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Learning Objectives
• To describe the unique properties of liquids.
Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to three unique properties of liquids that intimately depend on the nature of intermolecular interactions:
• surface tension,
• capillary action, and
• viscosity.
Surface Tension
If liquids tend to adopt the shapes of their containers, then why do small amounts of water on a freshly waxed car form raised droplets instead of a thin, continuous film? The answer lies in a property called surface tension, which depends on intermolecular forces. Surface tension is the energy required to increase the surface area of a liquid by a unit amount and varies greatly from liquid to liquid based on the nature of the intermolecular forces, e.g., water with hydrogen bonds has a surface tension of 7.29 x 10-2 J/m2 (at 20°C), while mercury with metallic bonds has as surface tension that is 15 times higher: 4.86 x 10-1 J/m2 (at 20°C).
Figure \(1\) presents a microscopic view of a liquid droplet. A typical molecule in the interior of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no net force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the surface experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads. A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads.
The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends above the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water. Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking. This is even observable in the zero gravity conditions of space as shown in Figure \(2\) (and more so in the video link) where water wrung from a wet towel continues to float along the towel's surface!
Such phenomena are manifestations of surface tension, which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m2) or dyne per centimeter (dyn/cm), where 1 dyn = 1 × 10−5 N. The values of the surface tension of some representative liquids are listed in Table \(1\). Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding.
Table \(1\): Surface Tension, Viscosity, Vapor Pressure (at 25°C Unless Otherwise Indicated), and Normal Boiling Points of Common Liquids
Substance Surface Tension (× 10−3 J/m2) Viscosity (mPa•s) Vapor Pressure (mmHg) Normal Boiling Point (°C)
Organic Compounds
diethyl ether 17 0.22 531 34.6
n-hexane 18 0.30 149 68.7
acetone 23 0.31 227 56.5
ethanol 22 1.07 59 78.3
ethylene glycol 48 16.1 ~0.08 198.9
Liquid Elements
bromine 41 0.94 218 58.8
mercury 486 1.53 0.0020 357
Water
0°C 75.6 1.79 4.6
20°C 72.8 1.00 17.5
60°C 66.2 0.47 149
100°C 58.9 0.28 760
Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or surfactants. In the 1960s, US Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids.
Capillary Action
Intermolecular forces also cause a phenomenon called capillary action, which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a capillary), as shown in Figure \(3\). When a glass capillary is is placed in liquid water, water rises up into the capillary. The height to which the water rises depends on the diameter of the tube and the temperature of the water but not on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises.
• Cohesive forces bind molecules of the same type together
• Adhesive forces bind a substance to a surface
Capillary action is the net result of two opposing sets of forces: cohesive forces, which are the intermolecular forces that hold a liquid together, and adhesive forces, which are the attractive forces between a liquid and the substance that composes the capillary. Water has both strong adhesion to glass, which contains polar SiOH groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (Figure \(4\)). The upper surface of a liquid in a tube is called the meniscus, and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (Figure \(4\)).
Polar substances are drawn up a glass capillary and generally have a concave meniscus.
Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body.
Viscosity
Viscosity (η) is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a low viscosity. Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a high viscosity. The two most common methods for evaluating the viscosity of a liquid are (1) to measure the time it takes for a quantity of liquid to flow through a narrow vertical tube and (2) to measure the time it takes steel balls to fall through a given volume of the liquid. The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa•s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in Table 11.3.1 and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH2CH2OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces.
There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous.
Viscosity increases as intermolecular interactions or molecular size increases.
Video Discussing Surface Tension and Viscosity. Video Link: Surface Tension, Viscosity, & Melting Point, YouTube(opens in new window) [youtu.be]
Application: Motor Oils
Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing. As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures.
The viscosity of motor oils is described by an SAE (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity (Figure \(5\)). So-called single-grade oils can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are multigrade, with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils.
Example \(1\)
Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.)
Given: substance and composition of the glass surface
Asked for: behavior of oil and the shape of meniscus
Strategy:
1. Identify the cohesive forces in the motor oil.
2. Determine whether the forces interact with the surface of glass. From the strength of this interaction, predict the behavior of the oil and the shape of the meniscus.
Solution
A Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains.
B Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury.
Exercise \(1\)
Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)?
Answer
Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave.
Summary
Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. Surfactants are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when cohesive forces, the intermolecular forces in the liquid, are weaker than adhesive forces, the attraction between a liquid and the surface of the capillary. The shape of the meniscus, the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The viscosity of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/16%3A_Liquids_and_Solids/16.02%3A_The_Liquid_State.txt |
Solid are characterized by structural rigidity and resistance to changes of shape or volume. Unlike a liquid, a solid object does not flow to take on the shape of its container, nor does expands to fill the entire volume available to it like a gas . The atoms in a solid are tightly bound to each other, either in a regular geometric lattice (crystalline solids, which include metals and ordinary water ice) or irregularly (an amorphous solid such as common window glass).
16.04: Structure and Bonding in Metals
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 $\sum \chi =2.94$. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/16%3A_Liquids_and_Solids/16.03%3A_An_Introduction_to_Structures_and_Types_of_Solids.txt |
Learning Objectives
• To understand the correlation between bonding and the properties of solids.
• To classify solids as ionic, molecular, covalent (network), or metallic, where the general order of increasing strength of interactions.
Crystalline solids fall into one of four categories. All four categories involve packing discrete molecules or atoms into a lattice or repeating array, though network solids are a special case. The categories are distinguished by the nature of the interactions holding the discrete molecules or atoms together. Based on the nature of the forces that hold the component atoms, molecules, or ions together, solids may be formally classified as ionic, molecular, covalent (network), or metallic. The variation in the relative strengths of these four types of interactions correlates nicely with their wide variation in properties.
Table $1$: Solids may be formally classified as ionic, molecular, covalent (network), or metallic
Type of Solid Interaction Properties Examples
Ionic Ionic High Melting Point, Brittle, Hard NaCl, MgO
Molecular Hydrogen Bonding,
Dipole-Dipole,
London Dispersion
Low Melting Point, Nonconducting H2, CO2
Metallic Metallic Bonding Variable Hardness and Melting Point (depending upon strength of metallic bonding), Conducting Fe, Mg
Network Covalent Bonding High Melting Point, Hard, Nonconducting C (diamond),
SiO2 (quartz)
In ionic and molecular solids, there are no chemical bonds between the molecules, atoms, or ions. The solid consists of discrete chemical species held together by intermolecular forces that are electrostatic or Coulombic in nature. This behavior is most obvious for an ionic solid such as $NaCl$, where the positively charged Na+ ions are attracted to the negatively charged $Cl^-$ ions. Even in the absence of ions, however, electrostatic forces are operational. For polar molecules such as $CH_2Cl_2$, the positively charged region of one molecular is attracted to the negatively charged region of another molecule (dipole-dipole interactions). For a nonpolar molecule such as $CO_2$, which has no permanent dipole moment, the random motion of electrons gives rise to temporary polarity (a temporary dipole moment). Electrostatic attractions between two temporarily polarized molecules are called London Dispersion Forces.
Hydrogen bonding is a term describing an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X–H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation. (See the IUPAC Provisional Recommendation on the definition of a hydrogen bond.) Dots are employed to indicate the presence of a hydrogen bond: X–H•••Y. The attractive interaction in a hydrogen bond typically has a strong electrostatic contribution, but dispersion forces and weak covalent bonding are also present.
In metallic solids and network solids, however, chemical bonds hold the individual chemical subunits together. The crystal is essential a single, macroscopic molecule with continuous chemical bonding throughout the entire structure. In metallic solids, the valence electrons are no longer exclusively associated with a single atom. Instead these electrons exist in molecular orbitals that are delocalized over many atoms, producing an electronic band structure. The metallic crystal essentially consists of a set of metal cations in a sea of electrons. This type of chemical bonding is called metallic bonding.
Ionic Solids
You learned previously that an ionic solid consists of positively and negatively charged ions held together by electrostatic forces. The strength of the attractive forces depends on the charge and size of the ions that compose the lattice and determines many of the physical properties of the crystal.
The lattice energy (i.e., the energy required to separate 1 mol of a crystalline ionic solid into its component ions in the gas phase) is directly proportional to the product of the ionic charges and inversely proportional to the sum of the radii of the ions. For example, NaF and CaO both crystallize in the face-centered cubic (fcc) sodium chloride structure, and the sizes of their component ions are about the same: Na+ (102 pm) versus Ca2+ (100 pm), and F (133 pm) versus O2− (140 pm). Because of the higher charge on the ions in CaO, however, the lattice energy of CaO is almost four times greater than that of NaF (3401 kJ/mol versus 923 kJ/mol). The forces that hold Ca and O together in CaO are much stronger than those that hold Na and F together in NaF, so the heat of fusion of CaO is almost twice that of NaF (59 kJ/mol versus 33.4 kJ/mol), and the melting point of CaO is 2927°C versus 996°C for NaF. In both cases, however, the values are large; that is, simple ionic compounds have high melting points and are relatively hard (and brittle) solids.
Molecular Solids
Molecular solids consist of atoms or molecules held to each other by dipole–dipole interactions, London dispersion forces, or hydrogen bonds, or any combination of these. The arrangement of the molecules in solid benzene is as follows:
Because the intermolecular interactions in a molecular solid are relatively weak compared with ionic and covalent bonds, molecular solids tend to be soft, low melting, and easily vaporized ($ΔH_{fus}$ and $ΔH_{vap}$ are low). For similar substances, the strength of the London dispersion forces increases smoothly with increasing molecular mass. For example, the melting points of benzene (C6H6), naphthalene (C10H8), and anthracene (C14H10), with one, two, and three fused aromatic rings, are 5.5°C, 80.2°C, and 215°C, respectively. The enthalpies of fusion also increase smoothly within the series: benzene (9.95 kJ/mol) < naphthalene (19.1 kJ/mol) < anthracene (28.8 kJ/mol). If the molecules have shapes that cannot pack together efficiently in the crystal, however, then the melting points and the enthalpies of fusion tend to be unexpectedly low because the molecules are unable to arrange themselves to optimize intermolecular interactions. Thus toluene (C6H5CH3) and m-xylene [m-C6H4(CH3)2] have melting points of −95°C and −48°C, respectively, which are significantly lower than the melting point of the lighter but more symmetrical analog, benzene.
Self-healing rubber is an example of a molecular solid with the potential for significant commercial applications. The material can stretch, but when snapped into pieces it can bond back together again through reestablishment of its hydrogen-bonding network without showing any sign of weakness. Among other applications, it is being studied for its use in adhesives and bicycle tires that will self-heal.
Covalent Network Solids
Covalent solids are formed by networks or chains of atoms or molecules held together by covalent bonds. A perfect single crystal of a covalent solid is therefore a single giant molecule. For example, the structure of diamond, shown in part (a) in Figure $1$, consists of sp3 hybridized carbon atoms, each bonded to four other carbon atoms in a tetrahedral array to create a giant network. The carbon atoms form six-membered rings.
The unit cell of diamond can be described as an fcc array of carbon atoms with four additional carbon atoms inserted into four of the tetrahedral holes. It thus has the zinc blende structure described in Section 12.3, except that in zinc blende the atoms that compose the fcc array are sulfur and the atoms in the tetrahedral holes are zinc. Elemental silicon has the same structure, as does silicon carbide (SiC), which has alternating C and Si atoms. The structure of crystalline quartz (SiO2), shown in Section 12.1, can be viewed as being derived from the structure of silicon by inserting an oxygen atom between each pair of silicon atoms.
All compounds with the diamond and related structures are hard, high-melting-point solids that are not easily deformed. Instead, they tend to shatter when subjected to large stresses, and they usually do not conduct electricity very well. In fact, diamond (melting point = 3500°C at 63.5 atm) is one of the hardest substances known, and silicon carbide (melting point = 2986°C) is used commercially as an abrasive in sandpaper and grinding wheels. It is difficult to deform or melt these and related compounds because strong covalent (C–C or Si–Si) or polar covalent (Si–C or Si–O) bonds must be broken, which requires a large input of energy.
Other covalent solids have very different structures. For example, graphite, the other common allotrope of carbon, has the structure shown in part (b) in Figure $1$. It contains planar networks of six-membered rings of sp2 hybridized carbon atoms in which each carbon is bonded to three others. This leaves a single electron in an unhybridized 2pz orbital that can be used to form C=C double bonds, resulting in a ring with alternating double and single bonds. Because of its resonance structures, the bonding in graphite is best viewed as consisting of a network of C–C single bonds with one-third of a π bond holding the carbons together, similar to the bonding in benzene.
To completely describe the bonding in graphite, we need a molecular orbital approach similar to the one used for benzene in Chapter 9. In fact, the C–C distance in graphite (141.5 pm) is slightly longer than the distance in benzene (139.5 pm), consistent with a net carbon–carbon bond order of 1.33. In graphite, the two-dimensional planes of carbon atoms are stacked to form a three-dimensional solid; only London dispersion forces hold the layers together. As a result, graphite exhibits properties typical of both covalent and molecular solids. Due to strong covalent bonding within the layers, graphite has a very high melting point, as expected for a covalent solid (it actually sublimes at about 3915°C). It is also very soft; the layers can easily slide past one another because of the weak interlayer interactions. Consequently, graphite is used as a lubricant and as the “lead” in pencils; the friction between graphite and a piece of paper is sufficient to leave a thin layer of carbon on the paper. Graphite is unusual among covalent solids in that its electrical conductivity is very high parallel to the planes of carbon atoms because of delocalized C–C π bonding. Finally, graphite is black because it contains an immense number of alternating double bonds, which results in a very small energy difference between the individual molecular orbitals. Thus light of virtually all wavelengths is absorbed. Diamond, on the other hand, is colorless when pure because it has no delocalized electrons. Table $2$ compares the strengths of the intermolecular and intramolecular interactions for three covalent solids, showing the comparative weakness of the interlayer interactions.
Table $2$: A Comparison of Intermolecular (ΔHsub) and Intramolecular Interactions
Substance ΔHsub (kJ/mol) Average Bond Energy (kJ/mol)
phosphorus (s) 58.98 201
sulfur (s) 64.22 226
iodine (s) 62.42 149
Carbon: An example of an Covalent Network Solid
In network solids, conventional chemical bonds hold the chemical subunits together. The bonding between chemical subunits, however, is identical to that within the subunits, resulting in a continuous network of chemical bonds. One common examples of network solids are diamond (a form of pure carbon) Carbon exists as a pure element at room temperature in three different forms: graphite (the most stable form), diamond, and fullerene.
Diamonds
The structure of diamond is shown at the right in a "ball-and-stick" format. The balls represent the carbon atoms and the sticks represent a covalent bond. Be aware that in the "ball-and-stick" representation the size of the balls do not accurately represent the size of carbon atoms. In addition, a single stick is drawn to represent a covalent bond irrespective of whether the bond is a single, double, or triple bond or requires resonance structures to represent. In the diamond structure, all bonds are single covalent bonds ($\sigma$ bonds). The "space-filling" format is an alternate representation that displays atoms as spheres with a radius equal to the van der Waals radius, thus providing a better sense of the size of the atoms.
Notice that diamond is a network solid. The entire solid is an "endless" repetition of carbon atoms bonded to each other by covalent bonds. (In the display at the right, the structure is truncated to fit in the display area.)
Questions to consider
• What is the bonding geometry around each carbon?
• What is the hybridization of carbon in diamond?
• The diamond structure consists of a repeating series of rings. How many carbon atoms are in a ring?
• Diamond are renowned for its hardness. Explain why this property is expected on the basis of the structure of diamond.
Graphite
The most stable form of carbon is graphite. Graphite consists of sheets of carbon atoms covalently bonded together. These sheets are then stacked to form graphite. Figure $3$ shows a ball-and-stick representation of graphite with sheets that extended "indefinitely" in the xy plane, but the structure has been truncated for display purposed. Graphite may also be regarded as a network solid, even though there is no bonding in the z direction. Each layer, however, is an "endless" bonded network of carbon atoms.
Questions to consider
• What is the bonding geometry around each carbon?
• What is the hybridization of carbon in graphite?
• The a layer of the graphite structure consists of a repeating series of rings. How many carbon atoms are in a ring?
• What force holds the carbon sheets together in graphite?
• Graphite is very slippery and is often used in lubricants. Explain why this property is expected on the basis of the structure of graphite.
• The slipperiness of graphite is enhanced by the introduction of impurities. Where would such impurities be located and why would they make graphite a better lubricant?
Fullerenes
Until the mid 1980's, pure carbon was thought to exist in two forms: graphite and diamond. The discovery of C60 molecules in interstellar dust in 1985 added a third form to this list. The existence of C60, which resembles a soccer ball, had been hypothesized by theoreticians for many years. In the late 1980's synthetic methods were developed for the synthesis of C60, and the ready availability of this form of carbon led to extensive research into its properties.
The C60 molecule (Figure $4$; left), is called buckminsterfullerene, though the shorter name fullerene is often used. The name is a tribute to the American architect R. Buckminster Fuller, who is famous for designing and constructing geodesic domes which bear a close similarity to the structure of C60. As is evident from the display, C60 is a sphere composed of six-member and five-member carbon rings. These balls are sometimes fondly referred to as "Bucky balls". It should be noted that fullerenes are an entire class of pure carbon compounds rather than a single compound. A distorted sphere containing more than 60 carbon atoms have also been found, and it is also possible to create long tubes (Figure $4$; right). All of these substances are pure carbon.
Questions to Consider
• What is the bonding geometry around each carbon? (Note that this geometry is distorted in $C_{60}$.)
• What is the hybridization of carbon in fullerene?
• A single crystal of C60 falls into which class of crystalline solids?
• It has been hypothesized that C60 would make a good lubricant. Why might C60 make a good lubricant?
Metallic Solids
Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure $5$. The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials.
Metals are characterized by their ability to reflect light, called luster, their high electrical and thermal conductivity, their high heat capacity, and their malleability and ductility. Every lattice point in a pure metallic element is occupied by an atom of the same metal. The packing efficiency in metallic crystals tends to be high, so the resulting metallic solids are dense, with each atom having as many as 12 nearest neighbors.
Bonding in metallic solids is quite different from the bonding in the other kinds of solids we have discussed. Because all the atoms are the same, there can be no ionic bonding, yet metals always contain too few electrons or valence orbitals to form covalent bonds with each of their neighbors. Instead, the valence electrons are delocalized throughout the crystal, providing a strong cohesive force that holds the metal atoms together.
Valence electrons in a metallic solid are delocalized, providing a strong cohesive force that holds the atoms together.
The strength of metallic bonds varies dramatically. For example, cesium melts at 28.4°C, and mercury is a liquid at room temperature, whereas tungsten melts at 3680°C. Metallic bonds tend to be weakest for elements that have nearly empty (as in Cs) or nearly full (Hg) valence subshells, and strongest for elements with approximately half-filled valence shells (as in W). As a result, the melting points of the metals increase to a maximum around group 6 and then decrease again from left to right across the d block. Other properties related to the strength of metallic bonds, such as enthalpies of fusion, boiling points, and hardness, have similar periodic trends.
A somewhat oversimplified way to describe the bonding in a metallic crystal is to depict the crystal as consisting of positively charged nuclei in an electron sea (Figure $6$). In this model, the valence electrons are not tightly bound to any one atom but are distributed uniformly throughout the structure. Very little energy is needed to remove electrons from a solid metal because they are not bound to a single nucleus. When an electrical potential is applied, the electrons can migrate through the solid toward the positive electrode, thus producing high electrical conductivity. The ease with which metals can be deformed under pressure is attributed to the ability of the metal ions to change positions within the electron sea without breaking any specific bonds. The transfer of energy through the solid by successive collisions between the metal ions also explains the high thermal conductivity of metals. This model does not, however, explain many of the other properties of metals, such as their metallic luster and the observed trends in bond strength as reflected in melting points or enthalpies of fusion. Some general properties of the four major classes of solids are summarized in Table $2$.
Table $2$: Properties of the Major Classes of Solids
Ionic Solids Molecular Solids Covalent Solids Metallic Solids
*Many exceptions exist. For example, graphite has a relatively high electrical conductivity within the carbon planes, and diamond has the highest thermal conductivity of any known substance.
poor conductors of heat and electricity poor conductors of heat and electricity poor conductors of heat and electricity* good conductors of heat and electricity
relatively high melting point low melting point high melting point melting points depend strongly on electron configuration
hard but brittle; shatter under stress soft very hard and brittle easily deformed under stress; ductile and malleable
relatively dense low density low density usually high density
dull surface dull surface dull surface lustrous
The general order of increasing strength of interactions in a solid is:
molecular solids < ionic solids ≈ metallic solids < covalent solids
Example $1$
Classify Ge, RbI, C6(CH3)6, and Zn as ionic, molecular, covalent, or metallic solids and arrange them in order of increasing melting points.
Given: compounds
Asked for: classification and order of melting points
Strategy:
1. Locate the component element(s) in the periodic table. Based on their positions, predict whether each solid is ionic, molecular, covalent, or metallic.
2. Arrange the solids in order of increasing melting points based on your classification, beginning with molecular solids.
Solution:
A Germanium lies in the p block just under Si, along the diagonal line of semimetallic elements, which suggests that elemental Ge is likely to have the same structure as Si (the diamond structure). Thus Ge is probably a covalent solid. RbI contains a metal from group 1 and a nonmetal from group 17, so it is an ionic solid containing Rb+ and I ions. The compound C6(CH3)6 is a hydrocarbon (hexamethylbenzene), which consists of isolated molecules that stack to form a molecular solid with no covalent bonds between them. Zn is a d-block element, so it is a metallic solid.
B Arranging these substances in order of increasing melting points is straightforward, with one exception. We expect C6(CH3)6 to have the lowest melting point and Ge to have the highest melting point, with RbI somewhere in between. The melting points of metals, however, are difficult to predict based on the models presented thus far. Because Zn has a filled valence shell, it should not have a particularly high melting point, so a reasonable guess is C6(CH3)6 < Zn ~ RbI < Ge. The actual melting points are C6(CH3)6, 166°C; Zn, 419°C; RbI, 642°C; and Ge, 938°C. This agrees with our prediction.
Exercise $1$
Classify C60, BaBr2, GaAs, and AgZn as ionic, covalent, molecular, or metallic solids and then arrange them in order of increasing melting points.
Answer
C60 (molecular) < AgZn (metallic) ~ BaBr2 (ionic) < GaAs (covalent). The actual melting points are C60, about 300°C; AgZn, about 700°C; BaBr2, 856°C; and GaAs, 1238°C.
Summary
The major types of solids are ionic, molecular, covalent, and metallic. Ionic solids consist of positively and negatively charged ions held together by electrostatic forces; the strength of the bonding is reflected in the lattice energy. Ionic solids tend to have high melting points and are rather hard. Molecular solids are held together by relatively weak forces, such as dipole–dipole interactions, hydrogen bonds, and London dispersion forces. As a result, they tend to be rather soft and have low melting points, which depend on their molecular structure. Covalent solids consist of two- or three-dimensional networks of atoms held together by covalent bonds; they tend to be very hard and have high melting points. Metallic solids have unusual properties: in addition to having high thermal and electrical conductivity and being malleable and ductile, they exhibit luster, a shiny surface that reflects light. An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements. Alloys can be formed by substituting one metal atom for another of similar size in the lattice (substitutional alloys), by inserting smaller atoms into holes in the metal lattice (interstitial alloys), or by a combination of both. Although the elemental composition of most alloys can vary over wide ranges, certain metals combine in only fixed proportions to form intermetallic compounds with unique properties. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/16%3A_Liquids_and_Solids/16.05%3A_Carbon_and_Silicon%3A_Network_Atomic_Solids.txt |
Contributors and Attributions
• Nathalie Interiano
16.10: Vapor Pressure and Changes of State
Learning Objectives
• To know how and why the vapor pressure of a liquid varies with temperature.
• To understand that the equilibrium vapor pressure of a liquid depends on the temperature and the intermolecular forces present.
• To understand that the relationship between pressure, enthalpy of vaporization, and temperature is given by the Clausius-Clapeyron equation.
Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid.
Evaporation and Condensation
Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy (KE) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure $1$), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy ($E_0$) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than $E_0$. The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than $E_0$ has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure.
To understand the causes of vapor pressure, consider the apparatus shown in Figure $2$. When a liquid is introduced into an evacuated chamber (part (a) in Figure $2$), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with $KE > E_0$ will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase.
