gas often contains relatively high concentrations of helium (up to 7%), however, and it is the only practical terrestrial source.
The elements of group 18 all have closed-shell valence electron configurations, either ns2np6 or 1s2 for He. Consistent with periodic trends in atomic properties, these elements have high ionization energies that decrease smoothly down the group. From their electron affinities, the data in Table $1$ indicate that the noble gases are unlikely to form compounds in negative oxidation states. A potent oxidant is needed to oxidize noble gases and form compounds in positive oxidation states. Like the heavier halogens, xenon and perhaps krypton should form covalent compounds with F, O, and possibly Cl, in which they have even formal oxidation states (+2, +4, +6, and possibly +8). These predictions actually summarize the chemistry observed for these elements.
Table $1$: Selected Properties of the Group 18 Elements
Property Helium Neon Argon Krypton Xenon Radon
*The configuration shown does not include filled d and f subshells. This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given.
atomic symbol He Ne Ar Kr Xe Rn
atomic number 2 10 18 36 54 86
atomic mass (amu) 4.00 20.18 39.95 83.80 131.29 222
valence electron configuration* 1s2 2s22p6 3s23p6 4s24p6 5s25p6 6s26p6
triple point/boiling point (°C) —/−269 −249 (at 43 kPa)/−246 −189 (at 69 kPa)/−189 −157/−153 −112 (at 81.6 kPa)/−108 −71/−62
density (g/L) at 25°C 0.16 0.83 1.63 3.43 5.37 9.07
atomic radius (pm) 31 38 71 88 108 120
first ionization energy (kJ/mol) 2372 2081 1521 1351 1170 1037
normal oxidation state(s) 0 0 0 0 (+2) 0 (+2, +4, +6, +8) 0 (+2)
electron affinity (kJ/mol) > 0 > 0 > 0 > 0 > 0 > 0
electronegativity 2.6
product of reaction with O2 none none none none not directly with oxygen, but $\ce{XeO3}$ can be formed by Equation \ref{Eq5}. none
type of oxide acidic
product of reaction with N2 none none none none none none
product of reaction with X2 none none none KrF2 XeF2, XeF4, XeF6 RnF2
product of reaction with H2 none none none none none none
Reactions and Compounds of the Noble Gases
For many years, it was thought that the only compounds the noble gases could form were clathrates. Clathrates are solid compounds in which a gas, the guest, occupies holes in a lattice formed by a less volatile, chemically dissimilar substance, the host (Figure $1$).
Because clathrate formation does not involve the formation of chemical bonds between the guest (Xe) and the host molecules (H2O, in the case of xenon hydrate), the guest molecules are immediately released when the clathrate is melted or dissolved.
Methane Clathrates
In addition to the noble gases, many other species form stable clathrates. One of the most interesting is methane hydrate, large deposits of which occur naturally at the bottom of the oceans. It is estimated that the amount of methane in such deposits could have a major impact on the world’s energy needs later in this century.
The widely held belief in the intrinsic lack of reactivity of the noble gases was challenged when Neil Bartlett, a British professor of chemistry at the University of British Columbia, showed that PtF6, a compound used in the Manhattan Project, could oxidize O2. Because the ionization energy of xenon (1170 kJ/mol) is actually lower than that of O2, Bartlett recognized that PtF6 should also be able to oxidize xenon. When he mixed colorless xenon gas with deep red PtF6 vapor, yellow-orange crystals immediately formed (Figure $3$). Although Bartlett initially postulated that they were $\ce{Xe^{+}PtF6^{−}}$, it is now generally agreed that the reaction also involves the transfer of a fluorine atom to xenon to give the $\ce{XeF^{+}}$ ion:
$\ce{Xe(g) + PtF6(g) -> [XeF^{+}][PtF5^{−}](s)} \label{Eq1}$
Subsequent work showed that xenon reacts directly with fluorine under relatively mild conditions to give XeF2, XeF4, or XeF6, depending on conditions; one such reaction is as follows:
$\ce{Xe(g) + 2F2(g) -> XeF4(s)} \label{Eq2}$
The ionization energies of helium, neon, and argon are so high (Table $1$) that no stable compounds of these elements are known. The ionization energies of krypton and xenon are lower but still very high; consequently only highly electronegative elements (F, O, and Cl) can form stable compounds with xenon and krypton without being oxidized themselves. Xenon reacts directly with only two elements: F2 and Cl2. Although $\ce{XeCl2}$ and $\ce{KrF2}$ can be prepared directly from the elements, they are substantially less stable than the xenon fluorides.
The ionization energies of helium, neon, and argon are so high that no stable compounds of these elements are known.
Because halides of the noble gases are powerful oxidants and fluorinating agents, they decompose rapidly after contact with trace amounts of water, and they react violently with organic compounds or other reductants. The xenon fluorides are also Lewis acids; they react with the fluoride ion, the only Lewis base that is not oxidized immediately on contact, to form anionic complexes. For example, reacting cesium fluoride with XeF6 produces CsXeF7, which gives Cs2XeF8 when heated:
$\ce{XeF6(s) + CsF(s) -> CsXeF7(s)} \label{Eq3}$
$\ce{2CsXeF7(s) ->[\Delta] Cs2XeF8(s) + XeF6(g)} \label{Eq4}$
The $\ce{XeF8^{2-}}$ ion contains eight-coordinate xenon and has the square antiprismatic structure, which is essentially identical to that of the IF8 ion. Cs2XeF8 is surprisingly stable for a polyatomic ion that contains xenon in the +6 oxidation state, decomposing only at temperatures greater than 300°C. Major factors in the stability of Cs2XeF8 are almost certainly the formation of a stable ionic lattice and the high coordination number of xenon, which protects the central atom from attack by other species. (Recall from that this latter effect is responsible for the extreme stability of SF6.)
For a previously “inert” gas, xenon has a surprisingly high affinity for oxygen, presumably because of π bonding between $O$ and $Xe$. Consequently, xenon forms an extensive series of oxides and oxoanion salts. For example, hydrolysis of either $XeF_4$ or $XeF_6$ produces $XeO_3$, an explosive white solid:
$\ce{XeF6(aq) + 3H2O(l) -> XeO3(aq) + 6HF(aq)} \label{Eq5}$
Treating a solution of XeO3 with ozone, a strong oxidant, results in further oxidation of xenon to give either XeO4, a colorless, explosive gas, or the surprisingly stable perxenate ion (XeO64−), both of which contain xenon in its highest possible oxidation state (+8). The chemistry of the xenon halides and oxides is best understood by analogy to the corresponding compounds of iodine. For example, XeO3 is isoelectronic with the iodate ion (IO3), and XeF82− is isoelectronic with the IF8 ion.
Xenon has a high affinity for both fluorine and oxygen.
Because the ionization energy of radon is less than that of xenon, in principle radon should be able to form an even greater variety of chemical compounds than xenon. Unfortunately, however, radon is so radioactive that its chemistry has not been extensively explored.
Example $1$
On a virtual planet similar to Earth, at least one isotope of radon is not radioactive. A scientist explored its chemistry and presented her major conclusions in a trailblazing paper on radon compounds, focusing on the kinds of compounds formed and their stoichiometries. Based on periodic trends, how did she summarize the chemistry of radon?
Given: nonradioactive isotope of radon
Asked for: summary of its chemistry
Strategy:
Based on the position of radon in the periodic table and periodic trends in atomic properties, thermodynamics, and kinetics, predict the most likely reactions and compounds of radon.
Solution
We expect radon to be significantly easier to oxidize than xenon. Based on its position in the periodic table, however, we also expect its bonds to other atoms to be weaker than those formed by xenon. Radon should be more difficult to oxidize to its highest possible oxidation state (+8) than xenon because of the inert-pair effect. Consequently, radon should form an extensive series of fluorides, including RnF2, RnF4, RnF6, and possibly RnF8 (due to its large radius). The ion RnF82− should also exist. We expect radon to form a series of oxides similar to those of xenon, including RnO3 and possibly RnO4. The biggest surprise in radon chemistry is likely to be the existence of stable chlorides, such as RnCl2 and possibly even RnCl4.
Exercise $1$
Predict the stoichiometry of the product formed by reacting XeF6 with a 1:1 stoichiometric amount of KF and propose a reasonable structure for the anion.
Answer
$\ce{KXeF7}$; the xenon atom in XeF7 has 16 valence electrons, which according to the valence-shell electron-pair repulsion model could give either a square antiprismatic structure with one fluorine atom missing or a pentagonal bipyramid if the 5s2 electrons behave like an inert pair that does not participate in bonding.
Summary
The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves. Xenon has a high affinity for both fluorine and oxygen, which form stable compounds that contain xenon in even oxidation states up to +8.
• Anonymous
22: The Chemistry of the Transition Elements
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
23: Carbon: Not Just Another Element
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
24: Biochemistry
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/21%3A_The_Chemistry_of_the_Main_Group_Elements/21.11%3A__The_Noble_Gases_Group_8A.txt |
Chemistry and Chemical Reactivity
John C. Kotz, Paul M. Treichel, and John Townsend
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XXIII
Nuclear chemistry deals with radioactivity, nuclear processes, such as nuclear transmutation, and nuclear properties.
25: Nuclear Chemistry
Learning Objectives
• Write and balance nuclear equations
• To know the different kinds of radioactive decay.
• To balance a nuclear reaction.
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of $\ce{^1_1H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ($Z$) of the element, and the sum of the number of protons and the number of neutrons is the mass number ($A$). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation:
$\large \ce{^{A}_{Z}X} \label{Eq1a}$
where
• $X$ is the symbol for the element,
• $A$ is the mass number, and
• $Z$ is the atomic number.
Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, $\ce{^{14}_6C}$ is called “carbon-14.”
Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about 10−15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10−10 meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm3. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately $6.4 \times 10^6$ meters, 30,000 times larger).
Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.
Nuclear Equations
A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:
1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
2. The sum of the charges of the reactants equals the sum of the charges of the products.
If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that $\ce{^{17}_8O}$ is a product of the nuclear reaction of $\ce{^{14}_7N}$ and $\ce{^4_2He}$ if we knew that a proton, $\ce{^1_1H}$, was one of the two products. Example $1$ shows how we can identify a nuclide by balancing the nuclear reaction.
Example $1$: Balancing Equations for Nuclear Reactions
The reaction of an α particle with magnesium-25 $(\ce{^{25}_{12}Mg})$ produces a proton and a nuclide of another element. Identify the new nuclide produced.
Solution
The nuclear reaction can be written as:
$\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X} \nonumber$
where
• $\ce A$ is the mass number and
• $\ce Z$ is the atomic number of the new nuclide, $\ce X$.
Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:
$\mathrm{25+4=A+1} \nonumber$
so
$\mathrm{A=28} \nonumber$
Similarly, the charges must balance, so:
$\mathrm{12+2=Z+1} \nonumber$
so
$\mathrm{Z=13} \nonumber$
Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is $\ce{^{28}_{13}Al}$.
Exercise $1$
The nuclide $\ce{^{125}_{53}I}$ combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?
Answer
$\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber$
The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions. In a nuclear decay reaction, also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a nuclear transmutation reaction, a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is more massive than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit.
Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced.
Nuclear Decay Reactions
Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions.
To describe nuclear decay reactions, chemists have extended the $^A _Z \textrm{X}$ notation for nuclides to include radioactive emissions. Table $1$ lists the name and symbol for each type of emitted radiation. The most notable addition is the positron, a particle that has the same mass as an electron but a positive charge rather than a negative charge.
Table $1$: Nuclear Decay Emissions and Their Symbols
Identity Symbol Charge Mass (amu)
helium nucleus $^4_2\alpha$ +2 4.001506
electron $^0_{-1}\beta$ or $\beta ^-$ −1 0.000549
photon $_0^0\gamma$
neutron $^1_0\textrm n$ 0 1.008665
proton $^1_1\textrm p$ +1 1.007276
positron $^0_{+1}\beta$ or $\beta ^+$ +1 0.000549
Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, A = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus.
Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, Z = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so Z = 0. In the case of an electron, Z = −1, and for a positron, Z = +1. Because γ rays are high-energy photons, both A and Z are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol $^0_{-1}\textrm e$, which is usually simplified to e, represents a free electron or an electron associated with an atom, whereas the symbol $^0_{-1}\beta$, which is often simplified to β, denotes an electron that originates from within the nucleus, which is a β particle. Similarly, $^4_{2}\textrm{He}^{2+}$ refers to the nucleus of a helium atom, and $^4_{2}\alpha$ denotes an identical particle that has been ejected from a heavier nucleus.
There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in Figure $1$. The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions.
Alpha $\alpha$ Decay
Many nuclei with mass numbers greater than 200 undergo alpha (α) decay, which results in the emission of a helium-4 nucleus as an alpha (α) particle, $^4_{2}\alpha$. The general reaction is as follows:
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A-4}_{Z-2} \textrm X'}+\underset{\textrm{alpha}\ \textrm{particle}}{^4_2 \alpha}\label{Eq1}$
The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number A − 4 and a nuclear charge Z − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222:
$^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm{Rn}+^{4}_{2}\alpha\label{Eq2}$
Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced.
Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction.
Beta $\beta^-$ Decay
Nuclei that contain too many neutrons often undergo beta (β) decay, in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle:
$\underset{\textrm{unstable} \ \textrm{neutron in} \ \textrm{nucleus}}{^1_0 \textrm n}\rightarrow \underset{\textrm{proton} \ \textrm{retained} \ \textrm{by nucleus}}{^{1}_{1} \textrm p}+\underset{\textrm{beta particle} \ \textrm{emitted by} \ \textrm{nucleus}}{^0_{-1} \beta}\label{Eq3}$
The general reaction for beta decay is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z+1} \textrm X'}+\underset{\textrm{beta particle}}{^0_{-1} \beta}\label{Eq4}$
Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14:
$^{14}_{6}\textrm{C}\rightarrow ^{14}_{7}\textrm{N}+\,^{0}_{-1}\beta \nonumber$
Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent.
Positron $\beta^+$ Emission
Because a positron has the same mass as an electron but opposite charge, positron emission is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron:
$^{1}_{1}\textrm{p}^+\rightarrow ^{1}_{0}\textrm{n}+\,^{0}_{+1}\beta^+\label{Eq6}$
The general reaction for positron emission is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\underset{\textrm{positron}}{^0_{+1} \beta^+} \nonumber$
Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11:
$^{11}_{6}\textrm{C}\rightarrow ^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta^+ \nonumber$
Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide.
Electron Capture
A neutron-poor nucleus can decay by either positron emission or electron capture (EC), in which an electron in an inner shell reacts with a proton to produce a neutron:
$^{1}_{1}\textrm{p} +\; ^{0}_{-1}\textrm{e}\rightarrow \, ^{1}_{0}\textrm n\label{Eq9}$
When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus
$\underset{\textrm{parent}}{^A_Z \textrm X}+\underset{\textrm{electron}}{^0_{-1} \textrm e}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\textrm{x-ray} \nonumber$
Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows:
$^{55}_{26}\textrm{Fe}\overset{\textrm{EC}}{\rightarrow}\, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber$
The atomic numbers of the parent and daughter nuclides differ in Equation 20.2.11, although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation:
$^{55}_{26}\textrm{Fe}+\,^{0}_{-1}\textrm{e}\rightarrow \, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber$
Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different.
Gamma $\gamma$ Emission
Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. These high-energy photons are γ rays. Gamma ($\gamma$) emission can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state:
$^{238}_{92}\textrm{U}\rightarrow \, \underset{\textrm{excited} \ \textrm{nuclear} \ \textrm{state}}{^{234}_{90}\textrm{Th*}} + ^{4}_{2}\alpha\xrightarrow {\textrm{relaxation}\,}\,^{234}_{90}\textrm{Th} + \ce{^0_0\gamma} \nonumber$
If we disregard the decay event that created the excited nucleus, then
$^{234}_{88}\textrm{Th*} \rightarrow\, ^{234}_{88}\textrm{Th} + ^{0}_{0}\gamma \nonumber$
or more generally,
$^{A}_{Z}\textrm{X*} \rightarrow\, ^{A}_{Z}\textrm{X} + ^{0}_{0}\gamma \nonumber$
Gamma emission can also occur after a significant delay. For example, technetium-99m has a half-life of about 6 hours before emitting a $γ$ ray to form technetium-99 (the m is for metastable). Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction.
Spontaneous Fission
Only very massive nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with Z ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of $^{254}_{98}\textrm{Cf}$, which gives a distribution of fission products; one possible set of products is shown in the following equation:
$^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n} \nonumber$
Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.
Example $2$
Write a balanced nuclear equation to describe each reaction.
1. the beta decay of $^{35}_{16}\textrm{S}$
2. the decay of $^{201}_{80}\textrm{Hg}$ by electron capture
3. the decay of $^{30}_{15}\textrm{P}$ by positron emission
Given: radioactive nuclide and mode of decay
Asked for: balanced nuclear equation
Strategy:
A Identify the reactants and the products from the information given.
B Use the values of A and Z to identify any missing components needed to balance the equation.
Solution
a.
A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as $^{A}_{Z}\textrm{X}$: $^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta \nonumber$
B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows: $^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber$
b.
A We know the identities of both reactants: $^{201}_{80}\textrm{Hg}$ and an inner electron, $^{0}_{-1}\textrm{e}$. The reaction is as follows: $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X} \nonumber$
B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au} \nonumber$
c.
A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore $^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta \nonumber$
B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: $^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta \nonumber$
Exercise $2$
Write a balanced nuclear equation to describe each reaction.
1. $^{11}_{6}\textrm{C}$ by positron emission
2. the beta decay of molybdenum-99
3. the emission of an α particle followed by gamma emission from $^{185}_{74}\textrm{W}$
Answer a
$^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta$
Answer d
$^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta$
Answer c
$^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma$
Example $3$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{45}_{22}\textrm{Ti}$
2. $^{242}_{94}\textrm{Pu}$
3. $^{12}_{5}\textrm{B}$
4. $^{256}_{100}\textrm{Fm}$
Given: nuclide
Asked for: type of nuclear decay
Strategy:
Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide.
Solution
1. This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
2. Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus $^{242}_{94}\textrm{Pu}$ is expected to decay by alpha emission.
3. This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that $^{12}_{5}\textrm{B}$ will undergo beta decay.
4. This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that $^{256}_{100}\textrm{Fm}$ will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.
Exercise $3$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{32}_{14}\textrm{Si}$
2. $^{43}_{21}\textrm{Sc}$
3. $^{231}_{91}\textrm{Pa}$
Answer a
beta decay
Answer d
positron emission or electron capture
Answer c
alpha decay
Radioactive Decay Series
The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases Z by only 2, and positron emission or electron capture decreases Z by only 1, it is impossible for any nuclide with Z > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with Z > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a radioactive decay series. The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions (Figure $2$). Although a radioactive decay series can be written for almost any isotope with Z > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic.
Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin.
Induced Nuclear Reactions
The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction.
The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process:
$^{4}_{2}\alpha + \, ^{14}_{7}\textrm{N} \rightarrow \,^{17}_{8}\textrm{O}+\,^{1}_{1}\textrm{p}\label{Eq17}$
Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows:
$^{4}_{2}\alpha + \, ^{9}_{4}\textrm{Be} \rightarrow \,^{12}_{6}\textrm{C}+\,^{1}_{0}\textrm{n}\label{Eq18}$
Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus (Figure $3$). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays:
$^{1}_{0}\textrm{n} + \, ^{14}_{7}\textrm{N} \rightarrow \,^{14}_{6}\textrm{C}+\,^{1}_{1}\textrm{p}\label{Eq19}$
Example $4$
In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each 27Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction.
Given: reactants in a nuclear transmutation reaction
Asked for: product nuclide and balanced nuclear equation
Strategy:
A Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of Z and A of the product nuclide and thus its identity.
B Write the balanced nuclear equation for the reaction.
Solution
A Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum (Z = 13) will be converted to phosphorus (Z = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, $^{30}_{15}\textrm{P}$.
B The balanced nuclear equation for the reaction is as follows:
$^{27}_{13}\textrm{Al} + \, ^{4}_{2}\alpha \rightarrow \,^{30}_{15}\textrm{P}+\,^{1}_{0}\textrm{n} \nonumber$
Exercise $4$
Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei $(^{2}_{1}\textrm{H})$ produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction.
Answer
neutron, $^{1}_{0}\textrm{n}$ ; $^{96}_{42}\textrm{Mo} + \, ^{2}_{1}\textrm{H} \rightarrow \,^{97}_{43}\textrm{Tc}+\,^{1}_{0}\textrm{n}$ :
We noted earlier in this section that very heavy nuclides, corresponding to Z ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of Z, such as the isotopes of uranium (Z = 92) and plutonium (Z = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as $^{235}_{92}\textrm{U}$ and $^{239}_{94}\textrm{Pu}$ undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in Figure $4$:
$^{235}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{236}_{92}\textrm{U}\rightarrow \,^{141}_{56}\textrm{Ba}+\,^{92}_{36}\textrm{Kr}+3^{1}_{0}\textrm{n}\label{Eq20}$
Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a fissile isotope.
During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in Figure $4$. Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in Figure $5$, a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2.
Synthesis of Transuranium Elements
Uranium (Z = 92) is the heaviest naturally occurring element. Consequently, all the elements with Z > 92, the transuranium elements, are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium (Z = 93), which was synthesized in 1940 by bombarding a 238U target with neutrons. As shown in Equation 20.21, this reaction occurs in two steps. Initially, a neutron combines with a 238U nucleus to form 239U, which is unstable and undergoes beta decay to produce 239Np:
$^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta\label{Eq21}$
Subsequent beta decay of 239Np produces the second transuranium element, plutonium (Z = 94):
$^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta\label{Eq22}$
Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus (Table $2$). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability."
Table $2$: Some Reactions Used to Synthesize Transuranium Elements
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{242}_{96}\textrm{Cm}+\,^{1}_{0}\textrm{n}$
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{241}_{95}\textrm{Am}+\,^{1}_{1}\textrm{p}+\,^{1}_{0}\textrm{n}$
$^{242}_{96}\textrm{Cm}+\,^{4}_{2}\alpha \rightarrow \,^{243}_{97}\textrm{Bk}+\,^{1}_{1}\textrm{p}+2^{1}_{0}\textrm{n}$
$^{253}_{99}\textrm{Es}+\,^{4}_{2}\alpha \rightarrow \,^{256}_{101}\textrm{Md}+\,^{1}_{0}\textrm{n}$
$^{238}_{92}\textrm{U}+\,^{12}_{6}\textrm{C} \rightarrow \,^{246}_{98}\textrm{Cf}+4^{1}_{0}\textrm{n}$
$^{252}_{98}\textrm{Cf}+\,^{10}_{5}\textrm{B} \rightarrow \,^{256}_{103}\textrm{Lr}+6^{1}_{0}\textrm{n}$
A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator (Figure $6$), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long.
To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target.
The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate.
Summary and Key Takeaway
• Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material.
In nuclear decay reactions (or radioactive decay), the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with A ≥ 200 and Z > 83) are unstable and tend to decay by emitting an α particle. When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. Alpha decay results in the emission of an α particle, $^4 _2 \alpha$, and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. Beta decay converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. Positron emission is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In electron capture (EC), an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In gamma emission, a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a radioactive decay series—a succession of some combination of alpha- and beta-decay reactions. In nuclear transmutation reactions, a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All transuranium elements—elements with Z > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons.
Key Equations
alpha decay
$^A_Z \textrm X\rightarrow \, ^{A-4}_{Z-2} \textrm X'+\,^4_2 \alpha \nonumber$
beta decay
$^A_Z \textrm X\rightarrow \, ^{A}_{Z+1} \textrm X'+\,^0_{-1} \beta \nonumber$
positron emission
$^A_Z \textrm X\rightarrow \, ^{A}_{Z-1} \textrm X'+\,^0_{+1} \beta \nonumber$
electron capture
$^A_Z \textrm X+\,^{0}_{-1} \textrm e\rightarrow \, ^{A}_{Z-1} \textrm X'+\textrm{x-ray} \nonumber$
gamma emission
$^A_Z \textrm{X*}\rightarrow \, ^{A}_{Z} \textrm X+\,^0_{0} \gamma \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.1%3A__Natural_Radioactivity.txt |
Learning Objectives
• To understand how nuclear transmutation reactions lead to the formation of the elements in stars and how they can be used to synthesize transuranium elements.
The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the 56Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen (1H), which accounts for about 90% of all atoms. In fact, 1H is the raw material from which all other elements are formed. In this section, we explain why 1H and 2He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements.
Relative Abundances of the Elements on Earth and in the Known Universe
The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in Figure $1$. The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in Figure $1$ illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the known trends in nuclear stability, but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table $1$ for some common elements.
Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH3, CH4, and H2O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as 40K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. This section explains some of the reasons for the great differences in abundances of the metallic elements.
Table $1$: Relative Abundances of Elements on Earth and in the Known Universe
Terrestrial/Universal Element Abundance Ratio
H 0.0020
He 2.4 × 10−8
C 0.36
N 0.02
O 46
Ne 1.9 × 10−6
Na 1200
Mg 48
Al 1600
Si 390
S 0.84
K 5000
Ca 710
Ti 2200
Fe 57
All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei.
Synthesis of the Elements in Stars
Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known (Figure $2$). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm3, and the temperature increases to about 1.5 × 107 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun.
In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium:
$_1^1\textrm H+\,_1^1\textrm H\rightarrow\,_1^2\textrm H+\,_{+1}^0\beta \_1^2\textrm H+\,_1^1\textrm H\rightarrow\,_2^3\textrm{He}+\,_{0}^0\gamma \_2^3\textrm{He}+\,_2^3\textrm{He}\rightarrow\,_2^4\textrm{He}+2_{1}^1\textrm H\label{Eq1}$
The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two $\gamma$ rays, and a great deal of energy:
$4_1^1\textrm H\rightarrow\,_2^4\textrm{He}+2_{+1}^0\beta+2_0^0\gamma\label{Eq2}$
These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 108 K, the helium-4 nuclei begin to fuse, producing beryllium-8:
$2_2^4\textrm{He}\rightarrow\,_4^8\textrm{Be}\label{Eq3}$
Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10−16 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24:
$_4^8\textrm{Be}\xrightarrow{_2^4\textrm{He}}\,_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\label{Eq4}$
So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star.
As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 108 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei:
$_6^{12}\textrm C+\,_6^{12}\textrm C\rightarrow \,_{11}^{23}\textrm{Na}+\,_1^1\textrm H\label{Eq5}$
$_6^{12}\textrm C+\,_8^{16}\textrm O\rightarrow \,_{14}^{28}\textrm{Si}+\,_0^0\gamma\label{Eq6}$
At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40:
$_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\xrightarrow{_2^4\textrm{He}}\,_{14}^{28}\textrm{Si}\xrightarrow{_2^4\textrm{He}}\,_{16}^{32}\textrm S\xrightarrow{_2^4\textrm{He}}\,_{18}^{36}\textrm{Ar}\xrightarrow{_2^4\textrm{He}}\,_{20}^{40}\textrm{Ca}\label{Eq7}$
The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 109 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known.
The Formation of Heavier Elements in Supernovas
None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called supernovas (Figure $2$). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula (Figure $3$). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120:
$_{26}^{56}\textrm{Fe}+64_0^1\textrm n\rightarrow \,_{26}^{120}\textrm{Fe}\rightarrow\,_{50}^{120}\textrm{Sn}+24_{-1}^0\beta\label{Eq8}$
Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past.
Example $1$: Carbon Burning Stars
The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of
1. magnesium-24.
2. neon-20 from two carbon-12 nuclei.
Given: reactant and product nuclides
Asked for: balanced nuclear equation
Strategy:
Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction.
Solution
1. A magnesium-24 nucleus (Z = 12, A = 24) has the same nucleons as two carbon-12 nuclei (Z = 6, A = 12). The reaction is therefore a fusion of two carbon-12 nuclei, and no other particles are produced: $_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{12}^{24}\textrm{Mg}$.
2. The neon-20 product has Z = 10 and A = 20. The conservation of mass requires that the other product have A = (2 × 12) − 20 = 4; because of conservation of charge, it must have Z = (2 × 6) − 10 = 2. These are the characteristics of an α particle. The reaction is therefore $_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{10}^{20}\textrm{Ne}+\,_2^4\alpha$.
Exercise $1$
How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction.
Answer
19 neutrons; $_{26}^{56}\textrm{Fe}+19_0^1\textrm n \rightarrow \,_{26}^{75}\textrm{Fe}\rightarrow \,_{33}^{75}\textrm{As}+7_{-1}^0\beta$
Summary
Hydrogen and helium are the most abundant elements in the universe. Heavier elements are formed in the interior of stars via multiple neutron-capture events. By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.2%3A__Nuclear_Reactions_and_Radioactive_Decay.txt |
Learning Objectives
• To understand the factors that affect nuclear stability.
The nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear force, an extremely powerful but very short-range attractive force between nucleons (Figure \(1\)). All stable nuclei except the hydrogen-1 nucleus (1H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability.
The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in Figure \(2\). The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, 1H and 3He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., \(^4_2 \textrm{He}\), \(^{10}_5 \textrm{B}\), and \(^{40}_{20} \textrm{Ca}\)). All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive.
As shown in Figure \(3\), more than half of the stable nuclei (166 out of 279) have even numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called magic numbers 2, 8, 20, 50, 82, and 126. For example, tin (Z = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium (Z = 49) and antimony (Z = 51), have only 2 stable isotopes each. Nuclei with magic numbers of both protons and neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are \(^4_2 \textrm{He}\), with 2 protons and 2 neutrons, and \(^{208}_{82} \textrm{Pb}\), with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element.
Note
Most stable nuclei contain even numbers of both neutrons and protons
The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in Figure \(2\), the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei.
Example \(1\)
Classify each nuclide as stable or radioactive.
1. \(_{15}^{30} \textrm P\)
2. \(_{43}^{98} \textrm{Tc}\)
3. tin-118
4. \(_{94}^{239} \textrm{Pu}\)
Given: mass number and atomic number
Asked for: predicted nuclear stability
Strategy:
Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide.
Solution:
a. This isotope of phosphorus has 15 neutrons and 15 protons, giving a neutron-to-proton ratio of 1.0. Although the atomic number, 15, is much less than the value of 83 above which all nuclides are unstable, the neutron-to-proton ratio is less than that expected for stability for an element with this mass. As shown in Figure \(2\), its neutron-to-proton ratio should be greater than 1. Moreover, this isotope has an odd number of both neutrons and protons, which also tends to make a nuclide unstable. Consequently, \(_{15}^{30} \textrm P\) is predicted to be radioactive, and it is.
b. This isotope of technetium has 55 neutrons and 43 protons, giving a neutron-to-proton ratio of 1.28, which places \(_{43}^{98} \textrm{Tc}\) near the edge of the band of stability. The atomic number, 55, is much less than the value of 83 above which all isotopes are unstable. These facts suggest that \(_{43}^{98} \textrm{Tc}\) might be stable. However, \(_{43}^{98} \textrm{Tc}\) has an odd number of both neutrons and protons, a combination that seldom gives a stable nucleus. Consequently, \(_{43}^{98} \textrm{Tc}\) is predicted to be radioactive, and it is.
c. Tin-118 has 68 neutrons and 50 protons, for a neutron-to-proton ratio of 1.36. As in part b, this value and the atomic number both suggest stability. In addition, the isotope has an even number of both neutrons and protons, which tends to increase nuclear stability. Most important, the nucleus has 50 protons, and 50 is one of the magic numbers associated with especially stable nuclei. Thus \(_{50}^{118} \textrm{Sn}\)should be particularly stable.
d. This nuclide has an atomic number of 94. Because all nuclei with Z > 83 are unstable, \(_{94}^{239} \textrm{Pu}\) must be radioactive.
Exercise \(1\)
Classify each nuclide as stable or radioactive.
1. \(_{90}^{232} \textrm{Th}\)
2. \(_{20}^{40} \textrm{Ca}\)
3. \(_8^{15} \textrm{O}\)
4. \(_{57}^{139} \textrm{La}\)
Answer:
1. radioactive
2. stable
3. radioactive
4. stable
Superheavy Elements
In addition to the “peninsula of stability” there is a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elements, with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with Z = 114 and Z = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with A = 292 that was found in 232Th. With an estimated half-life greater than 108 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with Z = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements.
Summary
Subatomic particles of the nucleus (protons and neutrons) are called nucleons. A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive, and its emissions are collectively called radioactivity. Isotopes that emit radiation are called radioisotopes. Each nucleon is attracted to other nucleons by the strong nuclear force. Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable. Superheavy elements, with atomic numbers near 126, may even be stable enough to exist in nature.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.3%3A__Stability_of_Atomic_Nuclei.txt |
Learning Objectives
• Recognize common modes of radioactive decay
• Identify common particles and energies involved in nuclear decay reactions
• Write and balance nuclear decay equations
• Calculate kinetic parameters for decay processes, including half-life
• Describe common radiometric dating techniques
Following the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences—chemistry and physics), who was the first to coin the term “radioactivity,” and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed.
The spontaneous change of an unstable nuclide into another is radioactive decay. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide. The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure $1$).
Although the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. Click here to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab.
Types of Radioactive Decay
Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure $2$) helped him determine that one type of radiation consisted of positively charged and relatively massive $α$ particles; a second type was made up of negatively charged and much less massive $β$ particles; and a third was uncharged electromagnetic waves, $γ$ rays. We now know that $α$ particles are high-energy helium nuclei, $β$ particles are high-energy electrons, and $γ$ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced.
Alpha ($α$) decay is the emission of an α particle from the nucleus. For example, polonium-210 undergoes α decay:
$\ce{^{210}_{84}Po⟶ ^4_2He + ^{206}_{82}Pb} \hspace{40px}\ce{or}\hspace{40px} \ce{^{210}_{84}Po ⟶ ^4_2α + ^{206}_{82}Pb}\nonumber$
Alpha decay occurs primarily in heavy nuclei (A > 200, Z > 83). Because the loss of an α particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing α decay lies below the band of stability, the daughter nuclide will lie closer to the band.
Beta (β) decay is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes β decay:
$\ce{^{131}_{53}I ⟶ ^0_{-1}e + ^{131}_{54}X} \hspace{40px}\ce{or}\hspace{40px} \ce{^{131}_{53}I ⟶ ^0_{-1}β + ^{131}_{54}Xe}\nonumber$
Beta decay, which can be thought of as the conversion of a neutron into a proton and a β particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Gamma emission (γ emission) is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a γ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (*). Cobalt-60 emits γ radiation and is used in many applications including cancer treatment:
$\mathrm{^{60}_{27}Co^* ⟶\, ^0_0γ +\, ^{60}_{27}Co}\nonumber$
There is no change in mass number or atomic number during the emission of a γ ray unless the γ emission accompanies one of the other modes of decay.
Positron emission ($β^+$ decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:
$\ce{^{15}_8O ⟶ ^0_{+1}e + ^{15}_7N} \hspace{40px}\ce{or}\hspace{40px} \ce{^{15}_8O ⟶ ^0_{+1}β + ^{15}_7N}\nonumber$
Positron emission is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Electron capture occurs when one of the inner electrons in an atom is captured by the atom’s nucleus. For example, potassium-40 undergoes electron capture:
$\ce{^{40}_{19}K + ^0_{-1}e ⟶ ^{40}_{18}Ar}\nonumber$
Electron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for “proton-rich” nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur. Figure $3$ summarizes these types of decay, along with their equations and changes in atomic and mass numbers.
PET Scan
Positron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient’s body function (Figure $4$). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This “tagged” compound, or radiotracer, is then put into the patient (injected via IV or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions.
For example, F-18 is produced by proton bombardment of 18O $(\ce{^{18}_8O + ^1_1p⟶ ^{18}_9F + ^1_0n})$ and incorporated into a glucose analog called fludeoxyglucose (FDG). How FDG is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The 18F emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient’s body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer’s disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and X-rays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan.
Radioactive Decay Series
The naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product—that is, a nuclide on the band of stability (Figure $5$). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.
Radioactive Half-Lives
Radioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life (t1/2), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem.
For example, cobalt-60, an isotope that emits gamma rays used to treat cancer, has a half-life of 5.27 years (Figure $6$). In a given cobalt-60 source, since half of the $\ce{^{60}_{27}Co}$ nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.
Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for first-order chemical reactions. We generally substitute the number of nuclei, N, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:
$\text{decay rate} = \lambda N\nonumber$
with $\lambda$ is the decay constant for the particular radioisotope.
The decay constant, $\lambda$, which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, t1/2:
$λ=\dfrac{\ln 2}{t_{1/2}}=\dfrac{0.693}{t_{1/2}} \hspace{40px}\ce{or}\hspace{40px} t_{1/2}=\dfrac{\ln 2}{λ}=\dfrac{0.693}{λ}\nonumber$
The first-order equations relating amount, N, and time are:
$N_t=N_0e^{−kt} \hspace{40px}\ce{or}\hspace{40px} t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)\nonumber$
where N0 is the initial number of nuclei or moles of the isotope, and Nt is the number of nuclei/moles remaining at time t. Example $1$ applies these calculations to find the rates of radioactive decay for specific nuclides.
Example $1$: Rates of Radioactive Decay
$\ce{^{60}_{27}Co}$ decays with a half-life of 5.27 years to produce $\ce{^{60}_{28}Ni}$.
1. What is the decay constant for the radioactive disintegration of cobalt-60?
2. Calculate the fraction of a sample of the $\ce{^{60}_{27}Co}$ isotope that will remain after 15 years.
3. How long does it take for a sample of $\ce{^{60}_{27}Co}$ to disintegrate to the extent that only 2.0% of the original amount remains?
Solution
(a) The value of the rate constant is given by:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{5.27\:y}=0.132\:y^{−1}} \nonumber$
(b) The fraction of $\ce{^{60}_{27}Co}$ that is left after time t is given by $\dfrac{N_t}{N_0}$. Rearranging the first-order relationship Nt = N0eλt to solve for this ratio yields:
$\dfrac{N_t}{N_0}=e^{-λt}=e^\mathrm{-(0.132/y)(15.0/y)}=0.138 \nonumber$
The fraction of $\ce{^{60}_{27}Co}$ that will remain after 15.0 years is 0.138. Or put another way, 13.8% of the $\ce{^{60}_{27}Co}$ originally present will remain after 15 years.
(c) 2.00% of the original amount of $\ce{^{60}_{27}Co}$ is equal to 0.0200 × N0. Substituting this into the equation for time for first-order kinetics, we have:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)=−\dfrac{1}{0.132\:\ce y^{−1}}\ln\left(\dfrac{0.0200×N_0}{N_0}\right)=29.6\:\ce y \nonumber$
Exercise $1$
Radon-222, $\ce{^{222}_{86}Rn}$, has a half-life of 3.823 days. How long will it take a sample of radon-222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222?
Answer
11.1 days
Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of $\ce{^{209}_{83}Bi}$ is 1.9 × 1019 years; $\ce{^{239}_{94}Ra}$ is 24,000 years; $\ce{^{222}_{86}Rn}$ is 3.82 days; and element-111 (Rg for roentgenium) is 1.5 × 10–3 seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table $1$, and others are listed in Appendix N1.
Table $1$: Half-lives of Radioactive Isotopes Important to Medicine
Type Decay Mode Half-Life Uses
F-18 β+ decay 110. minutes PET scans
Co-60 β decay, γ decay 5.27 years cancer treatment
Tc-99m1 γ decay 8.01 hours scans of brain, lung, heart, bone
I-131 β decay 8.02 days thyroid scans and treatment
Tl-201 electron capture 73 hours heart and arteries scans; cardiac stress tests
The “m” in Tc-99m stands for “metastable,” indicating that this is an unstable, high-energy state of Tc-99. Metastable isotopes emit $γ$ radiation to rid themselves of excess energy and become (more) stable.
Radiometric Dating
Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type.
Radioactive Dating Using Carbon-14
The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.
Naturally occurring carbon consists of three isotopes: $\ce{^{12}_6C}$, which constitutes about 99% of the carbon on earth; $\ce{^{13}_6C}$, about 1% of the total; and trace amounts of $\ce{^{14}_6C}$. Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:
$\ce{^{14}_7N + ^1_0n⟶ ^{14}_6C + ^1_1H}\nonumber$
All isotopes of carbon react with oxygen to produce CO2 molecules. The ratio of $\ce{^{14}_6CO2}$ to $\ce{^{12}_6CO2}$ depends on the ratio of $\ce{^{14}_6CO}$ to $\ce{^{12}_6CO}$ in the atmosphere. The natural abundance of $\ce{^{14}_6CO}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of $\ce{^{14}_6C ^{14}_6CO2}$ and $\ce{^{12}_6CO2}$ into plants is a regular part of the photosynthesis process, which means that the $\ce{^{14}_6C: ^{12}_6C}$ ratio found in a living plant is the same as the $\ce{^{14}_6C: ^{12}_6C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because $\ce{^{12}_6C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by β emission with a half-life of 5730 years:
$\ce{^{14}_6C⟶ ^{14}_7N + ^0_{-1}e}\nonumber$
Thus, the $\ce{^{14}_6C: ^{12}_6C}$ ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure $7$ visually depicts this process.
For example, with the half-life of $\ce{^{14}_6C}$ being 5730 years, if the $\ce{^{14}_6C : ^{12}_6C}$ ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of $\ce{^{14}_6C : ^{12}_6C}$ ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer.
Example $2$: Radiocarbon Dating
A tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations/min/g of C, estimate the age of the Dead Sea Scrolls.
Solution
The rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive C-14 left in the paper, so we can substitute the rates for the amounts, N, in the relationship:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)⟶t=−\dfrac{1}{λ}\ln\left(\dfrac{\ce{Rate}_t}{\ce{Rate}_0}\right) \nonumber$
where the subscript 0 represents the time when the plants were cut to make the paper, and the subscript t represents the current time.
The decay constant can be determined from the half-life of C-14, 5730 years:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{5730\: y}=1.21×10^{−4}\:y^{−1}} \nonumber$
Substituting and solving, we have:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{\ce{Rate}_t}{\ce{Rate}_0}\right)=\mathrm{−\dfrac{1}{1.21×10^{−4}\:y^{−1}}\ln\left(\dfrac{10.8\:dis/min/g\: C}{13.6\:dis/min/g\: C}\right)=1910\: y}\nonumber$
Therefore, the Dead Sea Scrolls are approximately 1900 years old (Figure $8$).
Exercise $2$
More accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun’s tomb have a C-14 decay rate of 9.07 disintegrations/min/g of C. How long ago did King Tut’s reign come to an end?
Answer
about 3350 years ago, or approximately 1340 BC
There have been some significant, well-documented changes to the $\ce{^{14}_6C : ^{12}_6C}$ ratio. The accuracy of a straightforward application of this technique depends on the $\ce{^{14}_6C : ^{12}_6C}$ ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of CO2 molecules (largely $\ce{^{12}_6CO2}$) in the atmosphere caused by combustion of fossil fuels (in which essentially all of the $\ce{^{14}_6C}$ has decayed), the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the atmosphere may be changing. This manmade increase in $\ce{^{12}_6CO2}$ in the atmosphere causes the $\ce{^{14}_6C : ^{12}_6C}$ ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years.
Radioactive Dating Using Nuclides Other than Carbon-14
Radioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original U-238 to decay into Pb-206. In a sample of rock that does not contain appreciable amounts of Pb-208, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form Ar-40 with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40 ratio yields the age of the rock. Other methods, such as rubidium-strontium dating (Rb-87 decays into Sr-87 with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth’s age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old.
Example $3$: Radioactive Dating of Rocks
An igneous rock contains 9.58 × 10–5 g of U-238 and 2.51 × 10–5 g of Pb-206, and much, much smaller amounts of Pb-208. Determine the approximate time at which the rock formed.
Solution
The sample of rock contains very little Pb-208, the most common isotope of lead, so we can safely assume that all the Pb-206 in the rock was produced by the radioactive decay of U-238. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay.
The amount of U-238 currently in the rock is:
$\mathrm{9.58×10^{−5}\cancel{g\: U}×\left( \dfrac{1\: mol\: U}{238\cancel{g\: U}}\right )=4.03×10^{−7}\:mol\: U}\nonumber$
Because when one mole of U-238 decays, it produces one mole of Pb-206, the amount of U-238 that has undergone radioactive decay since the rock was formed is:
$\mathrm{2.51×10^{-5}\cancel{g\: Pb}×\left( \dfrac{1\cancel{mol\: Pb}}{206\cancel{g\: Pb}}\right )×\left(\dfrac{1\: mol\: U}{1\cancel{mol\: Pb}}\right)=1.22×10^{-7}\:mol\: U}\nonumber$
The total amount of U-238 originally present in the rock is therefore:
$\mathrm{4.03×10^{−7}\:mol+1.22×10^{−7}\:mol=5.25×10^{−7}\:mol\: U}\nonumber$
The amount of time that has passed since the formation of the rock is given by:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)\nonumber$
with N0 representing the original amount of U-238 and Nt representing the present amount of U-238.
U-238 decays into Pb-206 with a half-life of 4.5 × 109 y, so the decay constant λ is:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{4.5×10^9\:y}=1.54×10^{−10}\:y^{−1}}\nonumber$
Substituting and solving, we have:
$t=\mathrm{−\dfrac{1}{1.54×10^{−10}\:y^{−1}}\ln\left(\dfrac{4.03×10^{−7}\cancel{mol\: U}}{5.25×10^{−7}\cancel{mol\: U}}\right)=1.7×10^9\:y}\nonumber$
Therefore, the rock is approximately 1.7 billion years old.
Exercise $3$
A sample of rock contains 6.14 × 10–4 g of Rb-87 and 3.51 × 10–5 g of Sr-87. Calculate the age of the rock. (The half-life of the β decay of Rb-87 is 4.7 × 1010 y.)
Answer
3.7 × 109 y
Summary
Nuclei that have unstable n:p ratios undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new nucleus with a more stable n:p ratio. Some substances undergo radioactive decay series, proceeding through multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics, and each radioisotope has its own characteristic half-life, the time that is required for half of its atoms to decay. Because of the large differences in stability among nuclides, there is a very wide range of half-lives of radioactive substances. Many of these substances have found useful applications in medical diagnosis and treatment, determining the age of archaeological and geological objects, and more.
Key Equations
• decay rate = λN
• $t_{1/2}=\dfrac{\ln 2}{λ}=\dfrac{0.693}{λ}$
Glossary
alpha (α) decay
loss of an alpha particle during radioactive decay
beta (β) decay
breakdown of a neutron into a proton, which remains in the nucleus, and an electron, which is emitted as a beta particle
daughter nuclide
nuclide produced by the radioactive decay of another nuclide; may be stable or may decay further
electron capture
combination of a core electron with a proton to yield a neutron within the nucleus
gamma (γ) emission
decay of an excited-state nuclide accompanied by emission of a gamma ray
half-life (t1/2)
time required for half of the atoms in a radioactive sample to decay
parent nuclide
unstable nuclide that changes spontaneously into another (daughter) nuclide
positron emission
(also, β+ decay) conversion of a proton into a neutron, which remains in the nucleus, and a positron, which is emitted
radioactive decay
spontaneous decay of an unstable nuclide into another nuclide
radioactive decay series
chains of successive disintegrations (radioactive decays) that ultimately lead to a stable end-product
radiocarbon dating
highly accurate means of dating objects 30,000–50,000 years old that were derived from once-living matter; achieved by calculating the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the object vs. the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the present-day atmosphere
radiometric dating
use of radioisotopes and their properties to date the formation of objects such as archeological artifacts, formerly living organisms, or geological formations | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.4%3A__Rates_of_Nuclear_Decay.txt |
Radioactivity is the process by which the nucleus of an unstable atom loses energy by emitting radiation, including alpha particles, beta particles, gamma rays and conversion electrons Although radioactivity is observed as a natural occurring process, it can also be artificially induced typically via the bombarding atoms of a specific element by radiating particles, thus creating new atoms.
Introduction
Ernest Rutherford was a prominent New Zealand scientist, and a winner of the Nobel Prize in chemistry in 1908. Amongst his vast list of discoveries, Rutherford was also the first to discover artificially induced radioactivity. Through the bombardment of alpha particles against nuclei of $\ce{^{14}N}$ with 7 protons/electrons, Rutherford produced $\ce{^{17}O}$ (8 protons/electrons) and protons (Figure $1$). Through this observation, Rutherford concluded that atoms of one specific element can be made into atoms of another element. If the resulting element is radioactive, then this process is called artificially induced radioactivity
Rutherford was the first researcher to create protons outside of the atomic nuclei and the $\ce{^{17}O}$ isotope of oxygen, which is nonradioactive. Similarly, other nuclei when bombarded with alpha particles will generate new elements (Figure $2$) that may be radioactive and decay naturally or that may be stable and persist like $\ce{^{17}O}$.
Before this discovery of artificial induction of radioactivity, it was a common belief that atoms of matter are unchangeable and indivisible. After the very first discoveries made by Ernest Rutherford, Irene Joliot-Curie and her husband, Frederic Joliot, a new point of view was developed. The point of view that although atoms appear to be stable, they can be transformed into new atoms with different chemical properties. Today over one thousand artificially created radioactive nuclides exist, which considerably outnumber the nonradioactive ones created.
Note: Irene Joliet-Curie and Frederic Joliot
Irene Joliet-Curie and her husband Frédéric both were French scientists who shared winning the Nobel Prize award in chemistry in 1935 for artificially synthesizing a radioactive isotope of phosphorus by bombarding aluminum with alpha particles. $\ce{^{30}P}$ with 15 protons was the first radioactive nuclide obtained through this method of artificially inducing radioactivity.
$\ce{^27_13Al + ^4_2He \rightarrow ^30_15P + ^1_0n}$
$\ce{^30_15P \rightarrow ^30_14Si + ^0_{-1}\beta}$
Activation (or radioactivation) involves making a radioactive isotope by neutron capture, e.g. the addition of a neutron to a nuclide resulting in an increase in isotope number by 1 while retaining the same atomic number (Figure $3$). Activation is often an inadvertent process occurring inside or near a nuclear reactor, where there are many neutrons flying around. For example, Coba in or near a nuclear reactor will capture a neutron forming the radioactive isotope Co-60.
$\ce{ ^1_0n + ^{59}Co \rightarrow ^{60}Co }$
The $\ce{ ^{60}Co}$ isotope is unstable (half life of 5.272 years) and disintegrates into $\ce{ ^{60}Ni }$ via the emission of $\beta$ particle and $\gamma$ radiation Figure $4$.
Example $1$: Neutron Bombardment
Write a nuclear equation for the creation of 56Mn through the bombardment of 59Co with neutrons.
Solution
A unknown particle is produced with 56Mn, in order to find the mass number (A) of the unknown we must subtract the mass number of the Manganese atom from the mass number of the Cobalt atom plus the neutron being thrown. In simpler terms,
Now, by referring to a periodic table to find the atomic numbers of Mn and Co, and then subtracting the atomic number of Mn from Co, we will receive the atomic number of the unknown particle
Thus, the unknown particle has A = 4, and Z = 2, which would make it a Helium particle, and the nuclear formula would be as follows:
$\ce{^{50}_{27}Co + ^1_0n \rightarrow ^{56}_{25}Mn + ^{4}_{2}\alpha } \nonumber$
Example $2$: Calcium Bombardment
Write a nuclear equation for the production of $\ce{^{147}Eu}$ by bombardment of $\ce{^{139}La}$ with $\ce{^{12}Ca}$.
Solution
Like the above example, you must first find the mass number of the unknown particle.
Thus, the mass number of the unknown particle is 4. Again by referring to a periodic table and finding the atomic numbers of Lanthanum, Carbon and Europium, we are able to calculate the atomic number of the unknown particle,
The atomic number for the unknown particle equals to zero, therefore 4 neutrons are emitted, and the nuclear equation is written as follows:
$\ce{^{139}_{57}La + ^{12}_6C \rightarrow ^{147}_{63}Eu + 4 ^{0}_{1}n } \nonumber$
Summary
Induced radioactivity occurs when a previously stable material has been made radioactive by exposure to specific radiation. Most radioactivity does not induce other material to become radioactive. This Induced radioactivity was discovered by Irène Curie and F. Joliot in 1934. This is also known as man-made radioactivity. The phenomenon by which even light elements are made radioactive by artificial or induced methods is called artificial radioactivity. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.5%3A__Artificial_Nuclear_Reactions.txt |
Learning Objectives
• Explain nuclear fission
• Relate the concepts of critical mass and nuclear chain reactions
• Summarize basic requirements for nuclear fission
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure \(1\).
Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure \(2\). Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium.
A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 × 1010 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal.
As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (Figure \(3\)). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.
Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure \(4\)).
An atomic bomb (Figure \(5\)) contains several pounds of fissionable material, \(\ce{^{235}_{92}U}\) or \(\ce{^{239}_{94}Pu}\), a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result.
Fission Reactors
Chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure \(6\)). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.
Nuclear Fuels
Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U3O8; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.
In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF6 (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF6 to pass through. The slightly lighter 235UF6 molecules diffuse through the barrier slightly faster than the heavier 238UF6 molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of 235UF6 to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF6 gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO2) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil.
Nuclear Moderators
Neutrons produced by nuclear reactions move too fast to cause fission (Figure \(4\)). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water \(\ce{( ^2_1H2O)}\) or light water (ordinary H2O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite.
Reactor Coolants
A nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts.
Control Rods
Nuclear reactors use control rods (Figure \(8\)) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:
\[\ce{^{10}_5B + ^1_0n⟶ ^7_3Li + ^4_2He}\]
When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.
Shield and Containment System
During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:
1. The reactor vessel, a steel shell that is 3–20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor
2. A main shield of 1–3 meters of high-density concrete
3. A personnel shield of lighter materials that protects operators from γ rays and X-rays
In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident.
Video \(1\): Click here to watch a 3-minute video from the Nuclear Energy Institute on how nuclear reactors work.
Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents.
Nuclear Accidents
The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima).
In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:
\[\ce{Zr}(s)+\ce{2H2O}(g)⟶\ce{ZrO2}(s)+\ce{2H2}(g)\]
The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process.
Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure \(8\)).
Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland.
In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events.
An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure \(1\)0).
The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate.
Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.6%3A__Nuclear_Fission.txt |
Learning Objectives
• Describe the nuclear reactions in a nuclear fusion reaction
• Quantify the energy released or absorbed in a fusion reaction
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:
$\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}n}$
A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 1011 kJ of energy per mole of $\ce{^4_2He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ).
It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, $^2_1H$ and a triton, $^3_1H$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:
$\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n}$
This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 109 kilojoules per mole of $\ce{^4_2He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide.
The most important fusion process in nature is the one that powers stars. In the 20th century, it was realized that the energy released from nuclear fusion reactions accounted for the longevity of the Sun and other stars as a source of heat and light. The fusion of nuclei in a star, starting from its initial hydrogen and helium abundance, provides that energy and synthesizes new nuclei as a byproduct of that fusion process. The prime energy producer in the Sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature of 14 million kelvin. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy (Figure $2$).
Example $1$
Calculate the energy released in each of the following hypothetical processes.
1. $\ce{3 ^4_2He \rightarrow ^{12}_6C}$
2. $\ce{6 ^1_1H + 6 ^1_0n \rightarrow ^{12}_6C}$
3. $\ce{6 ^2_1D \rightarrow ^{12}_6C}$
Solution
1. $Q_a = 3 \times 4.0026 - 12.000) \,amu \times (1.4924\times 10^{-10} \,J/amu) = 1.17 \times 10^{-12} \,J$
2. $Q_b = (6 \times (1.007825 + 1.008665) - 12.00000)\, amu \times (1.4924\times 10^{1-0} J/amu) = 1.476\times 10^{-11} \,J$
3. $Q_c = 6 \times 2.014102 - 12.00000 \, amu \times (1.4924\times 10^{-10} \, J/amu) = 1.263\times 10^{-11}\, J$
Fusion of $\ce{He}$ to give $\ce{C}$ releases the least amount of energy, because the fusion to produce He has released a large amount. The difference between the second and the third is the binding energy of deuterium. The conservation of mass-and-energy is well illustrated in these calculations. On the other hand, the calculation is based on the conservation of mass-and-energy.
Nuclear Reactors
Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.
Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure $3$). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.Contributors | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.7%3A__Nuclear_Fusion.txt |
Learning Objectives
• Describe the biological impact of ionizing radiation.
• Define units for measuring radiation exposure.
• Explain the operation of common tools for detecting radioactivity.
• List common sources of radiation exposure in the US.
The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure $1$).
Ionizing vs. Nonionizing Radiation
There is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation, emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure $2$).
Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H2O (the most abundant molecule in living organisms), which forms a H2O+ ion that reacts with water, forming a hydronium ion and a hydroxyl radical:
Biological Effects of Exposure to Radiation
Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy.
Different types of radiation have differing abilities to pass through material (Figure $4$). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of β particles, and about 20 times that of γ rays and X-rays.
For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238, which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure $5$).
Radon is found in buildings across the country, with amounts dependent on location. The average concentration of radon inside houses in the US (1.25 pCi/L) is about three times the level found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one’s risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year.
Measuring Radiation Exposure
Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure $6$). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator—a material that emits light (luminesces) when excited by ionizing radiation—and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters.
A variety of units are used to measure various aspects of radiation (Table $1$). The SI unit for rate of radioactive decay is the becquerel (Bq), with 1 Bq = 1 disintegration per second. The curie (Ci) and millicurie (mCi) are much larger units and are frequently used in medicine (1 curie = 1 Ci = $3.7 \times 10^{10}$ disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with 1 Gy = 1 J of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used (1 rad = 0.01 Gy; 1 rad results in the absorption of 0.01 J/kg of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose.
Table $1$: Units Used for Measuring Radiation
Measurement Purpose Unit Quantity Measured Description
activity of source becquerel (Bq) radioactive decays or emissions amount of sample that undergoes 1 decay/second
curie (Ci) amount of sample that undergoes $\mathrm{3.7 \times 10^{10}\; decays/second}$
absorbed dose gray (Gy) energy absorbed per kg of tissue 1 Gy = 1 J/kg tissue
radiation absorbed dose (rad) 1 rad = 0.01 J/kg tissue
biologically effective dose sievert (Sv) tissue damage Sv = RBE × Gy
roentgen equivalent for man (rem) Rem = RBE × rad
The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine (1 rem = 1 Sv). Note that the tissue damage units (rem or Sv) includes the energy of the radiation dose (rad or Gy), along with a biological factor referred to as the RBE (for relative biological effectiveness), that is an approximate measure of the relative damage done by the radiation. These are related by:
$\text{number of rems}=\text{RBE} \times \text{number of rads} \label{Eq2}$
with RBE approximately 10 for α radiation, 2(+) for protons and neutrons, and 1 for β and γ radiation.
Example $1$: Amount of Radiation
Cobalt-60 (t1/2 = 5.26 y) is used in cancer therapy since the $\gamma$ rays it emits can be focused in small areas where the cancer is located. A 5.00-g sample of Co-60 is available for cancer treatment.
1. What is its activity in Bq?
2. What is its activity in Ci?
Solution
The activity is given by:
$\textrm{Activity}=λN=\left( \dfrac{\ln 2}{t_{1/2} } \right) N=\mathrm{\left( \dfrac{\ln 2}{5.26\ y} \right) \times 5.00 \ g=0.659\ \dfrac{g}{y} \ of\ \ce{^{60}Co} \text{ that decay}} \nonumber$
And to convert this to decays per second:
$\mathrm{0.659\; \frac{g}{y} \times \dfrac{y}{365 \;day} \times \dfrac{1\; day}{ 24\; hours} \times \dfrac{1\; h}{3,600 \;s} \times \dfrac{1\; mol}{59.9\; g} \times \dfrac{6.02 \times 10^{23} \;atoms}{1 \;mol} \times \dfrac{1\; decay}{1\; atom}} \nonumber$
$\mathrm{=2.10 \times 10^{14} \; \frac{decay}{s}} \nonumber$
(a) Since $\mathrm{1\; Bq = 1\; \frac{ decay}{s}}$, the activity in Becquerel (Bq) is:
$\mathrm{2.10 \times 10^{14} \dfrac{decay}{s} \times \left(\dfrac{1\ Bq}{1 \; \frac{decay}{s}} \right)=2.10 \times 10^{14} \; Bq} \nonumber$
(b) Since $\mathrm{1\ Ci = 3.7 \times 10^{11}\; \frac{decay}{s}}$, the activity in curie (Ci) is:
$\mathrm{2.10 \times 10^{14} \frac{decay}{s} \times \left( \dfrac{1\ Ci}{3.7 \times 10^{11} \frac{decay}{s}} \right) =5.7 \times 10^2\;Ci} \nonumber$
Exercise $1$
Tritium is a radioactive isotope of hydrogen ($t_{1/2} = \mathrm{12.32\; years}$) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu. What is the activity of a sample containing 1.00mg of tritium (a) in Bq and (b) in Ci?
Answer a
$\mathrm{3.56 \times 10^{11} Bq}$
Answer b
$\mathrm{0.962\; Ci}$
Effects of Long-term Radiation Exposure on the Human Body
The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure $8$, the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131).
A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table $2$.
Table $2$: Health Effects of Radiation
Exposure (rem) Health Effect Time to Onset (Without Treatment)
5–10 changes in blood chemistry
50 nausea hours
55 fatigue
70 vomiting
75 hair loss 2–3 weeks
90 diarrhea
100 hemorrhage
400 possible death within 2 months
1000 destruction of intestinal lining
internal bleeding
death 1–2 weeks
2000 damage to central nervous system
loss of consciousness minutes
death hours to days
It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure.
Summary
We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to rapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating, but potentially most damaging, and gamma rays the most penetrating.
Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source of radiation, and limiting time of exposure.
Footnotes
1. 1 Source: US Environmental Protection Agency
Glossary
becquerel (Bq)
SI unit for rate of radioactive decay; 1 Bq = 1 disintegration/s.
curie (Ci)
Larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 1010 disintegrations/s.
Geiger counter
Instrument that detects and measures radiation via the ionization produced in a Geiger-Müller tube.
gray (Gy)
SI unit for measuring radiation dose; 1 Gy = 1 J absorbed/kg tissue.
ionizing radiation
Radiation that can cause a molecule to lose an electron and form an ion.
millicurie (mCi)
Larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 1010 disintegrations/s.
nonionizing radiation
Radiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules.
radiation absorbed dose (rad)
SI unit for measuring radiation dose, frequently used in medical applications; 1 rad = 0.01 Gy.
radiation dosimeter
Device that measures ionizing radiation and is used to determine personal radiation exposure.
relative biological effectiveness (RBE)
Measure of the relative damage done by radiation.
roentgen equivalent man (rem)
Unit for radiation damage, frequently used in medicine; 1 rem = 1 Sv.
scintillation counter
Instrument that uses a scintillator—a material that emits light when excited by ionizing radiation—to detect and measure radiation.
sievert (Sv)
SI unit measuring tissue damage caused by radiation; takes energy and biological effects of radiation into account. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.8%3A__Radiation_Health_and_Safety.txt |
Learning Objectives
• List common applications of radioactive isotopes
Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a radioactive tracer (or radioactive label). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more.
Radioisotopes have revolutionized medical practice, where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 $\ce{(^{99}_{43}Tc)}$, thallium-201 $\ce{(^{201}_{81}Tl)}$, iodine-131 $\ce{(^{131}_{53}I)}$, and sodium-24 $\ce{(^{24}_{11}Na)}$. Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the γ rays emitted by the Tc-99 isotope. Thallium-201 (Figure $1$) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave’s disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood.
Radioisotopes used in medicine typically have short half-lives—for example, the ubiquitous Tc-99m has a half-life of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes β decay with a half-life of 66 hours, and the Tc-99 is then chemically extracted (Figure $2$). The parent nuclide Mo-99 is part of a molybdate ion, $\ce{MoO4^2-}$; when it decays, it forms the pertechnetate ion, $\ce{TcO4-}$. These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests.
Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. Radiation therapy is the use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing (Figure $3$). A cancer patient may receive external beam radiation therapy delivered by a machine outside the body, or internal radiation therapy (brachytherapy) from a radioactive substance that has been introduced into the body. Note that chemotherapy is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells.
Cobalt-60 is a synthetic radioisotope produced by the neutron activation of Co-59, which then undergoes β decay to form Ni-60, along with the emission of γ radiation. The overall process is:
$\ce{^{59}_{27}Co + ^1_0n⟶ ^{60}_{27}Co⟶ ^{60}_{28}Ni + ^0_{−1}β + 2^0_0γ} \nonumber$
The overall decay scheme for this is shown graphically in Figure $4$.
Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants.
For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:
$\ce{6CO2}(g)+\ce{6H2O}(l)⟶\ce{C6H12O6}(s)+\ce{6O2}(g), \nonumber$
but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to CO2 containing a high concentration of $\ce{^{14}_6C}$. At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction.
Commercial applications of radioactive materials are equally diverse (Figure $5$). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with γ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil.
Americium-241, an α emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure $6$). The α emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm.
Summary
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally.
Glossary
chemotherapy
similar to internal radiation therapy, but chemical rather than radioactive substances are introduced into the body to kill cancer cells
external beam radiation therapy
radiation delivered by a machine outside the body
internal radiation therapy
(also, brachytherapy) radiation from a radioactive substance introduced into the body to kill cancer cells
radiation therapy
use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing
radioactive tracer
(also, radioactive label) radioisotope used to track or follow a substance by monitoring its radioactive emissions | textbooks/chem/General_Chemistry/Map%3A_Chemistry_and_Chemical_Reactivity_(Kotz_et_al.)/25%3A_Nuclear_Chemistry/25.9%3A__Applications_of_Nuclear_Chemistry.txt |
• 1.1: The Scientific Method
Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles
• 1.2: Properties of Matter
• 1.3: Classification of Matter
Matter is anything that occupies space and has mass. The basic building block of matter is the atom, the smallest unit of an element that can enter into combinations with atoms of the same or other elements. In many substances, atoms are combined into molecules. On earth, matter commonly exists in three states: solids, of fixed shape and volume; liquids, of variable shape but fixed volume; and gases, of variable shape and volume.
• 1.4: Measurement of Matter - SI (Metric) Units
• 1.5: Density and Percent Composition - Their Use in Problem Solving
Density and percent composition are important properties in chemistry. Each have basic components as well as broad applications. Components of density are: mass and volume, both of which can be more confusing than at first glance. An application of the concept of density is determining the volume of an irregular shape using a known mass and density. Determining Percent Composition requires knowing the mass of entire object or molecule and the mass of its components.
• 1.6: Uncertainties in Scientific Measurements
• 1.7: Significant Figures
• 1.E: Exercises
01: Matter- Its Properties And Measurement
Learning Objectives
• To identify the components of the scientific method
Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure \(1\)).
Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21 °C, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolve in 100 grams of water at 20 °C. An example of a quantitative observation was the initial observation leading to the modern theory of the dinosaurs’ extinction: iridium concentrations in sediments dating to 66 million years ago were found to be 20–160 times higher than normal. The development of this theory is a good exemplar of the scientific method in action (see Figure \(2\) below).
After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses:
1. Earth rotates on its axis every 24 hours, alternately exposing one side to the sun, or
2. The sun revolves around Earth every 24 hours.
Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists collected additional data that either support or refute it.
After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. For example, in the dinosaur extinction scenario, iridium concentrations were measured worldwide and compared. A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. Experiments often demonstrate that the hypothesis is incorrect or that it must be modified. More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why.
One example of a law, the Law of Definite Proportions, which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Some solid compounds do not strictly obey the law of definite proportions. The law of definite proportions should seem obvious—we would expect the composition of sodium chloride to be consistent—but the head of the US Patent Office did not accept it as a fact until the early 20th century.
Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. The theory developed to explain the extinction of the dinosaurs, for example, is that Earth occasionally encounters small- to medium-sized asteroids, and these encounters may have unfortunate implications for the continued existence of most species. This theory is by no means proven, but it is consistent with the bulk of evidence amassed to date. Figure \(2\) summarizes the application of the scientific method in this case.
Example \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Ice always floats on liquid water.
2. Birds evolved from dinosaurs.
3. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly.
4. When 10 g of ice were added to 100 mL of water at 25 °C, the temperature of the water decreased to 15.5 °C after the ice melted.
5. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised.
Given: components of the scientific method
Asked for: statement classification
Strategy: Refer to the definitions in this section to determine which category best describes each statement.
Solution
1. This is a general statement of a relationship between the properties of liquid and solid water, so it is a law.
2. This is a possible explanation for the origin of birds, so it is a hypothesis.
3. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory.
4. The temperature is measured before and after a change is made in a system, so these are quantitative observations.
5. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment.
Exercise \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.”
2. Heat always flows from hot objects to cooler ones, not in the opposite direction.
3. The universe was formed by a massive explosion that propelled matter into a vacuum.
4. Michael Jordan is the greatest pure shooter ever to play professional basketball.
5. Limestone is relatively insoluble in water but dissolves readily in dilute acid with the evolution of a gas.
6. Gas mixtures that contain more than 4% hydrogen in air are potentially explosive.
Answer a
experiment
Answer b
law
Answer c
theory
Answer d
hypothesis
Answer e
qualitative observation
Answer f
quantitative observation
Because scientists can enter the cycle shown in Figure \(1\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations.
It is important to remember that scientists have a tendency to formulate hypotheses in familiar terms simply because it is difficult to propose something that has never been encountered or imagined before. As a result, scientists sometimes discount or overlook unexpected findings that disagree with the basic assumptions behind the hypothesis or theory being tested. Fortunately, truly important findings are immediately subject to independent verification by scientists in other laboratories, so science is a self-correcting discipline. When the Alvarezes originally suggested that an extraterrestrial impact caused the extinction of the dinosaurs, the response was almost universal skepticism and scorn. In only 20 years, however, the persuasive nature of the evidence overcame the skepticism of many scientists, and their initial hypothesis has now evolved into a theory that has revolutionized paleontology and geology.
Summary
Chemists expand their knowledge by making observations, carrying out experiments, and testing hypotheses to develop laws to summarize their results and theories to explain them. In doing so, they are using the scientific method.
Fundamental Definitions in Chemistry: https://youtu.be/SBwjbkFNkdw | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.1%3A_The_Scientific_Method.txt |
Learning Objectives
• To separate physical from chemical properties and changes
All matter has physical and chemical properties. Physical properties are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical properties describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property).
Physical properties can be extensive or intensive. Extensive properties vary with the amount of the substance and include mass, weight, and volume. Intensive properties, in contrast, do not depend on the amount of the substance; they include color, melting point, boiling point, electrical conductivity, and physical state at a given temperature. For example, elemental sulfur is a yellow crystalline solid that does not conduct electricity and has a melting point of 115.2°C, no matter what amount is examined (Figure $1$). Scientists commonly measure intensive properties to determine a substance’s identity, whereas extensive properties convey information about the amount of the substance in a sample.
Although mass and volume are both extensive properties, their ratio is an important intensive property called density ($\rho$). Density is defined as mass per unit volume and is usually expressed in grams per cubic centimeter (g/cm3). As mass increases in a given volume, density also increases. For example, lead, with its greater mass, has a far greater density than the same volume of air, just as a brick has a greater density than the same volume of Styrofoam. At a given temperature and pressure, the density of a pure substance is a constant:
$density ={mass \over volume} \rightarrow \rho ={m \over v} \label{1.2.1}$
Pure water, for example, has a density of 0.998 g/cm3 at 25°C. The average densities of some common substances are in Table $1$. Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.”
Table $1$: Densities of Common Substances
Substance Density at 25°C (g/cm3)
blood 1.035
body fat 0.918
whole milk 1.030
corn oil 0.922
mayonnaise 0.910
honey 1.420
Physical Property and Change
Change in which the matter's physical appearance is altered, but composition remains unchanged, e.g., a change in state of matter. The three main states of matter are: Solid, Liquid, Gas
• Solid is distinguished by a fixed structure. Its shape and volume do not change. In a solid, atoms are tightly packed together in a fixed arrangement.
• Liquid is distinguished by its malleable shape (is able to form into the shape of its container), but constant volume. In a liquid, atoms are close together but not in a fixed arrangement.
• Gas is made up of atoms that are separate. However, unlike solid & liquid, a gas has no fixed shape and volume.
Different Definitions of Properties: https://youtu.be/n7UwjQJGh9Y
Example $1$: Physical Change
When liquid water ($H_2O$) freezes into a solid state (ice), it appears changed; However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen and 88.81% oxygen by mass.
Chemical Properties and Change
• Chemical Property is Any characteristic that gives a sample of matter the ability/inability to undergo a change that alters its composition. Examples: Alkali metals react with water; Paper's ability to burn.
• Chemical Change is a Change in which one or more kinds of matter are transformed to new kinds of matter with altered compositions (or Chemical Reaction).
Example $2$: Chemical Change
The combustion of magnetisum metal is a chemical change (Magnesium + Oxygen → Magnesium Oxide):
$2 Mg + O_2 \rightarrow 2 MgO$
The rusting of iron is a chemical change (Iron + Oxygen → Iron Oxide/ Rust):
$4 Fe + 3O_2 \rightarrow 2 Fe_2O_3$
Using the components of composition and properties, we have the ability to distinguish one sample of matter from the others.
Different Definitions of Changes: https://youtu.be/OiLaMHigCuo
Contributors and Attributions
• Samantha Ma (UC Davis) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.2%3A__Properties_of_Matter.txt |
Learning Objectives
• To classify matter.
Chemists study the structures, physical properties, and chemical properties of material substances. These consist of matter, which is anything that occupies space and has mass. Gold and iridium are matter, as are peanuts, people, and postage stamps. Smoke, smog, and laughing gas are matter. Energy, light, and sound, however, are not matter; ideas and emotions are also not matter.
The mass of an object is the quantity of matter it contains. Do not confuse an object’s mass with its weight, which is a force caused by the gravitational attraction that operates on the object. Mass is a fundamental property of an object that does not depend on its location.In physical terms, the mass of an object is directly proportional to the force required to change its speed or direction. A more detailed discussion of the differences between weight and mass and the units used to measure them is included in Essential Skills 1 (Section 1.9). Weight, on the other hand, depends on the location of an object. An astronaut whose mass is 95 kg weighs about 210 lb on Earth but only about 35 lb on the moon because the gravitational force he or she experiences on the moon is approximately one-sixth the force experienced on Earth. For practical purposes, weight and mass are often used interchangeably in laboratories. Because the force of gravity is considered to be the same everywhere on Earth’s surface, 2.2 lb (a weight) equals 1.0 kg (a mass), regardless of the location of the laboratory on Earth.
Under normal conditions, there are three distinct states of matter: solids, liquids, and gases. Solids are relatively rigid and have fixed shapes and volumes. A rock, for example, is a solid. In contrast, liquids have fixed volumes but flow to assume the shape of their containers, such as a beverage in a can. Gases, such as air in an automobile tire, have neither fixed shapes nor fixed volumes and expand to completely fill their containers. Whereas the volume of gases strongly depends on their temperature and pressure (the amount of force exerted on a given area), the volumes of liquids and solids are virtually independent of temperature and pressure. Matter can often change from one physical state to another in a process called a physical change. For example, liquid water can be heated to form a gas called steam, or steam can be cooled to form liquid water. However, such changes of state do not affect the chemical composition of the substance.
Pure Substances and Mixtures
A pure chemical substance is any matter that has a fixed chemical composition and characteristic properties. Oxygen, for example, is a pure chemical substance that is a colorless, odorless gas at 25°C. Very few samples of matter consist of pure substances; instead, most are mixtures, which are combinations of two or more pure substances in variable proportions in which the individual substances retain their identity. Air, tap water, milk, blue cheese, bread, and dirt are all mixtures. If all portions of a material are in the same state, have no visible boundaries, and are uniform throughout, then the material is homogeneous. Examples of homogeneous mixtures are the air we breathe and the tap water we drink. Homogeneous mixtures are also called solutions. Thus air is a solution of nitrogen, oxygen, water vapor, carbon dioxide, and several other gases; tap water is a solution of small amounts of several substances in water. The specific compositions of both of these solutions are not fixed, however, but depend on both source and location; for example, the composition of tap water in Boise, Idaho, is not the same as the composition of tap water in Buffalo, New York. Although most solutions we encounter are liquid, solutions can also be solid. The gray substance still used by some dentists to fill tooth cavities is a complex solid solution that contains 50% mercury and 50% of a powder that contains mostly silver, tin, and copper, with small amounts of zinc and mercury. Solid solutions of two or more metals are commonly called alloys.
If the composition of a material is not completely uniform, then it is heterogeneous (e.g., chocolate chip cookie dough, blue cheese, and dirt). Mixtures that appear to be homogeneous are often found to be heterogeneous after microscopic examination. Milk, for example, appears to be homogeneous, but when examined under a microscope, it clearly consists of tiny globules of fat and protein dispersed in water. The components of heterogeneous mixtures can usually be separated by simple means. Solid-liquid mixtures such as sand in water or tea leaves in tea are readily separated by filtration, which consists of passing the mixture through a barrier, such as a strainer, with holes or pores that are smaller than the solid particles. In principle, mixtures of two or more solids, such as sugar and salt, can be separated by microscopic inspection and sorting. More complex operations are usually necessary, though, such as when separating gold nuggets from river gravel by panning. First solid material is filtered from river water; then the solids are separated by inspection. If gold is embedded in rock, it may have to be isolated using chemical methods.
Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. Distillation makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. A simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask.
Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States.
Crystallization separates mixtures based on differences in solubility, a measure of how much solid substance remains dissolved in a given amount of a specified liquid. Most substances are more soluble at higher temperatures, so a mixture of two or more substances can be dissolved at an elevated temperature and then allowed to cool slowly. Alternatively, the liquid, called the solvent, may be allowed to evaporate. In either case, the least soluble of the dissolved substances, the one that is least likely to remain in solution, usually forms crystals first, and these crystals can be removed from the remaining solution by filtration.
Most mixtures can be separated into pure substances, which may be either elements or compounds. An element, such as gray, metallic sodium, is a substance that cannot be broken down into simpler ones by chemical changes; a compound, such as white, crystalline sodium chloride, contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed. With only a few exceptions, a particular compound has the same elemental composition (the same elements in the same proportions) regardless of its source or history. The chemical composition of a substance is altered in a process called a chemical change. The conversion of two or more elements, such as sodium and chlorine, to a chemical compound, sodium chloride, is an example of a chemical change, often called a chemical reaction. Currently, about 115 elements are known, but millions of chemical compounds have been prepared from these 115 elements. The known elements are listed in the periodic table.
In general, a reverse chemical process breaks down compounds into their elements. For example, water (a compound) can be decomposed into hydrogen and oxygen (both elements) by a process calledelectrolysis. In electrolysis, electricity provides the energy needed to separate a compound into its constituent elements (Figure \(5\)). A similar technique is used on a vast scale to obtain pure aluminum, an element, from its ores, which are mixtures of compounds. Because a great deal of energy is required for electrolysis, the cost of electricity is by far the greatest expense incurred in manufacturing pure aluminum. Thus recycling aluminum is both cost-effective and ecologically sound. The overall organization of matter and the methods used to separate mixtures are summarized in Figure \(6\).
Example \(1\)
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. filtered tea
2. freshly squeezed orange juice
3. a compact disc
4. aluminum oxide, a white powder that contains a 2:3 ratio of aluminum and oxygen atoms
5. selenium
Given: a chemical substance
Asked for: its classification
Strategy:
1. Decide whether a substance is chemically pure. If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound.
2. If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.
Solution
1. A Tea is a solution of compounds in water, so it is not chemically pure. It is usually separated from tea leaves by filtration. B Because the composition of the solution is uniform throughout, it is a homogeneous mixture.
2. A Orange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure. B Because its composition is not uniform throughout, orange juice is a heterogeneous mixture.
3. A A compact disc is a solid material that contains more than one element, with regions of different compositions visible along its edge. Hence a compact disc is not chemically pure. B The regions of different composition indicate that a compact disc is a heterogeneous mixture.
4. A Aluminum oxide is a single, chemically pure compound.
5. A Selenium is one of the known elements.
Exercise \(1\)
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. white wine
2. mercury
3. ranch-style salad dressing
4. table sugar (sucrose)
Answer
1. solution
2. element
3. heterogeneous mixture
4. compound
Different Definitions of Matter: https://youtu.be/qi_qLHc8wLk
Summary
Matter can be classified according to physical and chemical properties. Matter is anything that occupies space and has mass. The three states of matter are solid, liquid, and gas. A physical change involves the conversion of a substance from one state of matter to another, without changing its chemical composition. Most matter consists of mixtures of pure substances, which can be homogeneous (uniform in composition) or heterogeneous (different regions possess different compositions and properties). Pure substances can be either chemical compounds or elements. Compounds can be broken down into elements by chemical reactions, but elements cannot be separated into simpler substances by chemical means. The properties of substances can be classified as either physical or chemical. Scientists can observe physical properties without changing the composition of the substance, whereas chemical properties describe the tendency of a substance to undergo chemical changes (chemical reactions) that change its chemical composition. Physical properties can be intensive or extensive. Intensive properties are the same for all samples; do not depend on sample size; and include, for example, color, physical state, and melting and boiling points. Extensive properties depend on the amount of material and include mass and volume. The ratio of two extensive properties, mass and volume, is an important intensive property called density.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.3%3A_Classification_of_Matter.txt |
Learning Objectives
• To identify the basic units of measurement of the seven fundamental properties
• Describe the names and abbreviations of the SI base units and the SI decimal prefixes.
• Define the liter and the metric ton in these units.
• Explain the meaning and use of unit dimensions; state the dimensions of volume.
• State the quantities that are needed to define a temperature scale, and show how these apply to the Celsius, Kelvin, and Fahrenheit temperature scales.
• Explain how a Torricellian barometer works.
Have you ever estimated a distance by “stepping it off”— that is, by counting the number of steps required to take you a certain distance? Or perhaps you have used the width of your hand, or the distance from your elbow to a fingertip to compare two dimensions. If so, you have engaged in what is probably the first kind of measurement ever undertaken by primitive mankind. The results of a measurement are always expressed on some kind of a scale that is defined in terms of a particular kind of unit. The first scales of distance were likely related to the human body, either directly (the length of a limb) or indirectly (the distance a man could walk in a day).
As civilization developed, a wide variety of measuring scales came into existence, many for the same quantity (such as length), but adapted to particular activities or trades. Eventually, it became apparent that in order for trade and commerce to be possible, these scales had to be defined in terms of standards that would allow measures to be verified, and, when expressed in different units (bushels and pecks, for example), to be correlated or converted.
History of Units
Over the centuries, hundreds of measurement units and scales have developed in the many civilizations that achieved some literate means of recording them. Some, such as those used by the Aztecs, fell out of use and were largely forgotten as these civilizations died out. Other units, such as the various systems of measurement that developed in England, achieved prominence through extension of the Empire and widespread trade; many of these were confined to specific trades or industries. The examples shown here are only some of those that have been used to measure length or distance. The history of measuring units provides a fascinating reflection on the history of industrial development.
The most influential event in the history of measurement was undoubtedly the French Revolution and the Age of Rationality that followed. This led directly to the metric system that attempted to do away with the confusing multiplicity of measurement scales by reducing them to a few fundamental ones that could be combined in order to express any kind of quantity. The metric system spread rapidly over much of the world, and eventually even to England and the rest of the U.K. when that country established closer economic ties with Europe in the latter part of the 20th Century. The United States is presently the only major country in which “metrication” has made little progress within its own society, probably because of its relative geographical isolation and its vibrant internal economy.
Science, being a truly international endeavor, adopted metric measurement very early on; engineering and related technologies have been slower to make this change, but are gradually doing so. Even the within the metric system, however, a variety of units were employed to measure the same fundamental quantity; for example, energy could be expressed within the metric system in units of ergs, electron-volts, joules, and two kinds of calories. This led, in the mid-1960s, to the adoption of a more basic set of units, the Systeme Internationale (SI) units that are now recognized as the standard for science and, increasingly, for technology of all kinds.
The Seven SI Base Units and Decimal Prefixes
In principle, any physical quantity can be expressed in terms of only seven base units (Table $1$), with each base unit defined by a standard described in the NIST Web site.
Table $1$: The Seven Base Units
Property Unit Symbol
length meter m
mass kilogram kg
time second s
temperature (absolute) kelvin K
amount of substance mole mol
electric current ampere A
luminous intensity candela cd
A few special points about some of these units are worth noting:
• The base unit of mass is unique in that a decimal prefix (Table $2$) is built into it; i.e., the base SI unit is not the gram.
• The base unit of time is the only one that is not metric. Numerous attempts to make it so have never garnered any success; we are still stuck with the 24:60:60 system that we inherited from ancient times. The ancient Egyptians of around 1500 BC invented the 12-hour day, and the 60:60 part is a remnant of the base-60 system that the Sumerians used for their astronomical calculations around 100 BC.
• Of special interest to Chemistry is the mole, the base unit for expressing the quantity of matter. Although the number is not explicitly mentioned in the official definition, chemists define the mole as Avogadro’s number (approximately 6.02x1023) of anything.
Owing to the wide range of values that quantities can have, it has long been the practice to employ prefixes such as milli and mega to indicate decimal fractions and multiples of metric units. As part of the SI standard, this system has been extended and formalized (Table $2$).
Table $2$: Prefixes used to scale up or down base units
Prefix Abbreviation Multiplier Prefix Abbreviation Multiplier
peta P 1018 deci s 10–1
tera T 1012 centi c 10–2
giga G 109 milli m 10–3
mega M 106 micro μ 10–6
kilo k 103 nano n 10–9
hecto h 102 pico p 10–12
deca da 10 femto f 10–15
Pseudo-Si Units
There is a category of units that are “honorary” members of the SI in the sense that it is acceptable to use them along with the base units defined above. These include such mundane units as the hour, minute, and degree (of angle), etc., but the three shown here are of particular interest to chemistry, and you will need to know them.
liter (litre) L 1 L = 1 dm3 = 10–3 m3
metric ton t 1 t = 103 kg
united atomic mass unit (amu) u 1 u = 1.66054×10–27 kg
Derived Units and Dimensions
Most of the physical quantities we actually deal with in science and also in our daily lives, have units of their own: volume, pressure, energy and electrical resistance are only a few of hundreds of possible examples. It is important to understand, however, that all of these can be expressed in terms of the SI base units; they are consequently known as derived units. In fact, most physical quantities can be expressed in terms of one or more of the following five fundamental units:
• mass (M)
• length (L)
• time (T)
• electric charge (Q)
• temperature (Θ theta)
Consider, for example, the unit of volume, which we denote as V. To measure the volume of a rectangular box, we need to multiply the lengths as measured along the three coordinates:
$V = x · y · z$
We say, therefore, that volume has the dimensions of length-cubed:
$dim\{V\} = L^3$
Thus the units of volume will be m3 (in the SI) or cm3, ft3 (English), etc. Moreover, any formula that calculates a volume must contain within it the L3 dimension; thus the volume of a sphere is $4/3 πr^3$. The dimensions of a unit are the powers which M, L, t, Q and Q must be given in order to express the unit. Thus,
$dim\{V\} = M^0L^3T^0Q^0 Θ^0$
as given above.
There are several reasons why it is worthwhile to consider the dimensions of a unit.
1. Perhaps the most important use of dimensions is to help us understand the relations between various units of measure and thereby get a better understanding of their physical meaning. For example, a look at the dimensions of the frequently confused electrical terms resistance and resistivity should enable you to explain, in plain words, the difference between them.