As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure $2$). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a steady state will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure $3$.
Equilibrium Vapor Pressure
Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no net change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid.
If a liquid is in an open container, however, most of the molecules that escape into the vapor phase will not collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure $4$) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile.
The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does not depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure $4$. Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is much steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release.
Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions.
A Video Discussing Vapor Pressure and Boiling Points. Video Source: Vapor Pressure & Boiling Point(opens in new window) [youtu.be]
The exponential rise in vapor pressure with increasing temperature in Figure $4$ allows us to use natural logarithms to express the nonlinear relationship as a linear one.
$\boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1}$
where
• $\ln P$ is the natural logarithm of the vapor pressure,
• $ΔH_{vap}$ is the enthalpy of vaporization,
• $R$ is the universal gas constant [8.314 J/(mol•K)],
• $T$ is the temperature in kelvins, and
• $C$ is the y-intercept, which is a constant for any given line.
Plotting $\ln P$ versus the inverse of the absolute temperature ($1/T$) is a straight line with a slope of −ΔHvap/R. Equation $\ref{Eq1}$, called the Clausius–Clapeyron Equation, can be used to calculate the $ΔH_{vap}$ of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine $ΔH_{vap}$ is to measure the vapor pressure of a liquid at two temperatures and insert the values of $P$ and $T$ for these points into Equation $\ref{Eq2}$, which is derived from the Clausius–Clapeyron equation:
$\ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2}$
Conversely, if we know ΔHvap and the vapor pressure $P_1$ at any temperature $T_1$, we can use Equation $\ref{Eq2}$ to calculate the vapor pressure $P_2$ at any other temperature $T_2$, as shown in Example $1$.
A Video Discussing the Clausius-Clapeyron Equation. Video Link: The Clausius-Clapeyron Equation(opens in new window) [youtu.be]
Example $1$: Vapor Pressure of Mercury
The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table:
experimentally measured vapor pressures of liquid Hg at four temperatures
T (°C) 80.0 100 120 140
P (torr) 0.0888 0.2729 0.7457 1.845
From these data, calculate the enthalpy of vaporization (ΔHvap) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.)
Given: vapor pressures at four temperatures
Asked for: ΔHvap of mercury and vapor pressure at 160°C
Strategy:
1. Use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values in the table, making sure to convert all values to the appropriate units.
2. Substitute the calculated value of ΔHvap into Equation $\ref{Eq2}$ to obtain the unknown pressure (P2).
Solution:
A The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation $\ref{Eq1}$ and find the value of ΔHvap from the slope of the line, an alternative approach is to use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C (T1) and 120.0°C (T2) into Equation $\ref{Eq2}$ gives
\begin{align*} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align*} \nonumber
B We can now use this value of ΔHvap to calculate the vapor pressure of the liquid (P2) at 160.0°C (T2):
$\ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber$
Using the relationship $e^{\ln x} = x$, we have
\begin{align*} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \[4pt] P_{2} &= 4.21 Torr \end{align*} \nonumber
At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect.
Exercise $1$: Vapor Pressure of Nickel
The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr?
Answer
1896°C
Boiling Points
As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure $4$ are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the actual boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C.
Table $1$: The Boiling Points of Water at Various Locations on Earth
Place Altitude above Sea Level (ft) Atmospheric Pressure (mmHg) Boiling Point of Water (°C)
Mt. Everest, Nepal/Tibet 29,028 240 70
Bogota, Colombia 11,490 495 88
Denver, Colorado 5280 633 95
Washington, DC 25 759 100
Dead Sea, Israel/Jordan −1312 799 101.4
Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table $1$ lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked.
As pressure increases, the boiling point of a liquid increases and vice versa.
Example $2$: Boiling Mercury
Use Figure $4$ to estimate the following.
1. the boiling point of water in a pressure cooker operating at 1000 mmHg
2. the pressure required for mercury to boil at 250°C
Given: Data in Figure $4$, pressure, and boiling point
Asked for: corresponding boiling point and pressure
Strategy:
1. To estimate the boiling point of water at 1000 mmHg, refer to Figure $4$ and find the point where the vapor pressure curve of water intersects the line corresponding to a pressure of 1000 mmHg.
2. To estimate the pressure required for mercury to boil at 250°C, find the point where the vapor pressure curve of mercury intersects the line corresponding to a temperature of 250°C.
Solution:
1. A The vapor pressure curve of water intersects the P = 1000 mmHg line at about 110°C; this is therefore the boiling point of water at 1000 mmHg.
2. B The vertical line corresponding to 250°C intersects the vapor pressure curve of mercury at P ≈ 75 mmHg. Hence this is the pressure required for mercury to boil at 250°C.
Exercise $2$: Boiling Ethlyene Glycol
Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure $4$ to estimate the following.
1. the normal boiling point of ethylene glycol
2. the pressure required for diethyl ether to boil at 20°C.
Answer a
200°C
Answer b
450 mmHg
Summary
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of dynamic equilibrium. Under these conditions, a liquid exhibits a characteristic equilibrium vapor pressure that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the Clausius–Clapeyron equation. This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. Volatile liquids are liquids with high vapor pressures, which tend to evaporate readily from an open container; nonvolatile liquids have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its normal boiling point. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/16%3A_Liquids_and_Solids/16.07%3A_Ionic_Solids.txt |
Learning Objectives
• To understand the basics of a one-component phase diagram as a function of temperature and pressure in a closed system.
• To be able to identify the triple point, the critical point, and four regions: solid, liquid, gas, and a supercritical fluid.
The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system.
Introduction
A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance (Figure $1$). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of Figure $1$) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid.
The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure.
The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In Figure $1$, the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in Section 11.5. It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the solid phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple point; it is the only combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point.
Remember that a phase diagram, such as the one in Figure $1$, is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error.
The Phase Diagram of Water
Figure $2$ shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee).
The phase diagram for water illustrated in Figure $\PageIndex{2b}$ shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in Figure $1$; that is, the melting point of ice decreases with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is less dense than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice.
In Figure $\PageIndex{2b}$ point A is located at P = 1 atm and T = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines.
Ice Skating: An Incorrect Hypothesis of Phase Transitions
Until recently, many textbooks described ice skating as being possible because the pressure generated by the skater’s blade is high enough to melt the ice under the blade, thereby creating a lubricating layer of liquid water that enables the blade to slide across the ice. Although this explanation is intuitively satisfying, it is incorrect, as we can show by a simple calculation.
Recall that pressure (P) is the force (F) applied per unit area (A):
$P=\dfrac{F}{A} \nonumber$
To calculate the pressure an ice skater exerts on the ice, we need to calculate only the force exerted and the area of the skate blade. If we assume a 75.0 kg (165 lb) skater, then the force exerted by the skater on the ice due to gravity is
$F = mg \nonumber$
where m is the mass and g is the acceleration due to Earth’s gravity (9.81 m/s2). Thus the force is
$F = (75.0\; kg)(9.81\; m/s^2) = 736\; (kg•m)/s^2 = 736 N \nonumber$
If we assume that the skate blades are 2.0 mm wide and 25 cm long, then the area of the bottom of each blade is
$A = (2.0 \times 10^{−3}\; m)(25 \times 10^{−2}\; m) = 5.0 \times 10^{−4} m^2 \nonumber$
If the skater is gliding on one foot, the pressure exerted on the ice is
$P= \dfrac{736\;N}{5.0 \times 10^{-4} \; m^2} = 1.5 \times 10^6 \; N/m^2 = 1.5 \times 10^6\; Pa =15 \; atm \nonumber$
The pressure is much lower than the pressure needed to decrease the melting point of ice by even 1°C, and experience indicates that it is possible to skate even when the temperature is well below freezing. Thus pressure-induced melting of the ice cannot explain the low friction that enables skaters (and hockey pucks) to glide. Recent research indicates that the surface of ice, where the ordered array of water molecules meets the air, consists of one or more layers of almost liquid water. These layers, together with melting induced by friction as a skater pushes forward, appear to account for both the ease with which a skater glides and the fact that skating becomes more difficult below about −7°C, when the number of lubricating surface water layers decreases.
Example $1$: Water
Referring to the phase diagram of water in Figure $2$:
1. predict the physical form of a sample of water at 400°C and 150 atm.
2. describe the changes that occur as the sample in part (a) is slowly allowed to cool to −50°C at a constant pressure of 150 atm.
Given: phase diagram, temperature, and pressure
Asked for: physical form and physical changes
Strategy:
1. Identify the region of the phase diagram corresponding to the initial conditions and identify the phase that exists in this region.
2. Draw a line corresponding to the given pressure. Move along that line in the appropriate direction (in this case cooling) and describe the phase changes.
Solution:
1. A Locate the starting point on the phase diagram in part (a) in Figure $2$. The initial conditions correspond to point A, which lies in the region of the phase diagram representing water vapor. Thus water at T = 400°C and P = 150 atm is a gas.
2. B Cooling the sample at constant pressure corresponds to moving left along the horizontal line in part (a) in Figure $2$. At about 340°C (point B), we cross the vapor pressure curve, at which point water vapor will begin to condense and the sample will consist of a mixture of vapor and liquid. When all of the vapor has condensed, the temperature drops further, and we enter the region corresponding to liquid water (indicated by point C). Further cooling brings us to the melting curve, the line that separates the liquid and solid phases at a little below 0°C (point D), at which point the sample will consist of a mixture of liquid and solid water (ice). When all of the water has frozen, cooling the sample to −50°C takes us along the horizontal line to point E, which lies within the region corresponding to solid water. At P = 150 atm and T = −50°C, therefore, the sample is solid ice.
Exercise $2$
Referring to the phase diagram of water in Figure $2$, predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm.
Answer
The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid.
The Phase Diagram of Carbon Dioxide
In contrast to the phase diagram of water, the phase diagram of CO2 (Figure $3$) has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO2 cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO2 sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO2 is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed.
Also notice the critical point at 30.98°C and 72.79 atm. Supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps.
The Critical Point
As the phase diagrams above demonstrate, a combination of high pressure and low temperature allows gases to be liquefied. As we increase the temperature of a gas, liquefaction becomes more and more difficult because higher and higher pressures are required to overcome the increased kinetic energy of the molecules. In fact, for every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature (Tc), the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase. Instead, the substance forms a single phase that completely occupies the volume of the container. Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures. Each substance also has a critical pressure (Pc), the minimum pressure needed to liquefy it at the critical temperature. The combination of critical temperature and critical pressure is called the critical point. The critical temperatures and pressures of several common substances are listed in Figure $1$.
Figure $1$: Critical Temperatures and Pressures of Some Simple Substances
Substance Tc (°C) Pc (atm)
NH3 132.4 113.5
CO2 31.0 73.8
CH3CH2OH (ethanol) 240.9 61.4
He −267.96 2.27
Hg 1477 1587
CH4 −82.6 46.0
N2 −146.9 33.9
H2O 374.0 217.7
High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa.
Supercritical Fluids
A Video Discussing Phase Diagrams. Video Source: Phase Diagrams(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/16%3A_Liquids_and_Solids/16.11%3A_Phase_Diagrams.txt |
Solutions are all around us. Air, for example, is a solution. If you live near a lake, a river, or an ocean, that body of water is not pure H2O but most probably a solution. Much of what we drink—for example, soda, coffee, tea, and milk—is at least in part a solution. Solutions are a large part of everyday life. A lot of the chemistry occurring around us happens in solution. In fact, much of the chemistry that occurs in our own bodies takes place in solution, and many solutions—such as the Ringer’s lactate IV solution—are important for our health. In our understanding of chemistry, we need to understand a little bit about solutions. In this chapter, you will learn about the special characteristics of solutions, how solutions are characterized, and some of their properties.
17: Solutions
The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties. An ideal solution or ideal mixture is a solution in which the enthalpy of solution (\(\Delta{H_{solution}} = 0\)) is zero; with the closer to zero the enthalpy of solution, the more "ideal" the behavior of the solution becomes. Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing (\(\Delta{S_{solution}}\)).
• Raoult's Law
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing.
• Henry's Law
Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
17.2: The Energies of Solution Formation
Some forces that interact within pure liquids are also present during mixtures and solutions. Forces such as Cohesive as well as Adhesive forces still apply to mixtures; however, more importantly we focus on the interaction between different molecules. Why is oil only soluble in benzene and not water? Why do only "like" molecules dissolve in "like" molecules?
The process of Mixing
Before we go on to the more specific mechanisms of mixing, let's discuss its process. Mixing is a spontaneous process that increases the entropy of the solution. In order to form a mixture of homogenous solutions by distributing the solute molecules evenly within the solvent molecules, heat transfers are inevitable. This heat transfer is denoted ΔHsoln for our general comprehension. ΔH is the change in heat energy found by subtracting the enthalpy of the reactant from that of the product:
\[H_{products} - H_{reactants}= ΔH_{soln}.\]
Enthalpy of Solution
What then is the significance of \(ΔH_{soln}\)? It presents a clear indication of the magnitude as well as direction of the heat transfer so that when:
• ΔH>0 : Endothermic Reaction (positive), because the products encompass more energy than the reactants
• ΔH<0 : Exothermic Reaction (negative), because the reactants consist of more energy than the products.
What we have to supply for our understanding for this equation is that the extra energy is seized either from or give to the surrounding. And to ascertain the enthalpy of solution, we take the three step approach in enthalpy when a solute is mixed with the solvent.
Three Step Approach to Finding the Enthalpy of Solution: ΔHsoln = ΔH1 + ΔH2+ ΔH3
1. Each molecule of solute is Separated from each other (expand the solute), endothermic reaction. (ΔH1)
2. Each molecule of solvent is separated from each other (expand solvent), endothermic reaction. (ΔH2) Now the molecules of solute and molecules of solvent can be permitted to attract one another in solution.
3. The molecules of solute and solvent react with each other and a solution will result. exothermic reaction (ΔH3)
Note
note that usually \(ΔH_1\) and \(ΔH_2\) are opposite in sign as \(ΔH_3\). Separating the solute and the solvent solutions alone are usually endothermic reactions in that their cohesive forces are broken while letting the molecules to react freely is an exothermic reaction. The figure below explains pictorially how positive and negative \(ΔH\) can be obtained through the three step process of mixing solution.
ΔH's Relationship to the Behavior of the Solution
Ideal solution is the mixture that has little to no net intermolecular interactions that differentiates it from its ideal behavior. Thus if the intermolecular forces of attraction are the same and have the same strength, both the solvent and solute will mix at random. This solution is called an ideal solution, which means that \(ΔH_{soln} = 0\). If the intermolecular forces of attraction of different molecules are greater than the forces of attraction of like molecules, then it is called a nonideal solution. This will result in an exothermic process (\(ΔH_{soln}<0\)). If the intermolecular forces of attraction of different molecules are a bit weaker than the forces of attraction of like molecules. This solution is a nonideal solution, has bigger enthalpy value than pure components, and it goes through an endothermic process. Lastly, if the intermolecular forces of attraction of different molecules is a lot weaker than the forces of attraction of like molecules, the solution becomes a heterogeneous mixture (e.g., water and olive oil).
The Effects of Intermolecular Forces in Solution.
The epitome of intermolecular forces in solution is the miracle of solubility, because when a matter precipitates it no longer interacts with the solvent. So what is the attraction between "like" molecules that makes them attract to each other? Let's take a phospholipid, the building block of a cell's membrane, as an example. This molecule is amphipathic, meaning that it is both hydrophilic and hydrophobic. Beginning with the structure of a phospholipid, it has a polar head which is hydrophilic and a nonpolar tail which is hydrophobic as the picture below.
How can a single molecule be both polar molecule loving and polar molecule disliking at the same time? This is because at the polar head, the phosphate has a net negative charge thus attracting the partial positive charge of the hydrogen molecules of water. Its nonpolar tails on the other hand, is a very organized form of hydrocarbon, consisting of no net charges. The tail is then repelled by water as it struggles to fit between the partial positive and partial negative of the water molecule.
Another side effect of the interactions of molecules is reflected by the use of the activity coefficient during thermodynamic equilibrium constant calculations. This constant differentiates ideal and nonideal solutions so that interactions for solution equilibrium can be more accurately estimated. Most versions of the equilibrium constant K utilizes activity instead of concentration so that the units would disappear more fluently. For an ideal solution, the activity coefficient is 1 [x]/ oCelcius, thus when the concentration is dived by it to yield activity, it is unaltered.
Example \(1\)
Based on the concept of intermolecular interactions, ascertain the reason behind freezing-point depression and boiling-point elevation.
Solution
When an ion is added into solution, it exerts an intermolecular force which binds loosely to the water molecules in solution. This weak force then increases the energy necessary to break each molecule loose, thus increasing temperature in relationship to vapor pressure. It now takes more energy input to obtain the same vapor pressure, thus elevating the boiling point. For freezing-point depression, the same force that is holding the water molecules from evaporating is holding them against being placed into an organized solid form. It now takes more energy to form the weak bonds between each water molecule because the intermolecular forces between water and the ions first have to be overcome, hence reducing the freezing point.
Example \(2\)
Give examples that present the involvement of intermolecular forces thus differentiating ideal from nonideal solutions.
Solution
Gibbs free energy (relating ΔG with ΔGo), calculating thermodynamic equilibrium constants, boiling-point elevation, and freezing-point depression | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/17%3A_Solutions/17.1%3A_Solution_Composition.txt |
Learning Objectives
• To understand how Temperature, Pressure, and the presence of other solutes affect the solubility of solutes in solvents.
Solubility is defined as the upper limit of solute that can be dissolved in a given amount of solvent at equilibrium. In such an equilibrium, Le Chatelier's principle can be used to explain most of the main factors that affect solubility. Le Châtelier's principle dictates that the effect of a stress upon a system in chemical equilibrium can be predicted in that the system tends to shift in such a way as to alleviate that stress.
Solute-Solvent Interactions Affect Solubility
The relation between the solute and solvent is very important in determining solubility. Strong solute-solvent attractions equate to greater solubility while weak solute-solvent attractions equate to lesser solubility. In turn, polar solutes tend to dissolve best in polar solvents while non-polar solutes tend to dissolve best in non-polar solvents. In the case of a polar solute and non-polar solvent (or vice versa), it tends to be insoluble or only soluble to a miniscule degree. A general rule to remember is, "Like dissolves like."
Common-Ion Effect
The common-ion effect is a term that describes the decrease in solubility of an ionic compound when a salt that contains an ion that already exists in the chemical equilibrium is added to the mixture. This effect best be explained by Le Chatelier's principle. Imagine if the slightly soluble ionic compound calcium sulfate, CaSO4, is added to water. The net ionic equation for the resulting chemical equilibrium is the following:
\[ CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
Calcium sulfate is slightly soluble; at equilibrium, most of the calcium and sulfate exists in the solid form of calcium sulfate.
Suppose the soluble ionic compound copper sulfate (CuSO4) were added to the solution. Copper sulfate is soluble; therefore, its only important effect on the net ionic equation is the addition of more sulfate (SO42-) ions.
\[ CuSO_{4(s)} \rightleftharpoons Cu^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
The sulfate ions dissociated from copper sulfate are already present (common to) in the mixture from the slight dissociation of calcium sulfate. Thus, this addition of sulfate ions places stress on the previously established equilibrium. Le Chatelier's principle dictates that the additional stress on this product side of the equilibrium results in the shift of equilibrium towards the reactants side in order to alleviate this new stress. Because of the shift toward the reactant side, the solubility of the slightly soluble calcium sulfate is reduced even further.
Temperature Affects Solubility
Temperature changes affect the solubility of solids, liquids and gases differently. However, those effects are finitely determined only for solids and gases.
Solids
The effects of temperature on the solubility of solids differ depending on whether the reaction is endothermic or exothermic. Using Le Chatelier's principle, the effects of temperature in both scenarios can be determined.
1. First, consider an endothermic reaction (\(\Delta{H_{solvation}}>0\)): Increasing the temperature results in a stress on the reactants side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the product side in order to alleviate this stress. By shifting towards the product side, more of the solid is dissociated when equilibrium is again established, resulting in increased solubility.
2. Second, consider an exothermic reaction ((\(\Delta{H_{solvation}}<0\)): Increasing the temperature results in a stress on the products side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the reactant side in order to alleviate this stress. By shifting towards the reactant's side, less of the solid is dissociated when equilibrium is again established, resulting in decreased solubility.
Liquids
In the case of liquids, there is no defined trends for the effects of temperature on the solubility of liquids.
Gases
In understanding the effects of temperature on the solubility of gases, it is first important to remember that temperature is a measure of the average kinetic energy. As temperature increases, kinetic energy increases. The greater kinetic energy results in greater molecular motion of the gas particles. As a result, the gas particles dissolved in the liquid are more likely to escape to the gas phase and the existing gas particles are less likely to be dissolved. The converse is true as well. The trend is thus as follows: increased temperatures mean lesser solubility and decreased temperatures mean higher solubility.
Le Chatelier's principle allows better conceptualization of these trends. First, note that the process of dissolving gas in liquid is usually exothermic. As such, increasing temperatures result in stress on the product side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts towards the reactant side in order to alleviate this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase increases, resulting in lowered solubility.
Conversely, decreasing temperatures result in stress on the reactant side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts toward the product side in order to compensate for this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase would decrease, resulting in greater solubility.
Pressure Affects Solubility of Gases
The effects of pressure are only significant in affecting the solubility of gases in liquids.
• Solids & Liquids: The effects of pressure changes on the solubility of solids and liquids are negligible.
• Gases: The effects of pressure on the solubility of gases in liquids can best be described through a combination of Henry's law and Le Chatelier principle. Henry's law dictates that when temperature is constant, the solubility of the gas corresponds to it's partial pressure. Consider the following formula of Henry's law:
\[ p = k_h \; c \]
where:
• \(p\) is the partial pressure of the gas above the liquid,
• \(k_h\) is Henry's law constant, and
• \(c\) is the concentrate of the gas in the liquid.
This formula indicates that (at a constant temperature) when the partial pressure decreases, the concentration of gas in the liquid decreases as well, and consequently the solubility also decreases. Conversely, when the partial pressure increases in such a situation, the concentration of gas in the liquid will increase as well; the solubility also increases. Extending the implications from Henry's law, the usefulness of Le Chatelier's principle is enhanced in predicting the effects of pressure on the solubility of gases.
Consider a system consisting of a gas that is partially dissolved in liquid. An increase in pressure would result in greater partial pressure (because the gas is being further compressed). This increased partial pressure means that more gas particles will enter the liquid (there is therefore less gas above the liquid, so the partial pressure decreases) in order to alleviate the stress created by the increase in pressure, resulting in greater solubility.
The converse case in such a system is also true, as a decrease in pressure equates to more gas particles escaping the liquid to compensate.
Example 1
Consider the following exothermic reaction that is in equilibrium
\[ CO_2 (g) + H_2O (l) \rightleftharpoons H_2CO_3 (aq) \]
What will happen to the solubility of the carbon dioxide if:
1. Temperature is increased?
2. Pressure and temperature are increased?
3. Pressure is increased but temperature is decreased?
4. Pressure is increased?
Solution
1. The reaction is exothermic, so an increase in temperature means that solubility would decrease.
2. The change in solubility cannot be determined from the given information. Increasing pressure increased solubility, but increasing temperature decreases solubility
3. An increase in pressure and an increase in temperature in this reaction results in greater solubility.
4. An increase in pressure results in more gas particles entering the liquid in order to decrease the partial pressure. Therefore, the solubility would increase.
Example 2: The Common Ion Effect
Bob is in the business of purifying silver compounds to extract the actual silver. He is extremely frugal. One day, he finds a barrel containing a saturated solution of silver chloride. Bob has a bottle of water, a jar of table salt (NaCl(s)), and a bottle of vinegar (CH3COOH). Which of the three should Bob add to the solution to maximize the amount of solid silver chloride (minimizing the solubility of the silver chloride)?
Solution
Bob should add table salt to the solution. According to the common-ion effect, the additional Cl- ions would reduce the solubility of the silver chloride, which maximizes the amount of solid silver chloride.
Example 3:
Allison has always wanted to start her own carbonated drink company. Recently, she opened a factory to produce her drinks. She wants her drink to "out-fizz" all the competitors. That is, she wants to maximize the solubility of the gas in her drink. What conditions (high/low temperature, high/low pressure) would best allow her to achieve this goal?
Solution
She would be able to maximize the solubility of the gas, (\(CO_2\) in this case, in her drink (maximize fizz) when the pressure is high and temperature is low.
Example 4
Butters is trying to increase the solubility of a solid in some water. He begins to frantically stir the mixture. Should he continue stirring? Why or why not?
Solution
He stop stop stirring. Stirring only affects how fast the system will reach equilibrium and does not affect the solubility of the solid at all.
Example 5: Outgassing Soda
With respect to Henry's law, why is it a poor ideal to open a can of soda in a low pressure environment?
Solution
The fizziness of soda originates from dissolved \(CO_2\), partially in the form of carbonic acid. The concentration of \(CO_2\) dissolved in the soda depends on the amount of ambient pressure pressing down on the liquid. Hence, the soda can will be under pressure to maintain the desired \(CO_2\) concentration. When the can is opened to a lower pressure environment (e.g., the ambient atmosphere), the soda will quickly "outgas" (\(CO_2\) will come out of solution) at a rate depending on the surrounding atmospheric pressure. If a can of soda were opened under a lower pressure environment, this outgassing will be faster and hence more explosive (and dangerous) than under a high pressure environment.
Terms
• The solubility of a solute is the concentration of the saturated solution.
• A saturated solution a solution in which the maximum amount of solute has dissolved in the solvent at a given temperature.
• An unsaturated solution a solution in which the solute has completely dissolved in the solvent.
• A supersaturated solution is a solution in which the amount of solute dissolved under given conditions exceeds it's supposed upper limit.
• Le Châtelier's principle states that when a system in chemical equilibrium is stressed, the system will shift in a way that alleviates the stress.
• Endothermic reaction: a reaction in which heat is absorbed (ΔH>0)
• Exothermic reaction: a reaction in which heat is released (ΔH < 0) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/17%3A_Solutions/17.3%3A_Factors_Affecting_Solubility.txt |
Learning Objectives
• To understand how Temperature, Pressure, and the presence of other solutes affect the solubility of solutes in solvents.
Solubility is defined as the upper limit of solute that can be dissolved in a given amount of solvent at equilibrium. In such an equilibrium, Le Chatelier's principle can be used to explain most of the main factors that affect solubility. Le Châtelier's principle dictates that the effect of a stress upon a system in chemical equilibrium can be predicted in that the system tends to shift in such a way as to alleviate that stress.
Solute-Solvent Interactions Affect Solubility
The relation between the solute and solvent is very important in determining solubility. Strong solute-solvent attractions equate to greater solubility while weak solute-solvent attractions equate to lesser solubility. In turn, polar solutes tend to dissolve best in polar solvents while non-polar solutes tend to dissolve best in non-polar solvents. In the case of a polar solute and non-polar solvent (or vice versa), it tends to be insoluble or only soluble to a miniscule degree. A general rule to remember is, "Like dissolves like."
Common-Ion Effect
The common-ion effect is a term that describes the decrease in solubility of an ionic compound when a salt that contains an ion that already exists in the chemical equilibrium is added to the mixture. This effect best be explained by Le Chatelier's principle. Imagine if the slightly soluble ionic compound calcium sulfate, CaSO4, is added to water. The net ionic equation for the resulting chemical equilibrium is the following:
\[ CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
Calcium sulfate is slightly soluble; at equilibrium, most of the calcium and sulfate exists in the solid form of calcium sulfate.
Suppose the soluble ionic compound copper sulfate (CuSO4) were added to the solution. Copper sulfate is soluble; therefore, its only important effect on the net ionic equation is the addition of more sulfate (SO42-) ions.
\[ CuSO_{4(s)} \rightleftharpoons Cu^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
The sulfate ions dissociated from copper sulfate are already present (common to) in the mixture from the slight dissociation of calcium sulfate. Thus, this addition of sulfate ions places stress on the previously established equilibrium. Le Chatelier's principle dictates that the additional stress on this product side of the equilibrium results in the shift of equilibrium towards the reactants side in order to alleviate this new stress. Because of the shift toward the reactant side, the solubility of the slightly soluble calcium sulfate is reduced even further.
Temperature Affects Solubility
Temperature changes affect the solubility of solids, liquids and gases differently. However, those effects are finitely determined only for solids and gases.
Solids
The effects of temperature on the solubility of solids differ depending on whether the reaction is endothermic or exothermic. Using Le Chatelier's principle, the effects of temperature in both scenarios can be determined.
1. First, consider an endothermic reaction (\(\Delta{H_{solvation}}>0\)): Increasing the temperature results in a stress on the reactants side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the product side in order to alleviate this stress. By shifting towards the product side, more of the solid is dissociated when equilibrium is again established, resulting in increased solubility.
2. Second, consider an exothermic reaction ((\(\Delta{H_{solvation}}<0\)): Increasing the temperature results in a stress on the products side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the reactant side in order to alleviate this stress. By shifting towards the reactant's side, less of the solid is dissociated when equilibrium is again established, resulting in decreased solubility.
Liquids
In the case of liquids, there is no defined trends for the effects of temperature on the solubility of liquids.
Gases
In understanding the effects of temperature on the solubility of gases, it is first important to remember that temperature is a measure of the average kinetic energy. As temperature increases, kinetic energy increases. The greater kinetic energy results in greater molecular motion of the gas particles. As a result, the gas particles dissolved in the liquid are more likely to escape to the gas phase and the existing gas particles are less likely to be dissolved. The converse is true as well. The trend is thus as follows: increased temperatures mean lesser solubility and decreased temperatures mean higher solubility.
Le Chatelier's principle allows better conceptualization of these trends. First, note that the process of dissolving gas in liquid is usually exothermic. As such, increasing temperatures result in stress on the product side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts towards the reactant side in order to alleviate this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase increases, resulting in lowered solubility.
Conversely, decreasing temperatures result in stress on the reactant side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts toward the product side in order to compensate for this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase would decrease, resulting in greater solubility.
Pressure Affects Solubility of Gases
The effects of pressure are only significant in affecting the solubility of gases in liquids.
• Solids & Liquids: The effects of pressure changes on the solubility of solids and liquids are negligible.
• Gases: The effects of pressure on the solubility of gases in liquids can best be described through a combination of Henry's law and Le Chatelier principle. Henry's law dictates that when temperature is constant, the solubility of the gas corresponds to it's partial pressure. Consider the following formula of Henry's law:
\[ p = k_h \; c \]
where:
• \(p\) is the partial pressure of the gas above the liquid,
• \(k_h\) is Henry's law constant, and
• \(c\) is the concentrate of the gas in the liquid.
This formula indicates that (at a constant temperature) when the partial pressure decreases, the concentration of gas in the liquid decreases as well, and consequently the solubility also decreases. Conversely, when the partial pressure increases in such a situation, the concentration of gas in the liquid will increase as well; the solubility also increases. Extending the implications from Henry's law, the usefulness of Le Chatelier's principle is enhanced in predicting the effects of pressure on the solubility of gases.
Consider a system consisting of a gas that is partially dissolved in liquid. An increase in pressure would result in greater partial pressure (because the gas is being further compressed). This increased partial pressure means that more gas particles will enter the liquid (there is therefore less gas above the liquid, so the partial pressure decreases) in order to alleviate the stress created by the increase in pressure, resulting in greater solubility.
The converse case in such a system is also true, as a decrease in pressure equates to more gas particles escaping the liquid to compensate.
Example 1
Consider the following exothermic reaction that is in equilibrium
\[ CO_2 (g) + H_2O (l) \rightleftharpoons H_2CO_3 (aq) \]
What will happen to the solubility of the carbon dioxide if:
1. Temperature is increased?
2. Pressure and temperature are increased?
3. Pressure is increased but temperature is decreased?
4. Pressure is increased?
Solution
1. The reaction is exothermic, so an increase in temperature means that solubility would decrease.
2. The change in solubility cannot be determined from the given information. Increasing pressure increased solubility, but increasing temperature decreases solubility
3. An increase in pressure and an increase in temperature in this reaction results in greater solubility.
4. An increase in pressure results in more gas particles entering the liquid in order to decrease the partial pressure. Therefore, the solubility would increase.
Example 2: The Common Ion Effect
Bob is in the business of purifying silver compounds to extract the actual silver. He is extremely frugal. One day, he finds a barrel containing a saturated solution of silver chloride. Bob has a bottle of water, a jar of table salt (NaCl(s)), and a bottle of vinegar (CH3COOH). Which of the three should Bob add to the solution to maximize the amount of solid silver chloride (minimizing the solubility of the silver chloride)?
Solution
Bob should add table salt to the solution. According to the common-ion effect, the additional Cl- ions would reduce the solubility of the silver chloride, which maximizes the amount of solid silver chloride.
Example 3:
Allison has always wanted to start her own carbonated drink company. Recently, she opened a factory to produce her drinks. She wants her drink to "out-fizz" all the competitors. That is, she wants to maximize the solubility of the gas in her drink. What conditions (high/low temperature, high/low pressure) would best allow her to achieve this goal?
Solution
She would be able to maximize the solubility of the gas, (\(CO_2\) in this case, in her drink (maximize fizz) when the pressure is high and temperature is low.
Example 4
Butters is trying to increase the solubility of a solid in some water. He begins to frantically stir the mixture. Should he continue stirring? Why or why not?
Solution
He stop stop stirring. Stirring only affects how fast the system will reach equilibrium and does not affect the solubility of the solid at all.
Example 5: Outgassing Soda
With respect to Henry's law, why is it a poor ideal to open a can of soda in a low pressure environment?
Solution
The fizziness of soda originates from dissolved \(CO_2\), partially in the form of carbonic acid. The concentration of \(CO_2\) dissolved in the soda depends on the amount of ambient pressure pressing down on the liquid. Hence, the soda can will be under pressure to maintain the desired \(CO_2\) concentration. When the can is opened to a lower pressure environment (e.g., the ambient atmosphere), the soda will quickly "outgas" (\(CO_2\) will come out of solution) at a rate depending on the surrounding atmospheric pressure. If a can of soda were opened under a lower pressure environment, this outgassing will be faster and hence more explosive (and dangerous) than under a high pressure environment.
Terms
• The solubility of a solute is the concentration of the saturated solution.
• A saturated solution a solution in which the maximum amount of solute has dissolved in the solvent at a given temperature.
• An unsaturated solution a solution in which the solute has completely dissolved in the solvent.
• A supersaturated solution is a solution in which the amount of solute dissolved under given conditions exceeds it's supposed upper limit.
• Le Châtelier's principle states that when a system in chemical equilibrium is stressed, the system will shift in a way that alleviates the stress.
• Endothermic reaction: a reaction in which heat is absorbed (ΔH>0)
• Exothermic reaction: a reaction in which heat is released (ΔH < 0) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/17%3A_Solutions/17.4%3A_The_Vapor_Pressures_of_Solutions.txt |
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.
\begin{align*}\Delta{T_f} &= T_f(solvent) - T_f (solution) \[4pt] &= K_f \times m \end{align*}
where $\Delta{T_f}$ is the freezing point depression, $T_f$ (solution) is the freezing point of the solution, $T_f$ (solvent) is the freezing point of the solvent, $K_f$ is the freezing point depression constant, and m is the molality.
Introduction
Nonelectrolytes are substances with no ions, only molecules. Strong electrolytes, on the other hand, are composed mostly of ionic compounds, and essentially all soluble ionic compounds form electrolytes. Therefore, if we can establish that the substance that we are working with is uniform and is not ionic, it is safe to assume that we are working with a nonelectrolyte, and we may attempt to solve this problem using our formulas. This will most likely be the case for all problems you encounter related to freezing point depression and boiling point elevation in this course, but it is a good idea to keep an eye out for ions. It is worth mentioning that these equations work for both volatile and nonvolatile solutions. This means that for the sake of determining freezing point depression or boiling point elevation, the vapor pressure does not effect the change in temperature. Also, remember that a pure solvent is a solution that has had nothing extra added to it or dissolved in it. We will be comparing the properties of that pure solvent with its new properties when added to a solution.
Adding solutes to an ideal solution results in a positive $ΔS$, an increase in entropy. Because of this, the newly altered solution's chemical and physical properties will also change. The properties that undergo changes due to the addition of solutes to a solvent are known as colligative properties. These properties are dependent on the number of solutes added, not on their identity. Two examples of colligative properties are boiling point and freezing point: due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.
The freezing point and boiling point of a pure solvent can be changed when added to a solution. When this occurs, the freezing point of the pure solvent may become lower, and the boiling point may become higher. The extent to which these changes occur can be found using the formulas:
$\Delta{T}_f = -K_f \times m$
$\Delta{T}_b = K_b \times m$
where $m$ is the solute molality and $K$ values are proportionality constants; ($K_f$ and $K_b$ for freezing and boiling, respectively).
Molality
Molality is defined as the number of moles of solute per kilogram solvent. Be careful not to use the mass of the entire solution. Often, the problem will give you the change in temperature and the proportionality constant, and you must find the molality first in order to get your final answer.
If solving for the proportionality constant is not the ultimate goal of the problem, these values will most likely be given. Some common values for $K_f$ and $K_b$ respectively, are in Table $1$:
Table $1$: Ebullioscopic and cryoscopic constants for select solvents. Note that the nature of the solute does not affect colligative property relations.
Solvent $K_f$ $K_b$
Water 1.86 .512
Acetic acid 3.90 3.07
Benzene 5.12 2.53
Phenol 7.27 3.56
The solute, in order for it to exert any change on colligative properties, must fulfill two conditions. First, it must not contribute to the vapor pressure of the solution, and second, it must remain suspended in the solution even during phase changes. Because the solvent is no longer pure with the addition of solutes, we can say that the chemical potential of the solvent is lower. Chemical potential is the molar Gibb's energy that one mole of solvent is able to contribute to a mixture. The higher the chemical potential of a solvent is, the more it is able to drive the reaction forward. Consequently, solvents with higher chemical potentials will also have higher vapor pressures.
The boiling point is reached when the chemical potential of the pure solvent, a liquid, reaches that of the chemical potential of pure vapor. Because of the decrease in the chemical potential of mixed solvents and solutes, we observe this intersection at higher temperatures. In other words, the boiling point of the impure solvent will be at a higher temperature than that of the pure liquid solvent. Thus, boiling point elevation occurs with a temperature increase that is quantified using
$\Delta{T_b} = K_b m$
where $K_b$ is known as the ebullioscopic constant and $m$ is the molality of the solute.
Freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent. Again, since we are dealing with mixtures with decreased chemical potential, we expect the freezing point to change. Unlike the boiling point, the chemical potential of the impure solvent requires a colder temperature for it to reach the chemical potential of the pure solid solvent. Therefore, a freezing point depression is observed.
Example $1$
2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from 5.53 to 4.90 $^{\circ}C$. What is the molar mass of the compound?
Solution
First we must compute the molality of the benzene solution, which will allow us to find the number of moles of solute dissolved.
\begin{align*} m &= \dfrac{\Delta{T}_f}{-K_f} \[4pt] &= \dfrac{(4.90 - 5.53)^{\circ}C}{-5.12^{\circ}C / m} \[4pt] &= 0.123 m \end{align*}
\begin{align*} \text{Amount Solute} &= 0.07500 \; kg \; benzene \times \dfrac{0.123 \; m}{1 \; kg \; benzene} \[4pt] &= 0.00923 \; m \; solute \end{align*}
We can now find the molecular weight of the unknown compound:
\begin{align*} \text{Molecular Weight} =& \dfrac{2.00 \; g \; unknown}{0.00923 \; mol} \[4pt] &= 216.80 \; g/mol \end{align*}
The freezing point depression is especially vital to aquatic life. Since saltwater will freeze at colder temperatures, organisms can survive in these bodies of water.
Exercise $1$
Benzophenone has a freezing point of 49.00oC. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59oC. Find the freezing point depression constant for the solvent.
Answer
$9.80\,^oC/m$
Applications
Road salting takes advantage of this effect to lower the freezing point of the ice it is placed on. Lowering the freezing point allows the street ice to melt at lower temperatures. The maximum depression of the freezing point is about −18 °C (0 °F), so if the ambient temperature is lower, $\ce{NaCl}$ will be ineffective. Under these conditions, $\ce{CaCl_2}$ can be used since it dissolves to make three ions instead of two for $\ce{NaCl}$. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/17%3A_Solutions/17.5%3A_Boiling-Point_Elevation_and_Freezing-Point_Depression.txt |
Using the rule "like dissolves like" with the formation of ionic solutions, we must assess first assess two things: 1) the strength of the ion-dipole forces of attraction between water and the ionic compound and 2) the strength of the interionic bond of the ionic compound.
For an ionic compound to form a solution, the ion-dipole forces between water and ionic compound must be greater than the interionic bonds. Therefore, to form a compound: ion-dipole forces > interionic bonds
When the ionic compound is surrounded by water, the water dipoles surround the crystal's clustered structure. The water's negative ends of the dipole will be attracted to the positive dipoles of the ion and the positive ends of the water's dipole will be attracted to the negative dipoles of the ion. If the force of this attraction is stronger than the interionic bonds, the crystal's interionic bonds will be broken, then surrounded by the water molecules or hydrated .
There is a 3-step process that we can use to approach the energy involved in ionic solution formation. 1) Breaking apart the ionic compound is endothermic and requires energy. 2) Hydrating cation is exothermic and therefore releases energy. 3) Hydrating the anion is exothermic and also releases energy. The sum of these 3 steps will then give us the enthalpy of the solution.
Example \(1\): CaCl2
1) CaCl2 (s) -> Ca2+ (g) + Cl 2(g) energy > 0
2) Ca2+(g) H2O > Ca2+(aq) energy < 0
3) Cl2(g) H2O > Cl2(aq) energy < 0
CaCl2 (s) H2O> Ca2+(aq) + Cl2 (aq) energy > 0 The dissolution is endothermic because in the formation of ionic solutions, you must take into account entropy in addition to the enthalpy of the solution to determine whether it will occur spontaneously.
17.7: Osmotic Pressure
Introduction
Semipermiable membranes do not let the solute pass through (Think of the sugar example). A solvent will move to the side that is more concentrated to try to make each side more similar! Since there is a flow of solvents, the height of each side changes, which is osmotic pressure. When we work with aqueous solutions, we use mm of H2O to describe the difference.
Osmosis is the diffusion of a fluid through a semipermeable membrane. When a semipermeable membrane (animal bladders, skins of fruits and vegetables) separates a solution from a solvent, then only solvent molecules are able to pass through the membrane. The osmotic pressure of a solution is the pressure difference needed to stop the flow of solvent across a semipermeable membrane. The osmotic pressure of a solution is proportional to the molar concentration of the solute particles in solution.
$\Pi = i \dfrac{n}{V}RT = i M RT \label{eq1}$
where
• $\Pi$ is the osmotic pressure,
• $R$ is the ideal gas constant (0.0821 L atm / mol K),
• $T$ is the temperature in Kelvin,
• $i$ is the van 't Hoff factor
• $n$ is the number of moles of solute present,
• $V$ is the volume of the solution, and
• $M$ is the molar concentration of added solute (the $i$ factor accounts for how many species in solution are generated)
Exercise $1$
Calculate molarity of a sugar solution in water (300 K) has osmotic pressure of 3.00 atm.
Answer
Since it is sugar, we know it doesn't dissociate in water, so $i$ is 1. Then we use Equation \ref{eq1} directly
$M = \dfrac{\Pi}{RT} = \dfrac{3.00\, atm}{(0.0821\, atm.L/mol.K)(300\,K)} = 0.122\,M \nonumber$
Exercise $2$
Calculate osmotic pressure for 0.10 M $\ce{Na3PO4}$ aqueous solution at 20°C.
Answer
Since $\ce{Na3PO4}$ ionizes into four particles (3 Na+1 + $PO_4^{-3}$), then $i = 4$. We can then calculate the osmotic pressure via Equation \ref{eq1}
$\Pi = iMRT = (0.40)(0.0821)(293) = 9.6\, atm \nonumber$
Exercise $3$
Hemoglobin is a large molecule that carries oxygen in human blood. A water solution that contains 0.263 g of hemoglobin (Hb) in 10.0 mL of solution has an osmotic pressure of 7.51 torr at $25 ^oC$. What is the molar mass of the hemoglobin?
Answer
$6.51 \times 10^4 \; g/mol$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/17%3A_Solutions/17.7%3A_Colligative_Properties_of_Electrolyte_Solutions.txt |
A colloid is one of the three primary types of mixtures, with the other two being a solution and suspension. A colloid is a mixture that has particles ranging between 1 and 1000 nanometers in diameter, yet are still able to remain evenly distributed throughout the solution. These are also known as colloidal dispersions because the substances remain dispersed and do not settle to the bottom of the container. In colloids, one substance is evenly dispersed in another. The substance being dispersed is referred to as being in the dispersed phase, while the substance in which it is dispersed is in the continuous phase.
To be classified as a colloid, the substance in the dispersed phase must be larger than the size of a molecule but smaller than what can be seen with the naked eye. This can be more precisely quantified as one or more of the substance's dimensions must be between 1 and 1000 nanometers. If the dimensions are smaller than this the substance is considered a solution and if they are larger than the substance is a suspension.
Classifying Colloids
A common method of classifying colloids is based on the phase of the dispersed substance and what phase it is dispersed in. The types of colloids includes sol, emulsion, foam, and aerosol.
1. Sol is a colloidal suspension with solid particles in a liquid.
2. Emulsion is between two liquids.
3. Foam is formed when many gas particles are trapped in a liquid or solid.
4. Aerosol contains small particles of liquid or solid dispersed in a gas.
When the dispersion medium is water, the collodial system is often referred to as a hydrocolloid. The particles in the dispersed phase can take place in different phases depending on how much water is available. For example, Jello powder mixed in with water creates a hydrocolloid. A common use of hydrocolloids is in the creation of medical dressings.
Table 1: Examples of Colloids
Dispersion Medium Dispersed Phase Type of Colloid Example
Solid Solid Solid sol Ruby glass
Solid Liquid Solid emulsion/gel Pearl, cheese
Solid Gas Solid foam Lava, pumice
Liquid Solid Sol Paints, cell fluids
Liquid Liquid Emulsion Milk, oil in water
Liquid Gas Foam Soap suds, whipped cream
Gas Solid Aerosol Smoke
Gas Liquid Aerosol Fog, mist
An easy way of determining whether a mixture is colloidal or not is through use of the Tyndall Effect. When light is shined through a true solution, the light passes cleanly through the solution, however when light is passed through a colloidal solution, the substance in the dispersed phases scatters the light in all directions, making it readily seen. An example of this is shining a flashlight into fog. The beam of light can be easily seen because the fog is a colloid.
Another method of determining whether a mixture is a colloid is by passing it through a semipermeable membrane. The larger dispersed particles in a colloid would be unable to pass through the membrane, while the surrounding liquid molecules can. Dialysis takes advantage of the fact that colloids cannot diffuse through semipermeable membranes to filter them out of a medium.
Problems
1. Is dust a colloid? If so, what type is it?
2. Is whipped cream a colloid? if so, what type is it?
3. What does Sol mean?
4. When hit by light what happens to a colloidal mixture?
5. What is the mixture considered if the particles are larger than the particles of a colloidal substance
Answers
1. Dust is a colloid if suspended in air. It consists of a solid in a gas, so it is a aerosol.
2. Whipped cream is a colloid. It consists of a gas in a liquid, so it is a foam.
3. Sol is a colloidal suspension with solid particles in a liquid.
4. The light is reflected off the large particles and spread out.
5. It's considered a suspension if the particles are larger than 1000 nanometers.
Contributors and Attributions
• Jimmy Law (UCD), Abheetinder Brar (UCD)
• Thumbnail: pixabay.com/photos/milk-spra...-food-4755234/ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/17%3A_Solutions/17.8%3A_Colloids.txt |
Learning Objectives
• To know important periodic trends in several atomic properties.
As we begin our summary of periodic trends, recall that the single most important unifying principle in understanding the chemistry of the elements is the systematic increase in atomic number, accompanied by the orderly filling of atomic orbitals by electrons, which leads to periodicity in such properties as atomic and ionic size, ionization energy, electronegativity, and electron affinity. The same factors also lead to periodicity in valence electron configurations, which for each group results in similarities in oxidation states and the formation of compounds with common stoichiometries.
The most important periodic trends in atomic properties are summarized in Figure \(1\). Recall that these trends are based on periodic variations in a single fundamental property, the effective nuclear charge (\(Z_{eff}\)), which increases from left to right and from top to bottom in the periodic table.