2. By the same token, the dimensions essentially tell you how to calculate any of these quantities, using whatever specific units you wish. (Note here the distinction between dimensions and units.)
3. Just as you cannot add apples to oranges, an expression such as $a = b + cx^2$ is meaningless unless the dimensions of each side are identical. (Of course, the two sides should work out to the same units as well.)
4. Many quantities must be dimensionless— for example, the variable x in expressions such as $\log x$, $e^x$, and $\sin x$. Checking through the dimensions of such a quantity can help avoid errors.
The formal, detailed study of dimensions is known as dimensional analysis and is a topic in any basic physics course.
Example $1$
Find the dimensions of energy.
Solution
When mechanical work is performed on a body, its energy increases by the amount of work done, so the two quantities are equivalent and we can concentrate on work. The latter is the product of the force applied to the object and the distance it is displaced. From Newton’s law, force is the product of mass and acceleration, and the latter is the rate of change of velocity, typically expressed in meters per second per second. Combining these quantities and their dimensions yields the result shown in Table $1$.
Table $3$: Dimensions of units commonly used in Chemistry
Q
M
L
t quantity SI unit, other typical units
1 electric charge coulomb
1 mass kilogram, gram, metric ton, pound
1 length meter, foot, mile
1 time second, day, year
3 volume liter, cm3, quart, fluidounce
1 –3 density kg m–3, g cm–3
1 1 –2 force newton, dyne
1 –1 –2 pressure pascal, atmosphere, torr
1 2 –2 energy joule, erg, calorie, electron-volt
1 2 –3 power watt
1 1 2 –2 electric potential volt
1 –1 electric current ampere
1 1 1 –2 electric field intensity volt m–1
–2 1 2 –1 electric resistance ohm
2 1 3 –1 electric resistivity
2 –1 –2 1 electric conductance siemens, mho
Dimensional analysis is widely employed when it is necessary to convert one kind of unit into another, and chemistry students often use it in "chemical arithmetic" calculations, in which context it is also known as the "Factor-Label" method. In this section, we will look at some of the quantities that are widely encountered in Chemistry, and at the units in which they are commonly expressed. In doing so, we will also consider the actual range of values these quantities can assume, both in nature in general, and also within the subset of nature that chemistry normally addresses. In looking over the various units of measure, it is interesting to note that their unit values are set close to those encountered in everyday human experience
Mass is not weight
These two quantities are widely confused. Although they are often used synonymously in informal speech and writing, they have different dimensions: weight is the force exerted on a mass by the local gravational field:
$f = m a = m g \label{Eq1}$
where g is the acceleration of gravity. While the nominal value of the latter quantity is 9.80 m s–2 at the Earth’s surface, its exact value varies locally. Because it is a force, the SI unit of weight is properly the newton, but it is common practice (except in physics classes!) to use the terms "weight" and "mass" interchangeably, so the units kilograms and grams are acceptable in almost all ordinary laboratory contexts.
Please note that in this diagram and in those that follow, the numeric scale represents the logarithm of the number shown. For example, the mass of the electron is 10–30 kg.
The range of masses spans 90 orders of magnitude, more than any other unit. The range that chemistry ordinarily deals with has greatly expanded since the days when a microgram was an almost inconceivably small amount of material to handle in the laboratory; this lower limit has now fallen to the atomic level with the development of tools for directly manipulating these particles. The upper level reflects the largest masses that are handled in industrial operations, but in the recently developed fields of geochemistry and enivonmental chemistry, the range can be extended indefinitely. Flows of elements between the various regions of the environment (atmosphere to oceans, for example) are often quoted in teragrams.
Length
Chemists tend to work mostly in the moderately-small part of the distance range. Those who live in the lilliputian world of crystal- and molecular structures and atomic radii find the picometer a convenient currency, but one still sees the older non-SI unit called the Ångstrom used in this context; 1Å = 10–10 m = 100pm. Nanotechnology, the rage of the present era, also resides in this realm. The largest polymeric molecules and colloids define the top end of the particulate range; beyond that, in the normal world of doing things in the lab, the centimeter and occasionally the millimeter commonly rule.
Time
For humans, time moves by the heartbeat; beyond that, it is the motions of our planet that count out the hours, days, and years that eventually define our lifetimes. Beyond the few thousands of years of history behind us, those years-to-the-powers-of-tens that are the fare for such fields as evolutionary biology, geology, and cosmology, cease to convey any real meaning for us. Perhaps this is why so many people are not very inclined to accept their validity.
Most of what actually takes place in the chemist’s test tube operates on a far shorter time scale, although there is no limit to how slow a reaction can be; the upper limits of those we can directly study in the lab are in part determined by how long a graduate student can wait around before moving on to gainful employment. Looking at the microscopic world of atoms and molecules themselves, the time scale again shifts us into an unreal world where numbers tend to lose their meaning. You can gain some appreciation of the duration of a nanosecond by noting that this is about how long it takes a beam of light to travel between your two outstretched hands. In a sense, the material foundations of chemistry itself are defined by time: neither a new element nor a molecule can be recognized as such unless it lasts long enough to have its “picture” taken through measurement of its distinguishing properties.
Temperature
Temperature, the measure of thermal intensity, spans the narrowest range of any of the base units of the chemist’s measurement toolbox. The reason for this is tied into temperature’s meaning as a measure of the intensity of thermal kinetic energy. Chemical change occurs when atoms are jostled into new arrangements, and the weakness of these motions brings most chemistry to a halt as absolute zero is approached. At the upper end of the scale, thermal motions become sufficiently vigorous to shake molecules into atoms, and eventually, as in stars, strip off the electrons, leaving an essentially reaction-less gaseous fluid, or plasma, of bare nuclei (ions) and electrons.
The degree is really an increment of temperature, a fixed fraction of the distance between two defined reference points on a temperature scale.
Although rough means of estimating and comparing temperatures have been around since AD 170, the first mercury thermometer and temperature scale were introduced in Holland in 1714 by Gabriel Daniel Fahrenheit. Fahrenheit established three fixed points on his thermometer. Zero degrees was the temperature of an ice, water, and salt mixture, which was about the coldest temperature that could be reproduced in a laboratory of the time. When he omitted salt from the slurry, he reached his second fixed point when the water-ice combination stabilized at "the thirty-second degree." His third fixed point was "found at the ninety-sixth degree, and the spirit expands to this degree when the thermometer is held in the mouth or under the armpit of a living man in good health."
After Fahrenheit died in 1736, his thermometer was recalibrated using 212 degrees, the temperature at which water boils, as the upper fixed point. Normal human body temperature registered 98.6 rather than 96. In 1743, the Swedish astronomer Anders Celsius devised the aptly-named centigrade scale that places exactly 100 degrees between the two reference points defined by the freezing and boiling points of water.
When we say that the temperature is so many degrees, we must specify the particular scale on which we are expressing that temperature. A temperature scale has two defining characteristics, both of which can be chosen arbitrarily:
• The temperature that corresponds to 0° on the scale;
• The magnitude of the unit increment of temperature– that is, the size of the degree.
To express a temperature given on one scale in terms of another, it is necessary to take both of these factors into account. The key to temperature conversions is easy if you bear in mind that between the so-called ice- and steam-points of water there are 180 Fahrenheit degrees, but only 100 Celsius degrees, making the F° 100/180 = 5/9 the magnitude of the C°. Note the distinction between “°C” (a temperature) and “C°” (a temperature increment). Because the ice point is at 32°F, the two scales are offset by this amount. If you remember this, there is no need to memorize a conversion formula; you can work it out whenever you need it.
Near the end of the 19th Century when the physical significance of temperature began to be understood, the need was felt for a temperature scale whose zero really means zero— that is, the complete absence of thermal motion. This gave rise to the absolute temperature scale whose zero point is –273.15 °C, but which retains the same degree magnitude as the Celsius scale. This eventually got renamed after Lord Kelvin (William Thompson); thus the Celsius degree became the kelvin. Thus we can now express an increment such as five C° as “five kelvins”
The "other" Absolute Scale
In 1859 the Scottish engineer and physicist William J. M. Rankine proposed an absolute temperature scale based on the Fahrenheit degree. Absolute zero (0° Ra) corresponds to –459.67°F. The Rankine scale has been used extensively by those same American and English engineers who delight in expressing heat capacities in units of BTUs per pound per F°.
The importance of absolute temperature scales is that absolute temperatures can be entered directly in all the fundamental formulas of physics and chemistry in which temperature is a variable.
Pressure
Pressure is the measure of the force exerted on a unit area of surface. Its SI units are therefore newtons per square meter, but we make such frequent use of pressure that a derived SI unit, the pascal, is commonly used:
$1\; Pa = 1\; N \;m^{–2}$
The concept of pressure first developed in connection with studies relating to the atmosphere and vacuum that were carried out in the 17th century.
Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car below. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.
The molecules of a gas are in a state of constant thermal motion, moving in straight lines until experiencing a collision that exchanges momentum between pairs of molecules and sends them bouncing off in other directions. This leads to a completely random distribution of the molecular velocities both in speed and direction— or it would in the absence of the Earth’s gravitational field which exerts a tiny downward force on each molecule, giving motions in that direction a very slight advantage. In an ordinary container this effect is too small to be noticeable, but in a very tall column of air the effect adds up: the molecules in each vertical layer experience more downward-directed hits from those above it. The resulting force is quickly randomized, resulting in an increased pressure in that layer which is then propagated downward into the layers below.
At sea level, the total mass of the sea of air pressing down on each 1-cm2 of surface is about 1034 g, or 10340 kg m–2. The force (weight) that the Earth’s gravitional acceleration g exerts on this mass is
$f = ma = mg = (10340 \;kg)(9.81\; m\; s^{–2}) = 1.013 \times 10^5 \;kg \;m \;s^{–2} = 1.013 \times 10^5\; N$
resulting in a pressure of 1.013 × 105 n m–2 = 1.013 × 105 Pa. The actual pressure at sea level varies with atmospheric conditions, so it is customary to define standard atmospheric pressure as 1 atm = 1.01325 x 105 Pa or 101.325 kPa. Although the standard atmosphere is not an SI unit, it is still widely employed. In meteorology, the bar, exactly 1.000 × 105 = 0.967 atm, is often used.
The Barometer
In the early 17th century, the Italian physicist and mathematician Evangalisto Torricelli invented a device to measure atmospheric pressure. The Torricellian barometer consists of a vertical glass tube closed at the top and open at the bottom. It is filled with a liquid, traditionally mercury, and is then inverted, with its open end immersed in the container of the same liquid. The liquid level in the tube will fall under its own weight until the downward force is balanced by the vertical force transmitted hydrostatically to the column by the downward force of the atmosphere acting on the liquid surface in the open container. Torricelli was also the first to recognize that the space above the mercury constituted a vacuum, and is credited with being the first to create a vacuum.
One standard atmosphere will support a column of mercury that is 760 mm high, so the “millimeter of mercury”, now more commonly known as the torr, has long been a common pressure unit in the sciences: 1 atm = 760 torr.
Summary
The natural sciences begin with observation, and this usually involves numerical measurements of quantities such as length, volume, density, and temperature. Most of these quantities have units of some kind associated with them, and these units must be retained when you use them in calculations. Measuring units can be defined in terms of a very small number of fundamental ones that, through "dimensional analysis", provide insight into their derivation and meaning, and must be understood when converting between different unit systems. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.4%3A_Measurement_of_Matter_-_SI_%28Metric%29_Units.txt |
Learning Objectives
• To be introduced to the concepts of Density and Percent Composition as important properties of matter.
Density and percent composition are important properties in chemistry. Each have basic components as well as broad applications. Components of density are: mass and volume, both of which can be more confusing than at first glance. An application of the concept of density is determining the volume of an irregular shape using a known mass and density. Determining Percent Composition requires knowing the mass of entire object or molecule and the mass of its components.
Introduction
Which one weighs more, a kilogram of feathers or a kilogram of bricks? Though many people will say that a kilogram of bricks is heavier, they actually weigh the same! However, many people are caught up by the concept of density $\rho$, which causes them to answer the question incorrectly. A kilogram of feathers clearly takes up more space, but this is because it is less "dense." But what is density, and how can we determine it?
In the laboratory, density can be used to identify an element, while percent composition is used to determine the amount, by mass, of each element present in a chemical compound. In daily life, density explains everything from why boats float to why air bubbles will try to escape from soda. It even affects your health because bone density is very important. Similarly, percent composition is commonly used to make animal feed and compounds such as the baking soda found in your kitchen.
Density
Density is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary. The equation, as we already know, is as follows:
$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \label{1.5.4}$
or just
$\rho = \dfrac{m}{v} \label{1.5.5}$
Based on this equation, it's clear that density can, and does, vary from element to element and substance to substance due to differences in the relation of mass and volume. But let us break it down one step further. What are mass and volume? We cannot understand density until we know its parts: mass and volume. The following two sections will teach you all the information you need to know about volume and mass to properly solve and manipulate the density equation.
Mass
Mass concerns the quantity of matter in an object. The SI unit for mass is the kilogram (kg), although grams (g) are commonly used in the laboratory to measure smaller quantities. Often, people mistake weight for mass. Weight concerns the force exerted on an object as a function of mass and gravity. This can be written as
$\text{Weight} = \text{mass} \times \text{gravity} \label{1.5.6}$
$Weight = {m}{g} \label{1.5.7}$
Since on the earth, the $g$ in the equation is equal to one, weight and mass are considered equal on earth. It is important to note that although $g$ is equal to one in basic equations, it actually differs throughout earth by a small fraction depending on location; gravity at the equator is less than at the poles. On other astronomical objects, gravity and hence weight, highly differs. This is because weight changes due to variations in gravity and acceleration. The mass, however, of a 1 kg cube will continue to be 1 kg whether it is on the top of a mountain, the bottom of the sea, or on the moon.
Another important difference between mass and weight is how they are measured. Weight is measured with a scale, while mass must be measured with a balance. Just as people confuse mass and weight, they also confuse scales and balances. A balance counteracts the effects of gravity while a scale incorporates it. There are two types of balances found in the laboratory: electronic and manual. With a manual balance, you find the unknown mass of an object by adjusting or comparing known masses until equilibrium is reached. With an electronic balance, which is what you will work with in the UC Davis laboratory, the mass is found by electronic counterbalancing with little effort from the user. An electronic balance can be far more accurate than any other balance and is easier to use, but they are expensive. Also, keep in mind that all instruments used in the laboratory have systematic errors.
Volume
Volume describes the quantity of three dimensional space than an object occupies. The SI unit for volume is meters cubed (m3), but milliliters (mL), centimeters cubed (cm3), and liters (L) are more common in the laboratory. There are many equations to find volume. Here are just a few of the easy ones:
Volume = (length)3
or
Volume = (length)(width)(height)
or
Volume = (base area)(height)
Density: A Further Investigation
Units
We know all of density's components, so let's take a closer look at density itself. The unit most widely used to express density is g/cm3 or g/mL, though the SI unit for density is technically kg/m3. Grams per centimeter cubed is equivalent to grams per milliliter (g/cm3 = g/mL). To solve for density, simply follow the equation 1.5.1. For example, if you had a metal cube with mass 7.0 g and volume 5.0 cm3, the density would be $\rho = \frac{7\; g}{5\;cm^3}= 1.4\; g/cm^3$. Sometimes, you have to convert units to get the correct units for density, such as mg to g or in3 to cm3.
Density in the Periodic Table
Density can be used to help identify an unknown element. Of course, you have to know the density of an element with respect to other elements. Below is a table listing the density of a few elements from the Periodic Table at standard conditions for temperature and pressure, or STP. STP corresponds to a temperature of 273 K (0° Celsius) and 1 atmosphere of pressure.
Table $1$: Density of Elements at standard temperature and pressure
Element Name and Symbol Density (g/cm3) Atomic Number
Hydrogen (H) 0.09 1
Helium (He) 0.18 2
Aluminum (Al) 2.7 13
Zinc (Zn) 7.13 30
Tin (Sn) 7.31 50
Iron (Fe) 7.87 26
Nickel (Ni) 8.9 28
Cobalt (Co) 8.9 27
Copper (Cu) 8.96 29
Silver (Ag) 10.5 47
Lead (Pb) 11.35 82
Mercury (Hg) 11.55 80
Gold (Au) 19.32 79
Platinum (Pt) 21.45 78
Osmium (Os) 22.6 76
As can be seen from the table, the most dense element is Osmium (Os) with a density of 22.6 g/cm3. The least dense element is Hydrogen (H) with a density of 0.09 g/cm3.
Density and Temperature
Density generally decreases with increasing temperature and likewise increases with decreasing temperatures. This is because volume differs according to temperature. Volume increases with increasing temperature. Below is a table showing the density of pure water with differing temperatures.
Table $2$: Density fo water as a function of Temperature
Temperature (C) Density (g/cm3)
100 0.9584
80 0.9718
60 0.9832
40 0.9922
30 0.9957
25 0.997
22 0.9978
20 0.9982
15 0.9991
10 0.9997
4 1.000
0 (liquid) 0.9998
0 (solid) 0.9150
As can be seen from Table $2$, the density of water decreases with increasing temperature. Liquid water also shows an exception to this rule from 0 degrees Celsius to 4 degrees Celsius, where it increases in density instead of decreasing as expected. Looking at the table, you can also see that ice is less dense than water. This is unusual as solids are generally denser than their liquid counterparts. Ice is less dense than water due to hydrogen bonding. In the water molecule, the hydrogen bonds are strong and compact. As the water freezes into the hexagonal crystals of ice, these hydrogen bonds are forced farther apart and the volume increases. With this volume increase comes a decrease in density. This explains why ice floats to the top of a cup of water: the ice is less dense.
Even though the rule of density and temperature has its exceptions, it is still useful. For example, it explains how hot air balloons work.
Density and Pressure
Density increases with increasing pressure because volume decreases as pressure increases. And since density=mass/volume , the lower the volume, the higher the density. This is why all density values in the Periodic Table are recorded at STP, as mentioned in the section "Density and the Periodic Table." The decrease in volume as related to pressure is explained in Boyle's Law: $P_1V_1 = P_2V_2$ where P = pressure and V = volume. This idea is explained in the figure below.
Archimedes' Principle
The Greek scientist Archimedes made a significant discovery in 212 B.C. The story goes that Archimedes was asked to find out for the King if his goldsmith was cheating him by replacing his gold for the crown with silver, a cheaper metal. Archimedes did not know how to find the volume of an irregularly shaped object such as the crown, even though he knew he could distinguish between elements by their density. While meditating on this puzzle in a bath, Archimedes recognized that when he entered the bath, the water rose. He then realized that he could use a similar process to determine the density of the crown! He then supposedly ran through the streets naked shouting "Eureka," which means "I found it!" in Latin.
Archimedes then tested the king's crown by taking a genuine gold crown of equal mass and comparing the densities of the two. The king's crown displaced more water than the gold crown of the same mass, meaning that the king's crown had a greater volume and thus had a smaller density than the real gold crown. The king's "gold" crown, therefore, was not made of pure gold. Of course, this tale is disputed today because Archimedes was not precise in all his measurements, which would make it hard to determine accurately the differences between the two crowns.
Archimedes' Principle states that if an object has a greater density than the liquid that it is placed into, it will sink and displace a volume of liquid equal to its own. If it has a smaller density, it will float and displace a mass of liquid equal to its own. If the density is equal, it will not sink or float (Figure $2$). The principle explains why balloons filled with helium float. Balloons, as we learned in the section concerning density and temperature, float because they are less dense than the surrounding air. Helium is less dense than the atmospheric air, so it rises. Archimedes' Principle can also be used to explain why boats float. Boats, including all the air space, within their hulls, are far less dense than water. Boats made of steel can float because they displace their mass in water without submerging all the way. The table below gives the densities of a few liquids to put things into perspective.
Liquid
Density in kg/m3
Density in g/cm3
Table $3$: Density of Common Liquids
2-Methoxyethanol
964.60
0.9646
Acetic Acid
1049.10
1.049
Acetone
789.86
0.7898
Alcohol, ethyl
785.06
0.7851
Alcohol, methyl
786.51
0.7865
Ammonia
823.35
0.8234
Benzene
873.81
0.8738
Water, pure
1000.00
1.000
Percent Composition
Percent composition is very simple. Percent composition tells you by mass what percent of each element is present in a compound. A chemical compound is the combination of two or more elements. If you are studying a chemical compound, you may want to find the percent composition of a certain element within that chemical compound. The equation for percent composition is
$\text{Percent Composition} = \dfrac{\text{Total mass of element present}}{\text{Molecular mass}} \times 100\% \label{1.5.3}$
If you want to know the percent composition of the elements in an compound, follow these steps:
Steps to Solve:
1. Find the molar mass of all the elements in the compound in grams per mole.
2. Find the molecular mass of the entire compound.
3. Divide the component's molar mass by the entire molecular mass.
4. You will now have a number between 0 and 1. Multiply it by 100 to get percent composition!
Tips for solving:
1. The compounds will always add up to 100%, so in a binary compound, you can find the % of the first element, then do 100%-(% first element) to get (% second element)
2. If using a calculator, you can store the overall molar mass to a variable such as "A". This will speed up calculations, and reduce typographical errors.
These steps are outlined in the figure below.
For another example, if you wanted to know the percent composition of hydrochloric acid (HCl), first find the molar mass of Hydrogen. H = 1.00794g. Now find the molecular mass of HCl: 1.00794g + 35.4527g = 36.46064g. Follow steps 3 and 4: (1.00794g/36.46064g) x 100 = 2.76% Now just subtract to find the percent by mass of Chlorine in the compound: 100%-2.76% = 97.24% Therefore, HCl is 2.76% Hydrogen and 97.24% Chlorine by mass.
Percent Composition in Everyday Life
Percent composition plays an important role in everyday life. It is more than just the amount of chlorine in your swimming pool because it concerns everything from the money in your pocket to your health and how you live. The next two sections describe percent composition as it relates to you.
Example $1$: Nutritional Labels
The nutrition label found on the container of every bit of processed food sold by the local grocery store employs the idea of percent composition. On all nutrition labels, a known serving size is broken down in five categories: Total Fat, Cholesterol, Sodium, Total Carbohydrate, and Protein. These categories are broken down into further subcategories, including Saturated Fat and Dietary Fiber. The mass for each category, except Protein, is then converted to percent of Daily Value. Only two subcategories, Saturated Fat and Dietary Fiber are converted to percent of Daily Value. The Daily Value is based on a the mass of each category recommended per day per person for a 2000 calorie diet. The mass of protein is not converted to percent because their is no recommended daily value for protein. Following is a picture outlining these ideas.
For example, if you wanted to know the percent by mass of the daily value for sodium you are eating when you eat one serving of the food with this nutrition label, then go to the category marked sodium. Look across the same row and read the percent written. If you eat one serving of this food, then you will have consumed about 9% of your daily recommended value for sodium. To find the percent mass of fat in the whole food, you could divide 3.5 grams by 15 grams, and see that this snack is 23.33% fat.
Example $2$: The Lucky Penny
The penny should be called "the lucky copper coated coin." The penny has not been made of solid copper since part of 1857. After 1857, the US government started adding other cheaper metals to the mix. The penny, being only one cent, is literally not worth its weight in copper. People could melt copper pennies and sell the copper for more than the pennies were worth. After 1857, nickel was mixed with the more expensive copper. After 1864, the penny was made of bronze. Bronze is 95% copper and 5% zinc and tin. For one year, 1943, the penny had no copper in it due to the expense of the World War II. It was just zinc coated steel. After 1943 until 1982, the penny went through periods where it was brass or bronze.
The percent composition of a penny may actually affect health, particularly the health of small children and pets. Since the newer pennies are made mainly of zinc instead of copper, they are a danger to a child's health if ingested. Zinc is very susceptible to acid. If the thin copper coating is scratched and the hydrochloric acid present in the stomach comes into contact with the zinc core it could cause ulcers, anemia, kidney and liver damage, or even death in severe cases. Three important factors in penny ingestion are time, pH of the stomach, and amount of pennies ingested. Of course, the more pennies swallowed, the more danger of an overdose of zinc. The more acidic the environment, the more zinc will be released in less time. This zinc is then absorbed and sent to the liver where it begins to cause damage. In this kind of situation, time is of the essence. The faster the penny is removed, the less zinc is absorbed. If the penny or pennies are not removed, organ failure and death can occur.
Below is a picture of a scratched penny before and after it had been submerged in lemon juice. Lemon juice has a similar pH of 1.5-2.5 when compared to the normal human stomach after food has been consumed. Time elapsed: 36 hours.
As you can see, the copper is vastly unharmed by the lemon juice. That's why pennies made before 1982 with mainly copper (except the 1943 penny) are relatively safe to swallow. Chances are they would pass through the digestive system naturally before any damage could be done. Yet, it is clear that the zinc was partially dissolved even though it was in the lemon juice for only a limited amount of time. Therefore, the percent composition of post 1982 pennies is hazardous to your health and the health of your pets if ingested.
Problems
Following are examples of different types of percent composition and density problems.
Density Problems: These problems are meant to be easy in the beginning and then gradually become more challenging. Unless otherwise stated, answers should be in g/mL or the equivalent g/cm3.
1. If you have a 2.130 mL sample of acetic acid with mass 0.002234 kg, what is the density?
2. Calculate the density of a .03020 L sample of ethyl alcohol with a mass of 23.71002 g.
3. Find the density of a sample that has a volume of 36.5 L and a mass of 10.0 kg.
4. Find the volume in mL of an object that has a density of 10.2 g/L and a mass of 30.0 kg.
5. Calculate the mass in grams of an object with a volume of 23.5 mL and density of 10.0 g/L.
6. Calculate the density of a rectangular prism made of metal. The dimensions of the prism are: 5 cm by 4 cm by 5 cm. The metal has a mass of 50 grams.
7. Find the density of an unknown liquid in a beaker. The beaker's mass is 165 g when there is no liquid present. With the unknown liquid, the total mass is 309 g. The volume of the unknown is 125 mL.
8. Determine the density in g/L of an unknown with the following information. A 55 gallon tub weighs 137.5lb when empty and 500.0 lb when filled with the unknown.
9. A ring has a mass of 5.00g and a volume of 0.476 mL. Is it pure silver?
10. What is the density of the solid in the image if the mass is 40 g? Make your answer have 3 significant figures.
1. Below is a model of a pyramid found at an archeological dig made of an unknown substance. It is too large to find the volume by submerging it in water. Also, the scientists refuse to remove a piece to test because this pyramid is a part of history. Its height is 150.0m. The length of its base is 75.0m and the width is 50.0m. The mass of this pyramid is 5.50x105kg. What is the density?
Percent Composition Problems: These problems will follow the same pattern of difficulty as those of density.
1. Calculate the percent by mass of each element in Cesium Fluoride (CsF).
2. Calculate the percent by mass of each element present in carbon tetrachloride (CCl4)
3. A solution of salt and water is 33.0% salt by mass and has a density of 1.50 g/mL. What mass of the salt in grams is in 5.00L of this solution?
4. A solution of water and HCl contains 25% HCl by mass. The density of the solution is 1.05 g/mL. If you need 1.7g of HCl for a reaction, what volume of this solution will you use?
5. A solution containing 42% NaOH by mass has a density of 1.30 g/mL. What mass, in kilograms, of NaOH is in 6.00 L of this solution?
Answers
Here are the solutions to the listed practice problems.
Density Problem Solutions
1. 1.049 g/mL
2. 0.7851 g/mL
3. 0.274 g/mL
4. 2.94 x 106 mL
5. 0.3.27 kg
6. 0.5 g/cm3
7. 1.15 g/mL
8. 790 g/L
9. Yes
10. 0.195 g/cm3
11. 29.3 g/cm3
Percent Composition Problem Solutions
1. CsF is 87.5% Cs and 12.5% F by mass
2. CCl4 is 92.2% Cl and 7.8% C by mass
3. 2480 g
4. 6.5 mL
5. 3.26 kg
References
1. AUTOR , ARQUIMEDES , and Thomas Little . The Works of Archimedes . Courier Dover Publications, 2002.
2. Chande, D. and T. Fisher (2003). "Have a Penny? Need a Penny? Eliminating the One-Cent Coin from Circulation." Canadian Public Policy/Analyse de Politiques 29(4): 511-517.
3. Jefferson, T. (1999). "A Thought for Your Pennies." JAMA 281(2): 122.
4. Petrucci , Ralph , William Harwood , and Geoffrey Herring . Principles and Modern Application. ninth . New Jersey : Peason Eduation , 2007.
5. Rauch, F., H. Plotkin, et al. (2003). "Bone Mass, Size, and Density in Children and Adolescents with Osteogenesis Imperfecta: Effect of Intravenous Pamidronate Therapy." Journal of Bone and Mineral Research 18: 610-614.
6. Richardson, J., S. Gwaltney-Brant, et al. (2002). "Zinc Toxicosis from Penny Ingestion in Dogs." Vet Med 97(2): 96-99.
7. Tate, J. "Archimedes’ Discoveries: A Closer Look." | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.5%3A_Density_and_Percent_Composition_-_Their_Use_in_Problem_Solving.txt |
Learning Objectives
• To introduce the fundamental mathematical skills you will need to complete basic chemistry questions and problems
All measurements have a degree of uncertainty regardless of precision and accuracy. This is caused by two factors, the limitation of the measuring instrument (systematic error) and the skill of the experimenter making the measurements (random error).
Introduction
The graduated buret in Figure $1$ contains a certain amount of water (with yellow dye) to be measured. The amount of water is somewhere between 19 ml and 20 ml according to the marked lines. By checking to see where the bottom of the meniscus lies, referencing the ten smaller lines, the amount of water lies between 19.8 ml and 20 ml. The next step is to estimate the uncertainty between 19.8 ml and 20 ml. Making an approximate guess, the level is less than 20 ml, but greater than 19.8 ml. We then report that the measured amount is approximately 19.9 ml. The graduated cylinder itself may be distorted such that the graduation marks contain inaccuracies providing readings slightly different from the actual volume of liquid present.
Systematic vs. Random Error
Systematic errors: When we use tools meant for measurement, we assume that they are correct and accurate, however measuring tools are not always right. In fact, they have errors that naturally occur called systematic errors. Systematic errors tend to be consistent in magnitude and/or direction. If the magnitude and direction of the error is known, accuracy can be improved by additive or proportional corrections. Additive correction involves adding or subtracting a constant adjustment factor to each measurement; proportional correction involves multiplying the measurement(s) by a constant.
Random errors: Sometimes called human error, random error is determined by the experimenter's skill or ability to perform the experiment and read scientific measurements. These errors are random since the results yielded may be too high or low. Often random error determines the precision of the experiment or limits the precision. For example, if we were to time a revolution of a steadily rotating turnable, the random error would be the reaction time. Our reaction time would vary due to a delay in starting (an underestimate of the actual result) or a delay in stopping (an overestimate of the actual result). Unlike systematic errors, random errors vary in magnitude and direction. It is possible to calculate the average of a set of measured positions, however, and that average is likely to be more accurate than most of the measurements.
Accuracy and Precision
Measurements may be accurate, meaning that the measured value is the same as the true value; they may be precise, meaning that multiple measurements give nearly identical values (i.e., reproducible results); they may be both accurate and precise; or they may be neither accurate nor precise. The goal of scientists is to obtain measured values that are both accurate and precise. Figure $1$ help to understand the difference between precision (small expected difference between multiple measurements) and accuracy (difference between the result and a known value).
Suppose, for example, that the mass of a sample of gold was measured on one balance and found to be 1.896 g. On a different balance, the same sample was found to have a mass of 1.125 g. Which was correct? Careful and repeated measurements, including measurements on a calibrated third balance, showed the sample to have a mass of 1.895 g. The masses obtained from the three balances are in the following table:
Balance 1 Balance 2 Balance 3
1.896 g 1.125 g 1.893 g
1.895 g 1.158 g 1.895 g
1.894 g 1.067 g 1.895 g
Whereas the measurements obtained from balances 1 and 3 are reproducible (precise) and are close to the accepted value (accurate), those obtained from balance 2 are neither. Even if the measurements obtained from balance 2 had been precise (if, for example, they had been 1.125, 1.124, and 1.125), they still would not have been accurate. We can assess the precision of a set of measurements by calculating the average deviation of the measurements as follows:
1. Calculate the average value of all the measurements:
$\text{average} = \dfrac{ \text{sum of measurements} }{ \text{number of measurements}} \label{1.6.1}$
2. Calculate the deviation of each measurement, which is the absolute value of the difference between each measurement and the average value:
$deviation = |\text{measurement − average}| \label{1.6.2}$
where | | means absolute value (i.e., convert any negative number to a positive number).
3. Add all the deviations and divide by the number of measurements to obtain the average deviation:
$\text{average} = \dfrac{\text{sum of measurements} }{\text{number of measurements}} \label{Eq1}$
Then we can express the precision as a percentage by dividing the average deviation by the average value of the measurements and multiplying the result by 100. In the case of balance 2, the average value is
${1.125 \;g + 1.158 \;g + 1.067\; g \over 3} = 1.117 \;g$
The deviations are
• $|1.125\; g − 1.117 \;g| = 0.008\; g$
• $|1.158\; g − 1.117\; g| = 0.041 \:g$, and
• $|1.067\; g − 1.117\; g| = 0.050 \;g$.
So the average deviation is
${0.008 \:g + 0.041 \;g + 0.050 \;g \over 3} = 0.033\; g$
The precision of this set of measurements is therefore
${0.033\;g \over 1.117\;g} \times 100 = 3.0 \%$
When a series of measurements is precise but not accurate, the error is usually systematic. Systematic errors can be caused by faulty instrumentation or faulty technique.
When a series of measurements is precise but not accurate, the error is usually systematic. Systematic errors can be caused by faulty instrumentation or faulty technique.
Example $1$
The following archery targets show marks that represent the results of four sets of measurements. Which target shows
1. a precise but inaccurate set of measurements?
2. an accurate but imprecise set of measurements?
3. a set of measurements that is both precise and accurate?
4. a set of measurements that is neither precise nor accurate?
Solution
1. (B)
2. (a)
3. (c)
4. (d)
Example $2$
1. A 1-carat diamond has a mass of 200.0 mg. When a jeweler repeatedly weighed a 2-carat diamond, he obtained measurements of 450.0 mg, 459.0 mg, and 463.0 mg. Were the jeweler’s measurements accurate? Were they precise?
2. A single copper penny was tested three times to determine its composition. The first analysis gave a composition of 93.2% zinc and 2.8% copper, the second gave 92.9% zinc and 3.1% copper, and the third gave 93.5% zinc and 2.5% copper. The actual composition of the penny was 97.6% zinc and 2.4% copper. Were the results accurate? Were they precise?
Solution
a. The expected mass of a 2-carat diamond is 2 × 200.0 mg = 400.0 mg. The average of the three measurements is 457.3 mg, about 13% greater than the true mass. These measurements are not particularly accurate.
The deviations of the measurements are 7.3 mg, 1.7 mg, and 5.7 mg, respectively, which give an average deviation of 4.9 mg and a precision of
${4.9 mg \over 457.3 mg } \times 100 = 1.1 \% \nonumber$
These measurements are rather precise.
b. The average values of the measurements are 93.2% zinc and 2.8% copper versus the true values of 97.6% zinc and 2.4% copper. Thus these measurements are not very accurate, with errors of −4.5% and + 17% for zinc and copper, respectively. (The sum of the measured zinc and copper contents is only 96.0% rather than 100%, which tells us that either there is a significant error in one or both measurements or some other element is present.)
The deviations of the measurements are 0.0%, 0.3%, and 0.3% for both zinc and copper, which give an average deviation of 0.2% for both metals. We might therefore conclude that the measurements are equally precise, but that is not the case. Recall that precision is the average deviation divided by the average value times 100. Because the average value of the zinc measurements is much greater than the average value of the copper measurements (93.2% versus 2.8%), the copper measurements are much less precise.
$\text {precision (Zn)} = \dfrac {0.2 \%}{93.2 \% } \times 100 = 0.2 \% \nonumber$
$\text {precision (Cu)} = \dfrac {0.2 \%}{2.8 \% } \times 100 = 7 \% \nonumber$
Significant Figures
No measurement is free from error. Error is introduced by (1) the limitations of instruments and measuring devices (such as the size of the divisions on a graduated cylinder) and (2) the imperfection of human senses. Although errors in calculations can be enormous, they do not contribute to uncertainty in measurements. Chemists describe the estimated degree of error in a measurement as the uncertainty of the measurement, and they are careful to report all measured values using only significant figures, numbers that describe the value without exaggerating the degree to which it is known to be accurate. Chemists report as significant all numbers known with absolute certainty, plus one more digit that is understood to contain some uncertainty. The uncertainty in the final digit is usually assumed to be ±1, unless otherwise stated.
The following rules have been developed for counting the number of significant figures in a measurement or calculation:
1. Any nonzero digit is significant.
2. Any zeros between nonzero digits are significant. The number 2005, for example, has four significant figures.
3. Any zeros used as a placeholder preceding the first nonzero digit are not significant. So 0.05 has one significant figure because the zeros are used to indicate the placement of the digit 5. In contrast, 0.050 has two significant figures because the last two digits correspond to the number 50; the last zero is not a placeholder. As an additional example, 5.0 has two significant figures because the zero is used not to place the 5 but to indicate 5.0.
4. When a number does not contain a decimal point, zeros added after a nonzero number may or may not be significant. An example is the number 100, which may be interpreted as having one, two, or three significant figures. (Note: treat all trailing zeros in exercises and problems in this text as significant unless you are specifically told otherwise.)
5. Integers obtained either by counting objects or from definitions are exact numbers, which are considered to have infinitely many significant figures. If we have counted four objects, for example, then the number 4 has an infinite number of significant figures (i.e., it represents 4.000…). Similarly, 1 foot (ft) is defined to contain 12 inches (in), so the number 12 in the following equation has infinitely many significant figures:
$1 ft = 12 in \nonumber$
An effective method for determining the number of significant figures is to convert the measured or calculated value to scientific notation because any zero used as a placeholder is eliminated in the conversion. When 0.0800 is expressed in scientific notation as 8.00 × 10−2, it is more readily apparent that the number has three significant figures rather than five; in scientific notation, the number preceding the exponential (i.e., N) determines the number of significant figures.
Example $3$
Give the number of significant figures in each. Identify the rule for each.
1. 5.87
2. 0.031
3. 52.90
4. 00.2001
5. 500
6. 6 atoms
Solution
1. three (rule 1)
2. two (rule 3); in scientific notation, this number is represented as 3.1 × 10−2, showing that it has two significant figures.
3. four (rule 3)
4. four (rule 2); this number is 2.001 × 10−1 in scientific notation, showing that it has four significant figures.
5. one, two, or three (rule 4)
6. infinite (rule 5)
Example $4$
Which measuring apparatus would you use to deliver 9.7 mL of water as accurately as possible? To how many significant figures can you measure that volume of water with the apparatus you selected?
Solution
Use the 10 mL graduated cylinder, which will be accurate to two significant figures.
Mathematical operations are carried out using all the digits given and then rounding the final result to the correct number of significant figures to obtain a reasonable answer. This method avoids compounding inaccuracies by successively rounding intermediate calculations. After you complete a calculation, you may have to round the last significant figure up or down depending on the value of the digit that follows it. If the digit is 5 or greater, then the number is rounded up. For example, when rounded to three significant figures, 5.215 is 5.22, whereas 5.213 is 5.21. Similarly, to three significant figures, 5.005 kg becomes 5.01 kg, whereas 5.004 kg becomes 5.00 kg. The procedures for dealing with significant figures are different for addition and subtraction versus multiplication and division.
When we add or subtract measured values, the value with the fewest significant figures to the right of the decimal point determines the number of significant figures to the right of the decimal point in the answer. Drawing a vertical line to the right of the column corresponding to the smallest number of significant figures is a simple method of determining the proper number of significant figures for the answer:
3240.7 + 21.2
36
3261.9 36
The line indicates that the digits 3 and 6 are not significant in the answer. These digits are not significant because the values for the corresponding places in the other measurement are unknown (3240.7??). Consequently, the answer is expressed as 3261.9, with five significant figures. Again, numbers greater than or equal to 5 are rounded up. If our second number in the calculation had been 21.256, then we would have rounded 3261.956 to 3262.0 to complete our calculation.
When we multiply or divide measured values, the answer is limited to the smallest number of significant figures in the calculation; thus, 42.9 × 8.323 = 357.057 = 357. Although the second number in the calculation has four significant figures, we are justified in reporting the answer to only three significant figures because the first number in the calculation has only three significant figures. An exception to this rule occurs when multiplying a number by an integer, as in 12.793 × 12. In this case, the number of significant figures in the answer is determined by the number 12.973, because we are in essence adding 12.973 to itself 12 times. The correct answer is therefore 155.516, an increase of one significant figure, not 155.52.
When you use a calculator, it is important to remember that the number shown in the calculator display often shows more digits than can be reported as significant in your answer. When a measurement reported as 5.0 kg is divided by 3.0 L, for example, the display may show 1.666666667 as the answer. We are justified in reporting the answer to only two significant figures, giving 1.7 kg/L as the answer, with the last digit understood to have some uncertainty.
In calculations involving several steps, slightly different answers can be obtained depending on how rounding is handled, specifically whether rounding is performed on intermediate results or postponed until the last step. Rounding to the correct number of significant figures should always be performed at the end of a series of calculations because rounding of intermediate results can sometimes cause the final answer to be significantly in error.
In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate roundings need to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end.
In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step.
Significant Figures: https://youtu.be/E-OAkZglfO8
Problems
Complete the calculations and report your answers using the correct number of significant figures.
1. 87.25 mL + 3.0201 mL
2. 26.843 g + 12.23 g
3. 6 × 12.011
4. 2(1.008) g + 15.99 g
5. 137.3 + 2(35.45)
6. ${118.7 \over 2} g - 35.5 g$
7. $47.23 g - {207.2 \over 5.92 }g$
8. ${77.604 \over 6.467} −4.8$
9. ${24.86 \over 2.0 } - 3.26 (0.98 )$
10. $(15.9994 \times 9) + 2.0158$
Solution
1. 90.27 mL
2. 39.07 g
3. 72.066 (See rule 5 under “Significant Figures.”)
4. 2(1.008) g + 15.99 g = 2.016 g + 15.99 g = 18.01 g
5. 137.3 + 2(35.45) = 137.3 + 70.90 = 208.2
6. 59.35 g − 35.5 g = 23.9 g
7. 47.23 g − 35.0 g = 12.2 g
8. 12.00 − 4.8 = 7.2
9. 12 − 3.2 = 9
10. 143.9946 + 2.0158 = 146.0104 | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.6%3A_Uncertainties_in_Scientific_Measurements.txt |
Learning Objectives
• Give an example of a measurement whose number of significant digits is clearly too great, and explain why.
• State the purpose of rounding off, and describe the information that must be known to do it properly.
• Round off a number to a specified number of significant digits.
• Explain how to round off a number whose second-most-significant digit is 9.
• Carry out a simple calculation that involves two or more observed quantities, and express the result in the appropriate number of significant figures.
The numerical values we deal with in science (and in many other aspects of life) represent measurements whose values are never known exactly. Our pocket-calculators or computers don't know this; they treat the numbers we punch into them as "pure" mathematical entities, with the result that the operations of arithmetic frequently yield answers that are physically ridiculous even though mathematically correct. The purpose of this unit is to help you understand why this happens, and to show you what to do about it.
Digits: Significant and otherwise
Consider the two statements shown below:
• "The population of our city is 157,872."
• "The number of registered voters as of Jan 1 was 27,833.
Which of these would you be justified in dismissing immediately? Certainly not the second one, because it probably comes from a database which contains one record for each voter, so the number is found simply by counting the number of records. The first statement cannot possibly be correct. Even if a city’s population could be defined in a precise way (Permanent residents? Warm bodies?), how can we account for the minute-by minute changes that occur as people are born and die, or move in and move away?
What is the difference between the two population numbers stated above? The first one expresses a quantity that cannot be known exactly — that is, it carries with it a degree of uncertainty. It is quite possible that the last census yielded precisely 157,872 records, and that this might be the “population of the city” for legal purposes, but it is surely not the “true” population. To better reflect this fact, one might list the population (in an atlas, for example) as 157,900 or even 158,000. These two quantities have been rounded off to four and three significant figures, respectively, and the have the following meanings:
• 157900 (the significant digits are underlined here) implies that the population is believed to be within the range of about 157850 to about 157950. In other words, the population is 157900±50. The “plus-or-minus 50” appended to this number means that we consider the absolute uncertainty of the population measurement to be 50 – (–50) = 100. We can also say that the relative uncertainty is 100/157900, which we can also express as 1 part in 1579, or 1/1579 = 0.000633, or about 0.06 percent.
• The value 158000 implies that the population is likely between about 157500 and 158500, or 158000±500. The absolute uncertainty of 1000 translates into a relative uncertainty of 1000/158000 or 1 part in 158, or about 0.6 percent.
Which of these two values we would report as “the population” will depend on the degree of confidence we have in the original census figure; if the census was completed last week, we might round to four significant digits, but if it was a year or so ago, rounding to three places might be a more prudent choice. In a case such as this, there is no really objective way of choosing between the two alternatives.
This illustrates an important point: the concept of significant digits has less to do with mathematics than with our confidence in a measurement. This confidence can often be expressed numerically (for example, the height of a liquid in a measuring tube can be read to ±0.05 cm), but when it cannot, as in our population example, we must depend on our personal experience and judgment.
So, what is a significant digit? According to the usual definition, it is all the numerals in a measured quantity (counting from the left) whose values are considered as known exactly, plus one more whose value could be one more or one less:
• In “157900” (four significant digits), the left most three digits are known exactly, but the fourth digit, “9” could well be “8” if the “true value” is within the implied range of 157850 to 157950.
• In “158000” (three significant digits), the left most two digits are known exactly, while the third digit could be either “7” or “8” if the true value is within the implied range of 157500 to 158500.
Although rounding off always leads to the loss of numeric information, what we are getting rid of can be considered to be “numeric noise” that does not contribute to the quality of the measurement. The purpose in rounding off is to avoid expressing a value to a greater degree of precision than is consistent with the uncertainty in the measurement.
Implied Uncertainty
If you know that a balance is accurate to within 0.1 mg, say, then the uncertainty in any measurement of mass carried out on this balance will be ±0.1 mg. Suppose, however, that you are simply told that an object has a length of 0.42 cm, with no indication of its precision. In this case, all you have to go on is the number of digits contained in the data. Thus the quantity “0.42 cm” is specified to 0.01 unit in 0 42, or one part in 42 . The implied relative uncertainty in this figure is 1/42, or about 2%. The precision of any numeric answer calculated from this value is therefore limited to about the same amount.
Rounding Error
It is important to understand that the number of significant digits in a value provides only a rough indication of its precision, and that information is lost when rounding off occurs. Suppose, for example, that we measure the weight of an object as 3.28 g on a balance believed to be accurate to within ±0.05 gram. The resulting value of 3.28±.05 gram tells us that the true weight of the object could be anywhere between 3.23 g and 3.33 g. The absolute uncertainty here is 0.1 g (±0.05 g), and the relative uncertainty is 1 part in 32.8, or about 3 percent.
How many significant digits should there be in the reported measurement? Since only the left most “3” in “3.28” is certain, you would probably elect to round the value to 3.3 g. So far, so good. But what is someone else supposed to make of this figure when they see it in your report? The value “3.3 g” suggests an implied uncertainty of 3.3±0.05 g, meaning that the true value is likely between 3.25 g and 3.35 g. This range is 0.02 g below that associated with the original measurement, and so rounding off has introduced a bias of this amount into the result. Since this is less than half of the ±0.05 g uncertainty in the weighing, it is not a very serious matter in itself. However, if several values that were rounded in this way are combined in a calculation, the rounding-off errors could become significant.
Rules for Rounding
The standard rules for rounding off are well known. Before we set them out, let us agree on what to call the various components of a numeric value.
• The most significant digit is the left most digit (not counting any leading zeros which function only as placeholders and are never significant digits.)
• If you are rounding off to n significant digits, then the least significant digit is the nth digit from the most significant digit. The least significant digit can be a zero.
• The first non-significant digit is the n+1th digit.
Rounding-off rules
• If the first non-significant digit is less than 5, then the least significant digit remains unchanged.
• If the first non-significant digit is greater than 5, the least significant digit is incremented by 1.
• If the first non-significant digit is 5, the least significant digit can either be incremented or left unchanged (see below!)
• All non-significant digits are removed.
Fantasies about fives
Students are sometimes told to increment the least significant digit by 1 if it is odd, and to leave it unchanged if it is even. One wonders if this reflects some idea that even numbers are somehow “better” than odd ones! (The ancient superstition is just the opposite, that only the odd numbers are "lucky".)
In fact, you could do it equally the other way around, incrementing only the even numbers. If you are only rounding a single number, it doesn't really matter what you do. However, when you are rounding a series of numbers that will be used in a calculation, if you treated each first nonsignificant 5 in the same way, you would be over- or understating the value of the rounded number, thus accumulating round-off error. Since there are equal numbers of even and odd digits, incrementing only the one kind will keep this kind of error from building up. You could do just as well, of course, by flipping a coin!
Table \(1\): Examples of rounding-off
number to round
number of significant digits
result
comment
34.216 3 34.2 First non-significant digit (1) is less than 5, so number is simply truncated.
2.252 2 2.2 or 2.3 First non-significant digit is 5, so least sig. digit can either remain unchanged or be incremented.
39.99 3 40.0 Crossing "decimal boundary", so all numbers change.
85,381 3 85,400 The two zeros are just placeholders
0.04597 3 0.0460 The two leading zeros are not significant digits.
Rounding up the Nines
Suppose that an object is found to have a weight of 3.98 ± 0.05 g. This would place its true weight somewhere in the range of 3.93 g to 4.03 g. In judging how to round this number, you count the number of digits in “3.98” that are known exactly, and you find none! Since the “4” is the left most digit whose value is uncertain, this would imply that the result should be rounded to one significant figure and reported simply as 4 g. An alternative would be to bend the rule and round off to two significant digits, yielding 4.0 g. How can you decide what to do? In a case such as this, you should look at the implied uncertainties in the two values, and compare them with the uncertainty associated with the original measurement.
Table \(2\)
rounded value
implied max
implied min
absolute uncertainty
relative uncertainty
3.98 g 3.985 g 3.975 g ±.005 g or 0.01 g 1 in 400, or 0.25%
4 g 4.5 g 3.5 g ±.5 g or 1 g 1 in 4, 25%
4.0 g 4.05 g 3.95 g ±.05 g or 0.1 g 1 in 40, 2.5%
Clearly, rounding off to two digits is the only reasonable course in this example. Observed values should be rounded off to the number of digits that most accurately conveys the uncertainty in the measurement.
• Usually, this means rounding off to the number of significant digits in in the quantity; that is, the number of digits (counting from the left) that are known exactly, plus one more.
• When this cannot be applied (as in the example above when addition of subtraction of the absolute uncertainty bridges a power of ten), then we round in such a way that the relative implied uncertainty in the result is as close as possible to that of the observed value.
Rounding the Results of Calculations
When carrying out calculations that involve multiple steps, you should avoid doing any rounding until you obtain the final result. Suppose you use your calculator to work out the area of a rectangle:
rounded value
relative implied uncertainty
1.58 1 part in 158, or 0.6%
1.6 1 part in 16, or 6 %
Note
Your calculator is of course correct as far as the pure numbers go, but you would be wrong to write down "1.57676 cm2" as the answer. Two possible options for rounding off the calculator answer are shown at the right.
It is clear that neither option is entirely satisfactory; rounding to 3 significant digits overstates the precision of the answer, whereas following the rule and rounding to the two digits in ".42" has the effect of throwing away some precision. In this case, it could be argued that rounding to three digits is justified because the implied relative uncertainty in the answer, 0.6%, is more consistent with those of the two factors.
The "rules" for rounding off are generally useful, convenient guidelines, but they do not always yield the most desirable result. When in doubt, it is better to rely on relative implied uncertainties.
Addition and Subtraction
In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise that the least precise number used to get the answer. When adding or subtracting, we go by the number of decimal places (i.e., the number of digits on the right side of the decimal point) rather than by the number of significant digits. Identify the quantity having the smallest number of decimal places, and use this number to set the number of decimal places in the answer.
Multiplication and Division
The result must contain the same number of significant figures as in the value having the least number of significant figures.
Logarithms and antilogarithms
If a number is expressed in the form a × 10b ("scientific notation") with the additional restriction that the coefficient a is no less than 1 and less than 10, the number is in its normalized form. Express the base-10 logarithm of a value using the same number of significant figures as is present in the normalized form of that value. Similarly, for antilogarithms (numbers expressed as powers of 10), use the same number of significant figures as are in that power.
Examples \(1\)
The following examples will illustrate the most common problems you are likely to encounter in rounding off the results of calculations. They deserve your careful study!
calculator result
rounded remarks
1.6 Rounding to two significant figures yields an implied uncertainty of 1/16 or 6%, three times greater than that in the least-precisely known factor. This is a good illustration of how rounding can lead to the loss of information.
1.9E6 The "3.1" factor is specified to 1 part in 31, or 3%. In the answer 1.9, the value is expressed to 1 part in 19, or 5%. These precisions are comparable, so the rounding-off rule has given us a reasonable result.
A certain book has a thickness of 117 mm; find the height of a stack of 24 identical books:
2810 mm The “24” and the “1” are exact, so the only uncertain value is the thickness of each book, given to 3 significant digits. The trailing zero in the answer is only a placeholder.
10.4 In addition or subtraction, look for the term having the smallest number of decimal places, and round off the answer to the same number of places.
23 cm see below
The last of the examples shown above represents the very common operation of converting one unit into another. There is a certain amount of ambiguity here; if we take "9 in" to mean a distance in the range 8.5 to 9.5 inches, then the implied uncertainty is ±0.5 in, which is 1 part in 18, or about ± 6%. The relative uncertainty in the answer must be the same, since all the values are multiplied by the same factor, 2.54 cm/in. In this case we are justified in writing the answer to two significant digits, yielding an uncertainty of about ±1 cm; if we had used the answer "20 cm" (one significant digit), its implied uncertainty would be ±5 cm, or ±25%.
When the appropriate number of significant digits is in question, calculating the relative uncertainty can help you decide.
1.E: Exercises
1.1: The Scientific Method
Problems
1. What are the three components of the scientific method? Is it necessary for an individual to conduct experiments to follow the scientific method?
2. Identify each statement as a theory or a law and explain your reasoning.
a. The ratio of elements in a pure substance is constant.
b. An object appears black because it absorbs all the visible light that strikes it.
c. Energy is neither created nor destroyed.
d. Metals conduct electricity because their electrons are not tightly bound to a particular nucleus and are therefore free to migrate.
3. Identify each statement as a theory or a law and explain your reasoning.
a. A pure chemical substance contains the same proportion of elements by mass.
b. The universe is expanding.
c. Oppositely charged particles attract each other.
d. Life exists on other planets.
4. Classify each statement as a qualitative observation or a quantitative observation.
a. Mercury and bromine are the only elements that are liquids at room temperature.
b. An element is both malleable and ductile.
c. The density of iron is 7.87 g/cm3.
d. Lead absorbs sound very effectively.
e. A meteorite contains 20% nickel by mass.
5. Classify each statement as a quantitative observation or a qualitative observation.
a. Nickel deficiency in rats is associated with retarded growth.
b. Boron is a good conductor of electricity at high temperatures.
c. There are 1.4–2.3 g of zinc in an average 70 kg adult.
d. Certain osmium compounds found in air in concentrations as low as 10.7 µg/m3 can cause lung cancer.
3.
a. law
b. theory
c. law
d. theory
5.
a. qualitative
b. qualitative
c. quantitative
d. quantitative | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.7%3A_Significant_Figures.txt |
Thumbnail: Schematic of Rutherford's famous gold foil experiment.
02: Atoms and The Atomic Theory
Learning Objectives
• Explain how all matter is composed of atoms.
• Describe the modern atomic theory.
Take some aluminum foil. Cut it in half. Now there are two smaller pieces of aluminum foil. Cut one of the pieces in half again. Cut one of those smaller pieces in half again. Continue cutting, making smaller and smaller pieces of aluminum foil. It should be obvious that the pieces are still aluminum foil; they are just becoming smaller and smaller. But how far can this exercise be taken, at least in theory? Can one continue cutting the aluminum foil into halves forever, making smaller and smaller pieces? Or is there some limit, some absolute smallest piece of aluminum foil? Thought experiments like this—and the conclusions based on them—were debated as far back as the fifth century BC.
John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. The theory explains several concepts that are relevant in the observable world: the composition of a pure gold necklace, what makes the pure gold necklace different than a pure silver necklace, and what occurs when pure gold is mixed with pure copper. This article explains the theories that Dalton used as a basis for his theory:
Law 1: The Conservation of Mass
"Nothing comes from nothing" is an important idea in ancient Greek philosophy that argues that what exists now has always existed, since no new matter can come into existence where there was none before. Antoine Lavoisier (1743-1794) restated this principle for chemistry with the law of conservation of mass, which "means that the atoms of an object cannot be created or destroyed, but can be moved around and be changed into different particles." This law says that when a chemical reaction rearranges atoms into a new product, the mass of the reactants (chemicals before the chemical reaction) is the same as the mass of the products (the new chemicals made). More simply, whatever you do, you will still have the same amount of stuff (however, certain nuclear reactions like fusion and fission can convert a small part of the mass into energy.
The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction; in other words, mass is conserved. The law of conservation of mass was formulated by Lavoisier as a result of his combustion experiment, in which he observed that the mass of his original substance—a glass vessel, tin, and air—was equal to the mass of the produced substance—the glass vessel, “tin calx”, and the remaining air.
Historically, this was a difficult concept for scientists to grasp. If this law was true, then how could a large piece of wood be reduced to a small pile of ashes? The wood clearly has a greater mass than the ashes. From this observation scientists concluded that mass had been lost. However, Figure \(1\) shows that the burning of word does follow the law of conservation of mass. Scientists did not account for the gases that play a critical role in this reaction.
Law 2: Definite Proportions
Joseph Proust (1754-1826) formulated the law of definite proportions (also called the Law of Constant Composition or Proust's Law). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. The illustration below depicts this law; 31 grams of H2O and 8 grams of H2O are made up of the same percent of hydrogen and oxygen.
Example \(1\): water
Oxygen makes up 88.8% of the mass of any sample of pure water, while hydrogen makes up the remaining 11.2% of the mass. You can get water by melting ice or snow, by condensing steam, from river, sea, pond, etc. It can be from different places: USA, UK, Australia, or anywhere. It can be made by chemical reactions like burning hydrogen in oxygen.
However, if the water is pure, it will always consist of 88.8 % oxygen by mass and 11.2 % hydrogen by mass, irrespective of its source or method of preparation.
Law 3: Multiple Proportions
Many combinations of elements can react to form more than one compound. In such cases, this law states that the weights of one element that combine with a fixed weight of another of these elements are integer multiples of one another. It's easy to say this, but please make sure that you understand how it works. Nitrogen forms a very large number of oxides, five of which are shown here.
• Lineshows the ratio of the relative weights of the two elements in each compound. These ratios were calculated by simply taking the molar mass of each element, and multiplying by the number of atoms of that element per mole of the compound. Thus for NO2, we have (1 × 14) : (2 × 16) = 13:32. (These numbers were not known in the early days of Chemistry because atomic weights (i.e., molar masses) of most elements were not reliably known.)
• The numbers in Lineare just the mass ratios of O:N, found by dividing the corresponding ratios in line 1. But someone who depends solely on experiment would work these out by finding the mass of O that combines with unit mass (1 g) of nitrogen.
• Line is obtained by dividing the figures the previous line by the smallest O:N ratio in the line above, which is the one for N2O. Note that just as the law of multiple proportions says, the weight of oxygen that combines with unit weight of nitrogen work out to small integers.
• Of course we just as easily could have illustrated the law by considering the mass of nitrogen that combines with one gram of oxygen; it works both ways!
The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers.
Example \(2\): Oxides of Carbon
Consider two separate compounds are formed by only carbon and oxygen. The first compound contains 42.9% carbon and 57.1% oxygen (by mass) and the second compound contains 27.3% carbon and 72.7% oxygen (again by mass). Is this consistant with the law of multiple proportions?
Solution
The Law of Multiple Proportions states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers. Hence, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio.
Thus for every 1 g of the first compound there are 0.57 g of oxygen and 0.429 g of carbon. The mass of oxygen per gram carbon is:
\[ \dfrac{0.571\;g\; oxygen}{0.429 \;g \;carbon} = 1.33\; \dfrac{g oxygen}{g carbon}\]
Similarly, for 1 g of the second compound, there are 0.727 g oxygen and 0.273 g of carbon. The ration of mass of oxygen per gram of carbon is
\[ \dfrac{0.727\;g\; oxygen}{0.273 \;g \;carbon} = 2.66\; \dfrac{g oxygen}{g carbon}\]
Dividing the mass of oxygen per g of carbon of the second compound:
\[\dfrac{2.66}{1.33} = 2\]
Hence the masses of oxygen combine with carbon in a 2:1 ratio which s consistent with the Law of Multiple Proportions since they are whole numbers.
Dalton's Atomic Theory
The modern atomic theory, proposed about 1803 by the English chemist John Dalton (Figure \(4\)), is a fundamental concept that states that all elements are composed of atoms. Previously, an atom was defined as the smallest part of an element that maintains the identity of that element. Individual atoms are extremely small; even the largest atom has an approximate diameter of only 5.4 × 10−10 m. With that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of the human pinkie (about 1 cm).
Dalton’s ideas are called the modern atomic theory because the concept of atoms is very old. The Greek philosophers Leucippus and Democritus originally introduced atomic concepts in the fifth century BC. (The word atom comes from the Greek word atomos, which means “indivisible” or “uncuttable.”) Dalton had something that the ancient Greek philosophers didn’t have, however; he had experimental evidence, such as the formulas of simple chemicals and the behavior of gases. In the 150 years or so before Dalton, natural philosophy had been maturing into modern science, and the scientific method was being used to study nature. When Dalton announced a modern atomic theory, he was proposing a fundamental theory to describe many previous observations of the natural world; he was not just participating in a philosophical discussion.
Dalton's Theory was a powerful development as it explained the three laws of chemical combination (above) and recognized a workable distinction between the fundamental particle of an element (atom) and that of a compound (molecule). Six postulates are involved in Dalton's Atomic Theory:
1. All matter consists of indivisible particles called atoms.
2. Atoms of the same element are similar in shape and mass, but differ from the atoms of other elements.
3. Atoms cannot be created or destroyed.
4. Atoms of different elements may combine with each other in a fixed, simple, whole number ratios to form compound atoms.
5. Atoms of same element can combine in more than one ratio to form two or more compounds.
6. The atom is the smallest unit of matter that can take part in a chemical reaction.
In light of the current state of knowledge in the field of Chemistry, Dalton’s theory had a few drawbacks. According to Dalton’s postulates,
1. The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes part in chemical reactions.
2. According to Dalton, the atoms of same element are similar in all respects. However, atoms of some elements vary in their masses and densities. These atoms of different masses are called isotopes. For example, chlorine has two isotopes with mass numbers 35 and 37.
3. Dalton also claimed that atoms of different elements are different in all respects. This has been proven wrong in certain cases: argon and calcium atoms each have an same atomic mass (40 amu).
4. According to Dalton, atoms of different elements combine in simple whole number ratios to form compounds. This is not observed in complex organic compounds like sugar (\(C_{12}H_{22}O_{11}\)).
5. The theory fails to explain the existence of allotropes (different forms of pure elements); it does not account for differences in properties of charcoal, graphite, diamond.
Despite these drawbacks, the importance of Dalton’s theory should not be underestimated. He displayed exceptional insight into the nature of matter. and his ideas provided a framework that was later modified and expanded by other. Consequentiually, John Dalton is often considered to be the father of modern atomic theory.
Fundamental Experiments in Chemistry: https://youtu.be/IhqqLGKmah4
Summary
This article explains the theories that Dalton used as a basis for his theory:
• Law of Conservation of Mass
• Law of Constant Composition
• Law of Multiple Proportions | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.1%3A_Early_Chemical_Discoveries_and_the_Atomic_Theory.txt |
Learning Objectives
• To become familiar with the components and structure of the atom.
Long before the end of the 19th century, it was well known that applying a high voltage to a gas contained at low pressure in a sealed tube (called a gas discharge tube) caused electricity to flow through the gas, which then emitted light (Figure $1$). Researchers trying to understand this phenomenon found that an unusual form of energy was also emitted from the cathode, or negatively charged electrode; this form of energy was called a cathode ray.
In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure $2$). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Subsequently, the American scientist Robert Millikan (1868–1953) carried out a series of experiments using electrically charged oil droplets, which allowed him to calculate the charge on a single electron. With this information and Thomson’s mass-to-charge ratio, Millikan determined the mass of an electron:
$\dfrac {mass}{charge} \times {charge} ={mass}$
It was at this point that two separate lines of investigation began to converge, both aimed at determining how and why matter emits energy. The video below shows how JJ Thompson used such a tube to measure the ratio of charge over mass of an electron
Radioactivity
The second line of investigation began in 1896, when the French physicist Henri Becquerel (1852–1908) discovered that certain minerals, such as uranium salts, emitted a new form of energy. Becquerel’s work was greatly extended by Marie Curie (1867–1934) and her husband, Pierre (1854–1906); all three shared the Nobel Prize in Physics in 1903. Marie Curie coined the term radioactivity (from the Latin radius, meaning “ray”) to describe the emission of energy rays by matter. She found that one particular uranium ore, pitchblende, was substantially more radioactive than most, which suggested that it contained one or more highly radioactive impurities. Starting with several tons of pitchblende, the Curies isolated two new radioactive elements after months of work: polonium, which was named for Marie’s native Poland, and radium, which was named for its intense radioactivity. Pierre Curie carried a vial of radium in his coat pocket to demonstrate its greenish glow, a habit that caused him to become ill from radiation poisoning well before he was run over by a horse-drawn wagon and killed instantly in 1906. Marie Curie, in turn, died of what was almost certainly radiation poisoning.
Building on the Curies’ work, the British physicist Ernest Rutherford (1871–1937) performed decisive experiments that led to the modern view of the structure of the atom. While working in Thomson’s laboratory shortly after Thomson discovered the electron, Rutherford showed that compounds of uranium and other elements emitted at least two distinct types of radiation. One was readily absorbed by matter and seemed to consist of particles that had a positive charge and were massive compared to electrons. Because it was the first kind of radiation to be discovered, Rutherford called these substances α particles. Rutherford also showed that the particles in the second type of radiation, β particles, had the same charge and mass-to-charge ratio as Thomson’s electrons; they are now known to be high-speed electrons. A third type of radiation, γ rays, was discovered somewhat later and found to be similar to a lower-energy form of radiation called x-rays, now used to produce images of bones and teeth.
These three kinds of radiation—α particles, β particles, and γ rays—are readily distinguished by the way they are deflected by an electric field and by the degree to which they penetrate matter. As Figure $3$ illustrates, α particles and β particles are deflected in opposite directions; α particles are deflected to a much lesser extent because of their higher mass-to-charge ratio. In contrast, γ rays have no charge, so they are not deflected by electric or magnetic fields. Figure $5$ shows that α particles have the least penetrating power and are stopped by a sheet of paper, whereas β particles can pass through thin sheets of metal but are absorbed by lead foil or even thick glass. In contrast, γ-rays can readily penetrate matter; thick blocks of lead or concrete are needed to stop them.
Summary
• Atoms are the ultimate building blocks of all matter.
• The modern atomic theory establishes the concepts of atoms and how they compose matter.
Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy). | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.2%3A_Electrons_and_Other_Discoveries_in_Atomic_Physics.txt |
Learning Objectives
• To become familiar with the components and structure of the atom.
Dissecting the Atom
Once scientists concluded that all matter contains negatively charged electrons, it became clear that atoms, which are electrically neutral, must also contain positive charges to balance the negative ones. Thomson proposed that the electrons were embedded in a uniform sphere that contained both the positive charge and most of the mass of the atom, much like raisins in plum pudding or chocolate chips in a cookie (Figure $1$).
In a single famous experiment, however, Rutherford showed unambiguously that Thomson’s model of the atom was incorrect. Rutherford aimed a stream of α particles at a very thin gold foil target (part (a) in Figure $2$) and examined how the α particles were scattered by the foil. Gold was chosen because it could be easily hammered into extremely thin sheets, minimizing the number of atoms in the target. If Thomson’s model of the atom were correct, the positively-charged α particles should crash through the uniformly distributed mass of the gold target like cannonballs through the side of a wooden house. They might be moving a little slower when they emerged, but they should pass essentially straight through the target (part (b) in Figure $2$). To Rutherford’s amazement, a small fraction of the α particles were deflected at large angles, and some were reflected directly back at the source (part (c) in Figure $2$). According to Rutherford, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”
Rutherford’s results were not consistent with a model in which the mass and positive charge are distributed uniformly throughout the volume of an atom. Instead, they strongly suggested that both the mass and positive charge are concentrated in a tiny fraction of the volume of an atom, which Rutherford called the nucleus. It made sense that a small fraction of the α particles collided with the dense, positively charged nuclei in either a glancing fashion, resulting in large deflections, or almost head-on, causing them to be reflected straight back at the source.
Although Rutherford could not explain why repulsions between the positive charges in nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that repulsions between negatively charged electrons would cause the electrons to be uniformly distributed throughout the atom’s volume.Today it is known that strong nuclear forces, which are much stronger than electrostatic interactions, hold the protons and the neutrons together in the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he considered physics superior to chemistry. In his opinion, “All science is either physics or stamp collecting.”
The historical development of the different models of the atom’s structure is summarized in Figure $3$. Rutherford established that the nucleus of the hydrogen atom was a positively charged particle, for which he coined the name proton in 1920. He also suggested that the nuclei of elements other than hydrogen must contain electrically neutral particles with approximately the same mass as the proton. The neutron, however, was not discovered until 1932, when James Chadwick (1891–1974, a student of Rutherford; Nobel Prize in Physics, 1935) discovered it. As a result of Rutherford’s work, it became clear that an α particle contains two protons and neutrons, and is therefore the nucleus of a helium atom.
Rutherford’s model of the atom is essentially the same as the modern model, except that it is now known that electrons are not uniformly distributed throughout an atom’s volume. Instead, they are distributed according to a set of principles described by Quantum Mechanics. Figure $4$ shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today.
The Nuclear Atom
The precise physical nature of atoms finally emerged from a series of elegant experiments carried out between 1895 and 1915. The most notable of these achievements was Ernest Rutherford's famous 1911 alpha-ray scattering experiment, which established that
• Almost all of the mass of an atom is contained within a tiny (and therefore extremely dense) nucleus which carries a positive electric charge whose value identifies each element and is known as the atomic number of the element.
• Almost all of the volume of an atom consists of empty space in which electrons, the fundamental carriers of negative electric charge, reside. The extremely small mass of the electron (1/1840 the mass of the hydrogen nucleus) causes it to behave as a quantum particle, which means that its location at any moment cannot be specified; the best we can do is describe its behavior in terms of the probability of its manifesting itself at any point in space. It is common (but somewhat misleading) to describe the volume of space in which the electrons of an atom have a significant probability of being found as the electron cloud. The latter has no definite outer boundary, so neither does the atom. The radius of an atom must be defined arbitrarily, such as the boundary in which the electron can be found with 95% probability. Atomic radii are typically 30-300 pm.
The nucleus is itself composed of two kinds of particles. Protons are the carriers of positive electric charge in the nucleus; the proton charge is exactly the same as the electron charge, but of opposite sign. This means that in any [electrically neutral] atom, the number of protons in the nucleus (often referred to as the nuclear charge) is balanced by the same number of electrons outside the nucleus.
The other nuclear particle is the neutron. As its name implies, this particle carries no electrical charge. Its mass is almost the same as that of the proton. Most nuclei contain roughly equal numbers of neutrons and protons, so we can say that these two particles together account for almost all the mass of the atom.
Note
Because the electrons of an atom are in contact with the outside world, it is possible for one or more electrons to be lost, or some new ones to be added. The resulting electrically-charged atom is called an ion.
Atoms consist of electrons, protons, and neutrons. This is an oversimplification that ignores the other subatomic particles that have been discovered, but it is sufficient for discussion of chemical principles. Some properties of these subatomic particles are summarized in Table $1$ which illustrates three important points:
1. Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. Relative charges of −1 and +1 are assigned to the electron and proton, respectively.
2. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral.
3. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute the bulk of the mass of atoms.
The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science.
Table $1$: Properties of Subatomic Particles*
Particle Mass (g) Atomic Mass (amu) Electrical Charge (coulombs) Relative Charge
* For a review of using scientific notation and units of measurement, see Essential Skills 1 (Section 1.9).
electron $9.109 \times 10^{-28}$ 0.0005486 −1.602 × 10−19 −1
neutron $1.675 \times 10^{-24}$ 1.008665 0 0
proton $1.673 \times 10^{-24}$ 1.007276 +1.602 × 10−19 +1
The Nuclear Atom: https://youtu.be/eqoyZuv1tWA
Summary
• The atom consists of discrete particles that govern its chemical and physical behavior.
Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy).
Conceptual Problems
1. Describe the experiment that provided evidence that the proton is positively charged.
2. What observation led Rutherford to propose the existence of the neutron?
3. What is the difference between Rutherford’s model of the atom and the model chemists use today?
4. If cathode rays are not deflected when they pass through a region of space, what does this imply about the presence or absence of a magnetic field perpendicular to the path of the rays in that region?
5. Describe the outcome that would be expected from Rutherford’s experiment if the charge on α particles had remained the same but the nucleus were negatively charged. If the nucleus were neutral, what would have been the outcome?
6. Describe the differences between an α particle, a β particle, and a γ ray. Which has the greatest ability to penetrate matter?
Problems
Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9) before proceeding to the Numerical Problems.
1. Using the data in Table 1.3 and the periodic table (see Chapter 32), calculate the percentage of the mass of a silicon atom that is due to
a. electrons.
b. protons.
1. Using the data in Table 1.3 and the periodic table (see Chapter 32), calculate the percentage of the mass of a helium atom that is due to
a. electrons.
b. protons.
1. The radius of an atom is approximately 104 times larger than the radius of its nucleus. If the radius of the nucleus were 1.0 cm, what would be the radius of the atom in centimeters? in miles?
2. The total charge on an oil drop was found to be 3.84 × 10−18 coulombs. What is the total number of electrons contained in the drop? | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.3%3A_The_Nuclear_Atom.txt |
Learning Objectives
• To know the meaning of isotopes and atomic masses.
To date, about 115 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. In most cases, the symbols for the elements are derived directly from each element’s name, such as C for carbon, U for uranium, Ca for calcium, and Po for polonium. Elements have also been named for their properties [such as radium (Ra) for its radioactivity], for the native country of the scientist(s) who discovered them [polonium (Po) for Poland], for eminent scientists [curium (Cm) for the Curies], for gods and goddesses [selenium (Se) for the Greek goddess of the moon, Selene], and for other poetic or historical reasons. Some of the symbols used for elements that have been known since antiquity are derived from historical names that are no longer in use; only the symbols remain to indicate their origin. Examples are Fe for iron, from the Latin ferrum; Na for sodium, from the Latin natrium; and W for tungsten, from the German wolfram. Examples are in Table $1$.
Table $1$: Element Symbols Based on Names No Longer in Use
Element Symbol Derivation Meaning
antimony Sb stibium Latin for “mark”
copper Cu cuprum from Cyprium, Latin name for the island of Cyprus, the major source of copper ore in the Roman Empire
gold Au aurum Latin for “gold”
iron Fe ferrum Latin for “iron”
lead Pb plumbum Latin for “heavy”
mercury Hg hydrargyrum Latin for “liquid silver”
potassium K kalium from the Arabic al-qili, “alkali”
silver Ag argentum Latin for “silver”
sodium Na natrium Latin for “sodium”
tin Sn stannum Latin for “tin”
tungsten W wolfram German for “wolf stone” because it interfered with the smelting of tin and was thought to devour the tin
Atomic Number (Z)
What single parameter uniquely characterizes the atom of a given element? It is not the atom's relative mass, as we will see in the section on isotopes below. It is, rather, the number of protons in the nucleus, which we call the atomic number and denote by the symbol Z. Each proton carries an electric charge of +1, so the atomic number also specifies the electric charge of the nucleus. In the neutral atom, the Z protons within the nucleus are balanced by Z electrons outside it.
Henry Moseley
Atomic numbers were first worked out in 1913 by Henry Moseley, a young member of Rutherford's research group in Manchester. Moseley searched for a measurable property of each element that increases linearly with atomic number. He found this in a class of X-rays emitted by an element when it is bombarded with electrons. The frequencies of these X-rays are unique to each element, and they increase uniformly in successive elements. Moseley found that the square roots of these frequencies give a straight line when plotted against Z; this enabled him to sort the elements in order of increasing atomic number.
You can think of the atomic number as a kind of serial number of an element, commencing at 1 for hydrogen and increasing by one for each successive element. The chemical name of the element and its symbol are uniquely tied to the atomic number; thus the symbol "Sr" stands for strontium, whose atoms all have Z = 38.
Mass number (A)
This is just the sum of the numbers of protons and neutrons in the nucleus. It is sometimes represented by the symbol A, so
$A = Z + N \label{2.4.1}$
in which Z is the atomic number and N is the neutron number.
Nuclides and their Symbols
The term nuclide simply refers to any particular kind of nucleus. For example, a nucleus of atomic number 7 is a nuclide of nitrogen. Any nuclide is characterized by the pair of numbers (Z ,A). The element symbol depends on Z alone, so the symbol 26Mg is used to specify the mass-26 nuclide of manganese, whose name implies Z=12. A more explicit way of denoting a particular kind of nucleus is to add the atomic number as a subscript. Of course, this is somewhat redundant, since the symbol Mg always implies Z=12, but it is sometimes a convenience when discussing several nuclides.
Figure $2$: Formalism used for identifying specific nuclide (any particular kind of nucleus)
The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as $^A_Z X$, where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore $_6^{12} C$. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, $_6^{12} C$ is more often written as 12C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z.
Isotopes
Recall that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.
Two nuclides having the same atomic number but different mass numbers are known as isotopes. Most elements occur in nature as mixtures of isotopes, but twenty-three of them (including beryllium and fluorine, shown in the table) are monoisotopic. For example, there are three natural isotopes of magnesium: 24Mg (79% of all Mg atoms), 25Mg (10%), and 26Mg (11%); all three are present in all compounds of magnesium in about these same proportions.
Approximately 290 isotopes occur in nature. The two heavy isotopes of hydrogen are especially important— so much so that they have names and symbols of their own:
$\underset{\text{protium}}{\ce{^1_1H} } \label{2.4.2a}$
$\underset{\text{deuterium}}{\ce{^2_1H \equiv D}} \label{2.4.2b}$
$\underset{\text{tritium}}{\ce{^2_1H \equiv T}} \label{2.4.2c}$
Deuterium accounts for only about 15 out of every one million atoms of hydrogen. Tritium, which is radioactive, is even less abundant. All the tritium on the earth is a by-product of the decay of other radioactive elements.
For carbon, in addition to $^{12}C$, a typical sample of carbon contains 1.11% $_6^{13} C$ (13C), with 7 neutrons and 6 protons, and a trace of $_6^{14} C$ (14C), with 8 neutrons and 6 protons. The nucleus of 14C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Table $2$.
Table $2$: Properties of Selected Isotopes
Element Symbol Atomic Mass (amu) Isotope Mass Number Isotope Masses (amu) Percent Abundances (%)
hydrogen H 1.0079 1 1.007825 99.9855
2 2.014102 0.0115
boron B 10.81 10 10.012937 19.91
11 11.009305 80.09
carbon C 12.011 12 12 (defined) 99.89
13 13.003355 1.11
oxygen O 15.9994 16 15.994915 99.757
17 16.999132 0.0378
18 17.999161 0.205
iron Fe 55.845 54 53.939611 5.82
56 55.934938 91.66
57 56.935394 2.19
58 57.933276 0.33
uranium U 238.03 234 234.040952 0.0054
235 235.043930 0.7204
238 238.050788 99.274
Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991.
How Elements Are Represented on the Periodic Table: https://youtu.be/ik6ZsaSyISo
Example $1$
An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes.
Given: number of protons and neutrons
Asked for: element and atomic symbol
Strategy:
1. Refer to the periodic table and use the number of protons to identify the element.
2. Calculate the mass number of each isotope by adding together the numbers of protons and neutrons.
3. Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element.
Solution:
A The element with 82 protons (atomic number of 82) is lead: Pb.
B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are $^{206}_{82}Pb$, $^{207}_{82}Pb$, and $^{208}_{82}Pb$, which are usually abbreviated as $^{206}Pb$, $^{207}Pb$, and $^{208}Pb$.
Exercise $1$
Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons.
Answer: $\ce{^{79}_{35}Br}$ and $\ce{^{81}_{35}Br}$ or, more commonly, $\ce{^{79}Br}$ and $\ce{^{81}Br}$.
Summary
• The atom consists of discrete particles that govern its chemical and physical behavior.Contributors
Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.4%3A_Chemical_Elements.txt |
Learning Objectives
• to know the meaning of isotopes and atomic masses.
Atomic and Molecular Weights
The subscripts in chemical formulas, and the coefficients in chemical equations represent exact quantities. $\ce{H_2O}$, for example, indicates that a water molecule comprises exactly two atoms of hydrogen and one atom of oxygen. The following equation:
$\ce{C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H2O(l)} \label{Eq1}$
not only tells us that propane reacts with oxygen to produce carbon dioxide and water, but that 1 molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. Since counting individual atoms or molecules is a little difficult, quantitative aspects of chemistry rely on knowing the masses of the compounds involved.
Atoms of different elements have different masses. Early work on the separation of water into its constituent elements (hydrogen and oxygen) indicated that 100 grams of water contained 11.1 grams of hydrogen and 88.9 grams of oxygen:
$\text{100 grams Water} \rightarrow \text{11.1 grams Hydrogen} + \text{88.9 grams Oxygen} \label{Eq2}$
Later, scientists discovered that water was composed of two atoms of hydrogen for each atom of oxygen. Therefore, in the above analysis, in the 11.1 grams of hydrogen there were twice as many atoms as in the 88.9 grams of oxygen. Therefore, an oxygen atom must weigh about 16 times as much as a hydrogen atom:
$\dfrac{\dfrac{88.9\;g\;Oxygen}{1\; atom}}{\dfrac{111\;g\;Hydrogen}{2\;atoms}} = 16 \label{Eq3}$
Hydrogen, the lightest element, was assigned a relative mass of '1', and the other elements were assigned 'atomic masses' relative to this value for hydrogen. Thus, oxygen was assigned an atomic mass of 16. We now know that a hydrogen atom has a mass of 1.6735 x 10-24 grams, and that the oxygen atom has a mass of 2.6561 X 10-23 grams. As we saw earlier, it is convenient to use a reference unit when dealing with such small numbers: the atomic mass unit. The atomic mass unit (amu) was not standardized against hydrogen, but rather, against the 12C isotope of carbon (amu = 12).
Thus, the mass of the hydrogen atom (1H) is 1.0080 amu, and the mass of an oxygen atom (16O) is 15.995 amu. Once the masses of atoms were determined, the amu could be assigned an actual value:
1 amu = 1.66054 x 10-24 grams conversely: 1 gram = 6.02214 x 1023 amu
Average Atomic Mass
Although the masses of the electron, the proton, and the neutron are known to a high degree of precision (Table 2.3.1), the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1H (hydrogen) and 2H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions.
Because atoms are much too small to measure individually and do not have charges, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses (Figure $1$). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest.
The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu or u), defined as one-twelfth of the mass of one atom of 12C. Because the masses of all other atoms are calculated relative to the 12C standard, 12C is the only atom listed in Table 2.3.2 whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10−24 g.
Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2H to the mass of 12C, so the absolute mass of 2H is
$\rm{\text{mass of }^2H \over \text{mass of }^{12}C} \times \text{mass of }^{12}C = 0.167842 \times 12 \;amu = 2.104104\; amu \label{Eq4}$
The masses of the other elements are determined in a similar way.
The periodic table lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes in Table 2.3.2 reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes. Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12C (mass = 12 amu by definition) and 1.11% 13C (mass = 13.003355 amu). The percent abundance of 14C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as follows:
$\rm(0.9889 \times 12 \;amu) + (0.0111 \times 13.003355 \;amu) = 12.01 \;amu \label{Eq5}$
Carbon is predominantly 12C, so its average atomic mass should be close to 12 amu, which is in agreement with this calculation.
The value of 12.01 is shown under the symbol for C in the periodic table, although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes.
Finding the Averaged Atomic Weight of an Element: https://youtu.be/bmP6Gr9zJiQ
Example $1$: Bromine
Naturally occurring bromine consists of the two isotopes listed in the following table:
Isotope Exact Mass (amu) Percent Abundance (%)
79Br 78.9183 50.69
81Br 80.9163 49.31
Calculate the atomic mass of bromine.
Given: exact mass and percent abundance
Asked for: atomic mass
Strategy:
1. Convert the percent abundances to decimal form to obtain the mass fraction of each isotope.
2. Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass.
3. Add together the weighted masses to obtain the atomic mass of the element.
4. Check to make sure that your answer makes sense.
Solution:
A The atomic mass is the weighted average of the masses of the isotopes. In general, we can write
atomic mass of element = [(mass of isotope 1 in amu) (mass fraction of isotope 1)] + [(mass of isotope 2) (mass fraction of isotope 2)] + …
Bromine has only two isotopes. Converting the percent abundances to mass fractions gives
$\ce{^{79}Br}: {50.69 \over 100} = 0.5069$
$\ce{^{81}Br}: {49.31 \over 100} = 0.4931$
B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass:
$\ce{^{79}Br}: 79.9183 \;amu \times 0.5069 = 40.00\; amu$
$\ce{^{81}Br}: 80.9163 \;amu \times 0.4931 = 39.90 \;amu$
C The sum of the weighted masses is the atomic mass of bromine is
40.00 amu + 39.90 amu = 79.90 amu
D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%.
Exercise $1$
Magnesium has the three isotopes listed in the following table:
Isotope Exact Mass (amu) Percent Abundance (%)
24Mg 23.98504 78.70
25Mg 24.98584 10.13
26Mg 25.98259 11.17
Use these data to calculate the atomic mass of magnesium.
Answer
24.31 amu
Summary
The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.5%3A_Atomic_Mass.txt |
Learning Objectives
• To become familiar with the organization of the periodic table.
Rutherford’s nuclear model of the atom helped explain why atoms of different elements exhibit different chemical behavior. The identity of an element is defined by its atomic number (Z), the number of protons in the nucleus of an atom of the element. The atomic number is therefore different for each element. The known elements are arranged in order of increasing Z in the periodic table (Figure \(1\)). The rationale for the peculiar format of the periodic table is explained later. Each element is assigned a unique one-, two-, or three-letter symbol. The names of the elements are listed in the periodic table, along with their symbols, atomic numbers, and atomic masses. The chemistry of each element is determined by its number of protons and electrons. In a neutral atom, the number of electrons equals the number of protons.