The diagonal line in Figure \(1\) separates the metals (to the left of the line) from the nonmetals (to the right of the line). Because metals have relatively low electronegativities, they tend to lose electrons in chemical reactions to elements that have relatively high electronegativities, forming compounds in which they have positive oxidation states. Conversely, nonmetals have high electronegativities, and they therefore tend to gain electrons in chemical reactions to form compounds in which they have negative oxidation states. The semimetals lie along the diagonal line dividing metals and nonmetals. It is not surprising that they tend to exhibit properties and reactivities intermediate between those of metals and nonmetals. Because the elements of groups 13, 14, and 15 span the diagonal line separating metals and nonmetals, their chemistry is more complex than predicted based solely on their valence electron configurations.
Unique Chemistry of the Lightest Elements
The chemistry of the second-period element of each group (n = 2: Li, Be, B, C, N, O, and F) differs in many important respects from that of the heavier members, or congeners, of the group. Consequently, the elements of the third period (n = 3: Na, Mg, Al, Si, P, S, and Cl) are generally more representative of the group to which they belong. The anomalous chemistry of second-period elements results from three important characteristics: small radii, energetically unavailable d orbitals, and a tendency to form pi (π) bonds with other atoms.
Note
In contrast to the chemistry of the second-period elements, the chemistry of the third-period elements is more representative of the chemistry of the respective group.
Due to their small radii, second-period elements have electron affinities that are less negative than would be predicted from general periodic trends. When an electron is added to such a small atom, increased electron–electron repulsions tend to destabilize the anion. Moreover, the small sizes of these elements prevent them from forming compounds in which they have more than four nearest neighbors. Thus BF3 forms only the four-coordinate, tetrahedral BF4 ion, whereas under the same conditions AlF3 forms the six-coordinate, octahedral AlF63 ion. Because of the smaller atomic size, simple binary ionic compounds of second-period elements also have more covalent character than the corresponding compounds formed from their heavier congeners. The very small cations derived from second-period elements have a high charge-to-radius ratio and can therefore polarize the filled valence shell of an anion. As such, the bonding in such compounds has a significant covalent component, giving the compounds properties that can differ significantly from those expected for simple ionic compounds. As an example, LiCl, which is partially covalent in character, is much more soluble than NaCl in solvents with a relatively low dielectric constant, such as ethanol (ε = 25.3 versus 80.1 for H2O).
Because d orbitals are never occupied for principal quantum numbers less than 3, the valence electrons of second-period elements occupy 2s and 2p orbitals only. The energy of the 3d orbitals far exceeds the energy of the 2s and 2p orbitals, so using them in bonding is energetically prohibitive. Consequently, electron configurations with more than four electron pairs around a central, second-period element are simply not observed. You may recall that the role of d orbitals in bonding in main group compounds with coordination numbers of 5 or higher remains somewhat controversial. In fact, theoretical descriptions of the bonding in molecules such as SF6 have been published without mentioning the participation of d orbitals on sulfur. Arguments based on d-orbital availability and on the small size of the central atom, however, predict that coordination numbers greater than 4 are unusual for the elements of the second period, which is in agreement with experimental results.
One of the most dramatic differences between the lightest main group elements and their heavier congeners is the tendency of the second-period elements to form species that contain multiple bonds. For example, N is just above P in group 15: N2 contains an N≡N bond, but each phosphorus atom in tetrahedral P4 forms three P–P bonds. This difference in behavior reflects the fact that within the same group of the periodic table, the relative energies of the π bond and the sigma (σ) bond differ. A C=C bond, for example, is approximately 80% stronger than a C–C bond. In contrast, an Si=Si bond, with less p-orbital overlap between the valence orbitals of the bonded atoms because of the larger atomic size, is only about 40% stronger than an Si–Si bond. Consequently, compounds that contain both multiple and single C to C bonds are common for carbon, but compounds that contain only sigma Si–Si bonds are more energetically favorable for silicon and the other third-period elements.
The Inert-Pair Effect
The inert-pair effect refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. For example, although an oxidation state of +3 is common for group 13 elements, the heaviest element in group 13, thallium (Tl), is more likely to form compounds in which it has a +1 oxidation state. There appear to be two major reasons for the inert-pair effect: increasing ionization energies and decreasing bond strengths.
Note
In moving down a group in the p-block, increasing ionization energies and decreasing bond strengths result in an inert-pair effect.
The ionization energies increase because filled (n − 1)d or (n − 2)f subshells are relatively poor at shielding electrons in ns orbitals. Thus the two electrons in the ns subshell experience an unusually high effective nuclear charge, so they are strongly attracted to the nucleus, reducing their participation in bonding. It is therefore substantially more difficult than expected to remove these ns2electrons, as shown in Table \(1\) by the difference between the first ionization energies of thallium and aluminum. Because Tl is less likely than Al to lose its two ns2 electrons, its most common oxidation state is +1 rather than +3.
Table \(1\): Ionization Energies (I) and Average M–Cl Bond Energies for the Group 13 Elements
Element Electron Configuration I1 (kJ/mol) I1 + I2 + I3 (kJ/mol) Average M–Cl Bond Energy (kJ/mol)
B [He] 2s22p1 801 6828 536
Al [Ne] 3s23p1 578 5139 494
Ga [Ar] 3d104s24p1 579 5521 481
In [Kr] 4d105s2p1 558 5083 439
Tl [Xe] 4f145d106s2p1 589 5439 373
Source of data: John A. Dean, Lange’s Handbook of Chemistry, 15th ed. (New York: McGraw-Hill, 1999).
Going down a group, the atoms generally became larger, and the overlap between the valence orbitals of the bonded atoms decreases. Consequently, bond strengths tend to decrease down a column. As shown by the M–Cl bond energies listed in Table \(1\), the strength of the bond between a group 13 atom and a chlorine atom decreases by more than 30% from B to Tl. Similar decreases are observed for the atoms of groups 14 and 15.
The net effect of these two factors—increasing ionization energies and decreasing bond strengths—is that in going down a group in the p-block, the additional energy released by forming two additional bonds eventually is not great enough to compensate for the additional energy required to remove the two ns2 electrons.
Example \(1\)
Based on the positions of the group 13 elements in the periodic table and the general trends outlined in this section,
1. classify these elements as metals, semimetals, or nonmetals.
2. predict which element forms the most stable compounds in the +1 oxidation state.
3. predict which element differs the most from the others in its chemistry.
4. predict which element of another group will exhibit chemistry most similar to that of Al.
Given: positions of elements in the periodic table
Asked for: classification, oxidation-state stability, and chemical reactivity
Strategy:
From the position of the diagonal line in the periodic table separating metals and nonmetals, classify the group 13 elements. Then use the trends discussed in this section to compare their relative stabilities and chemical reactivities.
Solution
1. Group 13 spans the diagonal line separating the metals from the nonmetals. Although Al and B both lie on the diagonal line, only B is a semimetal; the heavier elements are metals.
2. All five elements in group 13 have an ns2np1 valence electron configuration, so they are expected to form ions with a +3 charge from the loss of all valence electrons. The inert-pair effect should be most important for the heaviest element (Tl), so it is most likely to form compounds in an oxidation state that is lower by 2. Thus the +1 oxidation state is predicted to be most important for thallium.
3. Among the main group elements, the lightest member of each group exhibits unique chemistry because of its small size resulting in a high concentration of charge, energetically unavailable d orbitals, and a tendency to form multiple bonds. In group 13, we predict that the chemistry of boron will be quite different from that of its heavier congeners.
4. Within the s and p blocks, similarities between elements in different groups are most marked between the lightest member of one group and the element of the next group immediately below and to the right of it. These elements exhibit similar electronegativities and charge-to-radius ratios. Because Al is the second member of group 13, we predict that its chemistry will be most similar to that of Be, the lightest member of group 2.
Exercise \(1\)
Based on the positions of the group 14 elements C, Si, Ge, Sn, and Pb in the periodic table and the general trends outlined in this section,
1. classify these elements as metals, semimetals, or nonmetals.
2. predict which element forms the most stable compounds in the +2 oxidation state.
3. predict which element differs the most from the others in its chemistry.
4. predict which element of group 14 will be chemically most similar to a group 15 element.
Answer
1. nonmetal: C; semimetals: Si and Ge; metals: Sn and Pb
2. Pb is most stable as M2+.
3. C is most different.
4. C and P are most similar in chemistry.
Summary
The most important unifying principle in describing the chemistry of the elements is that the systematic increase in atomic number and the orderly filling of atomic orbitals lead to periodic trends in atomic properties. The most fundamental property leading to periodic variations is the effective nuclear charge (Zeff). Because of the position of the diagonal line separating metals and nonmetals in the periodic table, the chemistry of groups 13, 14, and 15 is relatively complex. The second-period elements (n = 2) in each group exhibit unique chemistry compared with their heavier congeners because of their smaller radii, energetically unavailable d orbitals, and greater ability to form π bonds with other atoms. Increasing ionization energies and decreasing bond strengths lead to the inert-pair effect, which causes the heaviest elements of groups 13–17 to have a stable oxidation state that is lower by 2 than the maximum predicted for their respective groups.
Key Takeaway
• The chemistry of the third-period element in a group is most representative of the chemistry of the group because the chemistry of the second-period elements is dominated by their small radii, energetically unavailable d orbitals, and tendency to form π bonds with other atoms. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.01%3A_A_Survey_of_the_Representative_Elements.txt |
Li, Na, K, Rb, and Cs are all group IA elements, also known as the alkali metals. The seventh member of the group, francium (Fr) is radioactive and so rare that only 20 atoms of Fr may exist on Earth at any given moment[1]. The term alkali is derived from an Arabic word meaning “ashes.” Compounds of potassium as well as other alkali metals were obtained from wood ashes by early chemists. All the alkali metals are soft and, except for Cs which is yellow, are silvery-gray in color.
Lithium, sodium, potassium, rubidium, and cesium have a great many other properties in common. All are solids at 0°C and melt below 200°C. Each has metallic properties such as good conduction of heat and electricity, malleability (the ability to be hammered into sheets), and ductility (the ability to be drawn into wires). The high thermal (heat) conductivity and the relatively low melting point (for a metal) of sodium make it an ideal heat-transfer fluid. It is used to cool certain types of nuclear reactors (liquid-metal fast breeder reactors, LMFBRs) and to cool the valves of high-powered automobile engines for this reason.
Some general properties of the alkali metals are summarized in the table below. All these metal atoms contain a singles electron outside a noble-gas configuration, and so the valence electron is-well shielded from nuclear charge and the atomic radii are relatively large. The large volume of each atom results in a low density—small enough that Li, Na, and K float on water as they react with it.
Table $1$ Properties of the Group IA Alkali Metals
Element Symbol Electron Configuration Usual Oxidation State Atomic Radius/pm Ionic (M+) Radius/pm
Lithium Li [He]2s1 +1 122 60
Sodium Na [Ne]3s1 +1 157 95
Potassium K [Ar]4s1 +1 202 133
Rubidium Rb [Kr]5s1 +1 216 148
Cesium Cs [Xe]6s1 +1 235 169
Symbol Ionization Energy/MJ mol–1 Density/ g cm–3 Electronegativity Melting Point (in °C)
First Second
Li 0.526 7.305 0.534 1.0 179
Na 0.502 4.569 0.97 0.9 98
K 0.425 3.058 0.86 0.8 64
Rb 0.409 2.638 1.52 0.8 39
Cs 0.382 2.430 1.87 0.7 28
The atoms do not have a strong attraction for the single valence electron, and so it is easily lost (small first ionization energy) to from a +1 ion. Because they readily donate electrons in this way, all the alkali metals are strong reducing agents. They are quite reactive, even reducing water.
Weak attraction for the valence electron also results in weak metallic bonding, because it is attraction among nuclei and numerous valence electrons that holds metal atoms together. Weak metallic bonding results in low melting points, especially for the larger atoms toward the bottom of the group. Cs, for example, melts just above room temperature. Weak metallic bonding also accounts for the fact that all these metals are rather soft.
That the chemistry of alkali metals is confined to the +1 oxidation state is confirmed by the large second-ionization energies. Removing the first electron from the large, diffuses orbital is easy, but removing a second electron from an octet in an M+ ion is much too difficult for any oxidizing agent to do.
Two other elements are found in group IA. Hydrogen, although many of its compounds have formulas similar to the alkali metals, is a nonmetal and is almost unique in its chemical behavior. Therefore it is not usually included in this group. Francium (Fr) is quite radioactive, and only small quantities are available for study; so it too is usually omitted. Its properties, however, appear to be similar to those of Cs and the other alkali metals.
Chemical Reactions and Compounds
The element lithium combines violently and spectacularly with water. Hydrogen gas is given off, which propels the the lithium metal across the water as it reacts. If the excess water is evaporated, the compound lithium hydroxide (LiOH) remains behind. LiOH is visualized by phenolphthalein indicator, which turns pink as LiOH, a base, is produced. Thus the equation for this reaction is
$\text{2Li}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2LiOH}(aq) + \text{H}_\text{2}(g)\nonumber$
The elements sodium, potassium, rubidium, and cesium also combine violently with water to form hydroxides. The equations for their reactions are
$\text{2Na}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2NaOH}(aq) + \text{H}_\text{2} (g)\nonumber$
$\text{2K}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2KOH}(aq) + \text{H}_\text{2} (g)\nonumber$
$\text{2Rb}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2RbOH}(aq) + \text{H}_\text{2} (g)\nonumber$
$\text{2Cs}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2CsOH}(aq) + \text{H}_\text{2} (g)\nonumber$
Since the alkali metals all react with water in the same way, a general equation may be written:
$\text{2M}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2MOH}(aq) + \text{H}_\text{2} (g)\nonumber$
with M = K, Li, Na, Rb, or Cs.
The symbol M represents any one of the five elements.
In addition to their behavior when added to water, the alkali metals react directly with many elements. All combine swiftly with oxygen in air to form white oxide:
$\text{4M}(s) + \text{O}_2(g) \rightarrow \text{2M}_2 \text{O}(s)\nonumber$
with M = Li, Na, K, Rb, or Cs
(Li2O is lithium oxide, Na2O is sodium oxide, etc.)
All except lithium react further to form yellow peroxides, M2O2:
$\text{2M}_2 \text{O}(s) + \text{O}_2(g) \rightarrow \text{2M}_2 \text{O}_2(s)\nonumber$ M = Na, K, Rb, or Cs
(Na2O2 is sodium peroxide, etc.)
Potassium, rubidium, and cesium are sufficiently reactive that yellow superoxides (whose general formula is MO2) can be formed:
$\text{2M}_2 \text{O}_2(s) + \text{O}_2(g) \rightarrow \text{2MO}_2(s)\nonumber$
with M = K, Rb, or Cs
Unless the surface of a sample of an alkali metal is scraped clean, it will appear white or gray instead of having a silvery metallic luster. This is due to the oxide, peroxide, or superoxide coating that forms after a few seconds of exposure to air. The following movie shows how a freshly cut piece of lithium is shiny, but dulls to gray when exposed to oxygen in the air. The video also focuses on another important property of alkali metals: they are soft, and easy to cut, compared to other metals.
A dull gray oxidized cylinder of lithium metal is cut, revealing a shiny silvery surface. After 1 minute, the surface has dulled, and after 10 minutes, the cut surface has returned to the dull gray of the rest of the lithium metal. Since the alkali metal is lithium, the only reaction with oxygen that occurs is:
$\text{4Li}(s) + \text{O}_2(g) \rightarrow \text{2Li}_2\text{O}(s)\nonumber$
The alkali also combine directly with hydrogen gas to form compounds known as hydrides, MH:
$\text{2M}(s) + \text{H}_2(g) \rightarrow \text{2MH}(s)\nonumber$
with M = Li, Na, K, Rb, or Cs
They react with sulfur to form sulfides, M2S:
$\text{2M}(s) + \text{S}(g) \rightarrow \text{M}_2\text{S}(s)\nonumber$
with M = Li, Na, K, Rb, or Cs
These oxides, hydrides, hydroxides, and sulfides all dissolve in water to give basic solutions, and these compounds are among the strong bases.
The peroxides and superoxides formed when the heavier alkali metals react with O2 also dissolve to give basic solutions:
$\text{2NaO}_2(s) + \text{2H}_2\text{O}(l) \rightarrow \text{4Na}^{+}(aq) + \text{4OH}^{-}(aq) + \text{O}_2(g)\nonumber$
$\text{4K}_2\text{O}(s) + \text{2H}_2\text{O}(l) \rightarrow \text{4K}^{+}+ \text{4OH}^{-} + \text{3O}_2(g)\nonumber$
Both of the latter equations describe redox as well as acid-base processes, as you can confirm by assigning oxidation numbers. The peroxide and superoxide ions contain O atoms in the unusual (for O) –1 and –½ oxidation states:
Therefore disproportionation (simultaneous oxidation and reduction) of O22 or O2 to the more common oxidation states of 0 (in O2) and –2 (in OH) is possible.
The alkali metals also react directly with the halogens, for instance with chlorine, forming chlorides,
$\text{2M}(s) + \text{Cl}_2(g) \rightarrow \text{2MCl}(s)\nonumber$ M = Li, Na, K, Rb, or Cs
Below is an example of the reaction of Na with Cl2
A piece of sodium metal is added to a flask containing chlorine gas. Initially no reaction takes place, but when a drop of water is added, sodium and chlorine react, violently flaring up and producing so much heat that sand is needed in the bottom of the flask to absorb the heat and prevent the glass from cracking. This equation for this reaction is:
$\text{2Na}(s) + \text{Cl}_2(g) \rightarrow \text{2NaCl}(s)\nonumber$
with fluorine to form fluorides, MF:
$\text{2M}(s) + \text{F}_2(g) \rightarrow \text{2MF}(s)\nonumber$ M = Li, Na, K, Rb, or Cs
and with bromine to form bromides, MBr:
$\text{2M}(s) + \text{Br}_2(g) \rightarrow \text{2MBr}(s)\nonumber$ M = Li, Na, K, Rb, or Cs
Below is an example of K reacting with Br2
In this video, potassium, which is stored in inert mineral oil due to its high reactivity, is placed in a beaker of liquid bromine after the protective layer of mineral oil has been removed. The potassium reacts explosively with the bromine. The container is covered during the whole process to prevent reactants and products from entering the environment. The chemical equation for this reaction is:
$\text{2K}(s) + \text{Br}_2(g) \rightarrow \text{2KBr}(s)\nonumber$
Sodium and potassium are quite abundant, ranking sixth and seventh among all elements in the earth’s crust, but the other alkali metals are rare. Sodium and potassium ions are components of numerous silicate crystal lattices seen in the Earth's crust, but since most of their compounds are water soluble, they are also important constituents of seawater and underground deposits of brine. Sodium chloride obtained from such brines is the chief commercial source of sodium, while potassium can be obtained from the ores sylvite (KCl) or carnallite (KCl•MgCl2•6H2O).
Both sodium (Na+) and potassium (K+) ions are essential to living systems. Na+ is the main cation in fluids surrounding the cells, while K+is most important inside the cells. Na+ plays a role in muscle contraction, and both K+ and Na+ play a role in transmitting nerve impulses. K is more important than Na in plants, and it is one of three elements (K, P, N) which must be supplied in fertilizer to maintain high crop yields. K is especially abundant in trees—wood ashes from kitchen fires (potash) were the major source of this element as recently as a century ago, and they still make good fertilizer for your garden. Wood ashes contain a mixture of potassium oxide and potassium carbonate, the latter formed by combination of K2O with CO2 produced when C in the wood combines with O2:
$\text{K}_2\text{O} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3\nonumber$
Na compounds are obtained commercially from brine or from seawater. When an electrical current is passed through an NaCl solution (a process called electrolysis), Cl2(g), H2(g), and a concentrated solution of NaOH (caustic soda or lye) are obtained:
$\text{Na}^{+}(aq) + \text{2Cl}^{-}(aq) + \text{2H}_2\text{O}(l) \xrightarrow{\text{electrolysis}} \text{Cl}_2(g) + \text{H}_2(g) + \text{Na}^{+}(aq) + \text{2OH}^{-}(aq)\nonumber$
This process is described in more detail in the section on electrochemical cells, but you can see from the equation that the electrical current oxidizes Cl to Cl2 and reduces H2O to H2. NaOH(aq) is used as a strong base in numerous industrial processes to make soap, rayon, cellophane, paper, dyes, and many other products. Lye is also used in home drain cleaners. It must be handled with care because it is strongly basic, highly caustic, and can severely burn the skin. A second important industrial use of brine is the Solvay process:
$\text{CO}_2(g) + \text{NH}_3(aq) + \text{Na}^{+}(aq) + \text{Cl}^{-}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{NaHCO}_3(s) + \text{NH}_4^{+}(aq) + \text{Cl}^{-}(aq)\nonumber$
The Solvay process is an acid-base reaction combined with a precipitation. The acid anhydride, CO2, reacts with H2O to produce H2CO3. This weak acid donates a proton to NH3, yielding NH4+ and HCO3, and the latter ion precipitates with Na+. The weakly basic sodium hydrogen carbonate produced by the Solvay process can be purified for use as an antacid (bicarbonate of soda), but most of it is converted to sodium carbonate (soda ash) by heating:
$\text{2NaHCO}_3(s) \xrightarrow{\Delta } \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)\nonumber$
(The Δ in this equation indicates heating of the reactant.) Sodium carbonate (Na2CO3) is used in manufacturing glass and paper, and in some detergents. The carbonate ion is a rather strong base, however, and detergents containing Na2CO3 (washing soda) have resulted in severe chemical burns to some small children who, out of curiosity, have eaten them. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.02%3A_Group_1A_Metals.txt |
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_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.03%3A_The_Chemistry_of_Hydrogen.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_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.04%3A_Group_2A_Elements.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_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.05%3A_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.
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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_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.07%3A_The_Group_5A_Elements.txt |
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. Although nitrogen is the most abundant element in the atmosphere, it was the last of the pnicogens (Group 15 elements) 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 Nitrogen
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:
$\ce{2NaN3(s) ->[\Delta] 2Na(l) + 3N2(g)} \label{Eq1}$
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.
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):
$\ce{2NO2(g) + H2O(l) -> HNO2(aq) + HNO3(aq)} \label{Eq2}$
Nitrogen also forms $\ce{N2O}$ (dinitrogen monoxide, or nitrous oxide), a linear molecule that is isoelectronic with $\ce{CO2}$ 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 $\ce{Li3N}$ and $\ce{Ca3N2}$. These compounds consist of ionic lattices formed by $\ce{M^{n+}}$ and $\ce{N^{3−}}$ 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 $\ce{Si3N4}$ and $\ce{BN}$, 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 $\ce{O2}$ atmosphere:
$\ce{4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)} \label{Eq3}$
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 ($\ce{HN3}$), also called hydrogen azide, is a colorless, highly toxic, and explosive substance. Hydrazine ($\ce{N2H4}$) 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. $\ce{Sr(s) + N2O(g) ->[\Delta] SrO(s) + N2(g)}$
2. $\ce{NH4NO2(s) ->[\Delta] N2(g) + 2H2O(g)}$
3. $\ce{Pb(NO3)2(s) ->[\Delta] PbO2(s) + 2NO2(g)}$
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 $1$
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. $\ce{NO(g) + H2O(l) ->[\Delta]}$
2. $\ce{NH4NO3(s) ->[\Delta]}$
3. $\ce{Sr(s) + N2(g) ->}$
Answer
1. $\ce{NO(g) + H2O(l) ->[\Delta] no reaction}$
2. $\ce{NH4NO3(s) ->[\Delta] N2O(g) + 2H2O(g)}$
3. $\ce{3Sr(s) + N2(g) -> Sr3N2(s)}$
Summary
Nitrogen behaves chemically like nonmetals, Nitrogen forms compounds in nine different oxidation states. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.08%3A_The_Chemistry_of_Nitrogen.txt |
Learning Objectives
• Compare properties of Group 15 elements.
• Explain the major application of phosphate.
• Describe the equilibria of the ionization of phosphoric acid.
Phosphorus (P) is an essential part of life as we know it. Without the phosphates in biological molecules such as ATP, ADP and DNA, we would not be alive. Phosphorus compounds can also be found in the minerals in our bones and teeth. It is a necessary part of our diet. In fact, we consume it in nearly all of the foods we eat. Phosphorus is quite reactive. This quality of the element makes it an ideal ingredient for matches because it is so flammable. Phosphorus is a vital element for plants and that is why we put phosphates in our fertilizer to help them maximize their growth.