The elements are arranged in a periodic table, which is probably the single most important learning aid in chemistry. It summarizes huge amounts of information about the elements in a way that facilitates the prediction of many of their properties and chemical reactions. The elements are arranged in seven horizontal rows, in order of increasing atomic number from left to right and top to bottom. The rows are called periods, and they are numbered from 1 to 7. The elements are stacked in such a way that elements with similar chemical properties form vertical columns, called groups, numbered from 1 to 18 (older periodic tables use a system based on roman numerals). Groups 1, 2, and 13–18 are the main group elements, listed as A in older tables. Groups 3–12 are in the middle of the periodic table and are the transition elements, listed as B in older tables. The two rows of 14 elements at the bottom of the periodic table are the lanthanides and the actinides, whose positions in the periodic table are indicated in group 3.
Metals, Nonmetals, and Semimetals
The heavy orange zigzag line running diagonally from the upper left to the lower right through groups 13–16 in Figure \(1\) divides the elements into metals (in blue, below and to the left of the line) and nonmetals (in bronze, above and to the right of the line). Gold-colored lements that lie along the diagonal line exhibit properties intermediate between metals and nonmetals; they are called semimetals.
The distinction between metals and nonmetals is one of the most fundamental in chemistry. Metals—such as copper or gold—are good conductors of electricity and heat; they can be pulled into wires because they are ductile; they can be hammered or pressed into thin sheets or foils because they are malleable; and most have a shiny appearance, so they are lustrous. The vast majority of the known elements are metals. Of the metals, only mercury is a liquid at room temperature and pressure; all the rest are solids.
Nonmetals, in contrast, are generally poor conductors of heat and electricity and are not lustrous. Nonmetals can be gases (such as chlorine), liquids (such as bromine), or solids (such as iodine) at room temperature and pressure. Most solid nonmetals are brittle, so they break into small pieces when hit with a hammer or pulled into a wire. As expected, semimetals exhibit properties intermediate between metals and nonmetals.
Example \(1\)
Based on its position in the periodic table, do you expect selenium to be a metal, a nonmetal, or a semimetal?
Given: element
Asked for: classification
Strategy:
Find selenium in the periodic table shown in Figure \(1\) and then classify the element according to its location.
Solution:
The atomic number of selenium is 34, which places it in period 4 and group 16. In Figure \(1\), selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal. Note, however, that because selenium is close to the metal-nonmetal dividing line, it would not be surprising if selenium were similar to a semimetal in some of its properties.
Exercise \(1\)
Based on its location in the periodic table, do you expect indium to be a nonmetal, a metal, or a semimetal?
Answer: metal
Descriptive Names
As previously noted, the periodic table is arranged so that elements with similar chemical behaviors are in the same group. Chemists often make general statements about the properties of the elements in a group using descriptive names with historical origins. For example, the elements of Group 1 are known as the alkali metals, Group 2 are the alkaline earth metals, Group 17 are the halogens, and Group 18 are the noble gases.
Group 1: The Alkali Metals
The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Hydrogen is unique in that it is generally placed in Group 1, but it is not a metal.
The compounds of the alkali metals are common in nature and daily life. One example is table salt (sodium chloride); lithium compounds are used in greases, in batteries, and as drugs to treat patients who exhibit manic-depressive, or bipolar, behavior. Although lithium, rubidium, and cesium are relatively rare in nature, and francium is so unstable and highly radioactive that it exists in only trace amounts, sodium and potassium are the seventh and eighth most abundant elements in Earth’s crust, respectively.
Group 2: The Alkaline Earth Metals
The alkaline earth metals are beryllium, magnesium, calcium, strontium, barium, and radium. Beryllium, strontium, and barium are rare, and radium is unstable and highly radioactive. In contrast, calcium and magnesium are the fifth and sixth most abundant elements on Earth, respectively; they are found in huge deposits of limestone and other minerals.
Group 17: The Halogens
The halogens are fluorine, chlorine, bromine, iodine, and astatine. The name halogen is derived from the Greek words for “salt forming,” which reflects that all the halogens react readily with metals to form compounds, such as sodium chloride and calcium chloride (used in some areas as road salt).
Compounds that contain the fluoride ion are added to toothpaste and the water supply to prevent dental cavities. Fluorine is also found in Teflon coatings on kitchen utensils. Although chlorofluorocarbon propellants and refrigerants are believed to lead to the depletion of Earth’s ozone layer and contain both fluorine and chlorine, the latter is responsible for the adverse effect on the ozone layer. Bromine and iodine are less abundant than chlorine, and astatine is so radioactive that it exists in only negligible amounts in nature.
Group 18: The Noble Gases
The noble gases are helium, neon, argon, krypton, xenon, and radon. Because the noble gases are composed of only single atoms, they are called monatomic. At room temperature and pressure, they are unreactive gases. Because of their lack of reactivity, for many years they were called inert gases or rare gases. However, the first chemical compounds containing the noble gases were prepared in 1962. Although the noble gases are relatively minor constituents of the atmosphere, natural gas contains substantial amounts of helium. Because of its low reactivity, argon is often used as an unreactive (inert) atmosphere for welding and in light bulbs. The red light emitted by neon in a gas discharge tube is used in neon lights.
Note
The noble gases are unreactive at room temperature and pressure.
Periodic Law in the Periodic Table: https://youtu.be/ciJYvhRF5i4
Summary
• The periodic table is used as a predictive tool.
The periodic table is an arrangement of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. Some of the groups have widely-used common names, including the alkali metals (Group 1) and the alkaline earth metals (Group 2) on the far left, and the halogens (Group 17) and the noble gases (Group 18) on the far right. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals. Metals are located on the left of the periodic table, and nonmetals are located on the upper right. They are separated by a diagonal band of semimetals. Metals are lustrous, good conductors of electricity, and readily shaped (they are ductile and malleable), whereas solid nonmetals are generally brittle and poor electrical conductors. Other important groupings of elements in the periodic table are the main group elements, the transition metals, the lanthanides, and the actinides. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.6%3A_Introduction_to_the_Periodic_Table.txt |
$2 \times \text { atomic mass of carbon} = 2 \, atoms \left ( {12.011 \, amu \over atoms } \right ) = 24.022 \,amu$
$6 \times \text { atomic mass of hydrogen} = 2 \, atoms \left ( {1.0079 \, amu \over atoms } \right ) = 6.0474 \,amu$
$1 \times \text { atomic mass of oxygen} = 1 \, atoms \left ( {15.9994 \, amu \over atoms } \right ) = 15.994 \,amu$
C Adding together the masses gives the molecular mass:
$24.022 \,amu + 6.0474 \,amu + 15.9994 \,amu = 46.069 \,amu$
Alternatively, we could have used unit conversions to reach the result in one step:
$\left [ 2 \, atoms C \left ( {12.011 \, amu \over 1 \, atom C} \right ) \right ] + \left [ 6 \, atoms H \left ( {1.0079 \, amu \over 1 \, atom H} \right ) \right ] + \left [ 1 \, atoms C \left ( {15.9994 \, amu \over 1 \, atom 0} \right ) \right ] = 46.069 \, amu$
The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:
$2 C \, \, \, (2\, atoms) (12.011 \, amu/atom ) = 24.022 \, amu$
$6 H \, \, \, (6\, atoms) (1.0079 \, amu/atom ) = 6.0474 \, amu$
$+ 1O \, \, \, (1\, atoms) (15.9994 \, amu/atom ) = 15.9994 \, amu$
$C_2H_6O \, \, \, \, \, \text {molecular mass of ethanol} = 46.069 \, amu$
Exercise $1$: Molecular Mass of Freon
Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is $CCl_3F$. Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:
Answer: 137.368 \,amu
Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. The units are atomic mass units.
Note
Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.
Determining the Molar Mass of a Molecule: https://youtu.be/wOjQjZqX7l8
Example $2$: Formula Mass of Calcium Phosphate
Calculate the formula mass of Ca3(PO4)2, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.
Given: ionic compound
Asked for: formula mass
Strategy:
1. Determine the number of atoms of each element in the empirical formula.
2. Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
3. Add together the masses to give the formula mass.
Solution:
A The empirical formula—Ca3(PO4)2—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca2+ ions and two PO43 ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.
B Taking atomic masses from the periodic table, we obtain
$3 \times \text {atomic mass of calcium} = 3 \, atoms \left ( {40.078 \, amu \over atom } \right ) = 120.234 \, amu$
$2 \times \text {atomic mass of phosphorus} = 2 \, atoms \left ( {30.973761 \, amu \over atom } \right ) = 61.947522 \, amu$
$8 \times \text {atomic mass of oxygen} = 8 \, atoms \left ( {15.9994 \, amu \over atom } \right ) = 127.9952 \, amu$
C Adding together the masses gives the formula mass of Ca3(PO4)2:
$120.234 \,amu + 61.947522 \, amu + 127.9952 \, amu = 310.177 \, amu$
We could also find the formula mass of Ca3(PO4)2 in one step by using unit conversions or a tabular format:
$\left [ 3 \, atoms Ca \left ({40.078 \, amu \over 1 \, atom Ca } \right ) \right ] + \left [ 2 \, atoms P \left ({30.973761 \, amu \over 1 \, atom P } \right ) \right ] + \left [ 8 \, atoms O \left ({15.9994 \, amu \over 1 \, atom O } \right ) \right ]$
$= 310.177 \,amu$
$3Ca \, \, \, \, (3\, atoms)(40.078 \, amu/atom) = 120.234 \, amu$
$2P \, \, \, \, (2\, atoms)(30.973761 \, amu/atom) = 61.947522 \, amu$
$+ 8O \, \, \, \, (8\, atoms)(15.9994 \, amu/atom) = 127.9952 \, amu$
$Ca_3P_2O_8 \, \, \, \, \text {formula mass of Ca}_3(PO_4)_2 = 310.177 \, amu$
$(moles)(molar mass) \rightarrow mass \label{2.7.1}$
or, more specifically,
$moles \left ( {grams \over mole } \right ) = grams$
Conversely, to convert the mass of a substance to moles:
$\left ( {grams \over molar mass } \right ) \rightarrow moles \label{2.7.2A}$
$\left ( { grams \over grams/mole} \right ) = grams \left ( {mole \over grams } \right ) = moles \label{2.7.2B}$
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$
the production of two moles of water would require the consumption of 2 moles of $H_2$ and one mole of $O_2$. Therefore, when considering this particular reaction
• 2 moles of H2
• 1 mole of O2 and
• 2 moles of H2O
would be considered to be stoichiometrically equivalent quantitites.
These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of $H_2O$ would be produced from 1.57 moles of $O_2$?
$(1.57\; mol\; O_2) \left( \dfrac{2\; mol H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O$
The ratio $\left( \dfrac{2\; mol\; H_2O}{1\;mol\;O_2} \right)$ is the stoichiometric relationship between $H_2O$ and $O_2$ from the balanced equation for this reaction.
Example $3$
For the combustion of butane ($C_4H_{10}$) the balanced equation is:
$2C_4H_{10(l)} + 13O_{2(g)} \rightarrow 8CO_{2(g)} + 10H_2O_{(l)}$
Calculate the mass of $CO_2$ that is produced in burning 1.00 gram of $C_4H_{10}$.
Solution
Thus, the overall sequence of steps to solve this problem is:
First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:
$(1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10}$
Now, the stoichiometric relationship between $C_4H_{10}$ and $CO_2$ is:
$\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right)$
Therefore:
$\left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2$
The question called for the determination of the mass of $CO_2$ produced, thus we have to convert moles of $CO_2$ into grams (by using the molecular weight of $CO_2$):
$6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2$
Be sure to pay attention to the units when converting between mass and moles. Figure $1$ is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Example $3$ and Example $4$.
Conversions Between Grams, Mol, & Atoms: https://youtu.be/rOvErpAnoCg
Example $4$
For 35.00 g of ethylene glycol (HOCH2CH2OH), which is used in inks for ballpoint pens, calculate the number of
1. moles.
2. molecules.
Given: mass and molecular formula
Asked for: number of moles and number of molecules
Strategy:
1. Use the molecular formula of the compound to calculate its molecular mass in grams per mole.
2. Convert from mass to moles by dividing the mass given by the compound’s molar mass.
3. Convert from moles to molecules by multiplying the number of moles by Avogadro’s number.
Solution:
a. A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example $2$.7.1:
$2C (2 \,atoms )(12.011 \, amu/atom) = 24.022 \, amu$
$6H (6 \,atoms )(1.0079 \, amu/atom) = 6.0474 \, amu$
$2O (2 \,atoms )(15.9994 \, amu/atom) = 31.9988 \, amu$
$C_2H_6O_2 \text {molecular mass of ethylene glycol} = 62.068 \, amu$
The molar mass of ethylene glycol is 62.068 g/mol.
B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole):
${ \text {mass of ethylene glycol (g)} \over \text {molar mass (g/mol)} } = \text {moles ethylene glycol (mol) }$
So
$35.00 \, g \text {ethylene glycol} \left ( {1 \, mole \text {ethylene glycol} \over 62.068 \, g \text {ethylene glycol } } \right ) = 0.5639 \,mol \text {ethylene glycol}$
It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations.
b. C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:
$\text {molecules of ethylene glycol} = 0.5639 \, mol \left ( {6.022 \times 10^{23} \, molecules \over 1 \, mol } \right )$
$= 3.396 \times 10^{23} \, molecules$
Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 1023 molecules, which is indeed the case.
$2S (2 \, atoms)(32.065 \, amu/atom ) = 64.130 \, amu$
$+ 2Cl (2 \, atoms )(35.453 \, amu/atom ) = 70.906 \, amu$
$S_2 Cl_2 \text {molecular mass of } S_2Cl_2 = 135.036 \, amu$
The molar mass of S2Cl2 is 135.036 g/mol.
B The mass of 1.75 mol of S2Cl2 is calculated as follows:
$moles S_2Cl_2 \left [\text {molar mass}\left ({ g \over mol} \right )\right ] \rightarrow mass of S_2Cl_2 \, (g)$
$1.75 \, mol S_2Cl_2\left ({135.036 \, g S_2Cl_2 \over 1 \, mol S_2Cl_2 } \right ) = 236 \, g S_2Cl_2$
b.
A The formula mass of Ca(ClO)2 is obtained as follows:
$1Ca (1 \, atom)(40.078 \, amu/atom) = 40.078 \, amu$
$2Cl (2 \, atoms)(35.453 \, amu/atom) = 70.906 \, amu$
$+ 2O (2 \, atoms)(15.9994 \, amu/atom) = 31.9988 \, amu$
$Ca (ClO)_2 \text { formula mass of } Ca (ClO)_2 = 142.983 \, amu$
The molar mass of Ca(ClO)2 142.983 g/mol.
B The mass of 1.75 mol of Ca(ClO)2 is calculated as follows:
$moles Ca(ClO)_2 \left [{\text {molar mass} Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2} \right ]=mass Ca(ClO)_2$
$1.75 \, mol Ca(ClO)_2 \left [ {142.983 \, g Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2 } \right ] = 250 \, g Ca(ClO)_2$
Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is.
Exercise $5$
Calculate the mass of 0.0122 mol of each compound.
1. Si3N4 (silicon nitride), used as bearings and rollers
2. (CH3)3N (trimethylamine), a corrosion inhibitor
Answer:
1. 1.71 g
2. 0.721 g
The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the balanced equation:
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$
the production of two moles of water would require the consumption of 2 moles of $H_2$ and one mole of $O_2$. Therefore, when considering this particular reaction
• 2 moles of H2
• 1 mole of O2 and
• 2 moles of H2O
would be considered to be stoichiometrically equivalent quantitites.
These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of $H_2O$ would be produced from 1.57 moles of $O_2$?
$(1.57\; mol\; O_2) \left( \dfrac{2\; mol H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O$
The ratio $\left( \dfrac{2\; mol\l H_2O}{1\;mol\;O_2} \right)$ is the stoichiometric relationship between $H_2O$ and $O_2$ from the balanced equation for this reaction.
Example $6$
For the combustion of butane ($C_4H_{10}$) the balanced equation is:
$2C_4H_{10(l)} + 13O_{2(g)} \rightarrow 8CO_{2(g)} + 10H_2O_{(l)}$
Calculate the mass of $CO_2$ that is produced in burning 1.00 gram of $C_4H_{10}$.
Solution
First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:
$(1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10}$
Now, the stoichiometric relationship between $C_4H_{10}$ and $CO_2$ is:
$\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right)$
Therefore:
$\left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2$
The question called for the determination of the mass of $CO_2$ produced, thus we have to convert moles of $CO_2$ into grams (by using the molecular weight of $CO_2$):
$6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2$
Thus, the overall sequence of steps to solve this problem were:
In a similar way we could determine the mass of water produced, or oxygen consumed, etc.
Summary
• To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole.
The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12, Avogadro’s number (6.022 × 1023) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 1023 atoms, molecules, or formula units of that substance.
2.8: Using the Mole Concept in Calculations
under construction | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/02%3A_Atoms_and_The_Atomic_Theory/2.7%3A_The_Concept_of_Mole_and_the_Avogadro_Constant.txt |
Learning Objectives
• To understand the differences between covalent and ionic bonding.
The atoms in all substances that contain multiple atoms are held together by electrostatic interactions—interactions between electrically charged particles such as protons and electrons. Electrostatic attraction between oppositely charged species (positive and negative) results in a force that causes them to move toward each other, like the attraction between opposite poles of two magnets. In contrast, electrostatic repulsion between two species with the same charge (either both positive or both negative) results in a force that causes them to repel each other, as do the same poles of two magnets. Atoms form chemical compounds when the attractive electrostatic interactions between them are stronger than the repulsive interactions. Collectively, the attractive interactions between atoms are called chemical bonds.
Chemical bonds are generally divided into two fundamentally different types: ionic and covalent. In reality, however, the bonds in most substances are neither purely ionic nor purely covalent, but lie on a spectrum between these extremes. Although purely ionic and purely covalent bonds represent extreme cases that are seldom encountered in any but very simple substances, a brief discussion of these two extremes helps explain why substances with different kinds of chemical bonds have very different properties. Ionic compounds consist of positively and negatively charged ions held together by strong electrostatic forces, whereas covalent compounds generally consist of molecules, which are groups of atoms in which one or more pairs of electrons are shared between bonded atoms. In a covalent bond, atoms are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons they share. This discussion of structures and formulas begins by describing covalent compounds. The energetic factors involved in bond formation are described in more quantitative detail in later.
Note
Ionic compounds consist of ions of opposite charges held together by strong electrostatic forces, whereas pairs of electrons are shared between bonded atoms in covalent compounds.
Covalent Molecules and Compounds
Just as an atom is the simplest unit that has the fundamental chemical properties of an element, a molecule is the simplest unit that has the fundamental chemical properties of a covalent compound. Some pure elements exist as covalent molecules. Hydrogen, nitrogen, oxygen, and the halogens occur naturally as the diatomic (“two atoms”) molecules H2, N2, O2, F2, Cl2, Br2, and I2 (part (a) in Figure $1$). Similarly, a few pure elements exist as polyatomic (“many atoms”) molecules, such as elemental phosphorus and sulfur, which occur as P4 and S8 (part (b) in Figure $1$).
Each covalent compound is represented by a molecular formula, which gives the atomic symbol for each component element, in a prescribed order, accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number of atoms is greater than 1. For example, water, with two hydrogen atoms and one oxygen atom per molecule, is written as $H_2O$. Similarly, carbon dioxide, which contains one carbon atom and two oxygen atoms in each molecule, is written as $CO_2$.
Covalent compounds that predominantly contain carbon and hydrogen are called organic compounds. The convention for representing the formulas of organic compounds is to write carbon first, followed by hydrogen and then any other elements in alphabetical order (e.g., CH4O is methyl alcohol, a fuel). Compounds that consist primarily of elements other than carbon and hydrogen are called inorganic compounds; they include both covalent and ionic compounds. In inorganic compounds, the component elements are listed beginning with the one farthest to the left in the periodic table, as in CO2 or SF6. Those in the same group are listed beginning with the lower element and working up, as in ClF. By convention, however, when an inorganic compound contains both hydrogen and an element from groups 13–15, hydrogen is usually listed last in the formula. Examples are ammonia (NH3) and silane (SiH4). Compounds such as water, whose compositions were established long before this convention was adopted, are always written with hydrogen first: Water is always written as H2O, not OH2. The conventions for inorganic acids, such as hydrochloric acid (HCl) and sulfuric acid (H2SO4), are described elswhere.
Note
For organic compounds: write C first, then H, and then the other elements in alphabetical order. For molecular inorganic compounds: start with the element at far left in the periodic table; list elements in same group beginning with the lower element and working up.
Example $1$
Write the molecular formula of each compound.
1. The phosphorus-sulfur compound that is responsible for the ignition of so-called strike anywhere matches has 4 phosphorus atoms and 3 sulfur atoms per molecule.
2. Ethyl alcohol, the alcohol of alcoholic beverages, has 1 oxygen atom, 2 carbon atoms, and 6 hydrogen atoms per molecule.
3. Freon-11, once widely used in automobile air conditioners and implicated in damage to the ozone layer, has 1 carbon atom, 3 chlorine atoms, and 1 fluorine atom per molecule.
Given: identity of elements present and number of atoms of each
Asked for: molecular formula
Strategy:
A Identify the symbol for each element in the molecule. Then identify the substance as either an organic compound or an inorganic compound.
B If the substance is an organic compound, arrange the elements in order beginning with carbon and hydrogen and then list the other elements alphabetically. If it is an inorganic compound, list the elements beginning with the one farthest left in the periodic table. List elements in the same group starting with the lower element and working up.
C From the information given, add a subscript for each kind of atom to write the molecular formula.
Solution:
a.
• A The molecule has 4 phosphorus atoms and 3 sulfur atoms. Because the compound does not contain mostly carbon and hydrogen, it is inorganic.
• B Phosphorus is in group 15, and sulfur is in group 16. Because phosphorus is to the left of sulfur, it is written first.
• C Writing the number of each kind of atom as a right-hand subscript gives P4S3 as the molecular formula.
b.
• A Ethyl alcohol contains predominantly carbon and hydrogen, so it is an organic compound.
• B The formula for an organic compound is written with the number of carbon atoms first, the number of hydrogen atoms next, and the other atoms in alphabetical order: CHO.
• C Adding subscripts gives the molecular formula $\ce{C2H6O}$.
c.
• A Freon-11 contains carbon, chlorine, and fluorine. It can be viewed as either an inorganic compound or an organic compound (in which fluorine has replaced hydrogen). The formula for Freon-11 can therefore be written using either of the two conventions.
• B According to the convention for inorganic compounds, carbon is written first because it is farther left in the periodic table. Fluorine and chlorine are in the same group, so they are listed beginning with the lower element and working up: CClF. Adding subscripts gives the molecular formula CCl3F.
• C We obtain the same formula for Freon-11 using the convention for organic compounds. The number of carbon atoms is written first, followed by the number of hydrogen atoms (zero) and then the other elements in alphabetical order, also giving CCl3F.
Exercise $1$
Write the molecular formula for each compound.
1. Nitrous oxide, also called “laughing gas,” has 2 nitrogen atoms and 1 oxygen atom per molecule. Nitrous oxide is used as a mild anesthetic for minor surgery and as the propellant in cans of whipped cream.
2. Sucrose, also known as cane sugar, has 12 carbon atoms, 11 oxygen atoms, and 22 hydrogen atoms.
3. Sulfur hexafluoride, a gas used to pressurize “unpressurized” tennis balls and as a coolant in nuclear reactors, has 6 fluorine atoms and 1 sulfur atom per molecule.
Answer:
1. N2O
2. C12H22O11
3. SF6
Representations of Molecular Structures
Molecular formulas give only the elemental composition of molecules. In contrast, structural formulas show which atoms are bonded to one another and, in some cases, the approximate arrangement of the atoms in space. Knowing the structural formula of a compound enables chemists to create a three-dimensional model, which provides information about how that compound will behave physically and chemically.
The structural formula for H2 can be drawn as H–H and that for I2 as I–I, where the line indicates a single pair of shared electrons, a single bond. Two pairs of electrons are shared in a double bond, which is indicated by two lines—for example, O2 is O=O. Three electron pairs are shared in a triple bond, which is indicated by three lines—for example, N2 is N≡N (Figure $2$). Carbon is unique in the extent to which it forms single, double, and triple bonds to itself and other elements. The number of bonds formed by an atom in its covalent compounds is not arbitrary. Hydrogen, oxygen, nitrogen, and carbon have very strong tendencies to form substances in which they have one, two, three, and four bonds to other atoms, respectively (Table $1$).
Table $1$: The Number of Bonds That Selected Atoms Commonly Form to Other Atoms
Atom Number of Bonds
H (group 1) 1
O (group 16) 2
N (group 15) 3
C (group 14) 4
The structural formula for water can be drawn as follows:
Because the latter approximates the experimentally determined shape of the water molecule, it is more informative. Similarly, ammonia (NH3) and methane (CH4) are often written as planar molecules:
As shown in Figure $3$, however, the actual three-dimensional structure of NH3 looks like a pyramid with a triangular base of three hydrogen atoms. The structure of CH4, with four hydrogen atoms arranged around a central carbon atom as shown in Figure $3$, is tetrahedral: the hydrogen atoms are positioned at every other vertex of a cube. Many compounds—carbon compounds, in particular—have four bonded atoms arranged around a central atom to form a tetrahedron.
Figures $3$-$3$ illustrate different ways to represent the structures of molecules. It should be clear that there is no single “best” way to draw the structure of a molecule; the method used depends on which aspect of the structure should be emphasized and how much time and effort is required. Figure $4$ shows some of the different ways to portray the structure of a slightly more complex molecule: methanol. These representations differ greatly in their information content. For example, the molecular formula for methanol (part (a) in Figure $4$) gives only the number of each kind of atom; writing methanol as CH4O tells nothing about its structure. In contrast, the structural formula (part (b) in Figure $4$) indicates how the atoms are connected, but it makes methanol look as if it is planar (which it is not). Both the ball-and-stick model (part (c) in Figure $4$) and the perspective drawing (part (d) in Figure $4$) show the three-dimensional structure of the molecule. The latter (also called a wedge-and-dash representation) is the easiest way to sketch the structure of a molecule in three dimensions. It shows which atoms are above and below the plane of the paper by using wedges and dashes, respectively; the central atom is always assumed to be in the plane of the paper. The space-filling model (part (e) in Figure $4$) illustrates the approximate relative sizes of the atoms in the molecule, but it does not show the bonds between the atoms. In addition, in a space-filling model, atoms at the “front” of the molecule may obscure atoms at the “back.”
Although a structural formula, a ball-and-stick model, a perspective drawing, and a space-filling model provide a significant amount of information about the structure of a molecule, each requires time and effort. Consequently, chemists often use a condensed structural formula (part (f) in Figure $4$), which omits the lines representing bonds between atoms and simply lists the atoms bonded to a given atom next to it. Multiple groups attached to the same atom are shown in parentheses, followed by a subscript that indicates the number of such groups. For example, the condensed structural formula for methanol is CH3OH, which indicates that the molecule contains a CH3 unit that looks like a fragment of methane (CH4). Methanol can therefore be viewed either as a methane molecule in which one hydrogen atom has been replaced by an –OH group or as a water molecule in which one hydrogen atom has been replaced by a –CH3 fragment. Because of their ease of use and information content, we use condensed structural formulas for molecules throughout this text. Ball-and-stick models are used when needed to illustrate the three-dimensional structure of molecules, and space-filling models are used only when it is necessary to visualize the relative sizes of atoms or molecules to understand an important point.
Example $2$
Write the molecular formula for each compound. The condensed structural formula is given.
1. Sulfur monochloride (also called disulfur dichloride) is a vile-smelling, corrosive yellow liquid used in the production of synthetic rubber. Its condensed structural formula is ClSSCl.
2. Ethylene glycol is the major ingredient in antifreeze. Its condensed structural formula is HOCH2CH2OH.
3. Trimethylamine is one of the substances responsible for the smell of spoiled fish. Its condensed structural formula is (CH3)3N.
Given: condensed structural formula
Asked for: molecular formula
Strategy:
1. Identify every element in the condensed structural formula and then determine whether the compound is organic or inorganic.
2. As appropriate, use either organic or inorganic convention to list the elements. Then add appropriate subscripts to indicate the number of atoms of each element present in the molecular formula.
Solution:
The molecular formula lists the elements in the molecule and the number of atoms of each.
1. A Each molecule of sulfur monochloride has two sulfur atoms and two chlorine atoms. Because it does not contain mostly carbon and hydrogen, it is an inorganic compound. B Sulfur lies to the left of chlorine in the periodic table, so it is written first in the formula. Adding subscripts gives the molecular formula S2Cl2.
2. A Counting the atoms in ethylene glycol, we get six hydrogen atoms, two carbon atoms, and two oxygen atoms per molecule. The compound consists mostly of carbon and hydrogen atoms, so it is organic. B As with all organic compounds, C and H are written first in the molecular formula. Adding appropriate subscripts gives the molecular formula C2H6O2.
3. A The condensed structural formula shows that trimethylamine contains three CH3 units, so we have one nitrogen atom, three carbon atoms, and nine hydrogen atoms per molecule. Because trimethylamine contains mostly carbon and hydrogen, it is an organic compound. B According to the convention for organic compounds, C and H are written first, giving the molecular formula C3H9N.
Exercise $2$
Write the molecular formula for each molecule.
1. Chloroform, which was one of the first anesthetics and was used in many cough syrups until recently, contains one carbon atom, one hydrogen atom, and three chlorine atoms. Its condensed structural formula is CHCl3.
2. Hydrazine is used as a propellant in the attitude jets of the space shuttle. Its condensed structural formula is H2NNH2.
3. Putrescine is a pungent-smelling compound first isolated from extracts of rotting meat. Its condensed structural formula is H2NCH2CH2CH2CH2NH2. This is often written as H2N(CH2)4NH2 to indicate that there are four CH2 fragments linked together.
Answer a
CHCl3
Answer b
N2H4
Answer c
C4H12N2
Ionic Compounds
The substances described in the preceding discussion are composed of molecules that are electrically neutral; that is, the number of positively-charged protons in the nucleus is equal to the number of negatively-charged electrons. In contrast, ions are atoms or assemblies of atoms that have a net electrical charge. Ions that contain fewer electrons than protons have a net positive charge and are called cations. Conversely, ions that contain more electrons than protons have a net negative charge and are called anions. Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge.
Note
Ionic compounds contain both cations and anions in a ratio that results in zero electrical charge.
In covalent compounds, electrons are shared between bonded atoms and are simultaneously attracted to more than one nucleus. In contrast, ionic compounds contain cations and anions rather than discrete neutral molecules. Ionic compounds are held together by the attractive electrostatic interactions between cations and anions. In an ionic compound, the cations and anions are arranged in space to form an extended three-dimensional array that maximizes the number of attractive electrostatic interactions and minimizes the number of repulsive electrostatic interactions (Figure $5$). As shown in Equation 3.1.1, the electrostatic energy of the interaction between two charged particles is proportional to the product of the charges on the particles and inversely proportional to the distance between them:
$\text {electrostatic energy} \propto {Q_1Q_2 \over r} \label{3.1.1}$
where
• $Q_1$ and $Q_2$ are the electrical charges on particles 1 and 2, and
• $r$ is the distance between them.
When $Q_1$ and $Q_2$ are both positive, corresponding to the charges on cations, the cations repel each other and the electrostatic energy is positive. When $Q_1$ and $Q_2$ are both negative, corresponding to the charges on anions, the anions repel each other and the electrostatic energy is again positive. The electrostatic energy is negative only when the charges have opposite signs; that is, positively charged species are attracted to negatively charged species and vice versa. As shown in Figure $6$, the strength of the interaction is proportional to the magnitude of the charges and decreases as the distance between the particles increases. These energetic factors are discussed in greater quantitative detail later.
Note
If the electrostatic energy is positive, the particles repel each other; if the electrostatic energy is negative, the particles are attracted to each other.
One example of an ionic compound is sodium chloride (NaCl), formed from sodium and chlorine. In forming chemical compounds, many elements have a tendency to gain or lose enough electrons to attain the same number of electrons as the noble gas closest to them in the periodic table. When sodium and chlorine come into contact, each sodium atom gives up an electron to become a Na+ ion, with 11 protons in its nucleus but only 10 electrons (like neon), and each chlorine atom gains an electron to become a Cl ion, with 17 protons in its nucleus and 18 electrons (like argon), as shown in part (b) in Figure $5$. Solid sodium chloride contains equal numbers of cations (Na+) and anions (Cl), thus maintaining electrical neutrality. Each Na+ ion is surrounded by 6 Cl ions, and each Cl ion is surrounded by 6 Na+ ions. Because of the large number of attractive Na+Cl interactions, the total attractive electrostatic energy in NaCl is great.
Consistent with a tendency to have the same number of electrons as the nearest noble gas, when forming ions, elements in groups 1, 2, and 3 tend to lose one, two, and three electrons, respectively, to form cations, such as Na+ and Mg2+. They then have the same number of electrons as the nearest noble gas: neon. Similarly, K+, Ca2+, and Sc3+ have 18 electrons each, like the nearest noble gas: argon. In addition, the elements in group 13 lose three electrons to form cations, such as Al3+, again attaining the same number of electrons as the noble gas closest to them in the periodic table. Because the lanthanides and actinides formally belong to group 3, the most common ion formed by these elements is M3+, where M represents the metal. Conversely, elements in groups 17, 16, and 15 often react to gain one, two, and three electrons, respectively, to form ions such as Cl, S2−, and P3−. Ions such as these, which contain only a single atom, are called monatomic ions. The charges of most monatomic ions derived from the main group elements can be predicted by simply looking at the periodic table and counting how many columns an element lies from the extreme left or right. For example, barium (in Group 2) forms Ba2+ to have the same number of electrons as its nearest noble gas, xenon; oxygen (in Group 16) forms O2− to have the same number of electrons as neon; and cesium (in Group 1) forms Cs+, which has the same number of electrons as xenon. Note that this method is ineffective for most of the transition metals, as discussed in Section 2.3. Some common monatomic ions are listed in Table $2$.
Note
Elements in Groups 1, 2, and 3 tend to form 1+, 2+, and 3+ ions, respectively; elements in Groups 15, 16, and 17 tend to form 3−, 2−, and 1− ions, respectively.
Table $2$: Some Common Monatomic Ions and Their Names
Group 1 Group 2 Group 3 Group 13 Group 15 Group 16 Group 17
Li+
lithium
Be2+
beryllium
N3−
nitride
(azide)
O2−
oxide
F
fluoride
Na+
sodium
Mg2+
magnesium
Al3+
aluminum
P3−
phosphide
S2−
sulfide
Cl
chloride
K+
potassium
Ca2+
calcium
Sc3+
scandium
Ga3+
gallium
As3
arsenide
Se2
selenide
Br
bromide
Rb+
rubidium
Sr2+
strontium
Y3+
yttrium
In3+
indium
Te2
telluride
I
iodide
Cs+
cesium
Ba2+
barium
La3+
lanthanum
Example $3$
Predict the charge on the most common monatomic ion formed by each element.
1. aluminum, used in the quantum logic clock, the world’s most precise clock
2. selenium, used to make ruby-colored glass
3. yttrium, used to make high-performance spark plugs
Given: element
Asked for: ionic charge
Strategy:
A Identify the group in the periodic table to which the element belongs. Based on its location in the periodic table, decide whether the element is a metal, which tends to lose electrons; a nonmetal, which tends to gain electrons; or a semimetal, which can do either.
B After locating the noble gas that is closest to the element, determine the number of electrons the element must gain or lose to have the same number of electrons as the nearest noble gas.
Solution:
1. A Aluminum is a metal in group 13; consequently, it will tend to lose electrons. B The nearest noble gas to aluminum is neon. Aluminum will lose three electrons to form the Al3+ ion, which has the same number of electrons as neon.
2. A Selenium is a nonmetal in group 16, so it will tend to gain electrons. B The nearest noble gas is krypton, so we predict that selenium will gain two electrons to form the Se2 ion, which has the same number of electrons as krypton.
3. A Yttrium is in group 3, and elements in this group are metals that tend to lose electrons. B The nearest noble gas to yttrium is krypton, so yttrium is predicted to lose three electrons to form Y3+, which has the same number of electrons as krypton.
Exercise $3$
Predict the charge on the most common monatomic ion formed by each element.
1. calcium, used to prevent osteoporosis
2. iodine, required for the synthesis of thyroid hormones
3. zirconium, widely used in nuclear reactors
Answer:
1. Ca2+
2. I
3. Zr4+
Molecular and Ionic Compounds: https://youtu.be/zJejgCll1bw
Physical Properties of Ionic and Covalent Compounds
In general, ionic and covalent compounds have different physical properties. Ionic compounds form hard crystalline solids that melt at high temperatures and are resistant to evaporation. These properties stem from the characteristic internal structure of an ionic solid, illustrated schematically in part (a) in Figure $8$ which shows the three-dimensional array of alternating positive and negative ions held together by strong electrostatic attractions. In contrast, as shown in part (b) in Figure $8$ most covalent compounds consist of discrete molecules held together by comparatively weak intermolecular forces (the forces between molecules), even though the atoms within each molecule are held together by strong intramolecular covalent bonds (the forces within the molecule). Covalent substances can be gases, liquids, or solids at room temperature and pressure, depending on the strength of the intermolecular interactions. Covalent molecular solids tend to form soft crystals that melt at low temperatures and evaporate easily.Some covalent substances, however, are not molecular but consist of infinite three-dimensional arrays of covalently bonded atoms and include some of the hardest materials known, such as diamond. This topic will be addressed elsewhere. The covalent bonds that hold the atoms together in the molecules are unaffected when covalent substances melt or evaporate, so a liquid or vapor of independent molecules is formed. For example, at room temperature, methane, the major constituent of natural gas, is a gas that is composed of discrete CH4 molecules. A comparison of the different physical properties of ionic compounds and covalent molecular substances is given in Table $3$.
Table $3$: The Physical Properties of Typical Ionic Compounds and Covalent Molecular Substances
Ionic Compounds Covalent Molecular Substances
hard solids gases, liquids, or soft solids
high melting points low melting points
nonvolatile volatile
• The Learning Objective of this Module is to describe the composition of a chemical compound.
When chemists synthesize a new compound, they may not yet know its molecular or structural formula. In such cases, they usually begin by determining its empirical formula, the relative numbers of atoms of the elements in a compound, reduced to the smallest whole numbers. Because the empirical formula is based on experimental measurements of the numbers of atoms in a sample of the compound, it shows only the ratios of the numbers of the elements present. The difference between empirical and molecular formulas can be illustrated with butane, a covalent compound used as the fuel in disposable lighters. The molecular formula for butane is C4H10. The ratio of carbon atoms to hydrogen atoms in butane is 4:10, which can be reduced to 2:5. The empirical formula for butane is therefore C2H5. The formula unit is the absolute grouping of atoms or ions represented by the empirical formula of a compound, either ionic or covalent. Butane has the empirical formula C2H5, but it contains two C2H5 formula units, giving a molecular formula of C4H10.
Because ionic compounds do not contain discrete molecules, empirical formulas are used to indicate their compositions. All compounds, whether ionic or covalent, must be electrically neutral. Consequently, the positive and negative charges in a formula unit must exactly cancel each other. If the cation and the anion have charges of equal magnitude, such as Na+ and Cl, then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be NaCl. If the charges are not the same magnitude, then a cation:anion ratio other than 1:1 is needed to produce a neutral compound. In the case of Mg2+ and Cl, for example, two Cl ions are needed to balance the two positive charges on each Mg2+ ion, giving an empirical formula of MgCl2. Similarly, the formula for the ionic compound that contains Na+ and O2− ions is Na2O.
Note
Ionic compounds do not contain discrete molecules, so empirical formulas are used to indicate their compositions.
Binary Ionic Compounds
An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is MgCl2, a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows:
Crossing charges. One method for obtaining subscripts in the empirical formula is by crossing charges.
When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Mg2O2:
This simplifies to its correct empirical formula MgO. The empirical formula has one Mg2+ ion and one O2− ion.
Example $4$
Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair.
1. Ga3+ and As3
2. Eu3+ and O2−
3. calcium and chlorine
Given: ions or elements
Asked for: empirical formula for binary ionic compound
Strategy:
A If not given, determine the ionic charges based on the location of the elements in the periodic table.
B Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the lowest numbers
to write the empirical formula. Check to make sure the empirical formula is electrically neutral.
Solution
a. B Using the absolute values of the charges on the ions as the subscripts gives Ga3As3:
Reducing the subscripts to the smallest whole numbers gives the empirical formula GaAs, which is electrically neutral [+3 + (−3) = 0]. Alternatively, we could recognize that Ga3+ and As3 have charges of equal magnitude but opposite signs. One Ga3+ ion balances the charge on one As3 ion, and a 1:1 compound will have no net charge. Because we write subscripts only if the number is greater than 1, the empirical formula is GaAs. GaAs is gallium arsenide, which is widely used in the electronics industry in transistors and other devices.
b. B Because Eu3+ has a charge of +3 and O2− has a charge of −2, a 1:1 compound would have a net charge of +1. We must therefore find multiples of the charges that cancel. We cross charges, using the absolute value of the charge on one ion as the subscript for the other ion:
The subscript for Eu3+ is 2 (from O2−), and the subscript for O2− is 3 (from Eu3+), giving Eu2O3; the subscripts cannot be reduced further. The empirical formula contains a positive charge of 2(+3) = +6 and a negative charge of 3(−2) = −6, for a net charge of 0. The compound Eu2O3 is neutral. Europium oxide is responsible for the red color in television and computer screens.
c. A Because the charges on the ions are not given, we must first determine the charges expected for the most common ions derived from calcium and chlorine. Calcium lies in group 2, so it should lose two electrons to form Ca2+. Chlorine lies in group 17, so it should gain one electron to form Cl.
B Two Cl ions are needed to balance the charge on one Ca2+ ion, which leads to the empirical formula CaCl2. We could also cross charges, using the absolute value of the charge on Ca2+ as the subscript for Cl and the absolute value of the charge on Cl as the subscript for Ca:
The subscripts in CaCl2 cannot be reduced further. The empirical formula is electrically neutral [+2 + 2(−1) = 0]. This compound is calcium chloride, one of the substances used as “salt” to melt ice on roads and sidewalks in winter.
Exercise $4$
Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair.
1. Li+ and N3−
2. Al3+ and O2−
3. lithium and oxygen
Answer:
1. Li3N
2. Al2O3
3. Li2O
Polyatomic Ions
Polyatomic ions are groups of atoms that bear net electrical charges, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH4+) and the methylammonium (CH3NH3+) ions. Polyatomic anions are much more numerous than polyatomic cations; some common examples are in Table $4$.
Table $4$: Common Polyatomic Ions and Their Names
Formula Name of Ion
NH4+ ammonium
CH3NH3+ methylammonium
OH hydroxide
O22 peroxide
CN cyanide
SCN thiocyanate
NO2 nitrite
NO3 nitrate
CO32 carbonate
HCO3 hydrogen carbonate, or bicarbonate
SO32 sulfite
SO42 sulfate
HSO4 hydrogen sulfate, or bisulfate
PO43 phosphate
HPO42 hydrogen phosphate
H2PO4 dihydrogen phosphate
ClO hypochlorite
ClO2 chlorite
ClO3 chlorate
ClO4 perchlorate
MnO4 permanganate
CrO42 chromate
Cr2O72 dichromate
C2O42 oxalate
HCO2 formate
CH3CO2 acetate
C6H5CO2 benzoate
The method used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus, K+ and NO3 ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42 form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3 form NH4NO3 (ammonium nitrate), a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43 ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca2+ ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca3(PO4)2; the parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43 ions.
Example $5$
Write the empirical formula for the compound formed from each ion pair.
1. Na+ and HPO42
2. potassium cation and cyanide anion
3. calcium cation and hypochlorite anion
Given: ions
Asked for: empirical formula for ionic compound
Strategy:
A If it is not given, determine the charge on a monatomic ion from its location in the periodic table. Use Table $4$ "Common Polyatomic Ions and Their Names" to find the charge on a polyatomic ion.
B Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the smallest whole numbers when writing the empirical formula.
Solution:
a. B Because HPO42 has a charge of −2 and Na+ has a charge of +1, the empirical formula requires two Na+ ions to balance the charge of the polyatomic ion, giving Na2HPO4. The subscripts are reduced to the lowest numbers, so the empirical formula is Na2HPO4. This compound is sodium hydrogen phosphate, which is used to provide texture in processed cheese, puddings, and instant breakfasts.
b. A The potassium cation is K+, and the cyanide anion is CN. B Because the magnitude of the charge on each ion is the same, the empirical formula is KCN. Potassium cyanide is highly toxic, and at one time it was used as rat poison. This use has been discontinued, however, because too many people were being poisoned accidentally.
c. A The calcium cation is Ca2+, and the hypochlorite anion is ClO. B Two ClO ions are needed to balance the charge on one Ca2+ ion, giving Ca(ClO)2. The subscripts cannot be reduced further, so the empirical formula is Ca(ClO)2. This is calcium hypochlorite, the “chlorine” used to purify water in swimming pools.
Exercise $5$
Write the empirical formula for the compound formed from each ion pair.
1. Ca2+ and H2PO4
2. sodium cation and bicarbonate anion
3. ammonium cation and sulfate anion
Answer:
1. Ca(H2PO4)2: calcium dihydrogen phosphate is one of the ingredients in baking powder.
2. NaHCO3: sodium bicarbonate is found in antacids and baking powder; in pure form, it is sold as baking soda.
3. (NH4)2SO4: ammonium sulfate is a common source of nitrogen in fertilizers.
Summary
• There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties.
• The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit.Contributors
The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. Ionic compounds contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Most covalent compounds consist of molecules, groups of atoms in which one or more pairs of electrons are shared by at least two atoms to form a covalent bond. The atoms in molecules are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons shared by the nuclei. The molecular formula of a covalent compound gives the types and numbers of atoms present. Compounds that contain predominantly carbon and hydrogen are called organic compounds, whereas compounds that consist primarily of elements other than carbon and hydrogen are inorganic compounds. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two. A structural formula indicates the composition and approximate structure and shape of a molecule. Single bonds, double bonds, and triple bonds are covalent bonds in which one, two, and three pairs of electrons, respectively, are shared between two bonded atoms. Atoms or groups of atoms that possess a net electrical charge are called ions; they can have either a positive charge (cations) or a negative charge (anions). Ions can consist of one atom (monatomic ions) or several (polyatomic ions). The charges on monatomic ions of most main group elements can be predicted from the location of the element in the periodic table. Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure.
An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.1%3A_Types_of_Chemical_Compounds_and_their_Formulas.txt |
$2 \times \text { atomic mass of carbon} = 2 \, atoms \left ( {12.011 \, amu \over atoms } \right ) = 24.022 \,amu$
$6 \times \text { atomic mass of hydrogen} = 2 \, atoms \left ( {1.0079 \, amu \over atoms } \right ) = 6.0474 \,amu$
$1 \times \text { atomic mass of oxygen} = 1 \, atoms \left ( {15.9994 \, amu \over atoms } \right ) = 15.994 \,amu$
C Adding together the masses gives the molecular mass:
$24.022 \,amu + 6.0474 \,amu + 15.9994 \,amu = 46.069 \,amu$
Alternatively, we could have used unit conversions to reach the result in one step:
$\left [ 2 \, atoms C \left ( {12.011 \, amu \over 1 \, atom C} \right ) \right ] + \left [ 6 \, atoms H \left ( {1.0079 \, amu \over 1 \, atom H} \right ) \right ] + \left [ 1 \, atoms C \left ( {15.9994 \, amu \over 1 \, atom 0} \right ) \right ] = 46.069 \, amu$
The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:
$2 C \, \, \, (2\, atoms) (12.011 \, amu/atom ) = 24.022 \, amu$
$6 H \, \, \, (6\, atoms) (1.0079 \, amu/atom ) = 6.0474 \, amu$
$+ 1O \, \, \, (1\, atoms) (15.9994 \, amu/atom ) = 15.9994 \, amu$
$C_2H_6O \, \, \, \, \, \text {molecular mass of ethanol} = 46.069 \, amu$
Exercise $1$: Molecular Mass of Freon
Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is $CCl_3F$. Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:
Answer: 137.368 \,amu
Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. The units are atomic mass units.
Note
Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.
Example $2$: Formula Mass of Calcium Phosphate
Calculate the formula mass of Ca3(PO4)2, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.
Given: ionic compound
Asked for: formula mass
Strategy:
1. Determine the number of atoms of each element in the empirical formula.
2. Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
3. Add together the masses to give the formula mass.
Solution:
A The empirical formula—Ca3(PO4)2—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca2+ ions and two PO43 ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.
B Taking atomic masses from the periodic table, we obtain
$3 \times \text {atomic mass of calcium} = 3 \, atoms \left ( {40.078 \, amu \over atom } \right ) = 120.234 \, amu$
$2 \times \text {atomic mass of phosphorus} = 2 \, atoms \left ( {30.973761 \, amu \over atom } \right ) = 61.947522 \, amu$
$8 \times \text {atomic mass of oxygen} = 8 \, atoms \left ( {15.9994 \, amu \over atom } \right ) = 127.9952 \, amu$
C Adding together the masses gives the formula mass of Ca3(PO4)2:
$120.234 \,amu + 61.947522 \, amu + 127.9952 \, amu = 310.177 \, amu$
We could also find the formula mass of Ca3(PO4)2 in one step by using unit conversions or a tabular format:
$\left [ 3 \, atoms Ca \left ({40.078 \, amu \over 1 \, atom Ca } \right ) \right ] + \left [ 2 \, atoms P \left ({30.973761 \, amu \over 1 \, atom P } \right ) \right ] + \left [ 8 \, atoms O \left ({15.9994 \, amu \over 1 \, atom O } \right ) \right ]$
$= 310.177 \,amu$
$3Ca \, \, \, \, (3\, atoms)(40.078 \, amu/atom) = 120.234 \, amu$
$2P \, \, \, \, (2\, atoms)(30.973761 \, amu/atom) = 61.947522 \, amu$
$+ 8O \, \, \, \, (8\, atoms)(15.9994 \, amu/atom) = 127.9952 \, amu$
$Ca_3P_2O_8 \, \, \, \, \text {formula mass of Ca}_3(PO_4)_2 = 310.177 \, amu$
Molar Masses of Compounds: https://youtu.be/PhOqgNNv78s
$(moles)(molar mass) \rightarrow mass \label{3.2.1}$
or, more specifically,
$moles \left ( {grams \over mole } \right ) = grams$
Conversely, to convert the mass of a substance to moles:
$\left ( {grams \over molar mass } \right ) \rightarrow moles \label{3.2.2A}$
$\left ( { grams \over grams/mole} \right ) = grams \left ( {mole \over grams } \right ) = moles \label{3.2.2B}$
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$
the production of two moles of water would require the consumption of 2 moles of $H_2$ and one mole of $O_2$. Therefore, when considering this particular reaction
• 2 moles of H2
• 1 mole of O2 and
• 2 moles of H2O
would be considered to be stoichiometrically equivalent quantitites.
These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of $H_2O$ would be produced from 1.57 moles of $O_2$?
$(1.57\; mol\; O_2) \left( \dfrac{2\; mol H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O$
The ratio $\left( \dfrac{2\; mol\; H_2O}{1\;mol\;O_2} \right)$ is the stoichiometric relationship between $H_2O$ and $O_2$ from the balanced equation for this reaction.
Example $3$
For the combustion of butane ($C_4H_{10}$) the balanced equation is:
$2C_4H_{10(l)} + 13O_{2(g)} \rightarrow 8CO_{2(g)} + 10H_2O_{(l)}$
Calculate the mass of $CO_2$ that is produced in burning 1.00 gram of $C_4H_{10}$.
Solution
Thus, the overall sequence of steps to solve this problem is:
First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:
$(1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10}$
Now, the stoichiometric relationship between $C_4H_{10}$ and $CO_2$ is:
$\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right)$
Therefore:
$\left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2$
The question called for the determination of the mass of $CO_2$ produced, thus we have to convert moles of $CO_2$ into grams (by using the molecular weight of $CO_2$):
$6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2$
Conversions Between Grams, Mol, & Atoms: https://youtu.be/rOvErpAnoCg
Be sure to pay attention to the units when converting between mass and moles. Figure $1$ is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Example $3$ and Example $4$.
Example $4$
For 35.00 g of ethylene glycol (HOCH2CH2OH), which is used in inks for ballpoint pens, calculate the number of
1. moles.
2. molecules.
Given: mass and molecular formula
Asked for: number of moles and number of molecules
Strategy:
1. Use the molecular formula of the compound to calculate its molecular mass in grams per mole.
2. Convert from mass to moles by dividing the mass given by the compound’s molar mass.
3. Convert from moles to molecules by multiplying the number of moles by Avogadro’s number.
Solution:
a. A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example $3$.2.1:
$2C (2 \,atoms )(12.011 \, amu/atom) = 24.022 \, amu$
$6H (6 \,atoms )(1.0079 \, amu/atom) = 6.0474 \, amu$
$2O (2 \,atoms )(15.9994 \, amu/atom) = 31.9988 \, amu$
$C_2H_6O_2 \text {molecular mass of ethylene glycol} = 62.068 \, amu$
The molar mass of ethylene glycol is 62.068 g/mol.
B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole):
${ \text {mass of ethylene glycol (g)} \over \text {molar mass (g/mol)} } = \text {moles ethylene glycol (mol) }$
So
$35.00 \, g \text {ethylene glycol} \left ( {1 \, mole \text {ethylene glycol} \over 62.068 \, g \text {ethylene glycol } } \right ) = 0.5639 \,mol \text {ethylene glycol}$
It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations.
b. C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:
$\text {molecules of ethylene glycol} = 0.5639 \, mol \left ( {6.022 \times 10^{23} \, molecules \over 1 \, mol } \right )$
$= 3.396 \times 10^{23} \, molecules$
Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 1023 molecules, which is indeed the case.
$2S (2 \, atoms)(32.065 \, amu/atom ) = 64.130 \, amu$
$+ 2Cl (2 \, atoms )(35.453 \, amu/atom ) = 70.906 \, amu$
$S_2 Cl_2 \text {molecular mass of } S_2Cl_2 = 135.036 \, amu$
The molar mass of S2Cl2 is 135.036 g/mol.
B The mass of 1.75 mol of S2Cl2 is calculated as follows:
$moles S_2Cl_2 \left [\text {molar mass}\left ({ g \over mol} \right )\right ] \rightarrow mass of S_2Cl_2 \, (g)$
$1.75 \, mol S_2Cl_2\left ({135.036 \, g S_2Cl_2 \over 1 \, mol S_2Cl_2 } \right ) = 236 \, g S_2Cl_2$
b.
A The formula mass of Ca(ClO)2 is obtained as follows:
$1Ca (1 \, atom)(40.078 \, amu/atom) = 40.078 \, amu$
$2Cl (2 \, atoms)(35.453 \, amu/atom) = 70.906 \, amu$
$+ 2O (2 \, atoms)(15.9994 \, amu/atom) = 31.9988 \, amu$
$Ca (ClO)_2 \text { formula mass of } Ca (ClO)_2 = 142.983 \, amu$
The molar mass of Ca(ClO)2 142.983 g/mol.
B The mass of 1.75 mol of Ca(ClO)2 is calculated as follows:
$moles Ca(ClO)_2 \left [{\text {molar mass} Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2} \right ]=mass Ca(ClO)_2$
$1.75 \, mol Ca(ClO)_2 \left [ {142.983 \, g Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2 } \right ] = 250 \, g Ca(ClO)_2$
Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is.
Exercise $5$
Calculate the mass of 0.0122 mol of each compound.
1. Si3N4 (silicon nitride), used as bearings and rollers
2. (CH3)3N (trimethylamine), a corrosion inhibitor
Answer:
1. 1.71 g
2. 0.721 g
The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the balanced equation:
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$
the production of two moles of water would require the consumption of 2 moles of $H_2$ and one mole of $O_2$. Therefore, when considering this particular reaction
• 2 moles of H2
• 1 mole of O2 and
• 2 moles of H2O
would be considered to be stoichiometrically equivalent quantitites.
These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of $H_2O$ would be produced from 1.57 moles of $O_2$?
$(1.57\; mol\; O_2) \left( \dfrac{2\; mol H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O$
The ratio $\left( \dfrac{2\; mol\l H_2O}{1\;mol\;O_2} \right)$ is the stoichiometric relationship between $H_2O$ and $O_2$ from the balanced equation for this reaction.
Example $6$
For the combustion of butane ($C_4H_{10}$) the balanced equation is:
$2C_4H_{10(l)} + 13O_{2(g)} \rightarrow 8CO_{2(g)} + 10H_2O_{(l)}$
Calculate the mass of $CO_2$ that is produced in burning 1.00 gram of $C_4H_{10}$.
Solution
First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:
$(1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10}$
Now, the stoichiometric relationship between $C_4H_{10}$ and $CO_2$ is:
$\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right)$
Therefore:
$\left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2$
The question called for the determination of the mass of $CO_2$ produced, thus we have to convert moles of $CO_2$ into grams (by using the molecular weight of $CO_2$):
$6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2$
Thus, the overall sequence of steps to solve this problem were:
In a similar way we could determine the mass of water produced, or oxygen consumed, etc.
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): https://youtu.be/74mHV0CZcjw
Summary
• To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole.
The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12, Avogadro’s number (6.022 × 1023) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 1023 atoms, molecules, or formula units of that substance. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.2%3A_The_Mole_Concept_and_Chemical_Compounds.txt |
$2\times(1.0079\;amu) + 1 \times (15.9994 \;amu) = 18.01528 \;amu$
If a substance exists as discrete molecules (as with atoms that are chemically bonded together) then the chemical formula is the molecular formula, and the formula weight is the molecular weight. For example, carbon, hydrogen and oxygen can chemically bond to form a molecule of the sugar glucose with the chemical and molecular formula of C6H12O6. The formula weight and the molecular weight of glucose is thus:
$6\times(12\; amu) + 12\times(1.00794\; amu) + 6\times(15.9994\; amu) = 180.0 \;amu$
Ionic substances are not chemically bonded and do not exist as discrete molecules. However, they do associate in discrete ratios of ions. Thus, we can describe their formula weights, but not their molecular weights. Table salt (NaCl), for example, has a formula weight of:
$23.0\; amu + 35.5 \;amu = 58.5 \;amu$
Percentage Composition from Formulas
In some types of analyses of it is important to know the percentage by mass of each type of element in a compound. The law of definite proportions states that a chemical compound always contains the same proportion of elements by mass; that is, the percent composition—the percentage of each element present in a pure substance—is constant (although there are exceptions to this law). Take for example methane ($CH_4$) with a Formula and molecular weight:
$1\times (12.011 \;amu) + 4 \times (1.008) = 16.043 \;amu$
the relative (mass) percentages of carbon and hydrogen are
$\%C = \dfrac{1 \times (12.011\; amu)}{16.043 amu} = 0.749 = 74.9\%$
$\%H = \dfrac{4 \times (1.008 \;amu)}{16.043\; amu} = 0.251 = 25.1\%$
A more complex example is sucrose (table sugar) is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This means that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen. First the molecular formula of sucrose (C12H22O11) is used to calculate the mass percentage of the component elements; the mass percentage can then be used to determine an empirical formula.
According to its molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole of sucrose molecules therefore contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. This information can be used to calculate the mass of each element in 1 mol of sucrose, which gives the molar mass of sucrose. These masses can then be used to calculate the percent composition of sucrose. To three decimal places, the calculations are the following:
$\text {mass of C/mol of sucrose} = 12 \, mol \, C \times {12.011 \, g \, C \over 1 \, mol \, C} = 144.132 \, g \, C \label{3.3.1a}$
$\text {mass of H/mol of sucrose} = 22 \, mol \, H \times {1.008 \, g \, H \over 1 \, mol \, H} = 22.176 \, g \, C \label{3.3.1b}$
$\text {mass of O/mol of sucrose} = 11 \, mol \, O \times {15.999 \, g \, O \over 1 \, mol \, O} = 175.989 \, g \, O \label{3.3.1c}$
Thus 1 mol of sucrose has a mass of 342.297 g; note that more than half of the mass (175.989 g) is oxygen, and almost half of the mass (144.132 g) is carbon.
The mass percentage of each element in sucrose is the mass of the element present in 1 mol of sucrose divided by the molar mass of sucrose, multiplied by 100 to give a percentage. The result is shown to two decimal places:
$\text {mass % C in Sucrose} = {\text {mass of C/mol sucrose} \over \text {molar mass of sucrose} } \times100 = {144.132 \, g \, C \over 342.297 \, g/mol } \times 100 = 42.12 \%$
$\text {mass % H in Sucrose} = {\text {mass of H/mol sucrose} \over \text {molar mass of sucrose} } \times100 = {22.176 \, g \, H \over 342.297 \, g/mol } \times 100 = 6.48 \%$
$\text {mass % O in Sucrose} = {\text {mass of O/mol sucrose} \over \text {molar mass of sucrose} } \times100 = {175.989 \, g \, O \over 342.297 \, g/mol } \times 100 = 51.41 \%$
This can be checked by verifying that the sum of the percentages of all the elements in the compound is 100%:
$42.12\% + 6.48\% + 51.41\% = 100.01\%$
If the sum is not 100%, an error has been made in calculations. (Rounding to the correct number of decimal places can, however, cause the total to be slightly different from 100%.) Thus 100.00 g of sucrose contains 42.12 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to two decimal places, the percent composition of sucrose is indeed 42.12% carbon, 6.48% hydrogen, and 51.41% oxygen.
It is also possible to calculate mass percentages using atomic masses and molecular masses, with atomic mass units. Because the answer is a ratio, expressed as a percentage, the units of mass cancel whether they are grams (using molar masses) or atomic mass units (using atomic and molecular masses).
Example $1$
Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is C14H18N2O5.
1. Calculate the mass percentage of each element in aspartame.
2. Calculate the mass of carbon in a 1.00 g packet of Equal, assuming it is pure aspartame.
Given: molecular formula and mass of sample
Asked for: mass percentage of all elements and mass of one element in sample
Strategy:
1. Use atomic masses from the periodic table to calculate the molar mass of aspartame.
2. Divide the mass of each element by the molar mass of aspartame; then multiply by 100 to obtain percentages.
3. To find the mass of an element contained in a given mass of aspartame, multiply the mass of aspartame by the mass percentage of that element, expressed as a decimal.
Solution:
a.
A We calculate the mass of each element in 1 mol of aspartame and the molar mass of aspartame, here to three decimal places:
$14 \,C (14 \, mol \, C)(12.011 \, g/mol \, C) = 168.154 \, g$
$18 \,H (18 \, mol \, H)(1.008 \, g/mol \, H) = 18.114 \, g$
$2 \,N (2 \, mol \, N)(14.007 \, g/mol \, N) = 28.014 \, g$
$+5 \,O (5 \, mol \, O)(15.999 \, g/mol \, O) = 79.995 \, g$
$C_{14}H_{18}N_2O_5 \text {molar mass of aspartame} = 294.277 \, g/mol$
Thus more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g).
B To calculate the mass percentage of each element, we divide the mass of each element in the compound by the molar mass of aspartame and then multiply by 100 to obtain percentages, here reported to two decimal places:
$mass \% \, C = {168.154 \, g \, C \over 294.277 \, g \, aspartame } \times 100 = 57.14 \% C$
$mass \% \, H = {18.114 \, g \, H \over 294.277 \, g \, aspartame } \times 100 = 6.16 \% H$
$mass \% \, N = {28.014 \, g \, N \over 294.277 \, g \, aspartame } \times 100 = 9.52 \%$
$mass \% \, O = {79.995 \, g \, O \over 294.277 \, g \, aspartame } \times 100 = 27.18 \%$
As a check, we can add the percentages together:
$57.14\% + 6.16\% + 9.52\% + 27.18\% = 100.00\%$
If you obtain a total that differs from 100% by more than about ±1%, there must be an error somewhere in the calculation.
b. C The mass of carbon in 1.00 g of aspartame is calculated as follows:
$\text {mass of C} = 1.00 \, g \, aspartame \times {57.14 \, g \, C \over 100 \, g \, aspartame } = 0.571 \, g \, C$
Exercise $1$: Aluminum Oxide
Calculate the mass percentage of each element in aluminum oxide (Al2O3). Then calculate the mass of aluminum in a 3.62 g sample of pure aluminum oxide.
Answer: 52.93% aluminum; 47.08% oxygen; 1.92 g Al
Percent Composition: https://youtu.be/HNS6lItns10
Determining the Empirical Formula of Penicillin
Just as the empirical formula of a substance can be used to determine its percent composition, the percent composition of a sample can be used to determine its empirical formula, which can then be used to determine its molecular formula. Such a procedure was actually used to determine the empirical and molecular formulas of the first antibiotic to be discovered: penicillin.
Antibiotics are chemical compounds that selectively kill microorganisms, many of which cause diseases. Although antibiotics are often taken for granted today, penicillin was discovered only about 80 years ago. The subsequent development of a wide array of other antibiotics for treating many common diseases has contributed greatly to the substantial increase in life expectancy over the past 50 years. The discovery of penicillin is a historical detective story in which the use of mass percentages to determine empirical formulas played a key role.
In 1928, Alexander Fleming, a young microbiologist at the University of London, was working with a common bacterium that causes boils and other infections such as blood poisoning. For laboratory study, bacteria are commonly grown on the surface of a nutrient-containing gel in small, flat culture dishes. One day Fleming noticed that one of his cultures was contaminated by a bluish-green mold similar to the mold found on spoiled bread or fruit. Such accidents are rather common, and most laboratory workers would have simply thrown the cultures away. Fleming noticed, however, that the bacteria were growing everywhere on the gel except near the contaminating mold (part (a) in Figure $2$), and he hypothesized that the mold must be producing a substance that either killed the bacteria or prevented their growth. To test this hypothesis, he grew the mold in a liquid and then filtered the liquid and added it to various bacteria cultures. The liquid killed not only the bacteria Fleming had originally been studying but also a wide range of other disease-causing bacteria. Because the mold was a member of the Penicillium family (named for their pencil-shaped branches under the microscope) (part (b) in Figure $2$), Fleming called the active ingredient in the broth penicillin.
Although Fleming was unable to isolate penicillin in pure form, the medical importance of his discovery stimulated researchers in other laboratories. Finally, in 1940, two chemists at Oxford University, Howard Florey (1898–1968) and Ernst Chain (1906–1979), were able to isolate an active product, which they called penicillin G. Within three years, penicillin G was in widespread use for treating pneumonia, gangrene, gonorrhea, and other diseases, and its use greatly increased the survival rate of wounded soldiers in World War II. As a result of their work, Fleming, Florey, and Chain shared the Nobel Prize in Medicine in 1945.
As soon as they had succeeded in isolating pure penicillin G, Florey and Chain subjected the compound to a procedure called combustion analysis (described later in this section) to determine what elements were present and in what quantities. The results of such analyses are usually reported as mass percentages. They discovered that a typical sample of penicillin G contains 53.9% carbon, 4.8% hydrogen, 7.9% nitrogen, 9.0% sulfur, and 6.5% sodium by mass. The sum of these numbers is only 82.1%, rather than 100.0%, which implies that there must be one or more additional elements. A reasonable candidate is oxygen, which is a common component of compounds that contain carbon and hydrogen; do not assume that the “missing” mass is always due to oxygen. It could be any other element. For technical reasons, however, it is difficult to analyze for oxygen directly. Assuming that all the missing mass is due to oxygen, then penicillin G contains (100.0% − 82.1%) = 17.9% oxygen. From these mass percentages, the empirical formula and eventually the molecular formula of the compound can be determined.
To determine the empirical formula from the mass percentages of the elements in a compound such as penicillin G, the mass percentages must be converted to relative numbers of atoms. For convenience, assume a 100.0 g sample of the compound, even though the sizes of samples used for analyses are generally much smaller, usually in milligrams. This assumption simplifies the arithmetic because a 53.9% mass percentage of carbon corresponds to 53.9 g of carbon in a 100.0 g sample of penicillin G; likewise, 4.8% hydrogen corresponds to 4.8 g of hydrogen in 100.0 g of penicillin G; and so forth for the other elements. Each mass is then divided by the molar mass of the element to determine how many moles of each element are present in the 100.0 g sample:
${ mass \, (g) \over molar \,\, mass \,\, (g/mol)} = (g) \left ({mol \over g } \right ) = mol \label{3.3.2a}$
$53.9 \, g \, C \left ({1 \, mol \, C \over 12.011 \, g \, C} \right ) = 4.49 \, mol \, C \label{3.3.2b}$
$4.8 \, g \, H \left ({1 \, mol \, H \over 1.008 g \, H} \right ) = 4.8 \, mol \, H \label{3.3.2c}$
$7.9 \, g \, N \left ({1 \, mol \, N \over 14.007 \, g \, N} \right ) = 0.56 \, mol \, N \label{3.3.2d}$
$9 \, g \, S \left ({1 \, mol \, S \over 32.065 \, g \, S} \right ) = 0.28 \, mol \, S \label{3.3.2e}$
$6.5 \, g \, Na \left ({1 \, mol \, Na \over 22.990 \, g \, Na} \right ) = 0.28 \, mol \, Na \label{3.3.2f}$
Thus 100.0 g of penicillin G contains 4.49 mol of carbon, 4.8 mol of hydrogen, 0.56 mol of nitrogen, 0.28 mol of sulfur, 0.28 mol of sodium, and 1.12 mol of oxygen (assuming that all the missing mass was oxygen). The number of significant figures in the numbers of moles of elements varies between two and three because some of the analytical data were reported to only two significant figures.
These results give the ratios of the moles of the various elements in the sample (4.49 mol of carbon to 4.8 mol of hydrogen to 0.56 mol of nitrogen, and so forth), but they are not the whole-number ratios needed for the empirical formula—the empirical formula expresses the relative numbers of atoms in the smallest whole numbers possible. To obtain whole numbers, divide the numbers of moles of all the elements in the sample by the number of moles of the element present in the lowest relative amount, which in this example is sulfur or sodium. The results will be the subscripts of the elements in the empirical formula. To two significant figures, the results are as follows:
$C: {4.49 \over 0.28} = 16 \, \, \, \, \, H: {4.8 \over 0.28} = 17 \, \, \, \, \, N: {0.56 \over 0.28} = 2.0 \label{3.3.3a}$
$S: {0.28 \over 0.28 } = 1.0 \, \, \, \, \, Na: {0.28 \over 0.28 } = 1.0 \, \, \, \, \, O: {1.12 \over 0.28} = 4.0 \label{3.3.3b}$
The empirical formula of penicillin G is therefore C16H17N2NaO4S. Other experiments have shown that penicillin G is actually an ionic compound that contains Na+ cations and [C16H17N2O4S] anions in a 1:1 ratio. The complex structure of penicillin G (Figure $3$) was not determined until 1948.
In some cases, one or more of the subscripts in a formula calculated using this procedure may not be integers. Does this mean that the compound of interest contains a nonintegral number of atoms? No; rounding errors in the calculations as well as experimental errors in the data can result in nonintegral ratios. When this happens, judgment must be exercised in interpreting the results, as illustrated in Example 6. In particular, ratios of 1.50, 1.33, or 1.25 suggest that you should multiply all subscripts in the formula by 2, 3, or 4, respectively. Only if the ratio is within 5% of an integral value should one consider rounding to the nearest integer.
Example $2$: Calcium Phosphate in Toothpaste
Calculate the empirical formula of the ionic compound calcium phosphate, a major component of fertilizer and a polishing agent in toothpastes. Elemental analysis indicates that it contains 38.77% calcium, 19.97% phosphorus, and 41.27% oxygen.
Given: percent composition
Asked for: empirical formula
Strategy:
1. Assume a 100 g sample and calculate the number of moles of each element in that sample.
2. Obtain the relative numbers of atoms of each element in the compound by dividing the number of moles of each element in the 100 g sample by the number of moles of the element present in the smallest amount.
3. If the ratios are not integers, multiply all subscripts by the same number to give integral values.
4. Because this is an ionic compound, identify the anion and cation and write the formula so that the charges balance.
Solution:
A A 100 g sample of calcium phosphate contains 38.77 g of calcium, 19.97 g of phosphorus, and 41.27 g of oxygen. Dividing the mass of each element in the 100 g sample by its molar mass gives the number of moles of each element in the sample:
$\text {moles Ca} = 38.77 \, g \, Ca \times {1 \, mol \, Ca \over 40.078 \, g \, Ca} = 0.9674 \, mol \, Ca$
$\text {moles P} = 19.97 \, g \, P \times {1 \, mol \, P \over 30.9738 \, g \, P} = 0.6447 \, mol \, Ca$
$\text {moles O} = 41.27 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O} = 2.5800 \, mol \, O$
B To obtain the relative numbers of atoms of each element in the compound, divide the number of moles of each element in the 100-g sample by the number of moles of the element in the smallest amount, in this case phosphorus:
$P: {0.6447 \, mol\, P \over 0.6447 \, mol\, P} = 1.000 \, \, \, \, Ca: {0.9674 \over 0.6447} = 1.501\, \, \, \, O: {2.5800\over 0.6447}= 4.002$
C We could write the empirical formula of calcium phosphate as Ca1.501P1.000O4.002, but the empirical formula should show the ratios of the elements as small whole numbers. To convert the result to integral form, multiply all the subscripts by 2 to get Ca3.002P2.000O8.004. The deviation from integral atomic ratios is small and can be attributed to minor experimental errors; therefore, the empirical formula is Ca3P2O8.
D The calcium ion (Ca2+) is a cation, so to maintain electrical neutrality, phosphorus and oxygen must form a polyatomic anion. We know from Chapter 2 that phosphorus and oxygen form the phosphate ion (PO43; see Table 2.4). Because there are two phosphorus atoms in the empirical formula, two phosphate ions must be present. So we write the formula of calcium phosphate as Ca3(PO4)2.
Exercise $2$: Oklahoma City Bombing
Calculate the empirical formula of ammonium nitrate, an ionic compound that contains 35.00% nitrogen, 5.04% hydrogen, and 59.96% oxygen by mass. Although ammonium nitrate is widely used as a fertilizer, it can be dangerously explosive. For example, it was a major component of the explosive used in the 1995 Oklahoma City bombing.
The Alfred P. Murrah Federal Building destroyed in the Oklahoma City bombing via chemical explosives (rapid chemical reactions that generate massive quantities of gases).
Answer
N2H4O3 is NH4+NO3, written as NH4NO3
Determining Empirical and Molecular Formulas from % Composition: https://youtu.be/E-MxBYw1TSI
From Empirical Formula to Molecular Formula
The empirical formula gives only the relative numbers of atoms in a substance in the smallest possible ratio. For a covalent substance, chemists are usually more interested in the molecular formula, which gives the actual number of atoms of each kind present per molecule. Without additional information, however, it is impossible to know whether the formula of penicillin G, for example, is C16H17N2NaO4S or an integral multiple, such as C32H34N4Na2O8S2, C48H51N6Na3O12S3, or (C16H17N2NaO4S)n, where n is an integer. (The actual structure of penicillin G is shown in Figure $3$).
Consider glucose, the sugar that circulates in our blood to provide fuel for the body and brain. Results from combustion analysis of glucose report that glucose contains 39.68% carbon and 6.58% hydrogen. Because combustion occurs in the presence of oxygen, it is impossible to directly determine the percentage of oxygen in a compound by using combustion analysis; other more complex methods are necessary. Assuming that the remaining percentage is due to oxygen, then glucose would contain 53.79% oxygen. A 100.0 g sample of glucose would therefore contain 39.68 g of carbon, 6.58 g of hydrogen, and 53.79 g of oxygen. To calculate the number of moles of each element in the 100.0 g sample, divide the mass of each element by its molar mass:
$moles \, C = 39.68 \, g \, C \times {1 \, mol \, C \over 12.011 \, g \, C } = 3.304 \, mol \, C \label{3.3.4a}$
$moles \, H = 6.58 \, g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H } = 6.53 \, mol \, H \label{3.3.4b}$
$moles \, O = 53.79 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O } = 3.362 \, mol \, O \label{3.3.4c}$
Once again, the subscripts of the elements in the empirical formula are found by dividing the number of moles of each element by the number of moles of the element present in the smallest amount:
$C: {3.304 \over 3.304} = 1.000 \, \, \, \, H: {6.53 \over 3.304} = 1.98 \, \, \, \, O: {3.362 \over 3.304} = 1.018$
The oxygen:carbon ratio is 1.018, or approximately 1, and the hydrogen:carbon ratio is approximately 2. The empirical formula of glucose is therefore CH2O, but what is its molecular formula?
Many known compounds have the empirical formula CH2O, including formaldehyde, which is used to preserve biological specimens and has properties that are very different from the sugar circulating in the blood. At this point, it cannot be known whether glucose is CH2O, C2H4O2, or any other (CH2O)n. However, the experimentally determined molar mass of glucose (180 g/mol) can be used to resolve this dilemma.
First, calculate the formula mass, the molar mass of the formula unit, which is the sum of the atomic masses of the elements in the empirical formula multiplied by their respective subscripts. For glucose,
$\text {formula mass of} CH_2O = \left [ 1 \, mol C \left ( {12.011 \, g \over 1 \, mol \, C} \right ) \right ] + \left [ 2 \, mol \, H \left ({1.0079 \, g \over 1 \, mol \, H }\right )\right ] + \left [ 1 \, mole \, O \left ( {15.5994 \, mol \, O \over 1 \, mol \, O} \right ) \right ] = 30.026 g \label{3.3.5}$
This is much smaller than the observed molar mass of 180 g/mol.
Second, determine the number of formula units per mole. For glucose, calculate the number of (CH2O) units—that is, the n in (CH2O)n—by dividing the molar mass of glucose by the formula mass of CH2O:
$n={180 \, g \over 30.026\, g/CH_2O} = 5.99 \approx 6 CH_2O \, \text {formula units} \label{3.3.6}$
Each glucose contains six CH2O formula units, which gives a molecular formula for glucose of (CH2O)6, which is more commonly written as C6H12O6. The molecular structures of formaldehyde and glucose, both of which have the empirical formula CH2O, are shown in Figure $4$.
Example $3$: Caffeine
Calculate the molecular formula of caffeine, a compound found in coffee, tea, and cola drinks that has a marked stimulatory effect on mammals. The chemical analysis of caffeine shows that it contains 49.18% carbon, 5.39% hydrogen, 28.65% nitrogen, and 16.68% oxygen by mass, and its experimentally determined molar mass is 196 g/mol.
Given: percent composition and molar mass
Asked for: molecular formula
Strategy:
1. Assume 100 g of caffeine. From the percentages given, use the procedure given in Example 6 to calculate the empirical formula of caffeine.
2. Calculate the formula mass and then divide the experimentally determined molar mass by the formula mass. This gives the number of formula units present.
3. Multiply each subscript in the empirical formula by the number of formula units to give the molecular formula.
Solution:
A We begin by dividing the mass of each element in 100.0 g of caffeine (49.18 g of carbon, 5.39 g of hydrogen, 28.65 g of nitrogen, 16.68 g of oxygen) by its molar mass. This gives the number of moles of each element in 100 g of caffeine.
$moles \, C = 49.18 \, g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 4.095 \, mol \, C$
$moles \, H = 5.39 \, g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 5.35 \, mol \, H$
$moles \, N = 28.65 \, g \, N \times {1 \, mol \, N \over 14.0067 \, g \, N} = 2.045 \, mol \, N$
$moles \, O = 16.68 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O} = 1.043 \, mol \, O$
To obtain the relative numbers of atoms of each element present, divide the number of moles of each element by the number of moles of the element present in the least amount:
$O: {1.043 \over 1.043} = 1.000 \, \, \, \, C: {4.095 \over 1.043} = 3.926 \, \, \, \, H: {5.35 \over 1.043} = 5.13 \, \, \, \, N: {2.045 \over 1.043} = 1.960$
These results are fairly typical of actual experimental data. None of the atomic ratios is exactly integral but all are within 5% of integral values. Just as in Example 6, it is reasonable to assume that such small deviations from integral values are due to minor experimental errors, so round to the nearest integer. The empirical formula of caffeine is thus C4H5N2O.
B The molecular formula of caffeine could be C4H5N2O, but it could also be any integral multiple of this. To determine the actual molecular formula, we must divide the experimentally determined molar mass by the formula mass. The formula mass is calculated as follows:
$4C \, \, \, ( 4 \, atoms \, C) (12.011 \, g/ atom \, C) = 48.044 \, g$
$5H \, \, \, ( 5 \, atoms \, H ) (1.0079 \, g/ atom \, H) = 5.0395 \, g$
$2N \, \, \, (2 \, atoms \, N) (14.0067 \, g/ atom \, N) = 28.0134 \, g$
$+1O \, \, \, (1 \, atom \, O) (15.9994 \, g/ atom \, O) = 15.9994 \, g$
$C_4H_5N_2O \, \, \, \, \text {formula mass of caffeine} = 97.096 \, g$
Dividing the measured molar mass of caffeine (196 g/mol) by the calculated formula mass gives
${196 g/mol \over 97.096 g/C_4H_5N_2O } = 2.02 \approx 2\, C_4H_5N_2O \, \text {empirical formula units}$
C There are two C4H5N2O formula units in caffeine, so the molecular formula must be (C4H5N2O)2 = C8H10N4O2. The structure of caffeine is as follows:
Exercise $3$: Freon-114
Calculate the molecular formula of Freon-114, which has 13.85% carbon, 41.89% chlorine, and 44.06% fluorine. The experimentally measured molar mass of this compound is 171 g/mol. Like Freon-11, Freon-114 is a commonly used refrigerant that has been implicated in the destruction of the ozone layer.
Answer:
$C_2Cl_2F_4$
Combustion Analysis
One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in Figure $5$ and a typical combustion analysis is illustrated in Example $4$.
Example $4$
Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene.
Given: mass of sample and mass of combustion products
Asked for: empirical formula
Strategy:
1. Use the masses and molar masses of the combustion products, CO2 and H2O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.
2. Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.
Solution:
A Upon combustion, 1 mol of CO2 is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:
$mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C}$
$= 1.883 \times 10^{-2} \, g \, C$
$mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H}$
$= 1.264 \times 10^{-3} \, g \, H$
B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:
$moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C$
$moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H$
Dividing each number by the number of moles of the element present in the smaller amount gives
$H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250$
Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C10H8, which is consistent with our results.
Exercise $4$
1. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO2 and 14.53 mg of H2O. Determine the empirical formula of xylene.
2. The empirical formula of benzene is CH (its molecular formula is C6H6). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO2 and H2O will be produced?
Answer:
1. The empirical formula is C4H5. (The molecular formula of xylene is actually C8H10.)
2. 33.81 mg of CO2; 6.92 mg of H2O
Summary
• The empirical formula of a substance can be calculated from its percent composition, and the molecular formula can be determined from the empirical formula and the compound’s molar mass.
The empirical formula of a substance can be calculated from the experimentally determined percent composition, the percentage of each element present in a pure substance by mass. In many cases, these percentages can be determined by combustion analysis. If the molar mass of the compound is known, the molecular formula can be determined from the empirical formula.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.3%3A_Composition_of_Chemical_Compounds.txt |
Learning Objectives
• To identify oxidation–reduction reactions in solution.
The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, coherent, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O2−). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction.
Note
Any oxidation must ALWAYS be accompanied by a reduction and vice versa.
Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water). The reaction is as follows:
$Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g) \label{3.4.1}$
Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants, which follows the mnemonic oxidation is loss reduction is gain, or oil rig. The oxidation state of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O2 or H2, are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is
$4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s) \label{3.4.2}$
Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al2O3, electrons are transferred as follows (the superscript 0 emphasizes the oxidation state of the elements):
$4 Al^0 + 3 O_2^0 \rightarrow 4 Al^{3+} + 6 O^{2-} \label{3.4.3}$
Equation 3.4.1 and Equation 3.4.2 are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In Equation 3.4.3, for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:
$electrons \, lost = 4 \, Al \, atoms \times {3 \, e^- \, lost \over Al \, atom } = 12 \, e^- \, lost \label{3.4.4a}$
$electrons \, gained = 6 \, O \, atoms \times {2 \, e^- \, gained \over O \, atom} = 12 \, e^- \, gained \label{3.4.4a}$
The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained.
An additional example of a redox reaction, the reaction of sodium metal with oxygen in air, is illustrated in Figure $1$.
Note
In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.
Assigning Oxidation States
Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previosuly, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; Figure $1$), magnesium oxide (MgO), and calcium chloride (CaCl2). In covalent compounds, in contrast, atoms share electrons. Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.
A set of rules for assigning oxidation states to atoms in chemical compounds follows.
Rules for Assigning Oxidation States
1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
2. The oxidation state of a monatomic ion is the same as its charge—for example, Na+ = +1, Cl = −1.
3. The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens.
4. Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.
5. Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
6. The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.
Nonintegral oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states.
In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound.
Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6. Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H+ and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF2 but −½ in KO2. Note that an oxidation state of −½ for O in KO2 is perfectly acceptable.
The reduction of copper(I) oxide shown in Equation 3.4.5 demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H2 and Cu. From rule 4, hydrogen in H2O has an oxidation state of +1, and from rule 5, oxygen in both Cu2O and H2O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu2O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows:
$\overset {+1}{Cu_2} \underset {-2}{O} (s) + \overset {0}{H_2} (g) \rightarrow 2 \overset {0}{Cu} (s) + \overset {+1}{H_2} \underset {-2}{O} (g) \label{3.4.5}$
Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). So this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:
$electrons \, lost = 2 \, H \, atoms \times {1 \, e^- \, lost \over H \, atom } = 2 \, e^- \, lost \label{3.4.6a}$
$electrons \, gained = 2 \, Cu \, atoms \times {1 \, e^- \, gained \over Cu \, atom} = 2 \, e^- \, gained \label{3.4.6b}$
Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.
Example $1$
Assign oxidation states to all atoms in each compound.
1. sulfur hexafluoride (SF6)
2. methanol (CH3OH)
3. ammonium sulfate [(NH4)2SO4]
4. magnetite (Fe3O4)
5. ethanoic (acetic) acid (CH3CO2H)
Given: molecular or empirical formula
Asked for: oxidation states
Strategy:
Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.
Solution:
a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF6), the oxidation state of sulfur must be +6:
[(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0
b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2:
[(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0
c. Note that (NH4)2SO4 is an ionic compound that consists of both a polyatomic cation (NH4+) and a polyatomic anion (SO42−) (see Table 2.4). We assign oxidation states to the atoms in each polyatomic ion separately. For NH4+, hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:
[(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH4+ ion
For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:
[(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion
d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:
[(4 O atoms)(−2)]+[(3 Fe atoms)$\left (+{8 \over 3} \right )$]= 0
Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe3O4 can be viewed as having two Fe3+ ions and one Fe2+ ion per formula unit, giving a net positive charge of +8 per formula unit. Fe3O4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.”
e. Initially, we assign oxidation states to the components of CH3CO2H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of
[(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0
So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH3) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO2H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of
[(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3
To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus
$\underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H}$
Thus the sum of the oxidation states of the two carbon atoms is indeed zero.
Exercise $1$
Assign oxidation states to all atoms in each compound.
1. barium fluoride (BaF2)
2. formaldehyde (CH2O)
3. potassium dichromate (K2Cr2O7)
4. cesium oxide (CsO2)
5. ethanol (CH3CH2OH)
Answer:
1. Ba, +2; F, −1
2. C, 0; H, +1; O, −2
3. K, +1; Cr, +6; O, −2
4. Cs, +1; O, −½
5. C, −3; H, +1; C, −1; H, +1; O, −2; H, +1
Redox Reactions of Solid Metals in Aqueous Solution
A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure $1$). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:
$Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \label{3.4.5}$
In subsequent steps, FeCl2 undergoes oxidation to form a reddish-brown precipitate of Fe(OH)3.
Many metals dissolve through reactions of this type, which have the general form
$metal + acid \rightarrow salt + hydrogen \label{3.4.6}$
Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:
$Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \label{3.4.7}$
Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!
Single-Displacement Reactions
Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation 3.4.8) and the reduction of silver salts by copper (Equation 3.4.9 and Figure $2$):
$Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \label{3.4.8}$
$Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \label{3.4.9}$
The reaction in Equation 3.4.8 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.
The Activity Series
By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing Zn2+. Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:
$Zn(s) + Mg^{2+}(aq) \cancel{\rightarrow} Zn^{2+}(aq) + Mg(s) \label{3.4.10}$
$Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s) \label{3.4.11}$
Magnesium has a greater tendency to be oxidized than zinc does.
Pairwise reactions of this sort are the basis of the activity series (Figure $4$), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series).
When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series. Because magnesium is above zinc in Figure $4$, magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H2. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example $2$ demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions.
Example $2$
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
2. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
3. Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.
Given: reactants
Asked for: overall reaction and net ionic equation
Strategy:
1. Locate the reactants in the activity series in Figure .3.4.4 and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced.
2. Write the net ionic equation for the redox reaction.
Solution:
1. A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to their relative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionic equation is as follows:
$Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)$
Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.
2. A Mercury lies below lead in the activity series, so no reaction will occur.
3. A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H2. B From our discussion of solubilities, recall that Pb2+ and SO42− form insoluble lead(II) sulfate. In this case, the sulfate ions are not spectator ions, and the reaction is as follows:
$Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g)$
Lead(II) sulfate is the white solid that forms on corroded battery terminals.
Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.
Exercise $2$
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of chromium metal is placed in an aqueous solution of aluminum chloride.
2. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
3. A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).
Answer
1. $no\: reaction$
2. $3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)$
3. $2Al(s) + 6CH_3CO_2H(aq) \rightarrow 2Al^{3+}(aq) + 6CH_3CO_2^-(aq) + 3H_2(g)$
Summary
• Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation.
In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method (Table $1$), in which the overall reaction is separated into an oxidation equation and a reduction equation. Single-displacement reactions are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the activity series (Figure $3$), which arranges metals and H2 in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. Active metals lie at the top of the activity series, whereas inert metals are at the bottom of the activity series.
3.5: Importance of Nomenclature
The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning which chemical compound the name refers to: each chemical name should refer to a single substance. A less important aim is to ensure that each substance has a single name (although a limited number of alternative names is acceptable in some cases). The form of nomenclature used depends on the audience to which it is addressed. As such, no single correct form exists, but rather there are different forms that are more or less appropriate in different circumstances. A common name will often suffice to identify a chemical compound in a particular set of circumstances. However, in a few specific circumstances (such as the construction of large indices), it becomes necessary to ensure that each compound has a unique name.
In discussing chemistry nomenclatures is it necessary to identify the type of compound including stoichiometry and type of constituent atoms. The first separation of importance is to distinguish between inorganic and organic compounds. Unfortunately, this separation is not always clear. even to experienced chemists. Historically (two centuries ago) the definitions were that inorganic compounds were synthesized by geological systems and organic compounds were found in biological systems. The modern definitions argues that organic compounds are any molecule containing carbon and by default this means that inorganic chemistry deals with molecules lacking carbon. The confusion arise when chemists that study carbon based materials like diamond or graphite, but that argument is beyond the scope of this text. The following sections presents the current nomenclature rules for inorganic and organic molecules. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.4%3A_Oxidation_States%3A_A_Useful_Tool_in_Describing_Chemical_Compounds.txt |
Learning Objectives
• To name ionic compounds.
The empirical and molecular formulas discussed in the preceding section are precise and informative, but they have some disadvantages. First, they are inconvenient for routine verbal communication. For example, saying “C-A-three-P-O-four-two” for Ca3(PO4)2 is much more difficult than saying “calcium phosphate.” In addition, many compounds have the same empirical and molecular formulas but different arrangements of atoms, which differences result in very different chemical and physical properties. In such cases, it is necessary for the compounds to have different names that distinguish among the possible arrangements.
Many compounds, particularly those that have been known for a relatively long time, have more than one name: a common name (sometimes several), and a systematic name, which is the name assigned by adhering to specific rules. Like the names of most elements, the common names of chemical compounds generally have historical origins, although they often appear to be unrelated to the compounds of interest. For example, the systematic name for KNO3 is potassium nitrate, but its common name is saltpeter.
In this text, a systematic nomenclature is used to assign meaningful names to the millions of known substances. Unfortunately, some chemicals that are widely used in commerce and industry are still known almost exclusively by their common names; in such cases, familiarity with the common name as well as the systematic one is required. The objective of this and the next two sections is to teach how to write the formula for a simple inorganic compound from its name—and vice versa—and introduce some frequently-encountered common names.
Binary ionic compounds contain only two elements. The procedure for naming such compounds is outlined in Figure \(1\) and uses the following steps:
1. Place the ions in their proper order: cation and then anion.
2. Name the cation.
a. Metals that form only one cation. As noted previosuly, these metals are usually in Groups 1–3, 12, and 13. The name of the cation of a metal that forms only one cation is the same as the name of the metal (with the word ion added if the cation is by itself). For example, Na+ is the sodium ion, Ca2+ is the calcium ion, and Al3+ is the aluminum ion.
b. Metals that form more than one cation. As shown in Figure \(2\), many metals can form more than one cation. This behavior is observed for most transition metals, many actinides, and the heaviest elements of Groups 13–15. In such cases, the positive charge on the metal is indicated by a roman numeral in parentheses immediately following the name of the metal. Thus Cu+ is copper(I) (read as “copper one”), Fe2+ is iron(II), Fe3+ is iron(III), Sn2+ is tin(II), and Sn4+ is tin(IV).
An older system of nomenclature for such cations is still widely used, however. The name of the cation with the higher charge is formed from the root of the element’s Latin name with the suffix -ic attached, and the name of the cation with the lower charge has the same root with the suffix -ous. The names of Fe3+, Fe2+, Sn4+, and Sn2+ are therefore ferric, ferrous, stannic, and stannous, respectively. Even though this text uses the systematic names with roman numerals, it is important to recognize these common names because they are still often used. For example, on the label of dental fluoride rinse, the compound chemists call tin(II) fluoride is usually listed as stannous fluoride.
Some examples of metals that form more than one cation are listed in Table \(1\), along with the names of the ions. Note that the simple Hg+ cation does not occur in chemical compounds. Instead, all compounds of mercury(I) contain a dimeric cation, Hg22+, in which the two Hg atoms are bonded together.
Table \(1\): Common Cations of Metals That Form More Than One Ion
Cation Systematic Name Common Name Cation Systematic Name Common Name
* Not widely used.
†The isolated mercury(I) ion exists only as the gaseous ion.
Co2+ cobalt(II) cobaltous* Pb4+ lead(IV) plumbic*
Co3+ cobalt(III) cobaltic* Pb2+ lead(II) plumbous*
Cr2+ chromium(II) chromous Cu2+ copper(II) cupric
Cr3+ chromium(III) chromic Cu+ copper(I) cuprous
Fe2+ iron(II) ferrous Sn4+ tin(IV) stannic
Fe3+ iron(III) ferric Sn2+ tin(II) stannous
Mn2+ manganese(II) manganous* Hg2+ mercury(II) mercuric
Mn3+ manganese(III) manganic* Hg22+ mercury(I) mercurous†
c. Polyatomic cations. The names of the common polyatomic cations that are relatively important in ionic compounds (such as, the ammonium ion) are in Table 2.4
3. Name the anion.
a. Monatomic anions. Monatomic anions are named by adding the suffix -ide to the root of the name of the parent element; thus, Cl is chloride, O2− is oxide, P3− is phosphide, N3− is nitride (also called azide), and C4− is carbide. Because the charges on these ions can be predicted from their position in the periodic table, it is not necessary to specify the charge in the name. Examples of monatomic anions are in Table 2.2
b. Polyatomic anions. Polyatomic anions typically have common names that must be memorized; some examples are in Table 2.4. Polyatomic anions that contain a single metal or nonmetal atom plus one or more oxygen atoms are called oxoanions (or oxyanions). In cases where only two oxoanions are known for an element, the name of the oxoanion with more oxygen atoms ends in -ate, and the name of the oxoanion with fewer oxygen atoms ends in -ite. For example, NO3 is nitrate and NO2 is nitrite.
The halogens and some of the transition metals form more extensive series of oxoanions with as many as four members. In the names of these oxoanions, the prefix per- is used to identify the oxoanion with the most oxygen (so that ClO4 is perchlorate and ClO3 is chlorate), and the prefix hypo- is used to identify the anion with the fewest oxygen (ClO2 is chlorite and ClO is hypochlorite). The relationship between the names of oxoanions and the number of oxygen atoms present is diagrammed in Figure \(3\) "The Relationship between the Names of Oxoanions and the Number of Oxygen Atoms Present." Differentiating the oxoanions in such a series is no trivial matter; for example, the hypochlorite ion is the active ingredient in laundry bleach and swimming pool disinfectant, but compounds that contain the perchlorate ion can explode if they come into contact with organic substances.
4. Write the name of the compound as the name of the cation followed by the name of the anion.
It is not necessary to indicate the number of cations or anions present per formula unit in the name of an ionic compound because this information is implied by the charges on the ions. The charge of the ions must be considered when writing the formula for an ionic compound from its name, however. Because the charge on the chloride ion is −1 and the charge on the calcium ion is +2, for example, consistent with their positions in the periodic table, arithmetic indicates that calcium chloride must contain twice as many chloride ions as calcium ions to maintain electrical neutrality. Thus, the formula is CaCl2. Similarly, calcium phosphate must be Ca3(PO4)2 because the cation and the anion have charges of +2 and −3, respectively. The best way to learn how to name ionic compounds is to work through a few examples, referring to Figure \(1\), Table 2.2, Table 2.4, and Table \(1\) as needed.
With only a few exceptions, these metals are usually transition metals or actinides.
Note
Cations are always named before anions.
Most transition metals, many actinides, and the heaviest elements of groups 13–15 can form more than one cation.
Example \(1\)
Write the systematic name (and the common name if applicable) for each ionic compound.
1. LiCl
2. MgSO4
3. (NH4)3PO4
4. Cu2O
Given: empirical formula
Asked for: name
Strategy:
A If only one charge is possible for the cation, give its name, consulting Table 2.2 or Table 2.4 if necessary. If the cation can have more than one charge (Table \(1\)), specify the charge using roman numerals.
B If the anion does not contain oxygen, name it according to step 3a, using Table 2.2 and Table 2.4 if necessary. For polyatomic anions that contain oxygen, use Table 2.4 and the appropriate prefix and suffix listed in step 3b.
C Beginning with the cation, write the name of the compound.
Solution:
a. A B Lithium is in group 1, so we know that it forms only the Li+ cation, which is the lithium ion. Similarly, chlorine is in group 7, so it forms the Cl anion, which is the chloride ion. C Because we begin with the name of the cation, the name of this compound is lithium chloride, which is used medically as an antidepressant drug.
b. A B The cation is the magnesium ion, and the anion, which contains oxygen, is sulfate. C Because we list the cation first, the name of this compound is magnesium sulfate. A hydrated form of magnesium sulfate (MgSO4·7H2O) is sold in drugstores as Epsom salts, a harsh but effective laxative.
c. A B The cation is the ammonium ion (from Table 2.4), and the anion is phosphate. C The compound is therefore ammonium phosphate, which is widely used as a fertilizer. It is not necessary to specify that the formula unit contains three ammonium ions because three are required to balance the negative charge on phosphate.
d. A B The cation is a transition metal that often forms more than one cation (Table \(1\)). We must therefore specify the positive charge on the cation in the name: copper(I) or, according to the older system, cuprous. The anion is oxide. C The name of this compound is copper(I) oxide or, in the older system, cuprous oxide. Copper(I) oxide is used as a red glaze on ceramics and in antifouling paints to prevent organisms from growing on the bottoms of boats.
Cu2O. The bottom of a boat is protected with a red antifouling paint containing copper(I) oxide, Cu2O.
Exercise \(1\)
Exercise
Write the systematic name (and the common name if applicable) for each ionic compound.
1. CuCl2
2. MgCO3
3. FePO4
Answer:
1. copper(II) chloride (or cupric chloride)
2. magnesium carbonate
3. iron(III) phosphate (or ferric phosphate)
Example \(2\)
Write the formula for each compound.
1. calcium dihydrogen phosphate
2. aluminum sulfate
3. chromium(III) oxide
Given: systematic name
Asked for: formula
Strategy:
A Identify the cation and its charge using the location of the element in the periodic table and Table 2.2 and form cations with different charges, use the appropriate roman numeral or suffix to indicate its charge.
B Identify the anion using Table 2.2 and Table 2.4. Beginning with the cation, write the compound’s formula and then determine the number of cations and anions needed to achieve electrical neutrality.
Solution:
a. A Calcium is in group 2, so it forms only the Ca2+ ion. B Dihydrogen phosphate is the H2PO4− ion (Table 2.4). Two H2PO4 ions are needed to balance the positive charge on Ca2+, to give Ca(H2PO4)2. A hydrate of calcium dihydrogen phosphate, Ca(H2PO4)2·H2O, is the active ingredient in baking powder.
b. A Aluminum, near the top of group 13 in the periodic table, forms only one cation, Al3+ (Figure \(2\) "Metals That Form More Than One Cation and Their Locations in the Periodic Table"). B Sulfate is SO42 (Table 2.4). To balance the electrical charges, we need two Al3+ cations and three SO42 anions, giving Al2(SO4)3. Aluminum sulfate is used to tan leather and purify drinking water.
c. A Because chromium is a transition metal, it can form cations with different charges. The roman numeral tells us that the positive charge in this case is +3, so the cation is Cr3+. B Oxide is O2−. Thus two cations (Cr3+) and three anions (O2−) are required to give an electrically neutral compound, Cr2O3. This compound is a common green pigment that has many uses, including camouflage coatings.
Cr2O3. Chromium(III) oxide (Cr2O3) is a common pigment in dark green paints, such as camouflage paint.
Exercise \(2\)
Write the formula for each compound.
1. barium chloride
2. sodium carbonate
3. iron(III) hydroxide
Answer:
1. BaCl2
2. Na2CO3
3. Fe(OH)3
Summary
• There is a systematic method used to name ionic compounds.
Ionic compounds are named according to systematic procedures, although common names are widely used. Systematic nomenclature enables chemists to write the structure of any compound from its name and vice versa. Ionic compounds are named by writing the cation first, followed by the anion. If a metal can form cations with more than one charge, the charge is indicated by roman numerals in parentheses following the name of the metal. Oxoanions are polyatomic anions that contain a single metal or nonmetal atom and one or more oxygen atoms.
Learning Objectives
• To describe the composition of a chemical compound.
• To name covalent compounds that contain up to three elements.
As with ionic compounds, the system for naming covalent compounds enables chemists to write the molecular formula from the name and vice versa. This and the following section describe the rules for naming simple covalent compounds, beginning with inorganic compounds and then turning to simple organic compounds that contain only carbon and hydrogen.
When chemists synthesize a new compound, they may not yet know its molecular or structural formula. In such cases, they usually begin by determining its empirical formula, the relative numbers of atoms of the elements in a compound, reduced to the smallest whole numbers. Because the empirical formula is based on experimental measurements of the numbers of atoms in a sample of the compound, it shows only the ratios of the numbers of the elements present. The difference between empirical and molecular formulas can be illustrated with butane, a covalent compound used as the fuel in disposable lighters. The molecular formula for butane is C4H10. The ratio of carbon atoms to hydrogen atoms in butane is 4:10, which can be reduced to 2:5. The empirical formula for butane is therefore C2H5. The formula unit is the absolute grouping of atoms or ions represented by the empirical formula of a compound, either ionic or covalent. Butane has the empirical formula C2H5, but it contains two C2H5 formula units, giving a molecular formula of C4H10.
Because ionic compounds do not contain discrete molecules, empirical formulas are used to indicate their compositions. All compounds, whether ionic or covalent, must be electrically neutral. Consequently, the positive and negative charges in a formula unit must exactly cancel each other. If the cation and the anion have charges of equal magnitude, such as Na+ and Cl, then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be NaCl. If the charges are not the same magnitude, then a cation:anion ratio other than 1:1 is needed to produce a neutral compound. In the case of Mg2+ and Cl, for example, two Cl ions are needed to balance the two positive charges on each Mg2+ ion, giving an empirical formula of MgCl2. Similarly, the formula for the ionic compound that contains Na+ and O2− ions is Na2O.
Note
Ionic compounds do not contain discrete molecules, so empirical formulas are used to indicate their compositions.
Nomenclature of Metals: https://youtu.be/zVhGxYTgRk0
Nomenclature of Transition Metals: https://youtu.be/gIaRpko0A_A
Binary Ionic Compounds
An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is MgCl2, a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows:
Crossing charges. One method for obtaining subscripts in the empirical formula is by crossing charges.
When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Mg2O2:
This simplifies to its correct empirical formula MgO. The empirical formula has one Mg2+ ion and one O2− ion.
Example \(1\)
Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair.
1. Ga3+ and As3
2. Eu3+ and O2−
3. calcium and chlorine
Given: ions or elements
Asked for: empirical formula for binary ionic compound
Strategy:
1. If not given, determine the ionic charges based on the location of the elements in the periodic table.
2. Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the lowest numbers
to write the empirical formula. Check to make sure the empirical formula is electrically neutral.
Solution
a. B Using the absolute values of the charges on the ions as the subscripts gives Ga3As3:
Reducing the subscripts to the smallest whole numbers gives the empirical formula GaAs, which is electrically neutral [+3 + (−3) = 0]. Alternatively, we could recognize that Ga3+ and As3 have charges of equal magnitude but opposite signs. One Ga3+ ion balances the charge on one As3 ion, and a 1:1 compound will have no net charge. Because we write subscripts only if the number is greater than 1, the empirical formula is GaAs. GaAs is gallium arsenide, which is widely used in the electronics industry in transistors and other devices.
b. B Because Eu3+ has a charge of +3 and O2− has a charge of −2, a 1:1 compound would have a net charge of +1. We must therefore find multiples of the charges that cancel. We cross charges, using the absolute value of the charge on one ion as the subscript for the other ion:
The subscript for Eu3+ is 2 (from O2−), and the subscript for O2− is 3 (from Eu3+), giving Eu2O3; the subscripts cannot be reduced further. The empirical formula contains a positive charge of 2(+3) = +6 and a negative charge of 3(−2) = −6, for a net charge of 0. The compound Eu2O3 is neutral. Europium oxide is responsible for the red color in television and computer screens.
c. A Because the charges on the ions are not given, we must first determine the charges expected for the most common ions derived from calcium and chlorine. Calcium lies in group 2, so it should lose two electrons to form Ca2+. Chlorine lies in group 17, so it should gain one electron to form Cl.
B Two Cl ions are needed to balance the charge on one Ca2+ ion, which leads to the empirical formula CaCl2. We could also cross charges, using the absolute value of the charge on Ca2+ as the subscript for Cl and the absolute value of the charge on Cl as the subscript for Ca:
The subscripts in CaCl2 cannot be reduced further. The empirical formula is electrically neutral [+2 + 2(−1) = 0]. This compound is calcium chloride, one of the substances used as “salt” to melt ice on roads and sidewalks in winter.
Exercise \(1\)
Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair.
1. Li+ and N3−
2. Al3+ and O2−
3. lithium and oxygen
Answer:
1. Li3N
2. Al2O3
3. Li2O
Polyatomic Ions
Polyatomic ions are groups of atoms that bear net electrical charges, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH4+) and the methylammonium (CH3NH3+) ions. Polyatomic anions are much more numerous than polyatomic cations; some common examples are in Table \(1\).
Table \(1\): Common Polyatomic Ions and Their Names
Formula Name of Ion Formula Name of Ion
NH4+ ammonium HPO42 hydrogen phosphate
CH3NH3+ methylammonium H2PO4 dihydrogen phosphate
OH hydroxide ClO hypochlorite
O22 peroxide ClO2 chlorite
CN cyanide ClO3 chlorate
SCN thiocyanate ClO4 perchlorate
NO2 nitrite MnO4 permanganate
NO3 nitrate CrO42 chromate
CO32 carbonate Cr2O72 dichromate
HCO3 hydrogen carbonate, or bicarbonate C2O42 oxalate
SO32 sulfite HCO2 formate
SO42 sulfate CH3CO2 acetate
HSO4 hydrogen sulfate, or bisulfate C6H5CO2 benzoate
PO43 phosphate
The method used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus, K+ and NO3 ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42 form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3 form NH4NO3 (ammonium nitrate), a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43 ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca2+ ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca3(PO4)2; the parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43 ions.
Example \(2\)
Write the empirical formula for the compound formed from each ion pair.
1. Na+ and HPO42
2. potassium cation and cyanide anion
3. calcium cation and hypochlorite anion
Given: ions
Asked for: empirical formula for ionic compound
Strategy:
1. If it is not given, determine the charge on a monatomic ion from its location in the periodic table. Use Table \(1\) to find the charge on a polyatomic ion.
2. Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the smallest whole numbers when writing the empirical formula.
Solution:
1. B Because HPO42 has a charge of −2 and Na+ has a charge of +1, the empirical formula requires two Na+ ions to balance the charge of the polyatomic ion, giving Na2HPO4. The subscripts are reduced to the lowest numbers, so the empirical formula is Na2HPO4. This compound is sodium hydrogen phosphate, which is used to provide texture in processed cheese, puddings, and instant breakfasts.
2. A The potassium cation is K+, and the cyanide anion is CN. B Because the magnitude of the charge on each ion is the same, the empirical formula is KCN. Potassium cyanide is highly toxic, and at one time it was used as rat poison. This use has been discontinued, however, because too many people were being poisoned accidentally.
3. A The calcium cation is Ca2+, and the hypochlorite anion is ClO. B Two ClO ions are needed to balance the charge on one Ca2+ ion, giving Ca(ClO)2. The subscripts cannot be reduced further, so the empirical formula is Ca(ClO)2. This is calcium hypochlorite, the “chlorine” used to purify water in swimming pools.
Exercise \(2\)
Write the empirical formula for the compound formed from each ion pair.
1. Ca2+ and H2PO4
2. sodium cation and bicarbonate anion
3. ammonium cation and sulfate anion
Answer:
1. Ca(H2PO4)2: calcium dihydrogen phosphate is one of the ingredients in baking powder.
2. NaHCO3: sodium bicarbonate is found in antacids and baking powder; in pure form, it is sold as baking soda.
3. (NH4)2SO4: ammonium sulfate is a common source of nitrogen in fertilizers.
Polyatomics: https://youtu.be/kTSPkzDcntA
Hydrates
Many ionic compounds occur as hydrates, compounds that contain specific ratios of loosely bound water molecules, called waters of hydration. Waters of hydration can often be removed simply by heating. For example, calcium dihydrogen phosphate can form a solid that contains one molecule of water per Ca(H2PO4)2 unit and is used as a leavening agent in the food industry to cause baked goods to rise. The empirical formula for the solid is Ca(H2PO4)2·H2O. In contrast, copper sulfate usually forms a blue solid that contains five waters of hydration per formula unit, with the empirical formula CuSO4·5H2O. When heated, all five water molecules are lost, giving a white solid with the empirical formula CuSO4.
Compounds that differ only in the numbers of waters of hydration can have very different properties. For example, CaSO4·½H2O is plaster of Paris, which was often used to make sturdy casts for broken arms or legs, whereas CaSO4·2H2O is the less dense, flakier gypsum, a mineral used in drywall panels for home construction. When a cast would set, a mixture of plaster of Paris and water crystallized to give solid CaSO4·2H2O. Similar processes are used in the setting of cement and concrete.
Binary Inorganic Compounds
Binary covalent compounds—covalent compounds that contain only two elements—are named using a procedure similar to that used for simple ionic compounds, but prefixes are added as needed to indicate the number of atoms of each kind. The procedure, diagrammed in Figure \(2\) consists of the following steps:
1. Place the elements in their proper order.
1. The element farthest to the left in the periodic table is usually named first. If both elements are in the same group, the element closer to the bottom of the column is named first.
2. The second element is named as if it were a monatomic anion in an ionic compound (even though it is not), with the suffix -ide attached to the root of the element name.
2. Identify the number of each type of atom present.
1. Prefixes derived from Greek stems are used to indicate the number of each type of atom in the formula unit (Table \(2\)). The prefix mono- (“one”) is used only when absolutely necessary to avoid confusion, just as the subscript 1 is omitted when writing molecular formulas.
To demonstrate steps 1 and 2a, HCl is named hydrogen chloride (because hydrogen is to the left of chlorine in the periodic table), and PCl5 is phosphorus pentachloride. The order of the elements in the name of BrF3, bromine trifluoride, is determined by the fact that bromine lies below fluorine in Group 17.
Table \(2\): Prefixes for Indicating the Number of Atoms in Chemical Names
Prefix Number
mono- 1
di- 2
tri- 3
tetra- 4
penta- 5
hexa- 6
hepta- 7
octa- 8
nona- 9
deca- 10
undeca- 11
dodeca- 12
2. If a molecule contains more than one atom of both elements, then prefixes are used for both. Thus N2O3 is dinitrogen trioxide, as shown in Figure 2.13.
3. In some names, the final a or o of the prefix is dropped to avoid awkward pronunciation. Thus OsO4 is osmium tetroxide rather than osmium tetraoxide.
3. Write the name of the compound.
1. Binary compounds of the elements with oxygen are generally named as “element oxide,” with prefixes that indicate the number of atoms of each element per formula unit. For example, CO is carbon monoxide. The only exception is binary compounds of oxygen with fluorine, which are named as oxygen fluorides.
2. Certain compounds are always called by the common names that were assigned before formulas were used. For example, H2O is water (not dihydrogen oxide); NH3 is ammonia; PH3 is phosphine; SiH4 is silane; and B2H6, a dimer of BH3, is diborane. For many compounds, the systematic name and the common name are both used frequently, requiring familiarity with both. For example, the systematic name for NO is nitrogen monoxide, but it is much more commonly called nitric oxide. Similarly, N2O is usually called nitrous oxide rather than dinitrogen monoxide. Notice that the suffixes -ic and -ous are the same ones used for ionic compounds.
Note
Start with the element at the far left in the periodic table and work to the right. If two or more elements are in the same group, start with the bottom element and work up.
Example \(3\)
Write the name of each binary covalent compound.
1. SF6
2. N2O4
3. ClO2
Given: molecular formula
Asked for: name of compound
Strategy:
1. List the elements in order according to their positions in the periodic table. Identify the number of each type of atom in the chemical formula and then use Table \(2\) to determine the prefixes needed.
2. If the compound contains oxygen, follow step 3a. If not, decide whether to use the common name or the systematic name.
Solution:
1. A Because sulfur is to the left of fluorine in the periodic table, sulfur is named first. Because there is only one sulfur atom in the formula, no prefix is needed. B There are, however, six fluorine atoms, so we use the prefix for six: hexa- (Table \(2\)). The compound is sulfur hexafluoride.
2. A Because nitrogen is to the left of oxygen in the periodic table, nitrogen is named first. Because more than one atom of each element is present, prefixes are needed to indicate the number of atoms of each. According to Table \(2\) "Prefixes for Indicating the Number of Atoms in Chemical Names", the prefix for two is di-, and the prefix for four is tetra-. B The compound is dinitrogen tetroxide (omitting the a in tetra- according to step 2c) and is used as a component of some rocket fuels.
3. A Although oxygen lies to the left of chlorine in the periodic table, it is not named first because ClO2 is an oxide of an element other than fluorine (step 3a). Consequently, chlorine is named first, but a prefix is not necessary because each molecule has only one atom of chlorine. B Because there are two oxygen atoms, the compound is a dioxide. Thus the compound is chlorine dioxide. It is widely used as a substitute for chlorine in municipal water treatment plants because, unlike chlorine, it does not react with organic compounds in water to produce potentially toxic chlorinated compounds.
Example \(3\)
Write the name of each binary covalent compound.
1. IF7
2. N2O5
3. OF2
Answer
1. iodine heptafluoride
2. dinitrogen pentoxide
3. oxygen difluoride
Example \(4\)
Write the formula for each binary covalent compound.
1. sulfur trioxide
2. diiodine pentoxide
Given: name of compound
Asked for: formula
Strategy:
List the elements in the same order as in the formula, use Table \(2\) to identify the number of each type of atom present, and then indicate this quantity as a subscript to the right of that element when writing the formula.
Solution:
1. Sulfur has no prefix, which means that each molecule has only one sulfur atom. The prefix tri- indicates that there are three oxygen atoms. The formula is therefore SO3. Sulfur trioxide is produced industrially in huge amounts as an intermediate in the synthesis of sulfuric acid.
2. The prefix di- tells you that each molecule has two iodine atoms, and the prefix penta- indicates that there are five oxygen atoms. The formula is thus I2O5, a compound used to remove carbon monoxide from air in respirators.
Exercise \(4\)
Write the formula for each binary covalent compound.
1. silicon tetrachloride
2. disulfur decafluoride
Answer
1. SiCl4
2. S2F10
The structures of some of the compounds in Examples 3.6.3 and 3.6.4 are shown in Figure \(2\) along with the location of the “central atom” of each compound in the periodic table. It may seem that the compositions and structures of such compounds are entirely random, but this is not true. After mastering the material discussed later on this course, one is able to predict the compositions and structures of compounds of this type with a high degree of accuracy.
Nomenclature of Nonmetals: https://youtu.be/VgHCrtpDWJk
The Learning Objective of this Module is to identify and name some common acids and bases.
For our purposes at this point in the text, we can define an acid as a substance with at least one hydrogen atom that can dissociate to form an anion and an H+ ion (a proton) in aqueous solution, thereby forming an acidic solution. We can define bases as compounds that produce hydroxide ions (OH) and a cation when dissolved in water, thus forming a basic solution. Solutions that are neither basic nor acidic are neutral. We will discuss the chemistry of acids and bases in more detail later, but in this section we describe the nomenclature of common acids and identify some important bases so that you can recognize them in future discussions. Pure acids and bases and their concentrated aqueous solutions are commonly encountered in the laboratory. They are usually highly corrosive, so they must be handled with care.
Acids
The names of acids differentiate between (1) acids in which the H+ ion is attached to an oxygen atom of a polyatomic anion (these are called oxoacids, or occasionally oxyacids) and (2) acids in which the H+ ion is attached to some other element. In the latter case, the name of the acid begins with hydro- and ends in -ic, with the root of the name of the other element or ion in between. Recall that the name of the anion derived from this kind of acid always ends in -ide. Thus hydrogen chloride (HCl) gas dissolves in water to form hydrochloric acid (which contains H+ and Cl ions), hydrogen cyanide (HCN) gas forms hydrocyanic acid (which contains H+ and CN ions), and so forth (Table \(3\)). Examples of this kind of acid are commonly encountered and very important. For instance, your stomach contains a dilute solution of hydrochloric acid to help digest food. When the mechanisms that prevent the stomach from digesting itself malfunction, the acid destroys the lining of the stomach and an ulcer forms.
Note
Acids are distinguished by whether the H+ ion is attached to an oxygen atom of a polyatomic anion or some other element.
Table \(3\): Some Common Acids That Do Not Contain Oxygen
Formula Name in Aqueous Solution Name of Gaseous Species
HF hydrofluoric acid hydrogen fluoride
HCl hydrochloric acid hydrogen chloride
HBr hydrobromic acid hydrogen bromide
HI hydroiodic acid hydrogen iodide
HCN hydrocyanic acid hydrogen cyanide
H2S hydrosulfuric acid hydrogen sulfide
If an acid contains one or more H+ ions attached to oxygen, it is a derivative of one of the common oxoanions, such as sulfate (SO42) or nitrate (NO3). These acids contain as many H+ ions as are necessary to balance the negative charge on the anion, resulting in a neutral species such as H2SO4 and HNO3.
The names of acids are derived from the names of anions according to the following rules:
1. If the name of the anion ends in -ate, then the name of the acid ends in -ic. For example, because NO3 is the nitrate ion, HNO3 is nitric acid. Similarly, ClO4 is the perchlorate ion, so HClO4 is perchloric acid. Two important acids are sulfuric acid (H2SO4) from the sulfate ion (SO42) and phosphoric acid (H3PO4) from the phosphate ion (PO43). These two names use a slight variant of the root of the anion name: sulfate becomes sulfuric and phosphate becomes phosphoric.
2. If the name of the anion ends in -ite, then the name of the acid ends in -ous. For example, OCl is the hypochlorite ion, and HOCl is hypochlorous acid; NO2 is the nitrite ion, and HNO2 is nitrous acid; and SO32 is the sulfite ion, and H2SO3 is sulfurous acid. The same roots are used whether the acid name ends in -ic or -ous; thus, sulfite becomes sulfurous.
The relationship between the names of the oxoacids and the parent oxoanions is illustrated in Figure \(3\), and some common oxoacids are in Table \(4\).
Table \(4\): Some Common Oxoacids
Formula Name
\(HNO_2\) nitrous acid
\(HNO_3\) nitric acid
\(H_2SO_3\) sulfurous acid
\(H_2SO_4\) sulfuric acid
\(H_3PO_4\) phosphoric acid
\(H_2CO_3\) carbonic acid
\(HClO\) hypochlorous acid
\(HClO_2\) chlorous acid
\(HClO_3\) chloric acid
\(HClO_4\) perchloric acid
Example \(5\)
Name and give the formula for each acid.
1. the acid formed by adding a proton to the hypobromite ion (OBr)
2. the acid formed by adding two protons to the selenate ion (SeO42)
Given: anion
Asked for: parent acid
Strategy:
Refer to Table \(3\) and Table \(4\) to find the name of the acid. If the acid is not listed, use the guidelines given previously.
Solution:
Neither species is listed in Table \(3\) or Table \(4\), so we must use the information given previously to derive the name of the acid from the name of the polyatomic anion.
1. The anion name, hypobromite, ends in -ite, so the name of the parent acid ends in -ous. The acid is therefore hypobromous acid (HOBr).
2. Selenate ends in -ate, so the name of the parent acid ends in -ic. The acid is therefore selenic acid (H2SeO4).
Exercise \(5\)
Name and give the formula for each acid.
1. the acid formed by adding a proton to the perbromate ion (BrO4)
2. the acid formed by adding three protons to the arsenite ion (AsO33)
Answer:
1. perbromic acid; HBrO4
2. arsenous acid; H3AsO3
Nomenclature of Acids: https://youtu.be/in46UUzmSO4
Organic Acids
Many organic compounds contain the carbonyl group, in which there is a carbon–oxygen double bond. In carboxylic acids, an –OH group is covalently bonded to the carbon atom of the carbonyl group. Their general formula is RCO2H, sometimes written as RCOOH:
where R can be an alkyl group, an aryl group, or a hydrogen atom. The simplest example, HCO2H, is formic acid, so called because it is found in the secretions of stinging ants (from the Latin formica, meaning “ant”). Another example is acetic acid (CH3CO2H), which is found in vinegar. Like many acids, carboxylic acids tend to have sharp odors. For example, butyric acid (CH3CH2CH2CO2H), is responsible for the smell of rancid butter, and the characteristic odor of sour milk and vomit is due to lactic acid [CH3CH(OH)CO2H]. Some common carboxylic acids are shown in Table 3.6.5.
Table \(5\): Some Common Carboxylic Acids
Although carboxylic acids are covalent compounds, when they dissolve in water, they dissociate to produce H+ ions (just like any other acid) and RCO2 ions. Note that only the hydrogen attached to the oxygen atom of the CO2 group dissociates to form an H+ ion. In contrast, the hydrogen atom attached to the oxygen atom of an alcohol does not dissociate to form an H+ ion when an alcohol is dissolved in water. The reasons for the difference in behavior between carboxylic acids and alcohols will be discussed in Chapter 8.
Note
Only the hydrogen attached to the oxygen atom of the CO2 group dissociates to form an H+ ion.
Bases
We will present more comprehensive definitions of bases in later chapters, but virtually every base you encounter in the meantime will be an ionic compound, such as sodium hydroxide (NaOH) and barium hydroxide [Ba(OH)2], that contain the hydroxide ion and a metal cation. These have the general formula M(OH)n. It is important to recognize that alcohols, with the general formula ROH, are covalent compounds, not ionic compounds; consequently, they do not dissociate in water to form a basic solution (containing OH ions). When a base reacts with any of the acids we have discussed, it accepts a proton (H+). For example, the hydroxide ion (OH) accepts a proton to form H2O. Thus bases are also referred to as proton acceptors.
Concentrated aqueous solutions of ammonia (NH3) contain significant amounts of the hydroxide ion, even though the dissolved substance is not primarily ammonium hydroxide (NH4OH) as is often stated on the label. Thus aqueous ammonia solution is also a common base. Replacing a hydrogen atom of NH3 with an alkyl group results in an amine (RNH2), which is also a base. Amines have pungent odors—for example, methylamine (CH3NH2) is one of the compounds responsible for the foul odor associated with spoiled fish. The physiological importance of amines is suggested in the word vitamin, which is derived from the phrase vital amines. The word was coined to describe dietary substances that were effective at preventing scurvy, rickets, and other diseases because these substances were assumed to be amines. Subsequently, some vitamins have indeed been confirmed to be amines.
Note
Metal hydroxides (MOH) yield OH ions and are bases, alcohols (ROH) do not yield OH or H+ ions and are neutral, and carboxylic acids (RCO2H) yield H+ ions and are acids.
Summary
Common acids and the polyatomic anions derived from them have their own names and rules for nomenclature. The nomenclature of acids differentiates between oxoacids, in which the H+ ion is attached to an oxygen atom of a polyatomic ion, and acids in which the H+ ion is attached to another element. Carboxylic acids are an important class of organic acids. Ammonia is an important base, as are its organic derivatives, the amines.
Summary
• The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit.
• Covalent inorganic compounds are named using a procedure similar to that used for ionic compounds, whereas hydrocarbons use a system based on the number of bonds between carbon atoms.
Covalent inorganic compounds are named by a procedure similar to that used for ionic compounds, using prefixes to indicate the numbers of atoms in the molecular formula. An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.6%3A_Names_and_Formulas_of_Inorganic_Compounds.txt |
Learning Objectives
• Explain the following laws within the Ideal Gas Law:
• Avogadro's law of gases
Hydrocarbons
Approximately one-third of the compounds produced industrially are organic compounds. All living organisms are composed of organic compounds, as are most foods, medicines, clothing fibers, and plastics. The detection of organic compounds is useful in many fields. In one recently developed application, scientists have devised a new method called “material degradomics” to monitor the degradation of old books and historical documents. As paper ages, it produces a familiar “old book smell” from the release of organic compounds in gaseous form. The composition of the gas depends on the original type of paper used, a book’s binding, and the applied media. By analyzing these organic gases and isolating the individual components, preservationists are better able to determine the condition of an object and those books and documents most in need of immediate protection.
The simplest class of organic compounds is the hydrocarbons, which consist entirely of carbon and hydrogen. Petroleum and natural gas are complex, naturally occurring mixtures of many different hydrocarbons that furnish raw materials for the chemical industry. The four major classes of hydrocarbons are the following: the alkanes, which contain only carbon–hydrogen and carbon–carbon single bonds; the alkenes, which contain at least one carbon–carbon double bond; the alkynes, which contain at least one carbon–carbon triple bond; and the aromatic hydrocarbons, which usually contain rings of six carbon atoms that can be drawn with alternating single and double bonds. Alkanes are also called saturated hydrocarbons, whereas hydrocarbons that contain multiple bonds (alkenes, alkynes, and aromatics) are unsaturated.
Alkanes
The simplest alkane is methane (CH4), a colorless, odorless gas that is the major component of natural gas. In larger alkanes whose carbon atoms are joined in an unbranched chain (straight-chain alkanes), each carbon atom is bonded to at most two other carbon atoms. The structures of two simple alkanes are shown in Figure \(1\), and the names and condensed structural formulas for the first 10 straight-chain alkanes are in Table \(1\). The names of all alkanes end in -ane, and their boiling points increase as the number of carbon atoms increases.
Table \(1\): The First 10 Straight-Chain Alkanes
Name Number of Carbon Atoms Molecular Formula Condensed Structural Formula Boiling Point (°C) Uses
methane 1 CH4 CH4 −162 natural gas constituent
ethane 2 C2H6 CH3CH3 −89 natural gas constituent
propane 3 C3H8 CH3CH2CH3 −42 bottled gas
butane 4 C4H10 CH3CH2CH2CH3 or CH3(CH2)2CH3 0 lighters, bottled gas
pentane 5 C5H12 CH3(CH2)3CH3 36 solvent, gasoline
hexane 6 C6H14 CH3(CH2)4CH3 69 solvent, gasoline
heptane 7 C7H16 CH3(CH2)5CH3 98 solvent, gasoline
octane 8 C8H18 CH3(CH2)6CH3 126 gasoline
nonane 9 C9H20 CH3(CH2)7CH3 151 gasoline
decane 10 C10H22 CH3(CH2)8CH3 174 kerosene
Alkanes with four or more carbon atoms can have more than one arrangement of atoms. The carbon atoms can form a single unbranched chain, or the primary chain of carbon atoms can have one or more shorter chains that form branches. For example, butane (C4H10) has two possible structures. Normal butane (usually called n-butane) is CH3CH2CH2CH3, in which the carbon atoms form a single unbranched chain. In contrast, the condensed structural formula for isobutane is (CH3)2CHCH3, in which the primary chain of three carbon atoms has a one-carbon chain branching at the central carbon. Three-dimensional representations of both structures are as follows:
The systematic names for branched hydrocarbons use the lowest possible number to indicate the position of the branch along the longest straight carbon chain in the structure. Thus the systematic name for isobutane is 2-methylpropane, which indicates that a methyl group (a branch consisting of –CH3) is attached to the second carbon of a propane molecule. Similarly, Section 2.6 states that one of the major components of gasoline is commonly called isooctane; its structure is as follows:
The compound has a chain of five carbon atoms, so it is a derivative of pentane. There are two methyl group branches at one carbon atom and one methyl group at another. Using the lowest possible numbers for the branches gives 2,2,4-trimethylpentane for the systematic name of this compound.
Alkenes
The simplest alkenes are ethylene, C2H4 or CH2=CH2, and propylene, C3H6 or CH3CH=CH2 (part (a) in Figure \(2\)). The names of alkenes that have more than three carbon atoms use the same stems as the names of the alkanes (Table \(1\) "The First 10 Straight-Chain Alkanes") but end in -ene instead of -ane.
As with alkanes, more than one structure is possible for alkenes with four or more carbon atoms. For example, an alkene with four carbon atoms has three possible structures. One is CH2=CHCH2CH3 (1-butene), which has the double bond between the first and second carbon atoms in the chain. The other two structures have the double bond between the second and third carbon atoms and are forms of CH3CH=CHCH3 (2-butene). All four carbon atoms in 2-butene lie in the same plane, so there are two possible structures (part (a) in Figure \(2\)). If the two methyl groups are on the same side of the double bond, the compound is cis-2-butene (from the Latin cis, meaning “on the same side”). If the two methyl groups are on opposite sides of the double bond, the compound is trans-2-butene (from the Latin trans, meaning “across”). These are distinctly different molecules: cis-2-butene melts at −138.9°C, whereas trans-2-butene melts at −105.5°C.
Just as a number indicates the positions of branches in an alkane, the number in the name of an alkene specifies the position of the first carbon atom of the double bond. The name is based on the lowest possible number starting from either end of the carbon chain, so CH3CH2CH=CH2 is called 1-butene, not 3-butene. Note that CH2=CHCH2CH3 and CH3CH2CH=CH2 are different ways of writing the same molecule (1-butene) in two different orientations.
The name of a compound does not depend on its orientation. As illustrated for 1-butene, both condensed structural formulas and molecular models show different orientations of the same molecule. It is important to be able to recognize the same structure no matter what its orientation.
Note
The positions of groups or multiple bonds are always indicated by the lowest number possible.
Alkynes
The simplest alkyne is acetylene, C2H2 or HC≡CH (part (b) in Figure \(2\)). Because a mixture of acetylene and oxygen burns with a flame that is hot enough (>3000°C) to cut metals such as hardened steel, acetylene is widely used in cutting and welding torches. The names of other alkynes are similar to those of the corresponding alkanes but end in -yne. For example, HC≡CCH3 is propyne, and CH3C≡CCH3 is 2-butyne because the multiple bond begins on the second carbon atom.
Note
The number of bonds between carbon atoms in a hydrocarbon is indicated in the suffix:
• alkane: only carbon–carbon single bonds
• alkene: at least one carbon–carbon double bond
• alkyne: at least one carbon–carbon triple bond
Cyclic Hydrocarbons
In a cyclic hydrocarbon, the ends of a hydrocarbon chain are connected to form a ring of covalently bonded carbon atoms. Cyclic hydrocarbons are named by attaching the prefix cyclo- to the name of the alkane, the alkene, or the alkyne. The simplest cyclic alkanes are cyclopropane (C3H6) a flammable gas that is also a powerful anesthetic, and cyclobutane (C4H8) (part (c) in Figure \(2\)). The most common way to draw the structures of cyclic alkanes is to sketch a polygon with the same number of vertices as there are carbon atoms in the ring; each vertex represents a CH2 unit. The structures of the cycloalkanes that contain three to six carbon atoms are shown schematically in Figure \(3\).
Aromatic Hydrocarbons
Alkanes, alkenes, alkynes, and cyclic hydrocarbons are generally called aliphatic hydrocarbons. The name comes from the Greek aleiphar, meaning “oil,” because the first examples were extracted from animal fats. In contrast, the first examples of aromatic hydrocarbons, also called arenes, were obtained by the distillation and degradation of highly scented (thus aromatic) resins from tropical trees.
The simplest aromatic hydrocarbon is benzene (C6H6), which was first obtained from a coal distillate. The word aromatic now refers to benzene and structurally similar compounds. As shown in part (a) in Figure \(4\), it is possible to draw the structure of benzene in two different but equivalent ways, depending on which carbon atoms are connected by double bonds or single bonds. Toluene is similar to benzene, except that one hydrogen atom is replaced by a –CH3 group; it has the formula C7H8 (part (b) in Figure \(4\)). The chemical behavior of aromatic compounds differs from the behavior of aliphatic compounds. Benzene and toluene are found in gasoline, and benzene is the starting material for preparing substances as diverse as aspirin and nylon.
Figure \(4\): Two Aromatic Hydrocarbons: (a) Benzene and (b) Toluene
Figure \(5\) illustrates two of the molecular structures possible for hydrocarbons that have six carbon atoms. As shown, compounds with the same molecular formula can have very different structures.
Example \(1\)
Write the condensed structural formula for each hydrocarbon.
1. n-heptane
2. 2-pentene
3. 2-butyne
4. cyclooctene
Given: name of hydrocarbon
Asked for: condensed structural formula
Strategy:
1. Use the prefix to determine the number of carbon atoms in the molecule and whether it is cyclic. From the suffix, determine whether multiple bonds are present.
2. Identify the position of any multiple bonds from the number(s) in the name and then write the condensed structural formula.
Solution:
a. A The prefix hept- tells us that this hydrocarbon has seven carbon atoms, and n- indicates that the carbon atoms form a straight chain. The suffix -ane tells that it is an alkane, with no carbon–carbon double or triple bonds. B The condensed structural formula is CH3CH2CH2CH2CH2CH2CH3, which can also be written as \(CH_3(CH_2)_5CH_3\).
b. A The prefix pent- tells us that this hydrocarbon has five carbon atoms, and the suffix -ene indicates that it is an alkene, with a carbon–carbon double bond. B The 2- tells us that the double bond begins on the second carbon of the five-carbon atom chain. The condensed structural formula of the compound is therefore CH3CH=CHCH2CH3.
c. A The prefix but- tells us that the compound has a chain of four carbon atoms, and the suffix -yne indicates that it has a carbon–carbon triple bond. B The 2- tells us that the triple bond begins on the second carbon of the four-carbon atom chain. So the condensed structural formula for the compound is CH3C≡CCH3.
d. A The prefix cyclo- tells us that this hydrocarbon has a ring structure, and oct- indicates that it contains eight carbon atoms, which we can draw as
The suffix -ene tells us that the compound contains a carbon–carbon double bond, but where in the ring do we place the double bond? B Because all eight carbon atoms are identical, it doesn’t matter. We can draw the structure of cyclooctene as
Exercise \(1\)
Write the condensed structural formula for each hydrocarbon.
1. n-octane
2. 2-hexene
3. 1-heptyne
4. cyclopentane
Answer:
1. CH3(CH2)6CH3
2. CH3CH=CHCH2CH2CH3
3. HC≡C(CH2)4CH3
The general name for a group of atoms derived from an alkane is an alkyl group. The name of an alkyl group is derived from the name of the alkane by adding the suffix -yl. Thus the –CH3 fragment is a methyl group, the –CH2CH3 fragment is an ethyl group, and so forth, where the dash represents a single bond to some other atom or group. Similarly, groups of atoms derived from aromatic hydrocarbons are aryl groups, which sometimes have unexpected names. For example, the –C6H5 fragment is derived from benzene, but it is called a phenyl group. In general formulas and structures, alkyl and aryl groups are often abbreviated as R.
Structures of alkyl and aryl groups. The methyl group is an example of an alkyl group, and the phenyl group is an example of an aryl group.
Alcohols
Replacing one or more hydrogen atoms of a hydrocarbon with an –OH group gives an alcohol, represented as ROH. The simplest alcohol (CH3OH) is called either methanol (its systematic name) or methyl alcohol (its common name). Methanol is the antifreeze in automobile windshield washer fluids, and it is also used as an efficient fuel for racing cars, most notably in the Indianapolis 500. Ethanol (or ethyl alcohol, CH3CH2OH) is familiar as the alcohol in fermented or distilled beverages, such as beer, wine, and whiskey; it is also used as a gasoline additive (Section 2.6). The simplest alcohol derived from an aromatic hydrocarbon is C6H5OH, phenol (shortened from phenyl alcohol), a potent disinfectant used in some sore throat medications and mouthwashes.
Ethanol, which is easy to obtain from fermentation processes, has successfully been used as an alternative fuel for several decades. Although it is a “green” fuel when derived from plants, it is an imperfect substitute for fossil fuels because it is less efficient than gasoline. Moreover, because ethanol absorbs water from the atmosphere, it can corrode an engine’s seals. Thus other types of processes are being developed that use bacteria to create more complex alcohols, such as octanol, that are more energy efficient and that have a lower tendency to absorb water. As scientists attempt to reduce mankind’s dependence on fossil fuels, the development of these so-called biofuels is a particularly active area of research.
Summary
The simplest organic compounds are the hydrocarbons, which contain only carbon and hydrogen. Alkanes contain only carbon–hydrogen and carbon–carbon single bonds, alkenes contain at least one carbon–carbon double bond, and alkynes contain one or more carbon–carbon triple bonds. Hydrocarbons can also be cyclic, with the ends of the chain connected to form a ring. Collectively, alkanes, alkenes, and alkynes are called aliphatic hydrocarbons. Aromatic hydrocarbons, or arenes, are another important class of hydrocarbons that contain rings of carbon atoms related to the structure of benzene (C6H6). A derivative of an alkane or an arene from which one hydrogen atom has been removed is called an alkyl group or an aryl group, respectively. Alcohols are another common class of organic compound, which contain an –OH group covalently bonded to either an alkyl group or an aryl group (often abbreviated R). | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.7%3A__Names_of_Formulas_of_Organic_Compounds.txt |
3.1: Types of Chemical Compounds and their Formulas
Conceptual Problems
1. Ionic and covalent compounds are held together by electrostatic attractions between oppositely charged particles. Describe the differences in the nature of the attractions in ionic and covalent compounds. Which class of compounds contains pairs of electrons shared between bonded atoms?
2. Which contains fewer electrons than the neutral atom—the corresponding cation or the anion?
3. What is the difference between an organic compound and an inorganic compound?
4. What is the advantage of writing a structural formula as a condensed formula?
5. The majority of elements that exist as diatomic molecules are found in one group of the periodic table. Identify the group.
6. Discuss the differences between covalent and ionic compounds with regard to
a. the forces that hold the atoms together.
b. melting points.
c. physical states at room temperature and pressure.
7. Why do covalent compounds generally tend to have lower melting points than ionic compounds?
Answer 7. Covalent compounds generally melt at lower temperatures than ionic compounds because the intermolecular interactions that hold the molecules together in a molecular solid are weaker than the electrostatic attractions that hold oppositely charged ions together in an ionic solid.
Numerical Problems
1. The structural formula for chloroform (CHCl3) was shown in Example 2. Based on this information, draw the structural formula of dichloromethane (CH2Cl2).
2. What is the total number of electrons present in each ion?
a. F
b. Rb+
c. Ce3+
d. Zr4+
e. Zn2+
f. Kr2+
g. B3+
3. What is the total number of electrons present in each ion?
a. Ca2+
b. Se2
c. In3+
d. Sr2+
e. As3+
f. N3−
g. Tl+
4. Predict how many electrons are in each ion.
a. an oxygen ion with a −2 charge
b. a beryllium ion with a +2 charge
c. a silver ion with a +1 charge
d. a selenium ion with a +4 charge
e. an iron ion with a +2 charge
f. a chlorine ion with a −1 charge
5. Predict how many electrons are in each ion.
a. a copper ion with a +2 charge
b. a molybdenum ion with a +4 charge
c. an iodine ion with a −1 charge
d. a gallium ion with a +3 charge
e. an ytterbium ion with a +3 charge
f. a scandium ion with a +3 charge
6. Predict the charge on the most common monatomic ion formed by each element.
a. chlorine
b. phosphorus
c. scandium
d. magnesium
e. arsenic
f. oxygen
7. Predict the charge on the most common monatomic ion formed by each element.
a. sodium
b. selenium
c. barium
d. rubidium
e. nitrogen
f. aluminum
8. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found.
a. \(_4^9X^{2+} \)
b. \(_1^1X^-\)
c. \(_8^{16}X^{2-} \)
9. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found.
a. \(_3^7X^+ \)
b. \(_9^{19}X^-\)
c. \(_{13}^{27}X^{3+}\)
Answers
5.
a. 27
b. 38
c. 54
d. 28
e. 67
f. 18
9.
a. Li, Li+, 2nd period, group 1
b. F, F, 2nd period, group 17
c. Al, Al3+, 3nd period, group 13
Conceptual Problems2
1. What are the differences and similarities between a polyatomic ion and a molecule?
2. Classify each compound as ionic or covalent.
1. Zn3(PO4)2
2. C6H5CO2H
3. K2Cr2O7
4. CH3CH2SH
5. NH4Br
6. CCl2F2
3. Classify each compound as ionic or covalent. Which are organic compounds and which are inorganic compounds?
1. CH3CH2CO2H
2. CaCl2
3. Y(NO3)3
4. H2S
5. NaC2H3O2
4. Generally, one cannot determine the molecular formula directly from an empirical formula. What other information is needed?
5. Give two pieces of information that we obtain from a structural formula that we cannot obtain from an empirical formula.
6. The formulas of alcohols are often written as ROH rather than as empirical formulas. For example, methanol is generally written as CH3OH rather than CH4O. Explain why the ROH notation is preferred.
7. The compound dimethyl sulfide has the empirical formula C2H6S and the structural formula CH3SCH3. What information do we obtain from the structural formula that we do not get from the empirical formula? Write the condensed structural formula for the compound.
8. What is the correct formula for magnesium hydroxide—MgOH2 or Mg(OH)2? Why?
9. Magnesium cyanide is written as Mg(CN)2, not MgCN2. Why?
10. Does a given hydrate always contain the same number of waters of hydration?
Answers 2
7. The structural formula gives us the connectivity of the atoms in the molecule or ion, as well as a schematic representation of their arrangement in space. Empirical formulas tell us only the ratios of the atoms present. The condensed structural formula of dimethylsulfide is (CH3)2S.
Numerical Problems 2
1. Write the formula for each compound.
a. magnesium sulfate, which has 1 magnesium atom, 4 oxygen atoms, and 1 sulfur atom
b. ethylene glycol (antifreeze), which has 6 hydrogen atoms, 2 carbon atoms, and 2 oxygen atoms
c. acetic acid, which has 2 oxygen atoms, 2 carbon atoms, and 4 hydrogen atoms
d. potassium chlorate, which has 1 chlorine atom, 1 potassium atom, and 3 oxygen atoms
e. sodium hypochlorite pentahydrate, which has 1 chlorine atom, 1 sodium atom, 6 oxygen atoms, and 10 hydrogen atoms
2. Write the formula for each compound.
a. cadmium acetate, which has 1 cadmium atom, 4 oxygen atoms, 4 carbon atoms, and 6 hydrogen atoms
b. barium cyanide, which has 1 barium atom, 2 carbon atoms, and 2 nitrogen atoms
c. iron(III) phosphate dihydrate, which has 1 iron atom, 1 phosphorus atom, 6 oxygen atoms, and 4 hydrogen atoms
d. manganese(II) nitrate hexahydrate, which has 1 manganese atom, 12 hydrogen atoms, 12 oxygen atoms, and 2 nitrogen atoms
e. silver phosphate, which has 1 phosphorus atom, 3 silver atoms, and 4 oxygen atoms
3. Complete the following table by filling in the formula for the ionic compound formed by each cation-anion pair.
Ion K+ Fe3+ NH4+ Ba2+
Cl KCl
SO42−
PO43−
NO3
OH
4. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements.
a. zinc and sulfur
b. barium and iodine
c. magnesium and chlorine
d. silicon and oxygen
e. sodium and sulfur
5. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements.
a. lithium and nitrogen
b. cesium and chlorine
c. germanium and oxygen
d. rubidium and sulfur
e. arsenic and sodium
6. Write the empirical formula for each compound.
a. Na2S2O4
b. B2H6
c. C6H12O6
d. P4O10
e. KMnO4
7. Write the empirical formula for each compound.
a. Al2Cl6
b. K2Cr2O7
c. C2H4
d. (NH2)2CNH
e. CH3COOH
Answers
1.
a. MgSO4
b. C2H6O2
c. C2H4O2
d. KClO3
e. NaOCl·5H2O
3.
Ion K + Fe 3+ NH 4 + Ba 2+
Cl − KCl FeCl3 NH4Cl BaCl2
SO 4 2− K2SO4 Fe2(SO4)3 (NH4)2SO4 BaSO4
PO 4 3− K3PO4 FePO4 (NH4)3PO4 Ba3(PO4)2
NO 3 − KNO3 Fe(NO3)3 NH4NO3 Ba(NO3)2
OH − KOH Fe(OH)3 NH4OH Ba(OH)2
5.
a. Li3N
b. CsCl
c. GeO2
d. Rb2S
e. Na3As
7.
a. AlCl3
b. K2Cr2O7
c. CH2
d. CH5N3
e. CH2O
3.6: Names and Formulas of Inorganic Compounds
Conceptual Problems
1. Name each cation.
1. a. K+
2. b. Al3+
3. c. NH4+
4. d. Mg2+
5. e. Li+
2. Name each anion.
1. a. Br
2. b. CO32−
3. c. S2−
4. d. NO3
5. e. HCO2
6. f. F
7. g. ClO
8. h. C2O42−
3. Name each anion.
1. a. PO43−
2. b. Cl
3. c. SO32−
4. d. CH3CO2
5. e. HSO4
6. f. ClO4
7. g. NO2
8. h. O2−
4. Name each anion.
1. a. SO42−
2. b. CN
3. c. Cr2O72−
4. d. N3−
5. e. OH
6. f. I
7. g. O22−
5. Name each compound.
1. a. MgBr2
2. b. NH4CN
3. c. CaO
4. d. KClO3
5. e. K3PO4
6. f. NH4NO2
7. g. NaN3
6. Name each compound.
1. a. NaNO3
2. b. Cu3(PO4)2
3. c. NaOH
4. d. Li4C
5. e. CaF2
6. f. NH4Br
7. g. MgCO3
7. Name each compound.
1. a. RbBr
2. b. Mn2(SO4)3
3. c. NaClO
4. d. (NH4)2SO4
5. e. NaBr
6. f. KIO3
7. g. Na2CrO4
8. Name each compound.
1. a. NH4ClO4
2. b. SnCl4
3. c. Fe(OH)2
4. d. Na2O
5. e. MgCl2
6. f. K2SO4
7. g. RaCl2
9. Name each compound.
1. a. KCN
2. b. LiOH
3. c. CaCl2
4. d. NiSO4
5. e. NH4ClO2
6. f. LiClO4
7. g. La(CN)3
Answer
7.
a. rubidium bromide
b. manganese(III) sulfate
c. sodium hypochlorite
d. ammonium sulfate
e. sodium bromide
f. potassium iodate
g. sodium chromate
Numerical Problems
1. For each ionic compound, name the cation and the anion and give the charge on each ion.
1. a. BeO
2. b. Pb(OH)2
3. c. BaS
4. d. Na2Cr2O7
5. e. ZnSO4
6. f. KClO
7. g. NaH2PO4
2. For each ionic compound, name the cation and the anion and give the charge on each ion.
1. a. Zn(NO3)2
2. b. CoS
3. c. BeCO3
4. d. Na2SO4
5. e. K2C2O4
6. f. NaCN
7. g. FeCl2
3. Write the formula for each compound.
1. a. magnesium carbonate
2. b. aluminum sulfate
3. c. potassium phosphate
4. d. lead(IV) oxide
5. e. silicon nitride
6. f. sodium hypochlorite
7. g. titanium(IV) chloride
8. h. disodium ammonium phosphate
4. Write the formula for each compound.
1. a. lead(II) nitrate
2. b. ammonium phosphate
3. c. silver sulfide
4. d. barium sulfate
5. e. cesium iodide
6. f. sodium bicarbonate
7. g. potassium dichromate
8. h. sodium hypochlorite
5. Write the formula for each compound.
1. a. zinc cyanide
2. b. silver chromate
3. c. lead(II) iodide
4. d. benzene
5. e. copper(II) perchlorate
6. Write the formula for each compound.
1. a. calcium fluoride
2. b. sodium nitrate
3. c. iron(III) oxide
4. d. copper(II) acetate
5. e. sodium nitrite
7. Write the formula for each compound.
1. a. sodium hydroxide
2. d. calcium cyanide
3. c. magnesium phosphate
4. d. sodium sulfate
5. e. nickel(II) bromide
6. f. calcium chlorite
7. g. titanium(IV) bromide
8. Write the formula for each compound.
1. a. sodium chlorite
2. b. potassium nitrite
3. c. sodium nitride (also called sodium azide)
4. d. calcium phosphide
5. e. tin(II) chloride
6. f. calcium hydrogen phosphate
7. g. iron(II) chloride dihydrate
9. Write the formula for each compound.
1. a. potassium carbonate
2. b. chromium(III) sulfite
3. c. cobalt(II) phosphate
4. d. magnesium hypochlorite
5. e. nickel(II) nitrate hexahydrate
Conceptual Problems
1. Name each acid.
1. a. HCl
2. b. HBrO3
3. c. HNO3
4. d. H2SO4
5. e. HIO3
2. Name each acid.
1. a. HBr
2. b. H2SO3
3. c. HClO3
4. d. HCN
5. e. H3PO4
3. Name the aqueous acid that corresponds to each gaseous species.
1. a. hydrogen bromide
2. b. hydrogen cyanide
3. c. hydrogen iodide
4. For each structural formula, write the condensed formula and the name of the compound.
a.
b.
5. For each structural formula, write the condensed formula and the name of the compound.
a.
b.
6. When each compound is added to water, is the resulting solution acidic, neutral, or basic?
1. a. CH3CH2OH
2. b. Mg(OH)2
3. c. C6H5CO2H
4. d. LiOH
5. e. C3H7CO2H
6. f. H2SO4
7. Draw the structure of the simplest example of each type of compound.
1. a. alkane
2. b. alkene
3. c. alkyne
4. d. aromatic hydrocarbon
5. e. alcohol
6. f. carboxylic acid
7. g. amine
8. h. cycloalkane
8. Identify the class of organic compound represented by each compound.
a.
b. CH3CH2OH
c. HC≡CH
d.
e. C3H7NH2
f. CH3CH=CHCH2CH3
g.
h.
9. Identify the class of organic compound represented by each compound.
a.
b.
c.
d.
e.
f. CH3C≡CH
g.
h.
Numerical Problems
1. Write the formula for each compound.
1. a. hypochlorous acid
2. b. perbromic acid
3. c. hydrobromic acid
4. d. sulfurous acid
5. e. sodium perbromate
2. Write the formula for each compound.
1. a. hydroiodic acid
2. b. hydrogen sulfide
3. c phosphorous acid
4. d. perchloric acid
5. e. calcium hypobromite
3. Name each compound.
1. a. HBr
2. b. H2SO3
3. c. HCN
4. d. HClO4
5. e. NaHSO4
4. Name each compound.
1. a. H2SO4
2. b. HNO2
3. c. K2HPO4
4. d. H3PO3
5. e. Ca(H2PO4)2·H2O | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.E%3A_Homework_Problems.txt |
Thumbnail: Copper from a wire is displaced by silver in a silver nitrate solution it is dipped into, and solid silver precipitates out. (CC BY-SA 3.0; Toby Hudson)
04: Chemical Reactions
$2H_2 + O_2 \rightarrow 2H_2O$
Chemical formulas and other symbols are used to indicate the starting materials, or reactants, which by convention are written on the left side of the equation, and the final compounds, or products, which are written on the right. An arrow points from the reactant to the products. The chemical reaction for the ammonium dichromate volcano in Figure $1$ is
$(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + N_2 + 4H_2O \label{4.1.1}$
$reactant \, \, \, \, \,\, \, \, \, \, \, \,\,\, products$
The arrow is read as “yields” or “reacts to form.” Equation 4.1.1 indicates that ammonium dichromate (the reactant) yields chromium(III) oxide, nitrogen, and water (the products). The equation for this reaction is even more informative when written as follows:
$(NH_4)_2Cr_2O_7(s) \rightarrow Cr_2O_{3\;(s)} + N_{2\;(g)} + 4H_2O_{(g)} \label{4.1.2}$
Equation 4.1.2 is identical to Equation 4.1.1 except for the addition of abbreviations in parentheses to indicate the physical state of each species. The abbreviations are (s) for solid, (l) for liquid, (g) for gas, and (aq) for an aqueous solution, a solution of the substance in water.
Consistent with the law of conservation of mass, the numbers of each type of atom are the same on both sides of Equations 4.1.1 and 4.1.2. Each side of the reaction has two chromium atoms, seven oxygen atoms, two nitrogen atoms, and eight hydrogen atoms.
In a balanced chemical equation, both the numbers of each type of atom and the total charge are the same on both sides. Equations 4.1.1 and 4.1.2 are balanced chemical equations. What is different on each side of the equation is how the atoms are arranged to make molecules or ions. A chemical reaction represents a change in the distribution of atoms, but not in the number of atoms. In this reaction, and in most chemical reactions, bonds are broken in the reactants (here, Cr–O and N–H bonds), and new bonds are formed to create the products (here, O–H and N≡N bonds). If the numbers of each type of atom are different on the two sides of a chemical equation, then the equation is unbalanced, and it cannot correctly describe what happens during the reaction. To proceed, the equation must first be balanced.
of two substances in a chemical reaction is the ratio of their coefficients in the balanced chemical equation.
Introduction to Chemical Reaction Equations: https://youtu.be/5mjawuf7K2Q
4.2: Chemical Equations and Stoichiometry
$C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \label{4.2.1}$
Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain does not run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room?
The initial step in solving a problem of this type is to write the balanced chemical equation for the reaction. Inspection shows that it is balanced as written, so the strategy outlined in , can be adapted as follows:
1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:
$moles \, glucose = 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose } = 0.251 \, mol \, glucose$
2. According to the balanced chemical equation, 6 mol of CO2 is produced per mole of glucose; the mole ratio of CO2 to glucose is therefore 6:1. The number of moles of CO2 produced is thus
$moles \, CO_2 = mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose }$
$= 0.251 \, mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose }$
$= 1.51 \, mol \, CO_2$
3. Use the molar mass of CO2 (44.010 g/mol) to calculate the mass of CO2 corresponding to 1.51 mol of CO2:
$mass\, of\, CO_2 = 1.51 \, mol \, CO_2 \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.5 \, g \, CO_2$
$45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose} \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose} \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.4 \, g \, CO_2$
Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (For more information about rounding and significant digits, see Essential Skills 1 in , .) This amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. Similar methods can be used to calculate the amount of oxygen consumed or the amount of water produced.
The balanced chemical equation was used to calculate the mass of product that is formed from a certain amount of reactant. It can also be used to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example 11, the mass of one reactant that is required to consume a given mass of another reactant.
Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): https://youtu.be/74mHV0CZcjw | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/04%3A_Chemical_Reactions/4.1%3A_Chemical_Reactions_and_Chemical_Equations.txt |
Learning Objectives
• To describe the concentrations of solutions quantitatively
Many people have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions. Chemists use many different methods to define concentrations, some of which are described in this section.
Molarity
The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) is a common unit of concentration and is defined as the number of moles of solute present in exactly 1 L of solution. It is, equivalently, the number of millimoles of solute present in exactly 1 mL of solution:
$molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \label{4.3.1}$
The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as $M$. An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the molar concentration of a solute. Therefore,
$[\rm{sucrose}] = 1.00\: M$
is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either
$V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \label{4.3.2}$
or
$V_{mL} M_{mmol/mL} = \cancel{mL} \left( \dfrac{mmol} {\cancel{mL}} \right) = mmoles \label{4.3.3}$
Figure 4.3.1 illustrates the use of Equation 4.3.2 and Equation 4.3.3.
Example 4.3.1
Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.
Given: identity of solute and volume and molarity of solution
Asked for: amount of solute in moles
Strategy:
Use either Equation 4.3.2 or Equation 4.3.3, depending on the units given in the problem.
Solution:
Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation 4.3.2 is more useful:
$moles\: NaOH = V_L M_{mol/L} = (2 .50\: \cancel{L} ) \left( \dfrac{0.100\: mol } {\cancel{L}} \right) = 0 .250\: mol\: NaOH$
Exercise 4.3.1
Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.
Answer: 41.6 mmol
Concentrations are often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in parts per million (ppm), which is grams of solute per 106 g of solution, or in parts per billion (ppb), which is grams of solute per 109 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter (Table 4.3.1).
Table 4.3.1: Common Units of Concentration
Concentration Units
m/m g of solute/g of solution
m/v g of solute/mL of solution
ppm g of solute/106 g of solution
μg/mL
ppb g of solute/109 g of solution
ng/mL
Calculations Involving Molarity (M): https://youtu.be/TVTCvKoSR-Q
The Preparation of Solutions
To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Figure 4.3.1 illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always less than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure 4.3.2, for some substances this effect can be significant, especially for concentrated solutions.
Example 4.3.2
The solution in Figure 4.3.1 contains 10.0 g of cobalt(II) chloride dihydrate, CoCl2•2H2O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl2•2H2O?
Given: mass of solute and volume of solution
Asked for: concentration (M)
Strategy:
To find the number of moles of CoCl2•2H2O, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters.
Solution:
The molar mass of CoCl2•2H2O is 165.87 g/mol. Therefore,
$moles\: CoCl_2 \cdot 2H_2O = \left( \dfrac{10.0 \: \cancel{g}} {165 .87\: \cancel{g} /mol} \right) = 0 .0603\: mol$
The volume of the solution in liters is
$volume = 500\: \cancel{mL} \left( \dfrac{1\: L} {1000\: \cancel{mL}} \right) = 0 .500\: L$
Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is
$molarity = \dfrac{0.0603\: mol} {0.500\: L} = 0.121\: M = CoCl_2 \cdot H_2O$
Exercise 4.3.2
The solution shown in Figure 4.3.2 contains 90.0 g of (NH4)2Cr2O7 in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate?
Answer: $(NH_4)_2Cr_2O_7 = 1.43\: M$
To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation 4.3.2. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example 4.3.3.
Example 4.3.3
The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol.
Given: molarity, volume, and molar mass of solute
Asked for: mass of solute
Strategy:
1. Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity.
2. Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass.
Solution:
A We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:
$V_L M_{mol/L} = moles$
$500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .155\: mol\: glucose$
B We then convert the number of moles of glucose to the required mass of glucose:
$mass \: of \: glucose = 0.155 \: \cancel{mol\: glucose} \left( \dfrac{180.16 \: g\: glucose} {1\: \cancel{mol\: glucose}} \right) = 27.9 \: g \: glucose$
Exercise 4.3.3
Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution.
Answer: 2.3 g NaCl
A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution is a commercially prepared solution of known concentration and is a commercially prepared solution of known concentration, is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.
The procedure for preparing a solution of known concentration from a stock solution is shown in Figure 4.3.3. It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore
$(V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)\label{4.3.4}$
where the subscripts s and d indicate the stock and dilute solutions, respectively. Example 5.5.4 demonstrates the calculations involved in diluting a concentrated stock solution.
Example 4.3.4
What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example 4.3.3?
Given: volume and molarity of dilute solution
Asked for: volume of stock solution
Strategy:
1. Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity.
2. To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution.
Solution:
A The D5W solution in Example 4.3.3 was 0.310 M glucose. We begin by using Equation 4.3.4 to calculate the number of moles of glucose contained in 2500 mL of the solution:
$moles\: glucose = 2500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .775\: mol\: glucose$
B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:
$volume\: of\: stock\: soln = 0 .775\: \cancel{mol\: glucose} \left( \dfrac{1\: L} {3 .00\: \cancel{mol\: glucose}} \right) = 0 .258\: L\: or\: 258\: mL$
In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; only the amount of solvent has changed. The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M).
We could also have solved this problem in a single step by solving Equation 4.3.4 for Vs and substituting the appropriate values:
$V_s = \dfrac{( V_d )(M_d )}{M_s} = \dfrac{(2 .500\: L)(0 .310\: \cancel{M} )} {3 .00\: \cancel{M}} = 0 .258\: L$
As we have noted, there is often more than one correct way to solve a problem.
Exercise 4.3.4
What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)?
Answer: 16 mL
Calculating Moles from Volume
Quantitative calculations involving reactions in solution are carried out with masses, however, volumes of solutions of known concentration are used to determine the number of moles of reactants. Whether dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation give the number of moles of each reactant needed and the number of moles of each product that can be produced.
The flowchart for stoichiometric calculations illustrated in Figure 4.3.4 shows that in the balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used to determine the amounts of other species.
Note
The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations.
Example 4.3.5: Extraction of Gold
Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN)2] ion. Gold is then recovered by reduction with metallic zinc according to the following equation:
$Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s)$
What mass of gold can be recovered from 400.0 L of a 3.30 × 10−4 M solution of [Au(CN)2]?
Given: chemical equation and molarity and volume of reactant
Asked for: mass of product
Strategy:
1. Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN)2] present by multiplying the volume of the solution by its concentration.
2. From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.
Solution:
A The equation is balanced as written; proceed to the stoichiometric calculation. Figure 4.3.2 is adapted for this particular problem as follows:
As indicated in the strategy, start by calculating the number of moles of [Au(CN)2] present in the solution from the volume and concentration of the [Au(CN)2] solution:
\begin{align} moles\: [Au(CN)_2 ]^- & = V_L M_{mol/L} \ & = 400 .0\: \cancel{L} \left( \dfrac{3 .30 \times 10^{4-}\: mol\: [Au(CN)_2 ]^-} {1\: \cancel{L}} \right) = 0 .132\: mol\: [Au(CN)_2 ]^- \end{align}
B Because the coefficients of gold and the [Au(CN)2] ion are the same in the balanced chemical equation, assuming that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN)2] (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so the number of moles of gold must be converted to the corresponding mass using the molar mass of gold:
\begin{align} mass\: of\: Au &= (moles\: Au)(molar\: mass\: Au) \ &= 0 .132\: \cancel{mol\: Au} \left( \dfrac{196 .97\: g\: Au} {1\: \cancel{mol\: Au}} \right) = 26 .0\: g\: Au \end{align}
At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth$1170.
$26 .0\: \cancel{g\: Au} \times \dfrac{1\: \cancel{troy\: oz}} {31 .10\: \cancel{g}} \times \dfrac{\1400} {1\: \cancel{troy\: oz\: Au}} = \1170$
Exercise 4.3.5: Lanthanum Oxalate
What mass of solid lanthanum(III) oxalate nonahydrate [La2(C2O4)3•9H2O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl3 by adding a stoichiometric amount of sodium oxalate?
Answer: 3.89 g
Ion Concentrations in Solution
In Example 4.3.2, the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL were calculated to be 1.43 M. Let’s consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH4+ ions and one Cr2O72 ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH4+ and Cr2O72 ions:
$(NH_4 )_2 Cr_2 O_7 (s) \xrightarrow {H_2 O(l)} 2NH_4^+ (aq) + Cr_2 O_7^{2-} (aq)\label{4.3.5}$
Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr2O72 anions and 2 mol of NH4+ cations (see Figure 4.3.4).
When carrying out a chemical reaction using a solution of a salt such as ammonium dichromate, it is important to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH4)2Cr2O7, then the concentration of Cr2O72 must also be 1.43 M because there is one Cr2O72 ion per formula unit. However, there are two NH4+ ions per formula unit, so the concentration of NH4+ ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH4)2Cr2O7 produces three ions when dissolved in water (2NH4+ + 1Cr2O72), the total concentration of ions in the solution is 3 × 1.43 M = 4.29 M.
Example 4.3.6
What are the concentrations of all species derived from the solutes in these aqueous solutions?
1. 0.21 M NaOH
2. 3.7 M (CH3)CHOH
3. 0.032 M In(NO3)3
Given: molarity
Asked for: concentrations
Strategy:
A Classify each compound as either a strong electrolyte or a nonelectrolyte.
B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.
Solution:
1. Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution: $NaOH(s) \xrightarrow {H_2 O(l)} Na^+ (aq) + OH^- (aq)$
B Because each formula unit of NaOH produces one Na+ ion and one OH ion, the concentration of each ion is the same as the concentration of NaOH: [Na+] = 0.21 M and [OH] = 0.21 M.
2. A The formula (CH3)2CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 4.1 that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.
B The only solute species in solution is therefore (CH3)2CHOH molecules, so [(CH3)2CHOH] = 3.7 M.
3. A Indium nitrate is an ionic compound that contains In3+ ions and NO3 ions, so we expect it to behave like a strong electrolyte in aqueous solution:
$In(NO _3 ) _3 (s) \xrightarrow {H_ 2 O(l)} In ^{3+} (aq) + 3NO _3^- (aq)$
B One formula unit of In(NO3)3 produces one In3+ ion and three NO3 ions, so a 0.032 M In(NO3)3 solution contains 0.032 M In3+ and 3 × 0.032 M = 0.096 M NO3—that is, [In3+] = 0.032 M and [NO3] = 0.096 M.
Exercise 4.3.6
What are the concentrations of all species derived from the solutes in these aqueous solutions?
1. 0.0012 M Ba(OH)2
2. 0.17 M Na2SO4
3. 0.50 M (CH3)2CO, commonly known as acetone
Answer
1. $[Ba^{2+}] = 0.0012\: M; \: [OH^-] = 0.0024\: M$
2. $[Na^+] = 0.34\: M; \: [SO_4^{2-}] = 0.17\: M$
3. $[(CH_3)_2CO] = 0.50\: M$
Concentration of Ions in Solution from a Soluble Salt: https://youtu.be/qsekSJBLemc
Summary
• Solution concentrations are typically expressed as molarities and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution.
• definition of molarity: $molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution}$
• relationship among volume, molarity, and moles: $V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles$
• relationship between volume and concentration of stock and dilute solutions: $(V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)$
The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/04%3A_Chemical_Reactions/4.3%3A_Chemical_Reactions_in_Solution.txt |
Learning Objectives
• To understand the concept of limiting reactants and quantify incomplete reactions
In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant. The reactant that remains after a reaction has gone to completion is in excess.
Consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus
$1 \,\text{box mix} + 2 \,\text{eggs} \rightarrow 1 \, \text{batch brownies} \label{4.4.1}$
If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant. Even if you had a refrigerator full of eggs, you could make only two batches of brownies.
Now consider a chemical example of a limiting reactant: the production of pure titanium. This metal is fairly light (45% lighter than steel and only 60% heavier than aluminum) and has great mechanical strength (as strong as steel and twice as strong as aluminum). Because it is also highly resistant to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. Titanium is also used in medical implants and portable computer housings because it is light and resistant to corrosion. Although titanium is the ninth most common element in Earth’s crust, it is relatively difficult to extract from its ores. In the first step of the extraction process, titanium-containing oxide minerals react with solid carbon and chlorine gas to form titanium tetrachloride (TiCl4) and carbon dioxide. Titanium tetrachloride is then converted to metallic titanium by reaction with magnesium metal at high temperature:
$TiCl_4 (g) + 2 \, Mg (l) \rightarrow Ti (s) + 2 \, MgCl_2 (l) \label{4.4.2}$
Because titanium ores, carbon, and chlorine are all rather inexpensive, the high price of titanium (about \$100 per kilogram) is largely due to the high cost of magnesium metal. Under these circumstances, magnesium metal is the limiting reactant in the production of metallic titanium.
With 1.00 kg of titanium tetrachloride and 200 g of magnesium metal, how much titanium metal can be produced according to the equation above? Solving this type of problem requires that you carry out the following steps:
1. Determine the number of moles of each reactant.
2. Compare the mole ratio of the reactants with the ratio in the balanced chemical equation to determine which reactant is limiting.
3. Calculate the number of moles of product that can be obtained from the limiting reactant.
4. Convert the number of moles of product to mass of product.
1. To determine the number of moles of reactants present, calculate or look up their molar masses: 189.679 g/mol for titanium tetrachloride and 24.305 g/mol for magnesium. The number of moles of each is calculated as follows:
$moles \, TiCl_4 = {mass \, TiCl_4 \over molar \, mass \, TiCl_4}$
$= 1000 \, g \, TiCl_4 \times {1 \, mol \, TiCl_4 \over 189.679 \, g \, TiCl_4} = 5.272 \, mol \, TiCl_4$
$moles \, Mg = {mass \, Mg \over molar \, mass \, Mg}$
$= 200 \, g \, Mg \times {1 \, mol \, Mg \over 24.305 \, g \, Mg } = 8.23 \, mol \, Mg$
2. There are more moles of magnesium than of titanium tetrachloride, but the ratio is only the following:
${mol \, Mg \over mol \, TiCl_4} = {8.23 \, mol \over 5.272 \, mol } = 1.56$
Because the ratio of the coefficients in the balanced chemical equation is,
${ 2 \, mol \, Mg \over 1 \, mol \, TiCl_4} = 2$
there is not have enough magnesium to react with all the titanium tetrachloride. If this point is not clear from the mole ratio, calculate the number of moles of one reactant that is required for complete reaction of the other reactant. For example, there are 8.23 mol of Mg, so (8.23 ÷ 2) = 4.12 mol of TiCl4 are required for complete reaction. Because there are 5.272 mol of TiCl4, titanium tetrachloride is present in excess. Conversely, 5.272 mol of TiCl4 requires 2 × 5.272 = 10.54 mol of Mg, but there are only 8.23 mol. Therefore, magnesium is the limiting reactant.
3. Because magnesium is the limiting reactant, the number of moles of magnesium determines the number of moles of titanium that can be formed:
$moles \, Ti = 8.23 \, mol \, Mg = {1 \, mol \, Ti \over 2 \, mol \, Mg} = 4.12 \, mol \, Ti$
Thus only 4.12 mol of Ti can be formed.
4. To calculate the mass of titanium metal that can obtain, multiply the number of moles of titanium by the molar mass of titanium (47.867 g/mol):
$moles \, Ti = mass \, Ti \times molar \, mass \, Ti = 4.12 \, mol \, Ti \times {47.867 \, g \, Ti \over 1 \, mol \, Ti} = 197 \, g \, Ti$
Here is a simple and reliable way to identify the limiting reactant in any problem of this sort:
1. Calculate the number of moles of each reactant present: 5.272 mol of TiCl4 and 8.23 mol of Mg.
2. Divide the actual number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation: $TiCl_4 : { 5.272 \, mol \, (actual) \over 1 \, mol \, (stoich)} = 5.272 \, \, \, \, Mg: {8.23 \, mol \, (actual) \over 2 \, mol \, (stoich)} = 4.12$
3. The reactant with the smallest mole ratio is limiting. Magnesium, with a calculated stoichiometric mole ratio of 4.12, is the limiting reactant.
Density is the mass per unit volume of a substance. If we are given the density of a substance, we can use it in stoichiometric calculations involving liquid reactants and/or products, as Example $1$ demonstrates.
Determining the Limiting Reactant and Theoretical Yield for a Reaction: https://youtu.be/HmDm1qpNUD0
Example $1$: Fingernail Polish Remover
Ethyl acetate (CH3CO2C2H5) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethanol (C2H5OH) with acetic acid (CH3CO2H); the other product is water. A small amount of sulfuric acid is used to accelerate the reaction, but the sulfuric acid is not consumed and does not appear in the balanced chemical equation. Given 10.0 mL each of acetic acid and ethanol, how many grams of ethyl acetate can be prepared from this reaction? The densities of acetic acid and ethanol are 1.0492 g/mL and 0.7893 g/mL, respectively.
Given: reactants, products, and volumes and densities of reactants
Asked for: mass of product
Strategy:
1. Balance the chemical equation for the reaction.
2. Use the given densities to convert from volume to mass. Then use each molar mass to convert from mass to moles.
3. Using mole ratios, determine which substance is the limiting reactant. After identifying the limiting reactant, use mole ratios based on the number of moles of limiting reactant to determine the number of moles of product.