Introduction
Phosphorus plays a big role in our existence but it can also be dangerous. When fertilizers containing phosphorus enter the water, it produces rapid algae growth. This can lead to eutrophication of lakes and rivers; i.e., the ecosystem has an increase of chemical nutrients and this can led to negative environmental effects. With all the excess phosphorus, plants grow rapidly then die, causing a lack of oxygen in the water and an overall reduction of water quality. It is thus necessary to remove excess phosphorus from our wastewater. The process of removing the phosphorus is done chemically by reacting the phosphorus with compounds such as ferric chloride, ferric sulfate, and aluminum sulfate or aluminum chlorohydrate. Phosphorus, when combined with aluminum or iron, becomes an insoluble salt. The solubility equilibrium constants of $FePO_4$ and $AlPO_4$ are 1.3x10-22 and 5.8x10-19, respectively. With solubilitys this low, the resulting precipitates can then be filtered out.
Another example of the dangers of phosphorus is in the production of matches. The flammable nature and cheap manufacturing of white phosphorus made it possible to easily make matches around the turn of the 20th century. However, white phosphorus is highly toxic. Many workers in match factories developed brain damage and a disease called "phosphorus necrosis of the jaw" from exposure to toxic phosphorus vapors. Excess phosphorus accumulation caused their bone tissue to die and rot away. For this reason, we now use red phosphorus or phosphorus sesquisulfide in "safety" matches.
Discovery of Phosphorus
Named from the Greek word phosphoros ("bringer of light"), elemental phosphorus is not found in its elemental form because this form is quite reactive. Because of this factor it took a long period of time for it to be "discovered". The first recorded isolation of phosphorus was by alchemist Hennig Brand in 1669, involving about 60 pails of urine. After letting a large amount of urine putrefy for a long time, Brand distilled the liquid to a paste, heated the paste, discarded the salt formed, and put the remaining substance under cold water to form solid white phosphorus. Brand's process was not very efficient; the salt he discarded actually contained most of the phosphorus. Nevertheless, he obtained some pure, elemental phosphorus for his efforts. Others of the time improved the efficiency of the process by adding sand, but still continued to discard the salt. Later, phosphorus was manufactured from bone ash. Currently, the process for manufacturing phosphorus does not involve large amounts of putrefied urine or bone ash. Instead, manufacturers use calcium phosphate and coke (Emsley).
Allotropes of Phosphorus
Phosphorus is a nonmetal, solid at room temperature, and a poor conductor of heat and electricity. Phosphorus occurs in at least 10 allotropic forms, the most common (and reactive) of which is so-called white (or yellow) phosphorus, which looks like a waxy solid or plastic. It is very reactive and will spontaneously inflame in air, so it is stored under water. The other common form of phosphorus is red phosphorus, which is much less reactive and is one of the components on the striking surface of a match book. Red phosphorus can be converted to white phosphorus by careful heating.
White phosphorus consists of $\ce{P4}$ molecules, whereas the crystal structure of red phosphorus has a complicated network of bonding. White phosphorus has to be stored in water to prevent natural combustion, but red phosphorus is stable in air.
When burned, red phosphorus also forms the same oxides as those obtained in the burning of white phosphosrus, $\ce{P4O6}$ when air supply is limited, and $\ce{P4O10}$ when sufficient air is present.
Diphosphorus (P2)
Diphosphorus ($P_2$) is the gaseous form of phosphorus that is thermodynamically stable above 1200 °C and until 2000 °C. It can be generated by heating white phosphorus (see below) to 1100 K and is very reactive with a bond-dissociation energy (117 kcal/mol or 490 kJ/mol) half that of dinitrogen ($N_2$).
White Phosphorus (P4)
White phosphorus (P4) has a tetrahedral structure. It is soft and waxy, but insoluble in water. Its glow occurs as a result of its vapors slowly being oxidized by the air. It is so thermodynamically unstable that it combusts in air. It was once used in fireworks and the U.S. military still uses it in incendiary bombs.
This Youtube video link shows various experiments with white phosphorus, which help show the physical and chemical properties of it. It also shows white phosphorus combusting with air.
Red Phosphorus and Violet Phosphorus (Polymeric)
Red Phosphorus has more atoms linked together in a network than white phosphorus does, which makes it much more stable. It is not quite as flammable, but given enough energy it still reacts with air. For this reason, we now use red phosphorus in matches.
Violet phosphorus is obtained from heating and crystallizing red phosphorus in a certain way. The phosphorus forms pentagonal "tubes".
Black Phosphorus (Polymeric)
Black phosphorus is the most stable form; the atoms are linked together in puckered sheets, like graphite. Because of these structural similarities black phosphorus is also flaky like graphite and possesses other similar properties.
Isotopes of Phosphorus
There are many isotopes of phosphorus, only one of which is stable (31P). The rest of the isotopes are radioactive with generally very short half-lives, which vary from a few nanoseconds to a few seconds. Two of the radioactive phosphorus isotopes have longer half-lives: 32P has a half-life of 14 days and 33P has a half-life of 25 days. These half-lives are long enough to be useful for analysis, and for this reason the isotopes can be used to mark DNA.
32P played an important role in the 1952 Hershey-Chase Experiment. In this experiment, Alfred Hershey and Martha Chase used radioactive isotopes of phosphorus and sulfur to determine that DNA was genetic material and not proteins. Sulfur can be found in proteins but not DNA, and phosphorus can be found in DNA but not proteins. This made phosphorus and sulfur effective markers of DNA and protein, respectively. The experiment was set up as follows: Hershey and Chase grew one sample of a virus in the presence of radioactive 35S and another sample of a virus in the presence of 32P. Then, they allowed both samples to infect bacteria. They blended the 35S and the 32P samples separately and centrifuged the two samples. Centrifuging separated the genetic material from the non-genetic material. The genetic material penetrated the solid that contained the bacterial cells at the bottom of the tube while the non-genetic material remained in the liquid. By analyzing their radioactive markers, Hershey and Chase found that the 32P remained with the bacteria, and the 35S remained in the supernatant liquid. These results were confirmed by further tests involving the radioactive phosphorus..
Phosphorus and Life
We get most elements from nature in the form of minerals. In nature, phosphorus exists in the form of phosphates. Rocks containing phosphate are fluoroapatite ($\ce{3Ca3(PO4)2.CaF2}$), chloroapaptite, ($\ce{3Ca3(PO4)2. CaCl2}$), and hydroxyapatite ($\ce{3Ca3(PO4)2. Ca(OH)2}$). These minerals are very similar to the bones and teeth. The arrangements of atoms and ions of bones and teeth are similar to those of the phosphate-containing rocks. In fact, when the $\ce{OH-}$ ions of the teeth are replaced by $\ce{F-}$, the teeth resist decay. This discovery led to a series of social and economical issues.
Nitrogen, phosphorus and potassium are key ingredients for plants, and their contents are key in all forms of fertilizers. From an industrial and economical viewpoint, phosphorus-containing compounds are important commodities. Thus, the chemistry of phosphorus has academic, commercial and industrial interests.
Chemistry of Phosphorus
As a member of the Nitrogen Family, Group 15 on the Periodic Table, phosphorus has 5 valence shell electrons available for bonding. Its valence shell configuration is 3s23p3. Phosphorus forms mostly covalent bonds. Any phosphorus rock can be used for the production of elemental phosphorus. Crushed phosphate rocks and sand ($\ce{SiO2}$) react at 1700 K to give phosphorus oxide, $\ce{P4O10}$:
$\ce{2 Ca3(PO4)2 + 6 SiO2 \rightarrow P4O10 + 6 CaSiO3} \label{1}$
$\ce{P4O10}$ can be reduced by carbon:
$\ce{P4O10 + 10 C \rightarrow P4 + 10 CO}. \label{2}$
Waxy solids of white phosphorus are molecular crystals consisting of $\ce{P4}$ molecules. They have an interesting property in that they undergo spontaneous combustion in air:
$\ce{P4 + 5 O2 \rightarrow P4O10} \label{3}$
The structure of $\ce{P4}$ can be understood by thinking of the electronic configuration (s2 p3) of $\ce{P}$ in bond formation. Sharing three electrons with other $\ce{P}$ atoms gives rise to the 6 $\ce{P-P}$ bonds, leaving a lone pair occupying the 4th position in a distorted tetrahedron.
When burned with insufficient oxygen, $\ce{P4O6}$ is formed:
$\ce{P4 + 3 O2 \rightarrow P4O6} \label{4}$
Into each of the $\ce{P-P}$ bonds, an $\ce{O}$ atom is inserted.
Burning phosphorus with excess oxygen results in the formation of $\ce{P4O10}$. An additional $\ce{O}$ atom is attached to the $\ce{P}$ directly:
$\ce{P4 + 5 O2 \rightarrow P4O10} \label{5}$
Thus, the oxides $\ce{P4O6}$ and $\ce{P4O10}$ share interesting features. Oxides of phosphorus, $\ce{P4O10}$, dissolve in water to give phosphoric acid,
$\ce{P4O10 + 6 H2O \rightarrow 4 H3PO4} \label{6}$
Phosphoric acid is a polyprotic acid, and it ionizes in three stages:
$\ce{H3PO4 \rightleftharpoons H+ + H2PO4-} \label{7a}$
$\ce{H2PO4- \rightleftharpoons H+ + HPO4^2-} \label{7b}$
$\ce{HPO4^2- \rightarrow H+ + PO4^3-} \label{7c}$
Phosphoric Acid
Phosphoric acid is a polyprotic acid, which makes it an ideal buffer. It gets harder and harder to separate the hydrogen from the phosphate, making the pKa values increase in basicity: 2.12, 7.21, and 12.67. The conjugate bases H2PO4-, HPO42-, and PO43- can be mixed to form buffer solutions.
Reaction Dissociation Constant
Table 1: Ionization constants for the successive deprotonation of phosphoric acid states
$H_3PO_4 + H_2O \rightarrow H_3O^+ + H_2PO^{4-}$ Ka1=7.5x10-3
$H_2PO_4^{-} + H_2O \rightarrow H_3O^+ + HPO_4^{2-}$ Ka2=6.2x10-8
$H_2PO_4^{-} + H_2O \rightarrow H_3O^+ + PO_4^{3-}$ Ka3=2.14x10-13
Overall: $H_3PO_4 + 3H_2O \rightarrow 3 H_3O^+ + PO_4^{3-}$
Past and Present Uses of Phosphorus
Commercially, phosphorus compounds are used in the manufacture of phosphoric acid ($H_3PO_4$) (found in soft drinks and used in fertilizer compounding). Other compounds find applications in fireworks and, of course, phosphorescent compounds which glow in the dark. Phosphorus compounds are currently used in foods, toothpaste, baking soda, matches, pesticides, nerve gases, and fertilizers. Phosphoric acid is not only used in buffer solutions; it is also a key ingredient of Coca Cola and other sodas! Phosphorus compounds were once used in detergents as a water softener until they raised concerns about pollution and eutrophication. Pure phosphorus was once prescribed as a medicine and an aphrodisiac until doctors realized it was poisonous (Emsley).
Questions
1. About 85% of the total industrial output of phosphoric acid is used
a. in the detergent industry
b. to produce buffer solutions
c. in the paint industry
d. to produce superphosphate fertilizers
e. in the manufacture of plastics
2. What is the product when phosphorus pentoxide $\ce{P4O10}$ reacts with water? Give the formula of the product.
3. What is the phosphorus-containing product when $\ce{PCl3}$ reacts with water? Give the formula.
Solutions
1. Answer... d
The middle number, (for example, 6-5-8) specifies the percentage of phosphorus compound in a fertilizer. Phosphorus is an important element for plant life.
2. Answer $\ce{H3PO4}$
$\ce{P4O10 + 6 H2O \rightarrow 4 H3PO4} \nonumber$
3. Answer $\ce{H3PO3}$
$\ce{PCl3 + 3 H2O \rightarrow H3PO3 + 3 HCl} \nonumber$
This is a weaker acid than $\ce{H3PO4}$.
Contributors
• Aimee Kindel (UCD), Kirenjot Grewal (UCD), Tiffany Lui (UCD)
• Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.09%3A_The_Chemistry_of_Phosphorus.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_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.10%3A_The_Group_6A_Elements.txt |
Oxygen is an element that is widely known by the general public because of the large role it plays in sustaining life. Without oxygen, animals would be unable to breathe and would consequently die. Oxygen not only is important to supporting life, but also plays an important role in many other chemical reactions. Oxygen is the most common element in the earth's crust and makes up about 20% of the air we breathe. Historically the discovery of oxygen as an element essential for combustion stands at the heart of the phlogiston controversy (see below).
The Origin and History
Oxygen is found in the group 16 elements and is considered a chalcogen. Named from the Greek oxys + genes, "acid-former", oxygen was discovered in 1772 by Scheele and independently by Priestly in 1774. Oxygen was given its name by the French scientist, Antoine Lavoisier.
Scheele discovered oxygen through an experiment which involved burning manganese oxide. Scheele came to find that the hot manganese oxide produced a gas which he called "fire air". He also came to find that when this gas was able to come into contact with charcoal, it produced beautiful bright sparks. All of the other elements produced the same gas. Although Scheele discovered oxygen, he did not publish his work until three years after another chemist, Joseph Priestly, discovered oxygen. Joseph Priestly, an English chemist, repeated Scheele's experiment in 1774 using a slightly different setup. Priestly used a 12 in burning glass and aimed the sunlight directly towards the compound that he was testing, mercuric oxide. As a result, he was able to "discover better air" that was shown to expand a mouse's lifetime to four times as long and caused a flame to burn with higher intensity. Despite these experiments, both chemists were not able to pinpoint exactly what this element was. It was not until 1775 that Antoine Lavoisier, a French chemist, was able to recognize this unknown gas as an element.
Our atmosphere currently contains about 21% of free oxygen. Oxygen is produced in various ways. The process of photochemical dissociation in which water molecules are broken up by ultraviolet rays produces about 1-2% of our oxygen. Another process that produces oxygen is photosynthesis which is performed by plants and photosynthetic bacteria. Photosynthesis occurs through the following general reaction:
$\ce{CO2 + H2O + h\nu \rightarrow} \text{organic compounds} \ce{+ O2} \nonumber$
The Dangers of Phlogiston
Phlogiston theory is the outdated belief that a fire-like element called phlogiston is contained within combustible bodies and released during combustion. The name comes from the Ancient Greek φλογιστόν phlogistón (burning up), from φλόξ phlóx (flame). It was first stated in 1667 by Johann Joachim Becher, and then put together more formally by Georg Ernst Stahl. The theory attempted to explain burning processes such as combustion and rusting, which are now collectively known as oxidation.
Properties
• Element number: 8
• Atomic weight 15.9994
• Color: gas form- colorless, liquid- pale blue
• Melting point: 54.36K
• Boiling point: 90.2 K
• Density: .001429
• 21% of earth's atmosphere
• Third most abundant element in the universe
• Most abundant element in Earth's crust at 45.4%
• 3 Stable isotopes
• Ionization energy: 13.618 eV
• Oxygen is easily reduced and is a great oxidizing agent making it readily reactive with other elements
Magnetic Properties of Oxygen
Oxygen (O2) is paramagnetic. An oxygen molecule has six valence electrons, so the O2 molecule has 12 valence electrons with the electron configuration shown below:
As shown, there are two unpaired electrons, which causes O2 to be paramagnetic. There are also eight valence electrons in the bonding orbitals and four in antibonding orbitals, which makes the bond order 2. This accounts for the double covalent bond that is present in O2.
Video $1$: A chemical demonstration of the paramagnetism of molecular oxygen, as shown by the attraction of liquid oxygen to magnets.
As shown in Video $1$, since molecular oxygen ($O_2$) has unpaired electrons, it is paramagnetic and is attracted to the magnet. In contrast, molecular nitrogen ($N_2$) has no unpaired electrons and is not attracted to the magnet.
General Chemistry of Oxygen
Oxygen normally has an oxidation state of -2, but is capable of having oxidation states of -2, -1, -1/2, 0, +1, and +2. The oxidation states of oxides, peroxides and superoxides are as follows:
• Oxides: O-2 ,
• peroxides: O2-2 ,
• superoxide: O2-1.
Oxygen does not react with itself, nitrogen, or water under normal conditions. Oxygen does, however, dissolve in water at 20 degrees Celsius and 1 atmosphere. Oxygen also does not normally react with bases or acids. Group 1 metals (alkaline metals) are very reactive with oxygen and must be stored away from oxygen in order to prevent them from becoming oxidized. The metals at the bottom of the group are more reactive than those at the top. The reactions of a few of these metals are explored in more detail below.
Lithium: Reacts with oxygen to form white lithium oxide in the reaction below.
$\ce{4Li + O_2 \rightarrow 2Li_2O} \label{1}$
Sodium: Reacts with oxygen to form a white mixture of sodium oxide and sodium peroxide. The reactions are shown below.
• Sodium oxide: $\ce{4Na + O_2 \rightarrow 2Na_2O} \label{2}$
• Sodium peroxide: $\ce{2Na + O_2 \rightarrow Na_2O_2} \label{3}$
Potassium: Reacts with oxygen to form a mixture of potassium peroxide and potassium superoxide. The reactions are shown below.
• Potassium peroxide: $\ce{2K + O_2 \rightarrow 2K_2O_2} \label{4}$
• Potassium superoxide: $\ce{K + O_2 \rightarrow KO_2} \label{5}$
Rubidium and Cesium: Both metals react to produce superoxides through the same process as that of the potassium superoxide reaction.
The oxides of these metals form metal hydroxides when they react with water. These metal hydroxides make the solution basic or alkaline, hence the name alkaline metals.
Group 2 metals (alkaline earth metals) react with oxygen through the process of burning to form metal oxides but there are a few exceptions.
Beryllium is very difficult to burn because it has a layer of beryllium oxide on its surface which prevents further interaction with oxygen. Strontium and barium react with oxygen to form peroxides. The reaction of barium and oxygen is shown below, and the reaction with strontium would be the same.
$\ce{Ba(s) + O2 (g) \rightarrow BaO2 (s) }\label{6}$
Group 13 reacts with oxygen in order to form oxides and hydroxides that are of the form $X_2O_3$ and $X(OH)_3$. The variable X represents the various group 13 elements. As you go down the group, the oxides and hydroxides get increasingly basic.
Group 14 elements react with oxygen to form oxides. The oxides formed at the top of the group are more acidic than those at the bottom of the group. Oxygen reacts with silicon and carbon to form silicon dioxide and carbon dioxide. Carbon is also able to react with oxygen to form carbon monoxide, which is slightly acidic. Germanium, tin, and lead react with oxygen to form monoxides and dioxides that are amphoteric, which means that they react with both acids and bases.
Group 15 elements react with oxygen to form oxides. The most important are listed below.
• Nitrogen: N2O, NO, N2O3, N2O4, N2O5
• Phosphorus: P4O6, P4O8, P2O5
• Arsenic: As2O3, As2O5
• Antimony: Sb2O3, Sb2O5
• Bismuth: Bi2O3, Bi2O5
Group 16 elements react with oxygen to form various oxides. Some of the oxides are listed below.
• Sulfur: SO, SO2, SO3, S2O7
• Selenium: SeO2, SeO3
• Tellurium: TeO, TeO2, TeO3
• Polonium: PoO, PoO2, PoO3
Group 17 elements (halogens) fluorine, chlorine, bromine, and iodine react with oxygen to form oxides. Fluorine forms two oxides with oxygen: F2O and F2O2. Both fluorine oxides are called oxygen fluorides because fluorine is the more electronegative element. One of the fluorine reactions is shown below.
$\ce{O2 (g) + F2 (g) \rightarrow F2O2 (g)} \label{7}$
Group 18: Some would assume that the Noble Gases would not react with oxygen. However, xenon does react with oxygen to form $\ce{XeO_3}$ and $\ce{XeO_4}$. The ionization energy of xenon is low enough for the electronegative oxygen atom to "steal away" electrons. Unfortunately, $\ce{XeO_3}$ is HIGHLY unstable, and it has been known to spontaneously detonate in a clean, dry environment.
Transition metals react with oxygen to form metal oxides. However, gold, silver, and platinum do not react with oxygen. A few reactions involving transition metals are shown below:
$2Sn_{(s)} + O_{2(g)} \rightarrow 2SnO_{(s)} \label{8}$
$4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \label{9A}$
$4Al_{(s)} + 3O_{2(g)} \rightarrow 2Al_2O_{3(s)} \label{9B}$
Reaction of Oxides
We will be discussing metal oxides of the form $X_2O$. The variable $X$ represents any metal that is able to bond to oxygen to form an oxide.
• Reaction with water: The oxides react with water to form a metal hydroxide.
$X_2O + H_2O \rightarrow 2XOH \nonumber$
• Reaction with dilute acids: The oxides react with dilute acids to form a salt and water.
$X_2O + 2HCl \rightarrow 2XCl + H_2O \nonumber$
Reactions of Peroxides
The peroxides we will be discussing are of the form $X_2O_2$. The variable $X$ represents any metal that can form a peroxide with oxygen.
Reaction with water: If the temperature of the reaction is kept constant despite the fact that the reaction is exothermic, then the reaction proceeds as follows:
$X_2O_2+ 2H_2O \rightarrow 2XOH + H_2O_2 \nonumber$
If the reaction is not carried out at a constant temperature, then the reaction of the peroxide and water will result in decomposition of the hydrogen peroxide that is produced into water and oxygen.
Reaction with dilute acid: This reaction is more exothermic than that with water. The heat produced causes the hydrogen peroxide to decompose to water and oxygen. The reaction is shown below.
$X_2O_2 + 2HCl \rightarrow 2XCl + H_2O_2 \nonumber$
$2H_2O_2 \rightarrow 2H_2O + O_2 \nonumber$
Reaction of Superoxides
The superoxides we will be talking about are of the form $XO_2$, with $X$ representing any metal that forms a superoxide when reacting with oxygen.
Reaction with water: The superoxide and water react in a very exothermic reaction that is shown below. The heat that is produced in forming the hydrogen peroxide will cause the hydrogen peroxide to decompose to water and oxygen.
$2XO_2 + 2H_2O \rightarrow 2XOH + H_2O_2 + O_2 \nonumber$
Reaction with dilute acids: The superoxide and dilute acid react in a very exothermic reaction that is shown below. The heat produced will cause the hydrogen peroxide to decompose to water and oxygen.
$2XO_2 + 2HCl \rightarrow 2XCl + H_2O_2 + O_2 \nonumber$
Allotropes of Oxygen
There are two allotropes of oxygen; dioxygen (O2) and trioxygen (O3) which is called ozone. The reaction of converting dioxygen into ozone is very endothermic, causing it to occur rarely and only in the lower atmosphere. The reaction is shown below:
$3O_{2 (g)} \rightarrow 2O_{3 (g)} \;\;\; ΔH^o= +285 \;kJ \nonumber$
Ozone is unstable and quickly decomposes back to oxygen but is a great oxidizing agent.
Miscellaneous Reactions
Reaction with Alkanes: The most common reactions that involve alkanes occur with oxygen. Alkanes are able to burn and it is the process of oxidizing the hydrocarbons that makes them important as fuels. An example of an alkane reaction is the reaction of octane with oxygen as shown below.
C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(l) ∆Ho = -5.48 X 103 kJ
Reaction with ammonia: Oxygen is able to react with ammonia to produce dinitrogen (N2) and water (H2O) through the reaction shown below.
$4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O \nonumber$
Reaction with Nitrogen Oxide: Oxygen is able to react with nitrogen oxide in order to produce nitrogen dioxide through the reaction shown below.
$NO + O_2 \rightarrow NO_2 \nonumber$
Problems
1. Is oxygen reactive with noble gases?
2. Which transition metals does oxygen not react with?
3. What is produced when an oxide reacts with water?
4. Is oxygen reactive with alkali metals? Why are the alkali metals named that way?
5. If oxygen is reactive with alkali metals, are oxides, peroxides or superoxides produced?
Solutions
1. No, noble gases are unreactive with oxygen.
2. Oxygen is mostly unreactive with gold and platinum.
3. When an oxide reacts with water, a metal hydroxide is produced.
4. Oxygen is very reactive with alkali metals. Alkali metals are given the name alkali because the oxides of these metals react with water to form a metal hydroxide that is basic or alkaline.
5. Lithium produces an oxide, sodium produces a peroxide, and potassium, cesium, and rubidium produce superoxides.
Contributors and Attributions
• Phillip Ball (UCD), Katharine Williams (UCD)
18.12: The Chemistry of Sulfur
Learning Objectives
• Describe the properties, preparation, and uses of sulfur
Sulfur exists in nature as elemental deposits as well as sulfides of iron, zinc, lead, and copper, and sulfates of sodium, calcium, barium, and magnesium. Hydrogen sulfide is often a component of natural gas and occurs in many volcanic gases, like those shown in Figure $1$. Sulfur is a constituent of many proteins and is essential for life.
The Frasch process, illustrated in Figure $2$, is important in the mining of free sulfur from enormous underground deposits in Texas and Louisiana. Superheated water (170 °C and 10 atm pressure) is forced down the outermost of three concentric pipes to the underground deposit. The hot water melts the sulfur. The innermost pipe conducts compressed air into the liquid sulfur. The air forces the liquid sulfur, mixed with air, to flow up through the outlet pipe. Transferring the mixture to large settling vats allows the solid sulfur to separate upon cooling. This sulfur is 99.5% to 99.9% pure and requires no purification for most uses.
Larger amounts of sulfur also come from hydrogen sulfide recovered during the purification of natural gas.
Sulfur exists in several allotropic forms. The stable form at room temperature contains eight-membered rings, and so the true formula is S8. However, chemists commonly use S to simplify the coefficients in chemical equations; we will follow this practice in this book.
Like oxygen, which is also a member of group 16, sulfur exhibits a distinctly nonmetallic behavior. It oxidizes metals, giving a variety of binary sulfides in which sulfur exhibits a negative oxidation state (2−). Elemental sulfur oxidizes less electronegative nonmetals, and more electronegative nonmetals, such as oxygen and the halogens, will oxidize it. Other strong oxidizing agents also oxidize sulfur. For example, concentrated nitric acid oxidizes sulfur to the sulfate ion, with the concurrent formation of nitrogen(IV) oxide:
$\ce{S}(s)+\ce{6HNO3}(aq)⟶\ce{2H3O+}(aq)+\ce{SO4^2-}(aq)+\ce{6NO2}(g) \nonumber$
The chemistry of sulfur with an oxidation state of 2− is similar to that of oxygen. Unlike oxygen, however, sulfur forms many compounds in which it exhibits positive oxidation states.
Summary
Sulfur (group 16) reacts with almost all metals and readily forms the sulfide ion, S2−, in which it has as oxidation state of 2−. Sulfur reacts with most nonmetals.
Glossary
Frasch process
important in the mining of free sulfur from enormous underground deposits | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.11%3A_The_Chemistry_of_Oxygen.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_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.13%3A_The_Group_7A_Elements.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 | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/18%3A_The_Representative_Elements/18.14%3A_The_Group_8A_Elements.txt |
We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing.
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. As shown in Figure $2$, the d-block elements in groups 3–11 are transition elements. The f-block elements, also called inner transition metals (the lanthanides and actinides), also meet this criterion because the d orbital is partially occupied before the f orbitals. The d orbitals fill with the copper family (group 11); for this reason, the next family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals.
The d-block elements are divided into the first transition series (the elements Sc through Cu), the second transition series (the elements Y through Ag), and the third transition series (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the fourth transition series, which also includes Rf through Rg.
The f-block elements are the elements Ce through Lu, which constitute the lanthanide series (or lanthanoid series), and the elements Th through Lr, which constitute the actinide series (or actinoid series). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series
The elements of the second and third rows of the Periodic Table show gradual changes in properties across the table from left to right as expected. Electrons in the outer shells of the atoms of these elements have little shielding effects resulting in an increase in effective nuclear charge due to the addition of protons in the nucleus. Consequently, the effects on atomic properties are: smaller atomic radius, increased first ionization energy, enhanced electronegativity and more nonmetallic character. This trend continues until one reaches calcium (Z=20). There is an abrupt break at this point. The next ten elements called the first transition series are remarkably similar in their physical and chemical properties. This general similarity in properties has been explained in terms of their relatively small difference in effective nuclear charge over the series. This occurs because each additional electron enters the penultimate 3d shell providing an effective shield between the nucleus and the outer 4s shell.
What is a Transition Metal?
Thus, the transition elements can be defined as those in which the d electron shells are being filled and so we generally ignore Sc and Zn where Sc(III) is d0 and Zn(II) is d10.
It is useful, at the beginning, to identify the physical and chemical properties of transition elements which differ from main group elements (s-block). Properties of transition elements include:
• have large charge/radius ratio;
• are hard and have high densities;
• have high melting and boiling points;
• form compounds which are often paramagnetic;
• show variable oxidation states;
• form coloured ions and compounds;
• form compounds with profound catalytic activity;
• form stable complexes.
Table $1$ : Summary of select physical properties of transition elements:
Element Group density
/g cm-3
m. p.
/ °C
b.p.
/ °C
radius
/ pm
free atom
configuration
ionization energy
/ kJ mol-1
Uses
Sc 3 2.99 1541 2831 164 [Ar] 3d14s2 631
Ti 4 4.50 1660 3287 147 [Ar]3d24s2 658 -engines/aircraft industry-density is 60% of iron
V 5 5.96 1890 3380 135 [Ar]3d34s2 650 -stainless steel, 19% Cr, 9% Ni the rest Fe
Cr 6 7.20 1857 2670 129 [Ar]3d54s1 653 -alloys eg with C steel, the most significant use
Mn 7 7.20 1244 1962 137 [Ar]3d54s2 717 -alloys eg with Cu
Fe 8 7.86 1535 2750 126 [Ar]3d64s2 759 -alloys eg with C steel, the most significant use
Co 9 8.90 1495 2870 125 [Ar]3d74s2 758 -alloys eg with Cr and W for hardened drill bits
Ni 10 8.90 1455 2730 125 [Ar]3d84s2 737 -alloys Fe/Ni armor plating, resists corrosion
Cu 11 8.92 1083 2567 128 [Ar]3d104s1 746 -high electrical conductivity (2nd to Ag), wiring
Zn 12 7.14 420 907 137 [Ar]3d104s2 906
Densities and Metallic Radii
The transition elements are much denser than the s-block elements and show a gradual increase in density from scandium to copper. This trend in density can be explained by the small and irregular decrease in metallic radii coupled with the relative increase in atomic mass.
Melting and Boiling points
The melting points and the molar enthalpies of fusion of the transition metals are both high in comparison to main group elements. This arises from strong metallic bonding in transition metals which occurs due to delocalization of electrons facilitated by the availability of both d and s electrons.
Ionization Energies
In moving across the series of metals from scandium to zinc a small change in the values of the first and second ionization energies is observed. This is due to the build-up of electrons in the immediately underlying d-sub-shells that efficiently shields the 4s electrons from the nucleus and minimizing the increase in effective nuclear charge $Z_{eff}$ from element to element. The increases in third and fourth ionization energy values are more rapid. However, the trends in these values show the usual discontinuity half way along the series. The reason is that the five d electrons are all unpaired, in singly occupied orbitals. When the sixth and subsequent electrons enter, the electrons have to share the already occupied orbitals resulting in inter-electron repulsions, which would require less energy to remove an electron. Hence, the third ionization energy curve for the last five elements is identical in shape to the curve for the first five elements, but displaced upwards by 580 kJ mol-1.
Oxidation States
Transition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure $5$. As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+).
For the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the s and d orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two 3d and two 4s electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from 2+ to 6+, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air.
The elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable Cr3+ ions in water and, in the absence of air, less stable Cr2+ ions. The sulfide with the highest oxidation state for chromium is Cr2S3, which contains the Cr3+ ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of 4+ and 6+.
Electronic Configurations
The electronic configuration of the atoms of the first row transition elements are basically the same. It can be seen in the Table above that there is a gradual filling of the 3d orbitals across the series starting from scandium. This filling is, however, not regular, since at chromium and copper the population of 3d orbitals increase by the acquisition of an electron from the 4s shell. This illustrates an important generalization about orbital energies of the first row transition series. At chromium, both the 3d and 4s orbitals are occupied, but neither is completely filled in preference to the other. This suggests that the energies of the 3d and 4s orbitals are relatively close for atoms in this row.
In the case of copper, the 3d level is full, but only one electron occupies the 4s orbital. This suggests that in copper the 3d orbital energy is lower than the 4s orbital. Thus the 3d orbital energy has passed from higher to lower as we move across the period from potassium to zinc. However, the whole question of preference of an atom to adopt a particular electronic configuration is not determined by orbital energy alone. In chromium it can be shown that the 4s orbital energy is still below the 3d which suggests a configuration [Ar] 3d44s2. However due to the effect of electronic repulsion between the outer electrons the actual configuration becomes [Ar]3d54s1 where all the electrons in the outer orbitals are unpaired. It should be remembered that the factors that determine electronic configuration in this period are indeed delicately balanced.
Redox Couple E°/V
Mn2+(aq.)/Mn(s) -1.18
H+(aq.)/H2(g) 0.00
This shows that elemental Mn is a stronger reductant than molecular hydrogen and hence should be able to displace hydrogen gas from 1 mol dm-3 hydrochloric acid.
Valence Electrons in Transition Metals
Review how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the s electrons before the d or f electrons. Then, for each ion, give the electron configuration:
1. cerium(III)
2. lead(II)
3. Ti2+
4. Am3+
5. Pd2+
For the examples that are transition metals, determine to which series they belong.
Solution
For ions, the s-valence electrons are lost prior to the d or f electrons.
1. Ce3+[Xe]4f1; Ce3+ is an inner transition element in the lanthanide series.
2. Pb2+[Xe]6s25d104f14; the electrons are lost from the p orbital. This is a main group element.
3. titanium(II) [Ar]3d2; first transition series
4. americium(III) [Rn]5f6; actinide
5. palladium(II) [Kr]4d8; second transition series
Exercise $1$
Check Your Learning Give an example of an ion from the first transition series with no d electrons.
Answer
V5+ is one possibility. Other examples include Sc3+, Ti4+, Cr6+, and Mn7+.
Chemical Reactivity
Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (Table P1), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be2+ and Mg2+. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry.
Ions of the lighter d-block elements, such as Cr3+, Fe3+, and Co2+, form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo3+, Ru3+, and Ir2+) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier d-block elements are oxyanions such as $\ce{MoO4^2-}$ and $\ce{ReO4-}$.
Ruthenium, osmium, rhodium, iridium, palladium, and platinum are the "platinum metals". With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions.
Both the d- and f-block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from 1+ to 6+. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The f-block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts.
Activity of the Transition Metals
Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)?
Solution
First, we need to look up the reduction half reactions (Table P1) for each oxide in the specified oxidation state:
$\ce{Cr2O7^2- + 14H+ + 6e- ⟶ 2Cr^3+ + 7H2O} \hspace{20px} \mathrm{+1.33\: V}$
$\ce{MnO4- + 8H+ + 5e- ⟶ Mn^2+ + H2O} \hspace{20px} \mathrm{+1.51\: V}$
$\ce{TiO2 + 4H+ + 2e- ⟶ Ti^2+ + 2H2O} \hspace{20px} \mathrm{−0.50\: V}$
A larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set.
Exercise $2$
Predict what reaction (if any) will occur between HCl and Co(s), and between HBr and Pt(s). You will need to use the standard reduction potentials from (Table P1).
Answer
$\ce{Co}(s)+\ce{2HCl}⟶\ce{H2}+\ce{CoCl2}(aq)$; no reaction because Pt(s) will not be oxidized by H+ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.1%3A_The_Transition_Metals%3A_A_Survey.txt |
Learning Objectives
• To use periodic trends to understand the chemistry of the transition metals.
As we shall see, the two heaviest members of each group usually exhibit substantial similarities in chemical behavior and are quite different from the lightest member.
Group 3 (Sc, Y, La, and Ac)
As shown in Table $1$, the observed trends in the properties of the group 3 elements are similar to those of groups 1 and 2. Due to their ns2(n − 1)d1 valence electron configurations, the chemistry of all four elements is dominated by the +3 oxidation state formed by losing all three valence electrons. As expected based on periodic trends, these elements are highly electropositive metals and powerful reductants, with La (and Ac) being the most reactive. In keeping with their highly electropositive character, the group 3 metals react with water to produce the metal hydroxide and hydrogen gas:
$2M_{(s)} + 6H_2O_{(l)} \rightarrow 2M(OH)_{3(s)} + 3H_{2(g)}\label{Eq1}$
The chemistry of the group 3 metals is almost exclusively that of the M3+ ion; the elements are powerful reductants.
Moreover, all dissolve readily in aqueous acid to produce hydrogen gas and a solution of the hydrated metal ion: M3+(aq).
Table $1$: Some Properties of the Elements of Groups 3, 4, and 5
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
3 Sc 21 4s23d1 1.36 162 1541 2.99
Y 39 5s24d1 1.22 180 1522 4.47
La 57 6s25d1 1.10 183 918 6.15
Ac 89 7s26d1 1.10 188 1051 10.07
4 Ti 22 4s23d2 1.54 147 1668 4.51
Zr 40 5s24d2 1.33 160 1855 6.52
Hf 72 6s25d24f14 1.30 159 2233 13.31
5 V 23 4s23d3 1.63 134 1910 6.00
Nb 41 5s24d3 1.60 146 2477 8.57
Ta 73 6s25d34f14 1.50 146 3017 16.65
The group 3 metals react with nonmetals to form compounds that are primarily ionic in character. For example, reacting group 3 metals with the halogens produces the corresponding trihalides: MX3. The trifluorides are insoluble in water because of their high lattice energies, but the other trihalides are very soluble in water and behave like typical ionic metal halides. All group 3 elements react with air to form an oxide coating, and all burn in oxygen to form the so-called sesquioxides (M2O3), which react with H2O or CO2 to form the corresponding hydroxides or carbonates, respectively. Commercial uses of the group 3 metals are limited, but “mischmetal,” a mixture of lanthanides containing about 40% La, is used as an additive to improve the properties of steel and make flints for cigarette lighters.
Group 4 (Ti, Zr, and Hf)
Because the elements of group 4 have a high affinity for oxygen, all three metals occur naturally as oxide ores that contain the metal in the +4 oxidation state resulting from losing all four ns2(n − 1)d2 valence electrons. They are isolated by initial conversion to the tetrachlorides, as shown for Ti:
$2FeTiO_{3(s)} + 6C_{(s)} + 7Cl_{2(g)} \rightarrow 2TiCl_{4(g)} + 2FeCl_{3(g)} + 6CO_{(g)}\label{Eq2}$
followed by reduction of the tetrachlorides with an active metal such as Mg.
The chemistry of the group 4 metals is dominated by the +4 oxidation state. Only Ti has an extensive chemistry in lower oxidation states.
In contrast to the elements of group 3, the group 4 elements have important applications. Titanium (melting point = 1668°C) is often used as a replacement for aluminum (melting point = 660°C) in applications that require high temperatures or corrosion resistance. For example, friction with the air heats the skin of supersonic aircraft operating above Mach 2.2 to temperatures near the melting point of aluminum; consequently, titanium is used instead of aluminum in many aerospace applications. The corrosion resistance of titanium is increasingly exploited in architectural applications, as shown in the chapter-opening photo. Metallic zirconium is used in UO2-containing fuel rods in nuclear reactors, while hafnium is used in the control rods that modulate the output of high-power nuclear reactors, such as those in nuclear submarines.
Consistent with the periodic trends shown in Figure 23.2, the group 4 metals become denser, higher melting, and more electropositive down the column (Table $1$). Unexpectedly, however, the atomic radius of Hf is slightly smaller than that of Zr due to the lanthanide contraction. Because of their ns2(n − 1)d2 valence electron configurations, the +4 oxidation state is by far the most important for all three metals. Only titanium exhibits a significant chemistry in the +2 and +3 oxidation states, although compounds of Ti2+ are usually powerful reductants. In fact, the Ti2+(aq) ion is such a strong reductant that it rapidly reduces water to form hydrogen gas.
Reaction of the group 4 metals with excess halogen forms the corresponding tetrahalides (MX4), although titanium, the lightest element in the group, also forms dihalides and trihalides (X is not F). The covalent character of the titanium halides increases as the oxidation state of the metal increases because of increasing polarization of the anions by the cation as its charge-to-radius ratio increases. Thus TiCl2 is an ionic salt, whereas TiCl4 is a volatile liquid that contains tetrahedral molecules. All three metals react with excess oxygen or the heavier chalcogens (Y) to form the corresponding dioxides (MO2) and dichalcogenides (MY2). Industrially, TiO2, which is used as a white pigment in paints, is prepared by reacting TiCl4 with oxygen at high temperatures:
$TiCl_{4(g)} + O_{2(g)} \rightarrow TiO_{2(s)} + 2Cl_{2(g)}\label{Eq3}$
The group 4 dichalcogenides have unusual layered structures with no M–Y bonds holding adjacent sheets together, which makes them similar in some ways to graphite (Figure $1$). The group 4 metals also react with hydrogen, nitrogen, carbon, and boron to form hydrides (such as TiH2), nitrides (such as TiN), carbides (such as TiC), and borides (such as TiB2), all of which are hard, high-melting solids. Many of these binary compounds are nonstoichiometric and exhibit metallic conductivity.
Group 5 (V, Nb, and Ta)
Like the group 4 elements, all group 5 metals are normally found in nature as oxide ores that contain the metals in their highest oxidation state (+5). Because of the lanthanide contraction, the chemistry of Nb and Ta is so similar that these elements are usually found in the same ores.
Three-fourths of the vanadium produced annually is used in the production of steel alloys for springs and high-speed cutting tools. Adding a small amount of vanadium to steel results in the formation of small grains of V4C3, which greatly increase the strength and resilience of the metal, especially at high temperatures. The other major use of vanadium is as V2O5, an important catalyst for the industrial conversion of SO2 to SO3 in the contact process for the production of sulfuric acid. In contrast, Nb and Ta have only limited applications, and they are therefore produced in relatively small amounts. Although niobium is used as an additive in certain stainless steels, its primary application is in superconducting wires such as Nb3Zr and Nb3Ge, which are used in superconducting magnets for the magnetic resonance imaging of soft tissues. Because tantalum is highly resistant to corrosion, it is used as a liner for chemical reactors, in missile parts, and as a biologically compatible material in screws and pins for repairing fractured bones.
The chemistry of the two heaviest group 5 metals (Nb and Ta) is dominated by the +5 oxidation state. The chemistry of the lightest element (V) is dominated by lower oxidation states, especially +4.
As indicated in Table $1$, the trends in properties of the group 5 metals are similar to those of group 4. Only vanadium, the lightest element, has any tendency to form compounds in oxidation states lower than +5. For example, vanadium is the only element in the group that forms stable halides in the lowest oxidation state (+2). All three metals react with excess oxygen, however, to produce the corresponding oxides in the +5 oxidation state (M2O5), in which polarization of the oxide ions by the high-oxidation-state metal is so extensive that the compounds are primarily covalent in character. Vanadium–oxygen species provide a classic example of the effect of increasing metal oxidation state on the protonation state of a coordinated water molecule: vanadium(II) in water exists as the violet hydrated ion [V(H2O)6]2+; the blue-green [V(H2O)6]3+ ion is acidic, dissociating to form small amounts of the [V(H2O)5(OH)]2+ ion and a proton; and in water, vanadium(IV) forms the blue vanadyl ion [(H2O)4VO]2+, which contains a formal V=O bond (Figure $2$). Consistent with its covalent character, V2O5 is acidic, dissolving in base to give the vanadate ion ([VO4]3−), whereas both Nb2O5 and Ta2O5 are comparatively inert. Oxides of these metals in lower oxidation states tend to be nonstoichiometric.
Because vanadium ions with different oxidation states have different numbers of d electrons, aqueous solutions of the ions have different colors: in acid V(V) forms the pale yellow [VO2]+ ion; V(IV) is the blue vanadyl ion [VO]2+; and V(III) and V(II) exist as the hydrated V3+ (blue-green) and V2+ (violet) ions, respectively.
Although group 5 metals react with the heavier chalcogens to form a complex set of binary chalcogenides, the most important are the dichalcogenides (MY2), whose layered structures are similar to those of the group 4 dichalcogenides. The elements of group 5 also form binary nitrides, carbides, borides, and hydrides, whose stoichiometries and properties are similar to those of the corresponding group 4 compounds. One such compound, tantalum carbide (TiC), has the highest melting point of any compound known (3738°C); it is used for the cutting edges of high-speed machine tools.
Group 6 (Cr, Mo, and W)
As an illustration of the trend toward increasing polarizability as we go from left to right across the d block, in group 6 we first encounter a metal (Mo) that occurs naturally as a sulfide ore rather than as an oxide. Molybdenite (MoS2) is a soft black mineral that can be used for writing, like PbS and graphite. Because of this similarity, people long assumed that these substances were all the same. In fact, the name molybdenum is derived from the Greek molybdos, meaning “lead.” More than 90% of the molybdenum produced annually is used to make steels for cutting tools, which retain their sharp edge even when red hot. In addition, molybdenum is the only second- or third-row transition element that is essential for humans. The major chromium ore is chromite (FeCr2O4), which is oxidized to the soluble [CrO4]2− ion under basic conditions and reduced successively to Cr2O3 and Cr with carbon and aluminum, respectively. Pure chromium can be obtained by dissolving Cr2O3 in sulfuric acid followed by electrolytic reduction; a similar process is used for electroplating metal objects to give them a bright, shiny, protective surface layer. Pure tungsten is obtained by first converting tungsten ores to WO3, which is then reduced with hydrogen to give the metal.
The metals become increasing polarizable across the d block.
Consistent with periodic trends, the group 6 metals are slightly less electropositive than those of the three preceding groups, and the two heaviest metals are essentially the same size because of the lanthanide contraction (Table $2$). All three elements have a total of six valence electrons, resulting in a maximum oxidation state of +6. Due to extensive polarization of the anions, compounds in the +6 oxidation state are highly covalent. As in groups 4 and 5, the lightest element exhibits variable oxidation states, ranging from Cr2+, which is a powerful reductant, to CrO3, a red solid that is a powerful oxidant. For Mo and W, the highest oxidation state (+6) is by far the most important, although compounds in the +4 and +5 oxidation states are known.
Table $2$: Some Properties of the Elements of Groups 6 and 7
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
6 Cr 24 4s13d5 1.66 128 1907 7.15
Mo 42 5s14d5 2.16 139 2623 10.20
W 74 6s25d44f14 1.70 139 3422 19.30
7 Mn 25 4s23d5 1.55 127 1246 7.30
Tc 43 5s24d5 2.10 136 2157 11.50
Re 75 6s25d54f14 1.90 137 3186 20.80
The chemistry of the two heaviest group 6 metals (Mo and W) is dominated by the +6 oxidation state. The chemistry of the lightest element (Cr) is dominated by lower oxidation states.
As observed in previous groups, the group 6 halides become more covalent as the oxidation state of the metal increases: their volatility increases, and their melting points decrease. Recall that as the electronegativity of the halogens decreases from F to I, they are less able to stabilize high oxidation states; consequently, the maximum oxidation state of the corresponding metal halides decreases. Thus all three metals form hexafluorides, but CrF6 is unstable at temperatures above −100°C, whereas MoF6 and WF6 are stable. Consistent with the trend toward increased stability of the highest oxidation state for the second- and third-row elements, the other halogens can oxidize chromium to only the trihalides, CrX3 (X is Cl, Br, or I), while molybdenum forms MoCl5, MoBr4, and MoI3, and tungsten gives WCl6, WBr5, and WI4.
Both Mo and W react with oxygen to form the covalent trioxides (MoO3 and WO3), but Cr reacts to form only the so-called sesquioxide (Cr2O3). Chromium will form CrO3, which is a highly toxic compound that can react explosively with organic materials. All the trioxides are acidic, dissolving in base to form the corresponding oxoanions ([MO4]2−). Consistent with periodic trends, the sesquioxide of the lightest element in the group (Cr2O3) is amphoteric. The aqueous chemistry of molybdate and tungstate is complex, and at low pH they form a series of polymeric anions called isopolymetallates, such as the [Mo8O26]4− ion, whose structure is as follows:
An isopolymolybdate cluster. The [Mo8O26]4− ion, shown here in both side and top views, is typical of the oxygen-bridged clusters formed by Mo(VI) and W(VI) in aqueous solution.