4. Convert from moles of product to mass of product.
Solution:
A Always begin by writing the balanced chemical equation for the reaction:
$C_2H_5OH (l) + CH_3CO_2H (aq) \rightarrow CH_3CO_2C_2H_5 (aq) + H_2O (l)$
B We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. Recall from that the density of a substance is the mass divided by the volume:
$density = {mass \over volume }$
Rearranging this expression gives mass = (density)(volume). We can replace mass by the product of the density and the volume to calculate the number of moles of each substance in 10.0 mL (remember, 1 mL = 1 cm3):
$moles \, C_2H_5OH = { mass \, C_2H_5OH \over molar \, mass \, C_2H_5OH }$
$= {volume \, C_2H_5OH \times density \, C_2H_5OH \over molar \, mass \, C_2H_5OH}$
$= 10.0 \, ml \, C_2H_5OH \times {0.7893 \, g \, C_2H_5OH \over 1 \, ml \, C_2H_5OH} \times {1 \, mole \, C_2H_5OH \over 46.07 \, g\, C_2H_5OH}$
$= 0.171 \, mol \, C_2H_5OH$
$moles \, CH_3CO_2H = {mass \, CH_3CO_2H \over molar \, mass \, CH_3CO_2H}$
$= {volume \, CH_3CO_2H \times density \, CH_3CO_2H \over molar \, mass \, CH_3CO_2H}$
$= 10.0 \, ml \, CH_3CO_2H \times {1.0492 \, g \, CH_3CO_2H \over 1 \, ml \, CH_3CO_2H} \times {1 \, mol \, CH_3CO_2H \over 60.05 \, g \, CH_3CO_2H }$
$= 0.175 \, mol \, CH_3CO_2H$
C The number of moles of acetic acid exceeds the number of moles of ethanol. Because the reactants both have coefficients of 1 in the balanced chemical equation, the mole ratio is 1:1. We have 0.171 mol of ethanol and 0.175 mol of acetic acid, so ethanol is the limiting reactant and acetic acid is in excess. The coefficient in the balanced chemical equation for the product (ethyl acetate) is also 1, so the mole ratio of ethanol and ethyl acetate is also 1:1. This means that given 0.171 mol of ethanol, the amount of ethyl acetate produced must also be 0.171 mol:
$moles \, ethyl \, acetate = molethanol \times {1 \, mol \, ethyl \, acetate \over 1 \, mol \, ethanol }$
$= 0.171 \, mol \, C_2H_5OH \times {1 \, mol \, CH_3CO_2C_2H_5 \over 1 \, mol \, C_2H_5OH}$
$= 0.171 \, mol \, CH_3CO_2C_2H_5$
D The final step is to determine the mass of ethyl acetate that can be formed, which we do by multiplying the number of moles by the molar mass:
$mass \, of \, ethyl \, acetate = moleethyl \, acetate \times molar \, mass \, ethyl \, acetate$
$= 0.171 \, mol \, CH_3CO_2C_2H_5 \times {88.11 \, g \, CH_3CO_2C_2H_5 \over 1 \, mol \, CH_3CO_2C_2H_5}$
$= 15.1 \, g \, CH_3CO_2C_2H_5$
Thus 15.1 g of ethyl acetate can be prepared in this reaction. If necessary, you could use the density of ethyl acetate (0.9003 g/cm3) to determine the volume of ethyl acetate that could be produced:
$volume \, of \, ethyl \, acetate = 15.1 \, g \, CH_3CO_2C_2H_5 \times { 1 \, ml \, CH_3CO_2C_2H_5 \over 0.9003 \, g\, CH_3CO_2C_2H_5}$
$= 16.8 \, ml \, CH_3CO_2C_2H_5$
Exercise $1$
Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound P4S10. How much P4S10 can be prepared starting with 10.0 g of P4 and 30.0 g of S8?
Answer: 35.9 g
Limiting Reactants in Solutions
The concept of limiting reactants applies to reactions carried out in solution as well as to reactions involving pure substances. If all the reactants but one are present in excess, then the amount of the limiting reactant may be calculated as illustrated in Example $2$.
Example
Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion:
$3CH_3 CH_2 OH(aq) + \underset{yellow-orange}{2Cr_2 O_7^{2 -}}(aq) + 16H ^+ (aq) \underset{H_2 SO_4 (aq)}{\xrightarrow{\hspace{10px} Ag ^+\hspace{10px}} } 3CH_3 CO_2 H(aq) + \underset{green}{4Cr^{3+}} (aq) + 11H_2 O(l)$
When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the Cr2O72 ions are reduced from Cr6+ to Cr3+. In the presence of Ag+ ions that act as a catalyst, the reaction is complete in less than a minute. Because the Cr2O72 ion (the reactant) is yellow-orange and the Cr3+ ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol.
A Breathalyzer reaction with a test tube before (a) and after (b) ethanol is added. When a measured volume of a suspect’s breath is bubbled through the solution, the ethanol is oxidized to acetic acid, and the solution changes color from yellow-orange to green. The intensity of the green color indicates the amount of ethanol in the sample.
A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K2Cr2O7 in 50% H2SO4 as well as a fixed concentration of AgNO3 (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr6+ to Cr3+?
Given: volume and concentration of one reactant
Asked for: mass of other reactant needed for complete reaction
Strategy:
1. Calculate the number of moles of Cr2O72 ion in 1 mL of the Breathalyzer solution by dividing the mass of K2Cr2O7 by its molar mass.
2. Find the total number of moles of Cr2O72 ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL).
3. Use the mole ratios from the balanced chemical equation to calculate the number of moles of C2H5OH needed to react completely with the number of moles of Cr2O72 ions present. Then find the mass of C2H5OH needed by multiplying the number of moles of C2H5OH by its molar mass.
Solution:
A In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K2Cr2O7 in 1 mL of solution, which can be used to calculate the number of moles of K2Cr2O7 contained in 1 mL:
$\dfrac{moles\: K_2 Cr_2 O_7} {1\: mL} = \dfrac{(0 .25\: \cancel{mg}\: K_2 Cr_2 O_7 )} {mL} \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg}} \right) \left( \dfrac{1\: mol} {294 .18\: \cancel{g}\: K_2 Cr_2 O_7} \right) = 8.5 \times 10 ^{-7}\: moles$
B Because 1 mol of K2Cr2O7 produces 1 mol of Cr2O72 when it dissolves, each milliliter of solution contains 8.5 × 10−7 mol of Cr2O72. The total number of moles of Cr2O72 in a 3.0 mL Breathalyzer ampul is thus
$moles\: Cr_2 O_7^{2-} = \left( \dfrac{8 .5 \times 10^{-7}\: mol} {1\: \cancel{mL}} \right) ( 3 .0\: \cancel{mL} ) = 2 .6 \times 10^{-6}\: mol\: Cr_2 O_7^{2–}$
C The balanced chemical equation tells us that 3 mol of C2H5OH is needed to consume 2 mol of Cr2O72 ion, so the total number of moles of C2H5OH required for complete reaction is
$moles\: of\: C_2 H_5 OH = ( 2.6 \times 10 ^{-6}\: \cancel{mol\: Cr_2 O_7 ^{2-}} ) \left( \dfrac{3\: mol\: C_2 H_5 OH} {2\: \cancel{mol\: Cr _2 O _7 ^{2 -}}} \right) = 3 .9 \times 10 ^{-6}\: mol\: C _2 H _5 OH$
As indicated in the strategy, this number can be converted to the mass of C2H5OH using its molar mass:
$mass\: C _2 H _5 OH = ( 3 .9 \times 10 ^{-6}\: \cancel{mol\: C _2 H _5 OH} ) \left( \dfrac{46 .07\: g} {\cancel{mol\: C _2 H _5 OH}} \right) = 1 .8 \times 10 ^{-4}\: g\: C _2 H _5 OH$
Thus 1.8 × 10−4 g or 0.18 mg of C2H5OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal.
Exercise $2$
The compound para-nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction
Because the amount of para-nitrophenol is easily estimated from the intensity of the yellow color that results when excess NaOH is added, reactions that produce para-nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10−4 g of para-nitrophenol to ensure that formation of the yellow anion is complete?
Answer: 4.93 × 10−5 L or 49.3 μL
In Examples 4.4.1 and 4.4.2, the identities of the limiting reactants are apparent: [Au(CN)2], LaCl3, ethanol, and para-nitrophenol. When the limiting reactant is not apparent, it can be determined by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation. The only difference is that the volumes and concentrations of solutions of reactants, rather than the masses of reactants, are used to calculate the number of moles of reactants, as illustrated in Example $3$.
Example $3$
When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows:
$2AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2KNO_3(aq)$
What mass of Ag2Cr2O7 is formed when 500 mL of 0.17 M K2Cr2O7 are mixed with 250 mL of 0.57 M AgNO3?
Given: balanced chemical equation and volume and concentration of each reactant
Asked for: mass of product
Strategy:
1. Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity.
2. Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation.
3. Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product.
Solution:
A The balanced chemical equation tells us that 2 mol of AgNO3(aq) reacts with 1 mol of K2Cr2O7(aq) to form 1 mol of Ag2Cr2O7(s) (Figure $2$). The first step is to calculate the number of moles of each reactant in the specified volumes:
$moles\: K_2 Cr_2 O_7 = 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .17\: mol\: K_2 Cr_2 O_7} {1\: \cancel{L}} \right) = 0 .085\: mol\: K_2 Cr_2 O_7$
$moles\: AgNO_3 = 250\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .57\: mol\: AgNO_3} {1\: \cancel{L}} \right) = 0 .14\: mol\: AgNO_3$
B Now determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient:
$K_2 Cr_2 O_7: \: \dfrac{0 .085\: mol} {1\: mol} = 0 .085$
$AgNO_3: \: \dfrac{0 .14\: mol} {2\: mol} = 0 .070$
Because 0.070 < 0.085, we know that AgNO3 is the limiting reactant.
C Each mole of Ag2Cr2O7 formed requires 2 mol of the limiting reactant (AgNO3), so we can obtain only 0.14/2 = 0.070 mol of Ag2Cr2O7. Finally, convert the number of moles of Ag2Cr2O7 to the corresponding mass:
$mass\: of\: Ag_2 Cr_2 O_7 = 0 .070\: \cancel{mol} \left( \dfrac{431 .72\: g} {1 \: \cancel{mol}} \right) = 30\: g \: Ag_2 Cr_2 O_7$
The Ag+ and Cr2O72 ions form a red precipitate of solid Ag2Cr2O7, while the K+ and NO3 ions remain in solution. (Water molecules are omitted from molecular views of the solutions for clarity.)
Exercise $3$
Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation:
$2NaHCO_3(aq) + H_2SO_4(aq) \rightarrow 2CO_2(g) + Na_2SO_4(aq) + 2H_2O(l)$
If 13.0 mL of 3.0 M H2SO4 are added to 732 mL of 0.112 M NaHCO3, what mass of CO2 is produced?
Answer: 3.4 g
Limiting Reactant Problems Using Molarities: https://youtu.be/eOXTliL-gNw
Summary
• The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained.
The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a quantity of a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation. The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/04%3A_Chemical_Reactions/4.4%3A_Determining_the_Limiting_Reactant.txt |
Learning Objectives
• To understand the concept of quantify incomplete reactions, including sequential and parallel reactions.
Theoretical Yields
When reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yield, the amount obtained if the reaction occurred perfectly and the purification method were 100% efficient.
In reality, less product is always obtained than is theoretically possible because of mechanical losses (such as spilling), separation procedures that are not 100% efficient, competing reactions that form undesired products, and reactions that simply do not run to completion, resulting in a mixture of products and reactants; this last possibility is a common occurrence. Therefore, the actual yield, the measured mass of products obtained from a reaction, is almost always less than the theoretical yield (often much less). The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, multiplied by 100 to give a percentage:
$percent \, yield = {actual \, yield \, (g) \over theoretical \, yield \, (g) } \times 100 \tag{4.5.1}$
The method used to calculate the percent yield of a reaction is illustrated in Example $1$.
Example $1$: Novocain
Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H2SO4 (indicated above the arrow) by the reaction
$\underset {p-amino benzoic\,acid}{C_7H_7NO_2} + \underset {2-diethylaminoethanol}{C_6H_{15}NO}\,\,\underrightarrow {H_2SO_4} \,\, \underset {procaine}{C_{13}H_{20}N_2O_2} + H_2O$
If this reaction was carried out using 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and 15.7 g of procaine were isolated, what is the percent yield?
The preparation of procaine. A reaction of p-aminobenzoic acid with 2-diethylaminoethanol yields procaine and water.
Given: masses of reactants and product
Asked for: percent yield
Strategy:
1. Write the balanced chemical equation.
2. Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield.
3. Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.
Solution:
A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water.
B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C7H7NO2), 137.14 g/mol; 2-diethylaminoethanol (C6H15NO), 117.19 g/mol. Thus the reaction used the following numbers of moles of reactants:
$moles \, p-aminobenzoic \, acid = 10.0 \, g \, \times \, {1 \, mol \over 137.14 \, g } = 0.0729 \, mol \, p-aminbenzoic\, acid$
$moles \, 2-diethylaminoethanol = 10.0 \, g \times {1 \, mol \over 117.19 \, g} = 0.0853 \, mol \, 2-diethylaminoethanol$
The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine. We can therefore obtain only a maximum of 0.0729 mol of procaine. To calculate the corresponding mass of procaine, we use its structural formula (C13H20N2O2) to calculate its molar mass, which is 236.31 g/mol.
$theoretical \, yield \, of \, procaine = 0.0729 \, mol \times {236.31 \, g \over 1 \, mol } = 17.2 \, g$
C The actual yield was only 15.7 g of procaine, so the percent yield is
$percent \, yield = {15.7 \, g \over 17.2 \, g } \times 100 = 91.3 \%$
(If the product were pure and dry, this yield would indicate very good lab technique!)
Exercise $1$: Extraction of Lead
Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena (PbS), which is easily converted to lead oxide (PbO) in 100% yield by roasting in air via the following reaction:
$2PbS_{(s)} + 3O_2 \rightarrow 2PbO_{ (s)} + 2SO_{2 (g)}$
The resulting PbO is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows:
$PbO_{(s)} + C_{(s)} \rightarrow Pb_{(l)} + CO_{(g)}$
If 93.3 kg of PbO is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield?
Answer: 89.2%
Percent yield can range from 0% to 100%. In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments. A 100% yield means that everything worked perfectly, and the chemist obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows the unlikelihood of a 100% yield. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible.
Consecutive Reactions, Simultaneous Reactions, and Overall Reactions
Most of time it takes more than one steps to get desired product. Two more complex reactions are the consecutive and simultaneous reactions. Reactions that are carried out one after another in sequence to yield a final product are called consecutive reactions (e.g., occurring sequenceally):
$A \rightarrow Y \rightarrow Z \tag{4.5.2}$
Any substance that is produced in one step and is consumed in another step of a multistep process is called an intermediate (e.g., $Y$ in Equation 4.5.2). The Overall Reaction is the chemical equation that expresses all the reactions occurring in a single overall equation.
Example $2$:Purification of $TiO_2$
Step 1:
$2TiO2_{(\text{impure solid})} + 3C_{(s)} +4Cl_{2(g)} \rightarrow 2TiCl_{4(g)} + CO_{2(g)} + 2CO_{(g)}$
Step 2:
$2\times [ TiCl_{4(g)} + O_{2(g)} \rightarrow TiO_{2(s)} +2Cl_{2(g)}]$
Overall:
$3C_{(s)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2CO_{(g)}$
$TiCl_4$ is an intermediate in this reaction
Example $3$
How many gram of $NaBr$ is produced if 55.85g of $Fe$ is consumed the the following series of reactions?
$3Fe + 3Br_2 \rightarrow 3FeBr_2$
$3FeBr_2 + Br_2 \rightarrow Fe_3Br_8$
$Fe_3Br_8 + 4Na_2CO_3 \rightarrow 8NaBr + 4CO_2 + Fe_3O_4$
Overall reaction:
$3Fe + 4Br_2 + 4Na_2CO_3 \rightarrow 8NaBr + 4CO_2 + Fe_3O_4$
Solution
under construction
In simultaneous reactions, two or more substances react independently of one another in separate reactions occurring simultaneously
$A \rightarrow Y \tag{4.5.3a}$
and
$B \rightarrow B \tag{4.5.3b}$
Equations 4.5.3 involve two reactions evolving in parallel, however sometimes there exist a coupling between the two (or more) reactions with shared reactant, intermediates or products. is competition involved, like in the scheme:
$A + B \rightarrow Y \tag{4.5.4a}$
and
$A + C \rightarrow Z \tag{4.5.3b}$
In the above simultaneous reaction, both $B$ and $C$ compete with each another for reacting $A$. This is superficially similar to a common ion effect.
Example $4$: Oxidation of Magnalium
Magnalium is an aluminum alloy with magnesium that exhibits greater strength, greater corrosion resistance, and lower density than pure aluminum. How many grams of $H_{2(g)}$ are generated if a 0.710 g piece of magnalium (with a 70% Al and 30.0% Mg by mass composition) with excess $HCl_{(aq)}$?
Solution
under construction
Summary
• The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained.
The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a quantity of a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation. The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/04%3A_Chemical_Reactions/4.5%3A_Other_Practical_Matters_in_Reaction_Stoichiometry.txt |
Learning Objectives
• To understand how and why solutions form
The solvent in aqueous solutions is water, which makes up about 70% of the mass of the human body and is essential for life. Many of the chemical reactions that keep us alive depend on the interaction of water molecules with dissolved compounds. Moreover, the presence of large amounts of water on Earth’s surface helps maintain its surface temperature in a range suitable for life. In this section, we describe some of the interactions of water with various substances and introduce you to the characteristics of aqueous solutions.
Polar Substances
As shown in Figure $1$, the individual water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure. As is typical of group 16 elements, the oxygen atom in each O–H covalent bond attracts electrons more strongly than the hydrogen atom does. Consequently, the oxygen and hydrogen nuclei do not equally share electrons. Instead, hydrogen atoms are electron poor compared with a neutral hydrogen atom and have a partial positive charge, which is indicated by δ+. The oxygen atom, in contrast, is more electron rich than a neutral oxygen atom, so it has a partial negative charge. This charge must be twice as large as the partial positive charge on each hydrogen for the molecule to have a net charge of zero. Thus its charge is indicated by 2δ. This unequal distribution of charge creates a polar bond in which one portion of the molecule carries a partial negative charge, while the other portion carries a partial positive charge (Figure $1$). Because of the arrangement of polar bonds in a water molecule, water is described as a polar substance.
Because of the asymmetric charge distribution in the water molecule, adjacent water molecules are held together by attractive electrostatic (δ+…δ) interactions between the partially negatively charged oxygen atom of one molecule and the partially positively charged hydrogen atoms of adjacent molecules (Figure $2$). Energy is needed to overcome these electrostatic attractions. In fact, without them, water would evaporate at a much lower temperature, and neither Earth’s oceans nor we would exist!
As you learned previously,, ionic compounds such as sodium chloride (NaCl) are also held together by electrostatic interactions—in this case, between oppositely charged ions in the highly ordered solid, where each ion is surrounded by ions of the opposite charge in a fixed arrangement. In contrast to an ionic solid, the structure of liquid water is not completely ordered because the interactions between molecules in a liquid are constantly breaking and reforming.
The unequal charge distribution in polar liquids such as water makes them good solvents for ionic compounds. When an ionic solid dissolves in water, the ions dissociate. That is, the partially negatively charged oxygen atoms of the H2O molecules surround the cations (Na+ in the case of NaCl), and the partially positively charged hydrogen atoms in H2O surround the anions (Cl; Figure $3$). Individual cations and anions that are each surrounded by their own shell of water molecules are called hydrated ions. We can describe the dissolution of NaCl in water as
$NaCl(s) \xrightarrow{H_2O(l)} Na^+ (aq) + Cl^- (aq) \label{5.1.1}$
where (aq) indicates that Na+ and Cl are hydrated ions.
Note
Polar liquids are good solvents for ionic compounds.
Electrolytes
When electricity, in the form of an electrical potential, is applied to a solution, ions in solution migrate toward the oppositely charged rod or plate to complete an electrical circuit, whereas neutral molecules in solution do not (Figure $4$). Thus solutions that contain ions conduct electricity, while solutions that contain only neutral molecules do not. Electrical current will flow through the circuit shown in Figure $4$ and the bulb will glow only if ions are present. The lower the concentration of ions in solution, the weaker the current and the dimmer the glow. Pure water, for example, contains only very low concentrations of ions, so it is a poor electrical conductor.
Note
Solutions that contain ions conduct electricity.
An electrolyte is any compound that can form ions when dissolved in water (c.f. nonelectrolytes). Electrolytes may be strong or weak. is any compound that can form ions when it dissolves in water. When strong electrolytes dissolve, the constituent ions dissociate completely due to strong electrostatic interactions with the solvent, producing aqueous solutions that conduct electricity very well (Figure $4$). Examples include ionic compounds such as barium chloride ($BaCl_2$) and sodium hydroxide (NaOH), which are both strong electrolytes and dissociate as follows:
$BaCl_2 (s) \xrightarrow{H_2O(l)} Ba^{2+} (aq) + 2Cl^- (aq) \label{5.1.2}$
$NaOH(s) \xrightarrow{H_2O(l)} Na^+ (aq) + OH^- (aq) \label{5.1.3}$
The single arrows from reactant to products in Equation 5.1.2 and Equation 5.1.3 indicate that dissociation is complete.
When weak electrolytes dissolve, they produce relatively few ions in solution. This does not mean that the compounds do not dissolve readily in water; many weak electrolytes contain polar bonds and are therefore very soluble in a polar solvent such as water. They do not completely dissociate to form ions, however, because of their weaker electrostatic interactions with the solvent. Because very few of the dissolved particles are ions, aqueous solutions of weak electrolytes do not conduct electricity as well as solutions of strong electrolytes. One such compound is acetic acid (CH3CO2H), which contains the –CO2H unit. Although it is soluble in water, it is a weak acid and therefore also a weak electrolyte. Similarly, ammonia (NH3) is a weak base and therefore a weak electrolyte. The behavior of weak acids and weak bases will be described in more detail elsewhere.
General structure of an aldehyde and a ketone. Notice that both contain the C=O group.
Nonelectrolytes (a substance that dissolves in water to form neutral molecules and has essentially no effect on electrical conductivity) that dissolve in water do so as neutral molecules and thus have essentially no effect on conductivity. Examples of nonelectrolytes that are very soluble in water but that are essentially nonconductive are ethanol, ethylene glycol, glucose, and sucrose, all of which contain the –OH group that is characteristic of alcohols. The topic of why alcohols and carboxylic acids behave differently in aqueous solution is for a different Module; for now, however, you can simply look for the presence of the –OH and –CO2H groups when trying to predict whether a substance is a strong electrolyte, a weak electrolyte, or a nonelectrolyte. The distinctions between soluble and insoluble substances and between strong, weak, and nonelectrolytes are illustrated in Figure $5$.
Note
Ionic substances and carboxylic acids are electrolytes; alcohols, aldehydes, and ketones are nonelectrolytes.
Example $1$
Predict whether each compound is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in water.
1. formaldehyde
2. cesium chloride
Given: compound
Asked for: relative ability to form ions in water
Strategy:
A Classify the compound as ionic or covalent.
B If the compound is ionic and dissolves, it is a strong electrolyte that will dissociate in water completely to produce a solution that conducts electricity well. If the compound is covalent and organic, determine whether it contains the carboxylic acid group. If the compound contains this group, it is a weak electrolyte. If not, it is a nonelectrolyte.
Solution:
1. A Formaldehyde is an organic compound, so it is covalent. B It contains an aldehyde group, not a carboxylic acid group, so it should be a nonelectrolyte.
2. A Cesium chloride (CsCl) is an ionic compound that consists of Cs+ and Cl ions. B Like virtually all other ionic compounds that are soluble in water, cesium chloride will dissociate completely into Cs+(aq) and Cl(aq) ions. Hence it should be a strong electrolyte.
Exercise $1$
Predict whether each compound is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in water.
1. (CH3)2CHOH (2-propanol)
2. ammonium sulfate
Answer
1. nonelectrolyte
2. strong electrolyte
Predicting the Solubility of Ionic Compounds: https://youtu.be/U3QNwnfmvGU
Summary
• Aqueous solutions can be classified as polar or nonpolar depending on how well they conduct electricity.
Most chemical reactions are carried out in solutions, which are homogeneous mixtures of two or more substances. In a solution, a solute (the substance present in the lesser amount) is dispersed in a solvent (the substance present in the greater amount). Aqueous solutions contain water as the solvent, whereas nonaqueous solutions have solvents other than water.
Polar substances, such as water, contain asymmetric arrangements of polar bonds, in which electrons are shared unequally between bonded atoms. Polar substances and ionic compounds tend to be most soluble in water because they interact favorably with its structure. In aqueous solution, dissolved ions become hydrated; that is, a shell of water molecules surrounds them.
Substances that dissolve in water can be categorized according to whether the resulting aqueous solutions conduct electricity. Strong electrolytes dissociate completely into ions to produce solutions that conduct electricity well. Weak electrolytes produce a relatively small number of ions, resulting in solutions that conduct electricity poorly. Nonelectrolytes dissolve as uncharged molecules and have no effect on the electrical conductivity of water.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.1%3A_The_Nature_of_Aqueous_Solutions.txt |
Learning Objectives
• To identify a precipitation reaction and predict solubilities.
A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. We described a precipitation reaction in which a colorless solution of silver nitrate was mixed with a yellow-orange solution of potassium dichromate to give a reddish precipitate of silver dichromate:
$AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + KNO_3(aq)\label{5.2.1A}$
This equation has the general form of an exchange reaction:
$AC + BD \rightarrow \underset{insoluble}{AD} + BC \label{5.2.2B}$
Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Two important uses of precipitation reactions are to isolate metals that have been extracted from their ores and to recover precious metals for recycling.
Video: Mixing Potassium Chromate and Silver Nitrate together to initiate a precipitation reaction (Equation $\ref{5.2.1}$).
While chemical equations show the identities of the reactants and the products and gave the stoichiometries of the reactions, but they told us very little about what was occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent.
Let’s consider the reaction of silver nitrate with potassium dichromate. When aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall balanced chemical equation for the reaction shows each reactant and product as undissociated, electrically neutral compounds:
$2AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2KNO_3(aq) \label{5.2.1}$
Although Equation $\ref{5.2.1}$ gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as AgNO3 and K2Cr2O7 are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because Ag2Cr2O7 is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equation showing which ions and molecules are hydrated and which are present in other forms and phases:
$2Ag^+(aq) + 2NO_3^-(aq) + 2K^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2K^+(aq) + 2NO_3^-(aq) \label{5.2.2}$
Note that K+(aq) and NO3(aq) ions are present on both sides of the equation, and their coefficients are the same on both sides. These ions are called spectator ions because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation, which shows only those species that participate in the chemical reaction:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) \label{5.2.3}$
Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation $\ref{5.2.3}$, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag2Cr2O7 formula unit.
By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows:
$2AgF(aq) + (NH_4)_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4F(aq)\label{5.2.4}$
The complete ionic equation for this reaction is as follows:
$2Ag^+(aq) + 2F^-(aq) + 2NH_4^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4^+(aq) + 2F^-(aq)\label{5.2.5}$
Because two NH4+(aq) and two F(aq) ions appear on both sides of Equation $\ref{5.2.5}$, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation $\ref{5.2.6}$), which is identical to Equation $\ref{5.2.3}$:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{5.2.6}$
If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.
Example $1$
Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate.
Given: reactants and products
Asked for: overall, complete ionic, and net ionic equations
Strategy:
Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation.
Solution:
From the information given, we can write the unbalanced chemical equation for the reaction:
$Ba(NO_3)_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaNO_3(aq)$
Because the product is Ba3(PO4)2, which contains three Ba2+ ions and two PO43 ions per formula unit, we can balance the equation by inspection:
$3Ba(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaNO_3(aq)$
This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form:
$3Ba^{2+}(aq) + 6NO_3^-(aq) + 6Na^+(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s) + 6Na^+(aq) + 6NO_3^-(aq)$
The six NO3(aq) ions and the six Na+(aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation:
$3Ba^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s)$
Exercise $1$: Mixing Silver Fluoride with Sodium Phosphate
Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride.
Answer
overall chemical equation:
$3AgF(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaF(aq) \nonumber$
complete ionic equation:
$3Ag^+(aq) + 3F^-(aq) + 3Na^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^+(aq) + 3F^-(aq) \nonumber$
net ionic equation:
$3Ag^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) \nonumber$
So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome.
The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually present in the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions.
Determining the Products for Precipitation Reactions: https://youtu.be/r0kYeZVuTAM
Predicting Solubilities
Table $2$ gives guidelines for predicting the solubility of a wide variety of ionic compounds. To determine whether a precipitation reaction will occur, we identify each species in the solution and then refer to Table $2$ to see which, if any, combination(s) of cation and anion are likely to produce an insoluble salt. In doing so, it is important to recognize that soluble and insoluble are relative terms that span a wide range of actual solubilities. We will discuss solubilities in more detail later, where you will learn that very small amounts of the constituent ions remain in solution even after precipitation of an “insoluble” salt. For our purposes, however, we will assume that precipitation of an insoluble salt is complete.
Table $2$: Guidelines for Predicting the Solubility of Ionic Compounds in Water
Soluble Exceptions
Rule 1 most salts that contain an alkali metal (Li+, Na+, K+, Rb+, and Cs+) and ammonium (NH4+)
Rule 2 most salts that contain the nitrate (NO3) anion
Rule 3 most salts of anions derived from monocarboxylic acids (e.g., CH3CO2) but not silver acetate and salts of long-chain carboxylates
Rule 4 most chloride, bromide, and iodide salts but not salts of metal ions located on the lower right side of the periodic table (e.g., Cu+, Ag+, Pb2+, and Hg22+).
Insoluble Exceptions
Rule 5 most salts that contain the hydroxide (OH) and sulfide (S2−) anions but not salts of the alkali metals (group 1), the heavier alkaline earths (Ca2+, Sr2+, and Ba2+ in group 2), and the NH4+ ion.
Rule 6 most carbonate (CO32) and phosphate (PO43) salts but not salts of the alkali metals or the NH4+ ion.
Rule 7 most sulfate (SO42) salts that contain main group cations with a charge ≥ +2 but not salts of +1 cations, Mg2+, and dipositive transition metal cations (e.g., Ni2+)
Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of a 1.0 M aqueous NaCl solution is mixed with 500 mL of a 1.0 M aqueous KBr solution, the final solution has a volume of 1.00 L and contains 0.50 M Na+(aq), 0.50 M Cl(aq), 0.50 M K+(aq), and 0.50 M Br(aq). As you will see in the following sections, none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other (Figure $1$).
Example $2$
Using the information in Table $2$, predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.
1. Aqueous solutions of barium chloride and lithium sulfate are mixed.
2. Aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.
3. Aqueous solutions of strontium bromide and aluminum nitrate are mixed.
4. Solid lead(II) acetate is added to an aqueous solution of ammonium iodide.
Given: reactants
Asked for: reaction and net ionic equation
Strategy:
1. Identify the ions present in solution and write the products of each possible exchange reaction.
2. Refer to Table $2$ to determine which, if any, of the products is insoluble and will therefore form a precipitate. If a precipitate forms, write the net ionic equation for the reaction.
Solution:
A Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba2+, Cl, Li+, and SO42 ions. The only possible exchange reaction is to form LiCl and BaSO4:
B We now need to decide whether either of these products is insoluble. Table $2$ shows that LiCl is soluble in water (rules 1 and 4), but BaSO4 is not soluble in water (rule 5). Thus BaSO4 will precipitate according to the net ionic equation
$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$
Although soluble barium salts are toxic, BaSO4 is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO4 particles in water.
1. A Rubidium hydroxide and cobalt(II) chloride are strong electrolytes, so when aqueous solutions of these compounds are mixed, the resulting solution initially contains Rb+, OH, Co2+, and Cl ions. The possible products of an exchange reaction are rubidium chloride and cobalt(II) hydroxide):
B According to Table $2$, RbCl is soluble (rules 1 and 4), but Co(OH)2 is not soluble (rule 5). Hence Co(OH)2 will precipitate according to the following net ionic equation:
$Co^{2+}(aq) + 2OH^-(aq) \rightarrow Co(OH)_2(s)$
2. A When aqueous solutions of strontium bromide and aluminum nitrate are mixed, we initially obtain a solution that contains Sr2+, Br, Al3+, and NO3 ions. The two possible products from an exchange reaction are aluminum bromide and strontium nitrate:
B According to Table $2$, both AlBr3 (rule 4) and Sr(NO3)2 (rule 2) are soluble. Thus no net reaction will occur.
1. A According to Table $2$, lead acetate is soluble (rule 3). Thus solid lead acetate dissolves in water to give Pb2+ and CH3CO2 ions. Because the solution also contains NH4+ and I ions, the possible products of an exchange reaction are ammonium acetate and lead(II) iodide:
B According to Table $2$, ammonium acetate is soluble (rules 1 and 3), but PbI2 is insoluble (rule 4). Thus Pb(C2H3O2)2 will dissolve, and PbI2 will precipitate. The net ionic equation is as follows:
$Pb^{2+} (aq) + 2I^-(aq) \rightarrow PbI_2(s)$
Exercise $2$
Using the information in Table $2$, predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.
1. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.
2. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.
3. Solid sodium fluoride is added to an aqueous solution of ammonium formate.
4. Aqueous solutions of calcium bromide and cesium carbonate are mixed.
Answer
1. $Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s)$
2. $2PO_4^{3-}(aq) + 3Hg^{2+}(aq) \rightarrow Hg_3(PO_4)_2(s)$
3. $NaF(s)$ dissolves; no net reaction
4. $Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s)$
Precipitation Reactions in Photography
Precipitation reactions can be used to recover silver from solutions used to develop conventional photographic film. Although largely supplanted by digital photography, conventional methods are often used for artistic purposes. Silver bromide is an off-white solid that turns black when exposed to light, which is due to the formation of small particles of silver metal. Black-and-white photography uses this reaction to capture images in shades of gray, with the darkest areas of the film corresponding to the areas that received the most light. The first step in film processing is to enhance the black/white contrast by using a developer to increase the amount of black. The developer is a reductant: because silver atoms catalyze the reduction reaction, grains of silver bromide that have already been partially reduced by exposure to light react with the reductant much more rapidly than unexposed grains.
After the film is developed, any unexposed silver bromide must be removed by a process called “fixing”; otherwise, the entire film would turn black with additional exposure to light. Although silver bromide is insoluble in water, it is soluble in a dilute solution of sodium thiosulfate (Na2S2O3; photographer’s hypo) because of the formation of [Ag(S2O3)2]3− ions. Thus washing the film with thiosulfate solution dissolves unexposed silver bromide and leaves a pattern of metallic silver granules that constitutes the negative. This procedure is summarized in Figure $2$. The negative image is then projected onto paper coated with silver halides, and the developing and fixing processes are repeated to give a positive image. (Color photography works in much the same way, with a combination of silver halides and organic dyes superimposed in layers.) “Instant photo” operations can generate more than a hundred gallons of dilute silver waste solution per day. Recovery of silver from thiosulfate fixing solutions involves first removing the thiosulfate by oxidation and then precipitating Ag+ ions with excess chloride ions.
Example $3$
A silver recovery unit can process 1500 L of photographic silver waste solution per day. Adding excess solid sodium chloride to a 500 mL sample of the waste (after removing the thiosulfate as described previously) gives a white precipitate that, after filtration and drying, consists of 3.73 g of AgCl. What mass of NaCl must be added to the 1500 L of silver waste to ensure that all the Ag+ ions precipitate?
Given: volume of solution of one reactant and mass of product from a sample of reactant solution
Asked for: mass of second reactant needed for complete reaction
Strategy:
1. Write the net ionic equation for the reaction. Calculate the number of moles of AgCl obtained from the 500 mL sample and then determine the concentration of Ag+ in the sample by dividing the number of moles of AgCl formed by the volume of solution.
2. Determine the total number of moles of Ag+ in the 1500 L solution by multiplying the Ag+ concentration by the total volume.
3. Use mole ratios to calculate the number of moles of chloride needed to react with Ag+. Obtain the mass of NaCl by multiplying the number of moles of NaCl needed by its molar mass.
Solution:
We can use the data provided to determine the concentration of Ag+ ions in the waste, from which the number of moles of Ag+ in the entire waste solution can be calculated. From the net ionic equation, we can determine how many moles of Cl are needed, which in turn will give us the mass of NaCl necessary.
A The first step is to write the net ionic equation for the reaction:
$Cl^-(aq) + Ag^+(aq) \rightarrow AgCl(s)$
We know that 500 mL of solution produced 3.73 g of AgCl. We can convert this value to the number of moles of AgCl as follows:
$moles\: AgCl = \dfrac{grams\: AgCl} {molar\: mass\: AgCl} = 3 .73\: \cancel{g\: AgCl} \left( \dfrac{1\: mol\: AgCl} {143 .32\: \cancel{g\: AgCl}} \right) = 0 .0260\: mol\: AgCl$
Therefore, the 500 mL sample of the solution contained 0.0260 mol of Ag+. The Ag+ concentration is determined as follows:
$[Ag^+ ] = \dfrac{moles\: Ag^+} {liters\: soln} = \dfrac{0 .0260\: mol\: AgCl} {0 .500\: L} = 0 .0520\: M$
B The total number of moles of Ag+ present in 1500 L of solution is as follows:
$moles\: Ag^+ = 1500\: \cancel{L} \left( \dfrac{0 .520\: mol} {1\: \cancel{L}} \right) = 78 .1\: mol\: Ag^+$
C According to the net ionic equation, one Cl ion is required for each Ag+ ion. Thus 78.1 mol of NaCl are needed to precipitate the silver. The corresponding mass of NaCl is
$mass\: NaCl = 78 .1 \: \cancel{mol\: NaCl} \left( \dfrac{58 .44\: g\: NaCl} {1\: \cancel{mol\: NaCl}} \right) = 4560\: g\: NaCl = 4 .56\: kg\: NaCl$
Note that 78.1 mol of AgCl correspond to 8.43 kg of metallic silver, which is worth about $7983 at 2011 prices ($32.84 per troy ounce). Silver recovery may be economically attractive as well as ecologically sound, although the procedure outlined is becoming nearly obsolete for all but artistic purposes with the growth of digital photography.
Exercise $3$
Because of its toxicity, arsenic is the active ingredient in many pesticides. The arsenic content of a pesticide can be measured by oxidizing arsenic compounds to the arsenate ion (AsO43), which forms an insoluble silver salt (Ag3AsO4). Suppose you are asked to assess the purity of technical grade sodium arsenite (NaAsO2), the active ingredient in a pesticide used against termites. You dissolve a 10.00 g sample in water, oxidize it to arsenate, and dilute it with water to a final volume of 500 mL. You then add excess AgNO3 solution to a 50.0 mL sample of the arsenate solution. The resulting precipitate of Ag3AsO4 has a mass of 3.24 g after drying. What is the percentage by mass of NaAsO2 in the original sample?
Answer: 91.0%
Note
Precipitation reactions are a subclass of double displacement reactions.
Determining the Net Ionic Equation for a Precipitation Reaction: https://youtu.be/AMJz1Sdz8IA
Summary
A complete ionic equation consists of the net ionic equation and spectator ions. Predicting the solubility of ionic compounds in water can give insight into whether or not a reaction will occur. The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances present in their undissociated forms; the complete ionic equation shows all the substances present in the form in which they actually exist in solution; and the net ionic equation is derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equation with the same coefficients. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction.
In a precipitation reaction, a subclass of exchange reactions, an insoluble material (a precipitate) forms when solutions of two substances are mixed. To predict the product of a precipitation reaction, all species initially present in the solutions are identified, as are any combinations likely to produce an insoluble salt. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.2%3A_Precipitation_Reactions.txt |
Learning Objectives
• To know the characteristic properties of acids and bases.
Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in our homes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids are bases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carried out an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids and bases.
Definitions of Acids and Bases
We defined acids as substances that dissolve in water to produce H+ ions, whereas bases were defined as substances that dissolve in water to produce OH ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases have been known for more than a thousand years, the definitions of acid and base have changed dramatically as scientists have learned more about them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent color changes in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and a solution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO3) with the evolution of carbon dioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyes that differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions were useful, they were entirely descriptive.
Although we include their tastes among the common characteristics of acids and bases, we never advocate tasting an unknown chemical!
The Arrhenius Definition of Acids and Bases
The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry, 1903). According to the Arrhenius definition, an acid is a substance like hydrochloric acid that dissolves in water to produce H+ ions (protons; Equation 5.3.1), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH) ions (Equation 5.3.2):
$\underset{an\: Arrhenius\: acid}{HCl_{(g)}} \xrightarrow {H_2 O_{(l)}} H^+_{(aq)} + Cl^-_{(aq)} \label{5.3.1}$
$\underset{an\: Arrhenius\: base}{NaOH_{(s)}} \xrightarrow {H_2O_{(l)}} Na^+_{(aq)} + OH^-_{(aq)} \label{5.3.2}$
According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H+ and OH ions, respectively, in solution. Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations:
1. First, because acids and bases were defined in terms of ions obtained from water, the Arrhenius concept applied only to substances in aqueous solution.
2. Second, and more important, the Arrhenius definition predicted that only substances that dissolve in water to produce $H^+$ and $OH^−$ ions should exhibit the properties of acids and bases, respectively. For example, according to the Arrhenius definition, the reaction of ammonia (a base) with gaseous HCl (an acid) to give ammonium chloride (Equation $\ref{5.3.3}$) is not an acid–base reaction because it does not involve $H^+$ and $OH^−$:
$NH_{3\;(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \label{5.3.3}$
The Brønsted–Lowry Definition of Acids and Bases
Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposed independently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936), who defined acid–base reactions in terms of the transfer of a proton (H+ ion) from one substance to another.
According to Brønsted and Lowry, an acid (A substance with at least one hydrogen atom that can dissociate to form an anion and an $H^+$ ion (a proton) in aqueous solution, thereby forming an acidic solution) is any substance that can donate a proton, and a base (a substance that produces one or more hydroxide ions ($OH^-$ and a cation when dissolved in aqueous solution, thereby forming a basic solution) is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrhenius definition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more general because the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form $NH_4^+$, so in Equation $\ref{5.3.3}$, $NH_3$ is a Brønsted–Lowry base and $HCl$ is a Brønsted–Lowry acid. Because of its more general nature, the Brønsted–Lowry definition is used throughout this text unless otherwise specified.
Definitions of Acids and Bases: https://youtu.be/r8reN0CSIHw
Polyprotic Acids
Acids differ in the number of protons they can donate. For example, monoprotic acids (a compound that is capable of donating one proton per molecule) are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl, HBr, HI, HNO3, and HNO2. All carboxylic acids that contain a single −CO2H group, such as acetic acid (CH3CO2H), are monoprotic acids, dissociating to form RCO2 and H+. A compound that can donate more than one proton per molecule. can donate more than one proton per molecule. For example, H2SO4 can donate two H+ ions in separate steps, so it is a diprotic acid (a compound that can donate two protons per molecule in separate steps) and H3PO4, which is capable of donating three protons in successive steps, is a triprotic acid (a compound that can donate three protons per molecule in separate steps), (Equations $\ref{5.3.4}$, $\ref{5.3.5}$, and $\ref{5.3.6}$):
$H_3 PO_4 (l) \overset{H_2 O(l)}{\rightleftharpoons} H ^+ ( a q ) + H_2 PO_4 ^- (aq) \label{5.3.4}$
$H_2 PO_4 ^- (aq) \rightleftharpoons H ^+ (aq) + HPO_4^{2-} (aq) \label{5.3.5}$
$HPO_4^{2-} (aq) \rightleftharpoons H^+ (aq) + PO_4^{3-} (aq) \label{5.3.6}$
In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occur simultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactants and products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. The reaction is then said to be in equilibrium (the point at which the rates of the forward and reverse reactions become the same, so that the net composition of the system no longer changes with time).
Strengths of Acids and Bases
We will not discuss the strengths of acids and bases quantitatively until next semester. Qualitatively, however, we can state that strong acids (An acid that reacts essentially completely with water) to give $H^+$ and the corresponding anion. react essentially completely with water to give $H^+$ and the corresponding anion. Similarly, strong bases (A base that dissociates essentially completely in water) to give $OH^-$ and the corresponding cation) dissociate essentially completely in water to give $OH^−$ and the corresponding cation. Strong acids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acids (An acid in which only a fraction of the molecules react with water) to producee $H^+$ and the corresponding anion. and weak bases (A base in which only a fraction of the molecules react with water to produce $OH^-$ and the corresponding cation) react with water to produce ions, so weak acids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereas more than 95% is present in undissociated form.
In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO3, HClO4, and H2SO4 (H3PO4 is only moderately strong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH, KOH, and Ca(OH)2. Common weak acids include HCN, H2S, HF, oxoacids such as HNO2 and HClO, and carboxylic acids such as acetic acid. The ionization reaction of acetic acid is as follows:
$CH_3 CO_2 H(l) \overset{H_2 O(l)}{\rightleftharpoons} H^+ (aq) + CH_3 CO_2^- (aq) \label{5.3.7}$
Although acetic acid is very soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than 1% dissociates), as we stated in section 8.1. Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation 5.3.8) but a weak acid when it donates its second proton (Equation 5.3.9) as indicated by the single and double arrows, respectively:
$\underset{strong\: acid}{H_2 SO_4 (l)} \xrightarrow {H_2 O(l)} H ^+ (aq) + HSO_4 ^- (aq) \label{5.3.8}$
$\underset{weak\: acid}{HSO_4^- (aq)} \rightleftharpoons H^+ (aq) + SO_4^{2-} (aq) \label{5.3.9}$
Consequently, an aqueous solution of sulfuric acid contains $H^+_{(aq)}$ ions and a mixture of $HSO^-_{4\;(aq)}$ and $SO^{2−}_{4\;(aq)}$ ions, but no $H_2SO_4$ molecules.
The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion:
$NH_3 (g) + H_2 O(l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq) \label{5.3.10}$
Most of the ammonia (>99%) is present in the form of NH3(g). Amines, which are organic analogues of ammonia, are also weak bases, as are ionic compounds that contain anions derived from weak acids (such as S2−).
There is no correlation between the solubility of a substance and whether it is a strong electrolyte, a weak electrolyte, or a nonelectrolyte.
Table $1$ lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other bases you encounter are most likely weak. Remember that there is no correlation between solubility and whether a substance is a strong or a weak electrolyte! Many weak acids and bases are extremely soluble in water.
Table $1$: Common Strong Acids and Bases
Strong Acids Strong Bases
Hydrogen Halides Oxoacids Group 1 Hydroxides Hydroxides of the Heavier Group 2 Elements
HCl HNO3 LiOH Ca(OH)2
HBr H2SO4 NaOH Sr(OH)2
HI HClO4 KOH Ba(OH)2
RbOH
CsOH
Example $1$
Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.
1. CH3CH2CO2H
2. CH3OH
3. Sr(OH)2
4. CH3CH2NH2
5. HBrO4
Given: compound
Asked for: acid or base strength
Strategy:
A Determine whether the compound is organic or inorganic.
B If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H+ or OH ions, respectively. If organic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group, respectively. Recall that all polyprotic acids except H2SO4 are weak acids.
Solution:
1. A This compound is propionic acid, which is organic. B It contains a carboxylic acid group analogous to that in acetic acid, so it must be a weak acid.
2. A CH3OH is methanol, an organic compound that contains the −OH group. B As a covalent compound, it does not dissociate to form the OH ion. Because it does not contain a carboxylic acid (−CO2H) group, methanol also cannot dissociate to form H+(aq) ions. Thus we predict that in aqueous solution methanol is neither an acid nor a base.
3. A Sr(OH)2 is an inorganic compound that contains one Sr2+ and two OH ions per formula unit. B We therefore expect it to be a strong base, similar to Ca(OH)2.
4. A CH3CH2NH2 is an amine (ethylamine), an organic compound in which one hydrogen of ammonia has been replaced by an R group. B Consequently, we expect it to behave similarly to ammonia (Equation 5.3.7), reacting with water to produce small amounts of the OH ion. Ethylamine is therefore a weak base.
5. A HBrO4 is perbromic acid, an inorganic compound. B It is not listed in Table $1$ as one of the common strong acids, but that does not necessarily mean that it is a weak acid. If you examine the periodic table, you can see that Br lies directly below Cl in group 17. We might therefore expect that HBrO4 is chemically similar to HClO4, a strong acid—and, in fact, it is.
Exercise $1$
Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.
1. Ba(OH)2
2. HIO4
3. CH3CH2CH2CO2H
4. (CH3)2NH
5. CH2O
Answer
1. strong base
2. strong acid
3. weak acid
4. weak base
5. none of these; formaldehyde is a neutral molecule
Definition of Strong/Weak Acids & Bases: https://youtu.be/gN2l8H_AWGU
The Hydronium Ion
Because isolated protons are very unstable and hence very reactive, an acid never simply “loses” an H+ ion. Instead, the proton is always transferred to another substance, which acts as a base in the Brønsted–Lowry definition. Thus in every acid–base reaction, one species acts as an acid and one species acts as a base. Occasionally, the same substance performs both roles, as you will see later. When a strong acid dissolves in water, the proton that is released is transferred to a water molecule that acts as a proton acceptor or base, as shown for the dissociation of sulfuric acid:
$\underset{acid\: (proton\: donor)}{H_2 SO_4 (l)} + \underset{base\: (proton\: acceptor)} {H_2 O(l)} \rightarrow \underset{acid}{H _3 O^+ (aq)} + \underset{base}{HSO_4^- (aq)} \label{5.3.11}$
Technically, therefore, it is imprecise to describe the dissociation of a strong acid as producing $H^+_{(aq)}$ ions, as we have been doing. The resulting $H_3O^+$ ion, called the hydronium ion and is a more accurate representation of $H^+_{(aq)}$. For the sake of brevity, however, in discussing acid dissociation reactions, we often show the product as $H^+_{(aq)}$ (as in Equation 5.3.7) with the understanding that the product is actually the$H_3O^+ _{(aq)}$ ion.
Conversely, bases that do not contain the hydroxide ion accept a proton from water, so small amounts of OH are produced, as in the following:
$\underset{base}{NH_3 (g)} + \underset{acid}{H_2 O(l)} \rightleftharpoons \underset{acid}{NH_4^+ (aq)} + \underset{base}{OH^- (aq)} \label{5.3.12}$
Again, the double arrow indicates that the reaction does not go to completion but rather reaches a state of equilibrium. In this reaction, water acts as an acid by donating a proton to ammonia, and ammonia acts as a base by accepting a proton from water. Thus water can act as either an acid or a base by donating a proton to a base or by accepting a proton from an acid. Substances that can behave as both an acid and a base are said to be amphotericWhen substances can behave as both an acid and a base..
The products of an acid–base reaction are also an acid and a base. In Equation $\ref{5.3.11}$, for example, the products of the reaction are the hydronium ion, here an acid, and the hydrogen sulfate ion, here a weak base. In Equation $\ref{5.3.12}$, the products are NH4+, an acid, and OH, a base. The product NH4+ is called the conjugate acid of the base NH3, and the product OH is called the conjugate baseThe substance formed when a Brønsted–Lowry acid donates a proton. of the acid H2O. Thus all acid–base reactions actually involve two conjugate acid–base pairs. All acid–base reactions involve two conjugate acid–base pairs, the Brønsted–Lowry acid and the base it forms after donating its proton, and the Brønsted–Lowry base and the acid it forms after accepting a proton.; in Equation $\ref{5.3.12}$, they are NH4+/NH3 and H2O/OH.
Neutralization Reactions
A neutralization reaction (a chemical reaction in which an acid and a base react in stoichiometric amounts to produce water and a salt) is one in which an acid and a base react in stoichiometric amounts to produce water and a salt (the general term for any ionic substance that does not have OH− as the anion or H+ as the cation), the general term for any ionic substance that does not have OH as the anion or H+ as the cation. If the base is a metal hydroxide, then the general formula for the reaction of an acid with a base is described as follows: Acid plus base yields water plus salt. For example, the reaction of equimolar amounts of HBr and NaOH to give water and a salt (NaBr) is a neutralization reaction:
$\underset{acid}{HBr(aq)} + \underset{base}{NaOH(aq)} \rightarrow \underset{water}{H_2 O(l)} + \underset{salt}{NaBr(aq)} \label{5.3.13}$
Acid plus base yields water plus salt.
If we write the complete ionic equation for the reaction in Equation $\ref{5.3.13}$, we see that $Na^+_{(aq)}$ and $Br^−_{(aq)}$ are spectator ions and are not involved in the reaction:
$H^+ (aq) + \cancel{Br^- (aq)} + \cancel{Na^+ (aq)} + OH^- (aq) \rightarrow H_2 O(l) + \cancel{Na^+ (aq)} + \cancel{Br^- (aq)} \label{5.3.14}$
The overall reaction is therefore simply the combination of H+(aq) and OH(aq) to produce H2O, as shown in the net ionic equation:
$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)\label{5.3.15}$
The net ionic equation for the reaction of any strong acid with any strong base is identical to Equation $\ref{5.3.15}$.
The strengths of the acid and the base generally determine whether the reaction goes to completion. The reaction of any strong acid with any strong base goes essentially to completion, as does the reaction of a strong acid with a weak base, and a weak acid with a strong base. Examples of the last two are as follows:
$\underset{strong\: acid}{HCl(aq)} + \underset{weak\: base}{NH_3 (aq)} \rightarrow \underset{salt}{NH_4 Cl(aq)} \label{5.3.16}$
$\underset{weak\: acid} {CH_3 CO _2 H(aq)} + \underset{strong\: base}{NaOH(aq)} \rightarrow \underset{salt}{CH _3 CO _2 Na(aq)} + H_2 O(l) \label{5.3.17}$
Sodium acetate is written with the organic component first followed by the cation, as is usual for organic salts. Most reactions of a weak acid with a weak base also go essentially to completion. One example is the reaction of acetic acid with ammonia:
$\underset{weak\: acid}{CH _3 CO _2 H(aq)} + \underset{weak\: base}{NH_3 (aq)} \rightarrow \underset{salt}{CH_3 CO_2 NH_4 (aq)} \label{5.3.18}$
An example of an acid–base reaction that does not go to completion is the reaction of a weak acid or a weak base with water, which is both an extremely weak acid and an extremely weak base.
Note
Except for the reaction of a weak acid or a weak base with water, acid–base reactions essentially go to completion.
In some cases, the reaction of an acid with an anion derived from a weak acid (such as HS) produces a gas (in this case, H2S). Because the gaseous product escapes from solution in the form of bubbles, the reverse reaction cannot occur. Therefore, these reactions tend to be forced, or driven, to completion. Examples include reactions in which an acid is added to ionic compounds that contain the HCO3, CN, or S2− anions, all of which are driven to completion:
$HCO_3^- (aq) + H^+ (aq) \rightarrow H_2 CO_3 (aq) \label{5.3.19}$
$H_2 CO_3 (aq) \rightarrow CO_2 (g) + H_2 O(l)$
$CN^- (aq) + H^+ (aq) \rightarrow HCN(g) \label{5.3.20}$
$S ^{2-} (aq) + H^+ (aq) \rightarrow HS^- (aq) \label{5.3.21}$
$HS^- (aq) + H^+ (aq) \rightarrow H_2 S(g)$
The reactions in Equations $\ref{5.3.19}$-$\ref{5.3.21}$ are responsible for the rotten egg smell that is produced when metal sulfides come in contact with acids.
Acid/Base Neutralization Reactions & Net Ionic Equations: https://youtu.be/gDS93ySeF80
Example $2$: Calcium Propionate
Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemical equation for the reaction of aqueous propionic acid (CH3CH2CO2H) with aqueous calcium hydroxide [Ca(OH)2] to give calcium propionate. Do you expect this reaction to go to completion, making it a feasible method for the preparation of calcium propionate?
Given: reactants and product
Asked for: balanced chemical equation and whether the reaction will go to completion
Strategy
Write the balanced chemical equation for the reaction of propionic acid with calcium hydroxide. Based on their acid and base strengths, predict whether the reaction will go to completion.
Solution
Propionic acid is an organic compound that is a weak acid, and calcium hydroxide is an inorganic compound that is a strong base. The balanced chemical equation is as follows:
$2CH_3CH_2CO_2H(aq) + Ca(OH)_2(aq) \rightarrow (CH_3CH_2CO_2)_2Ca(aq) + 2H_2O(l) \nonumber$
The reaction of a weak acid and a strong base will go to completion, so it is reasonable to prepare calcium propionate by mixing solutions of propionic acid and calcium hydroxide in a 2:1 mole ratio.
Exercise $2$
Write a balanced chemical equation for the reaction of solid sodium acetate with dilute sulfuric acid to give sodium sulfate.
Answer:
$2CH_3CO_2Na(s) + H_2SO_4(aq) \rightarrow Na_2SO_4(aq) + 2CH_3CO_2H(aq) \nonumber$
One of the most familiar and most heavily advertised applications of acid–base chemistry is antacids, which are bases that neutralize stomach acid. The human stomach contains an approximately 0.1 M solution of hydrochloric acid that helps digest foods. If the protective lining of the stomach breaks down, this acid can attack the stomach tissue, resulting in the formation of an ulcer. Because one factor that is believed to contribute to the formation of stomach ulcers is the production of excess acid in the stomach, many individuals routinely consume large quantities of antacids. The active ingredients in antacids include sodium bicarbonate and potassium bicarbonate (NaHCO3 and KHCO3; Alka-Seltzer); a mixture of magnesium hydroxide and aluminum hydroxide [Mg(OH)2 and Al(OH)3; Maalox, Mylanta]; calcium carbonate (CaCO3; Tums); and a complex salt, dihydroxyaluminum sodium carbonate [NaAl(OH)2CO3; original Rolaids]. Each has certain advantages and disadvantages. For example, Mg(OH)2 is a powerful laxative (it is the active ingredient in milk of magnesia), whereas Al(OH)3 causes constipation. When mixed, each tends to counteract the unwanted effects of the other. Although all antacids contain both an anionic base (OH, CO32, or HCO3) and an appropriate cation, they differ substantially in the amount of active ingredient in a given mass of product.
Example $3$
Assume that the stomach of someone suffering from acid indigestion contains 75 mL of 0.20 M HCl. How many Tums tablets are required to neutralize 90% of the stomach acid, if each tablet contains 500 mg of CaCO3? (Neutralizing all of the stomach acid is not desirable because that would completely shut down digestion.)
Given: volume and molarity of acid and mass of base in an antacid tablet
Asked for: number of tablets required for 90% neutralization
Strategy:
1. Write the balanced chemical equation for the reaction and then decide whether the reaction will go to completion.
2. Calculate the number of moles of acid present. Multiply the number of moles by the percentage to obtain the quantity of acid that must be neutralized. Using mole ratios, calculate the number of moles of base required to neutralize the acid.
3. Calculate the number of moles of base contained in one tablet by dividing the mass of base by the corresponding molar mass. Calculate the number of tablets required by dividing the moles of base by the moles contained in one tablet.
Solution:
A We first write the balanced chemical equation for the reaction:
$2HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2CO_3(aq)$
Each carbonate ion can react with 2 mol of H+ to produce H2CO3, which rapidly decomposes to H2O and CO2. Because HCl is a strong acid and CO32 is a weak base, the reaction will go to completion.
B Next we need to determine the number of moles of HCl present:
$75\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .20\: mol\: HCl} {\cancel{L}} \right) = 0. 015\: mol\: HCl$
Because we want to neutralize only 90% of the acid present, we multiply the number of moles of HCl by 0.90:
$(0.015\: mol\: HCl)(0.90) = 0.014\: mol\: HCl$
We know from the stoichiometry of the reaction that each mole of CaCO3 reacts with 2 mol of HCl, so we need
$moles\: CaCO_3 = 0 .014\: \cancel{mol\: HCl} \left( \dfrac{1\: mol\: CaCO_3}{2\: \cancel{mol\: HCl}} \right) = 0 .0070\: mol\: CaCO_3$
C Each Tums tablet contains
$\left( \dfrac{500\: \cancel{mg\: CaCO_3}} {1\: Tums\: tablet} \right) \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg\: CaCO_3}} \right) \left( \dfrac{1\: mol\: CaCO_3} {100 .1\: \cancel{g}} \right) = 0 .00500\: mol\: CaCO_ 3$
Thus we need $\dfrac{0.0070\: \cancel{mol\: CaCO_3}}{0.00500\: \cancel{mol\: CaCO_3}}= 1.4$ Tums tablets.
Exercise $3$
Assume that as a result of overeating, a person’s stomach contains 300 mL of 0.25 M HCl. How many Rolaids tablets must be consumed to neutralize 95% of the acid, if each tablet contains 400 mg of NaAl(OH)2CO3? The neutralization reaction can be written as follows:
$NaAl(OH)_2CO_3(s) + 4HCl(aq) \rightarrow AlCl_3(aq) + NaCl(aq) + CO_2(g) + 3H_2O(l)$
Answer: 6.4 tablets
The pH Scale
One of the key factors affecting reactions that occur in dilute solutions of acids and bases is the concentration of H+ and OH ions. The pH scale provides a convenient way of expressing the hydrogen ion (H+) concentration of a solution and enables us to describe acidity or basicity in quantitative terms.
Pure liquid water contains extremely low, but measurable concentrations of H3O+(aq) and OH(aq) ions produced via an autoionization reaction, in which water acts simultaneously as an acid and as a base:
$H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)\label{5.3.22}$
The concentration of hydrogen ions in pure water is only 1.0 × 10−7 M at 25°C. Because the autoionization reaction produces both a proton and a hydroxide ion, the OH concentration in pure water is also 1.0 × 10−7 M. Pure water is a neutral solution in which the total positive charge from all the cations is matched by an identical total negative charge from all the anions. At room temperature this is [H+] = [OH] = 1.0 × 10−7 M.
The pH scale describes the hydrogen ion concentration of a solution in a way that avoids the use of exponential notation. The negative base-10 logarithm of the hydrogen ion concentration: pH is often defined as the negative base-10 logarithm of the hydrogen ion concentration; pH is actually defined as the negative base-10 logarithm of hydrogen ion activity. As you will learn in a more advanced course, the activity of a substance in solution is related to its concentration. For dilute solutions such as those we are discussing, the activity and the concentration are approximately the same.
$pH = -log[H^+]\label{5.3.23}$
Conversely,
$[H^+] = 10^{-pH}\label{5.3.24}$
(If you are not familiar with logarithms or using a calculator to obtain logarithms and antilogarithms)
Because the hydrogen ion concentration is 1.0 × 10−7 M in pure water at 25°C, the pH of pure liquid water (and, by extension, of any neutral solution) is
$pH = -\log[1.0 \times 10^{-7}] = 7.00\label{5.3.25}$
Adding an acid to pure water increases the hydrogen ion concentration and decreases the hydroxide ion concentration because a neutralization reaction occurs, such as that shown in Equation 5.3.15. Because the negative exponent of [H+] becomes smaller as [H+] increases, the pH decreases with increasing [H+]. For example, a 1.0 M solution of a strong monoprotic acid such as HCl or HNO3 has a pH of 0.00:
$pH = -\log[1.0] = 0.00\label{5.3.26}$
Note
pH decreases with increasing [H+].
Conversely, adding a base to pure water increases the hydroxide ion concentration and decreases the hydrogen ion concentration. Because the autoionization reaction of water does not go to completion, neither does the neutralization reaction. Even a strongly basic solution contains a detectable amount of H+ ions. For example, a 1.0 M OH solution has [H+] = 1.0 × 10−14 M. The pH of a 1.0 M NaOH solution is therefore
$pH = -log[1.0 \times 10^{-14}] = 14.00\label{5.3.27}$
For practical purposes, the pH scale runs from pH = 0 (corresponding to 1 M H+) to pH 14 (corresponding to 1 M OH), although pH values less than 0 or greater than 14 are possible.
We can summarize the relationships between acidity, basicity, and pH as follows:
• If pH = 7.0, the solution is neutral.
• If pH < 7.0, the solution is acidic.
• If pH > 7.0, the solution is basic.
Keep in mind that the pH scale is logarithmic, so a change of 1.0 in the pH of a solution corresponds to a tenfold change in the hydrogen ion concentration. The foods and consumer products we encounter daily represent a wide range of pH values, as shown in Figure $2$.
Example $4$
1. What is the pH of a 2.1 × 10−2 M aqueous solution of HClO4?
2. The pH of a vinegar sample is 3.80. What is its hydrogen ion concentration?
Given: molarity of acid or pH
Asked for: pH or [H+]
Strategy:
Using the balanced chemical equation for the acid dissociation reaction and Equation 5.3.25 or 5.3.24, determine [H+] and convert it to pH or vice versa.
Solution:
1. HClO4 (perchloric acid) is a strong acid, so it dissociates completely into H+ ions and ClO4 ions:
$HClO_4(l) \rightarrow H^+(aq) + ClO_4^-(aq)$
The H+ ion concentration is therefore the same as the perchloric acid concentration. The pH of the perchloric acid solution is thus
$pH = -log[H^+] = -log(2.1 \times 10^{-2}) = 1.68$
The result makes sense: the H+ ion concentration is between 10−1 M and 10−2 M, so the pH must be between 1 and 2.
Note: The assumption that [H+] is the same as the concentration of the acid is valid for only strong acids. Because weak acids do not dissociate completely in aqueous solution, a more complex procedure is needed to calculate the pH of their solutions.
2. We are given the pH and asked to calculate the hydrogen ion concentration. From Equation 5.3.24,
$10^{-pH} = [H^+]$
Thus $[H^+] = 10^{-3.80} = 1.6 \times 10^{-4}\: M$.
Exercise $4$
1. What is the pH of a 3.0 × 10−5 M aqueous solution of HNO3?
2. What is the hydrogen ion concentration of turnip juice, which has a pH of 5.41?
Answer
1. $pH = 4.52$
2. $[H^+] = 3.9 \times 10^{-6}\: M$
Tools have been developed that make the measurement of pH simple and convenient (Figures 5.3.3 and 5.3.4). For example, pH paper consists of strips of paper impregnated with one or more acid–base indicators, which are intensely colored organic molecules whose colors change dramatically depending on the pH of the solution. Placing a drop of a solution on a strip of pH paper and comparing its color with standards give the solution’s approximate pH. A more accurate tool, the pH meter, uses a glass electrode, a device whose voltage depends on the H+ ion concentration (Figure $4$).
Introduction to pH: https://youtu.be/pQOa3bb5YEE
Summary
An acidic solution and a basic solution react together in a neutralization reaction that also forms a salt. Acid–base reactions require both an acid and a base. In Brønsted–Lowry terms, an acid is a substance that can donate a proton (H+), and a base is a substance that can accept a proton. All acid–base reactions contain two acid–base pairs: the reactants and the products. Acids can donate one proton (monoprotic acids), two protons (diprotic acids), or three protons (triprotic acids). Compounds that are capable of donating more than one proton are generally called polyprotic acids. Acids also differ in their tendency to donate a proton, a measure of their acid strength. Strong acids react completely with water to produce H3O+(aq) (the hydronium ion), whereas weak acids dissociate only partially in water. Conversely, strong bases react completely with water to produce the hydroxide ion, whereas weak bases react only partially with water to form hydroxide ions. The reaction of a strong acid with a strong base is a neutralization reaction, which produces water plus a salt.
The acidity or basicity of an aqueous solution is described quantitatively using the pH scale. The pH of a solution is the negative logarithm of the H+ ion concentration and typically ranges from 0 for strongly acidic solutions to 14 for strongly basic ones. Because of the autoionization reaction of water, which produces small amounts of hydronium ions and hydroxide ions, a neutral solution of water contains 1 × 10−7 M H+ ions and has a pH of 7.0. An indicator is an intensely colored organic substance whose color is pH dependent; it is used to determine the pH of a solution. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.3%3A_Acid-Base_Reactions.txt |
Learning Objectives
• To identify oxidation–reduction reactions in solution.
The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, coherent, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O2−). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction.
Note
Any oxidation must ALWAYS be accompanied by a reduction and vice versa.
Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water). The reaction is as follows:
$Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g) \label{5.4.1}$
Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants, which follows the mnemonic oxidation is loss reduction is gain, or oil rig. The oxidation state of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O2 or H2, are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is
$4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s) \label{5.4.2}$
Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al2O3, electrons are transferred as follows (the superscript 0 emphasizes the oxidation state of the elements):
$4 Al^0 + 3 O_2^0 \rightarrow 4 Al^{3+} + 6 O^{2-} \label{5.4.3}$
Equation 5.4.1 and Equation 5.4.2 are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In Equation 5.4.3, for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:
$electrons \, lost = 4 \, Al \, atoms \times {3 \, e^- \, lost \over Al \, atom } = 12 \, e^- \, lost \label{5.4.4a}$
$electrons \, gained = 6 \, O \, atoms \times {2 \, e^- \, gained \over O \, atom} = 12 \, e^- \, gained \label{5.4.4a}$
The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained.
An additional example of a redox reaction, the reaction of sodium metal with oxygen in air, is illustrated in Figure $1$.
Note
In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.
Redox Reactions: https://youtu.be/1v3yaaR_nHc
Assigning Oxidation States
Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previosuly, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; Figure $1$), magnesium oxide (MgO), and calcium chloride (CaCl2). In covalent compounds, in contrast, atoms share electrons. Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.
A set of rules for assigning oxidation states to atoms in chemical compounds follows.
Rules for Assigning Oxidation States
1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
2. The oxidation state of a monatomic ion is the same as its charge—for example, Na+ = +1, Cl = −1.
3. The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens.
4. Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.
5. Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
6. The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.
Nonintegral oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states.
In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound.
Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6. Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H+ and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF2 but −½ in KO2. Note that an oxidation state of −½ for O in KO2 is perfectly acceptable.
The reduction of copper(I) oxide shown in Equation 5.4.5 demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H2 and Cu. From rule 4, hydrogen in H2O has an oxidation state of +1, and from rule 5, oxygen in both Cu2O and H2O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu2O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows:
$\overset {+1}{Cu_2} \underset {-2}{O} (s) + \overset {0}{H_2} (g) \rightarrow 2 \overset {0}{Cu} (s) + \overset {+1}{H_2} \underset {-2}{O} (g) \label{5.4.5}$
Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). So this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:
$electrons \, lost = 2 \, H \, atoms \times {1 \, e^- \, lost \over H \, atom } = 2 \, e^- \, lost \label{5.4.6a}$
$electrons \, gained = 2 \, Cu \, atoms \times {1 \, e^- \, gained \over Cu \, atom} = 2 \, e^- \, gained \label{5.4.6b}$
Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.
Example $1$
Assign oxidation states to all atoms in each compound.
1. sulfur hexafluoride (SF6)
2. methanol (CH3OH)
3. ammonium sulfate [(NH4)2SO4]
4. magnetite (Fe3O4)
5. ethanoic (acetic) acid (CH3CO2H)
Given: molecular or empirical formula
Asked for: oxidation states
Strategy:
Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.
Solution:
a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF6), the oxidation state of sulfur must be +6:
[(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0
b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2:
[(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0
c. Note that (NH4)2SO4 is an ionic compound that consists of both a polyatomic cation (NH4+) and a polyatomic anion (SO42) (see Table 2.4). We assign oxidation states to the atoms in each polyatomic ion separately. For NH4+, hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:
[(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH4+ ion
For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:
[(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion
d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:
[(4 O atoms)(−2)]+[(3 Fe atoms)$\left (+{8 \over 3} \right )$]= 0
Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe3O4 can be viewed as having two Fe3+ ions and one Fe2+ ion per formula unit, giving a net positive charge of +8 per formula unit. Fe3O4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.”
e. Initially, we assign oxidation states to the components of CH3CO2H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of
[(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0
So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH3) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO2H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of
[(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3
To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus
$\underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H}$
Thus the sum of the oxidation states of the two carbon atoms is indeed zero.
Exercise $1$
Assign oxidation states to all atoms in each compound.
1. barium fluoride (BaF2)
2. formaldehyde (CH2O)
3. potassium dichromate (K2Cr2O7)
4. cesium oxide (CsO2)
5. ethanol (CH3CH2OH)
Answer:
1. Ba, +2; F, −1
2. C, 0; H, +1; O, −2
3. K, +1; Cr, +6; O, −2
4. Cs, +1; O, −½
5. C, −3; H, +1; C, −1; H, +1; O, −2; H, +1
Redox Reactions of Solid Metals in Aqueous Solution
A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure $1$). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:
$Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \label{5.4.5}$
In subsequent steps, FeCl2 undergoes oxidation to form a reddish-brown precipitate of Fe(OH)3.
Many metals dissolve through reactions of this type, which have the general form
$metal + acid \rightarrow salt + hydrogen \label{5.4.6}$
Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:
$Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \label{5.4.7}$
Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!
Single-Displacement Reactions
Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation 5.4.8) and the reduction of silver salts by copper (Equation 5.4.9 and Figure $2$):
$Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \label{5.4.8}$
$Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \label{5.4.9}$
The reaction in Equation 5.4.8 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.
The Activity Series
By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing Zn2+. Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:
$Zn(s) + Mg^{2+}(aq) \cancel{\rightarrow} Zn^{2+}(aq) + Mg(s) \label{5.4.10}$
$Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s) \label{5.4.11}$
Magnesium has a greater tendency to be oxidized than zinc does.
Pairwise reactions of this sort are the basis of the activity series (Figure $4$), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series).
When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series. Because magnesium is above zinc in Figure $4$, magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H2. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example $2$ demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions.
Example $2$
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
2. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
3. Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.
Given: reactants
Asked for: overall reaction and net ionic equation
Strategy:
1. Locate the reactants in the activity series in Figure .5.4.4 and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced.
2. Write the net ionic equation for the redox reaction.
Solution:
1. A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to their relative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionic equation is as follows:
$Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)$
Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.
2. A Mercury lies below lead in the activity series, so no reaction will occur.
3. A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H2. B From our discussion of solubilities, recall that Pb2+ and SO42 form insoluble lead(II) sulfate. In this case, the sulfate ions are not spectator ions, and the reaction is as follows:
$Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g)$
Lead(II) sulfate is the white solid that forms on corroded battery terminals.
Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.
Exercise $2$
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of chromium metal is placed in an aqueous solution of aluminum chloride.
2. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
3. A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).
Answer
1. $no\: reaction$
2. $3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)$
3. $2Al(s) + 6CH_3CO_2H(aq) \rightarrow 2Al^{3+}(aq) + 6CH_3CO_2^-(aq) + 3H_2(g)$
Summary
• Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation.
In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method (Table $1$), in which the overall reaction is separated into an oxidation equation and a reduction equation. Single-displacement reactions are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the activity series (Figure $4$), which arranges metals and H2 in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. Active metals lie at the top of the activity series, whereas inert metals are at the bottom of the activity series. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.4%3A_Oxidation-Reduction%3A_Some_General_Principles.txt |
The Learning Objectives of this Module are:
1. To identify oxidation–reduction reactions in solution.
We described the defining characteristics of oxidation–reduction, or redox, reactions already. Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses oxidation states, and a second is referred to as the half-reaction method. We show you how to balance redox equations using oxidation states in this section; the half-reaction method is discussed in the second semester
Balancing Redox Equations Using Oxidation States
To balance a redox equation using the , we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of Cr2+(aq) with manganese dioxide (MnO2) in the presence of dilute acid. Equation 8.8.1 is the net ionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using the procedure described in Section 7.5:
$\begin{matrix} +2 & +4 & +1 & +3 & +2 & +1\; \; \ Cr^{2+}\left ( aq \right ) &+\; \; \; \; \; \; \; \;MnO_{2}\left ( aq \right ) &+\; \; H^{+}\left ( aq \right ) &\rightarrow\; Cr^{3+}\left ( aq \right ) &+\;Mn^{2+} \left ( aq \right ) &+\; \; \; H_{2}O\left ( l \right ) \ & \; \; \; \; \; \; \;-2 & & & & \; \; \; \; -2 \end{matrix} \tag{8.8.1}$
Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced:
$Cr^{2+} + Mn^{4+} \rightarrow Cr^{3+} + Mn^{2+} \tag{8.8.2}$
The oxidation can be written as
$Cr^{2+} \rightarrow Cr^{3+} + e^- \tag{8.8.3}$
and the reduction as
$Mn^{4+} + 2e^- \rightarrow Mn^{2+} \tag{8.8.4}$
For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by the oxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same number of electrons in both. In this example, we must multiply the oxidation equation by 2 to give
$2Cr^{2+} \rightarrow 2Cr^{3+} + 2e^- \tag{8.8.5}$
Note the Patterns
In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.
The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction:
$2Cr^{2+} \rightarrow 2Cr^{3+} + 2e^- \tag{8.8.6}$
$Mn^{4+} + 2e^- \rightarrow Mn^{2+}$
We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical forms of the reactants and products:
$\begin{matrix} 2Cr^{2+}\left ( aq \right ) & \rightarrow & 2Cr^{3+}\left ( aq \right )+\cancel{2e^{-}}\ & & & \ 2Cr^{2+}\left ( aq \right )+\cancel{2e^{-}}& \rightarrow & Mn^{2+}\left ( aq \right )\ & & & \ 2Cr^{2+}\left ( aq \right )+2Cr^{2+}\left ( aq \right )&\rightarrow & 2Cr^{3+}\left ( aq \right ) + Mn^{2+}\left ( aq \right ) \end{matrix} \tag{8.8.7}$
Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of the equation (+4) does not equal the charge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add H+ as necessary to either side of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, we could balance the charge by adding OH as necessary to either side of the equation to balance the charges. In this case, adding four H+ ions to the left side of the equation gives
$2Cr^{2+}(aq) + MnO_2(s) + 4H^+(aq) \rightarrow 2Cr^{3+}(aq) + Mn^{2+}(aq) \tag{8.8.8}$
Although the charges are now balanced, we have two oxygen atoms on the left side of the equation and none on the right. We can balance the oxygen atoms without affecting the overall charge balance by adding H2O as necessary to either side of the equation. Here, we need to add two H2O molecules to the right side:
$2Cr^{2+}(aq) + MnO_2(s) + 4H^+(aq) \rightarrow 2Cr^{3+}(aq) + Mn^{2+}(aq) + 2H_2O(l) \tag{8.8.9}$
Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is now balanced. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing reactions is summarized in Table 8.8.1 and illustrated in Example 17.
Table 8.8.1 Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method
1. Write the unbalanced chemical equation for the reaction, showing the reactants and the products.
2. Assign oxidation states to all atoms in the reactants and the products (see Section 3.5 ) and determine which atoms change oxidation state.
3. Write separate equations for oxidation and reduction, showing (a) the atom(s) that is (are) oxidized and reduced and (b) the number of electrons accepted or donated by each.
4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons.
5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting the coefficients as necessary to give the numbers of atoms in step 4.
6. Add the two equations and cancel the electrons.
7. Balance the charge by adding H+ or OH ions as necessary for reactions in acidic or basic solution, respectively.
8. Balance the oxygen atoms by adding H2O molecules to one side of the equation.
9. Check to make sure that the equation is balanced in both atoms and total charges.
Example 17
Arsenic acid (H3AsO4) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine (AsH3, a highly toxic and unstable gas) and Zn2+(aq). Balance the equation for this reaction using oxidation states:
$H_3AsO_4(aq) + Zn(s) \rightarrow AsH_3(g) + Zn^{2+}(aq)$
Given: reactants and products in acidic solution
Asked for: balanced chemical equation using oxidation states
Strategy:
Follow the procedure given in Table 8.8.1 for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced.
Solution:
1. Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip this step.
2. Assign oxidation states using the procedure described in Section 3.5 and determine which atoms change oxidation state. The oxidation state of arsenic in arsenic acid is +6, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in Zn2+(aq) is +2:
3. Write separate equations for oxidation and reduction. The arsenic atom in H3AsO4 is reduced from the +5 to the −3 oxidation state, which requires the addition of eight electrons:
$Reduction \textrm: \: As^{5+} + 8e^- \rightarrow \underset{-3}{As ^{3-}}$
Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:
$Oxidation \textrm:\: Zn \rightarrow Zn^{2+} + 2e^-$
4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain
\begin{align} & Reduction\: (\times 1) \textrm : \: As^{5+} + 8e^- \rightarrow As^{3-} \ & Oxidation\: (\times 4) \textrm : \: 4Zn^0 \rightarrow 4Zn^{2+} + 8e^- \end{align}
5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting coefficients as necessary to give the numbers of atoms shown in step 4. Inserting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives
\begin{align} & Reduction \textrm : \: H_3AsO_4(aq) + 8e^- \rightarrow AsH_3(g) \ & Oxidation \textrm : \: 4Zn(s) \rightarrow 4Zn^{2+}(aq) + 8e^- \end{align}
6. Add the two equations and cancel the electrons. The sum of the two equations in step 5 is
$H_3 AsO_4 (aq) + 4Zn(s) + \cancel{8e^-} \rightarrow AsH_3 (g) + 4Zn^{2+} (aq) + \cancel{8e^-}$
which then yields
$H_3AsO_4(aq) + 4Zn(s) \rightarrow AsH_3(g) + 4Zn^{2+}(aq)$
7. Balance the charge by adding H+or OHions as necessary for reactions in acidic or basic solution, respectively. Because the reaction is carried out in acidic solution, we can add H+ ions to whichever side of the equation requires them to balance the charge. The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H+ ions to the left side gives a charge of +8 on both sides of the equation:
$H_3AsO_4(aq) + 4Zn(s) + 8H^+(aq) \rightarrow AsH_3(g) + 4Zn^{2+}(aq)$
8. Balance the oxygen atoms by adding H2O molecules to one side of the equation. There are 4 O atoms on the left side of the equation. Adding 4 H2O molecules to the right side balances the O atoms:
$H_3AsO_4(aq) + 4Zn(s) + 8H^+(aq) \rightarrow AsH_3(g) + 4Zn^{2+}(aq) + 4H_2O(l)$
Although we have not explicitly balanced H atoms, each side of the equation has 11 H atoms.
9. Check to make sure that the equation is balanced in both atoms and total charges. To guard against careless errors, it is important to check that both the total number of atoms of each element and the total charges are the same on both sides of the equation:
\begin{align} & Atoms \textrm : \: 1As + 4Zn + 4O + 11H = 1As + 4Zn + 4O + 11H \ & Total\: charge \textrm : \: 8(+1) = 4(+2) = +8 \end{align}
The balanced chemical equation for the reaction is therefore:
$H_3AsO_4(aq) + 4Zn(s) + 8H^+(aq) \rightarrow AsH_3(g) + 4Zn^{2+}(aq) + 4H_2O(l)$
Exercise
Copper commonly occurs as the sulfide mineral CuS. The first step in extracting copper from CuS is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to NO. Balance the equation for this reaction using oxidation states:
$CuS(s) + H^+(aq) + NO_3^-(aq) \rightarrow Cu^{2+}(aq) + NO(g) + SO_4^{2-}(aq)$
Answer: $3CuS(s) + 8H^+(aq) + 8NO_3^-(aq) \rightarrow 3Cu^{2+}(aq) + 8NO(g) + 3SO_4^{2-}(aq) + 4H_2O(l)$
Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example 18.
Example 18
The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows:
$Al(s) + H_2O(aq) \rightarrow [Al(OH)_4]^-(aq) + H_2(g)$
Balance this equation using oxidation states.
Given: reactants and products in a basic solution
Asked for: balanced chemical equation
Strategy:
Follow the procedure given in Table 8.8.1 for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced.
Solution:
We will apply the same procedure used in Example 17 but in a more abbreviated form.
1. The equation for the reaction is given, so we can skip this step.
2. The oxidation state of aluminum changes from 0 in metallic Al to +3 in [Al(OH)4]. The oxidation state of hydrogen changes from +1 in H2O to 0 in H2. Aluminum is oxidized, while hydrogen is reduced:
$\overset{0}{Al} (s) + \overset{+1}{H}_2 O(aq) \rightarrow [ \overset{+3}{Al} (OH)_4 ]^- (aq) + \overset{0}{H_2} (g)$
3. \begin{align} & Reduction\textrm : \: Al^0 \rightarrow Al^{3+} + 3e^- \ & Oxidation\textrm : \: H^+ + e^- \rightarrow H^0 \: (in\: H_2 ) \end{align}
4. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation:
\begin{align} & Reduction \textrm : \: 3H^+ + 3e^- \rightarrow 3H^0\: (in\: H_2) \ & Oxidation \textrm : \: Al^0 \rightarrow Al^{3+} + 3e^- \end{align}
5. Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbers of atoms as in step 4. Because a molecule of H2O contains two protons, in this case, 3H+ corresponds to 3/2H2O. Similarly, each molecule of hydrogen gas contains two H atoms, so 3H corresponds to 3/2H2.
\begin{align} & Reduction\textrm : \: \dfrac{3}{2} H_2 O + 3e^- \rightarrow \dfrac{3}{2} H_2 \ & Oxidation \textrm : \: Al \rightarrow [ Al ( OH )_4 ]^- + 3e^- \end{align}
6. Adding the equations and canceling the electrons gives
$Al + \dfrac{3}{2} H_2 O + \cancel{3e^-} \rightarrow [Al(OH)_4 ]^- + \dfrac{3}{2} H_2 + \cancel{3e^-}$
$Al + \dfrac{3}{2} H_2 O \rightarrow [Al(OH)_4 ]^- + \dfrac{3}{2} H_2$
To remove the fractional coefficients, multiply both sides of the equation by 2:
$2Al + 3H_2O \rightarrow 2[Al(OH)_4]^- + 3H_2$
7. The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two OH ions to the left side:
$2Al + 2OH^- + 3H_2O \rightarrow 2[Al(OH)_4]^- + 3H_2$
8. The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H2O molecules to the left side:
$2Al + 2OH^- + 6H_2O \rightarrow 2[Al(OH)_4]^- + 3H_2$
9. Be sure the equation is balanced:
\begin{align} & Atoms \textrm : \: 2Al + 8O + 14H = 2Al + 8O + 14H \ & Total\: charge \textrm : \: (2)(0) + (2)(-1) + (6)(0) = (2)(-1) + (3)(0) \end{align}
$-2 = -2$
10. The balanced chemical equation is therefore
$2Al(s) + 2OH^-(aq) + 6H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq) + 3H_2(g)$
Thus 3 mol of H2 gas are produced for every 2 mol of Al.
Exercise
The permanganate ion reacts with nitrite ion in basic solution to produce manganese(IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction.
Answer: $2MnO_4^-(aq) + 3NO_2^-(aq) + H_2O(l) \rightarrow 2MnO_2(s) + 3NO_3^-(aq) + 2OH^-(aq)$
As suggested in Example 17 and Example 18, a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines for predicting whether a redox reaction will occur and the identity of the products:
1. Compounds of elements in high oxidation states (such as ClO4, NO3, MnO4, Cr2O72−, and UF6) tend to act as oxidants and become reduced in chemical reactions.
2. Compounds of elements in low oxidation states (such as CH4, NH3, H2S, and HI) tend to act as reductants and become oxidized in chemical reactions.
Note the Pattern
Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.
When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.
Redox Reactions of Solid Metals in Aqueous Solution
A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure 8.8.1). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:
$Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \tag{8.8.10}$
In subsequent steps, FeCl2 undergoes oxidation to form a reddish-brown precipitate of Fe(OH)3.
Figure 8.8.1 Rust Formation. The corrosion process involves an oxidation–reduction reaction in which metallic iron is converted to Fe(OH)3, a reddish-brown solid.
Many metals dissolve through reactions of this type, which have the general form
$metal + acid \rightarrow salt + hydrogen \tag{8.8.11}$
Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:
$Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \tag{8.8.12}$
Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!
Single-Displacement Reactions
Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation 8.8.13) and the reduction of silver salts by copper (Equation 8.8.14 and Figure 8.8.2):
$Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \tag{8.8.13}$
$Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \tag{8.8.14}$
The reaction in Equation 8.8.13 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.
Figure 8.8.2 The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate. When a copper coil is placed in a solution of silver nitrate, silver ions are reduced to metallic silver on the copper surface, and some of the copper metal dissolves. Note the formation of a metallic silver precipitate on the copper coil and a blue color in the surrounding solution due to the presence of aqueous Cu2+ ions. Figure used with permission of Wikipedia
Summary
In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method (Table 8.8.1), in which the overall reaction is separated into an oxidation equation and a reduction equation.
Key Takeaway
• Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation.
Conceptual Problems
1. Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?
2. If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why?
3. In each redox reaction, determine which species is oxidized and which is reduced:
1. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
2. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
3. BrO3(aq) + 2MnO2(s) + H2O(l) → Br(aq) + 2MnO4(aq) + 2H+(aq)
4. Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is not a single-displacement reaction.