Reacting molybdenum or tungsten with heavier chalcogens gives binary chalcogenide phases, most of which are nonstoichiometric and electrically conducting. One of the most stable is MoS2; it has a layered structure similar to that of TiS2 (Figure $1$), in which the layers are held together by only weak van der Waals forces, which allows them to slide past one another rather easily. Consequently, both MoS2 and WS2 are used as lubricants in a variety of applications, including automobile engines. Because tungsten itself has an extraordinarily high melting point (3380°C), lubricants described as containing “liquid tungsten” actually contain a suspension of very small WS2 particles.
As in groups 4 and 5, the elements of group 6 form binary nitrides, carbides, and borides whose stoichiometries and properties are similar to those of the preceding groups. Tungsten carbide (WC), one of the hardest compounds known, is used to make the tips of drill bits.
Group 7 (Mn, Tc, and Re)
Continuing across the periodic table, we encounter the group 7 elements (Table $2$). One group 7 metal (Mn) is usually combined with iron in an alloy called ferromanganese, which has been used since 1856 to improve the mechanical properties of steel by scavenging sulfur and oxygen impurities to form MnS and MnO. Technetium is named after the Greek technikos, meaning “artificial,” because all its isotopes are radioactive. One isotope, 99mTc (m for metastable), has become an important biomedical tool for imaging internal organs. Because of its scarcity, Re is one of the most expensive elements, and its applications are limited. It is, however, used in a bimetallic Pt/Re catalyst for refining high-octane gasoline.
All three group 7 elements have seven valence electrons and can form compounds in the +7 oxidation state. Once again, the lightest element exhibits multiple oxidation states. Compounds of Mn in oxidation states ranging from −3 to +7 are known, with the most common being +2 and +4 (Figure $3$). In contrast, compounds of Tc and Re in the +2 oxidation state are quite rare. Because the electronegativity of Mn is anomalously low, elemental manganese is unusually reactive. In contrast, the chemistry of Tc is similar to that of Re because of their similar size and electronegativity, again a result of the lanthanide contraction. Due to the stability of the half-filled 3d5 electron configuration, the aqueous Mn3+ ion, with a 3d4 valence electron configuration, is a potent oxidant that is able to oxidize water. It is difficult to generalize about other oxidation states for Tc and Re because their stability depends dramatically on the nature of the compound.
Like vanadium, compounds of manganese in different oxidation states have different numbers of d electrons, which leads to compounds with different colors: the Mn2+(aq) ion is pale pink; Mn(OH)3, which contains Mn(III), is a dark brown solid; MnO2, which contains Mn(IV), is a black solid; and aqueous solutions of Mn(VI) and Mn(VII) contain the green manganate ion [MnO4]2− and the purple permanganate ion [MnO4], respectively.
Consistent with higher oxidation states being more stable for the heavier transition metals, reacting Mn with F2 gives only MnF3, a high-melting, red-purple solid, whereas Re reacts with F2 to give ReF7, a volatile, low-melting, yellow solid. Again, reaction with the less oxidizing, heavier halogens produces halides in lower oxidation states. Thus reaction with Cl2, a weaker oxidant than F2, gives MnCl2 and ReCl6. Reaction of Mn with oxygen forms only Mn3O4, a mixed-valent compound that contains two Mn(II) and one Mn(III) per formula unit and is similar in both stoichiometry and structure to magnetite (Fe3O4). In contrast, Tc and Re form high-valent oxides, the so-called heptoxides (M2O7), consistent with the increased stability of higher oxidation states for the second and third rows of transition metals. Under forced conditions, manganese will form Mn2O7, an unstable, explosive, green liquid. Also consistent with this trend, the permanganate ion [MnO4]2− is a potent oxidant, whereas [TcO4] and [ReO4] are much more stable. Both Tc and Re form disulfides and diselenides with layered structures analogous to that of MoS2, as well as more complex heptasulfides (M2S7). As is typical of the transition metals, the group 7 metals form binary nitrides, carbides, and borides that are generally stable at high temperatures and exhibit metallic properties.
The chemistry of the group 7 metals (Mn, Tc, and Re) is dominated by lower oxidation states. Compounds in the maximum possible oxidation state (+7) are readily reduced.
Groups 8, 9, and 10
In many older versions of the periodic table, groups 8, 9, and 10 were combined in a single group (group VIII) because the elements of these three groups exhibit many horizontal similarities in their chemistry, in addition to the similarities within each column. In part, these horizontal similarities are due to the fact that the ionization potentials of the elements, which increase slowly but steadily across the d block, have now become so large that the oxidation state corresponding to the formal loss of all valence electrons is encountered only rarely (group 8) or not at all (groups 9 and 10). As a result, the chemistry of all three groups is dominated by intermediate oxidation states, especially +2 and +3 for the first-row metals (Fe, Co, and Ni). The heavier elements of these three groups are called precious metals because they are rather rare in nature and mostly chemically inert.
The chemistry of groups 8, 9, and 10 is dominated by intermediate oxidation states such as +2 and +3.
Group 8 (Fe, Ru, and Os)
The chemistry of group 8 is dominated by iron, whose high abundance in Earth’s crust is due to the extremely high stability of its nucleus. Ruthenium and osmium, on the other hand, are extremely rare elements, with terrestrial abundances of only about 0.1 ppb and 5 ppb, respectively, and they were not discovered until the 19th century. Because of the high melting point of iron (1538°C), early humans could not use it for tools or weapons. The advanced techniques needed to work iron were first developed by the Hittite civilization in Asia Minor sometime before 2000 BC, and they remained a closely guarded secret that gave the Hittites military supremacy for almost a millennium. With the collapse of the Hittite civilization around 1200 BC, the technology became widely distributed, however, leading to the Iron Age.
Group 9 (Co, Rh, and Ir)
Cobalt is one of the least abundant of the first-row transition metals. Its oxide ores, however, have been used in glass and pottery for thousands of years to produce the brilliant color known as “cobalt blue,” and its compounds are consumed in large quantities in the paint and ceramics industries. The heavier elements of group 9 are also rare, with terrestrial abundances of less than 1 ppb; they are generally found in combination with the heavier elements of groups 8 and 10 in Ni–Cu–S ores.
Group 10 (Ni, Pd, and Pt)
Nickel silicates are easily processed; consequently, nickel has been known and used since antiquity. In fact, a 75:25 Cu:Ni alloy was used for more than 2000 years to mint “silver” coins, and the modern US nickel uses the same alloy. In contrast to nickel, palladium and platinum are rare (their terrestrial abundance is about 10–15 ppb), but they are at least an order of magnitude more abundant than the heavier elements of groups 8 and 9. Platinum and palladium are used in jewelry, the former as the pure element and the latter as the Pd/Au alloy known as white gold.
Over 2000 years ago, the Bactrian civilization in Western Asia used a 75:25 alloy of copper and nickel for its coins. A modern US nickel has the same composition, but a modern Canadian nickel is nickel-plated steel and contains only 2.5% nickel by mass.
Some properties of the elements in groups 8–10 are summarized in Table $3$. As in earlier groups, similarities in size and electronegativity between the two heaviest members of each group result in similarities in chemistry. We are now at the point in the d block where there is no longer a clear correlation between the valence electron configuration and the preferred oxidation state. For example, all the elements of group 8 have eight valence electrons, but only Ru and Os have any tendency to form compounds in the +8 oxidation state, and those compounds are powerful oxidants. The predominant oxidation states for all three group 8 metals are +2 and +3. Although the elements of group 9 possess a total of nine valence electrons, the +9 oxidation state is unknown for these elements, and the most common oxidation states in the group are +3 and +1. Finally, the elements of group 10 all have 10 valence electrons, but all three elements are normally found in the +2 oxidation state formed by losing the ns2 valence electrons. In addition, Pd and Pt form numerous compounds and complexes in the +4 oxidation state.
Table $3$: Some Properties of the Elements of Groups 8, 9, and 10
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
8 Fe 26 4s23d6 1.83 126 1538 7.87
Ru 44 5s14d7 2.20 134 2334 12.10
Os 76 6s25d64f14 2.20 135 3033 22.59
9 Co 27 4s23d7 1.88 125 1495 8.86
Rh 45 5s14d8 2.28 134 1964 12.40
Ir 77 6s25d74f14 2.20 136 2446 22.50
10 Ni 28 4s23d8 1.91 124 1455 8.90
Pd 46 4d10 2.20 137 1555 12.00
Pt 78 6s25d84f14 2.20 139 1768 21.50
We stated that higher oxidation states become less stable as we go across the d-block elements and more stable as we go down a group. Thus Fe and Co form trifluorides, but Ni forms only the difluoride NiF2. In contrast to Fe, Ru and Os form a series of fluorides up to RuF6 and OsF7. The hexafluorides of Rh and Ir are extraordinarily powerful oxidants, and Pt is the only element in group 10 that forms a hexafluoride. Similar trends are observed among the oxides. For example, Fe forms only FeO, Fe2O3, and the mixed-valent Fe3O4 (magnetite), all of which are nonstoichiometric. In contrast, Ru and Os form the dioxides (MO2) and the highly toxic, volatile, yellow tetroxides, which contain formal M=O bonds. As expected for compounds of metals in such high oxidation states, the latter are potent oxidants. The tendency of the metals to form the higher oxides decreases rapidly as we go farther across the d block.
Higher oxidation states become less stable across the d-block, but more stable down a group.
Reactivity with the heavier chalcogens is rather complex. Thus the oxidation state of Fe, Ru, Os, Co, and Ni in their disulfides is +2 because of the presence of the disulfide ion (S22−), but the disulfides of Rh, Ir, Pd, and Pt contain the metal in the +4 oxidation state together with sulfide ions (S2−). This combination of highly charged cations and easily polarized anions results in substances that are not simple ionic compounds and have significant covalent character.
The groups 8–10 metals form a range of binary nitrides, carbides, and borides. By far the most important of these is cementite (Fe3C), which is used to strengthen steel. At high temperatures, Fe3C is soluble in iron, but slow cooling causes the phases to separate and form particles of cementite, which gives a metal that retains much of its strength but is significantly less brittle than pure iron. Palladium is unusual in that it forms a binary hydride with the approximate composition PdH0.5. Because the H atoms in the metal lattice are highly mobile, thin sheets of Pd are highly permeable to H2 but essentially impermeable to all other gases, including He. Consequently, diffusion of H2 through Pd is an effective method for separating hydrogen from other gases.
Group 11 (Cu, Ag, and Au)
The coinage metals—copper, silver, and gold—occur naturally (like the gold nugget shown here); consequently, these were probably the first metals used by ancient humans. For example, decorative gold artifacts dating from the late Stone Age are known, and some gold Egyptian coins are more than 5000 yr old. Copper is almost as ancient, with objects dating to about 5000 BC. Bronze, an alloy of copper and tin that is harder than either of its constituent metals, was used before 3000 BC, giving rise to the Bronze Age. Deposits of silver are much less common than deposits of gold or copper, yet by 3000 BC, methods had been developed for recovering silver from its ores, which allowed silver coins to be widely used in ancient times.
This 1 kg gold nugget was found in Australia; in 2005, it was for sale in Hong Kong at an asking price of more than US\$64,000.
Deposits of gold and copper are widespread and numerous, and for many centuries it was relatively easy to obtain large amounts of the pure elements. For example, a single gold nugget discovered in Australia in 1869 weighed more than 150 lb. Because the demand for these elements has outstripped their availability, methods have been developed to recover them economically from even very low-grade ores (as low as 1% Cu content for copper) by operating on a vast scale, as shown in the photo of an open-pit copper mine. Copper is used primarily to manufacture electric wires, but large quantities are also used to produce bronze, brass, and alloys for coins. Much of the silver made today is obtained as a by-product of the manufacture of other metals, especially Cu, Pb, and Zn. In addition to its use in jewelry and silverware, silver is used in Ag/Zn and Ag/Cd button batteries. Gold is typically found either as tiny particles of the pure metal or as gold telluride (AuTe2). It is used as a currency reserve, in jewelry, in the electronics industry for corrosion-free contacts, and, in very thin layers, as a reflective window coating that minimizes heat transfer.
The Chuquicamata copper mine in northern Chile, the world’s largest open-pit copper mine, is 4.3 km long, 3 km wide, and 825 m deep. Each gigantic truck in the foreground (and barely visible in the lower right center) can hold 330 metric tn (330,000 kg) of copper ore.
Some properties of the coinage metals are listed in Table $4$. The electronegativity of gold (χ = 2.40) is close to that of the nonmetals sulfur and iodine, which suggests that the chemistry of gold should be somewhat unusual for a metal. The coinage metals have the highest electrical and thermal conductivities of all the metals, and they are also the most ductile and malleable. With an ns1(n − 1)d10 valence electron configuration, the chemistry of these three elements is dominated by the +1 oxidation state due to losing the single ns electron. Higher oxidation states are also known, however: +2 is common for Cu and, to a lesser extent, Ag, and +3 for Au because of the relatively low values of the second and (for Au) third ionization energies. All three elements have significant electron affinities due to the half-filled ns orbital in the neutral atoms. As a result, gold reacts with powerful reductants like Cs and solutions of the alkali metals in liquid ammonia to produce the gold anion Au with a 6s25d10 valence electron configuration.
Table $4$: Some Properties of the Elements of Groups 11 and 12
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
11 Cu 29 4s13d10 1.90 128 1085 8.96
Ag 47 5s14d10 1.93 144 962 10.50
Au 79 6s15d104f14 2.40 144 1064 19.30
12 Zn 30 4s23d10 1.65 134 420 7.13
Cd 48 5s24d10 1.69 149 321 8.69
Hg 80 6s25d104f14 1.90 151 −38.8 13.53
All group 11 elements are relatively unreactive, and their reactivity decreases from Cu to Au. Hence they are noble metals that are particularly well suited for use in coins and jewelry. Copper reacts with O2 at high temperatures to produce Cu2O and with sulfur to form Cu2S. Neither silver nor gold reacts directly with oxygen, although oxides of these elements can be prepared by other routes. Silver reacts with sulfur compounds to form the black Ag2S coating known as tarnish. Gold is the only metal that does not react with sulfur; it also does not react with nitrogen, carbon, or boron. All the coinage metals do, however, react with oxidizing acids. Thus both Cu and Ag dissolve in HNO3 and in hot concentrated H2SO4, while Au dissolves in the 3:1 HCl:HNO3 mixture known as aqua regia. Furthermore, all three metals dissolve in basic cyanide solutions in the presence of oxygen to form very stable [M(CN)2] ions, a reaction that is used to separate gold from its ores.
Although the most important oxidation state for group 11 is +1, the elements are relatively unreactive, with reactivity decreasing from Cu to Au.
All the monohalides except CuF and AuF are known (including AgF). Once again, iodine is unable to stabilize the higher oxidation states (Au3+ and Cu2+). Thus all the copper(II) halides except the iodide are known, but the only dihalide of silver is AgF2. In contrast, all the gold trihalides (AuX3) are known, again except the triiodide. No binary nitrides, borides, or carbides are known for the group 11 elements.
Group 12 (Zn, Cd, and Hg)
We next encounter the group 12 elements. Because none of the elements in group 12 has a partially filled (n − 1)d subshell, they are not, strictly speaking, transition metals. Nonetheless, much of their chemistry is similar to that of the elements that immediately precede them in the d block. The group 12 metals are similar in abundance to those of group 11, and they are almost always found in combination with sulfur. Because zinc and cadmium are chemically similar, virtually all zinc ores contain significant amounts of cadmium. All three metals are commercially important, although the use of Cd is restricted because of its toxicity. Zinc is used for corrosion protection, in batteries, to make brass, and, in the form of ZnO, in the production of rubber and paints. Cadmium is used as the cathode in rechargeable NiCad batteries. Large amounts of mercury are used in the production of chlorine and NaOH by the chloralkali process, while smaller amounts are consumed in mercury-vapor streetlights and mercury batteries.
As shown in Table $4$, the group 12 metals are significantly more electropositive than the elements of group 11, and they therefore have less noble character. They also have much lower melting and boiling points than the preceding transition metals. In contrast to trends in the preceding groups, Zn and Cd are similar to each other, but very different from the heaviest element (Hg). In particular, Zn and Cd are rather active metals, whereas mercury is not. Because mercury, the only metal that is a liquid at room temperature, can dissolve many metals by forming amalgams, medieval alchemists especially valued it when trying to transmute base metals to gold and silver. All three elements in group 12 have ns2(n − 1)d10 valence electron configurations; consequently, the +2 oxidation state, corresponding to losing the two ns electrons, dominates their chemistry. In addition, mercury forms a series of compounds in the +1 oxidation state that contain the diatomic mercurous ion Hg22+.
The most important oxidation state for group 12 is +2; the metals are significantly more electropositive than the group 11 elements, so they are less noble.
All the possible group 12 dihalides (MX2) are known, and they range from ionic (the fluorides) to highly covalent (such as HgCl2). The highly covalent character of many mercuric and mercurous halides is surprising given the large size of the cations, and this has been attributed to the existence of an easily distorted 5d10 subshell. Zinc and cadmium react with oxygen to form amphoteric MO, whereas mercury forms HgO only within a narrow temperature range (350–400°C). Whereas zinc and cadmium dissolve in mineral acids such as HCl with the evolution of hydrogen, mercury dissolves only in oxidizing acids such as HNO3 and H2SO4. All three metals react with sulfur and the other chalcogens to form the binary chalcogenides; mercury also has an extraordinarily high affinity for sulfur.
Example $1$
For each reaction, explain why the indicated products form.
1. TiCl4(l) + 2H2O(l) → TiO2(s) + 4HCl(aq)
2. WO3(s) + 3C(s) + 3Cl2(g) $\xrightarrow{\Delta}$ WCl6(s) + 3CO(g)
3. Sc2O3(s) + 2OH(aq) + 3H2O(l) → 2[Sc(OH)4](aq)
4. 2KMnO4(aq) + 2H2SO4(l) → Mn2O7(l) + 2KHSO4(soln) + H2O(soln)
5. 4CrCl2(aq) + O2(g) + 4H+(aq) → 4Cr3+(aq) + 8Cl(aq) + 2H2O(l)
Given: balanced chemical equation
Asked for: why the indicated products form
Strategy:
Refer to the periodic trends in this section.
Solution:
1. The most stable oxidation state for Ti is +4, and neither reactant is a particularly strong oxidant or reductant; hence a redox reaction is unlikely. Similarly, neither reactant is a particularly strong acid or base, so an acid–base reaction is unlikely. Because TiCl4 contains Ti in a relatively high oxidation state (+4), however, it is likely to be rather covalent in character, with reactivity similar to that of a semimetal halide such as SiCl4. Covalent halides tend to hydrolyze in water to produce the hydrohalic acid and either the oxide of the other element or a species analogous to an oxoacid.
2. This reaction involves the oxide of a group 6 metal in its highest oxidation state (WO3) and two elements, one of which is a reductant (C) and the other an oxidant (Cl2). Consequently, some sort of redox reaction will occur. Carbon can certainly react with chlorine to form CCl4, and WO3 is a potential source of oxygen atoms that can react with carbon to produce CO, which is very stable. If CO is one of the products, then it seems likely that the other product will contain the metal and chlorine. The most likely combination is WCl6 (leaving the oxidation state of the metal unchanged).
3. One of the reactants is a strong base (OH), so an acid–base reaction is likely if the other reactant is acidic. Because oxides like Sc2O3, in which the metal is in an intermediate oxidation state, are often amphoteric, we expect Sc2O3 to dissolve in base to form a soluble hydroxide complex.
4. Concentrated sulfuric acid is both an oxidant and a strong acid that tends to protonate and dehydrate other substances. The permanganate ion already contains manganese in its highest possible oxidation state (+7), so it cannot be oxidized further. A redox reaction is impossible, which leaves an acid–base reaction as the most likely alternative. Sulfuric acid is likely to protonate the terminal oxygen atoms of permanganate, allowing them to be lost as water.
5. Molecular oxygen is an oxidant, so a redox reaction is likely if the other reactant can be oxidized. Because chromous chloride contains chromium in its lowest accessible oxidation state, a redox reaction will occur in which Cr2+ ions are oxidized and O2 is reduced. In the presence of protons, the reduction product of O2 is water, so we need to determine only the identity of the oxidation product of Cr2+. Chromium forms compounds in two common higher oxidation states: the Cr3+ ion, which is the most stable, and the [Cr2O7]2− ion, which is a more powerful oxidant than O2. We therefore predict that the reaction will form Cr3+(aq) and water.
Exercise $1$
Predict the products of each reactions and then balance each chemical equation.
1. Cr2+(aq) + Fe3+(aq) $\xrightarrow{\mathrm{H^+}}$
2. Na2Cr2O7(aq) + H2SO4(l) →
3. FeBr2(aq) + O2(g) $\xrightarrow{\mathrm{H^+}}$
4. VBr4(l) + H2O(l) →
5. ZrO2(s) + C(s) + Cl2(g) $\xrightarrow{\Delta}$
Answer
1. Cr2+(aq) + Fe3+(aq) $\xrightarrow{\mathrm{H^+}}$ Cr3+(aq) + Fe2+(aq)
2. Na2Cr2O7(aq) + 2H2SO4(l) → 2NaHSO4(soln) + H2O(soln) + 2CrO3(s)
3. 4FeBr2(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq) + 2H2O(l) + 8Br(aq)
4. VBr4(l) + H2O(l) → VO2+(aq) + 4Br(aq) + 2H+(aq)
5. ZrO2(s) + 2C(s) + 2Cl2(g) $\xrightarrow{\Delta}$ ZrCl4(s) + 2CO(g)
Summary
The elements tend to become more polarizable going across the d block and higher oxidation states become less stable; higher oxidation states become more stable going down a group. The group 3 transition metals are highly electropositive metals and powerful reductants. They react with nonmetals to form compounds that are largely ionic and with oxygen to form sesquioxides (M2O3). The group 4 metals also have a high affinity for oxygen. In their reactions with halogens, the covalent character of the halides increases as the oxidation state of the metal increases because the high charge-to-radius ratio causes extensive polarization of the anions. The dichalcogenides have layered structures similar to graphite, and the hydrides, nitrides, carbides, and borides are all hard, high-melting-point solids with metallic conductivity. The group 5 metals also have a high affinity for oxygen. Consistent with periodic trends, only the lightest (vanadium) has any tendency to form compounds in oxidation states lower than +5. The oxides are sufficiently polarized to make them covalent in character. These elements also form layered chalcogenides, as well as nitrides, carbides, borides, and hydrides that are similar to those of the group 4 elements. As the metals become more polarizable across the row, their affinity for oxygen decreases. The group 6 metals are less electropositive and have a maximum oxidation state of +6, making their compounds in high oxidation states largely covalent in character. As the oxidizing strength of the halogen decreases, the maximum oxidation state of the metal also decreases. All three trioxides are acidic, but Cr2O3 is amphoteric. The chalcogenides of the group 6 metals are generally nonstoichiometric and electrically conducting, and these elements also form nitrides, carbides, and borides that are similar to those in the preceding groups. The metals of group 7 have a maximum oxidation state of +7, but the lightest element, manganese, exhibits an extensive chemistry in lower oxidation states. As with the group 6 metals, reaction with less oxidizing halogens produces metals in lower oxidation states, and disulfides and diselenides of Tc and Re have layered structures. The group 7 metals also form nitrides, carbides, and borides that are stable at high temperatures and have metallic properties. In groups 8, 9, and 10, the ionization potentials of the elements are so high that the oxidation state corresponding to the formal loss of all valence electrons is encountered rarely (group 8) or not at all (groups 9 and 10). Compounds of group 8 metals in their highest oxidation state are powerful oxidants. The reaction of metals in groups 8, 9, and 10 with the chalcogens is complex, and these elements form a range of binary nitrides, carbides, and borides. The coinage metals (group 11) have the highest electrical and thermal conductivities and are the most ductile and malleable of the metals. Although they are relatively unreactive, they form halides but not nitrides, borides, or carbides. The group 12 elements, whose chemistry is dominated by the +2 oxidation state, are almost always found in nature combined with sulfur. Mercury is the only metal that is a liquid at room temperature, and it dissolves many metals to form amalgams. The group 12 halides range from ionic to covalent. These elements form chalcogenides and have a high affinity for soft ligands.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.2%3A_The_First-Row_Transition_Metals.txt |
Learning Objectives
• To know the most common structures observed for metal complexes.