Numerical Problems
1. Balance each redox reaction under the conditions indicated.
1. CuS(s) + NO3(aq) → Cu2+(aq) + SO42−(aq) + NO(g); acidic solution
2. Ag(s) + HS(aq) + CrO42−(aq) → Ag2S(s) + Cr(OH)3(s); basic solution
3. Zn(s) + H2O(l) → Zn2+(aq) + H2(g); acidic solution
4. O2(g) + Sb(s) → H2O2(aq) + SbO2(aq); basic solution
5. UO22+(aq) + Te(s) → U4+(aq) + TeO42−(aq); acidic solution
2. Balance each redox reaction under the conditions indicated.
1. MnO4(aq) + S2O32−(aq) → Mn2+(aq) + SO42−(aq); acidic solution
2. Fe2+(aq) + Cr2O72−(aq) → Fe3+(aq) + Cr3+(aq); acidic solution
3. Fe(s) + CrO42−(aq) → Fe2O3(s) + Cr2O3(s); basic solution
4. Cl2(aq) → ClO3(aq) + Cl(aq); acidic solution
5. CO32−(aq) + N2H4(aq) → CO(g) + N2(g); basic solution
3. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.
1. Platinum wire is dipped in hydrochloric acid.
2. Manganese metal is added to a solution of iron(II) chloride.
3. Tin is heated with steam.
4. Hydrogen gas is bubbled through a solution of lead(II) nitrate.
4. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.
1. A few drops of NiBr2 are dropped onto a piece of iron.
2. A strip of zinc is placed into a solution of HCl.
3. Copper is dipped into a solution of ZnCl2.
4. A solution of silver nitrate is dropped onto an aluminum plate.
5. Dentists occasionally use metallic mixtures called amalgams for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction.
6. Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu2+(aq) and nitric oxide gas.
1. What has been oxidized? What has been reduced?
2. Balance the chemical equation.
7. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:
1. Pt2+(aq) + Ag(s) →
2. HCN(aq) + NaOH(aq) →
3. Fe(NO3)3(aq) + NaOH(aq) →
4. CH4(g) + O2(g) →
8. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:
1. Zn(s) + HCl(aq) →
2. HNO3(aq) + AlCl3(aq) →
3. K2CrO4(aq) + Ba(NO3)2(aq) →
4. Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.5%3A_Balancing_Oxidation-Reduction_Equations.txt |
Compounds that are capable of accepting electrons, such as O2 or F2, are called oxidants (or oxidizing agents) because they can oxidize other compounds. In the process of accepting electrons, an oxidant is reduced. Compounds that are capable of donating electrons, such as sodium metal or cyclohexane (C6H12), are called reductants (or reducing agents) because they can cause the reduction of another compound. In the process of donating electrons, a reductant is oxidized. These relationships are summarized in Equation 3.30:
oxidant + reductant → oxidation−reduction
$\underset {gains\, e \, (is \, reduced)}{O_2 (g)} + \underset {loses \, e \, (is \, oxidized)}{4Na (s)} \rightarrow 2 \underset {redox \, reaction}{Na_2O (s)} \tag{3.30}$
Some oxidants have a greater ability than others to remove electrons from other compounds. Oxidants can range from very powerful, capable of oxidizing most compounds with which they come in contact, to rather weak. Both F2 and Cl2 are powerful oxidants: for example, F2 will oxidize H2O in a vigorous, potentially explosive reaction. In contrast, S8 is a rather weak oxidant, and O2 falls somewhere in between. Conversely, reductants vary in their tendency to donate electrons to other compounds. Reductants can also range from very powerful, capable of giving up electrons to almost anything, to weak. The alkali metals are powerful reductants, so they must be kept away from atmospheric oxygen to avoid a potentially hazardous redox reaction.
A combustion reaction, first introduced in Section 3.2 "Determining Empirical and Molecular Formulas", is an oxidation–reduction reaction in which the oxidant is O2. One example of a combustion reaction is the burning of a candle, shown in Figure 3.9 "An Example of a Combustion Reaction". Consider, for example, the combustion of cyclohexane, a typical hydrocarbon, in excess oxygen. The balanced chemical equation for the reaction, with the oxidation state shown for each atom, is as follows:
$\underset {-2}{C_6} \overset {+1}{H_{12}} + 9 \overset {0}{O_2} \rightarrow 6 \overset {+4}{C} \underset {-2}{O_2} + 6 \overset {+1}{H_2} \underset {-2}{O} \tag{3.31}$
If we compare the oxidation state of each element in the products and the reactants, we see that hydrogen is the only element whose oxidation state does not change; it remains +1. Carbon, however, has an oxidation state of −2 in cyclohexane and +4 in CO2; that is, each carbon atom changes its oxidation state by six electrons during the reaction. Oxygen has an oxidation state of 0 in the reactants, but it gains electrons to have an oxidation state of −2 in CO2 and H2O. Because carbon has been oxidized, cyclohexane is the reductant; because oxygen has been reduced, it is the oxidant. All combustion reactions are therefore oxidation–reduction reactions.
We described the defining characteristics of oxidation–reduction, or redox, reactions in Chapter 7. Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses oxidation states, and a second is referred to as the half-reaction method. We show you how to balance redox equations using oxidation states in this section; the half-reaction method is discussed in the second semester
Note the Pattern
Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.
When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.6%3A_Oxidizing_and_Reducing_Agents.txt |
The Learning Objective of this Module is to use titration methods to analyze solutions quantitatively.
To determine the amounts or concentrations of substances present in a sample, chemists use a combination of chemical reactions and stoichiometric calculations in a methodology called quantitative analysis (A methodology that combines chemical reactions and stoichiometric calculations to determine the amounts or concentrations of substances present in a sample). Suppose, for example, we know the identity of a certain compound in a solution but not its concentration. If the compound reacts rapidly and completely with another reactant, we may be able to use the reaction to determine the concentration of the compound of interest. In a titrationAn experimental procedure in which a carefully measured volume of a solution of known concentration is added to a measured volume of a solution containing a compound whose concentration is to be determined., a carefully measured volume of a solution of known concentration, called the titrantThe solution of known concentration that is reacted with a compound in a solution of unknown concentration in a titration., is added to a measured volume of a solution containing a compound whose concentration is to be determined (the unknown). The reaction used in a titration can be an acid–base reaction, a precipitation reaction, or an oxidation–reduction reaction. In all cases, the reaction chosen for the analysis must be fast, complete, and specific; that is, only the compound of interest should react with the titrant. The equivalence point (The point in a titration where a stoichiometric amount (i.e., the amount required to react completely with the unknown) of the titrant has been added) is reached when a stoichiometric amount of the titrant has been added—the amount required to react completely with the unknown.
Determining the Concentration of an Unknown Solution Using a Titration
The chemical nature of the species present in the unknown dictates which type of reaction is most appropriate and also how to determine the equivalence point. The volume of titrant added, its concentration, and the coefficients from the balanced chemical equation for the reaction allow us to calculate the total number of moles of the unknown in the original solution. Because we have measured the volume of the solution that contains the unknown, we can calculate the molarity of the unknown substance. This procedure is summarized graphically here:
Example 20
The calcium salt of oxalic acid [Ca(O2CCO2)] is found in the sap and leaves of some vegetables, including spinach and rhubarb, and in many ornamental plants. Because oxalic acid and its salts are toxic, when a food such as rhubarb is processed commercially, the leaves must be removed, and the oxalate content carefully monitored.
The reaction of MnO4 with oxalic acid (HO2CCO2H) in acidic aqueous solution produces Mn2+ and CO2:
$\begin{pmatrix} MnO_{4}^{-}\left ( aq \right )+HO_{2}CCO_{2}H\left ( aq \right ) &\rightarrow Mn^{2+}\left ( aq \right )&+CO_{2}\left ( g \right )+H_{2}O\left ( l \right ) \ purple & colorless & \end{pmatrix}$
Because this reaction is rapid and goes to completion, potassium permanganate (KMnO4) is widely used as a reactant for determining the concentration of oxalic acid. The following video (chemistry Nakajima) demonstrates the reaction
Figure 8.9.1
Suppose you stirred a 10.0 g sample of canned rhubarb with enough dilute H2SO4(aq) to obtain 127 mL of colorless solution. Because the added permanganate is rapidly consumed, adding small volumes of a 0.0247 M KMnO4 solution, which has a deep purple color, to the rhubarb extract does not initially change the color of the extract. When 15.4 mL of the permanganate solution have been added, however, the solution becomes a faint purple due to the presence of a slight excess of permanganate (Figure 8.3.1). If we assume that oxalic acid is the only species in solution that reacts with permanganate, what percentage of the mass of the original sample was calcium oxalate? The video below demonstrates the titration when small, measured amounts of a known permaganate solution are added. At the endpoint, the number of moles of permaganage added equals the number of moles of oxalate in the solution, thus determining how many moles of oxalate we started with
Figure 8.9.2 Titration of oxalate using a known permaganate solution As permanganate is added to the oxalate solution the purple color appears and then disappears as the permanganate is consumed. As more permanganate is added, eventually all the oxalate is oxidized, and a faint purple color from the presence of excess permanganate appears, marking the endpoint.
Given: equation, mass of sample, volume of solution, and molarity and volume of titrant
Asked for: mass percentage of unknown in sample
Strategy:
1. Balance the chemical equation for the reaction using oxidation states.
2. Calculate the number of moles of permanganate consumed by multiplying the volume of the titrant by its molarity. Then calculate the number of moles of oxalate in the solution by multiplying by the ratio of the coefficients in the balanced chemical equation. Because calcium oxalate contains a 1:1 ratio of Ca2+:O2CCO2, the number of moles of oxalate in the solution is the same as the number of moles of calcium oxalate in the original sample.
3. Find the mass of calcium oxalate by multiplying the number of moles of calcium oxalate in the sample by its molar mass. Divide the mass of calcium oxalate by the mass of the sample and convert to a percentage to calculate the percentage by mass of calcium oxalate in the original sample.
Solution:
A As in all other problems of this type, the first requirement is a balanced chemical equation for the reaction. Using oxidation states gives
$2MnO_4^-(aq) + 5HO_2CCO_2H(aq) + 6H^+(aq) \rightarrow 2Mn^{2+}(aq) + 10CO_2(g) + 8H_2O(l)$
Thus each mole of MnO4 added consumes 2.5 mol of oxalic acid.
B Because we know the concentration of permanganate (0.0247 M) and the volume of permanganate solution that was needed to consume all the oxalic acid (15.4 mL), we can calculate the number of moles of MnO4 consumed. To do this we first convert the volume in mL to a volume in liters. Then simply multiplying the molarity of the solution by the volume in liters we find the number of moles of MnO4
$15.4\; \cancel{mL}\left ( \dfrac{1 \: \cancel{L}}{1000\; \cancel{mL}} \right )\left ( \dfrac{0.0247\; mol\; MnO_{4}^{-}}{1\; \cancel{L}} \right )=3.80\times 10^{4\;} mol\; MnO_{4}^{-}$
The number of moles of oxalic acid, and thus oxalate, present can be calculated from the mole ratio of the reactants in the balanced chemical equation. We can abbreviate the table needed to calculate the number of moles of oxalic acid in the
\begin{align} moles\: HO_2 CCO_2 H & = 3 .80 \times 10^{-4}\: \cancel{mol\: MnO_4^-} \left( \dfrac{5\: mol\: HO_2 CCO_2 H} {2\:\cancel{mol\: MnO_4^-}} \right) \ &= 9 .50 \times 10^{-4}\: mol\: HO_2 CCO_2 H \end{align}
C The problem asks for the percentage of calcium oxalate by mass in the original 10.0 g sample of rhubarb, so we need to know the mass of calcium oxalate that produced 9.50 × 10−4 mol of oxalic acid. Because calcium oxalate is Ca(O2CCO2), 1 mol of calcium oxalate gave 1 mol of oxalic acid in the initial acid extraction:
$Ca(O_2CCO_2)(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + HO_2CCO_2H(aq)$
The mass of calcium oxalate originally present was thus
\begin{align} mass\: of\: CaC_2 O_4 &= 9 .50 \times 10^{-4}\: \cancel{mol\: HO_2 CCO_2 H} \left( \dfrac{1\: \cancel{mol\: CaC_2 O_4}} {1\: \cancel{mol\: HO_2 CCO_2 H}} \right) \left( \dfrac{128 .10\: g\: CaC_2 O_4} {1\: \cancel{mol\: CaC_2 O_4}} \right) \ &= 0 .122\: g\: CaC_2 O_4 \end{align}
The original sample contained 0.122 g of calcium oxalate per 10.0 g of rhubarb. The percentage of calcium oxalate by mass was thus
$\% CaC_2 O_4 = \dfrac{0 .122\: g} {10 .0\: g} \times 100 = 1 .22\%$
Because the problem asked for the percentage by mass of calcium oxalate in the original sample rather than for the concentration of oxalic acid in the extract, we do not need to know the volume of the oxalic acid solution for this calculation.
Exercise 20: Glutathione
Glutathione is a low-molecular-weight compound found in living cells that is produced naturally by the liver. Health-care providers give glutathione intravenously to prevent side effects of chemotherapy and to prevent kidney problems after heart bypass surgery. Its structure is as follows:
Glutathione is found in two forms: one abbreviated as (left) GSH (indicating the presence of an –SH group) and the other (right) as GSSG (the disulfide form, in which an S–S bond links two glutathione units). The GSH form is easily oxidized to GSSG with elemental iodine:
$2GSH(aq) + I_2(aq) \rightarrow GSSG(aq) + 2HI(aq)$
A small amount of soluble starch is added as an indicator. Because starch reacts with excess I2 to give an intense blue color, the appearance of a blue color indicates that the equivalence point of the reaction has been reached.
Adding small volumes of a 0.0031 M aqueous solution of I2 to 194 mL of a solution that contains glutathione and a trace of soluble starch initially causes no change. After 16.3 mL of iodine solution have been added, however, a permanent pale blue color appears because of the formation of the starch-iodine complex. What is the concentration of glutathione in the original solution?
Answer: 5.2 × 10−4 M
Standard Solutions
In Example 20, the concentration of the titrant (I2) was accurately known. The accuracy of any titration analysis depends on an accurate knowledge of the concentration of the titrant. Most titrants are first standardized; that is, their concentration is measured by titration with a standard solutionA solution whose concentration is precisely known., which is a solution whose concentration is known precisely. Only pure crystalline compounds that do not react with water or carbon dioxide are suitable for use in preparing a standard solution. One such compound is potassium hydrogen phthalate (KHP), a weak monoprotic acid suitable for standardizing solutions of bases such as sodium hydroxide. The reaction of KHP with NaOH is a simple acid–base reaction. If the concentration of the KHP solution is known accurately and the titration of a NaOH solution with the KHP solution is carried out carefully, then the concentration of the NaOH solution can be calculated precisely. The standardized NaOH solution can then be used to titrate a solution of an acid whose concentration is unknown.
The reaction of KHP with NaOH. As with all acid-base reactions, a salt is formed.
Acid–Base Titrations
Because most common acids and bases are not intensely colored, a small amount of an acid–base indicator is usually added to detect the equivalence point in an acid–base titration. The point in the titration at which an indicator changes color is called the endpoint (the point in a titration at which an indicator changes color). The procedure is illustrated in Example 21.
Example 21: Vitamin C
The structure of vitamin C (ascorbic acid, a monoprotic acid) is as follows:
Ascorbic acid. The upper figure shows the three-dimensional representation of ascorbic acid. Hatched lines indicate bonds that are behind the plane of the paper, and wedged lines indicate bonds that are out of the plane of the paper.
An absence of vitamin C in the diet leads to the disease known as scurvy, a breakdown of connective tissue throughout the body and of dentin in the teeth. Because fresh fruits and vegetables rich in vitamin C are readily available in developed countries today, scurvy is not a major problem. In the days of slow voyages in wooden ships, however, scurvy was common. Ferdinand Magellan, the first person to sail around the world, lost more than 90% of his crew, many to scurvy. Although a diet rich in fruits and vegetables contains more than enough vitamin C to prevent scurvy, many people take supplemental doses of vitamin C, hoping that the extra amounts will help prevent colds and other illness.
Suppose a tablet advertised as containing 500 mg of vitamin C is dissolved in 100.0 mL of distilled water that contains a small amount of the acid–base indicator bromothymol blue, an indicator that is yellow in acid solution and blue in basic solution, to give a yellow solution. The addition of 53.5 mL of a 0.0520 M solution of NaOH results in a change to green at the endpoint, due to a mixture of the blue and yellow forms of the indicator (Figure 8.9.3). What is the actual mass of vitamin C in the tablet? (The molar mass of ascorbic acid is 176.13 g/mol.)
Figure 8.9.3: Bromothymol Blue is used as an indicator. If placed in a acidic solution it will turn the solution yellow, and if placed in a basic solution it will turn the solution blue. The equivalence point is seen when the solution turns green. pH is important for a variety of reasons. For example if you are trying to discard an acidic solution, you could add bromothymol blue to the solution and then titrate in some base until the solution turned green. This indicates that it is neutral
Given: reactant, volume of sample solution, and volume and molarity of titrant
Asked for: mass of unknown
Strategy:
A Write the balanced chemical equation for the reaction and calculate the number of moles of base needed to neutralize the ascorbic acid.
B Using mole ratios, determine the amount of ascorbic acid consumed. Calculate the mass of vitamin C by multiplying the number of moles of ascorbic acid by its molar mass.
Solution:
A Because ascorbic acid acts as a monoprotic acid, we can write the balanced chemical equation for the reaction as
$HAsc(aq) + OH^-(aq) \rightarrow Asc^-(aq) + H_2O(l)$
where HAsc is ascorbic acid and Asc is ascorbate. The number of moles of OH ions needed to neutralize the ascorbic acid is
$moles\: OH^- = 53 .5\: \cancel{mL} \left( \dfrac{\cancel{1\: L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .0520\: mol\: OH^-} {\cancel{1\: L}} \right) = 2 .78 \times 10^{-3}\: mol\: OH^-$
B The mole ratio of the base added to the acid consumed is 1:1, so the number of moles of OH added equals the number of moles of ascorbic acid present in the tablet:
$mass\: ascorbic\: acid = 2 .78 \times 10^{-3}\: \cancel{mol\: HAsc} \left( \dfrac{176 .13\: g\: HAsc} {1 \: \cancel{mol\: HAsc}} \right) = 0 .490\: g\: HAsc$
Because 0.490 g equals 490 mg, the tablet contains about 2% less vitamin C than advertised.
Exercise 20: Vinegar
Vinegar is essentially a dilute solution of acetic acid in water. Vinegar is usually produced in a concentrated form and then diluted with water to give a final concentration of 4%–7% acetic acid; that is, a 4% m/v solution contains 4.00 g of acetic acid per 100 mL of solution. If a drop of bromothymol blue indicator is added to 50.0 mL of concentrated vinegar stock and 31.0 mL of 2.51 M NaOH are needed to turn the solution from yellow to green, what is the percentage of acetic acid in the vinegar stock? (Assume that the density of the vinegar solution is 1.00 g/mL.)
Answer: 9.35%
Summary
The concentration of a species in solution can be determined by quantitative analysis. One such method is a titration, in which a measured volume of a solution of one substance, the titrant, is added to a solution of another substance to determine its concentration. The equivalence point in a titration is the point at which exactly enough reactant has been added for the reaction to go to completion. A standard solution, a solution whose concentration is known precisely, is used to determine the concentration of the titrant. Many titrations, especially those that involve acid–base reactions, rely on an indicator. The point at which a color change is observed is the endpoint, which is close to the equivalence point if the indicator is chosen properly.
Key Takeaway
• Quantitative analysis of an unknown solution can be achieved using titration methods.
Conceptual Problems
1. The titration procedure is an application of the use of limiting reactants. Explain why this is so.
2. Explain how to determine the concentration of a substance using a titration.
3. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M acetic acid with 0.10 M NaOH, and the other corresponds to the titration of 100 mL 0.10 M NaOH with 0.10 M acetic acid. Which graph corresponds to which titration? Justify your answer.
4. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M ammonia with 0.10 M HCl, and the other corresponds to the titration of 100 mL 0.10 M NH4Cl with 0.10 M NaOH. Which graph corresponds to which titration? Justify your answer.
5. Following are two graphs that illustrate how the electrical conductivity of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M Ba(OH)2 with 0.10 M H2SO4 , and the other corresponds to the titration of 100 mL of 0.10 M NaOH with 0.10 M H2SO4. Which graph corresponds to which titration? Justify your answer.
Answers
1. titration of NaOH with acetic acid
2. titration of acetic acid with NaOH
1. titration of Ba(OH)2 with sulfuric acid
2. titration of NaOH with sulfuric acid
Numerical Problems
1. A 10.00 mL sample of a 1.07 M solution of potassium hydrogen phthalate (KHP, formula mass = 204.22 g/mol) is diluted to 250.0 mL. What is the molarity of the final solution? How many grams of KHP are in the 10.00 mL sample?
2. What volume of a 0.978 M solution of NaOH must be added to 25.0 mL of 0.583 M HCl to completely neutralize the acid? How many moles of NaOH are needed for the neutralization?
3. A student was titrating 25.00 mL of a basic solution with an HCl solution that was 0.281 M. The student ran out of the HCl solution after having added 32.46 mL, so she borrowed an HCl solution that was labeled as 0.317 M. An additional 11.5 mL of the second solution was needed to complete the titration. What was the concentration of the basic solution?
Contributors
• Anonymous
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.7%3A_Stoichiometry_of_Reactions_in_Aqueous_Solutions%3A_Titrations.txt |
Learning Objectives
• To describe the characteristics of a gas.
The three common phases (or states) of matter are gases, liquids, and solids. Gases have the lowest density of the three, are highly compressible, and completely fill any container in which they are placed. Gases behave this way because their intermolecular forces are relatively weak, so their molecules are constantly moving independently of the other molecules present. Solids, in contrast, are relatively dense, rigid, and incompressible because their intermolecular forces are so strong that the molecules are essentially locked in place. Liquids are relatively dense and incompressible, like solids, but they flow readily to adapt to the shape of their containers, like gases. We can therefore conclude that the sum of the intermolecular forces in liquids are between those of gases and solids. Figure \(1\) compares the three states of matter and illustrates the differences at the molecular level.
The state of a given substance depends strongly on conditions. For example, H2O is commonly found in all three states: solid ice, liquid water, and water vapor (its gaseous form). Under most conditions, we encounter water as the liquid that is essential for life; we drink it, cook with it, and bathe in it. When the temperature is cold enough to transform the liquid to ice, we can ski or skate on it, pack it into a snowball or snow cone, and even build dwellings with it. Water vapor (the term vapor refers to the gaseous form of a substance that is a liquid or a solid under normal conditions so nitrogen (N2) and oxygen (O2) are referred to as gases, but gaseous water in the atmosphere is called water vapor) is a component of the air we breathe, and it is produced whenever we heat water for cooking food or making coffee or tea. Water vapor at temperatures greater than 100°C is called steam. Steam is used to drive large machinery, including turbines that generate electricity. The properties of the three states of water are summarized in Table \(1\).
Table \(1\): Properties of Water at 1.0 atm
Temperature State Density (g/cm3)
≤0°C solid (ice) 0.9167 (at 0.0°C)
0°C–100°C liquid (water) 0.9997 (at 4.0°C)
≥100°C vapor (steam) 0.005476 (at 127°C)
The geometric structure and the physical and chemical properties of atoms, ions, and molecules usually do not depend on their physical state; the individual water molecules in ice, liquid water, and steam, for example, are all identical. In contrast, the macroscopic properties of a substance depend strongly on its physical state, which is determined by intermolecular forces and conditions such as temperature and pressure.
Figure \(2\) shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states. Except for hydrogen, the elements that occur naturally as gases are on the right side of the periodic table. Of these, all the noble gases (group 18) are monatomic gases, whereas the other gaseous elements are diatomic molecules (H2, N2, O2, F2, and Cl2). Oxygen can also form a second allotrope, the highly reactive triatomic molecule ozone (O3), which is also a gas. In contrast, bromine (as Br2) and mercury (Hg) are liquids under normal conditions (25°C and 1.0 atm, commonly referred to as “room temperature and pressure”). Gallium (Ga), which melts at only 29.76°C, can be converted to a liquid simply by holding a container of it in your hand or keeping it in a non-air-conditioned room on a hot summer day. The rest of the elements are all solids under normal conditions.
All of the gaseous elements (other than the monatomic noble gases) are molecules. Within the same group (1, 15, 16 and 17), the lightest elements are gases. All gaseous substances are characterized by weak interactions between the constituent molecules or atoms.
Defining Gas Pressure: https://youtu.be/_CRn3cFs2CI
Summary
Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N2, O2). | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.1%3A_Properties_of_Gases%3A_Gas_Pressure.txt |
Learning Objectives
• To understand the relationships among pressure, temperature, volume, and the amount of a gas.
Early scientists explored the relationships among the pressure of a gas (P) and its temperature (T), volume (V), and amount (n) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the scientific method.
The Relationship between Pressure and Volume: Boyle's Law
As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.
The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure $1$. In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure $2$). A simple plot of $V$ versus $P$ gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:
$PV = \rm constant \label{6.2.1}$
Dividing both sides by $P$ gives an equation illustrating the inverse relationship between $P$ and $V$:
$V=\dfrac{\rm const.}{P} = {\rm const.}\left(\dfrac{1}{P}\right) \label{6.2.2}$
or
$V \propto \dfrac{1}{P} \label{6.2.3}$
where the ∝ symbol is read “is proportional to.” A plot of V versus 1/P is thus a straight line whose slope is equal to the constant in Equation 6.2.1 and Equation 6.2.3. Dividing both sides of Equation 6.2.1 by V instead of P gives a similar relationship between P and 1/V. The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure.
Boyle’s Law: https://youtu.be/lu86VSupPO4
The Relationship between Temperature and Volume: Charles's Law
Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air.
The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure $3$).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure $3$ that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure $3$, plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.
The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K).
We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure $3$ is often referred to as Charles’s law and is stated mathematically as
$V ={\rm const.}\; T \label{6.2.4}$
or
$V \propto T \label{6.2.5}$
with temperature expressed in kelvins, not in degrees Celsius. Charles’s law is valid for virtually all gases at temperatures well above their boiling points.
Charles’s Law: https://youtu.be/NBf510ZnlR0
The Relationship between Amount and Volume: Avogadro's Law
We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure $4$). This is the historic “Avogadro’s hypothesis.”
A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: At constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample. Stated mathematically,
$V ={\rm const.} \; (n) \label{6.2.6}$
or
$V \propto.n \text{@ constant T and P} \label{6.2.7}$
This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures.
Note
For a sample of gas,
• V increases as P decreases (and vice versa)
• V increases as T increases (and vice versa)
• V increases as n increases (and vice versa)
The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure $5$. Volume increases with increasing temperature or amount but decreases with increasing pressure.
Avogadro’s Law: https://youtu.be/dRY3Trl4T24
Summary
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure (Charles’s law), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present (Avogadro’s law). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K), the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.2%3A_The_Simple_Gas_Laws.txt |
Learning Objectives
• To use the ideal gas law to describe the behavior of a gas.
In this module, the relationship between Pressure, Temperature, Volume, and Amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas.
Deriving the Ideal Gas Law
Any set of relationships between a single quantity (such as V) and several other variables ($P$, $T$, and $n$) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions are as follows:
• Boyle’s law
$V \propto \dfrac{1}{P} \;\; \text{@ constant n and T}$
• Charles’s law
$V \propto T \;\; \text{@ constant n and P}$
• Avogadro’s law
$V \propto n \;\; \text{@ constant T and P}$
Combining these three expressions gives
$V \propto \dfrac{nT}{P} \tag{6.3.1}$
which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as
$V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \tag{6.3.2}$
By convention, the proportionality constant in Equation 6.3.1 is called the gas constant, which is represented by the letter $R$. Inserting R into Equation 6.3.2 gives
$V = \dfrac{Rnt}{P} = \dfrac{nRT}{P} \tag{6.3.3}$
Clearing the fractions by multiplying both sides of Equation 6.3.4 by $P$ gives
$PV = nRT \tag{6.3.4}$
This equation is known as the ideal gas law.
An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed.
Note
Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures.
Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then
$R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \tag{6.3.5}$
Because the product PV has the units of energy, R can also have units of J/(K•mol):
$R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\tag{6.3.6}$
Standard Conditions of Temperature and Pressure
Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and $\rm1\; bar = 100 \;kPa = 10^5\;Pa$ pressure, referred to as standard temperature and pressure (STP).
$\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa$
Please note that STP was defined differently in the part. The old definition was based on a standard pressure of 1 atm.
We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation 6.3.4:
$V=\dfrac{nRT}{P}\tag{6.3.7}$
Thus the volume of 1 mol of an ideal gas is 22.71 L at STP and 22.41 L at 0°C and 1 atm, approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed.
Table $1$: Molar Volumes of Selected Gases at 0°C and 1 atm
Gas Molar Volume (L)
He 22.434
Ar 22.397
H2 22.433
N2 22.402
O2 22.397
CO2 22.260
NH3 22.079
Applying the Ideal Gas Law
The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables (P, V, T, and n). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters (P, V, T, and n) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample.
Example $1$
The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft3), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?
Given: volume, temperature, and pressure
Asked for: amount of gas
Strategy:
1. Solve the ideal gas law for the unknown quantity, in this case n.
2. Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.
Solution:
A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law (Equation 6.3.4) for n, we obtain
$\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm$
B P and T are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:
$T=273+30=303{\rm K}$
Substituting these values into the expression we derived for n, we obtain
$n=\dfrac{PV}{RT}=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K}=1.23\times10^3\;mol$
Exercise $1$
Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. What is the pressure of the gas at 25°C?
Answer: 1.5 atm
In Example $1$, we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example $5$.
The Ideal Gas Law: https://youtu.be/rHGs23368mE
General Gas Equation
When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is:
$\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f)\P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array}$
Both equations can be rearranged to give:
$R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f}$
The two equations are equal to each other since each is equal to the same constant $R$. Therefore, we have:
$\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\tag{6.3.8}$
The equation is called the general gas equation. The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties.
Example $2$
Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example $1$?
Given: temperature, pressure, amount, and volume in August; temperature in January
Asked for: volume in January
Strategy:
1. Use the results from Example $1$ for August as the initial conditions and then calculate the change in volume due to the change in temperature from 30°C to −10°C. Begin by constructing a table showing the initial and final conditions.
2. Simplify the general gas equation by eliminating the quantities that are held constant between the initial and final conditions, in this case $P$ and $n$.
3. Solve for the unknown parameter.
Solution:
A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions:
Initial (August) Final (January)
$T_i=30$°C = 303 K $T_f=$−10°C = 263 K
$P_i=$0.980 atm $P_f=$0.980 atm
$n_i=$1.23 × 103 mol $n_f=$1.23 × 103 mol
$V_i=31150$ L $V_f=?$
B Both $n$ and $P$ are the same in both cases ($n_i=n_f,P_i=P_f$). Therefore, Equation can be simplified to:
$\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}$
This is the relationship first noted by Charles.
C Solving the equation for $V_f$, we get:
$V_f=V_i\times\dfrac{T_f}{T_i}=\rm31150\;L\times\dfrac{263\;K}{303\;K}=2.70\times10^4\;L$
It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change.
Exercise $2$
At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon?
Answer: 0.52 L
Example $1$ illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle (PV = constant) and the relationship between volume and amount observed by Avogadro (V/n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example $1$ can be applied in any such case, as we demonstrate in Example $2$ (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion).
Example $3$
Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise 5 (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)?
Given: initial volume, amount, temperature, and pressure; final temperature
Asked for: final pressure
Strategy:
Follow the strategy outlined in Example $5$.
Solution:
Prepare a table to determine which parameters change and which are held constant:
Initial Final
$V_i=0.406\;\rm L$ $V_f=0.406\;\rm L$
$n_i=0.025\;\rm mol$ $n_f=0.025\;\rm mol$
$T_i=\rm25\;^\circ C=298\;K$ $T_i=\rm750\;^\circ C=1023\;K$
$P_i=1.5\;\rm atm$ $P_f=?$
Both $V$ and $n$ are the same in both cases ($V_i=V_f,n_i=n_f$). Therefore, Equation can be simplified to:
$P_iT_i=P_fT_f$
By solving the equation for $P_f$, we get:
$P_f=P_i\times\dfrac{T_i}{T_f}=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K}=5.1\;atm$
This pressure is more than enough to rupture a thin sheet metal container and cause an explosion!
Exercise $3$
Suppose that a fire extinguisher, filled with CO2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher?
Answer: 23.4 atm
In Example $1$ and Example $2$, two of the four parameters (P, V, T, and n) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P, V, and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions.
Example $4$
We saw in Example $1$ that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 103 mol of H2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude?
Given: initial pressure, temperature, amount, and volume; final pressure and temperature
Asked for: final volume
Strategy:
Follow the strategy outlined in Example $5$.
Solution:
Begin by setting up a table of the two sets of conditions:
Initial Final
$P_i=745\;\rm mmHg=0.980\;atm$ $P_f=312\;\rm mmHg=0.411\;atm$
$T_i=\rm30\;^\circ C=303\;K$ $T_f=\rm750-30\;^\circ C=243\;K$
$n_i=\rm1.2\times10^3\;mol$ $n_i=\rm1.2\times10^3\;mol$
$V_i=\rm31150\;L$ $V_f=?$
By eliminating the constant property ($n$) of the gas, Equation 6.3.8 is simplified to:
$\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}$
By solving the equation for $V_f$, we get:
$V_f=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i}=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K}=5.96\times10^4\;L$
Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/P, the variable that changes the most will have the greatest effect on V. In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation.
We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft:
Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems.
Exercise $4$
A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.)
Answer: 4.07 × 103
Second Type of Ideal Gas Law Problems: https://youtu.be/WQDJOqddPI0
Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses
The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain
$\dfrac{n}{V}=\dfrac{P}{RT}\tag{6.3.9}$
The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass ($m$, in grams) divided by its molar mass ($M$, in grams per mole):
$n=\dfrac{m}{M}\tag{6.3.10}$
Substituting this expression for $n$ into Equation 6.3.9 gives
$\dfrac{m}{MV}=\dfrac{P}{RT}\tag{6.3.11}$
Because $m/V$ is the density $d$ of a substance, we can replace $m/V$ by $d$ and rearrange to give
$\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\tag{6.3.12}$
The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL).
Example $5$
Calculate the density of butane at 25°C and a pressure of 750 mmHg.
Given: compound, temperature, and pressure
Asked for: density
Strategy:
1. Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.
2. Substitute these values into Equation 6.3.12 to obtain the density.
Solution:
A The molar mass of butane (C4H10) is
$M=(4)(12.011) + (10)(1.0079) = 58.123 \rm g/mol$
Using 0.08206 (L•atm)/(K•mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres:
$P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm$
B Substituting these values into Equation 6.3.12 gives
$\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L$
Exercise $5$
Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.
Answer: radon, 9.23 g/L; N2, 1.17 g/L
A common use of Equation 6.3.12 is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example $6$.
Example $6$
The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.
Given: pressure, temperature, mass, and volume
Asked for: molar mass and chemical formula
Strategy:
1. Solve Equation 6.3.12 for the molar mass of the gas and then calculate the density of the gas from the information given.
2. Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.
3. Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.
Solution:
A Solving Equation 6.3.12 for the molar mass gives
$M=\dfrac{mRT}{PV}=\dfrac{dRT}{P}$
Density is the mass of the gas divided by its volume:
$\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.17\rm L}=1.84 \rm g/L$
B We must convert the other quantities to the appropriate units before inserting them into the equation:
$T=18+273=291 K$
$P=727\rm mmHg\times\dfrac{1\rm atm}{760\rm mmHg}=0.957\rm atm$
The molar mass of the unknown gas is thus
$\rho=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol$
C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:
$M({\rm NO})=14 + 16=30 \rm\; g/mol$
$M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol$
$M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol$
The most likely choice is NO2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO2 gas.
Exercise $6$
You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.
Answer: 44 g/mol; $CO_2$
Density and the Molar Mass of Gases: https://youtu.be/gnkGBsvUFVk
Summary
The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known.
Ideal gas equation: $PV = nRT$,
where $R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}$
General gas equation: $\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}$
Density of a gas: $\rho=\dfrac{MP}{RT}$
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume. All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity (P, V, T, or n) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P, V, T, and n) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.3%3A_Combining_the_Gas_Laws%3A_The_Ideal_Gas_Equation_and_the_General_Gas_Equation.txt |
Learning Objectives
• To relate the amount of gas consumed or released in a chemical reaction to the stoichiometry of the reaction.
• To understand how the ideal gas equation and the stoichiometry of a reaction can be used to calculate the volume of gas produced or consumed in a reaction.
With the ideal gas law, we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing.
Gas Densities and Molar Mass
For gases the density varies with the number of gas molecules in a constant volume. The ideal-gas equation can be manipulated to solve a variety of different types of problems. To determine the density, $\rho$, of a gas, we rearrange the equation to
$\rho=\dfrac{n}{V}=\dfrac{P}{RT}\label{6.4.1}$
Density of a gas is generally expressed in g/L. Multiplication of the left and right sides of the equation by the molar mass ($M$) of the gas gives
$\dfrac{g}{L}=\dfrac{PM}{RT} \label{6.4.2}$
This allows us to determine the density of a gas when we know the molar mass, or vice versa.
Example $1$
What is the density of nitrogen gas ($N_2$) at 248.0 Torr and 18º C?
Solution
Step 1: Write down your given information
• P = 248.0 Torr
• V = ?
• n = ?
• R = 0.0820574 L•atm•mol-1 K-1
• T = 18º C
Step 2: Convert as necessary.
$(248 \; \rm{Torr}) \times \dfrac{1 \; \rm{atm}}{760 \; \rm{Torr}} = 0.3263 \; \rm{atm}$
$18ºC + 273 = 291 K$
Step 3: This one is tricky. We need to manipulate the Ideal Gas Equation to incorporate density into the equation. *Write down all known equations:
$PV = nRT$
$\rho=\dfrac{m}{V}$
where $\rho$=density, m=mass, V=Volume
$m=M \times n$
where m=mass, M=molar mass, n=moles
*Now take the density equation.
$\rho=\dfrac{m}{V}$
*Keeping in mind $m=M \times n$...replace $(M \times n)$ for $mass$ within the density formula.
$\rho=\dfrac{M \times n}{V}$
$\dfrac{\rho}{M} = \dfrac{n}{V}$
*Now manipulate the Ideal Gas Equation
$PV = nRT$
$\dfrac{n}{V} = \dfrac{P}{RT}$
*$(n/V)$ is in both equations.
$\dfrac{n}{V} = \dfrac{\rho}{M}$
$\dfrac{n}{V} = \dfrac{P}{RT}$
*Now combine them please.
$\dfrac{\rho}{M} = \dfrac{P}{RT}$
*Isolate density.
$\rho = \dfrac{PM}{RT}$
Step 4: Now plug in the information you have.
$\rho = \dfrac{PM}{RT}$
$\rho = \dfrac{(0.3263\; \rm{atm})(2*14.01 \; \rm{g/mol})}{(0.08206 L atm/K mol)(291 \; \rm{K})}$
$\rho = 0.3828 \; g/L$
An example of varying density for a useful purpose is the hot air balloon, which consists of a bag (called the envelope) that is capable of containing heated air. As the air in the envelope is heated, it becomes less dense than the surrounding cooler air (Equation 6.4.1), which is has enough lifting power (due to buoyancy) to cause the balloon to float and rise into the air. Constant heating of the air is required to keep the balloon aloft. As the air in the balloon cools, it contracts, allowing outside cool air to enter, and the density increases. When this is carefully controlled by the pilot, the balloon can land as gently as it rose.
Note
• The density of a gas INCREASES with increasing pressure (Equation 6.4.1)
• The density of a gas DECREASES with increasing temperature (Equation 6.4.1)
Density and the Molar Mass of Gases: https://youtu.be/gnkGBsvUFVk
Determining Gas Volumes in Chemical Reactions
The ideal gas law can be used to calculate volume of gases consumed or produced. The ideal-gas equation frequently is used to interconvert between volumes and molar amounts in chemical equations.
Example $2$
What volume of carbon dioxide gas is produced at STP by the decomposition of 0.150 g $CaCO_3$ via the equation:
$CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$
Solution
Begin by converting the mass of calcium carbonate to moles.
$\dfrac{0.150\;g}{100.1\;g/mol} = 0.0015\; mol$
The stoichiometry of the reaction dictates that the number of moles $CaCO_3$ decomposed equals the number of moles $CO_2$ produced. Use the ideal-gas equation to convert moles of $CO_2$ to a volume.
$V = \dfrac{nRT}{R} = \dfrac{(0.0015\;mol)\left( 0.08206\; \frac{L \cdot atm}{mol \cdot K} \right) ( 273.15\;K)}{1\;atm} = 0.0336\;L \; or \; 33.6\;mL$
Example $3$
A 3.00 L container is filled with $Ne_{(g)}$ at 770 mmHg at 27oC. A $0.633\;\rm{g}$ sample of $CO_2$ vapor is then added. What is the partial pressure of $CO_2$ and $Ne$ in atm? What is the total pressure in the container in atm?
Solution
Step 1: Write down all given information, and convert as necessary.
Before:
• P = 770mmHg --> 1.01 atm
• V = 3.00L
• nNe=?
• T = 27oC --> $300\; K$
Other Unknowns: $n_{CO_2}$= ?
$n_{CO_2} = 0.633\; \rm{g} \;CO_2 \times \dfrac{1 \; \rm{mol}}{44\; \rm{g}} = 0.0144\; \rm{mol} \; CO_2$
Step 2: After writing down all your given information, find the unknown moles of Ne.
$n_{Ne} = \dfrac{PV}{RT}$
$n_{Ne} = \dfrac{(1.01\; \rm{atm})(3.00\; \rm{L})}{(0.08206\;atm\;L/mol\;K)(300\; \rm{K})}$
$n_{Ne} = 0.123 \; \rm{mol}$
Because the pressure of the container before the $CO_2$ was added contained only $Ne$, that is your partial pressure of $Ne$. After converting it to atm, you have already answered part of the question!
$P_{Ne} = 1.01\; \rm{atm}$
Step 3: Now that have pressure for Ne, you must find the partial pressure for $CO_2$. Use the ideal gas equation.
$\dfrac{P_{Ne}V}{n_{Ne}RT} = \dfrac{P_{CO_2}V}{n_{CO_2}RT}$
but because both gases share the same Volume ($V$) and Temperature ($T$) and since the Gas Constant ($R$) is constants, all three terms cancel and can be removed them from the equation.
$\dfrac{P}{n_{Ne}} = \dfrac{P}{n_{CO_2}}$
$\dfrac{1.01 \; \rm{atm}}{0.123\; \rm{mol} \;Ne} = \dfrac{P_{CO_2}}{0.0144\; \rm{mol} \;CO_2}$
$P_{CO_2} = 0.118 \; \rm{atm}$
This is the partial pressure $CO_2$.
Step 4: Now find total pressure.
$P_{total}= P_{Ne} + P_{CO_2}$
$P_{total}= 1.01 \; \rm{atm} + 0.118\; \rm{atm}$
$P_{total}= 1.128\; \rm{atm} \approx 1.13\; \rm{atm} \; \text{(with appropriate significant figures)}$
Example $4$
Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give SO2, followed by the reaction of SO2 with O2 in the presence of a catalyst to give SO3, which reacts with water to give H2SO4. The overall chemical equation is as follows:
$\rm 2S_{(s)}+3O_{2(g)}+2H_2O_{(l)}\rightarrow 2H_2SO_{4(aq)}$
What volume of O2 (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H2SO4?
Given: reaction, temperature, pressure, and mass of one product
Asked for: volume of gaseous reactant
Strategy:
A Calculate the number of moles of H2SO4 in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of O2 required.
B Use the ideal gas law to determine the volume of O2 required under the given conditions. Be sure that all quantities are expressed in the appropriate units.
Solution:
mass of H2SO4 → moles H2SO4 → moles O2 → liters O2
A We begin by calculating the number of moles of H2SO4 in 1.00 ton:
$\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4$
We next calculate the number of moles of O2 required:
$\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2$
B After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O2:
$V=\dfrac{nRT}{P}=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}}=3.43\times10^5\;L$
The answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry.
Exercise $4$
Charles used a balloon containing approximately 31,150 L of H2 for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation:
$Fe_{(s)} + 2 HCl_{(aq)} \rightarrow H_{2(g)} + FeCl_{2(aq)}$
How much iron (in kilograms) was needed to produce this volume of H2 if the temperature was 30°C and the atmospheric pressure was 745 mmHg?
Answer: 68.6 kg of Fe (approximately 150 lb)
Example $5$
Sodium azide ($NaN_3$) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation:
$2NaN_3 \rightarrow 2Na_{(s)} + 3N_{2\; (g)}$
This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the $N_2$ gas that results from the decomposition of a 5.00 g sample of $NaN_3$ could be collected by displacing water from an inverted flask, as in Figure $4$, what volume of gas would be produced at 21°C and 762 mmHg?
Given: reaction, mass of compound, temperature, and pressure
Asked for: volume of nitrogen gas produced
Strategy:
A Calculate the number of moles of N2 gas produced. From the data in Table $4$, determine the partial pressure of N2 gas in the flask.
B Use the ideal gas law to find the volume of N2 gas produced.
Solution:
A Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of N2 gas produced:
$\rm\dfrac{5.00\;g\;NaN_3}{(22.99+3\times14.01)\;g/mol}\times\dfrac{3mol\;N_2}{2mol\;NaN_3}=0.115\;mol\; N_2$
The pressure given (762 mmHg) is the total pressure in the flask, which is the sum of the pressures due to the N2 gas and the water vapor present. Table $4$ tells us that the vapor pressure of water is 18.65 mmHg at 21°C (294 K), so the partial pressure of the N2 gas in the flask is only
$\rm(762 − 18.65)\;mmHg \times\dfrac{1\;atm}{760\;atm}= 743.4\; mmHg \times\dfrac{1\;atm}{760\;atm}= 0.978\; atm.$
B Solving the ideal gas law for V and substituting the other quantities (in the appropriate units), we get
$V=\dfrac{nRT}{P}=\rm\dfrac{0.115\;mol\times0.08206\dfrac{atm\cdot L}{mol\cdot K}\times294\;K}{0.978\;atm}=2.84\;L$
Exercise $5$
A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce H2 gas according to the equation Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq). The resulting H2 gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy?
Answer: 0.397 L
Ideal Gas law Equation and Reaction Stoichiometry: https://youtu.be/8pPlW8MRhgI
Summary
The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.4%3A_Applications_of_the_Ideal_Gas_Equation.txt |
Learning Objectives
• To relate the amount of gas consumed or released in a chemical reaction to the stoichiometry of the reaction.
• To understand how the ideal gas equation and the stoichiometry of a reaction can be used to calculate the volume of gas produced or consumed in a reaction.
Introduction
With the ideal gas law, we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing. As a chemical engineer said to one of the authors, “Gases always go where you want them to, liquids sometimes do, but solids almost never do.”
Example $1$
Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give SO2, followed by the reaction of SO2 with O2 in the presence of a catalyst to give SO3, which reacts with water to give H2SO4. The overall chemical equation is as follows:
$\rm 2S_{(s)}+3O_{2(g)}+2H_2O_{(l)}\rightarrow 2H_2SO_{4(aq)}$
What volume of O2 (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H2SO4?
Given: reaction, temperature, pressure, and mass of one product
Asked for: volume of gaseous reactant
Strategy:
A Calculate the number of moles of H2SO4 in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of O2 required.
B Use the ideal gas law to determine the volume of O2 required under the given conditions. Be sure that all quantities are expressed in the appropriate units.
Solution:
mass of H2SO4 → moles H2SO4 → moles O2 → liters O2
A We begin by calculating the number of moles of H2SO4 in 1.00 ton:
$\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4$
We next calculate the number of moles of O2 required:
$\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2$
B After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O2:
$V=\dfrac{nRT}{P}=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}}=3.43\times10^5\;L$
The answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry.
Exercise $1$
In Example 5, we saw that Charles used a balloon containing approximately 31,150 L of H2 for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation:
$Fe_{(s)} + 2 HCl_{(aq)} \rightarrow H_{2(g)} + FeCl_{2(aq)}$
How much iron (in kilograms) was needed to produce this volume of H2 if the temperature was 30°C and the atmospheric pressure was 745 mmHg?
Answer: 68.6 kg of Fe (approximately 150 lb)
Ideal Gas law Equation and Reaction Stoichiometry: https://youtu.be/8pPlW8MRhgI
Collecting gases over water
As shown in Figure $1$, a common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. Because the gas is less dense than liquid water, it bubbles to the top of the bottle, displacing the water. Eventually, all the water is forced out and the bottle contains only gas. If a calibrated bottle is used (i.e., one with markings to indicate the volume of the gas) and the bottle is raised or lowered until the level of the water is the same both inside and outside, then the pressure within the bottle will exactly equal the atmospheric pressure measured separately with a barometer ($P_{\rm bar.}$).
Remember, however, when calculating the amount of gas formed in the reaction, the gas collected inside the bottle is not pure. Instead, it is a mixture of the product gas and water vapor. All liquids (including water) have a measurable amount of vapor in equilibrium with the liquid because molecules of the liquid are continuously escaping from the liquid’s surface, while other molecules from the vapor phase collide with the surface and return to the liquid. The vapor thus exerts a pressure above the liquid, which is called the liquid’s vapor pressure. In the case shown in Figure $1$, the bottle is therefore actually filled with a mixture of O2 and water vapor, and the total pressure is, by Dalton’s law of partial pressures, the sum of the pressures of the two components:
$P_{\rm tot}=P_{\rm gas}+P_{\rm H_2O}=P_{\rm bar.} \label{6.6.1}$
If we want to know the pressure of the gas generated in the reaction to calculate the amount of gas formed, we must first subtract the pressure due to water vapor from the total pressure. This is done by referring to tabulated values of the vapor pressure of water as a function of temperature (Table $1$).
Table $1$: Vapor Pressure of Water at Various Temperatures
T (°C) P (in mmHg)
0 4.58
15 12.79
17 14.53
19 16.48
21 18.65
23 21.07
25 23.76
30 31.82
50 92.51
70 233.8
100 760.0
As shown in Figure $2$, the vapor pressure of water increases rapidly with increasing temperature, and at the normal boiling point (100°C), the vapor pressure is exactly 1 atm. The methodology is illustrated in Example $2$. The only gases that cannot be collected using this technique are those that readily dissolve in water (e.g., NH3, H2S, and CO2) and those that react rapidly with water (such as F2 and NO2).
Figure $2$: A Plot of the Vapor Pressure of Water versus Temperature. The vapor pressure is very low (but not zero) at 0°C and reaches 1 atm = 760 mmHg at the normal boiling point, 100°C.
Example $2$
Sodium azide ($NaN_3$) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation:
$2NaN_3 \rightarrow 2Na_{(s)} + 3N_{2\; (g)}$
This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the $N_2$ gas that results from the decomposition of a 5.00 g sample of $NaN_3$ could be collected by displacing water from an inverted flask, as in Figure $1$, what volume of gas would be produced at 21°C and 762 mmHg?
Given: reaction, mass of compound, temperature, and pressure
Asked for: volume of nitrogen gas produced
Strategy:
A Calculate the number of moles of N2 gas produced. From the data in Table $1$, determine the partial pressure of N2 gas in the flask.
B Use the ideal gas law to find the volume of N2 gas produced.
Solution:
A Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of N2 gas produced:
$\rm\dfrac{5.00\;g\;NaN_3}{(22.99+3\times14.01)\;g/mol}\times\dfrac{3mol\;N_2}{2mol\;NaN_3}=0.115\;mol\; N_2$
The pressure given (762 mmHg) is the total pressure in the flask, which is the sum of the pressures due to the N2 gas and the water vapor present. Table $1$ tells us that the vapor pressure of water is 18.65 mmHg at 21°C (294 K), so the partial pressure of the N2 gas in the flask is only
$\rm(762 − 18.65)\;mmHg \times\dfrac{1\;atm}{760\;atm}= 743.4\; mmHg \times\dfrac{1\;atm}{760\;atm}= 0.978\; atm.$
B Solving the ideal gas law for V and substituting the other quantities (in the appropriate units), we get
$V=\dfrac{nRT}{P}=\rm\dfrac{0.115\;mol\times0.08206\dfrac{atm\cdot L}{mol\cdot K}\times294\;K}{0.978\;atm}=2.84\;L$
Exercise $2$
A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce H2 gas according to the equation Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq). The resulting H2 gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy?
Answer: 0.397 L
Collecting a Product Gas over Water: https://youtu.be/zFuy3t81vjQ
Summary
The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature.
Conceptual Problems
1. Why are so many industrially important reactions carried out in the gas phase?
2. The volume of gas produced during a chemical reaction can be measured by collecting the gas in an inverted container filled with water. The gas forces water out of the container, and the volume of liquid displaced is a measure of the volume of gas. What additional information must be considered to determine the number of moles of gas produced? The volume of some gases cannot be measured using this method. What property of a gas precludes the use of this method?
3. Equal masses of two solid compounds (A and B) are placed in separate sealed flasks filled with air at 1 atm and heated to 50°C for 10 hours. After cooling to room temperature, the pressure in the flask containing A was 1.5 atm. In contrast, the pressure in the flask containing B was 0.87 atm. Suggest an explanation for these observations. Would the masses of samples A and B still be equal after the experiment? Why or why not?
Numerical Problems
1. Balance each chemical equation and then determine the volume of the indicated reactant at STP that is required for complete reaction. Assuming complete reaction, what is the volume of the products?
1. SO2(g) + O2(g) → SO3(g) given 2.4 mol of O2
2. H2(g) + Cl2(g) → HCl(g) given 0.78 g of H2
3. C2H6(g) + O2(g) → CO2(g) + H2O(g) given 1.91 mol of O2
• During the smelting of iron, carbon reacts with oxygen to produce carbon monoxide, which then reacts with iron(III) oxide to produce iron metal and carbon dioxide. If 1.82 L of CO2 at STP is produced,
1. what mass of CO is consumed?
2. what volume of CO at STP is consumed?
3. how much O2 (in liters) at STP is used?
4. what mass of carbon is consumed?
5. how much iron metal (in grams) is produced?
• Complete decomposition of a sample of potassium chlorate produced 1.34 g of potassium chloride and oxygen gas.
1. What is the mass of KClO3 in the original sample?
2. What mass of oxygen is produced?
3. What is the volume of oxygen produced at STP?
• The combustion of a 100.0 mg sample of an herbicide in excess oxygen produced 83.16 mL of CO2 and 72.9 mL of H2O vapor at STP. A separate analysis showed that the sample contained 16.44 mg of chlorine. If the sample is known to contain only C, H, Cl, and N, determine the percent composition and the empirical formula of the herbicide.
• The combustion of a 300.0 mg sample of an antidepressant in excess oxygen produced 326 mL of CO2 and 164 mL of H2O vapor at STP. A separate analysis showed that the sample contained 23.28% oxygen. If the sample is known to contain only C, H, O, and N, determine the percent composition and the empirical formula of the antidepressant.
Answers
1. 2.20 g KClO3
2. 0.863 g O2
3. 604 mL O2
1. Percent composition: 58.3% C, 4.93% H, 23.28% O, and 13.5% N; empirical formula: C10H10O3N2 | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.5%3A_Gases_in_Chemical_Reactions.txt |
Learning Objectives
• To determine the contribution of each component gas to the total pressure of a mixture of gases
In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture.
Partial Pressures
The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present:
$P=n \bigg(\dfrac{RT}{V}\bigg) = n \times \rm const. \label{6.6.1}$
Nothing in the equation depends on the nature of the gas—only the amount.
With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressure, the pressure the gas would exert if it were the only one present (at the same temperature and volume).
To summarize, the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressures. We can write it mathematically as
$P_{tot}= P_1+P_2+P_3+P_4 \; ... = \sum_{i=1}^n{P_i} \label{6.6.2}$
where $P_{tot}$ is the total pressure and the other terms are the partial pressures of the individual gases (up to $n$ component gases).
For a mixture of two ideal gases, $A$ and $B$, we can write an expression for the total pressure:
$P_{tot}=P_A+P_B=n_A\bigg(\dfrac{RT}{V}\bigg) + n_B\bigg(\dfrac{RT}{V}\bigg)=(n_A+n_B)\bigg(\dfrac{RT}{V}\bigg) \label{6.6.3}$
More generally, for a mixture of $n$ component gases, the total pressure is given by
$P_{tot}=(P_1+P_2+P_3+ \; \cdots +P_n)\bigg(\dfrac{RT}{V}\bigg)\label{6.6.2a}$
$P_{tot}=\sum_{i=1}^n{n_i}\bigg(\dfrac{RT}{V}\bigg)\label{6.6.2b}$
Equation 6.6.4 restates Equation 6.6.3 in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For Equation 6.6.4 to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example $1$.
Example $1$
Deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of $O_2$ and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature?
Given: masses of components, total volume, and temperature
Asked for: partial pressures and total pressure
Strategy:
1. Calculate the number of moles of $He$ and $O_2$ present.
2. Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture.
Solution:
A The number of moles of $He$ is
$n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol$
The number of moles of $O_2$ is
$n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol$
B We can now use the ideal gas law to calculate the partial pressure of each:
$P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm$
$P_{\rm O_2}=\dfrac{n_{\rm O_2} RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm$
The total pressure is the sum of the two partial pressures:
$P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm$
Exercise $1$
A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder.
Answer: $P_{CH_4}=137 \; atm$; $P_{C_2H_6}=13.4\; atm$; $P_{tot}=151\; atm$
Mole Fractions of Gas Mixtures
The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction ($X$) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture ($n_{tot}$):
$x_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\label{6.6.5}$
The mole fraction is a dimensionless quantity between 0 and 1. If $x_A = 1.0$, then the sample is pure $A$, not a mixture. If $x_A = 0$, then no $A$ is present in the mixture. The sum of the mole fractions of all the components present must equal 1.
To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas $A$ to the total pressure of a gas mixture that contains $A$. We can use the ideal gas law to describe the pressures of both gas $A$ and the mixture: $P_A = n_ART/V$ and $P_{tot} = n_tRT/V$. The ratio of the two is thus
$\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=x_A \label{6.6.6}$
Rearranging this equation gives
$P_A = x_AP_{tot} \label{6.6.7}$
That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. By volume, Earth’s atmosphere is about 78% $N_2$, 21% $O_2$, and 0.9% $Ar$, with trace amounts of gases such as $CO_2$, $H_2O$, and others. This means that 78% of the particles present in the atmosphere are $N_2$; hence the mole fraction of $N_2$ is 78%/100% = 0.78. Similarly, the mole fractions of $O_2$ and $Ar$ are 0.21 and 0.009, respectively. Using Equation 6.6.7, we therefore know that the partial pressure of N2 is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of $O_2$ and $Ar$ are 0.21 and 0.009 atm, respectively.
Example $2$
We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we exhale yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air.
Inhaled Air / mmHg Exhaled Air / mmHg
$P_{\rm N_2}$ 597 568
$P_{\rm O_2}$ 158 116
$P_{\rm H_2O}$ 0.3 28
$P_{\rm CO_2}$ 5 48
$P_{\rm Ar}$ 8 8
$P_{tot}$ 767 767
Given: pressures of gases in inhaled and exhaled air
Asked for: mole fractions of gases in exhaled air
Strategy:
Calculate the mole fraction of each gas using Equation 6.6.7.
Solution:
The mole fraction of any gas $A$ is given by
$x_A=\dfrac{P_A}{P_{tot}}$
where $P_A$ is the partial pressure of $A$ and $P_{tot}$ is the total pressure. For example, the mole fraction of $CO_2$ is given as:
$x_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063$
The following table gives the values of $x_A$ for the gases in the exhaled air.
Gas Mole Fraction
${\rm N_2}$ 0.741
${\rm O_2}$ 0.151
${\rm H_2O}$ 0.037
${\rm CO_2}$ 0.063
${\rm Ar}$ 0.010
Exercise $2$
Venus is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO2 and 3% N2, with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO2 and N2.
Answer
$P_{\rm CO_2}=\rm86\; atm$, $P_{\rm N_2}=\rm2.7\;atm$
Dalton’s Law of Partial Pressures: https://youtu.be/y5-SbspyvBA
Summary
• The partial pressure of each gas in a mixture is proportional to its mole fraction.
The pressure exerted by each gas in a gas mixture (its partial pressure) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The mole fraction of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.6%3A_Mixtures_of_Gases.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases.
The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles.
A Molecular Description
The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates:
1. A gas is composed of a large number of particles called molecules (whether monatomic or polyatomic) that are in constant random motion.
2. Because the distance between gas molecules is much greater than the size of the molecules, the volume of the molecules is negligible.
3. Intermolecular interactions, whether repulsive or attractive, are so weak that they are also negligible.
4. Gas molecules collide with one another and with the walls of the container, but these collisions are perfectly elastic; that is, they do not change the average kinetic energy of the molecules.
5. The average kinetic energy of the molecules of any gas depends on only the temperature, and at a given temperature, all gaseous molecules have exactly the same average kinetic energy.
Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In Section 10.8, we explain how this theory must be modified to account for the behavior of real gases.
Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a momentum transfer (impulse) from molecules to the walls (Figure $2$).
The momentum transfer to the wall perpendicular to $x$ axis as a molecule with an initial velocity $u_x$ in $x$ direction hits is expressed as:
$\Delta p_x=2mu_x \label{6.7.1}$
The collision frequency, a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume.
$f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{6.7.2}$
The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency.
$P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{6.7.3}$
At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace $u_x^2$ in the expression above with the average value of $u_x^2$, which is denoted by $\overline{u_x^2}$. The overbar designates the average value over all molecules.
The exact expression for pressure is given as :
$P=\dfrac{N}{V}m\overline{u_x^2} \label{6.7.4}$
Finally, we must consider that there is nothing special about $x$ direction. We should expect that $\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}$. Here the quantity $\overline{u^2}$ is called the mean-square speed defined as the average value of square-speed ($u^2$) over all molecules. Since $u^2=u_x^2+u_y^2+u_z^2$ for each molecule, $\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}$. By substituting $\dfrac{1}{3}\overline{u^2}$ for $\overline{u_x^2}$ in the expression above, we can get the final expression for the pressure:
$P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{6.7.5}$
Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules.
Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the average translational kinetic energy of the molecules of a gas $(\overline{e_K})$, which can be represented as and states that at a given Kelvin temperature $(T)$, all gases have the same value of
$\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{6.7.6}$
where $N_A$ is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by $N_A$:
$N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{6.7.7}$
where $M$ is the molar mass of the gas molecules and is related to the molecular mass by $M=N_Am$.
By rearranging the equation, we can get the relationship between the root-mean square speed ($u_{\rm rms}$) and the temperature.
The rms speed ($u_{\rm rms}$) is the square root of the sum of the squared speeds divided by the number of particles:
$u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{6.7.8}$
where $N$ is the number of particles and $u_i$ is the speed of particle $i$.
The relationship between $u_{\rm rms}$ and the temperature is given by:
$u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{6.7.9}$
In this equation, $u_{\rm rms}$ has units of meters per second; consequently, the units of molar mass $M$ are kilograms per mole, temperature $T$ is expressed in kelvins, and the ideal gas constant $R$ has the value 8.3145 J/(K•mol).
The equation shows that $u_{\rm rms}$ of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones.
The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, not the same average speed. In contrast, the most probable speed (vp) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then Equation 6.7.8 tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and vp increases.
Note
At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed.
Example $1$
The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed ($v_{\rm av}$) root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$).
Given: particle speeds
Asked for: average speed ($v_{\rm av}$), root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$)
Strategy:
Use Equation 6.7.6 to calculate the average speed and Equation 6.7.8 to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving.
Solution:
The average speed is the sum of the speeds divided by the number of particles:
$v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s$
The rms speed is the square root of the sum of the squared speeds divided by the number of particles:
$v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s$
The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence $v_{\rm m}=6.0$ m/s. The $v_{\rm rms}$ of the particles, which is related to the average kinetic energy, is greater than their average speed.
Boltzmann Distributions
At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction, as you will learn in Chapter 14. This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure $1$. Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the value of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure $1$ were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases.
The Relationships among Pressure, Volume, and Temperature
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
• Pressure versus Volume: At constant temperature, the kinetic energy of the molecules of a gas and hence the rms speed remain unchanged. If a given gas sample is allowed to occupy a larger volume, then the speed of the molecules does not change, but the density of the gas (number of particles per unit volume) decreases, and the average distance between the molecules increases. Hence the molecules must, on average, travel farther between collisions. They therefore collide with one another and with the walls of their containers less often, leading to a decrease in pressure. Conversely, increasing the pressure forces the molecules closer together and increases the density, until the collective impact of the collisions of the molecules with the container walls just balances the applied pressure.
• Volume versus Temperature: Raising the temperature of a gas increases the average kinetic energy and therefore the rms speed (and the average speed) of the gas molecules. Hence as the temperature increases, the molecules collide with the walls of their containers more frequently and with greater force. This increases the pressure, unless the volume increases to reduce the pressure, as we have just seen. Thus an increase in temperature must be offset by an increase in volume for the net impact (pressure) of the gas molecules on the container walls to remain unchanged.
• Pressure of Gas Mixtures: Postulate 3 of the kinetic molecular theory of gases states that gas molecules exert no attractive or repulsive forces on one another. If the gaseous molecules do not interact, then the presence of one gas in a gas mixture will have no effect on the pressure exerted by another, and Dalton’s law of partial pressures holds.
Example $2$
The temperature of a 4.75 L container of N2 gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the
1. average kinetic energy of the N2 molecules?
2. rms speed of the N2 molecules?
3. average speed of the N2 molecules?
4. impact of each N2 molecule on the wall of the container during a collision with the wall?
5. total number of collisions per second of N2 molecules with the walls of the entire container?
6. number of collisions per second of N2 molecules with each square centimeter of the container wall?
7. pressure of the N2 gas?
Given: temperatures and volume
Asked for: effect of increase in temperature
Strategy:
Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas.
Solution:
1. Increasing the temperature increases the average kinetic energy of the N2 molecules.
2. An increase in average kinetic energy can be due only to an increase in the rms speed of the gas particles.
3. If the rms speed of the N2 molecules increases, the average speed also increases.
4. If, on average, the particles are moving faster, then they strike the container walls with more energy.
5. Because the particles are moving faster, they collide with the walls of the container more often per unit time.
6. The number of collisions per second of N2 molecules with each square centimeter of container wall increases because the total number of collisions has increased, but the volume occupied by the gas and hence the total area of the walls are unchanged.
7. The pressure exerted by the N2 gas increases when the temperature is increased at constant volume, as predicted by the ideal gas law.
Exercise $2$
A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the
1. average kinetic energy of the He atoms?
2. rms speed of the He atoms?
3. average speed of the He atoms?
4. impact of each He atom on the wall of the container during a collision with the wall?
5. total number of collisions per second of He atoms with the walls of the entire container?
6. number of collisions per second of He atoms with each square centimeter of the container wall?
7. pressure of the He gas?
Answer: a. no change; b. no change; c. no change; d. no change; e. decreases; f. decreases; g. decreases
Kinetic-Molecular Theory of Gases: https://youtu.be/9f83XAYfXAg
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T,$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}},$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}.$
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square (rms) speed (vrms). The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a Boltzmann distribution, in which some molecules have higher or lower speeds (and kinetic energies) than average. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.7%3A_Kinetic-Molecular_Theory_of_Gases.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
Diffusion and Effusion
As you have learned, the molecules of a gas are not stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment.
Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space.
The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as Graham’s law, after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses:
$\dfrac{\text{rate of effusion A}}{\text{rate of effusion B}}=\sqrt{\dfrac{M_B}{M_A}} \label{6.8.1}$
Note
At a given temperature, heavier molecules move more slowly than lighter molecules.
Example $1$
During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of $\ce{^{235}U}$. Naturally occurring uranium is only 0.720% $\ce{^{235}U}$, whereas most of the rest (99.275%) is $\ce{^{238}U}$, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound $UF_6$ (boiling point = 56°C).
1. Calculate the ratio of the rates of effusion of 235UF6 and 238UF6 for a single step in which UF6 is allowed to pass through a porous barrier. (The atomic mass of 235U is 235.04, and the atomic mass of 238U is 238.05.)
2. If n identical successive separation steps are used, the overall separation is given by the separation in a single step (in this case, the ratio of effusion rates) raised to the nth power. How many effusion steps are needed to obtain 99.0% pure 235UF6?
Given: isotopic content of naturally occurring uranium and atomic masses of 235U and 238U
Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure 235UF6
Strategy:
1. Calculate the molar masses of 235UF6 and 238UF6, and then use Graham’s law to determine the ratio of the effusion rates. Use this value to determine the isotopic content of 235UF6 after a single effusion step.
2. Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity. Use a logarithmic expression to compute the number of separation steps required.
Solution:
1. The molar mass of 238UF6 is 238.05 + (6)(18.998) = 352.04 g/mol
The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation 6.8.1:
$\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043$
2. rate U235F6rate U238F6=352.04349.03=1.0043
Thus passing UF6 containing a mixture of the two isotopes through a single porous barrier gives an enrichment of 1.0043, so after one step the isotopic content is (0.720%)(1.0043) = 0.723% 235UF6.
3. B To obtain 99.0% pure 235UF6 requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated: final purity = (initial purity)(separation)n
In this case, 0.990 = (0.00720)(1.0043)n, which can be rearranged to give
$1.0043^n=\dfrac{0.990}{0.00720}=137.50$
4. Taking the logarithm of both sides gives
$n\ln(1.0043)=\ln(137.50)$
$n=\dfrac{\ln(137.50)}{\ln(1.0043)}=1148$
Thus at least a thousand effusion steps are necessary to obtain highly enriched 235U. Figure $2$ shows a small part of a system that is used to prepare enriched uranium on a large scale.
Exercise $1$
Helium consists of two isotopes: 3He (natural abundance = 0.000134%) and 4He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant 4He by a process of gaseous effusion.
1. Calculate the ratio of the effusion rates of 3He and 4He and thus the enrichment possible in a single effusion step.
2. How many effusion steps are necessary to yield 99.0% pure 3He?
Answer: a. ratio of effusion rates = 1.15200; one step gives 0.000154% 3He; b. 96 steps
Rates of Diffusion or Effusion
Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses:
$KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{6.8.2}$
Multiplying both sides by 2 and rearranging give
$\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{6.8.3}$
Taking the square root of both sides gives
$\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{6.8.4}$
Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation 6.8.4 shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy.
Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure $3$ for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution, molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds.
The lightest gases have a wider distribution of speeds and the highest average speeds.
Note
Molecules with lower masses have a wider distribution of speeds and a higher average speed.
Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 1010 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure $4$.
The average distance traveled by a molecule between collisions is the mean free path. The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10−8 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 1010 m (about 6 million miles).
Note
The denser the gas, the shorter the mean free path.
Example $2$
Calculate the rms speed of a sample of cis-2-butene (C4H8) at 20°C.
Given: compound and temperature
Asked for: rms speed
Strategy:
Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 6.8.5 to calculate the rms speed of the gas.
Solution:
To use Equation 6.8.4, we need to calculate the molar mass of cis-2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C4H8, so its molar mass is 56.11 g/mol. Thus
$u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}=361\;m/s$
or approximately 810 mi/h.
Exercise $2$
Calculate the rms speed of a sample of radon gas at 23°C.
Answer: 1.82 × 102 m/s (about 410 mi/h)
The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. In Section 10.8, we explore how the behavior of real gases differs from that of ideal gases.
Graham’s law of Diffusion and Effusion: https://youtu.be/9HO-qgh-iGI
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T,$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}},$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}.$
• Graham’s law for effusion: $\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}$
Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The mean free path of a molecule is the average distance it travels between collisions. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.8%3A_Gas_Properties_Relating_to_the_Kinetic-Molecular_Theory.txt |
Learning Objectives
• To recognize the differences between the behavior of an ideal gas and a real gas.
• To understand how molecular volumes and intermolecular attractions cause the properties of real gases to deviate from those predicted by the ideal gas law.
The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases.
Pressure, Volume, and Temperature Relationships in Real Gases
For an ideal gas, a plot of $PV/nRT$ versus $P$ gives a horizontal line with an intercept of 1 on the $PV/nRT$ axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (part (a) in Figure $1$). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (part (b) in Figure $1$).
Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure $2$ for $N_2$. Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid.
Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure $3$). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.
Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is less than expected. Thus as shown in Figure $2$, at low temperatures, the ratio of $PV/nRT$ is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the $PV/nRT$ versus $P$ plot for many gases.
Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted.
At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid).
The van der Waals Equation
The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equation,
$\left(P + \dfrac{an^2}{V^2}\right) (V − nb)=nRT \label{6.9.1}$
a and b are empirical constants that are different for each gas. The values of $a$ and $b$ are listed in Table $1$ for several common gases.
Table $1$: van der Waals Constants for Some Common Gases (see Table A8 for more complete list)
Gas a ( (L2·atm)/mol2) b (L/mol)
He 0.03410 0.0238
Ne 0.205 0.0167
Ar 1.337 0.032
H2 0.2420 0.0265
N2 1.352 0.0387
O2 1.364 0.0319
Cl2 6.260 0.0542
NH3 4.170 0.0371
CH4 2.273 0.0430
CO2 3.610 0.0429
The pressure term in Equation $\ref{6.9.1}$ —$P + (an^2/V^2$)—corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, $n^2/V^2$ represents the concentration of the gas ($n/V$) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure $4$. The volume term—$V − nb$—corrects for the volume occupied by the gaseous molecules.
The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on V and P, respectively. Because nonzero molecular volumes produce a measured volume that is larger than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is less than that expected based on the ideal gas law, so the an2/V2 term must be added to the measured pressure to correct for these effects.
Example $1$
You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)?
Given: volume of cylinder, mass of compound, pressure, and temperature
Asked for: safety
Strategy:
A Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law.
B Obtain a and b values for Cl2 from Table $1$. Use the van der Waals equation to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture.
Solution:
A We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol):
$n=\dfrac{m}{M}=\rm\dfrac{500\;g}{70.906\;g/mol}=7.052\;mol$
Using the ideal gas law and the temperature in kelvins (298 K), we calculate the pressure:
$P=\dfrac{nRT}{V}=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L}=43.1\;atm$
If chlorine behaves like an ideal gas, you have a real problem!
B Now let’s use the van der Waals equation with the a and b values for Cl2 from Table $1$. Solving for P gives
$\begin{split}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\&=\rm28.2\;atm\end{split}$
This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation.
Exercise $1$
A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the
1. ideal gas law.
2. van der Waals equation.
Answer: a. 77 atm; b. 67 atm
Liquefaction of Gases
Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure).
Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals $a$ coefficients are relatively easy to liquefy because large a coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small a coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O2, N2, Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points.
A large value of a indicates the presence of relatively strong intermolecular attractive interactions.
The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek kryo, meaning “cold,” and genes, meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called cryosurgery, which selectively destroys tissues with a minimal loss of blood by the use of extreme cold.
Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure $5$). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles.
Summary
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of PV/nRT versus P at a given temperature; for an ideal gas, PV/nRT versus P = 1 under all conditions. At high pressures, most real gases exhibit larger PV/nRT values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit PV/nRT values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, liquefaction occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold cryogenic liquids. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.9%3A_Non-ideal_%28Real%29_Gases.txt |
• 7.1: Getting Started: Some Terminology
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work.
• 7.2: Heat
Thermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of “hot” or “cold.” When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is “hot.” When the atoms and molecules are moving slowly, they have lower KE, and we say that the object is “cold”
• 7.3: Heats of Reactions and Calorimetry
Calorimetry is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called calorimeters, which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system.
• 7.4: Work
One definition of energy is the capacity to do work. There are many kinds of work, including mechanical work, electrical work, and work against a gravitational or a magnetic field. Here we will consider only mechanical work and focus on the work done during changes in the pressure or the volume of a gas.
• 7.5: The First Law of Thermodynamics
The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (U) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal g
• 7.6: Heats of Reactions - ΔU and ΔH
Enthalpy is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the change in enthalpy (ΔH) can be measured. A negative ΔH means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings. For a chemical reaction, the enthalpy of reaction (ΔHrxn) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Revers
• 7.7: Indirect Determination of ΔH - Hess's Law
Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This principle is called Hess’s law, which allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known ΔH values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps.
• 7.8: Standard Enthalpies of Formation
The enthalpy of formation (ΔHf) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation (ΔHof) are determined under standard conditions: a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their standard states (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is de
• 7.9: Fuels as Sources of Energy
According to the law of conservation of energy, energy can never actually be “consumed”; it can only be changed from one form to another. What is consumed on a huge scale, however, are resources that can be readily converted to a form of energy that is useful for doing work. energy that is not used to perform work is either stored as potential energy for future use or transferred to the surroundings as heat.
07: Thermochemistry
Learning Objectives
• To understand the concept of energy and its various forms.
• To know the relationship between energy, work, and heat.
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work.
Forms of Energy
The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure $1$). Thermal energy results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperature of an object is a measure of its thermal energy content. Radiant energy is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energy results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energy is stored in the nucleus of an atom, and chemical energy is stored within a chemical compound because of a particular arrangement of atoms.
Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE), which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE) is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other.
Energy can be converted from one form to another (Figure $2$) or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to mechanical work to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction.
Although energy can be converted from one form to another, the total amount of energy in the universe remains constant. This is known as the law of conservation of energy: Energy cannot be created or destroyed.
Kinetic and Potential Energy
The kinetic energy of an object is related to its mass $m$ and velocity $v$:
$KE=\dfrac{1}{2}mv^2 \tag{7.1.4}$
For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is
$KE=\dfrac{1}{2}(1360 kg)(26.8 ms)^2= 4.88 \times 10^5 g \cdot m^2 \tag{7.1.5}$
Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J), is named after the British physicist James Joule (1818–1889), an early worker in the field of energy. is defined as 1 kilogram·meter2/second2 (kg·m2/s2). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 103 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 105 J or 4.88 × 102 kJ. It is important to remember that the units of energy are the same regardless of the form of energy, whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same.
To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is w = Fd. According to Equation 7.1.2, the force (F) exerted by gravity on any object is equal to its mass (m, in this case, 1360 kg) times the acceleration (a) due to gravity (g, 9.81 m/s2 at Earth’s surface). The distance (d) is the height (h) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows:
$PE= F\;d = m\,a\;d = m\,g\,h \tag{7.1.6a}$
$PE=(1360, Kg)\left(\dfrac{9.81\, m}{s^2}\right)(36.6\;m) = 4.88 \times 10^5\; \frac{Kg \cdot m}{s^2} \tag{7.1.6b}$
$=4.88 \times 10^5 J = 488\; kJ \tag{7.1.6c}$
The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h.
If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero.
Potential energy is usually defined relative to an arbitrary standard position (in this case, the street was assigned an elevation of zero). As a result, we usually calculate only differences in potential energy: in this case, the difference between the potential energy of the car on the top level of the parking garage and the potential energy of the same car on the street at the base of the garage.
Units of Energy
The units of energy are the same for all forms of energy. Energy can also be expressed in the non-SI units of calories (cal), where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5°C to 15.5°C.We specify the exact temperatures because the amount of energy needed to raise the temperature of 1 g of water 1°C varies slightly with elevation. To three significant figures, however, this amount is 1.00 cal over the temperature range 0°C–100°C. The name is derived from the Latin calor, meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule:
$1 \;cal = 4.184 \;J \;\text{exactly} \tag{7.1.7a}$
$1 \;J = 0.2390\; cal \tag{7.1.7b}$
In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information.
Example $1$
1. If the mass of a baseball is 149 g, what is the kinetic energy of a fastball clocked at 100 mi/h?
2. A batter hits a pop fly, and the baseball (with a mass of 149 g) reaches an altitude of 250 ft. If we assume that the ball was 3 ft above home plate when hit by the batter, what is the increase in its potential energy?
Given: mass and velocity or height
Asked for: kinetic and potential energy
Strategy:
Use Equation 7.1.4 to calculate the kinetic energy and Equation 7.1.6 to calculate the potential energy, as appropriate.
Solution:
1. The kinetic energy of an object is given by $\frac{1}{2} mv^2$ In this case, we know both the mass and the velocity, but we must convert the velocity to SI units: $v= \left(\dfrac{100\; \cancel{mi}}{1\;\cancel{h}} \right) \left(\dfrac{1 \;\cancel{h}}{60 \;\cancel{min}} \right) \left(\dfrac{1 \; \cancel{min}}{60 \;s} \right)\left(\dfrac{1.61\; \cancel{km}}{1 \;\cancel{mi}} \right) (\dfrac{1000\; m}{1\; \cancel{km}})= 44.7 \;m/s$
The kinetic energy of the baseball is therefore $KE= 1492 \;\cancel{g} \left(\dfrac{1\; kg}{1000 \;\cancel{g}} \right) \left(\dfrac{44.7 \;m}{s} \right)^2= 1.49 \times 10^2 \dfrac{kg⋅m^2}{s^2}= 1.49 \times10^2\; J$
2. The increase in potential energy is the same as the amount of work required to raise the ball to its new altitude, which is (250 − 3) = 247 feet above its initial position. Thus $PE= 149\;\cancel{g} \left(\dfrac{1\; kg}{1000\; \cancel{g}} \right)\left(\dfrac{9.81\; m}{s^2} \right) \left(247\; \cancel{ft} \right) \left(\dfrac{0.3048\; m}{1 \;\cancel{ft}} \right)= 1.10 \times 10^2 \dfrac{kg⋅m^2}{s^2}= 1.10 \times 10^2\; J$
Exercise
1. In a bowling alley, the distance from the foul line to the head pin is 59 ft, 10 13/16 in. (18.26 m). If a 16 lb (7.3 kg) bowling ball takes 2.0 s to reach the head pin, what is its kinetic energy at impact? (Assume its speed is constant.)
2. What is the potential energy of a 16 lb bowling ball held 3.0 ft above your foot?
Answer
1. 3.10 × 102 J
2. 65 J
Summary
• All forms of energy can be interconverted. Three things can change the energy of an object: the transfer of heat, work performed on or by an object, or some combination of heat and work.
Thermochemistry is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. Energy is the capacity to do work. Mechanical work is the amount of energy required to move an object a given distance when opposed by a force. Thermal energy is due to the random motions of atoms, molecules, or ions in a substance. The temperature of an object is a measure of the amount of thermal energy it contains. Heat (q) is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of potential energy (PE), energy caused by the relative position or orientation of an object. Kinetic energy (KE) is the energy an object possesses due to its motion. Energy can be converted from one form to another, but the law of conservation of energy states that energy can be neither created nor destroyed. The most common units of energy are the joule (J), defined as 1 (kg·m2)/s2, and the calorie, defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J). | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.1%3A_Getting_Started%3A_Some_Terminology.txt |
Learning Objectives
• Distinguish the related properties of heat, thermal energy, and temperature
• Define and distinguish specific heat and heat capacity, and describe the physical implications of both
• Perform calculations involving heat, specific heat, and temperature change
Thermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of “hot” or “cold.” When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is “hot.” When the atoms and molecules are moving slowly, they have lower KE, and we say that the object is “cold” (Figure $1$). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease.
Most substances expand as their temperature increases and contract as their temperature decreases. This property can be used to measure temperature changes, as shown in Figure $2$. The operation of many thermometers depends on the expansion and contraction of substances in response to temperature changes.
Heat (q) is the transfer of thermal energy between two bodies at different temperatures. Heat flow (a redundant term, but one commonly used) increases the thermal energy of one body and decreases the thermal energy of the other. Suppose we initially have a high temperature (and high thermal energy) substance (H) and a low temperature (and low thermal energy) substance (L). The atoms and molecules in H have a higher average KE than those in L. If we place substance H in contact with substance L, the thermal energy will flow spontaneously from substance H to substance L. The temperature of substance H will decrease, as will the average KE of its molecules; the temperature of substance L will increase, along with the average KE of its molecules. Heat flow will continue until the two substances are at the same temperature (Figure $3$).
Matter undergoing chemical reactions and physical changes can release or absorb heat. A change that releases heat is called an exothermic process. For example, the combustion reaction that occurs when using an oxyacetylene torch is an exothermic process—this process also releases energy in the form of light as evidenced by the torch’s flame (Figure $4$a). A reaction or change that absorbs heat is an endothermic process. A cold pack used to treat muscle strains provides an example of an endothermic process. When the substances in the cold pack (water and a salt like ammonium nitrate) are brought together, the resulting process absorbs heat, leading to the sensation of cold.
Historically, energy was measured in units of calories (cal). A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to 1 kg m2/s2, which is also called 1 newton–meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules.
Heat Capacity
We now introduce two concepts useful in describing heat flow and temperature change. The heat capacity (C) of a body of matter is the quantity of heat (q) it absorbs or releases when it experiences a temperature change (ΔT) of 1 degree Celsius (or equivalently, 1 kelvin)
$C=\dfrac{q}{ΔT} \label{7.2.1}$
Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property—its value is proportional to the amount of the substance.
For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast iron frying pan is found by observing that it takes 18,150 J of energy to raise the temperature of the pan by 50.0 °C
$C_{\text{small pan}}=\dfrac{18,140 J}{50.0\; °C} =363\; J/°C \label{7.2.2}$
The larger cast iron frying pan, while made of the same substance, requires 90,700 J of energy to raise its temperature by 50.0 °C. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change:
$C_{\text{large pan}}=\dfrac{90,700\; J}{50.0\;°C}=1814\; J/°C \label{7.2.3}$
The specific heat capacity (c) of a substance, commonly called its “specific heat,” is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin):
$c = \dfrac{q}{m\Delta T} \label{7.2.4}$
Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property—the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g. The specific heat of iron (the material used to make the pan) is therefore:
$c_{iron}=\dfrac{18,140\; J}{(808\; g)(50.0\;°C)} = 0.449\; J/g\; °C \label{7.2.5}$
The large frying pan has a mass of 4040 g. Using the data for this pan, we can also calculate the specific heat of iron:
$c_{iron}=\dfrac{90,700 J}{(4,040\; g)(50.0\;°C)}=0.449\; J/g\; °C \label{7.2.6}$
Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of J/mol °C (Figure $5$).
The heat capacity of an object depends on both its mass and its composition. For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The molar heat capacity (Cp) is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of Cp are thus J/(mol•°C).The subscript p indicates that the value was measured at constant pressure. The specific heat (Cs) is the amount of energy needed to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C).
We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change either via moles (Equation 7.3.2) or mass (Equation 7.3.3):
$q = nC_pΔT \label{7.3.2}$
where
• $n$ is the number of moles of substance and
• $C_p$ is the molar heat capacity (i.e., heat capacity per mole of substance)
$q = mC_sΔT \label{7.3.3}$
where
The specific heats of some common substances are given in Table $1$. Note that the specific heat values of most solids are less than 1 J/(g•°C), whereas those of most liquids are about 2 J/(g•°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g•°C), is one of the highest known.
Liquid water has a relatively high specific heat (about 4.2 J/g °C); most metals have much lower specific heats (usually less than 1 J/g °C). The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table $1$.
Table $1$: Specific Heats of Common Substances at 25 °C and 1 bar
Substance Symbol (state) Specific Heat (J/g °C)
helium He(g) 5.193
water H2O(l) 4.184
ethanol C2H6O(l) 2.376
ice H2O(s) 2.093 (at −10 °C)
water vapor H2O(g) 1.864
nitrogen N2(g) 1.040
air mixture 1.007
oxygen O2(g) 0.918
aluminum Al(s) 0.897
carbon dioxide CO2(g) 0.853
argon Ar(g) 0.522
iron Fe(s) 0.449
copper Cu(s) 0.385
lead Pb(s) 0.130
gold Au(s) 0.129
silicon Si(s) 0.712
The value of C is intrinsically a positive number, but ΔT and q can be either positive or negative, and they both must have the same sign. If ΔT and q are positive, then heat flows from the surroundings into an object. If ΔT and q are negative, then heat flows from an object into its surroundings.
If we know the mass of a substance and its specific heat, we can determine the amount of heat, q, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost:
\begin{align}q&=\ce{(specific\: heat)×(mass\: of\: substance)×(temperature\: change)}\label{7.2.7}\q&=c×m×ΔT=c×m×(T_\ce{final}−T_\ce{initial})\end{align}
In this equation, $c$ is the specific heat of the substance, m is its mass, and ΔT (which is read “delta T”) is the temperature change, TfinalTinitial. If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, TfinalTinitial has a positive value, and the value of q is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, TfinalTinitial has a negative value, and the value of q is negative.
The Movement of Heat in a Substance: https://youtu.be/gaJQYke-lVY
Example $1$: Measuring Heat
A flask containing $8.0 \times 10^2\; g$ of water is heated, and the temperature of the water increases from 21 °C to 85 °C. How much heat did the water absorb?
Solution
To answer this question, consider these factors:
• the specific heat of the substance being heated (in this case, water)
• the amount of substance being heated (in this case, 800 g)
• the magnitude of the temperature change (in this case, from 21 °C to 85 °C).
The specific heat of water is 4.184 J/g °C, so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C).
This can be summarized using the equation:
\begin{align} q &=c×m×ΔT=c\times m \times (T_\ce{final}−T_\ce{initial}) \ &=\mathrm{(4.184\:J/\cancel{g}°C)×(800\:\cancel{g})×(85−20)°C}\ &=\mathrm{(4.184\:J/\cancel{g}°\cancel{C})×(800\:\cancel{g})×(65)°\cancel{C}}\ &=\mathrm{210,000\: J(=210\: kJ)} \end{align}
Because the temperature increased, the water absorbed heat and $q$ is positive.
Exercise $1$
How much heat, in joules, must be added to a $5.00 \times 10^2 \;g$ iron skillet to increase its temperature from 25 °C to 250 °C? The specific heat of iron is 0.451 J/g °C.
Answer:
$5.05 \times 10^4\; J$
Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced.
Example $2$: Determining Other Quantities
A piece of unknown metal weighs 348 g. When the metal piece absorbs 6.64 kJ of heat, its temperature increases from 22.4 °C to 43.6 °C. Determine the specific heat of this metal (which might provide a clue to its identity).
Solution
Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship:
$q=c \times m \times \Delta T=c \times m \times (T_{final}−T_{initial})$
Substituting the known values:
$6,640\; J=c \times (348\; g) \times (43.6 − 22.4)\; °C$
Solving:
$c=\dfrac{6,640\; J}{(348\; g) \times (21.2°C)} =0.900\; J/g\; °C$
Comparing this value with the values in Table $1$, this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum.
Exercise $2$
A piece of unknown metal weighs 217 g. When the metal piece absorbs 1.43 kJ of heat, its temperature increases from 24.5 °C to 39.1 °C. Determine the specific heat of this metal, and predict its identity.
Answer
$c = 0.45 \;J/g \;°C$; the metal is likely to be iron from checking Tabel 7.2.1
Example $3$: Solar Heating
A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0°C. During the course of the day, the temperature of the water rises to 38.0°C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0°C is 0.998 g/mL.)
Given: volume and density of water and initial and final temperatures
Asked for: amount of energy stored
Strategy:
1. Use the density of water at 22.0°C to obtain the mass of water (m) that corresponds to 400 L of water. Then compute ΔT for the water.
2. Determine the amount of heat absorbed by substituting values for m, Cs, and ΔT into Equation 7.3.1.
Solution:
A The mass of water is
$mass \; of \; H_{2}O=400 \; \cancel{L}\left ( \dfrac{1000 \; \cancel{mL}}{1 \; \cancel{L}} \right ) \left ( \dfrac{0.998 \; g}{1 \; \cancel{mL}} \right ) = 3.99\times 10^{5}g\; H_{2}O$
The temperature change (ΔT) is 38.0°C − 22.0°C = +16.0°C.
B From Table $1$, the specific heat of water is 4.184 J/(g•°C). From Equation 7.3.3, the heat absorbed by the water is thus
$q=mC_{s}\Delta T=\left ( 3.99X10^{5} \; \cancel{g} \right )\left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right ) \left ( 16.0 \; \cancel{^{o}C} \right ) = 2.67 \times 10^{7}J = 2.67 \times 10^{4}kJ$
Both q and ΔT are positive, consistent with the fact that the water has absorbed energy.
Exercise $3$: Solar Heating
Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0°C to 34.5°C during the day? Assume that the specific heat of sandstone is the same as that of quartz (SiO2) in Table $1$.
Answer
2.7 × 104 kJ (Even though the mass of sandstone is more than six times the mass of the water in Example $1$, the amount of thermal energy stored is the same to two significant figures.)
When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process:
$q_{cold} + q_{hot} = 0 \label{7.3.4}$
The equation implies that the amount of heat that flows from a warmer object is the same as the amount of heat that flows into a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite:
$q_{cold} = −q_{hot} \label{7.3.5}$
Thus heat is conserved in any such process, consistent with the law of conservation of energy.
The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object.
Substituting for q from Equation 7.3.2 gives
$\left [ mC_{s} \Delta T \right ] _{cold} + \left [ mC_{s} \Delta T \right ] _{hot}=0 \label{7.3.6}$
which can be rearranged to give
$\left [ mC_{s} \Delta T \right ] _{cold} = - \left [ mC_{s} \Delta T \right ] _{hot} \label{7.3.7}$
When two objects initially at different temperatures are placed in contact, we can use Equation 7.3.7 to calculate the final temperature if we know the chemical composition and mass of the objects.
Example $2$: Thermal Equilibration of Copper and Water
If a 30.0 g piece of copper pipe at 80.0°C is placed in 100.0 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings.
Given: mass and initial temperature of two objects
Asked for: final temperature
Strategy: Using Equation 7.3.7 and writing ΔT as TfinalTinitial for both the copper and the water, substitute the appropriate values of m, Cs, and Tinitial into the equation and solve for Tfinal.
Solution
We can adapt Equation 7.3.7 to solve this problem, remembering that ΔT is defined as TfinalTinitial:
$\left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0$
Substituting the data provided in the problem and Table $1$ gives
$\left [ \left (30 \; g \right ) \left (0.385 \; J \right ) \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0$
$T_{final}\left ( 11.6 \; J/ ^{o}C \right ) -924 \; J + T_{final}\left ( 418.4 \; J/ ^{o}C \right ) -11,300 \; J$
$T_{final}\left ( 430 \; J/\left ( g\cdot ^{o}C \right ) \right ) = -12,224 \; J$
$T_{final} = -28.4 \; ^{o}C$
Exercise $2$a: Thermal Equilibration of Gold and Water
If a 14.0 g chunk of gold at 20.0°C is dropped into 25.0 g of water at 80.0°C, what is the final temperature if no heat is transferred to the surroundings?
Answer: 80.0°C
Exercise $2$b: Thermal Equilibration of Aluminum and Water
A 28.0 g chunk of aluminum is dropped into 100.0 g of water with an initial temperature of 20.0°C. If the final temperature of the water is 24.0°C, what was the initial temperature of the aluminum? (Assume that no heat is transferred to the surroundings.)
Answer: 90.6°C
Conservation of Energy: The Movement of Heat between Substances: https://youtu.be/pGEYy-pNHBg | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.2%3A_Heat.txt |
Learning Objectives
• Explain the technique of calorimetry
• Calculate and interpret heat and related properties using typical calorimetry data
• To use calorimetric data to calculate enthalpy changes.
Measuring Heat Flow
One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section.
A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process.
The thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat (qrxn < 0), then heat is absorbed by the calorimeter (qcalorimeter > 0) and its temperature increases. Conversely, if the reaction absorbs heat (qrxn > 0), then heat is transferred from the calorimeter to the system (qcalorimeter < 0) and the temperature of the calorimeter decreases. In both cases, the amount of heat absorbed or released by the calorimeter is equal in magnitude and opposite in sign to the amount of heat produced or consumed by the reaction. The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants.
Note
The amount of heat absorbed or released by the calorimeter is equal in magnitude and opposite in sign to the amount of heat produced or consumed by the reaction.
Constant-Pressure Calorimetry
Because ΔH is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter (a device used to measure enthalpy changes in chemical processes at constant pressure) give ΔH values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (Figure $2$), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10−6°C).
Before we practice calorimetry problems involving chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure $4$). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero:
$q_\mathrm{\,substance\: M} + q_\mathrm{\,substance\: W}=0 \label{7.3.1}$
This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:
$q_\mathrm{\,substance\: M}=-q_\mathrm{\,substance\: W} \label{7.3.2}$
The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that $q_{substance\; M}$ and $q_{substance\; W}$ are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, qsubstance M is a negative value and qsubstance W is positive, since heat is transferred from M to W.
Heat Transfer between Substances at Different Temperatures
A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table T4), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).
Solution
The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or:
$q_\ce{rebar}=−q_\ce{water} \label{5.3.3}$
Since we know how heat is related to other measurable quantities, we have:
$(c×m×ΔT)_\ce{rebar}=−(c×m×ΔT)_\ce{water} \label{5.3.4}$
Letting f = final and i = initial, in expanded form, this becomes:
$c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \label{5.3.5}$
The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:
$\mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar})=(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} \label{5.3.6a}$
$\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C} \label{5.3.6b}$
Solving this gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C.
Exercise $\PageIndex{1A}$
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.
Answer:
The initial temperature of the copper was 335.6 °C.
Exercise $\PageIndex{1B}$
A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.
Answer:
The final temperature (reached by both copper and water) is 38.8 °C.
This method can also be used to determine other quantities, such as the specific heat of an unknown metal.
Identifying a Metal by Measuring Specific Heat
A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.
Solution
Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or:
$q_\ce{metal}=−q_\ce{water} \label{5.3.7}$
In expanded form, this is:
$c_\ce{metal}×m_\ce{metal}×(T_\mathrm{f,metal}−T_\mathrm{i, metal})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \label{5.3.8}$
Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:
$\mathrm{(\mathit c_{metal})(59.7\:g)(28.5°C−100.0°C)=−(4.18\:J/g\: °C)(60.0\:g)(28.5°C−22.0°C)} \label{5.3.9}$
Solving this:
$\mathrm{\mathit c_{metal}=\dfrac{−(4.184\:J/g\: °C)(60.0\:g)(6.5°C)}{(59.7\:g)(−71.5°C)}=0.38\:J/g\: °C}$
Comparing this with values in Table T4, our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.
Exercise $2$
A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).
Answer
$c_{metal}= 0.13 \;J/g\; °C$
This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.
When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), qreaction, plus the heat absorbed or lost by the solution (the “surroundings”), qsolution, must add up to zero:
$q_\ce{reaction}+q_\ce{solution}=0\ \label{ $3$}$
This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:
$q_\ce{reaction}=−q_\ce{solution} \label{$4$}$
This concept lies at the heart of all calorimetry problems and calculations. Because the heat released or absorbed at constant pressure is equal to ΔH, the relationship between heat and ΔHrxn is
$\Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mC_{s} \Delta T \label{$5$}$
The use of a constant-pressure calorimeter is illustrated in Example $3$.
Example $3$
When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm3. What is ΔHsoln (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water.
Given: mass of substance, volume of solvent, and initial and final temperatures
Asked for: ΔHsoln
Strategy:
1. Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution.
2. Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation $1$.
3. Use the molar mass of KOH to calculate ΔHsoln.
Solution:
A To calculate ΔHsoln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is
$\left (100.0 \; mL\; H2O \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g$
The temperature change is (34.7°C − 23.0°C) = +11.7°C.
B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus
$q_{calorimater}=mC_{s} \Delta T =\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right )=5130 \; J =5.13 \; lJ$
The temperature of the solution increased because heat was absorbed by the solution (q > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation $1$, we see that
ΔHrxn = −qcalorimeter = −5.13 kJ
This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic.
C The last step is to use the molar mass of KOH to calculate ΔHsoln—the heat released when dissolving 1 mol of KOH:
$\Delta H_{soln}= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right )=-57.2 \; kJ/mol$
Exercise $3$
A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example $3$, find ΔHsoln for NH4Br (in kilojoules per mole).
Answer: 16.6 kJ/mol
Conservation of Energy: Coffee Cup Calorimetry: https://youtu.be/FwQcc17PN0k
Constant-Volume Calorimetry
Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter (A device used to measure energy changes in chemical processes. shown schematically in Figure $3$). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated.
Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. The heat released by a reaction carried out at constant volume is identical to the change in internal energy (ΔU) rather than the enthalpy change (ΔH); ΔU is related to ΔH by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that ΔU < ΔH, the relationship between the measured temperature change and ΔHcomb is given in Equation \ref{7.3.6}, where Cbomb is the total heat capacity of the steel bomb and the water surrounding it:
$\Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{7.3.6}$
To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C6H5CO2H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its ΔHcomb = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used in Equation \ref{5.42} to determine Cbomb. The use of a bomb calorimeter to measure the ΔHcomb of a substance is illustrated in Example $4$.
Video $1$: Video of view how a bomb calorimeter is prepared for action.
Example $4$: Combustion of Glucose
The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the ΔHcomb of glucose?
Given: mass and ΔT for combustion of standard and sample
Asked for: ΔHcomb of glucose
Strategy:
1. Calculate the value of qrxn for benzoic acid by multiplying the mass of benzoic acid by its ΔHcomb. Then use Equation $2$ to determine the heat capacity of the calorimeter (Cbomb) from qcomb and ΔT.
2. Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the ΔHcomb of glucose by multiplying the amount of heat released per gram by the molar mass of glucose.
Solution:
The first step is to use Equation $2$ and the information obtained from the combustion of benzoic acid to calculate Cbomb. We are given ΔT, and we can calculate qcomb from the mass of benzoic acid:
$q_{comb} = \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) = - 15.3 \; kJ$
From Equation $2$,
$-C_{bomb} = \dfrac{q_{comb}}{\Delta T} = \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} =- 7.34 \; kJ/^{o}C$
B According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose:
$q_{comb}=-C_{bomb}\Delta T = \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right )=- 26.7 \; kJ$
Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the ΔHcomb of glucose is
$\Delta H_{comb}=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right )=-2780 \; kJ/mol =2.78 \times 10^{3} \; kJ/mol$
This result is in good agreement (< 1% error) with the value of ΔHcomb = −2803 kJ/mol that calculated using enthalpies of formation.
Exercise $4$: Combustion of Benzoic Acid
When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH3NHNH2) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the ΔHcomb of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle.
Answer: −1.30 × 103 kJ/mol
Conservation of Energy: Bomb Calorimetry: https://youtu.be/SSNZGgwYBsQ
Summary
• Calorimetry measures enthalpy changes during chemical processes, where the magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system.
Calorimetry is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called calorimeters, which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The heat capacity (C) of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The specific heat (Cs) of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the molar heat capacity (Cp) is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a constant-pressure calorimeter, which gives ΔH values directly, or a bomb calorimeter, which operates at constant volume and is particularly useful for measuring enthalpies of combustion.
Thermal energy itself cannot be measured easily, but the temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry describes a set of techniques employed to measure enthalpy changes in chemical processes using devices called calorimeters. To have any meaning, the quantity that is actually measured in a calorimetric experiment, the change in the temperature of the device, must be related to the heat evolved or consumed in a chemical reaction. We begin this section by explaining how the flow of thermal energy affects the temperature of an object. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.3%3A_Heats_of_Reactions_and_Calorimetry.txt |
Learning Objectives
• To know the relationship between energy, work, and heat.
One definition of energy is the capacity to do work. There are many kinds of work, including mechanical work, electrical work, and work against a gravitational or a magnetic field. Here we will consider only mechanical work and focus on the work done during changes in the pressure or the volume of a gas.
Mechanical Work
The easiest form of work to visualize is mechanical work (Figure $1$), which is the energy required to move an object a distance d when opposed by a force F, such as gravity:
$w=F\,d \label{7.4.1}$
with
• $w$ is work
• $F$ is opposing force
• $d$ is distance
Because the force (F) that opposes the action is equal to the mass (m) of the object times its acceleration ($a$), Equation \ref{7.4.1} can be rewritten to:
$w = m\,a\,d \label{7.4.2}$
with
• $w$ is work
• $m$ is mass
• $a$ is a acceleration, and
• $d$ is distance
Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth. Hence for works against gravity (on Earth), $a$ can be set to $g=9.8\; m/s^2)$. Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the opposing force of gravity. The amount of work done (w) and thus the energy required depends on three things:
1. the height of the second floor (the distance $d$);
2. your mass, which must be raised that distance against the downward acceleration due to gravity; and
3. your path.
Pressure-Volume (PV) Work
To describe this pressure–volume work (PV work), we will use such imaginary oddities as frictionless pistons, which involve no component of resistance, and ideal gases, which have no attractive or repulsive interactions. Imagine, for example, an ideal gas, confined by a frictionless piston, with internal pressure Pint and initial volume Vi (Figure $2$). If $P_{ext} = P_{int}$, the system is at equilibrium; the piston does not move, and no work is done. If the external pressure on the piston (Pext) is less than Pint, however, then the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings; that is, the final volume (Vf) will be greater than $V_i$. If $P_{ext} > P_{int}$, then the gas will be compressed, and the surroundings will perform work on the system.
If the piston has cross-sectional area $A$, the external pressure exerted by the piston is, by definition, the force per unit area:
$P_{ext} = \dfrac{F}{A}$
The volume of any three-dimensional object with parallel sides (such as a cylinder) is the cross-sectional area times the height (V = Ah). Rearranging to give F = PextA and defining the distance the piston moves (d) as Δh, we can calculate the magnitude of the work performed by the piston by substituting into Equation 7.4.1:
$w = F d = P_{ext}AΔh \label{7.4.3}$
The change in the volume of the cylinder (ΔV) as the piston moves a distance d is ΔV = AΔh, as shown in Figure $3$. The work performed is thus
$w = P_{ext}ΔV \label{7.4.4}$
The units of work obtained using this definition are correct for energy: pressure is force per unit area (newton/m2) and volume has units of cubic meters, so
$w=\left(\dfrac{F}{A}\right)_{\textrm{ext}}(\Delta V)=\dfrac{\textrm{newton}}{\textrm m^2}\times \textrm m^3=\mathrm{newton\cdot m}=\textrm{joule}$
If we use atmospheres for P and liters for V, we obtain units of L·atm for work. These units correspond to units of energy, as shown in the different values of the ideal gas constant R:
$R=\dfrac{0.08206\;\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}=\dfrac{8.314\textrm{ J}}{\mathrm{mol\cdot K}}$
Thus 0.08206 L·atm = 8.314 J and 1 L·atm = 101.3 J.
Whether work is defined as having a positive sign or a negative sign is a matter of convention. Heat flow is defined from a system to its surroundings as negative; using that same sign convention, we define work done by a system on its surroundings as having a negative sign because it results in a transfer of energy from a system to its surroundings. This is an arbitrary convention and one that is not universally used. Some engineering disciplines are more interested in the work done on the surroundings than in the work done by the system and therefore use the opposite convention. Because ΔV > 0 for an expansion, Equation 7.4.4 must be written with a negative sign to describe PV work done by the system as negative:
$w = −P_{ext}ΔV \label{7.4.5}$
The work done by a gas expanding against an external pressure is therefore negative, corresponding to work done by a system on its surroundings. Conversely, when a gas is compressed by an external pressure, ΔV < 0 and the work is positive because work is being done on a system by its surroundings.
Note: A Matter of Convention
• Heat flow is defined from the system to its surroundings as negative
• Work is defined as by the system on its surroundings as negative
Suppose, for example, that the system under study is a mass of steam heated by the combustion of several hundred pounds of coal and enclosed within a cylinder housing a piston attached to the crankshaft of a large steam engine. The gas is not ideal, and the cylinder is not frictionless. Nonetheless, as steam enters the engine chamber and the expanding gas pushes against the piston, the piston moves, so useful work is performed. In fact, PV work launched the Industrial Revolution of the 19th century and powers the internal combustion engine on which most of us still rely for transportation.
In contrast to internal energy, work is not a state function. We can see this by examining Figure $4$, in which two different, two-step pathways take a gaseous system from an initial state to a final state with corresponding changes in temperature. In pathway A, the volume of a gas is initially increased while its pressure stays constant (step 1); then its pressure is decreased while the volume remains constant (step 2). In pathway B, the order of the steps is reversed. The temperatures, pressures, and volumes of the initial and final states are identical in both cases, but the amount of work done, indicated by the shaded areas in the figure, is substantially different. As we can see, the amount of work done depends on the pathway taken from ($V_1$, $P_1$) to ($V_2$, $P_2$), which means that work is not a state function.
Note
Internal energy is a state function, whereas work is not.
Example $1$
A small high-performance internal combustion engine has six cylinders with a total nominal displacement (volume) of 2.40 L and a 10:1 compression ratio (meaning that the volume of each cylinder decreases by a factor of 10 when the piston compresses the air–gas mixture inside the cylinder prior to ignition). How much work in joules is done when a gas in one cylinder of the engine expands at constant temperature against an opposing pressure of 40.0 atm during the engine cycle? Assume that the gas is ideal, the piston is frictionless, and no energy is lost as heat.
Given: final volume, compression ratio, and external pressure
Asked for: work done
Strategy:
1. Calculate the final volume of gas in a single cylinder. Then compute the initial volume of gas in a single cylinder from the compression ratio.
2. Use Equation 7.4.5 to calculate the work done in liter-atmospheres. Convert from liter-atmospheres to joules.
Solution:
A To calculate the work done, we need to know the initial and final volumes. The final volume is the volume of one of the six cylinders with the piston all the way down: Vf = 2.40 L/6 = 0.400 L. With a 10:1 compression ratio, the volume of the same cylinder with the piston all the way up is Vi = 0.400 L/10 = 0.0400 L. Work is done by the system on its surroundings, so work is negative.
w = −PextΔV = −(40.0 atm)(0.400 L − 0.0400 L) = −14.4 L·atm
Converting from liter-atmospheres to joules,
$w=-(14.4\;\mathrm{L\cdot atm})[101.3\;\mathrm{J/(L\cdot atm)}]=-1.46\times10^3\textrm{ J}$
In the following exercise, you will see that the concept of work is not confined to engines and pistons. It is found in other applications as well.
Exercise $1$
Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm. How much work in liter-atmospheres and joules was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work in joules did he require to take a breath while exercising?
Answer: −0.500 L·atm, or −50.7 J; −304 J; if he takes a breath every three seconds, this corresponds to 1.4 Calories per minute (1.4 kcal).
Work and Chemical Reactions
We have stated that the change in energy (ΔU) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called pressure-volume work, (or just PV work). Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:
$Cu_{(s)} + 4HNO_{3(aq)} \rightarrow Cu(NO_{3})_{2(aq)} + 2H_{2}O_{(l)} + 2NO_{2(g)}$
If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure $5$). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by movement of the piston (ΔV). At a constant external pressure (here, atmospheric pressure)
$w = −PΔV \label{7.4.6}$
The negative sign associated with PV work done indicates that the system loses energy. If the volume increases at constant pressure (ΔV > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (ΔV < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.
The symbol $U$ in Equation 5.2.2 represents the internal energy of a system, which is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy (H) (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy $U$ plus the product of its pressure P and volume V:
$H =U + PV \label{7.4.7}$
Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function.
If a chemical change occurs at constant pressure (i.e., for a given P, ΔP = 0), the change in enthalpy (ΔH) is
$ΔH = Δ(U + PV) = ΔU + ΔPV = ΔU + PΔV \label{7.4.8}$
Substituting q + w for ΔU (Equation 5.2.2) and −w for PΔV (Equation 7.4.6), we obtain
$ΔH = ΔU + PΔV = q_p + w − w = q_p \label{7.4.9}$
The subscript $p$ is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation 7.4.9 we see that at constant pressure the change in enthalpy, ΔH of the system, defined as HfinalHinitial, is equal to the heat gained or lost.
$ΔH = H_{final} − H_{initial} = q_p \label{7.4.10}$
Just as with ΔU, because enthalpy is a state function, the magnitude of ΔH depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.
Note
To find $ΔH$ for a reaction, measure $q_p$.
Summary
• All forms of energy can be interconverted. Three things can change the energy of an object: the transfer of heat, work performed on or by an object, or some combination of heat and work.
Problems
1. How much work is done by a gas that expands from 2 liters to 5 liters against an external pressure of 750 mmHg?
2. How much work is done by 0.54 moles of a gas that has an initial volume of 8 liters and expands under the following conditions: 30 oC and 1.3 atm?
3. How much work is done by a gas (p=1.7 atm, V=1.56 L) that expands against an external pressure of 1.8 atm?
Solutions
1. W = − pΔV
ΔV = Vfinal - VInitial = 5 L - 2 L = 3 L
Convert 750 mmHg to atm: 750 mmHg * 1/760 (atm/mmHg) = 0.9868 atm.
W = − pΔV = -(.9868 atm)(3 Liters) = -2.96 L atm.
2. First we must find the final volume using the idela gas law: pv = nRT or v = (nRT)/P = [(.54 moles)(.082057(L atm)/ (mol K))(303K)] / (1.3 atm) = 10.33 L
ΔV = Vfinal - Vinitial = 10.3 Liters - 8 Liters = 2.3 Liters
W = − pΔV = - (1.3 atm)(2.3 Liters) = -3 L atm.
3. $W = - p * ΔV$ = - 1.8 atm * ΔV.
Given $p_1$,$V_1$, and $p_2$, find $V_2$: $p_1V_1=p_2V_2$ (at constant $T$ and $n$)
$V_2= (V_1* P_1) / P_2$ = (1.56 L * 1.7 atm) / 1.8 atm = 1.47 L
Now, $ΔV = V_2 - V_1=1.47 L - 1.56 L = -0.09$
W = - (1.8 atm) * (-0.09 L) = 0.162 L atm.
Outside Links
• Gasparro, Frances P. "Remembering the sign conventions for q and w in deltaU = q - w." J. Chem. Educ. 1976: 53, 389.
• Koubek, E. "PV work demonstration (TD)." J. Chem. Educ. 1980: 57, 374. ' | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.4%3A_Work.txt |
Learning Objectives
• To calculate changes in internal energy
To study the flow of energy during a chemical reaction, we need to distinguish between a system, the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings, the rest of the universe, including the container in which the reaction is carried out (Figure $1$). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa.
Three kinds of systems are important in chemistry. An open system can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed system can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated system exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, the total energy of a system plus its surroundings is constant, which must be true if energy is conserved.
The state of a system is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are path dependent. For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure $2$). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose.
Direction of Heat Flow
The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (see chapter opening image).
The balanced chemical equation for the reaction is as follows:
$2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s) \tag{7.5.1}$
We can also write this chemical equation as
$2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s) + \text{heat} \tag{7.5.2}$
to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations. In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings.A process in which heat (q) is transferred from a system to its surroundings is described as exothermic. By convention, $q < 0$ for an exothermic reaction.
When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation:
$heat + H_2O_{(s)} \rightarrow H_2O_{(l)} \tag{7.5.3}$
When heat is transferred to a system from its surroundings, the process is endothermic. By convention, $q > 0$ for an endothermic reaction.
Note: Heat is technically not a component in Chemical Reactions
Technically, it is poor form to have a $heat$ term in the chemical reaction like in Equations 7.5.2 and 7.5.3 since is it not a true species in the reaction. However, this is a convenient approach to represent exothermic and endothermic behavior and is commonly used by chemists.
The First Law
The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamics, which states that the energy of the universe is constant. We can express this law mathematically as follows:
$U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \tag{7.5.4a}$
$\Delta{U_{sys}}=−ΔU_{surr} \tag{7.5.4b}$
where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings.
Note
The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy.
An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the $CO_2/H_2O$ product mixture is less than that of the isooctane $O_2$ reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w):
$ΔU_{sys} = q + w \tag{7.5.5}$
Although $q$ and $w$ are not state functions on their own, their sum ($ΔU_{sys}$) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed. Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings.
Note
Although $q$ and $w$ are not state functions, their sum ($ΔU_{sys}$) is independent of the path taken and therefore is a state function.
Example $1$
A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔU) of the gas in joules?
Given: initial volume, final volume, external pressure, and quantity of energy transferred as heat
Asked for: total change in internal energy
Strategy:
1. Determine the sign of $q$ to use in Equation 7.5.5.
2. From Equation 7.5.5 calculate $w$ from the values given. Substitute this value into Equation 7.5.5 to calculate $ΔU$.
Solution
A From Equation 7.5.5, we know that ΔU = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention.
B Because the gas is being compressed, we know that work is being done on the system, so $w$ must be positive. From Equation 7.5.5,
$w=-P_{\textrm{ext}}\Delta V=-8.00\textrm{ atm}(\textrm{0.0500 L} - \textrm{0.400 L})\left(\dfrac{\textrm{101.3 J}}{\mathrm{L\cdot atm}} \right)=284\textrm{ J}$
Thus
ΔU = q + w = −140 J + 284 J = 144 J
In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs.
Exercise $1$
A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔU) of the gas in joules?
Answer: −216 J
Note
By convention, both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa.
Summary
• Enthalpy is a state function, and the change in enthalpy of a system is equal to the sum of the change in the internal energy of the system and the PV work done.
The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (U) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.5%3A_The_First_Law_of_Thermodynamics.txt |
Learning Objectives
• To understand how enthalpy pertains to chemical reactions
When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction ($ΔH_{rxn}$), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so $ΔH_{rxn}$ is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so $ΔH_{rxn}$ is positive. Thus $ΔH_{rxn} < 0$ for an exothermic reaction, and $ΔH_{rxn} > 0$ for an endothermic reaction. In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table:
Reaction Type q $ΔH_{rxn}$
exothermic < 0 < 0 (heat flows from a system to its surroundings)
endothermic > 0 > 0 (heat flows from the surroundings to a system)
If $ΔH_{rxn}$ is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (Figure $\PageIndex{1a}$). Conversely, if $ΔH_{rxn}$ is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (Figure $\PageIndex{1b}$).
Bond breaking ALWAYS requires an input of energy; bond making ALWAYS releases energy.
Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.
• Reversing a reaction or a process changes the sign of ΔH. Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed):
\begin{align} \text{heat} + \ce{H_{2}O(s)} & \ce{ -> H_{2}O(l)} & \Delta H > 0 \label{7.6.7} \[4pt] \ce{H2O (l)} & \ce{-> H2O(s) + heat} & \Delta H < 0 \end{align} \label{7.6.8}
In both cases, the magnitude of the enthalpy change is the same; only the sign is different.
• Enthalpy is an extensive property (like mass). The magnitude of $ΔH$ for a reaction is proportional to the amounts of the substances that react. For example, a large fire produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy change for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider Equation $\ref{7.6.9}$, which describes the reaction of aluminum with iron(III) oxide (Fe2O3) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al2O3, and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe2O3 consumed:
$\ce{ 2Al (s ) + Fe2O3 (s ) -> 2Fe (s) + Al2O3 (s )} + 851.5 \; kJ \label{7.6.9}$
Thus $ΔH = −851.5 \,kJ/mol$ of $\ce{Fe2O3}$. We can also describe $ΔH$ for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for $ΔH$, in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation \ref{7.6.9}, it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:
$\ce{ 2Al (s) + Fe2O3 (s) -> 2Fe (s) + Al2O3 (s)} \quad\quad \Delta H_{rxn}= - 851.5 \; kJ \label{7.6.10}$
If 4 mol of Al and 2 mol of Fe2O3 react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:
$- \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{7.6.6}$
The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example $1$.
Example $1$
Certain parts of the world, such as southern California and Saudi Arabia, are short of fresh water for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If $ΔH$ is 6.01 kJ/mol for the reaction $\ce{H2O(s) → H2O(l)}$ at 0°C and constant pressure, how much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 106 metric tons)? (A metric ton is 1000 kg.)
Given: energy per mole of ice and mass of iceberg
Asked for: energy required to melt iceberg
Strategy:
1. Calculate the number of moles of ice contained in 1 million metric tons (1.00 × 106 metric tons) of ice.
2. Calculate the energy needed to melt the ice by multiplying the number of moles of ice in the iceberg by the amount of energy required to melt 1 mol of ice.
Solution:
A Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given ΔH for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by ΔH (+6.01 kJ/mol):
\begin{align*} moles \; H_{2}O & = 1.00\times 10^{6} \; \cancel{metric \; tons}\, H_{2}O \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{metric \; ton}} \right ) \left ( \dfrac{1000 \; \cancel{g}}{1 \; \cancel{kg}} \right ) \left ( \dfrac{1 \; mol \; H_{2}O}{18.015 \; \cancel{g \; H_{2}O}} \right )\[4pt] & = 5.55\times 10^{10} \; mol H_{2}O \end{align*}
B The energy needed to melt the iceberg is thus
$\left ( \dfrac{6.01 \; kJ}{\cancel{mol \; H_{2}O}} \right )\left ( 5.55 \times 10^{10} \; \cancel{mol \; H_{2}O} \right )= 3.34 \times 10^{11} \; kJ \nonumber$
Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown in the table below.
Possible sources of the approximately 3.34 × 1011 kJ needed to melt a 1.00 × 106 metric ton iceberg
• Combustion of 3.8 × 103 ft3 of natural gas
• Combustion of 68,000 barrels of oil
• Combustion of 15,000 tons of coal
• 1.1 × 108 kilowatt-hours of electricity
Exercise $1$
If 17.3 g of powdered aluminum are allowed to react with excess $\ce{Fe2O3}$, how much heat is produced?
Answer
• 273 kJ
Types of Enthalpies of Reactions
One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following:
• Enthalpy of combustion ($ΔH_{comb}$) is the change in enthalpy that occurs during a combustion reaction. Enthalpy changes have been measured for the combustion of virtually any substance that will burn in oxygen; these values are usually reported as the enthalpy of combustion per mole of substance.
• Enthalpy of fusion ($ΔH_{fus}$) is the enthalpy change that accompanies the melting (fusion) of 1 mol of a substance. The enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance; these values have been measured for almost all the elements and for most simple compounds.
• Enthalpy of vaporization ($ΔH_{vap}$) is the enthalpy change that accompanies the vaporization of 1 mol of a substance. The enthalpy change that accompanies the vaporization of 1 mol of a substance; these values have also been measured for nearly all the elements and for most volatile compounds.
• Enthalpy of solution ($ΔH_{soln}$) is the change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent.
Table $1$: Enthalpies of Vaporization and Fusion for Selected Substances at Their Boiling Points and Melting Points
Substance ΔHvap (kJ/mol) ΔHfus (kJ/mol)
argon (Ar) 6.3 1.3
methane (CH4) 9.2 0.84
ethanol (CH3CH2OH) 39.3 7.6
benzene (C6H6) 31.0 10.9
water (H2O) 40.7 6.0
mercury (Hg) 59.0 2.29
iron (Fe) 340 14
The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.
Enthalpy of Reaction: https://youtu.be/z2KUaIEF9qI
Summary
Enthalpy is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the change in enthalpy ($ΔH$) can be measured. A negative $ΔH$ means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings. For a chemical reaction, the enthalpy of reaction ($ΔH_{rxn}$) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of $ΔH_{rxn}$.
The magnitude of ΔHrxn also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the enthalpy of fusion ($ΔH_{fus}$) and the enthalpy of vaporization ($ΔH_{vap}$), respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is Hess’s law. The enthalpy of combustion ($ΔH_{comb}$) is the enthalpy change that occurs when a substance is burned in excess oxygen. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.6%3A_Heats_of_Reactions_-_U_and_H.txt |
Learning Objectives
• To use Hess’s law and thermochemical cycles to calculate enthalpy changes of chemical reactions.
Because enthalpy is a state function, the enthalpy change for a reaction depends on only two things:
1. the masses of the reacting substances and
2. the physical states of the reactants and products.
It does not depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, $ΔH$ for the net reaction is the sum of the $ΔH$ values for the individual reactions. This principle is called Hess’s law, after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known $ΔH$ values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps.
We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation $\ref{12.7.1}$ can be viewed as occurring in three distinct steps with known $ΔH$ values. As shown in Figure $1$:
1. The first reaction produces 1 mol of solid aluminum oxide ($\ce{Al2O3}$) and 2 mol of liquid iron at its melting point of 1758°C (Equation \ref{12.7.1}a); the enthalpy change for this reaction is −732.5 kJ/mol of $\ce{Fe2O3}$.
2. The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (Equation \ref{12.7.1}b); the enthalpy change for this reaction is −13.8 kJ/mol of $\ce{Fe}$ (−27.6 kJ per 2 mol $\ce{Fe}$).
3. In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (Equation \ref{12.7.1}c); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol $\ce{Fe}$).
As you can see in Figure $1$, the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding Equations \ref{12.7.1}a-\ref{12.7.1}c} gives the overall reaction, shown Equation \ref{12.7.1}d:
$\begin{matrix} 2Al\left ( s, \; 25 ^{o}C \right ) + Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right )& \rightarrow & 2Fe\left ( l, \; 1758 ^{o}C \right ) + Al_2O_3 \left ( s, \; 1758 ^{o}C \right ) & \Delta H=-732.5 \; kJ& \left ( reaction \,a \right ) \ 2Fe\left ( l, \; 1758 ^{o}C \right ) & \rightarrow & 2Fe\left ( s, \; 1758 ^{o}C \right ) & \Delta H=-\;\; 27.6 \; kJ & \left (reaction\, b \right )\ 2Fe\left ( s, \; 1758 ^{o}C \right ) + 2Al\left ( s, \; 1758 ^{o}C \right ) & \rightarrow & 2Fe\left ( s, \; 25 ^{o}C \right ) + 2Al\left ( s, \; 25 ^{o}C \right ) & \Delta H=-\;\; 91.0 \; kJ & \left ( reaction \,c \right )\ \hline 2Al\left ( s, \; 25 ^{o}C \right ) + Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right ) & \rightarrow & 2Al\left ( s, \; 25 ^{o}C \right ) + 2Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right ) & \Delta H=-852.2 \; kJ & \left ( total \,reaction \, d \right ) \end{matrix} \label{12.7.1}$
By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in Figure $1$.
Comparing parts (a) and (d) in Equation $\ref{12.7.1}$ also illustrates an important point: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution). When the product is liquid iron at its melting point (part (a) in Equation $\ref{12.7.1}$), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in Equation $\ref{12.7.1}$). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation.
When using Hess’s law to calculate the value of ΔH for a reaction, follow this procedure:
1. Identify the equation whose $ΔH$ value is unknown and write individual reactions with known $ΔH$ values that, when added together, will give the desired equation. We illustrate how to use this procedure in Example $1$.
2. Arrange the chemical equations so that the reaction of interest is the sum of the individual reactions.
3. If a reaction must be reversed, change the sign of $ΔH$ for that reaction. Additionally, if a reaction must be multiplied by a factor to obtain the correct number of moles of a substance, multiply its $ΔH$ value by that same factor.
4. Add together the individual reactions and their corresponding $ΔH$ values to obtain the reaction of interest and the unknown ΔH.
Example $1$
When carbon is burned with limited amounts of oxygen gas ($\ce{O2}$), carbon monoxide ($\ce{CO}$) is the main product:
$\ce{ 2C (s) + O2 (g) -> 2CO (g)} \quad\quad \Delta H=-221.0 \; kJ \label{reaction 1} \tag{reaction 1}$
When carbon is burned in excess O2, carbon dioxide (CO2) is produced:
$\ce{ C (s) + O2(g) \rightarrow CO2 (g)} \quad\quad \Delta H=-393.5 \; kJ \label{reaction 2} \tag{reaction 2}$
Use this information to calculate the enthalpy change per mole of $\ce{CO}$ for the reaction of $\ce{CO}$ with $\ce{O2}$ to give $\ce{CO2}$.
Given: two balanced chemical equations and their $ΔH$ values
Asked for: enthalpy change for a third reaction
Strategy:
1. After balancing the chemical equation for the overall reaction, write two equations whose ΔH values are known and that, when added together, give the equation for the overall reaction. (Reverse the direction of one or more of the equations as necessary, making sure to also reverse the sign of $ΔH$.)
2. Multiply the equations by appropriate factors to ensure that they give the desired overall chemical equation when added together. To obtain the enthalpy change per mole of CO, write the resulting equations as a sum, along with the enthalpy change for each.
Solution:
A We begin by writing the balanced chemical equation for the reaction of interest:
$\ce{ CO (g) + 1/2O2 (g) -> CO2 (g)} \quad\quad \Delta H_{rxn}=? \label{reaction 3} \tag{reaction 3}$
There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added together to give the desired equation and for which the enthalpy changes are known. Observing that $\ce{CO}$, a reactant in \ref{reaction 2} and a product in Equation \ref{reaction 1}, we can reverse \ref{reaction 1} to give
$2CO\left ( g \right ) \rightarrow 2C\left ( s \right ) + O_{2}\left ( g \right ) \; \;\ \; \Delta H=+221.0 \; kJ \nonumber$
Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use \ref{reaction 2}, as written because its product, CO2, is the product we want in \ref{reaction 3},:
$\ce{C (s) + O2(g) -> CO2 (s)} \quad \quad \Delta H=-393.5 \; kJ$
B Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant:
$\ce{ 2C(s) + 2O2 (g) -> 2CO2 (s)} \quad\quad \Delta H=-787.0 \; kJ$
Writing the resulting equations as a sum, along with the enthalpy change for each, gives
$\begin{matrix} 2CO\left ( g \right ) & \rightarrow & \cancel{2C\left ( s \right )}+\cancel{O_{2}\left ( g \right )} & \Delta H & = & -\Delta H_{1} & = & +221.0 \; kJ \ \cancel{2C\left ( s \right )}+\cancel{2}O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = & -\Delta 2H_{2} & = & -787.0 \; kJ \ 2CO\left ( g \right ) + O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = & & -566.0 \; kJ \end{matrix}$
Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of $\ce{CO}$ with $\ce{O2}$, and the problem asks for the amount per mole of $\ce{CO}$. Consequently, we must divide both sides of the final equation and the magnitude of $ΔH$ by 2:
$\begin{matrix} CO\left ( g \right ) + \frac{1}{2}O_{2}\left ( g \right ) & \rightarrow & CO_{2} \left ( g \right ) & \Delta H & = & & -283.0 \; kJ \end{matrix} \nonumber$
An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms:
$\small \begin{matrix} \left ( A \right ) & 2C\left ( s \right ) + O_{2}\left ( g \right ) & \rightarrow & \cancel{2CO\left ( g \right )} & \Delta H_{A} & = & \Delta H_{1} & = & +221.0 \; kJ \ \left ( B \right ) &\cancel{2CO\left ( g \right )} + O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H_{B} & & & = & ? \ \left ( C \right ) & 2C\left ( s \right ) + 2O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = 2\Delta H_{2} & =2\times \left ( -393.5 \; kJ \right ) & =-787.0 \; kJ \end{matrix}$
The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO2. From Hess’s law, ΔHA + ΔHB = ΔHC, and we are given ΔH for reactions A and C. Substituting the appropriate values gives
\begin{align*} -221.0 \; kJ + \Delta H_{B} &= -787.0 \; kJ \[4pt] \Delta H_{B} &= -566.0 \end{align*}
This is again the enthalpy change for the conversion of 2 mol of CO to CO2. The enthalpy change for the conversion of 1 mol of CO to CO2 is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, there may be more than one correct way to solve a problem.
Exercise $1$
The reaction of acetylene (C2H2) with hydrogen (H2) can produce either ethylene (C2H4) or ethane (C2H6):
$\begin{matrix} C_{2}H_{2}\left ( g \right ) + H_{2}\left ( g \right ) \rightarrow C_{2}H_{4}\left ( g \right )& \Delta H = -175.7 \; kJ/mol \; C_{2}H_{2} \ C_{2}H_{2}\left ( g \right ) + 2H_{2}\left ( g \right ) \rightarrow C_{2}H_{6}\left ( g \right ) & \Delta H = -312.0 \; kJ/mol \; C_{2}H_{2} \end{matrix} \nonumber$
What is ΔH for the reaction of C2H4 with H2 to form C2H6?
Answer
−136.3 kJ/mol of C2H4
Summary
Hess's law is that the overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions. For a chemical reaction, the enthalpy of reaction ($ΔH_{rxn}$) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of $ΔH_{rxn}$. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.7%3A_Indirect_Determination_of_H_-_Hess%27s_Law.txt |
Learning Objectives
• To understand Enthalpies of Formation and be able to use them to calculate Enthalpies of Reaction
One way to report the heat absorbed or released by chemical reactions would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, starting from the elemental forms of each atom at 25 oC and 1 atm pressure.
Enthalpy of formation ($ΔH_f$) is the enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The formation of any chemical can be as a reaction from the corresponding elements:
$\text{elements} \rightarrow \text{compound} \nonumber$
which in terms of the the Enthalpy of formation becomes
$\Delta H_{rxn} = \Delta H_{f} \label{7.8.1}$
For example, consider the combustion of carbon:
$\ce{ C(s) + O2 (g) -> CO2 (g)} \nonumber$
then
$\Delta H_{rxn} = \Delta H_{f}\left [CO_{2}\left ( g \right ) \right ] \nonumber$
The sign convention for ΔHf is the same as for any enthalpy change: $ΔH_f < 0$ if heat is released when elements combine to form a compound and $ΔH_f > 0$ if heat is absorbed.
The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.
Standard Enthalpies of Formation
The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane.
The standard conditions for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state, which is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation ($ΔH^o_f$) The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition.
The standard enthalpy of formation of any element in its standard state is zero by definition.
For example, although oxygen can exist as ozone (O3), atomic oxygen (O), and molecular oxygen (O2), O2 is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H2(g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure $1$). Therefore, $\ce{O2(g)}$, $\ce{H2(g)}$, and graphite have $ΔH^o_f$ values of zero.
The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction:
$6C\left (s, graphite \right ) + 6H_{2}\left (g \right ) + 3O_{2}\left (g \right ) \rightarrow C_{6}H_{12}O_{6}\left (s \right )\; \; \; \Delta H_{f}^{o} = - 1273.3 \; kJ \label{7.8.2}$
It is not possible to measure the value of $ΔH^oo_f$ for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, $\ce{O2}$, and $\ce{H2}$ and measuring the heat evolved as glucose is formed since the reaction shown in Equation $\ref{7.8.2}$ does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of $ΔH^oo_f$ are obtained using Hess’s law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of $ΔH^o_f$ for an extensive list of compounds are given in Table T1. Note that $ΔH^o_f$ values are always reported in kilojoules per mole of the substance of interest. Also notice in Table T1 that the standard enthalpy of formation of O2(g) is zero because it is the most stable form of oxygen in its standard state.
Example $1$: Enthalpy of Formation
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
1. $\ce{HCl(g)}$
2. $\ce{MgCO3(s)}$
3. $\ce{CH3(CH2)14CO2H(s)}$ (palmitic acid)
Given:
compound formula and phase.
Asked for:
balanced chemical equation for its formation from elements in standard states
Strategy:
Use Table T1 to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made.
Solution:
To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in Table T1: by a $ΔH^o_f$ value of 0 kJ/mol.
Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are $\ce{H2(g)}$ and $\ce{Cl2(g)}$, respectively, the unbalanced chemical equation is
$\ce{H2(g) + Cl2(g) \rightarrow HCl(g)} \nonumber$
Fractional coefficients are required in this case because ΔHof values are reported for 1 mol of the product, $\ce{HCl}$. Multiplying both $\ce{H2(g)}$ and $\ce{Cl2(g)}$ by 1/2 balances the equation:
$\ce{1/2 H_{2} (g) + 1/2 Cl_{2} (g) \rightarrow HCl (g)} \nonumber$
The standard states of the elements in this compound are $\ce{Mg(s)}$, $\ce{C(s, graphite)}$, and $\ce{O2(g)}$. The unbalanced chemical equation is thus
$\ce{Mg(s) + C (s, graphite) + O2 (g) \rightarrow MgCO3 (s)} \nonumber$
This equation can be balanced by inspection to give
$\ce{Mg (s) + C (s, graphite ) + 3/2 O2 (g)\rightarrow MgCO3 (s)} \nonumber$
Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows:
$\ce{C(s, graphite) + H2(g) + O2(g) \rightarrow CH3(CH2)14CO2H(s)} \nonumber$
There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is
$\ce{16C (s, graphite) + 16 H2(g) + O2(g) -> CH3(CH2)14CO2H(s) } \nonumber$
Exercise $1$
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
1. $\ce{NaCl(s)}$
2. $\ce{H2SO4(l)}$
3. $\ce{CH3CO2H(l)}$ (acetic acid)
Answer a
$\ce{ Na (s) + 1/2 Cl2 (g) \rightarrow NaCl (s)} \nonumber$
Answer b
$\ce{H_{2} (g) + 1/8 S8 (s) + 2O2 ( g) \rightarrow H2 SO4( l) } \nonumber$
Answer c
$\ce{2C(s) + O2(g) + 2H2(g) -> CH3CO2H(l)} \nonumber$
Definition of Heat of Formation Reactions: https://youtu.be/A20k0CK4doI
Standard Enthalpies of Reaction
Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for any reaction involving substances whose $\Delta{H_f^o}$ values are known. The standard enthalpy of reaction $\Delta{H_{rxn}^o}$ is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction
$aA + bB \rightarrow cC + dD \label{7.8.3}$
where $A$, $B$, $C$, and $D$ are chemical substances and $a$, $b$, $c$, and $d$ are their stoichiometric coefficients. The magnitude of $ΔH^ο$ is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:
$\Delta H_{rxn}^{o} = \underbrace{ \left [c\Delta H_{f}^{o}\left ( C \right ) + d\Delta H_{f}^{o}\left ( D \right ) \right ] }_{\text{products} } - \underbrace{ \left [a\Delta H_{f}^{o}\left ( A \right ) + b\Delta H_{f}^{o}\left ( B \right ) \right ]}_{\text{reactants }} \label{7.8.4}$
More generally, we can write
$\Delta H_{rxn}^{o} = \sum m\Delta H_{f}^{o}\left ( products \right ) - \sum n\Delta H_{f}^{o}\left ( reactants \right ) \label{7.8.5}$
where the symbol $\sum$ means “sum of” and $m$ and $n$ are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation $\ref{7.8.5}$ arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters.
"Products minus reactants" summations are typical of state functions.
To demonstrate the use of tabulated ΔHο values, we will use them to calculate $ΔH_{rxn}$ for the combustion of glucose, the reaction that provides energy for your brain:
$\ce{ C6H12O6 (s) + 6O2 (g) \rightarrow 6CO2 (g) + 6H2O (l)} \label{7.8.6}$
Using Equation $\ref{7.8.5}$, we write
$\Delta H_{f}^{o} =\left \{ 6\Delta H_{f}^{o}\left [ CO_{2}\left ( g \right ) \right ] + 6\Delta H_{f}^{o}\left [ H_{2}O\left ( g \right ) \right ] \right \} - \left \{ \Delta H_{f}^{o}\left [ C_{6}H_{12}O_{6}\left ( s \right ) \right ] + 6\Delta H_{f}^{o}\left [ O_{2}\left ( g \right ) \right ] \right \} \label{7.8.7}$
From Table T1, the relevant ΔHοf values are ΔHοf [CO2(g)] = -393.5 kJ/mol, ΔHοf [H2O(l)] = -285.8 kJ/mol, and ΔHοf [C6H12O6(s)] = -1273.3 kJ/mol. Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol. Inserting these values into Equation $\ref{7.8.7}$ and changing the subscript to indicate that this is a combustion reaction, we obtain
\begin{align} \Delta H_{comb}^{o} &= \left [ 6\left ( -393.5 \; kJ/mol \right ) + 6 \left ( -285.8 \; kJ/mol \right ) \right ] - \left [-1273.3 + 6\left ( 0 \; kJ\;mol \right ) \right ] \label{7.8.8} \[4pt] &= -2802.5 \; kJ/mol \end{align}
As illustrated in Figure $2$, we can use Equation $\ref{7.8.8}$ to calculate $ΔH^ο_f$ for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled $ΔH^ο_{comb}$. The alternative hypothetical pathway consists of four separate reactions that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the $ΔH^ο_f$ values of the reactants. Consequently, the enthalpy changes are
\begin{align} \Delta H_{1}^{o} &= \Delta H_{f}^{o} \left [ glucose \left ( s \right ) \right ] \nonumber \[4pt] &= -1 \; \cancel{mol \; glucose}\left ( \dfrac{1273.3 \; kJ}{1 \; \cancel{mol \; glucose}} \right ) \nonumber \[4pt] &= +1273.3 \; kJ \nonumber \[4pt] \Delta H_{2}^{o} &= 6 \Delta H_{f}^{o} \left [ O_{2} \left ( g \right ) \right ] \nonumber \[4pt] & =6 \; \cancel{mol \; O_{2}}\left ( \dfrac{0 \; kJ}{1 \; \cancel{mol \; O_{2}}} \right ) \nonumber \[4pt] &= 0 \; kJ \end{align} \label{7.8.9}
Recall that when we reverse a reaction, we must also reverse the sign of the accompanying enthalpy change (Equation \ref{7.8.4} since the products are now reactants and vice versa.
The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O2) to the elements is therefore +1273.3 kJ.
The reactions that convert the elements to final products (downward purple arrows in Figure $2$) are identical to those used to define the ΔHοf values of the products. Consequently, the enthalpy changes (from Table T1) are
$\begin{matrix} \Delta H_{3}^{o} = \Delta H_{f}^{o} \left [ CO_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; CO_{2}}\left ( \dfrac{393.5 \; kJ}{1 \; \cancel{mol \; CO_{2}}} \right ) = -2361.0 \; kJ \ \Delta H_{4}^{o} = 6 \Delta H_{f}^{o} \left [ H_{2}O \left ( l \right ) \right ] = 6 \; \cancel{mol \; H_{2}O}\left ( \dfrac{-285.8 \; kJ}{1 \; \cancel{mol \; H_{2}O}} \right ) = -1714.8 \; kJ \end{matrix}$
The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using any pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ):
$\Delta H_{comb}^{o} = +1273.3 \; kJ +\left ( -4075.8 \; kJ \right ) = -2802.5 \; kJ \label{7.8.10}$
This is the same result we obtained using the “products minus reactants” rule (Equation $\ref{7.8.5}$) and ΔHοf values. The two results must be the same because Equation $\ref{7.8.10}$ is just a more compact way of describing the thermochemical cycle shown in Figure $1$.
Example $2$: Heat of Combustion
Long-chain fatty acids such as palmitic acid ($\ce{CH3(CH2)14CO2H}$) are one of the two major sources of energy in our diet ($ΔH^o_f$ =−891.5 kJ/mol). Use the data in Table T1 to calculate ΔHοcomb for the combustion of palmitic acid. Based on the energy released in combustion per gram, which is the better fuel — glucose or palmitic acid?
Given: compound and $ΔH^ο_{f}$ values
Asked for: $ΔH^ο_{comb}$ per mole and per gram
Strategy:
1. After writing the balanced chemical equation for the reaction, use Equation $\ref{7.8.5}$ and the values from Table T1 to calculate $ΔH^ο_{comb}$ the energy released by the combustion of 1 mol of palmitic acid.
2. Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of palmitic acid. Compare this value with the value calculated in Equation $\ref{7.8.8}$ for the combustion of glucose to determine which is the better fuel.
Solution:
A To determine the energy released by the combustion of palmitic acid, we need to calculate its $ΔH^ο_f$. As always, the first requirement is a balanced chemical equation:
$C_{16}H_{32}O_{2(s)} + 23O_{2(g)} \rightarrow 16CO_{2(g)} + 16H_2O_{(l)} \nonumber$
Using Equation $\ref{7.8.5}$ (“products minus reactants”) with ΔHοf values from Table T1 (and omitting the physical states of the reactants and products to save space) gives
\begin{align*} \Delta H_{comb}^{o} &= \sum m \Delta H^o_f\left( {products} \right) - \sum n \Delta H^o_f \left( {reactants} \right) \[4pt] &= \left [ 16\left ( -393.5 \; kJ/mol \; CO_{2} \right ) + 16\left ( -285.8 \; kJ/mol \; H_{2}O \; \right ) \right ] \[4pt] & - \left [ -891.5 \; kJ/mol \; C_{16}H_{32}O_{2} + 23\left ( 0 \; kJ/mol \; O_{2} \; \right ) \right ] \[4pt] &= -9977.3 \; kJ/mol \nonumber \end{align*}
This is the energy released by the combustion of 1 mol of palmitic acid.
B The energy released by the combustion of 1 g of palmitic acid is
$\Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{9977.3 \; kJ}{\cancel{1 \; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{256.42 \; g} \right )= -38.910 \; kJ/g \nonumber$
As calculated in Equation $\ref{7.8.8}$, $ΔH^o_f$ of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore
$\Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{-2802.5 \; kJ}{\cancel{1\; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{180.16\; g} \right ) = -15.556 \; kJ/g \nonumber$
The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight.
Exercise $2$: Water–gas shift reaction
Use Table T1 to calculate $ΔH^o_{rxn}$ for the water–gas shift reaction, which is used industrially on an enormous scale to obtain H2(g):
$\ce{ CO ( g ) + H2O (g ) -> CO2 (g) + H2 ( g )} \nonumber$
Answer
−41.2 kJ/mol
We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s $ΔH^ο_f$ which we cannot obtain otherwise. This procedure is illustrated in Example $3$.
Example $3$: Tetraethyllead
Beginning in 1923, tetraethyllead [$\ce{(C2H5)4Pb}$] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are $\ce{CO2(g)}$, $\ce{H2O(l)}$, and red $\ce{PbO(s)}$. What is the standard enthalpy of formation of tetraethyllead, given that $ΔH^ο_f$ is −19.29 kJ/g for the combustion of tetraethyllead and $ΔH^ο_f$ of red PbO(s) is −219.0 kJ/mol?
Given: reactant, products, and $ΔH^ο_{comb}$ values
Asked for: $ΔH^ο_f$ of the reactants
Strategy:
1. Write the balanced chemical equation for the combustion of tetraethyl lead. Then insert the appropriate quantities into Equation $\ref{7.8.5}$ to get the equation for ΔHοf of tetraethyl lead.
2. Convert $ΔH^ο_{comb}$ per gram given in the problem to $ΔH^ο_{comb}$ per mole by multiplying $ΔH^ο_{comb}$ per gram by the molar mass of tetraethyllead.
3. Use Table T1 to obtain values of $ΔH^ο_f$ for the other reactants and products. Insert these values into the equation for $ΔH^ο_f$ of tetraethyl lead and solve the equation.
Solution:
A The balanced chemical equation for the combustion reaction is as follows:
$\ce{2(C2H5)4Pb(l) + 27O2(g) → 2PbO(s) + 16CO2(g) + 20H2O(l)} \nonumber$
Using Equation $\ref{7.8.5}$ gives
$\Delta H_{comb}^{o} = \left [ 2 \Delta H_{f}^{o}\left ( PbO \right ) + 16 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 20 \Delta H_{f}^{o}\left ( H_{2}O \right )\right ] - \left [2 \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) + 27 \Delta H_{f}^{o}\left ( O_{2} \right ) \right ] \nonumber$
Solving for $ΔH^o_f [\ce{(C2H5)4Pb}]$ gives
$\Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) = \Delta H_{f}^{o}\left ( PbO \right ) + 8 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 10 \Delta H_{f}^{o}\left ( H_{2}O \right ) - \dfrac{27}{2} \Delta H_{f}^{o}\left ( O_{2} \right ) - \dfrac{\Delta H_{comb}^{o}}{2} \nonumber$
The values of all terms other than $ΔH^o_f [\ce{(C2H5)4Pb}]$ are given in Table T1.
B The magnitude of $ΔH^o_{comb}$ is given in the problem in kilojoules per gram of tetraethyl lead. We must therefore multiply this value by the molar mass of tetraethyl lead (323.44 g/mol) to get $ΔH^o_{comb}$ for 1 mol of tetraethyl lead:
\begin{align*} \Delta H_{comb}^{o} &= \left ( \dfrac{-19.29 \; kJ}{\cancel{g}} \right )\left ( \dfrac{323.44 \; \cancel{g}}{mol} \right ) \[4pt] &= -6329 \; kJ/mol \end{align*}
Because the balanced chemical equation contains 2 mol of tetraethyllead, $ΔH^o_{rxn}$ is
\begin{align*} \Delta H_{rxn}^{o} &= 2 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb} \left ( \dfrac{-6329 \; kJ}{1 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb }} \right ) \[4pt] &= -12,480 \; kJ \end{align*}
C Inserting the appropriate values into the equation for $ΔH^o_f [\ce{(C2H5)4Pb}]$ gives
\begin{align*} \Delta H_{f}^{o} \left [ \left (C_{2}H_{4} \right )_{4}Pb \right ] & = \left [1 \; mol \;PbO \;\times 219.0 \;kJ/mol \right ]+\left [8 \; mol \;CO_{2} \times \left (-393.5 \; kJ/mol \right )\right ] +\left [10 \; mol \; H_{2}O \times \left ( -285.8 \; kJ/mol \right )\right ] + \left [-27/2 \; mol \; O_{2}) \times 0 \; kJ/mol \; O_{2}\right ] \left [12,480.2 \; kJ/mol \; \left ( C_{2}H_{5} \right )_{4}Pb \right ]\[4pt] &= -219.0 \; kJ -3148 \; kJ - 2858 kJ - 0 kJ + 6240 \; kJ = 15 kJ/mol \end{align*}
Exercise $3$
Ammonium sulfate, $\ce{(NH4)2SO4}$, is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid:
$\ce{2NH3(g) + H2SO4(aq) \rightarrow (NH4)2SO4(s)} \nonumber$
The value of $ΔH^o_{rxn}$ is -179.4 kJ/mole $\ce{H2SO4}$. Use the data in Table T1 to calculate the standard enthalpy of formation of ammonium sulfate (in kilojoules per mole).
Answer
−1181 kJ/mol
Calculating DH° using DHf°: https://youtu.be/Y3aJJno9W2c
Summary
• The standard state for measuring and reporting enthalpies of formation or reaction is 25 oC and 1 atm.
• The elemental form of each atom is that with the lowest enthalpy in the standard state.
• The standard state heat of formation for the elemental form of each atom is zero.
The enthalpy of formation ($ΔH_{f}$) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation ($ΔH^o_{f}$) are determined under standard conditions: a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their standard states (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The standard enthalpy of reaction ($ΔH^o_{rxn}$) can be calculated from the sum of the standard enthalpies of formation of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The enthalpy of solution ($ΔH_{soln}$) is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.8%3A_Standard_Enthalpies_of_Formation.txt |
Learning Objectives
• To use thermochemical concepts to solve environmental issues.
Our contemporary society requires the constant expenditure of huge amounts of energy to heat our homes, provide telephone and cable service, transport us from one location to another, provide light when it is dark outside, and run the machinery that manufactures material goods. The United States alone consumes almost 106 kJ per person per day, which is about 100 times the normal required energy content of the human diet. This figure is about 30% of the world’s total energy usage, although only about 5% of the total population of the world lives in the United States. In contrast, the average energy consumption elsewhere in the world is about 105 kJ per person per day, although actual values vary widely depending on a country’s level of industrialization. In this section, we describe various sources of energy and their impact on the environment.
Fuels
According to the law of conservation of energy, energy can never actually be “consumed”; it can only be changed from one form to another. What is consumed on a huge scale, however, are resources that can be readily converted to a form of energy that is useful for doing work. Energy that is not used to perform work is either stored as potential energy for future use or transferred to the surroundings as heat.
A major reason for the huge consumption of energy by our society is the low efficiency of most machines in transforming stored energy into work. Efficiency can be defined as the ratio of useful work accomplished to energy expended. Automobiles, for example, are only about 20% efficient in converting the energy stored in gasoline to mechanical work; the rest of the energy is released as heat, either emitted in the exhaust or produced by friction in bearings and tires. The production of electricity by coal- or oil-powered steam turbines (Figure $1$) is can be more than 50% efficient.
In general, it is more efficient to use primary sources of energy directly (such as natural gas or oil) than to transform them to a secondary source such as electricity prior to their use. For example, if a furnace is well maintained, heating a house with natural gas is about 70% efficient. In contrast, burning the natural gas in a remote power plant, converting it to electricity, transmitting it long distances through wires, and heating the house by electric baseboard heaters have an overall efficiency of less than 35%.
The total expenditure of energy in the world each year is about 3 × 1017 kJ. 80% of this energy is provided by the combustion of fossil fuels: oil, coal, and natural gas (the sources of the energy consumed in the United States in 2019 are shown in Figure $2$). Natural gas and petroleum are the preferred fuels because many of the products derived from them are gases or liquids that are readily transported, stored, and burned. Natural gas and petroleum are derived from the remains of marine creatures that died hundreds of millions of years ago and were buried beneath layers of sediment. As the sediment turned to rock, the tremendous heat and pressure inside Earth transformed the organic components of the buried sea creatures to petroleum and natural gas.
Coal
Coal is a complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. Because plants contain large amounts of cellulose, derived from linked glucose units, the structure of coal is more complex than that of petroleum (Figure $3$). In particular, coal contains a large number of oxygen atoms that link parts of the structure together, in addition to the basic framework of carbon–carbon bonds. It is impossible to draw a single structure for coal; however, because of the prevalence of rings of carbon atoms (due to the original high cellulose content), coal is more similar to an aromatic hydrocarbon than an aliphatic one.
Table $1$: Properties of Different Types of Coal
Type % Carbon Hydrogen:Carbon Mole Ratio % Oxygen % Sulfur Heat Content US Deposits
anthracite 92 0.5 3 1 high Pennsylvania, New York
bituminous 80 0.6 8 5 medium Appalachia, Midwest, Utah
subbituminous 77 0.9 16 1 medium Rocky Mountains
lignite 71 1.0 23 1 low Montana
There are four distinct classes of coal (Table $1$); their hydrogen and oxygen contents depend on the length of time the coal has been buried and the pressures and temperatures to which it has been subjected. Lignite, with a hydrogen:carbon ratio of about 1.0 and a high oxygen content, has the lowest ΔHcomb. Anthracite, in contrast, with a hydrogen:carbon ratio of about 0.5 and the lowest oxygen content, has the highest ΔHcomb and is the highest grade of coal. The most abundant form in the United States is bituminous coal, which has a high sulfur content because of the presence of small particles of pyrite (FeS2). The combustion of coal releases the sulfur in FeS2 as SO2, which is a major contributor to acid rain. Table $2$ compares the ΔHcomb per gram of oil, natural gas, and coal with those of selected organic compounds.
Table $2$: Enthalpies of Combustion of Common Fuels and Selected Organic Compounds
Fuel ΔHcomb (kJ/g)
dry wood −15
peat −20.8
bituminous coal −28.3
charcoal −35
kerosene −37
C6H6 (benzene) −41.8
crude oil −43
natural gas −50
C2H2 (acetylene) −50.0
CH4 (methane) −55.5
gasoline −84
hydrogen −143
Peat, a precursor to coal, is the partially decayed remains of plants that grow in swampy areas. It is removed from the ground in the form of soggy bricks of mud that will not burn until they have been dried. Even though peat is a smoky, poor-burning fuel that gives off relatively little heat, humans have burned it since ancient times (Figure $4$). If a peat bog were buried under many layers of sediment for a few million years, the peat could eventually be compressed and heated enough to become lignite, the lowest grade of coal; given enough time and heat, lignite would eventually become anthracite, a much better fuel.
Converting Coal to Gaseous and Liquid Fuels
Oil and natural gas resources are limited. Current estimates suggest that the known reserves of petroleum will be exhausted in about 60 years, and supplies of natural gas are estimated to run out in about 120 years. Coal, on the other hand, is relatively abundant, making up more than 90% of the world’s fossil fuel reserves. As a solid, coal is much more difficult to mine and ship than petroleum (a liquid) or natural gas. Consequently, more than 75% of the coal produced each year is simply burned in power plants to produce electricity. A great deal of current research focuses on developing methods to convert coal to gaseous fuels (coal gasification) or liquid fuels (coal liquefaction). In the most common approach to coal gasification, coal reacts with steam to produce a mixture of CO and H2 known as synthesis gas, or syngas:Because coal is 70%–90% carbon by mass, it is approximated as C in Equation $\ref{7.9.1}$.
$\mathrm{C_{(s)} +H_2O_{(g)} → CO_{(g)}+H_{2(g)}} \;\;\; ΔH= \mathrm{131\: kJ} \label{7.9.1}$
Converting coal to syngas removes any sulfur present and produces a clean-burning mixture of gases.
Syngas is also used as a reactant to produce methane and methanol. A promising approach is to convert coal directly to methane through a series of reactions:
$\mathrm{2C(s)+2H_2O(g)→\cancel{2CO(g)}+\cancel{2H_2(g)}}\hspace{20px}ΔH_1= \mathrm{262\: kJ}\ \mathrm{\cancel{CO(g)}+\cancel{H_2O(g)}→CO_2(g)+\cancel{H_2(g)}}\hspace{20px}ΔH_2=\mathrm{−41\: kJ}\ \mathrm{\cancel{CO(g)}+\cancel{3H_2(g)}→CH_4(g)+\cancel{H_2O(g)}}\hspace{20px}ΔH_3=\mathrm{−206\: kJ}\ \overline{\mathrm{Overall:\hspace{10px}2C(s)+2H_2O(g)→CH_4(g)+CO_2(g)}\hspace{20px}ΔH_\ce{comb}= \mathrm{15\: kJ}}\hspace{40px}\label{7.9.2}$
Burning a small amount of coal or methane provides the energy consumed by these reactions. Unfortunately, methane produced by this process is currently significantly more expensive than natural gas. As supplies of natural gas become depleted, however, this coal-based process may well become competitive in cost.
Similarly, the techniques available for converting coal to liquid fuels are not yet economically competitive with the production of liquid fuels from petroleum. Current approaches to coal liquefaction use a catalyst to break the complex network structure of coal into more manageable fragments. The products are then treated with hydrogen (from syngas or other sources) under high pressure to produce a liquid more like petroleum. Subsequent distillation, cracking, and reforming can be used to create products similar to those obtained from petroleum. The total yield of liquid fuels is about 5.5 bbl of crude liquid per ton of coal (1 bbl is 42 gal or 160 L). Although the economics of coal liquefaction are currently even less attractive than for coal gasification, liquid fuels based on coal are likely to become economically competitive as supplies of petroleum are consumed.
Example $1$
If bituminous coal is converted to methane by the process in Equation $\ref{7.9.1}$, what is the ratio of the ΔHcomb of the methane produced to the enthalpy of the coal consumed to produce the methane? (Note that 1 mol of CH4 is produced for every 2 mol of carbon in coal.)
Given: chemical reaction and ΔHcomb (Table $2$)
Asked for: ratio of ΔHcomb of methane produced to coal consumed
Strategy:
A Write a balanced chemical equation for the conversion of coal to methane. Referring to Table $2$, calculate the ΔHcomb of methane and carbon.
B Calculate the ratio of the energy released by combustion of the methane to the energy released by combustion of the carbon.
Solution:
A The balanced chemical equation for the conversion of coal to methane is as follows:
$\ce{2C (s) + 2H2O(g) → CH4(g) + CO2(g)} \nonumber$
Thus 1 mol of methane is produced for every 2 mol of carbon consumed. The ΔHcomb of 1 mol of methane is
$\mathrm{1 \cancel{mol\: CH_4}\left(\dfrac{16.043\cancel{g}}{1 \cancel{mol\: CH_4}}\right)\left(\dfrac{−55.5\: kJ}{\cancel{g}}\right)=−890\: kJ}$
The ΔHcomb of 2 mol of carbon (as coal) is
$\mathrm{2 \cancel{mol\: C}\left(\dfrac{12.011 \cancel{g}}{1 \cancel{mol\: C}}\right)\left(\dfrac{−28.3\: kJ}{\cancel{g}}\right)=−680\: kJ}$
B The ratio of the energy released from the combustion of methane to the energy released from the combustion of carbon is
$\mathrm{\dfrac{−890 \cancel{kJ}}{−680 \cancel{kJ}}= 1.31}$
The energy released from the combustion of the product (methane) is 131% of that of the reactant (coal). The fuel value of coal is actually increased by the process!
How is this possible when the law of conservation of energy states that energy cannot be created? The reaction consumes 2 mol of water ($ΔH^\circ_\ce{f}=\mathrm{−285.8\: kJ/mol}$) but produces only 1 mol of CO2 ($ΔH^\circ_\ce{f}=\mathrm{−393.5\: kJ/mol}$). Part of the difference in potential energy between the two (approximately 180 kJ/mol) is stored in CH4 and can be released during combustion.
Exercise $1$
Using the data in Table $2$, calculate the mass of hydrogen necessary to provide as much energy during combustion as 1 bbl of crude oil (density approximately 0.75 g/mL).
Answer
36 kg
The Carbon Cycle and the Greenhouse Effect
Even if carbon-based fuels could be burned with 100% efficiency, producing only CO2(g) and H2O(g), doing so could still potentially damage the environment when carried out on the vast scale required by an industrial society. The amount of CO2 released is so large and is increasing so rapidly that it is apparently overwhelming the natural ability of the planet to remove CO2 from the atmosphere. In turn, the elevated levels of CO2 are thought to be affecting the temperature of the planet through a mechanism known as the greenhouse effect. As you will see, there is little doubt that atmospheric CO2 levels are increasing, and the major reason for this increase is the combustion of fossil fuels. There is substantially less agreement, however, on whether the increased CO2 levels are responsible for a significant increase in temperature.
The Global Carbon Cycle
Figure $5$ illustrates the global carbon cycle, the distribution and flow of carbon on Earth. Normally, the fate of atmospheric CO2 is to either (1) dissolve in the oceans and eventually precipitate as carbonate rocks or (2) be taken up by plants. The rate of uptake of CO2 by the ocean is limited by its surface area and the rate at which gases dissolve, which are approximately constant. The rate of uptake of CO2 by plants, representing about 60 billion metric tons of carbon per year, partly depends on how much of Earth’s surface is covered by vegetation. Unfortunately, the rapid deforestation for agriculture is reducing the overall amount of vegetation, and about 60 billion metric tons of carbon are released annually as CO2 from animal respiration and plant decay. The amount of carbon released as CO2 every year by fossil fuel combustion is estimated to be about 5.5 billion metric tons. The net result is a system that is slightly out of balance, experiencing a slow but steady increase in atmospheric CO2 levels (Figure $6$). As a result, average CO2 levels have increased by about 30% since 1850.
Most of Earth’s carbon is found in the crust, where it is stored as calcium and magnesium carbonate in sedimentary rocks. The oceans also contain a large reservoir of carbon, primarily as the bicarbonate ion (HCO3). Green plants consume about 60 billion metric tons of carbon per year as CO2 during photosynthesis, and about the same amount of carbon is released as CO2 annually from animal and plant respiration and decay. The combustion of fossil fuels releases about 5.5 billion metric tons of carbon per year as CO2.
The Atmospheric Greenhouse Effect
The increasing levels of atmospheric CO2 are of concern because CO2 absorbs thermal energy radiated by the Earth, as do other gases such as water vapor, methane, and chlorofluorocarbons. Collectively, these substances are called greenhouse gases; they mimic the effect of a greenhouse by trapping thermal energy in the Earth’s atmosphere, a phenomenon known as the greenhouse effect (Figure $7$).
Venus is an example of a planet that has a runaway greenhouse effect. The atmosphere of Venus is about 95 times denser than that of Earth and contains about 95% CO2. Because Venus is closer to the sun, it also receives more solar radiation than Earth does. The result of increased solar radiation and high CO2 levels is an average surface temperature of about 450°C, which is hot enough to melt lead.
Data such as those in Figure Figure $6$ indicate that atmospheric levels of greenhouse gases have increased dramatically over the past 100 years, and it seems clear that the heavy use of fossil fuels by industry is largely responsible. It is not clear, however, how large an increase in temperature (global warming) may result from a continued increase in the levels of these gases. Estimates of the effects of doubling the preindustrial levels of CO2 range from a 0°C to a 4.5°C increase in the average temperature of Earth’s surface, which is currently about 14.4°C. Even small increases, however, could cause major perturbations in our planet’s delicately balanced systems. For example, an increase of 5°C in Earth’s average surface temperature could cause extensive melting of glaciers and the Antarctic ice cap. It has been suggested that the resulting rise in sea levels could flood highly populated coastal areas, such as New York City, Calcutta, Tokyo, Rio de Janeiro, and Sydney. An analysis conducted in 2009 by leading climate researchers from the US National Oceanic and Atmospheric Administration, Switzerland, and France shows that CO2 in the atmosphere will remain near peak levels far longer than other greenhouse gases, which dissipate more quickly. The study predicts a rise in sea levels of approximately 3 ft by the year 3000, excluding the rise from melting glaciers and polar ice caps. According to the analysis, southwestern North America, the Mediterranean, and southern Africa are projected to face droughts comparable to that of the Dust Bowl of the 1930s as a result of global climate changes.
The increase in CO2 levels is only one of many trends that can affect Earth’s temperature. In fact, geologic evidence shows that the average temperature of Earth has fluctuated significantly over the past 400,000 years, with a series of glacial periods (during which the temperature was 10°C–15°C lower than it is now and large glaciers covered much of the globe) interspersed with relatively short, warm interglacial periods (Figure $8$). Although average temperatures appear to have increased by 0.5°C in the last century, the statistical significance of this increase is open to question, as is the existence of a cause-and-effect relationship between the temperature change and CO2 levels. Despite the lack of incontrovertible scientific evidence, however, many people believe that we should take steps now to limit CO2 emissions and explore alternative sources of energy, such as solar energy, geothermal energy from volcanic steam, and nuclear energy, to avoid even the possibility of creating major perturbations in Earth’s environment. In 2010, international delegates met in Cancún, Mexico, and agreed on a broad array of measures that would advance climate protection. These included the development of low-carbon technologies, providing a framework to reduce deforestation, and aiding countries in assessing their own vulnerabilities. They avoided, however, contentious issues of assigning emissions reductions commitments.
Example $2$
A student at UCLA decided to fly home to New York for Christmas. The round trip was 4500 air miles, and part of the cost of her ticket went to buy the 100 gal of jet fuel necessary to transport her and her baggage. Assuming that jet fuel is primarily n-dodecane (C12H26) with a density of 0.75 g/mL, how much energy was expended and how many tons of $\ce{CO2}$ were emitted into the upper atmosphere to get her home and back?
Given: volume and density of reactant in combustion reaction
Asked for: energy expended and mass of CO2 emitted
Strategy:
A After writing a balanced chemical equation for the reaction, calculate $ΔH^\circ_\ce{comb}$
B Determine the number of moles of dodecane in 100 gal by using the density and molar mass of dodecane and the appropriate conversion factors.
C Obtain the amount of energy expended by multiplying $ΔH^\circ_\ce{comb}$ by the number of moles of dodecane. Calculate the amount of CO2 emitted in tons by using mole ratios from the balanced chemical equation and the appropriate conversion factors.
Solution
A We first need to write a balanced chemical equation for the reaction:
$\ce{2C12H26 (l) + 37O2(g) -> 24CO2(g) + 26H2O(l)} \nonumber$
We can calculate $ΔH^\circ_\ce{comb}$ using the $ΔH^\circ_\ce{f}$ values corresponding to each substance in the specified phase (phases are not shown for simplicity):
\begin{align*} ΔH^\circ_\ce{comb}&=ΣmΔH^\circ_\ce{f}(\ce{products})−ΣnΔH^\circ_\ce{f}(\ce{reactants})\ &= [24ΔH^\circ_\ce{f}(\ce{CO2}) + 26ΔH^\circ_\ce{f}(\ce{H2O})]−[37ΔH^\circ_\ce{f}(\ce{O2}) + 2ΔH^\circ_\ce{f}(\ce{C12H26})]\ &= \mathrm{[24(−393.5\: kJ/mol\: CO_2) + 26(−285.8\: kJ/mol\: H_2O)]} \ &\:\:\:\:\:\mathrm{−[37(0\: kJ/mol\: O_2) + 2(−350.9\: kJ/mol\: C_{12}H_{26})]}\ &=\mathrm{−16,173.0\: kJ} \end{align*}
According to the balanced chemical equation for the reaction, this value is $ΔH^\circ_\ce{comb}$ for the combustion of 2 mol of n-dodecane. So we must divide by 2 to obtain $ΔH^\circ_\ce{comb}$ per mole of n-dodecane:
$ΔH^\circ_\ce{comb}=\mathrm{−8,086.5\; kJ/mol\; C_{12}H_{26}} \nonumber$
B The number of moles of dodecane in 100 gal can be calculated as follows, using density, molar mass, and appropriate conversion factors:
$\mathrm{100\: \cancel{gal} \left( \dfrac{3.785 \cancel{L}}{1 \cancel{gal}}\right )\left(\dfrac{1000 \cancel{mL}}{\cancel{L}}\right)\left(\dfrac{0.75 \cancel{g}}{\cancel{mL}}\right)\left(\dfrac{1\: mol}{170.34 \cancel{g}}\right)= 1.7×10^3\: mol\: C_{12}H_{26}}$
C The total energy released is
$ΔH^\circ_\ce{comb}= \mathrm{(−8086.5\: kJ/\cancel{mol}) (1.7×10^3 \cancel{mol}) =−1.4×10^7\: kJ}$
From the balanced chemical equation for the reaction, we see that each mole of dodecane forms 12 mol of $\ce{CO2}$ upon combustion. Hence the amount of $\ce{CO2}$ emitted is
$\mathrm{1.7×10^3 \cancel{mol\: C_{12}H_{26}}\left(\dfrac{\dfrac{24}{2} \cancel{mol\: CO_2}}{1 \cancel{mol\: C_{12}H_{26}}}\right)\left(\dfrac{44.0 \cancel{g}}{1 \cancel{mol\: CO_2}}\right)\left(\dfrac{1 \cancel{lb}}{454 \cancel{g}}\right)\left(\dfrac{1\: tn}{2000 \cancel{lb}}\right)= 0.99\: tn}$
Exercise $2$
Suppose the student in Example $2$ couldn’t afford the plane fare, so she decided to drive home instead. Assume that the round-trip distance by road was 5572 miles, her fuel consumption averaged 31 mpg, and her fuel was pure isooctane (C8H18, density = 0.6919 g/mL). How much energy was expended and how many tons of CO2 were produced during her trip?
Answer
2.2 × 107 kJ; 1.6 tons of CO2 (about twice as much as is released by flying)
Summary
Thermochemical concepts can be used to calculate the efficiency of various forms of fuel, which can then be applied to environmental issues. More than 80% of the energy used by modern society (about 3 × 1017 kJ/yr) is from the combustion of fossil fuels. Because of their availability, ease of transport, and facile conversion to convenient fuels, natural gas and petroleum are currently the preferred fuels. Supplies of coal, a complex solid material derived from plants that lived long ago, are much greater, but the difficulty in transporting and burning a solid makes it less attractive as a fuel. Coal releases the smallest amount of energy per gram of any fossil fuel, and natural gas the greatest amount. The combustion of fossil fuels releases large amounts of CO2 that upset the balance of the carbon cycle and result in a steady increase in atmospheric CO2 levels. Because CO2 is a greenhouse gas, which absorbs heat before it can be radiated from Earth into space, CO2 in the atmosphere can result in increased surface temperatures (the greenhouse effect). The temperature increases caused by increased CO2 levels because of human activities are, however, superimposed on much larger variations in Earth’s temperature that have produced phenomena such as the ice ages and are still poorly understood. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.9%3A_Fuels_as_Sources_of_Energy.txt |
In this chapter, we describe how electrons are arranged in atoms and how the spatial arrangements of electrons are related to their energies. We also explain how knowing the arrangement of electrons in an atom enables chemists to predict and explain the chemistry of an element. As you study the material presented in this chapter, you will discover how the shape of the periodic table reflects the electronic arrangements of elements. In this and subsequent chapters, we build on this information to explain why certain chemical changes occur and others do not. After reading this chapter, you will know enough about the theory of the electronic structure of atoms to explain what causes the characteristic colors of neon signs, how laser beams are created, and why gemstones and fireworks have such brilliant colors. In later chapters, we will develop the concepts introduced here to explain why the only compound formed by sodium and chlorine is NaCl, an ionic compound, whereas neon and argon do not form any stable compounds, and why carbon and hydrogen combine to form an almost endless array of covalent compounds, such as CH4, C2H2, C2H4, and C2H6. You will discover that knowing how to use the periodic table is the single most important skill you can acquire to understand the incredible chemical diversity of the elements.
08: Electrons in Atoms
Learning Objectives
• to learn about the characteristics of electromagnetic waves. Light, X-Rays, infrared and microwaves among the types of electromagnetic waves.
Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light we detect with our eyes, the infrared radiation we feel as heat, the ultraviolet light that causes sunburn, and the x-rays that produce images of our teeth or bones. All these forms of radiant energy should be familiar to you. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation.
Properties of Waves
A wave is a periodic oscillation that transmits energy through space. Anyone who has visited a beach or dropped a stone into a puddle has observed waves traveling through water (Figure $1$). These waves are produced when wind, a stone, or some other disturbance, such as a passing boat, transfers energy to the water, causing the surface to oscillate up and down as the energy travels outward from its point of origin. As a wave passes a particular point on the surface of the water, anything floating there moves up and down.
Waves have characteristic properties (Figure $2$). As you may have noticed in Figure $1$, waves are periodic, that is, they repeat regularly in both space and time. The distance between two corresponding points in a wave—between the midpoints of two peaks, for example, or two troughs—is the wavelength (λ), distance between two corresponding points in a wave—between the midpoints of two peaks or two troughs. $\lambda$ is the lowercase Greek lambda, and $u$ is the lowercase Greek nu. Wavelengths are described by a unit of distance, typically meters. The frequency (ν), the number of oscillations (i.e., of a wave) that pass a particular point in a given period of time. of a wave is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations persecond (1/s = s−1), which in the SI system is called the hertz (Hz). It is named after German physicist Heinrich Hertz (1857–1894),a pioneer in the field of electromagnetic radiation. The amplitude (the vertical height of a wave, which is defined as half the peak-to-trough height), or vertical height, of a wave is defined as half the peak-to-trough height; as the amplitude of a wave with a given frequency increases, so does its energy. As you can see in Figure $2$, two waves can have the same amplitude but different wavelengths and vice versa. The distance traveled by a wave per unit time is its speed (v), the distance traveled by a wave per unit time, which is typically measured in meters per second (m/s). The speed of a wave is equal to the product of its wavelength and frequency:
\begin{align} \text{(wavelength)(frequency)} &= speed \[4pt] \lambda u &=v \label{eq1a} \[4pt] \left ( \dfrac{meters}{\cancel{wave}} \right )\left ( \dfrac{\cancel{wave}}{second} \right ) &=\dfrac{meters}{second} \label{eq1b} \end{align}
Be careful not to confuse the symbols for the speed, $v$, with the frequency, $u$. Water waves are slow compared to sound waves, which can travel through solids, liquids, and gases. Whereas water waves may travel a few meters per second, the speed of sound in dry air at 20°C is 343.5 m/s. Ultrasonic waves, which travel at an even higher speed (>1500 m/s) and have a greater frequency, are used in such diverse applications as locating underwater objects and the medical imaging of internal organs.
Electromagnetic Radiation
Water waves transmit energy through space by the periodic oscillation of matter (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as electromagnetic radiation, which is energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields. (Figure $3$). Some forms of electromagnetic radiation are shown in Figure $4$. In a vacuum, all forms of electromagnetic radiation—whether microwaves, visible light, or gamma rays—travel at the speed of light (c), which is the speed with which all forms of electromagnetic radiation travel in a vacuum, a fundamental physical constant with a value of 2.99792458 × 108 m/s (which is about 3.00 ×108 m/s or 1.86 × 105 mi/s). This is about a million times faster than the speed of sound.
Because the various kinds of electromagnetic radiation all have the same speed (c), they differ in only wavelength and frequency. As shown in Figure $4$ and Table $1$, the wavelengths of familiar electromagnetic radiation range from 101 m for radio waves to 10−12 m for gamma rays, which are emitted by nuclear reactions. By replacing v with c in Equation $1$, we can show that the frequency of electromagnetic radiation is inversely proportional to its wavelength:
$\begin{array}{cc} c=\lambda u \ u =\dfrac{c}{\lambda } \end{array} \label{eq2}$
For example, the frequency of radio waves is about 108 Hz, whereas the frequency of gamma rays is about 1020 Hz. Visible light, which is electromagnetic radiation that can be detected by the human eye, has wavelengths between about 7 × 10−7 m (700 nm, or 4.3 × 1014 Hz) and 4 × 10−7 m (400 nm, or 7.5 × 1014 Hz). Note that when frequency increases, wavelength decreases; c being a constant stays the same. Similarly when frequency decreases, the wavelength increases.
Within this visible range our eyes perceive radiation of different wavelengths (or frequencies) as light of different colors, ranging from red to violet in order of decreasing wavelength. The components of white light—a mixture of all the frequencies of visible light—can be separated by a prism, as shown in part (b) in Figure $4$. A similar phenomenon creates a rainbow, where water droplets suspended in the air act as tiny prisms.
Table $1$: Common Wavelength Units for Electromagnetic Radiation
Unit Symbol Wavelength (m) Type of Radiation
picometer pm 10−12 gamma ray
angstrom Å 10−10 x-ray
nanometer nm 10−9 x-ray
micrometer μm 10−6 infrared
millimeter mm 10−3 infrared
centimeter cm 10−2 microwave
meter m 100 radio
As you will soon see, the energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength:
$E\; \propto\; u \label{$3$}$
$E\; \propto\; \dfrac{1}{\lambda } \label{$4$}$
Whereas visible light is essentially harmless to our skin, ultraviolet light, with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the ozone layer absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation.
The energy of electromagnetic radiation increases with increasing frequency and decreasing wavelength.
Example $1$
Your favorite FM radio station, WXYZ, broadcasts at a frequency of 101.1 MHz. What is the wavelength of this radiation?
Given: frequency
Asked for: wavelength
Strategy:
Substitute the value for the speed of light in meters per second into Equation $2$ to calculate the wavelength in meters.
Solution:
From Equation \ref{eq2}, we know that the product of the wavelength and the frequency is the speed of the wave, which for electromagnetic radiation is 2.998 × 108 m/s:
$λν = c = 2.998 \times 10^8 m/s$
Thus the wavelength λ is given by
$\lambda =\dfrac{c}{ u }=\left ( \dfrac{2.988\times 10^{8}\; m/\cancel{s}}{101.1\; \cancel{MHz}} \right )\left ( \dfrac{1\; \cancel{MHz}}{10^{6}\; \cancel{s^{-1}}} \right )=2.965\; m$
Exercise $1$
As the police officer was writing up your speeding ticket, she mentioned that she was using a state-of-the-art radar gun operating at 35.5 GHz. What is the wavelength of the radiation emitted by the radar gun?
Answer: 8.45 mm
Electromagnetic Radiation: https://youtu.be/TZy7a69pP-w
Summary
Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties: wavelength (λ), the distance between successive waves; frequency (ν), the number of waves that pass a fixed point per unit time; speed (v), the rate at which the wave propagates through space; and amplitude, the magnitude of the oscillation about the mean position. The speed of a wave is equal to the product of its wavelength and frequency. Electromagnetic radiation consists of two perpendicular waves, one electric and one magnetic, propagating at the speed of light (c). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.01%3A_Electromagnetic_Radiation.txt |
Learning Objectives
• To know the relationship between atomic spectra and the electronic structure of atoms.
The photoelectric effect provided indisputable evidence for the existence of the photon and thus the particle-like behavior of electromagnetic radiation. The concept of the photon, however, emerged from experimentation with thermal radiation, electromagnetic radiation emitted as the result of a source’s temperature, which produces a continuous spectrum of energies. More direct evidence was needed to verify the quantized nature of electromagnetic radiation. In this section, we describe how experimentation with visible light provided this evidence.
Line Spectra
Although objects at high temperature emit a continuous spectrum of electromagnetic radiation, a different kind of spectrum is observed when pure samples of individual elements are heated. For example, when a high-voltage electrical discharge is passed through a sample of hydrogen gas at low pressure, the resulting individual isolated hydrogen atoms caused by the dissociation of H2 emit a red light. Unlike blackbody radiation, the color of the light emitted by the hydrogen atoms does not depend greatly on the temperature of the gas in the tube. When the emitted light is passed through a prism, only a few narrow lines, called a line spectrum, which is a spectrum in which light of only a certain wavelength is emitted or absorbed, rather than a continuous range of wavelengths (Figure $1$), rather than a continuous range of colors. The light emitted by hydrogen atoms is red because, of its four characteristic lines, the most intense line in its spectrum is in the red portion of the visible spectrum, at 656 nm. With sodium, however, we observe a yellow color because the most intense lines in its spectrum are in the yellow portion of the spectrum, at about 589 nm.
Such emission spectra were observed for many other elements in the late 19th century, which presented a major challenge because classical physics was unable to explain them. Part of the explanation is provided by the observation of only a few values of λ (or ν) in the line spectrum meant that only a few values of E were possible. Thus the energy levels of a hydrogen atom had to be quantized; in other words, only states that had certain values of energy were possible, or allowed. If a hydrogen atom could have any value of energy, then a continuous spectrum would have been observed, similar to blackbody radiation.
In 1885, a Swiss mathematics teacher, Johann Balmer (1825–1898), showed that the frequencies of the lines observed in the visible region of the spectrum of hydrogen fit a simple equation that can be expressed as follows:
$u=constant\; \left ( \dfrac{1}{2^{2}}-\dfrac{1}{n^{^{2}}} \right ) \label{$1$}$
where n = 3, 4, 5, 6. As a result, these lines are known as the Balmer series. The Swedish physicist Johannes Rydberg (1854–1919) subsequently restated and expanded Balmer’s result in the Rydberg equation:
$\dfrac{1}{\lambda }=\Re\; \left ( \dfrac{1}{n^{2}_{1}}-\dfrac{1}{n^{2}_{2}} \right ) \label{$2$}$
where $n_1$ and $n_2$ are positive integers, $n_2 > n_1$, and $\Re$ the Rydberg constant, has a value of 1.09737 × 107 m−1.
Johann Balmer (1825–1898)
A mathematics teacher at a secondary school for girls in Switzerland, Balmer was 60 years old when he wrote the paper on the spectral lines of hydrogen that made him famous.
Balmer published only one other paper on the topic, which appeared when he was 72 years old.
Like Balmer’s equation, Rydberg’s simple equation described the wavelengths of the visible lines in the emission spectrum of hydrogen (with n1 = 2, n2 = 3, 4, 5,…). More important, Rydberg’s equation also described the wavelengths of other series of lines that would be observed in the emission spectrum of hydrogen: one in the ultraviolet (n1 = 1, n2 = 2, 3, 4,…) and one in the infrared (n1 = 3, n2 = 4, 5, 6). Unfortunately, scientists had not yet developed any theoretical justification for an equation of this form.
The Bohr Atom: https://youtu.be/GuFQEOzFOgA
Summary
There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy excited state to a higher-energy excited state. This produces an absorption spectrum, which has dark lines in the same position as the bright lines in the emission spectrum of an element.
Atoms of individual elements emit light at only specific wavelengths, producing a line spectrum rather than the continuous spectrum of all wavelengths produced by a hot object. Niels Bohr explained the line spectrum of the hydrogen atom by assuming that the electron moved in circular orbits and that orbits with only certain radii were allowed. Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states. Transitions from an excited state to a lower-energy state resulted in the emission of light with only a limited number of wavelengths. Bohr’s model could not, however, explain the spectra of atoms heavier than hydrogen.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.02%3A_Atomic_Spectra.txt |
Learning Objectives
• To understand how energy is quantized.
By the late 19th century, many physicists thought their discipline was well on the way to explaining most natural phenomena. They could calculate the motions of material objects using Newton’s laws of classical mechanics, and they could describe the properties of radiant energy using mathematical relationships known as Maxwell’s equations, developed in 1873 by James Clerk Maxwell, a Scottish physicist. The universe appeared to be a simple and orderly place, containing matter, which consisted of particles that had mass and whose location and motion could be accurately described, and electromagnetic radiation, which was viewed as having no mass and whose exact position in space could not be fixed. Thus matter and energy were considered distinct and unrelated phenomena. Soon, however, scientists began to look more closely at a few inconvenient phenomena that could not be explained by the theories available at the time.
Blackbody Radiation
One phenomenon that seemed to contradict the theories of classical physics was blackbody radiation, which is electromagnetic radiation whose wavelength and color that depends on the temperature of the object. The wavelength of energy emitted by an object depends on only its temperature, not its surface or composition. Hence an electric stove burner or the filament of a space heater glows dull red or orange when heated, whereas the much hotter tungsten wire in an incandescent light bulb gives off a yellowish light.
The intensity of radiation is a measure of the energy emitted per unit area. A plot of the intensity of blackbody radiation as a function of wavelength for an object at various temperatures is shown in Figure $2$. One of the major assumptions of classical physics was that energy increased or decreased in a smooth, continuous manner. For example, classical physics predicted that as wavelength decreased, the intensity of the radiation an object emits should increase in a smooth curve without limit at all temperatures, as shown by the broken line for 6000 K in Figure $2$. Thus classical physics could not explain the sharp decrease in the intensity of radiation emitted at shorter wavelengths (primarily in the ultraviolet region of the spectrum), which was referred to as the “ultraviolet catastrophe.” In 1900, however, the German physicist Max Planck (1858–1947) explained the ultraviolet catastrophe by proposing that the energy of electromagnetic waves is quantized rather than continuous. This means that for each temperature, there is a maximum intensity of radiation that is emitted in a blackbody object, corresponding to the peaks in Figure $2$, so the intensity does not follow a smooth curve as the temperature increases, as predicted by classical physics. Thus energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantum (the smallest possible unit of energy). Energy can be gained or lost only in integral multiples of a quantum..
Max Planck (1858–1947)
In addition to being a physicist, Planck was a gifted pianist, who at one time considered music as a career. During the 1930s, Planck felt it was his duty to remain in Germany, despite his open opposition to the policies of the Nazi government.
One of his sons was executed in 1944 for his part in an unsuccessful attempt to assassinate Hitler, and bombing during the last weeks of World War II destroyed Planck’s home. After WWII, the major German scientific research organization was renamed the Max Planck Society.
Although quantization may seem to be an unfamiliar concept, we encounter it frequently. For example, US money is integral multiples of pennies. Similarly, musical instruments like a piano or a trumpet can produce only certain musical notes, such as C or F sharp. Because these instruments cannot produce a continuous range of frequencies, their frequencies are quantized. Even electrical charge is quantized: an ion may have a charge of −1 or −2 but not −1.33 electron charges.
Planck postulated that the energy of a particular quantum of radiant energy could be described Max Planck (1858–1947) by the equation
$E=h u \label{8.3.1}$
where the proportionality constant h is called Planck’s constant, one of the most accurately known fundamental constants in science. For our purposes, its value to four significant figures is generally sufficient:
h = 6.626 × 10−34 J•s (joule-seconds)
As the frequency of electromagnetic radiation increases, the magnitude of the associated quantum of radiant energy increases. By assuming that energy can be emitted by an object only in integral multiples of hν, Planck devised an equation that fit the experimental data shown in Figure $2$. We can understand Planck’s explanation of the ultraviolet catastrophe qualitatively as follows: At low temperatures, radiation with only relatively low frequencies is emitted, corresponding to low-energy quanta. As the temperature of an object increases, there is an increased probability of emitting radiation with higher frequencies, corresponding to higher-energy quanta. At any temperature, however, it is simply more probable for an object to lose energy by emitting a large number of lower-energy quanta than a single very high-energy quantum that corresponds to ultraviolet radiation. The result is a maximum in the plot of intensity of emitted radiation versus wavelength, as shown in Figure $2$, and a shift in the position of the maximum to lower wavelength (higher frequency) with increasing temperature.
At the time he proposed his radical hypothesis, Planck could not explain why energies should be quantized. Initially, his hypothesis explained only one set of experimental data—blackbody radiation. If quantization were observed for a large number of different phenomena, then quantization would become a law. In time, a theory might be developed to explain that law. As things turned out, Planck’s hypothesis was the seed from which modern physics grew.
The Photoelectric Effect
Only five years after he proposed it, Planck’s quantization hypothesis was used to explain a second phenomenon that conflicted with the accepted laws of classical physics. When certain metals are exposed to light, electrons are ejected from their surface (Figure $3$). Classical physics predicted that the number of electrons emitted and their kinetic energy should depend on only the intensity of the light, not its frequency. In fact, however, each metal was found to have a characteristic threshold frequency of light; below that frequency, no electrons are emitted regardless of the light’s intensity. Above the threshold frequency, the number of electrons emitted was found to be proportional to the intensity of the light, and their kinetic energy was proportional to the frequency. This phenomenon was called the photoelectric effect (A phenomenon in which electrons are ejected from the surface of a metal that has been exposed to light).
Albert Einstein (1879–1955; Nobel Prize in Physics, 1921) quickly realized that Planck’s hypothesis about the quantization of radiant energy could also explain the photoelectric effect. The key feature of Einstein’s hypothesis was the assumption that radiant energy arrives at the metal surface in particles that we now call photons (a quantum of radiant energy, each of which possesses a particular energy energy E given by Equation 8.3.1 Einstein postulated that each metal has a particular electrostatic attraction for its electrons that must be overcome before an electron can be emitted from its surface (Eo = hνo). If photons of light with energy less than Eo strike a metal surface, no single photon has enough energy to eject an electron, so no electrons are emitted regardless of the intensity of the light. If a photon with energy greater than Eo strikes the metal, then part of its energy is used to overcome the forces that hold the electron to the metal surface, and the excess energy appears as the kinetic energy of the ejected electron:
$kinetic\; energy\; of\; ejected\; electron=E-E_{o}=h u -h u _{o}=h\left ( u - u _{o} \right ) \label{8.3.2}$
When a metal is struck by light with energy above the threshold energy Eo, the number of emitted electrons is proportional to the intensity of the light beam, which corresponds to the number of photons per square centimeter, but the kinetic energy of the emitted electrons is proportional to the frequency of the light. Thus Einstein showed that the energy of the emitted electrons depended on the frequency of the light, contrary to the prediction of classical physics.
Albert Einstein (1879–1955)
In 1900, Einstein was working in the Swiss patent office in Bern. He was born in Germany and throughout his childhood his parents and teachers had worried that he might be developmentally disabled. The patent office job was a low-level civil service position that was not very demanding, but it did allow Einstein to spend a great deal of time reading and thinking about physics.
In 1905, his "miracle year" he published four papers that revolutionized physics. One was on the special theory of relativity, a second on the equivalence of mass and energy, a third on Brownian motion, and the fourth on the photoelectric effect, for which he received the Nobel Prize in 1921, the theory of relativity and energy-matter equivalence being still controversial at the time
Planck’s and Einstein’s postulate that energy is quantized is in many ways similar to Dalton’s description of atoms. Both theories are based on the existence of simple building blocks, atoms in one case and quanta of energy in the other. The work of Planck and Einstein thus suggested a connection between the quantized nature of energy and the properties of individual atoms.
Example $1$
A ruby laser, a device that produces light in a narrow range of wavelengths emits red light at a wavelength of 694.3 nm (Figure $4$). What is the energy in joules of a single photon?
Given: wavelength
Asked for: energy of single photon.
Strategy:
1. Use Equation 6.1.2 and Equation 8.3.1 to calculate the energy in joules.
Solution:
The energy of a single photon is given by E = hν = hc/λ.
Exercise $1$
An x-ray generator, such as those used in hospitals, emits radiation with a wavelength of 1.544 Å.
1. What is the energy in joules of a single photon?
2. How many times more energetic is a single x-ray photon of this wavelength than a photon emitted by a ruby laser?
Answer
1. $1.287 \times 10^{-15}\; J/photon$
2. 4497 times
The Photoelectric Effect: https://youtu.be/mxBMxJLauQk
Summary
• The fundamental building blocks of energy are quanta and of matter are atoms.
The properties of blackbody radiation, the radiation emitted by hot objects, could not be explained with classical physics. Max Planck postulated that energy was quantized and could be emitted or absorbed only in integral multiples of a small unit of energy, known as a quantum. The energy of a quantum is proportional to the frequency of the radiation; the proportionality constant h is a fundamental constant (Planck’s constant). Albert Einstein used Planck’s concept of the quantization of energy to explain the photoelectric effect, the ejection of electrons from certain metals when exposed to light. Einstein postulated the existence of what today we call photons, particles of light with a particular energy, E = hν. Both energy and matter have fundamental building blocks: quanta and atoms, respectively.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.03%3A_Quantum_Theory.txt |
Learning Objectives
• To know the relationship between atomic spectra and the electronic structure of atoms.
In 1913, a Danish physicist, Niels Bohr (1885–1962; Nobel Prize in Physics, 1922), proposed a theoretical model for the hydrogen atom that explained its emission spectrum. Bohr’s model required only one assumption: The electron moves around the nucleus in circular orbits that can have only certain allowed radii. Rutherford’s earlier model of the atom had also assumed that electrons moved in circular orbits around the nucleus and that the atom was held together by the electrostatic attraction between the positively charged nucleus and the negatively charged electron. Although we now know that the assumption of circular orbits was incorrect, Bohr’s insight was to propose that the electron could occupy only certain regions of space.
Using classical physics, Niels Bohr showed that the energy of an electron in a particular orbit is given by
$E_{n}=\dfrac{-\Re hc}{n^{2}} \label{$3$}$
where $\Re$ is the Rydberg constant, h is Planck’s constant, c is the speed of light, and n is a positive integer corresponding to the number assigned to the orbit, with n = 1 corresponding to the orbit closest to the nucleus. In this model n = ∞ corresponds to the level where the energy holding the electron and the nucleus together is zero. In that level, the electron is unbound from the nucleus and the atom has been separated into a negatively charged (the electron) and a positively charged (the nucleus) ion. In this state the radius of the orbit is also infinite. The atom has been ionized.
Niels Bohr (1885–1962)
During the Nazi occupation of Denmark in World War II, Bohr escaped to the United States, where he became associated with the Atomic Energy Project.
In his final years, he devoted himself to the peaceful application of atomic physics and to resolving political problems arising from the development of atomic weapons.
As n decreases, the energy holding the electron and the nucleus together becomes increasingly negative, the radius of the orbit shrinks and more energy is needed to ionize the atom. The orbit with n = 1 is the lowest lying and most tightly bound. The negative sign in Equation $3$ indicates that the electron-nucleus pair is more tightly bound when they are near each other than when they are far apart. Because a hydrogen atom with its one electron in this orbit has the lowest possible energy, this is the ground state (the most stable arrangement of electrons for an element or a compound), the most stable arrangement for a hydrogen atom. As n increases, the radius of the orbit increases; the electron is farther from the proton, which results in a less stable arrangement with higher potential energy (Figure 2.10). A hydrogen atom with an electron in an orbit with n > 1 is therefore in an excited state. Any arrangement of electrons that is higher in energy than the ground state.: its energy is higher than the energy of the ground state. When an atom in an excited state undergoes a transition to the ground state in a process called decay, it loses energy by emitting a photon whose energy corresponds to the difference in energy between the two states (Figure $1$ ).
So the difference in energy (ΔE) between any two orbits or energy levels is given by $\Delta E=E_{n_{1}}-E_{n_{2}}$ where n1 is the final orbit and n2 the initial orbit. Substituting from Bohr’s equation (Equation $3$) for each energy value gives
$\Delta E=E_{final}-E_{initial}=-\dfrac{\Re hc}{n_{2}^{2}}-\left ( -\dfrac{\Re hc}{n_{1}^{2}} \right )=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{$4$}$
If n2 > n1, the transition is from a higher energy state (larger-radius orbit) to a lower energy state (smaller-radius orbit), as shown by the dashed arrow in part (a) in Figure $3$. Substituting hc/λ for ΔE gives
$\Delta E = \dfrac{hc}{\lambda }=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{$5$}$
Canceling hc on both sides gives
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{$6$}$
Except for the negative sign, this is the same equation that Rydberg obtained experimentally. The negative sign in Equation $5$ and Equation $6$ indicates that energy is released as the electron moves from orbit n2 to orbit n1 because orbit n2 is at a higher energy than orbit n1. Bohr calculated the value of $\Re$ from fundamental constants such as the charge and mass of the electron and Planck's constant and obtained a value of 1.0974 × 107 m−1, the same number Rydberg had obtained by analyzing the emission spectra.
We can now understand the physical basis for the Balmer series of lines in the emission spectrum of hydrogen (part (b) in Figure 2.9 ). As shown in part (b) in Figure $3$, the lines in this series correspond to transitions from higher-energy orbits (n > 2) to the second orbit (n = 2). Thus the hydrogen atoms in the sample have absorbed energy from the electrical discharge and decayed from a higher-energy excited state (n > 2) to a lower-energy state (n = 2) by emitting a photon of electromagnetic radiation whose energy corresponds exactly to the difference in energy between the two states (part (a) in Figure $3$ ). The n = 3 to n = 2 transition gives rise to the line at 656 nm (red), the n = 4 to n = 2 transition to the line at 486 nm (green), the n = 5 to n = 2 transition to the line at 434 nm (blue), and the n = 6 to n = 2 transition to the line at 410 nm (violet). Because a sample of hydrogen contains a large number of atoms, the intensity of the various lines in a line spectrum depends on the number of atoms in each excited state. At the temperature in the gas discharge tube, more atoms are in the n = 3 than the n ≥ 4 levels. Consequently, the n = 3 to n = 2 transition is the most intense line, producing the characteristic red color of a hydrogen discharge (part (a) in Figure $1$ ). Other families of lines are produced by transitions from excited states with n > 1 to the orbit with n = 1 or to orbits with n ≥ 3. These transitions are shown schematically in Figure $4$
In contemporary applications, electron transitions are used in timekeeping that needs to be exact. Telecommunications systems, such as cell phones, depend on timing signals that are accurate to within a millionth of a second per day, as are the devices that control the US power grid. Global positioning system (GPS) signals must be accurate to within a billionth of a second per day, which is equivalent to gaining or losing no more than one second in 1,400,000 years. Quantifying time requires finding an event with an interval that repeats on a regular basis. To achieve the accuracy required for modern purposes, physicists have turned to the atom. The current standard used to calibrate clocks is the cesium atom. Supercooled cesium atoms are placed in a vacuum chamber and bombarded with microwaves whose frequencies are carefully controlled. When the frequency is exactly right, the atoms absorb enough energy to undergo an electronic transition to a higher-energy state. Decay to a lower-energy state emits radiation. The microwave frequency is continually adjusted, serving as the clock’s pendulum. In 1967, the second was defined as the duration of 9,192,631,770 oscillations of the resonant frequency of a cesium atom, called the cesium clock. Research is currently under way to develop the next generation of atomic clocks that promise to be even more accurate. Such devices would allow scientists to monitor vanishingly faint electromagnetic signals produced by nerve pathways in the brain and geologists to measure variations in gravitational fields, which cause fluctuations in time, that would aid in the discovery of oil or minerals.
Example $1$: The Lyman Series
The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. In what region of the electromagnetic spectrum does it occur?
Given: lowest-energy orbit in the Lyman series
Asked for: wavelength of the lowest-energy Lyman line and corresponding region of the spectrum
Strategy:
1. Substitute the appropriate values into Equation $2$ (the Rydberg equation) and solve for $\lambda$.
2. Use Figure $1$ to locate the region of the electromagnetic spectrum corresponding to the calculated wavelength.
Solution:
We can use the Rydberg equation to calculate the wavelength:
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right )$
A For the Lyman series, n1 = 1. The lowest-energy line is due to a transition from the n = 2 to n = 1 orbit because they are the closest in energy.
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right )=1.097\times m^{-1}\left ( \dfrac{1}{1}-\dfrac{1}{4} \right )=8.228 \times 10^{6}\; m^{-1}$
It turns out that spectroscopists (the people who study spectroscopy) use cm-1 rather than m-1 as a common unit. Wavelength is inversely proportional to energy but frequency is directly proportional as shown by Planck's formula, E=h$u$.
Spectroscopists often talk about energy and frequency as equivalent. The cm-1 unit is particularly convenient. The infrared range is roughly 200 - 5,000 cm-1, the visible from 11,000 to 25.000 cm-1 and the UV between 25,000 and 100,000 cm-1. The units of cm-1 are called wavenumbers, although people often verbalize it as inverse centimeters. We can convert the answer in part A to cm-1.
$\varpi =\dfrac{1}{\lambda }=8.228\times 10^{6}\cancel{m^{-1}}\left (\dfrac{\cancel{m}}{100\;cm} \right )=82,280\: cm^{-1}$
and
$\lambda = 1.215 \times 10^{−7}\; m = 122\; nm$
This emission line is called Lyman alpha. It is the strongest atomic emission line from the sun and drives the chemistry of the upper atmosphere of all the planets producing ions by stripping electrons from atoms and molecules. It is completely absorbed by oxygen in the upper stratosphere, dissociating O2 molecules to O atoms which react with other O2 molecules to form stratospheric ozone
B This wavelength is in the ultraviolet region of the spectrum.
Exercise $1$: The Pfund Series
The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the n = 5 orbit. Calculate the wavelength of the second line in the Pfund series to three significant figures. In which region of the spectrum does it lie?
Answer: 4.65 × 103 nm; infrared
Bohr’s model of the hydrogen atom gave an exact explanation for its observed emission spectrum. The following are his key contributions to our understanding of atomic structure:
• Electrons can occupy only certain regions of space, called orbits.
• Orbits closer to the nucleus are lower in energy.
• Electrons can move from one orbit to another by absorbing or emitting energy, giving rise to characteristic spectra.
Unfortunately, Bohr could not explain why the electron should be restricted to particular orbits. Also, despite a great deal of tinkering, such as assuming that orbits could be ellipses rather than circles, his model could not quantitatively explain the emission spectra of any element other than hydrogen (Figure $5$). In fact, Bohr’s model worked only for species that contained just one electron: H, He+, Li2+, and so forth. Scientists needed a fundamental change in their way of thinking about the electronic structure of atoms to advance beyond the Bohr model.
Thus far we have explicitly considered only the emission of light by atoms in excited states, which produces an emission spectrum (a spectrum produced by the emission of light by atoms in excited states). The converse, absorption of light by ground-state atoms to produce an excited state, can also occur, producing an absorption spectrum (a spectrum produced by the absorption of light by ground-state atoms). Because each element has characteristic emission and absorption spectra, scientists can use such spectra to analyze the composition of matter.
When an atom emits light, it decays to a lower energy state; when an atom absorbs light, it is excited to a higher energy state.
The Energy States of the Hydrogen Atom
If white light is passed through a sample of hydrogen, hydrogen atoms absorb energy as an electron is excited to higher energy levels (orbits with n ≥ 2). If the light that emerges is passed through a prism, it forms a continuous spectrum with black lines (corresponding to no light passing through the sample) at 656, 468, 434, and 410 nm. These wavelengths correspond to the n = 2 to n = 3, n = 2 to n = 4, n = 2 to n = 5, and n = 2 to n = 6 transitions. Any given element therefore has both a characteristic emission spectrum and a characteristic absorption spectrum, which are essentially complementary images.
Emission and absorption spectra form the basis of spectroscopy, which uses spectra to provide information about the structure and the composition of a substance or an object. In particular, astronomers use emission and absorption spectra to determine the composition of stars and interstellar matter. As an example, consider the spectrum of sunlight shown in Figure $7$ Because the sun is very hot, the light it emits is in the form of a continuous emission spectrum. Superimposed on it, however, is a series of dark lines due primarily to the absorption of specific frequencies of light by cooler atoms in the outer atmosphere of the sun. By comparing these lines with the spectra of elements measured on Earth, we now know that the sun contains large amounts of hydrogen, iron, and carbon, along with smaller amounts of other elements. During the solar eclipse of 1868, the French astronomer Pierre Janssen (1824–1907) observed a set of lines that did not match those of any known element. He suggested that they were due to the presence of a new element, which he named helium, from the Greek helios, meaning “sun.” Helium was finally discovered in uranium ores on Earth in 1895. Alpha particles are helium nuclei. Alpha particles emitted by the radioactive uranium, pick up electrons from the rocks to form helium atoms.
The familiar red color of “neon” signs used in advertising is due to the emission spectrum of neon shown in part (b) in Figure $5$. Similarly, the blue and yellow colors of certain street lights are caused, respectively, by mercury and sodium discharges. In all these cases, an electrical discharge excites neutral atoms to a higher energy state, and light is emitted when the atoms decay to the ground state. In the case of mercury, most of the emission lines are below 450 nm, which produces a blue light (part (c) in Figure $5$). In the case of sodium, the most intense emission lines are at 589 nm, which produces an intense yellow light.
The Bohr Atom: https://youtu.be/GuFQEOzFOgA
The Chemistry of Fireworks
The colors of fireworks are also due to atomic emission spectra. As shown in part (a) in Figure $9$, a typical shell used in a fireworks display contains gunpowder to propel the shell into the air and a fuse to initiate a variety of reactions that produce heat and small explosions. Thermal energy excites the atoms to higher energy states; as they decay to lower energy states, the atoms emit light that gives the familiar colors.
When oxidant/reductant mixtures listed in Table $1$ are ignited, a flash of white or yellow light is produced along with a loud bang. Achieving the colors shown in part (b) in Figure $9$ requires adding a small amount of a substance that has an emission spectrum in the desired portion of the visible spectrum. For example, sodium is used for yellow because of its 589 nm emission lines. The intense yellow color of sodium would mask most other colors, so potassium and ammonium salts, rather than sodium salts, are usually used as oxidants to produce other colors, which explains the preponderance of such salts in Table $1$. Strontium salts, which are also used in highway flares, emit red light, whereas barium gives a green color. Blue is one of the most difficult colors to achieve. Copper(II) salts emit a pale blue light, but copper is dangerous to use because it forms highly unstable explosive compounds with anions such as chlorate. As you might guess, preparing fireworks with the desired properties is a complex, challenging, and potentially hazardous process. If you have the time here is a NOVA program about how fireworks are made.
Table $1$: Common Chemicals Used in the Manufacture of Fireworks*
Oxidizers Fuels (reductants) Special effects
ammonium perchlorate aluminum blue flame: copper carbonate, copper sulfate, or copper oxide
barium chlorate antimony sulfide red flame: strontium nitrate or strontium carbonate
barium nitrate charcoal white flame: magnesium or aluminum
potassium chlorate magnesium yellow flame: sodium oxalate or cryolite (Na3AlF6)
potassium nitrate sulfur green flame: barium nitrate or barium chlorate
potassium perchlorate titanium white smoke: potassium nitrate plus sulfur
strontium nitrate colored smoke: potassium chlorate and sulfur, plus organic dye
whistling noise: potassium benzoate or sodium salicylate
white sparks: aluminum, magnesium, or titanium
gold sparks: iron fillings or charcoal
*Almost any combination of an oxidizer and a fuel may be used along with the compounds needed to produce a desired special effect.
Summary
There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Atoms of individual elements emit light at only specific wavelengths, producing a line spectrum rather than the continuous spectrum of all wavelengths produced by a hot object. Niels Bohr explained the line spectrum of the hydrogen atom by assuming that the electron moved in circular orbits and that orbits with only certain radii were allowed. Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states. Transitions from an excited state to a lower-energy state resulted in the emission of light with only a limited number of wavelengths. Bohr’s model could not, however, explain the spectra of atoms heavier than hydrogen.
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.04%3A_The_Bohr_Atom.txt |
Learning Objectives
• To understand the wave–particle duality of matter.
Einstein’s photons of light were individual packets of energy having many of the characteristics of particles. Recall that the collision of an electron (a particle) with a sufficiently energetic photon can eject a photoelectron from the surface of a metal. Any excess energy is transferred to the electron and is converted to the kinetic energy of the ejected electron. Einstein’s hypothesis that energy is concentrated in localized bundles, however, was in sharp contrast to the classical notion that energy is spread out uniformly in a wave. We now describe Einstein’s theory of the relationship between energy and mass, a theory that others built on to develop our current model of the atom.
The Wave Character of Matter
Einstein initially assumed that photons had zero mass, which made them a peculiar sort of particle indeed. In 1905, however, he published his special theory of relativity, which related energy and mass according to the following equation:
$E=h\nu=h\dfrac{c}{\lambda }=mc^{2} \label{8.5.1}$
According to this theory, a photon of wavelength λ and frequency ν has a nonzero mass, which is given as follows:
$m=\dfrac{E}{c^{2}}=\dfrac{h\nu }{c^{2}}=\dfrac{h}{\lambda c} \label{8.5.2}$
That is, light, which had always been regarded as a wave, also has properties typical of particles, a condition known as wave–particle duality (a principle that matter and energy have properties typical of both waves and particles). Depending on conditions, light could be viewed as either a wave or a particle.
In 1922, the American physicist Arthur Compton (1892–1962) reported the results of experiments involving the collision of x-rays and electrons that supported the particle nature of light. At about the same time, a young French physics student, Louis de Broglie (1892–1972), began to wonder whether the converse was true: Could particles exhibit the properties of waves? In his PhD dissertation submitted to the Sorbonne in 1924, de Broglie proposed that a particle such as an electron could be described by a wave whose wavelength is given by
$\lambda =\dfrac{h}{mv} \label{8.5.3}$
where
• is Planck’s constant,
• m is the mass of the particle, and
• v is the velocity of the particle.
This revolutionary idea was quickly confirmed by American physicists Clinton Davisson (1881–1958) and Lester Germer (1896–1971), who showed that beams of electrons, regarded as particles, were diffracted by a sodium chloride crystal in the same manner as x-rays, which were regarded as waves. It was proven experimentally that electrons do exhibit the properties of waves. For his work, de Broglie received the Nobel Prize in Physics in 1929.
If particles exhibit the properties of waves, why had no one observed them before? The answer lies in the numerator of de Broglie’s equation, which is an extremely small number. As you will calculate in Example $1$, Planck’s constant (6.63 × 10−34 J•s) is so small that the wavelength of a particle with a large mass is too short (less than the diameter of an atomic nucleus) to be noticeable.
Example $1$
Calculate the wavelength of a baseball, which has a mass of 149 g and a speed of 100 mi/h.
Given: mass and speed of object
Asked for: wavelength
Strategy:
1. Convert the speed of the baseball to the appropriate SI units: meters per second.
2. Substitute values into Equation 8.5.3 and solve for the wavelength.
Solution:
The wavelength of a particle is given by λ = h/mv. We know that m = 0.149 kg, so all we need to find is the speed of the baseball:
$v=\left ( \dfrac{100\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )$
B Recall that the joule is a derived unit, whose units are (kg•m2)/s2. Thus the wavelength of the baseball is
$\lambda =\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )}= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )}=9.95\times 10^{-35}\; m$
(You should verify that the units cancel to give the wavelength in meters.) Given that the diameter of the nucleus of an atom is approximately 10−14 m, the wavelength of the baseball is almost unimaginably small.
Exercise $1$
Calculate the wavelength of a neutron that is moving at 3.00 × 103 m/s.
Answer: 1.32 Å, or 132 pm
As you calculated in Example $1$, objects such as a baseball or a neutron have such short wavelengths that they are best regarded primarily as particles. In contrast, objects with very small masses (such as photons) have large wavelengths and can be viewed primarily as waves. Objects with intermediate masses, such as electrons, exhibit the properties of both particles and waves. Although we still usually think of electrons as particles, the wave nature of electrons is employed in an electron microscope, which has revealed most of what we know about the microscopic structure of living organisms and materials. Because the wavelength of an electron beam is much shorter than the wavelength of a beam of visible light, this instrument can resolve smaller details than a light microscope can (Figure $1$).
An Important Wave Property: Phase
A wave is a disturbance that travels in space. The magnitude of the wave at any point in space and time varies sinusoidally. While the absolute value of the magnitude of one wave at any point is not very important, the relative displacement of two waves called the phase difference, is vitally important because it determines whether the waves reinforce or interfere with each other. Figure $2$a, on the right shows an arbitrary phase difference between two wave. Figure $2$b shows what happens when the two waves are 180 degrees out of phase. The green line is their sum. Figure $2$c shows what happens when the two lines are in phase, exactly superimposed on each other. Again, the green line is the sum of the intensities.
Standing Waves
De Broglie also investigated why only certain orbits were allowed in Bohr’s model of the hydrogen atom. He hypothesized that the electron behaves like a standing wave (a wave that does not travel in space). An example of a standing wave is the motion of a string of a violin or guitar. When the string is plucked, it vibrates at certain fixed frequencies because it is fastened at both ends (Figure $3$ ). If the length of the string is L, then the lowest-energy vibration (the fundamental (the lowest-energy standing wave) has wavelength
$\begin{array}{ll} \dfrac{\lambda }{2}=L \ \lambda =2L \end{array} \label{8.5.4}$
Higher-energy vibrations are called overtones (the vibration of a standing wave that is higher in energy than the fundamental vibration) and are produced when the string is plucked more strongly; they have wavelengths given by
$\lambda=\dfrac{2L}{n} \label{8.5.5}$
where n has any integral value. Thus the resonant vibrational energies of the string are quantized. When plucked, all other frequencies die out immediately. Only the resonant frequencies survive and are heard. By analogy we can think of the resonant frequencies as being quantized. Notice in Figure $3$ that all overtones have one or more nodes (the points where the amplitude of a wave is zero), points where the string does not move. The amplitude of the wave at a node is zero.
Quantized vibrations and overtones containing nodes are not restricted to one-dimensional systems, such as strings. A two-dimensional surface, such as a drumhead, also has quantized vibrations. Similarly, when the ends of a string are joined to form a circle, the only allowed vibrations are those with wavelength
$2πr = nλ \label{8.5.6}$
where r is the radius of the circle. De Broglie argued that Bohr’s allowed orbits could be understood if the electron behaved like a standing circular wave (Figure $4$). The standing wave could exist only if the circumference of the circle was an integral multiple of the wavelength such that the propagated waves were all in phase, thereby increasing the net amplitudes and causing constructive interference. Otherwise, the propagated waves would be out of phase, resulting in a net decrease in amplitude and causing destructive interference. The non resonant waves interfere with themselves! De Broglie’s idea explained Bohr’s allowed orbits and energy levels nicely: in the lowest energy level, corresponding to n = 1 in Equation 8.5.2, one complete wavelength would close the circle. Higher energy levels would have successively higher values of n with a corresponding number of nodes.
Standing waves are often observed on rivers, reservoirs, ponds, and lakes when seismic waves from an earthquake travel through the area. The waves are called seismic seiches, a term first used in 1955 when lake levels in England and Norway oscillated from side to side as a result of the Assam earthquake of 1950 in Tibet. They were first described in the Proceedings of the Royal Society in 1755 when they were seen in English harbors and ponds after a large earthquake in Lisbon, Portugal. Seismic seiches were also observed in many places in North America after the Alaska earthquake of March 28, 1964. Those occurring in western reservoirs lasted for two hours or longer, and amplitudes reached as high as nearly 6 ft along the Gulf Coast. The height of seiches is approximately proportional to the thickness of surface sediments; a deeper channel will produce a higher seiche.
Still, as all analogies, although the standing wave model helps us understand much about why Bohr's theory worked, it also, if pushed too far can mislead.
As you will see, some of de Broglie’s ideas are retained in the modern theory of the electronic structure of the atom: the wave behavior of the electron and the presence of nodes that increase in number as the energy level increases. Unfortunately, his (and Bohr's) explanation also contains one major feature that we know to be incorrect: in the currently accepted model, the electron in a given orbit is not always at the same distance from the nucleus.
The de Broglie Equation: https://youtu.be/pz0zMHWtK7Q
The Heisenberg Uncertainty Principle
Because a wave is a disturbance that travels in space, it has no fixed position. One might therefore expect that it would also be hard to specify the exact position of a particle that exhibits wavelike behavior. A characteristic of light is that is can be bent or spread out by passing through a narrow slit as shown in the video below. You can literally see this by half closing your eyes and looking through your eye lashes. This reduces the brightness of what you are seeing and somewhat fuzzes out the image, but the light bends around your lashes to provide a complete image rather than a bunch of bars across the image. This is called diffraction.
This behavior of waves is captured in Maxwell's equations (1870 or so) for electromagnetic waves and was and is well understood. Heisenberg's uncertainty principle for light is, if you will, merely a conclusion about the nature of electromagnetic waves and nothing new. DeBroglie's idea of wave particle duality means that particles such as electrons which all exhibit wave like characteristics, will also undergo diffraction from slits whose size is of the order of the electron wavelength.
This situation was described mathematically by the German physicist Werner Heisenberg (1901–1976; Nobel Prize in Physics, 1932), who related the position of a particle to its momentum. Referring to the electron, Heisenberg stated that “at every moment the electron has only an inaccurate position and an inaccurate velocity, and between these two inaccuracies there is this uncertainty relation.” Mathematically, the Heisenberg uncertainty principle is greater than or equal to Planck’s constant h divided by 4π: states that the uncertainty in the position of a particle (Δx) multiplied by the uncertainty in its momentum [Δ(mv)] is greater than or equal to Planck’s constant divided by 4π:
$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \label{8.5.7}$
Because Planck’s constant is a very small number, the Heisenberg uncertainty principle is important only for particles such as electrons that have very low masses. These are the same particles predicted by de Broglie’s equation to have measurable wavelengths.
If the precise position x of a particle is known absolutely (Δx = 0), then the uncertainty in its momentum must be infinite:
$\left ( \Delta \left [ mv \right ] \right )= \dfrac{h}{4\pi \left ( \Delta x \right ) }=\dfrac{h}{4\pi \left ( 0 \right ) }=\infty \label{8.5.8}$
Because the mass of the electron at rest (m) is both constant and accurately known, the uncertainty in Δ(mv) must be due to the Δv term, which would have to be infinitely large for Δ(mv) to equal infinity. That is, according to Equation 8.5.8, the more accurately we know the exact position of the electron (as Δx → 0), the less accurately we know the speed and the kinetic energy of the electron (1/2 mv2) because Δ(mv) → ∞. Conversely, the more accurately we know the precise momentum (and the energy) of the electron [as Δ(mv) → 0], then Δx → ∞ and we have no idea where the electron is.
Bohr’s model of the hydrogen atom violated the Heisenberg uncertainty principle by trying to specify simultaneously both the position (an orbit of a particular radius) and the energy (a quantity related to the momentum) of the electron. Moreover, given its mass and wavelike nature, the electron in the hydrogen atom could not possibly orbit the nucleus in a well-defined circular path as predicted by Bohr. You will see, however, that the most probable radius of the electron in the hydrogen atom is exactly the one predicted by Bohr’s model.
Example $2$
Calculate the minimum uncertainty in the position of the pitched baseball from Example $1$ that has a mass of exactly 149 g and a speed of 100 ± 1 mi/h.
Given: mass and speed of object
Asked for: minimum uncertainty in its position
Strategy:
1. Rearrange the inequality that describes the Heisenberg uncertainty principle (Equation 8.5.7) to solve for the minimum uncertainty in the position of an object (Δx).
2. Find Δv by converting the velocity of the baseball to the appropriate SI units: meters per second.
3. Substitute the appropriate values into the expression for the inequality and solve for Δx.
Solution:
A The Heisenberg uncertainty principle tells us that (Δx)[Δ(mv)] = h/4π. Rearranging the inequality gives
$\Delta x \ge \left( {\dfrac{h}{4\pi }} \right)\left( {\dfrac{1}{\Delta (mv)}} \right)$
B We know that h = 6.626 × 10−34 J•s and m = 0.149 kg. Because there is no uncertainty in the mass of the baseball, Δ(mv) = mΔv and Δv = ±1 mi/h. We have
$\Delta \nu =\left ( \dfrac{1\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1\; \cancel{min}}{60\; s} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )=0.4469\; m/s$
C Therefore,
$\Delta x \geqslant \left ( \dfrac{6.626\times 10^{-34}\; J\cdot s}{4\left ( 3.1416 \right )} \right ) \left ( \dfrac{1}{\left ( 0.149\; kg \right )\left ( 0.4469\; m\cdot s^{-1} \right )} \right )$
Inserting the definition of a joule (1 J = 1 kg•m2/s2) gives
$\Delta x \geqslant \left ( \dfrac{6.626\times 10^{-34}\; \cancel{kg} \cdot m^{\cancel{2}} \cdot s}{4\left ( 3.1416 \right )\left ( \cancel{s^{2}} \right )} \right ) \left ( \dfrac{1\; \cancel{s}}{\left ( 0.149\; \cancel{kg} \right )\left ( 0.4469\; \cancel{m} \right )} \right )$
$\Delta x \geqslant 7.92 \pm \times 10^{-34}\; m$
This is equal to 3.12 × 10−32 inches. We can safely say that if a batter misjudges the speed of a fastball by 1 mi/h (about 1%), he will not be able to blame Heisenberg’s uncertainty principle for striking out.
Exercise $2$
Calculate the minimum uncertainty in the position of an electron traveling at one-third the speed of light, if the uncertainty in its speed is ±0.1%. Assume its mass to be equal to its mass at rest.
Answer: 6 × 10−10 m, or 0.6 nm (about the diameter of a benzene molecule)
Videos and Examples
Answers for these quizzes are included.
Summary
• An electron possesses both particle and wave properties.
The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called wave–particle duality. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle.
$\lambda =\dfrac{h}{mv} \tag{8.5.3}$
The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the fundamental vibration, and higher-energy vibrations are overtones and have successively more nodes, points where the amplitude of the wave is always zero. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.
$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \tag{8.5.7}$
Contributors and Attributions
Modified by Joshua Halpern (Howard University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.05%3A_Two_Ideas_Leading_to_a_New_Quantum_Mechanics.txt |
Learning Objectives
• To apply the results of quantum mechanics to chemistry.
The paradox described by Heisenberg’s uncertainty principle and the wavelike nature of subatomic particles such as the electron made it impossible to use the equations of classical physics to describe the motion of electrons in atoms. Scientists needed a new approach that took the wave behavior of the electron into account. In 1926, an Austrian physicist, Erwin Schrödinger (1887–1961; Nobel Prize in Physics, 1933), developed wave mechanics, a mathematical technique that describes the relationship between the motion of a particle that exhibits wavelike properties (such as an electron) and its allowed energies. In doing so, Schrödinger developed the theory of quantum mechanics, which is used today to describe the energies and spatial distributions of electrons in atoms and molecules.
Schrödinger (1887–1961)
Schrödinger’s unconventional approach to atomic theory was typical of his unconventional approach to life. He was notorious for his intense dislike of memorizing data and learning from books. When Hitler came to power in Germany, Schrödinger escaped to Italy. He then worked at Princeton University in the United States but eventually moved to the Institute for Advanced Studies in Dublin, Ireland, where he remained until his retirement in 1955.
Although quantum mechanics uses sophisticated mathematics, you do not need to understand the mathematical details to follow our discussion of its general conclusions. We focus on the properties of the wave functions that are the solutions of Schrödinger’s equations.
Wave Functions
A wave function (Ψ) is a mathematical function that relates the location of an electron at a given point in space (identified by x, y, and z coordinates) to the amplitude of its wave, which corresponds to its energy. Thus each wave function is associated with a particular energy E. The properties of wave functions derived from quantum mechanics are summarized here:
• A wave function uses three variables to describe the position of an electron. A fourth variable is usually required to fully describe the location of objects in motion. Three specify the position in space (as with the Cartesian coordinates x, y, and z), and one specifies the time at which the object is at the specified location. For example, if you wanted to intercept an enemy submarine, you would need to know its latitude, longitude, and depth, as well as the time at which it was going to be at this position (Figure $1$). For electrons, we can ignore the time dependence because we will be using standing waves, which by definition do not change with time, to describe the position of an electron.
If you are the captain of a ship trying to intercept an enemy submarine, you need to deliver your depth charge to the right location at the right time.
• The magnitude of the wave function at a particular point in space is proportional to the amplitude of the wave at that point. Many wave functions are complex functions, which is a mathematical term indicating that they contain $\sqrt{-1}$, represented as $i$. Hence the amplitude of the wave has no real physical significance. In contrast, the sign of the wave function (either positive or negative) corresponds to the phase of the wave, which will be important in our discussion of chemical bonding. The sign of the wave function should not be confused with a positive or negative electrical charge.
• The square of the wave function at a given point is proportional to the probability of finding an electron at that point, which leads to a distribution of probabilities in space. The square of the wave function ($\Psi^2$) is always a real quantity [recall that that $\sqrt{-1}^2=-1$] is proportional to the probability of finding an electron at a given point. More accurately, the probability is given by the product of the wave function Ψ and its complex conjugate Ψ*, in which all terms that contain i are replaced by $−i$. We use probabilities because, according to Heisenberg’s uncertainty principle, we cannot precisely specify the position of an electron. The probability of finding an electron at any point in space depends on several factors, including the distance from the nucleus and, in many cases, the atomic equivalent of latitude and longitude. As one way of graphically representing the probability distribution, the probability of finding an electron is indicated by the density of colored dots, as shown for the ground state of the hydrogen atom in Figure $2$.
• Describing the electron distribution as a standing wave leads to sets of quantum numbers that are characteristic of each wave function. From the patterns of one- and two-dimensional standing waves shown in Figure 6.18 and Figure 6.19, you might expect (correctly) that the patterns of three-dimensional standing waves would be complex. Fortunately, however, in the 18th century, a French mathematician, Adrien Legendre (1752–1783), developed a set of equations to describe the motion of tidal waves on the surface of a flooded planet. Schrödinger incorporated Legendre’s equations into his wave functions. The requirement that the waves must be in phase with one another to avoid cancellation and produce a standing wave results in a limited number of solutions (wave functions), each of which is specified by a set of numbers called quantum numbers.
• Each wave function is associated with a particular energy. As in Bohr’s model, the energy of an electron in an atom is quantized; it can have only certain allowed values. The major difference between Bohr’s model and Schrödinger’s approach is that Bohr had to impose the idea of quantization arbitrarily, whereas in Schrödinger’s approach, quantization is a natural consequence of describing an electron as a standing wave.
Quantum Numbers
Schrödinger’s approach uses three quantum numbers (n, l, and ml) to specify any wave function. The quantum numbers provide information about the spatial distribution of an electron. Although n can be any positive integer, only certain values of l and ml are allowed for a given value of n.
Introduction to Quantum Numbers: https://youtu.be/07JpBeaPxL8
The Principal Quantum Number
The principal quantum number (n) tells the average relative distance of an electron from the nucleus:
$n = 1, 2, 3, 4,… \label{8.6.1}$
As n increases for a given atom, so does the average distance of an electron from the nucleus. A negatively charged electron that is, on average, closer to the positively charged nucleus is attracted to the nucleus more strongly than an electron that is farther out in space. This means that electrons with higher values of n are easier to remove from an atom. All wave functions that have the same value of n are said to constitute a principal shell because those electrons have similar average distances from the nucleus. As you will see, the principal quantum number n corresponds to the n used by Bohr to describe electron orbits and by Rydberg to describe atomic energy levels.
The Azimuthal Quantum Number
The second quantum number is often called the azimuthal quantum number (l). The value of l describes the shape of the region of space occupied by the electron. The allowed values of l depend on the value of n and can range from 0 to n − 1:
$l = 0, 1, 2,…, n − 1 \label{8.6.2}$
For example, if n = 1, l can be only 0; if n = 2, l can be 0 or 1; and so forth. For a given atom, all wave functions that have the same values of both n and l form a subshell. The regions of space occupied by electrons in the same subshell usually have the same shape, but they are oriented differently in space.
Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: https://youtu.be/ms7WR149fAY
The Magnetic Quantum Number
The third quantum number is the magnetic quantum number (ml). The value of $m_l$ describes the orientation of the region in space occupied by an electron with respect to an applied magnetic field. The allowed values of $m_l$ depend on the value of l: ml can range from −l to l in integral steps:
$m_l = −l, −l + 1,…, 0,…, l − 1, l \label{8.6.3}$
For example, if $l = 0$, $m_l$ can be only 0; if l = 1, ml can be −1, 0, or +1; and if l = 2, ml can be −2, −1, 0, +1, or +2.
Each wave function with an allowed combination of n, l, and ml values describes an atomic orbital, a particular spatial distribution for an electron. For a given set of quantum numbers, each principal shell has a fixed number of subshells, and each subshell has a fixed number of orbitals.
Example $1$: n=4 Shell Structure
How many subshells and orbitals are contained within the principal shell with n = 4?
Given: value of n
Asked for: number of subshells and orbitals in the principal shell
Strategy:
1. Given n = 4, calculate the allowed values of l. From these allowed values, count the number of subshells.
2. For each allowed value of l, calculate the allowed values of ml. The sum of the number of orbitals in each subshell is the number of orbitals in the principal shell.
Solution:
A We know that l can have all integral values from 0 to n − 1. If n = 4, then l can equal 0, 1, 2, or 3. Because the shell has four values of l, it has four subshells, each of which will contain a different number of orbitals, depending on the allowed values of ml.
B For l = 0, ml can be only 0, and thus the l = 0 subshell has only one orbital. For l = 1, ml can be 0 or ±1; thus the l = 1 subshell has three orbitals. For l = 2, ml can be 0, ±1, or ±2, so there are five orbitals in the l = 2 subshell. The last allowed value of l is l = 3, for which ml can be 0, ±1, ±2, or ±3, resulting in seven orbitals in the l = 3 subshell. The total number of orbitals in the n = 4 principal shell is the sum of the number of orbitals in each subshell and is equal to n2:
Exercise $1$: n=3 Shell Structure
How many subshells and orbitals are in the principal shell with n = 3?
Answer: three subshells; nine orbitals
Rather than specifying all the values of n and l every time we refer to a subshell or an orbital, chemists use an abbreviated system with lowercase letters to denote the value of l for a particular subshell or orbital:
l = 0 1 2 3
Designation s p d f
The principal quantum number is named first, followed by the letter s, p, d, or f as appropriate. These orbital designations are derived from corresponding spectroscopic characteristics: sharp, principle, diffuse, and fundamental. A 1s orbital has n = 1 and l = 0; a 2p subshell has n = 2 and l = 1 (and has three 2p orbitals, corresponding to ml = −1, 0, and +1); a 3d subshell has n = 3 and l = 2 (and has five 3d orbitals, corresponding to ml = −2, −1, 0, +1, and +2); and so forth.
We can summarize the relationships between the quantum numbers and the number of subshells and orbitals as follows (Table $1$):
• Each principal shell has n subshells. For n = 1, only a single subshell is possible (1s); for n = 2, there are two subshells (2s and 2p); for n = 3, there are three subshells (3s, 3p, and 3d); and so forth. Every shell has an ns subshell, any shell with n ≥ 2 also has an np subshell, and any shell with n ≥ 3 also has an nd subshell. Because a 2d subshell would require both n = 2 and l = 2, which is not an allowed value of l for n = 2, a 2d subshell does not exist.
• Each subshell has 2l + 1 orbitals. This means that all ns subshells contain a single s orbital, all np subshells contain three p orbitals, all nd subshells contain five d orbitals, and all nf subshells contain seven f orbitals.
Note
Each principal shell has n subshells, and each subshell has 2l + 1 orbitals.
Table $1$ Values of n, l, and ml through n = 4
n l Subshell Designation ml Number of Orbitals in Subshell Number of Orbitals in Shell
1 0 1s 0 1 1
2 0 2s 0 1 4
1 2p −1, 0, 1 3
3 0 3s 0 1 9
1 3p −1, 0, 1 3
2 3d −2, −1, 0, 1, 2 5
4 0 4s 0 1 16
1 4p −1, 0, 1 3
2 4d −2, −1, 0, 1, 2 5
3 4f −3, −2, −1, 0, 1, 2, 3 7
Magnetic Quantum Number (ml) & Spin Quantum Number (ms): https://youtu.be/gbmGVUXBOBk
Summary
• There is a relationship between the motions of electrons in atoms and molecules and their energies that is described by quantum mechanics.
Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wave functions (Ψ) to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies. Wave functions have five important properties:
1. the wave function uses three variables (Cartesian axes x, y, and z) to describe the position of an electron;
2. the magnitude of the wave function is proportional to the intensity of the wave;
3. the probability of finding an electron at a given point is proportional to the square of the wave function at that point, leading to a distribution of probabilities in space that is often portrayed as an electron density plot;
4. describing electron distributions as standing waves leads naturally to the existence of sets of quantum numbers characteristic of each wave function; and
5. each spatial distribution of the electron described by a wave function with a given set of quantum numbers has a particular energy.
Quantum numbers provide important information about the energy and spatial distribution of an electron. The principal quantum number n can be any positive integer; as n increases for an atom, the average distance of the electron from the nucleus also increases. All wave functions with the same value of n constitute a principal shell in which the electrons have similar average distances from the nucleus. The azimuthal quantum number l can have integral values between 0 and n − 1; it describes the shape of the electron distribution. Wave functions that have the same values of both n and l constitute a subshell, corresponding to electron distributions that usually differ in orientation rather than in shape or average distance from the nucleus. The magnetic quantum number ml can have 2l + 1 integral values, ranging from −l to +l, and describes the orientation of the electron distribution. Each wave function with a given set of values of n, l, and ml describes a particular spatial distribution of an electron in an atom, an atomic orbital. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.06%3A_Wave_Mechanics.txt |
Now that we have introduced the basic concepts of quantum mechanics, we can start to apply these concepts to build up matter, starting from its most elementary constituents, namely atoms, up to molecules, supramolecular complexes (complexes built from weak interactions such as hydrogen bonds and van der Waals interactions), networks, and bulk condensed phases, including liquids, glasses, solids,... As explore the structure of matter, itself, we should stand back and wonder at how the mathematically elegant, yet somehow not quite tangible, structure of quantum theory is able to describe so accurately all phases of matter and types of substances, ranging from metallic and semiconducting crystals to biological macromolecules, to morphologically complex polymeric materials.
We begin with the simplest system, the hydrogen atom (and hydrogen-like single-electron cations), which is the only atomic system (thus far) for which the Schrödinger equation can be solved exactly for the energy levels and wave functions. To this end, we consider a nucleus of charge $+Ze$ at the origin and a single electron a distance $r$ away from it. If we just naïvely start by writing the (total) classical energy
$\underbrace{\dfrac{p_{x}^{2}}{2m}+\dfrac{p_{y}^{2}}{2m}+\dfrac{p_{z}^{2}}{2m}}_{\text{Kinetic Energy}} \underbrace{-\dfrac{Ze^2}{4\pi \epsilon_0 \sqrt{x^2 +y^2 +z^2}}}_{\text{Potential Energy}}=E \label{8.1.1}$
where the constant $k$ in Coulomb's Law has been rewritten as $1/4\pi \epsilon_0$, where $\epsilon_0$ is the permittivity of free space, and where we have written $r=\sqrt{x^2 +y^2 +z^2}$ in terms of its Cartesian components, then we see immediately that the energy is not a simple sum of terms involving $(x,p_x)$, $(y, p_y)$, and $(z,p_z )$, and therefore, the wave function is not a simple product of a function of $x$, a function of $y$ and a function of $z$. Thus, it would seem that we have already hit a mathematical wall. Fortunately, we are not confined to work in Cartesian coordinates. In fact, there are numerous ways to locate a point $r$ in space, and for this problem, there is a coordinate system, known as spherical coordinates , that is particularly convenient because it uses the distance $r$ of a point from the origin as one of its explicit coordinates.
Spherical Coordinates
In Cartesian coordinates, a point in a three-dimensional space requires three numbers to locate it $r=(x,y,z)$. Thus, if we change to a different coordinate system, we still need three numbers to full locate the point. Figure $1$ shows how to locate a point in the system of spherical coordinates:
Thus, we use the distance $r$ of the point from the origin, which is also the magnitude of the vector $r$, the angle $\theta$ the vector $r$ makes with the positive z -axis, and the angle $\phi$ between the projection of the vector $r$ into the $xy$ plane and the positive x -axis. The angle $\theta$ is called the polar angle and the angle $\phi$ is called the azimuthal angle . These three variables have the following ranges
$r \in [0,\infty ) \;\;\;\; \theta \in [0,\pi ] \;\;\;\; \phi \in [0,2\pi ] \label{8.1.2}$
To map a point from the Cartesian representation $r=(x,y,z)$ to spherical coordinates $(r, \theta ,\phi )$, a coordinate transformation is needed tells us how points in the two coordinate representations are connected. To determine $x$ and $y$, we need the component of $r$ in the $xy$ plane, which is $r\sin\theta$, and its projection into the $x$ and $y$ directions, which are
\begin{align*}x &= r\sin\theta \cos\phi \ y &= r\sin \theta \sin \phi \end{align*} \label{8.1.3}
Finally, to determine $z$, we simply need the component of $r$ along the z-axis, which is $r\cos\theta$, so
$z=r\cos\theta \label{8.1.4}$
These three relations tell us how to map the point $(r,\theta ,\phi )$ into the original Cartesian frame $(x,y,z)$. The inverse of this transformation tells us how to map the Cartesian numbers $(x,y,z)$ back into spherical coordinates. It is simple algebra to show that the inverse is
\begin{align*}r &= \sqrt{x^2 +y^2 +z^2}\ \phi &= \tan^{-1}\dfrac{y}{x}\ \theta &= \tan^{-1}\dfrac{\sqrt{x^2+y^2}}{z}\end{align*} \label{8.1.5}
A note about vectors
Consider a two-dimensional vector $v$ in the $xy$ plane. The vector can be represented in terms of its two components $v=(v_x ,v_y)$. That is, just two numbers $v_x$ and $v_y$ are needed, and we know everything about the vector. However, if we know the angle $\theta$ that $v$ makes with the x-axis, then we can also use the magnitude $v=|v|$ and the angle as two alternative numbers that can be used to represent the vector. In this case, since $v_x =v\cos\theta =|v|\cos\theta$ and $v_y =v\sin \theta =|v|\sin\theta$, then $v$ is specified as $v=(v\cos\theta ,v\sin\theta )$. There is still another alternative scheme for representing $v$. We can use the magnitude $v=|v|$ and just one of the components of $v$. Suppose we wish to represent $v$ in terms of $v$ and $v_x$. since $v=v_{x}^{2}+v_{y}^{2}$, the other component $v_y$ is given by $v_{y}^{2}=v^2 -v_{x}^{2}$, then $v_y =\sqrt{v^2 -v_{x}^{2}}$, we can write $v=(v_x ,\sqrt{v^2-v_{x}^{2}})$, which means we only need to know the magnitude $v$ and one of the components to specify $v$ entirely. This is the scheme that will be used below to represent the angular momentum, as discussed in the next section.
Angular momentum
In spherical coordinates, the momentum $p$ of the electron has a radial component $p_r$, corresponding to motion radially outward from the origin, and an angular component $L$, corresponding to motion along the surface of a sphere of radius $r$, i.e. motion perpendicular to the radial direction:
$p=\left ( p_r, \dfrac{L}{r} \right ) \label{8.1.6}$
Note that $L$ is, itself, still a vector, since motion along the surface of a sphere still has more than one component. Its components include motion along the $\theta$ and along the $\phi$ directions.
Let us step back briefly into Cartesian components to look at the angular momentum vector $L$. Its definition is
$L=r \times p \label{8.1.7}$
i.e. the vector cross product of $r$ and $p$, which is why $L$ is perpendicular to $r$. In Cartesian components, $L$ actually has three components, which are given by
\begin{align*}L_x &= yp_z -zp_y \ L_y &= zp_x -xp_z \ L_z &= xp_y -yp_x\end{align*} \label{8.1.8}
using the definition of the cross product.
The important thing about angular momentum is that if the potential energy $V$ depends only on $r$, then angular momentum is conserved. For example, suppose $V(r)$ is the Coulomb potential, which we will write compactly as
$V(r)=-\dfrac{k}{r} \label{8.1.9}$
where $k=e^2 /(4\pi \epsilon_0 )$. We will show that angular momentum is conserved within the classical mechanical description of the system. If it is true classically, it is also true in quantum mechanics.
The force on the classical electron is
$F=-\dfrac{k}{r^3}r \label{8.1.10}$
Therefore, the three force components are
$F_x =-\dfrac{kx}{r^3}\;\;\;\; F_y =-\dfrac{ky}{r^3}\;\;\;\; F_z =-\dfrac{kz}{r^3} \label{8.1.11}$
From Newton's second law:
$F=ma=m\dfrac{dv}{dt}=\dfrac{d(mv)}{dt}=\dfrac{dp}{dt} \label{8.1.12}$
Now consider one of the components of the angular momentum vector, e.g. $L_z$. Its time derivative is
\begin{align*}\dfrac{dL_z}{dt} &= \dfrac{dx}{dt}p_y +x\dfrac{dp_y}{dt}-\dfrac{dy}{dt}p_x -y\dfrac{dp_x}{dt}\ &= \dfrac{p_x}{m}p_y +xF_y -\dfrac{p_y}{m}p_x -yF_x \ &=\dfrac{p_x p_y }{m} -\dfrac{kxy}{r^3} -\dfrac{p_y p_x}{m}+\dfrac{kyx}{r^3} \ &= 0\end{align*} \label{8.1.13}
The same can be shown about $L_x$ and $L_y$. Moreover, since $L^2 =L_{x}^{2}+L_{y}^{2}+L_{z}^{2}$, and the three components $L_x$, $L_y$ and $L_z$ are all constant, $L^2$ is constant as well.
Thus, both the energy and the angular momentum are conserved in classical mechanics, but classically, both quantities can take on any values. In quantum mechanics, the energy can have only certain allowed values, and because angular momentum is conserved, it too can only have certain allowed values, as Bohr predicted in his model of the hydrogen atom. The allowed energies are
$E_n = -\dfrac{Z^2 e^4 m_e}{8\epsilon_{0}^{2}h^2}\dfrac{1}{n^2} \label{8.1.14}$
where $n=1,2,3,...$
What about angular momentum? In spherical coordinates, the momentum is $p=(p_r ,L/r)$, and $L$ only has two components, one along $\theta$ and one along $\phi$. We can assign allowed values to these components separately, however, there is a compelling reason to do things a little differently. In spherical coordinates, the classical energy is given by
$\dfrac{p_{r}^{2}}{2m_e}+\dfrac{L^2}{2m_e r^2}-\dfrac{Ze^2}{4\pi \epsilon_0 r}=E \label{8.1.15}$
Thus, $E$ separates into a purely radial part and an angular part:
$E=\varepsilon_r +\dfrac{L^2}{2m_e r^2}$
Note that the energy depends on $L^2$. since $L^2$ and each of the three components $L_x$, $L_y$ and $L_z$ are conserved, we can use the representation of $L$ in which we choose the magnitude $L$ and one of the components. Why just one component? Note that even though $L$ has three Cartesian components, it really is only a two dimensional vector since it describes the motion on a two-dimensional surface. Just like the $xy$ plane is a two-dimensional surface, the surface of a sphere is also two-dimensional, using the angles $\theta$ and $\phi$ to locate points on the surface.
The difference is that the surface of a sphere is not flat but curved, so we have to embed it in three-dimensional space. So as a Cartesian representation, three components are needed, but in spherical coordinates, we only need two, i.e. two numbers to represent $L$. Thus, we can use the magnitude $L$ and one of its components, which by convention is taken to be the z -component $L_z$. $L_z$ describes motion in the $xy$ plane, i.e. motion in the azimuthal ($\phi$) di rection only. However, this is just motion on a ring, so one part of the wave function will just be the particle-on-a-ring wave functions
$\dfrac{1}{\sqrt{2\pi }}e^{im\phi} \ \label{8.1.17}$
where $m=0,\pm 1,\pm 2,...$, and so the allowed values of $L_z$ are $m\hbar$.
For $L^2$, it turns out that the allowed values are
$L^2 \rightarrow l(l+1)\hbar^2 \label{8.1.18}$
where $l$ is an integer that has the range $l=0,1,2,...,n-1$ in a given energy level characterized by $n$. Moreover, because $L_z \leq |L|$, once the value of $l$ is specified, $m$ cannot exceed $l$, so the range of $m$ is $m=-l ,-l+1 ,...,l-1,l$.
Schrödinger Equation for the Hydrogen atom and hydrogen-like Cations
The Schrödinger equation for single-electron Coulomb systems in spherical coordinates is
$-\dfrac{\hbar^2}{2m_e r^2}\left [ r\dfrac{\partial^2}{\partial r^2}(r\psi (r,\theta ,\phi))+\dfrac{1}{\sin\theta }\dfrac{\partial }{\partial \theta}\sin\theta \dfrac{\partial }{\partial \theta}\psi (r,\theta ,\phi )+\dfrac{1}{\sin^2 \theta}\dfrac{\partial^2}{\partial \phi^2}\psi (r, \theta ,\phi) \right ] -\dfrac{Ze^2}{4\pi \epsilon_0 r}\psi (r, \theta ,\phi ) \=E\psi (r, \theta ,\phi ) \label{Seq}$
This type of equation is an example of a partial differential equation , which is no simple task to solve. However, solving it gives both the allowed values of the angular momentum discussed above and the allowed energies $E_n$, which agree with the simpler Bohr model. Thus, we do not need to assume anything except the validity of the Schrödinger equation, and the allowed values of energy and angular momentum, together with the corresponding wavefunctions, all emerge from the solution.
Obviously, we are not going to go through the solution of the Schrödinger equation, but we can understand something about its mechanics and the solutions from a few simple considerations. Remember that the Schrödinger equation is set up starting from the classical energy, which we said takes the form
$\dfrac{p_{r}^{2}}{2m_e}+\dfrac{L^2}{2m_e r^2}-\dfrac{Ze^2}{4\pi \epsilon_0 r}=E$
which we can write as
$E=\varepsilon_r +\dfrac{L^2}{2m_e r^2}$
where
$\varepsilon_r =\dfrac{p_{r}^{2}}{2m_e}-\dfrac{Ze^2}{4\pi \epsilon_0 r}$
The term $L^2/(2m_e r^2)$ is actually dependent only on $\theta$ and $\phi$, so it is purely angular. Given the separability of the energy into radial and angular terms, the wavefunction can be decomposed into a product of the form
$\psi (r,\theta ,\phi )=R(r)Y(\theta ,\phi )$
Solution of the angular part for the function $Y(\theta ,\phi )$ yields the allowed values of the angular momentum $L^2$ and the z -component $L_z$. The functions $Y(\theta ,\phi )$ are then characterized by the integers $l$ and $m$, and are denoted $Y_{lm}(\theta ,\phi )$. They are known as spherical harmonics . Here we present just a few of them for a few values of $l$.
For $l=0$, there is just one value of $m$, $m=0$, and, therefore, one spherical harmonic, which turns out to be a simple constant:
$Y_{00}(\theta ,\phi )=\dfrac{1}{\sqrt{4\pi}}$
For $l=1$, there are three values of $m$, $m=-1,0,1$, and, therefore, three functions $Y_{1m}(\theta ,\phi )$. These are given by
\begin{align*}Y_{11}(\theta ,\phi ) &= -\left ( \dfrac{3}{8\pi} \right )^{1/2}\sin\theta e^{i\phi }\ Y_{1-1}(\theta ,\phi ) &= \left ( \dfrac{3}{8\pi} \right )^{1/2} \sin\theta e^{-i\phi }\ Y_{10}(\theta ,\phi ) &= \left ( \dfrac{3}{4\pi } \right )^{1/2}\cos\theta \end{align*}
Note: Exponentials of Complex Numbers
Remember that
$e^{i\phi } = \cos\phi +i\sin\phi$
$e^{-i\phi } = \cos\phi -i\sin\phi$
For $l=2$, there are five values of $m$, $m=-2,-1,0,1,2$, and, therefore, five spherical harmonics, given by
\begin{align*}Y_{22}(\theta ,\phi ) &=\left ( \dfrac{15}{32\pi } \right )^{1/2}\sin^2 \theta e^{2i\phi }\ Y_{2-2}(\theta ,\phi ) &= \left ( \dfrac{15}{32\pi} \right )^{1/2}\sin^2\theta e^{-2i\phi }\ Y_{21}(\theta ,\phi ) &= -\left ( \dfrac{15}{8\pi } \right )^{1/2}\sin\theta \cos\theta e^{i\phi }\ Y_{2-1}(\theta ,\phi ) &= \left ( \dfrac{15}{8\pi } \right )^{1/2}\sin\theta \cos\theta e^{-i\phi }\ Y_{20}(\theta ,\phi ) &= \left ( \dfrac{5}{16\pi} \right )^{1/2} (3\cos^2 \theta -1)\end{align*}
The remaining function $R(r)$ is characterized by the integers $n$ and $l$, as this function satisfies the radial part of the Schrödinger equation, also known as the radial Schrödinger equation:
$-\dfrac{\hbar^2}{2m_e}\dfrac{1}{r}\dfrac{d^2}{dr^2}(rR_{nl}(r))-\dfrac{l(l+1)\hbar^2}{2m_e r^2}R_{nl}(r)-\dfrac{Ze^2}{4\pi \epsilon_0 r}R_{nl}(r)=E_n R_{nl}(r)$
Note that, while the functions $Y_{lm}(\theta ,\phi )$ are not particular to the potential $V(r)$, the radial functions $R_{nl}(r)$ are particular for the Coulomb potential. It is the solution of the radial Schrödinger equation that leads to the allowed energy levels. The boundary conditions that lead to the quantized energies are $rR_{nl}(0)=0$ and $rR_{nl}(\infty )=0$. The radial parts of the wavefunctions that emerge are given by (for the first few values of $n$ and $l$ ):
\begin{align*}R_{10}(r) &= 2\left ( \dfrac{Z}{a_0} \right )^{3/2}e^{-Zr/a_0}\ R_{20}(r) &= \dfrac{1}{2\sqrt{2}}\left ( \dfrac{Z}{a_0} \right )^{3/2} \left ( 2-\dfrac{Zr}{a_0} \right ) e^{-Zr/2a_0}\ R_{21}(r) &= \dfrac{1}{2\sqrt{6}}\left ( \dfrac{Z}{a_0} \right )^{3/2} \dfrac{Zr}{a_0}e^{-Zr/2a_0}\ R_{30}(r) &= \dfrac{2}{81\sqrt{3}}\left ( \dfrac{Z}{a_0} \right )^{3/2}\left [ 27-18\dfrac{Zr}{a_0}+2 \left ( \dfrac{Zr}{a_0} \right )^2 \right ]e^{-Zr/3a_0}\end{align*}
where $a_0$ is the Bohr radius
$a_0 = \dfrac{4\pi \epsilon_0 \hbar^2}{e^2 m_e}=0.529177 \times 10^{-10} \ m$
The full wavefunctions are then composed of products of the radial and angular parts as
$\psi_{nlm}(r,\theta ,\phi)=R_{nl}(r)Y_{lm}(\theta ,\phi )$
At this points, several comments are in order. First, the integers $n,l,m$ that characterize each state are known as the quantum numbers of the system. Each of them corresponds to a quantity that is classically conserved. The number $n$ is known as the principal quantum number, the number $l$ is known as the angular quantum number, and the number $m$ is known as the magnetic quantum number.
As with any quantum system, the wavefunctions $\psi_{nlm}(r,\theta ,\phi )$ give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. The probability of finding the electron in a small volume element $dV$ of space around the point $r=(r,\theta ,\phi )$ is
\begin{align*}P(electron \ in \ dV \ about \ r,\theta ,\phi ) &= |\psi_{nlm}(r,\theta ,\phi )|^2dV\ &= R_{nl}^{2}(r)|Y_{lm}(\theta ,\phi)|^2 dV \end{align*}
What is $dV$ ? In Cartesian coordinates, $dV$ is the volume of a small box of dimensions $dx$, $dy$, and $dz$ in the $x$, $y$, and $z$ directions. That is,
$dV=dx\,dy\,dz$
In spherical coordinates, the volume element $dV$ is a small element of a spherical volume and is given by
$dV=r^2 \sin \theta\, dr\, d\theta \,d\phi \label{dV}$
which is derivable from the transformation equations.
Note: 3-D Integration
If we integrate $dV$ from Equation $\ref{dV}$ over a sphere of radius $R$, we should obtain the volume of the sphere:
$V= \left[\int_0^R r^2 dr\right]\left[\int_0^{\pi}\sin\theta d\theta\right]\left[\int_0^{2\pi}d\phi\right]$
$= \left[\left.{r^3 \over 3}\right\vert _0^R\right]\left[\left.-\cos\theta\right\vert _0^{\pi}\right]\left[\left.\phi\right\vert _0^{2\pi}\right]$
$= {R^3 \over 3}\times 2\times 2\pi$
$= {4 \over 3}\pi R^3$
which is the formula for the volume of a sphere of radius $R$.
Example $1$
The electron in a hydrogen atom $(Z=1)$ is in the state with quantum numbers $n=1$, $l=0$ and $m=0$. What is the probability that a measurement of the electron's position will yield a value $r\geq 2a_0$?
Solution
The wavefunction $\psi_{100}(r,\theta ,\phi )$ is
$\psi_{100}(r,\theta ,\phi )=\dfrac{2}{a_{0}^{3/2}} \left ( \dfrac{1}{4\pi} \right )^{1/2}e^{-r/a_0}$
Therefore, the probability we seek is
$P(r \geq 2a_0 ) = \int_0^{2\pi}\int_0^{\pi}\int_{2a_0}^{\infty} |\psi_{100}(r,\theta ,\phi)|^2 r^2 \sin\,\theta dr\, d\theta \,d\phi\$
$= \dfrac{4}{a_0^{3}}\dfrac{1}{4\pi }\left [ \int_0^{2\pi}d\phi \right ] \left [ \int_0^{\pi} \sin \theta d\theta \right ] \left [ \int_{2a_0}^{\infty}r^2 e^{-2r/a_0}dr \right]$
$= \dfrac{1}{a_{0}^3\pi}\cdot 2\pi \cdot 2\int_{2a_0}^{\infty}r^2e^{-2r/a_0}dr$
Let $x=2r/a_0$. Then
$P(r\geq 2a_0)=\dfrac{1}{2}\int_{4}^{\infty}x^2 e^{-x}dx$
After integrating by parts, we find
$P(r\geq 2a_0)=13e^{-4}\approx 0.24=24\%$
which is relatively large given that this is at least two Bohr radii away from the nucleus!
Radial Probability Distribution Functions
The part of the probability involving the product
$P_{nl}(r)=r^2 R_{nl}^{2}(r) \label{radeq}$
is known as the radial probability distribution function or simply the radial distribution function . $P_{nl}(r)dr$ is the probability that a measurement of the electron's position yields a value in a radial shell of thickness $dr$ and radius $r$ as shown in the Figure $2$:
\What the radial probability distribution shows is that the electron cannot be sucked into the nucleus because $P_{nl}(0)=0$. Hence, as we shrink the radial shell into the nucleus, the probability of finding the electron in that shell goes to 0.
Another point concerns the number of allowed states for each allowed energy. Remember that each wavefunction corresponds to a probability distribution in which the electron can be found for each energy. The more possible states there are, the more varied the electronic properties and behavior of the system will be.
For $n=1$, there is one energy $E_1$ and only one wavefunction $\psi_{100}$.
For $n=2$, there is one energy $E_2$ and four possible states, corresponding to the following allowable values of $l$ and $m$
\begin{align*} l &= 0 \;\;\;\; m=0\ l &= 1 \;\;\;\; m=-1,0,1\end{align*}
Thus, there are four wavefunctions $\psi_{200}$, $\psi_{21-1}$, $\psi_{210}$, and $\psi_{211}$.
Note: Degeneracy
Whenever there is more than one wavefunction corresponding to a given energy level, then that energy level is said to be degenerate . In the above example, the $n=2$ energy level of the hydrogen atom is fourfold degenerate .
Physical Character of the Wavefunctions
The wavefunctions $\psi_{nlm}(r,\theta ,\phi)$ of the electron are called orbitals, but should not be confused these with trajectories or orbits. These are complete static objects that only give static probabilities when the electron has a well-defined allowable energy. The shapes of the orbitals are largely determined by the values of angular momentum, so we will characterize the orbitals this way.
l=0 orbitals
The $l=0$ orbitals are called $s$ ($s$ for "sharp") orbitals. When $l=0$, $m=0$ as well, and the wavefunctions are of the form
$\psi_{n00}(r,\theta ,\phi )=R_{n0}(r)Y_{00}(\theta ,\phi )=\left ( \dfrac{1}{4\pi} \right )^{1/2}R_{n0}(r)$
There is no dependence on $\theta$ or $\phi$ because $Y_{00}$ is a constant. Thus, all of these orbitals are spherically symmetric (Figure $3$).
Note several things about these orbitals. First, the density of points dies off exponentially as $r$ increases, consistent with the exponential dependence of the functions $R_{n0}(r)$ (Figure $4$a) .
As $n$ increases, the exponentials decay more slowly as $e^{-r/a_0}$ for $n=1$, $e^{-r/2a_0}$ for $n=2$ and $e^{-r/3a_0}$ for $n=3$. Note, also, that the wavefunctions are peaked at $r=0$, which would suggest that the amplitude is maximal to find the electron right on top of the nucleus! In fact, we need to be careful about this interpretation, since the radial probability density (Equation $\ref{radeq}$) contains an extra $r^2$ factor from the volume element which goes to $0$ as $r \rightarrow 0$ ( Figure $4$b ).
Figure $5$ compares the radial probability densities of the first three $l=0$ radial wavefunctions.
We also see that the wavefunctions have radial nodes where the electron will never be found (Figure $5$). The number of nodes for $R_{n0}(r)$ is $n-1$.
Note: Time Dependence
The Schrödinger Equation ($\ref{Seq}$) is often called the time-independent Schrödinger Equation and is limiting case of the more complex time-dependent Schrödinger Equation that includes an explicit dependence to the wavefunction (where the "wave" part is most applicable). This is demonstrated in the simplified "drum" models below for the first three $l=0$ orbitals.
Simplified 2D drum models of the 1s (left), 2s (middle) and 3s (left) orbitals.
In all of the modes analogous to s orbitals, it can be seen that the center of the $r=0$ drum vibrates most strongly.
l=1 orbitals
The $l=1$ orbitals are known as $p$ ($"p"$ for principal'') orbitals. The orbitals take the form
$\psi_{nlm}(r,\theta ,\phi )=R_{n1}(r)Y_{1m}(\theta ,\phi )$
where
\begin{align*}Y_{1\pm 1}(\theta ,\phi ) &= \pm \left ( \dfrac{3}{8\pi} \right )^{1/2} \sin\theta e^{\pm i\phi }\ Y_{10}(\theta ,\phi ) &= \left ( \dfrac{3}{4\pi } \right )^{1/2} \cos\theta \end{align*}
Thus, these orbitals are not spherically symmetric.
The $m=0$ orbital is known as the $p_z$ orbital because of the $\cos\theta$ dependence and lack of $\phi$ dependence. This resembles the spherical coordinate transformation for $z$, $z=r\cos\theta$. Figure $6$ shows the $2p$ orbitals in more detail
The nodal plane in the $p$ orbital at $\theta =\pi /2$ arises because $\cos(\pi /2)=0$ for all $\phi$, meaning that the entire $xy$ plane is a nodal plane.
The orbitals $\psi_{n11}(r,\theta ,\phi )$ and $\psi_{n1-1}(r,\theta ,\phi )$ are not real because of the $exp(\pm i\phi )$ dependence of $Y_{1\pm 1}(\theta ,\phi )$. Thus, these orbitals are not entire convenient to work with. Fortunately, because $\psi_{n11}$ and $\psi_{n1-1}$ are solutions of the Schrödinger equation with the same energy $E_n$, we can take any combination of these two functions we wish, and we still have a solution of the Schrödinger equation with the same energy. Thus, it is useful to take two combinations that give us two real orbitals. Consider, for example:
\begin{align*}\tilde{\psi}_{p_x} &= \dfrac{1}{\sqrt{2}}[\psi_{n1-1}(r,\theta ,\phi )-\psi_{n11}(r,\theta ,\phi )]\ \tilde{\psi}_{p_y} &= \dfrac{i}{\sqrt{2}}[\psi_{n1-1}(r,\theta ,\phi )+\psi_{n11}(r,\theta ,\phi )]\end{align*}
which corresponds to defining new spherical harmonics:
\begin{align*}Y_{p_x}(\theta ,\phi ) &= \dfrac{1}{\sqrt{2}}[Y_{1-1}(\theta ,\phi )-Y_{11}(\theta ,\phi )]=\left ( \dfrac{3}{4\pi} \right )^{1/2}\sin\theta \cos\phi \ Y_{p_y}(\theta ,\phi ) &= \dfrac{i}{\sqrt{2}}[Y_{1-1}(\theta ,\phi )+Y_{11}(\theta ,\phi )]=\left ( \dfrac{3}{4\pi} \right )^{1/2}\sin\theta \sin\phi \end{align*}
Again, the notation $p_x$ and $p_y$ is used because of the similarity to the spherical coordinate transformations $x=r\sin\theta \cos\phi$ and $y=r\sin\theta \sin\phi$. These orbitals have the same shape as the $p_z$ orbital but are rotated to be oriented along the x -axis for the $p_x$ orbital and along the y -axis for the $p_y$ orbital.
l=2 orbitals
The $l=2$ orbitals are known as $d$ ($"d"$ for "diffuse") orbitals. Again, we seek combinations of the spherical harmonics that give us real orbitals. The combinations we arrive at are known as $Y_{xy}$, $Y_{xz}$, $Y_{yz}$, $Y_{z^2}$, $Y_{x^2 -y^2 }$, which gives us the required five orbitals we need for $m=-2,-1,0,1,2$. The notation again reflects the angular dependence we would have if we took products $xy$, $xz$, $yz$, $z^2$, and $x^2 -y^2$ using the spherical coordinate transformation equations.
Figure $7$ compared the five $3d$ orbitals. Note the presence of two nodal planes in most of the orbitals. The exception is the $3d_{z^2}$ orbital which has a nodal cone. Here, the number of radial nodes is still $n-l-1$, and, as noted earlier, the overall number of nodes remains the same, so as $l$ increases, radial nodes are exchanged for angular nodes.
For any of the wavefunctions $\psi_{nlm}(r,\theta ,\phi )$, the result of measuring the distance of the electron from the nucleus many times yields the average value or expectation value of $r$, which can be shown to be
\begin{align*}\langle r\rangle &= \int_{0}^{2\pi }\int_{0}^{\pi }\int_{0}^{\infty }|\psi_{nlm}(r,\theta ,\phi )|^2 r^3 \sin\theta drd\theta d\phi \ &= \dfrac{n^2 a_0}{Z}\left [ 1+\dfrac{1}{2}\left ( 1-\dfrac{l(l+1)}{n^2} \right ) \right ]\end{align*} | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.07%3A_Quantum_Numbers_and_Electron_Orbitals.txt |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.