• To predict the relative stabilities of metal complexes with different ligands
One of the most important properties of metallic elements is their ability to act as Lewis acids that form complexes with a variety of Lewis bases. A metal complex consists of a central metal atom or ion that is bonded to one or more ligands (from the Latin ligare, meaning “to bind”), which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, such as Co(NH3)3Cl3; positively charged, such as [Nd(H2O)9]3+; or negatively charged, such as [UF8]4−. Electrically charged metal complexes are sometimes called complex ions. A coordination compound contains one or more metal complexes.
Coordination compounds are important for at least three reasons. First, most of the elements in the periodic table are metals, and almost all metals form complexes, so metal complexes are a feature of the chemistry of more than half the elements. Second, many industrial catalysts are metal complexes, and such catalysts are steadily becoming more important as a way to control reactivity. For example, a mixture of a titanium complex and an organometallic compound of aluminum is the catalyst used to produce most of the polyethylene and polypropylene “plastic” items we use every day. Finally, transition-metal complexes are essential in biochemistry. Examples include hemoglobin, an iron complex that transports oxygen in our blood; cytochromes, iron complexes that transfer electrons in our cells; and complexes of Fe, Zn, Cu, and Mo that are crucial components of certain enzymes, the catalysts for all biological reactions.
History of the Coordination Compounds
Coordination compounds have been known and used since antiquity; probably the oldest is the deep blue pigment called Prussian blue: $\ce{KFe2(CN)6}$. The chemical nature of these substances, however, was unclear for a number of reasons. For example, many compounds called “double salts” were known, such as $\ce{AlF3·3KF}$, $\ce{Fe(CN)2·4KCN}$, and $\ce{ZnCl2·2CsCl}$, which were combinations of simple salts in fixed and apparently arbitrary ratios. Why should $\ce{AlF3·3KF}$ exist but not $\ce{AlF3·4KF}$ or $\ce{AlF3·2KF}$? And why should a 3:1 KF:AlF3 mixture have different chemical and physical properties than either of its components? Similarly, adducts of metal salts with neutral molecules such as ammonia were also known—for example, $\ce{CoCl3·6NH3}$, which was first prepared sometime before 1798. Like the double salts, the compositions of these adducts exhibited fixed and apparently arbitrary ratios of the components. For example, $\ce{CoCl3·6NH3}$, $\ce{CoCl3·5NH3}$, $\ce{CoCl3·4NH3}$, and $\ce{CoCl3·3NH3}$ were all known and had very different properties, but despite all attempts, chemists could not prepare $\ce{CoCl3·2NH3}$ or $\ce{CoCl3·NH3}$.
Although the chemical composition of such compounds was readily established by existing analytical methods, their chemical nature was puzzling and highly controversial. The major problem was that what we now call valence (i.e., the oxidation state) and coordination number were thought to be identical. As a result, highly implausible (to modern eyes at least) structures were proposed for such compounds, including the “Chattanooga choo-choo” model for CoCl3·4NH3 shown here.
The modern theory of coordination chemistry is based largely on the work of Alfred Werner (1866–1919; Nobel Prize in Chemistry in 1913). In a series of careful experiments carried out in the late 1880s and early 1890s, he examined the properties of several series of metal halide complexes with ammonia. For example, five different “adducts” of ammonia with PtCl4 were known at the time: PtCl4·nNH3 (n = 2–6). Some of Werner’s original data on these compounds are shown in Table $1$. The electrical conductivity of aqueous solutions of these compounds was roughly proportional to the number of ions formed per mole, while the number of chloride ions that could be precipitated as AgCl after adding Ag+(aq) was a measure of the number of “free” chloride ions present. For example, Werner’s data on PtCl4·6NH3 in Table $1$ showed that all the chloride ions were present as free chloride. In contrast, PtCl4·2NH3 was a neutral molecule that contained no free chloride ions.
Alfred Werner (1866–1919)
Werner, the son of a factory worker, was born in Alsace. He developed an interest in chemistry at an early age, and he did his first independent research experiments at age 18. While doing his military service in southern Germany, he attended a series of chemistry lectures, and he subsequently received his PhD at the University of Zurich in Switzerland, where he was appointed professor of chemistry at age 29. He won the Nobel Prize in Chemistry in 1913 for his work on coordination compounds, which he performed as a graduate student and first presented at age 26. Apparently, Werner was so obsessed with solving the riddle of the structure of coordination compounds that his brain continued to work on the problem even while he was asleep. In 1891, when he was only 25, he woke up in the middle of the night and, in only a few hours, had laid the foundation for modern coordination chemistry.
Table $1$: Werner’s Data on Complexes of Ammonia with $PtCl_4$
Complex Conductivity (ohm−1) Number of Ions per Formula Unit Number of Cl Ions Precipitated by Ag+
PtCl4·6NH3 523 5 4
PtCl4·5NH3 404 4 3
PtCl4·4NH3 299 3 2
PtCl4·3NH3 97 2 1
PtCl4·2NH3 0 0 0
These data led Werner to postulate that metal ions have two different kinds of valence: (1) a primary valence (oxidation state) that corresponds to the positive charge on the metal ion and (2) a secondary valence (coordination number) that is the total number of ligand-metal bonds bound to the metal ion. If $\ce{Pt}$ had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the $\ce{PtCl4·NH3}$ adducts by the following reactions, where the metal complex is enclosed in square brackets:
\begin{align*} \mathrm{[Pt(NH_3)_6]Cl_4} &\rightarrow \mathrm{[Pt(NH_3)_6]^{4+}(aq)+4Cl^-(aq)} \[4pt] \mathrm{[Pt(NH_3)_5Cl]Cl_3} &\rightarrow \mathrm{[Pt(NH_3)_5Cl]^{3+}(aq) +3Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_4Cl_2]Cl_2} &\rightarrow \mathrm{[Pt(NH_3)_4Cl_2]^{2+}(aq) +2Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_3Cl_3]Cl} &\rightarrow \mathrm{[Pt(NH_3)_3Cl_3]^+(aq) + Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_2Cl_4]} &\rightarrow \mathrm{[Pt(NH_3)_2Cl_4]^0(aq)} \end{align*} \label{23.9}
Further work showed that the two missing members of the series—[Pt(NH3)Cl5] and [PtCl6]2−—could be prepared as their mono- and dipotassium salts, respectively. Similar studies established coordination numbers of 6 for Co3+ and Cr3+ and 4 for Pt2+ and Pd2+.
Werner’s studies on the analogous Co3+ complexes also allowed him to propose a structural model for metal complexes with a coordination number of 6. Thus he found that [Co(NH3)6]Cl3 (yellow) and [Co(NH3)5Cl]Cl2 (purple) were 1:3 and 1:2 electrolytes. Unexpectedly, however, two different [Co(NH3)4Cl2]Cl compounds were known: one was red, and the other was green (Figure $\PageIndex{1a}$). Because both compounds had the same chemical composition and the same number of groups of the same kind attached to the same metal, there had to be something different about the arrangement of the ligands around the metal ion. Werner’s key insight was that the six ligands in [Co(NH3)4Cl2]Cl had to be arranged at the vertices of an octahedron because that was the only structure consistent with the existence of two, and only two, arrangements of ligands (Figure $\PageIndex{1b}$. His conclusion was corroborated by the existence of only two different forms of the next compound in the series: Co(NH3)3Cl3.
Example $1$
In Werner’s time, many complexes of the general formula MA4B2 were known, but no more than two different compounds with the same composition had been prepared for any metal. To confirm Werner’s reasoning, calculate the maximum number of different structures that are possible for six-coordinate MA4B2 complexes with each of the three most symmetrical possible structures: a hexagon, a trigonal prism, and an octahedron. What does the fact that no more than two forms of any MA4B2 complex were known tell you about the three-dimensional structures of these complexes?
Given: three possible structures and the number of different forms known for MA4B2 complexes
Asked for: number of different arrangements of ligands for MA4B2 complex for each structure
Strategy:
Sketch each structure, place a B ligand at one vertex, and see how many different positions are available for the second B ligand.
Solution
The three regular six-coordinate structures are shown here, with each coordination position numbered so that we can keep track of the different arrangements of ligands. For each structure, all vertices are equivalent. We begin with a symmetrical MA6 complex and simply replace two of the A ligands in each structure to give an MA4B2 complex:
For the hexagon, we place the first B ligand at position 1. There are now three possible places for the second B ligand: at position 2 (or 6), position 3 (or 5), or position 4. These are the only possible arrangements. The (1, 2) and (1, 6) arrangements are chemically identical because the two B ligands are adjacent to each other. The (1, 3) and (1, 5) arrangements are also identical because in both cases the two B ligands are separated by an A ligand.
Turning to the trigonal prism, we place the first B ligand at position 1. Again, there are three possible choices for the second B ligand: at position 2 or 3 on the same triangular face, position 4 (on the other triangular face but adjacent to 1), or position 5 or 6 (on the other triangular face but not adjacent to 1). The (1, 2) and (1, 3) arrangements are chemically identical, as are the (1, 5) and (1, 6) arrangements.
In the octahedron, however, if we place the first B ligand at position 1, then we have only two choices for the second B ligand: at position 2 (or 3 or 4 or 5) or position 6. In the latter, the two B ligands are at opposite vertices of the octahedron, with the metal lying directly between them. Although there are four possible arrangements for the former, they are chemically identical because in all cases the two B ligands are adjacent to each other.
The number of possible MA4B2 arrangements for the three geometries is thus: hexagon, 3; trigonal prism, 3; and octahedron, 2. The fact that only two different forms were known for all MA4B2 complexes that had been prepared suggested that the correct structure was the octahedron but did not prove it. For some reason one of the three arrangements possible for the other two structures could have been less stable or harder to prepare and had simply not yet been synthesized. When combined with analogous results for other types of complexes (e.g., MA3B3), however, the data were best explained by an octahedral structure for six-coordinate metal complexes.
Exercise $1$
Determine the maximum number of structures that are possible for a four-coordinate MA2B2 complex with either a square planar or a tetrahedral symmetrical structure.
Answer
square planar, 2; tetrahedral, 1
Structures of Metal Complexes
The coordination numbers of metal ions in metal complexes can range from 2 to at least 9. In general, the differences in energy between different arrangements of ligands are greatest for complexes with low coordination numbers and decrease as the coordination number increases. Usually only one or two structures are possible for complexes with low coordination numbers, whereas several different energetically equivalent structures are possible for complexes with high coordination numbers (n > 6). The following presents the most commonly encountered structures for coordination numbers 2–9. Many of these structures should be familiar to you from our discussion of the valence-shell electron-pair repulsion (VSEPR) model because they correspond to the lowest-energy arrangements of n electron pairs around a central atom.
Compounds with low coordination numbers exhibit the greatest differences in energy between different arrangements of ligands.
Coordination Number 2
Although it is rare for most metals, this coordination number is surprisingly common for d10 metal ions, especially Cu+, Ag+, Au+, and Hg2+. An example is the [Au(CN)2] ion, which is used to extract gold from its ores. As expected based on VSEPR considerations, these complexes have the linear L–M–L structure shown here.
Coordination Number 3
Although it is also rare, this coordination number is encountered with d10 metal ions such as Cu+ and Hg2+. Among the few known examples is the HgI3 ion. Three-coordinate complexes almost always have the trigonal planar structure expected from the VSEPR model.
Coordination Number 4
Two common structures are observed for four-coordinate metal complexes: tetrahedral and square planar. The tetrahedral structure is observed for all four-coordinate complexes of nontransition metals, such as [BeF4]2−, and d10 ions, such as [ZnCl4]2−. It is also found for four-coordinate complexes of the first-row transition metals, especially those with halide ligands (e.g., [FeCl4] and [FeCl4]2−). In contrast, square planar structures are routinely observed for four-coordinate complexes of second- and third-row transition metals with d8 electron configurations, such as Rh+ and Pd2+, and they are also encountered in some complexes of Ni2+ and Cu2+.
Coordination Number 5
This coordination number is less common than 4 and 6, but it is still found frequently in two different structures: trigonal bipyramidal and square pyramidal. Because the energies of these structures are usually rather similar for most ligands, many five-coordinate complexes have distorted structures that lie somewhere between the two extremes.
Coordination Number 6
This coordination number is by far the most common. The six ligands are almost always at the vertices of an octahedron or a distorted octahedron. The only other six-coordinate structure is the trigonal prism, which is very uncommon in simple metal complexes.
Coordination Number 7
This relatively uncommon coordination number is generally encountered for only large metals (such as the second- and third-row transition metals, lanthanides, and actinides). At least three different structures are known, two of which are derived from an octahedron or a trigonal prism by adding a ligand to one face of the polyhedron to give a “capped” octahedron or trigonal prism. By far the most common, however, is the pentagonal bipyramid.
Coordination Number 8
This coordination number is relatively common for larger metal ions. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Common structures are the square antiprism and the dodecahedron, both of which can be generated from the cube.
Coordination Number 9
This coordination number is found in larger metal ions, and the most common structure is the tricapped trigonal prism, as in [Nd(H2O)9]3+.
Key Takeaways
• Coordination compounds are a major feature of the chemistry of over half the elements.
• Coordination compounds have important roles as industrial catalysts in controlling reactivity, and they are essential in biochemical processes.
Summary
Transition metals form metal complexes, polyatomic species in which a metal ion is bound to one or more ligands, which are groups bound to a metal ion. Complex ions are electrically charged metal complexes, and a coordination compound contains one or more metal complexes. Metal complexes with low coordination numbers generally have only one or two possible structures, whereas those with coordination numbers greater than six can have several different structures. Coordination numbers of two and three are common for d10 metal ions. Tetrahedral and square planar complexes have a coordination number of four; trigonal bipyramidal and square pyramidal complexes have a coordination number of five; and octahedral complexes have a coordination number of six. At least three structures are known for a coordination number of seven, which is generally found for only large metal ions. Coordination numbers of eight and nine are also found for larger metal ions. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.3%3A_Coordination_Compounds.txt |
Learning Objectives
• To understand that there may be more than one way to arrange the same groups around the same atom with the same geometry (stereochemistry).
Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimensional space. These two, different molecules are called stereoisomers.
One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride, (NH3)2PtCl2. This important compound is sometimes called "platin" for short. As the formula implies, it contains a platinum ion that is coordinated to two ammonia ligands and two chloride ligands (remember, a ligand in inorganic chemistry is an electron donor that is attached to a metal atom, donating a pair of electrons to form a bond).
Platin is an example of a coordination compound. The way the different pieces of coordination compounds bond together is discussed in the chapter of Lewis acids and bases. For reasons arising from molecular orbital interactions, platin has a square planar geometry at the platinum atom. That arrangement results in two possible ways the ligands could be connected. The two sets of like ligands could be connected on the same side of the square or on opposite corners.
These two arrangements result in two different compounds; they are isomers that differ only in three-dimensional space.
• The one with the two amines beside each other is called cis-platin.
• These two ligands are 90 degrees from each other.
• The one with the amines across from each other is trans-platin.
• These two ligands are 180 degrees from each other.
CIS/TRANS isomers have different physical properties
Although these two compounds are very similar, they have slightly different physical properties. Both are yellow compounds that decompose when heated to 270 degrees C, but trans-platin forms pale yellow crystals and is more soluble than cis-platin in water.
CIS/TRANS isomers have different biological properties
Cis-platin has clinical importance in the treatment of ovarian and testicular cancers. The biological mechanism of the drug's action was long suspected to involve binding of the platinum by DNA. Further details were worked out by MIT chemist Steve Lippard and graduate student Amy Rosenzweig in the 1990's. Inside the cell nucleus, the two ammines in cis-platin can be replaced by nitrogen donors from a DNA strand. To donate to the Lewis acidic platinum, the DNA molecule must bend slightly. Normally that bend is detected and repaired by proteins in the cell. However, ovarian and testicular cells happen to contain a protein that is just the right shape to fit around this slightly bent DNA strand. The DNA strand becomes lodged in the protein and can't be displaced, and so it is unable to bind with other proteins used in DNA replication. The cell becomes unable to replicate, and so cancerous growth is stopped.
Exercise \(1\)
Draw the cis and trans isomers of the following compounds:
1. \(\ce{(NH3)2IrCl(CO)}\)
2. \(\ce{(H3P)2PtHBr}\)
3. \(\ce{(AsH3)2PtH(CO)}\)
Exercise \(2\)
Only one isomer of (tmeda)PtCl2 is possible [tmeda = (CH3)2NCH2CH2N(CH3)2; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible.
Geometric Isomers
The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomers. Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. Recall that in many cases more than one structure is possible for organic compounds with the same molecular formula; examples discussed previously include n-butane versus isobutane and cis-2-butene versus trans-2-butene. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique.
Planar Isomers
Metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal are called geometrical isomers. They are most important for square planar and octahedral complexes.
Because all vertices of a square are equivalent, it does not matter which vertex is occupied by the ligand B in a square planar MA3B complex; hence only a single geometrical isomer is possible in this case (and in the analogous MAB3 case). All four structures shown here are chemically identical because they can be superimposed simply by rotating the complex in space:
For an MA2B2 complex, there are two possible isomers: either the A ligands can be adjacent to one another (cis), in which case the B ligands must also be cis, or the A ligands can be across from one another (trans), in which case the B ligands must also be trans. Even though it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways, all members of each set are chemically equivalent:
The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs, which primarily affect individuals in their 20s and were notoriously difficult to cure. For example, after being diagnosed with metastasized testicular cancer in 1991 and given only a 50% chance of survival, Lance Armstrong was cured by treatment with cisplatin.
Square planar complexes that contain symmetrical bidentate ligands, such as [Pt(en)2]2+, have only one possible structure, in which curved lines linking the two N atoms indicate the ethylenediamine ligands:
Octahedral Isomers
Octahedral complexes also exhibit cis and trans isomers. Like square planar complexes, only one structure is possible for octahedral complexes in which only one ligand is different from the other five (MA5B). Even though we usually draw an octahedron in a way that suggests that the four “in-plane” ligands are different from the two “axial” ligands, in fact all six vertices of an octahedron are equivalent. Consequently, no matter how we draw an MA5B structure, it can be superimposed on any other representation simply by rotating the molecule in space. Two of the many possible orientations of an MA5B structure are as follows:
If two ligands in an octahedral complex are different from the other four, giving an MA4B2 complex, two isomers are possible. The two B ligands can be cis or trans. Cis- and trans-[Co(NH3)4Cl2]Cl are examples of this type of system:
Replacing another A ligand by B gives an MA3B3 complex for which there are also two possible isomers. In one, the three ligands of each kind occupy opposite triangular faces of the octahedron; this is called the fac isomer (for facial). In the other, the three ligands of each kind lie on what would be the meridian if the complex were viewed as a sphere; this is called the mer isomer (for meridional):
Example \(1\)
Draw all the possible geometrical isomers for the complex [Co(H2O)2(ox)BrCl], where ox is O2CCO2, which stands for oxalate.
Given: formula of complex
Asked for: structures of geometrical isomers
Solution
This complex contains one bidentate ligand (oxalate), which can occupy only adjacent (cis) positions, and four monodentate ligands, two of which are identical (H2O). The easiest way to attack the problem is to go through the various combinations of ligands systematically to determine which ligands can be trans. Thus either the water ligands can be trans to one another or the two halide ligands can be trans to one another, giving the two geometrical isomers shown here:
In addition, two structures are possible in which one of the halides is trans to a water ligand. In the first, the chloride ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens. Exchanging the chloride and bromide ligands gives the other, in which the bromide ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens:
This complex can therefore exist as four different geometrical isomers.
Exercise \(1\)
Draw all the possible geometrical isomers for the complex [Cr(en)2(CN)2]+.
Answer
Two geometrical isomers are possible: trans and cis.
Summary
Many metal complexes form isomers, which are two or more compounds with the same formula but different arrangements of atoms. Structural isomers differ in which atoms are bonded to one another, while geometrical isomers differ only in the arrangement of ligands around the metal ion. Ligands adjacent to one another are cis, while ligands across from one another are trans. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.4%3A_Isomerism.txt |
Learning Objectives
• To understand how crystal field theory explains the electronic structures and colors of metal complexes.
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy. The Learning Objective of this Module is to understand how crystal field theory explains the electronic structures and colors of metal complexes.
d-Orbital Splittings
CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion (VSEPR) model, the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands.
We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (Figure $\PageIndex{1a}$). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in Figure $\PageIndex{1b}$, the dz2 and dx2−y2 orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the eg orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (dxy, dxz, and dyz, collectively called the t2g orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge.
The difference in energy between the two sets of d orbitals is called the crystal field splitting energy (Δo), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δt.) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two eg orbitals increase in energy by 0.6Δo, whereas the three t2g orbitals decrease in energy by 0.4Δo. Thus the total change in energy is
$2(0.6Δ_o) + 3(−0.4Δ_o) = 0. \nonumber$
Crystal field splitting does not change the total energy of the d orbitals.
Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in Figure $\PageIndex{1a}$).
Electronic Structures of Metal Complexes
We can use the d-orbital energy-level diagram in Figure $1$ to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti3+ ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, while keeping their spins parallel as required by Hund’s rule. As shown in Figure 24.6.2, for d1–d3 systems—such as [Ti(H2O)6]3+, [V(H2O)6]3+, and [Cr(H2O)6]3+, respectively—the electrons successively occupy the three degenerate t2g orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex [Cr(H2O)6]3+, for example, by saying that the chromium ion has a d3 electron configuration or, more succinctly, Cr3+ is a d3 ion.
When we reach the d4 configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty eg orbitals or one of the singly occupied t2g orbitals. Recall that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the spin-pairing energy (P). If Δo is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty eg orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron configuration, such as the [Cr(H2O)6]2+ ion, is called a high-spin complex. Conversely, if Δo is greater than P, then the lowest-energy arrangement has the fourth electron in one of the occupied t2g orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the [Mn(CN)6]3− ion, is called a low-spin complex. Similarly, metal ions with the d5, d6, or d7 electron configurations can be either high spin or low spin, depending on the magnitude of Δo.
In contrast, only one arrangement of d electrons is possible for metal ions with d8–d10 electron configurations. For example, the [Ni(H2O)6]2+ ion is d8 with two unpaired electrons, the [Cu(H2O)6]2+ ion is d9 with one unpaired electron, and the [Zn(H2O)6]2+ ion is d10 with no unpaired electrons.
If Δo is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δo is greater, a low-spin configuration forms.
Factors That Affect the Magnitude of Δo
The magnitude of Δo dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δo (i.e., Δo > P) yield a low-spin complex, whereas small values of Δo (i.e., Δo < P) produce a high-spin complex. As we noted, the magnitude of Δo depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δo for some representative transition-metal complexes are given in Table $1$.
Table $1$: Crystal Field Splitting Energies for Some Octahedral (Δo)* and Tetrahedral (Δt) Transition-Metal Complexes
Octahedral Complexes Δo (cm−1) Octahedral Complexes Δo (cm−1) Tetrahedral Complexes Δt (cm−1)
*Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol.
[Ti(H2O)6]3+ 20,300 [Fe(CN)6]4− 32,800 VCl4 9010
[V(H2O)6]2+ 12,600 [Fe(CN)6]3− 35,000 [CoCl4]2− 3300
[V(H2O)6]3+ 18,900 [CoF6]3− 13,000 [CoBr4]2− 2900
[CrCl6]3− 13,000 [Co(H2O)6]2+ 9300 [CoI4]2− 2700
[Cr(H2O)6]2+ 13,900 [Co(H2O)6]3+ 27,000
[Cr(H2O)6]3+ 17,400 [Co(NH3)6]3+ 22,900
[Cr(NH3)6]3+ 21,500 [Co(CN)6]3− 34,800
[Cr(CN)6]3− 26,600 [Ni(H2O)6]2+ 8500
Cr(CO)6 34,150 [Ni(NH3)6]2+ 10,800
[MnCl6]4− 7500 [RhCl6]3− 20,400
[Mn(H2O)6]2+ 8500 [Rh(H2O)6]3+ 27,000
[MnCl6]3− 20,000 [Rh(NH3)6]3+ 34,000
[Mn(H2O)6]3+ 21,000 [Rh(CN)6]3− 45,500
[Fe(H2O)6]2+ 10,400 [IrCl6]3− 25,000
[Fe(H2O)6]3+ 14,300 [Ir(NH3)6]3+ 41,000
Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994).
Charge on the Metal Ion
Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δo increases as the charge on the metal ion increases. Typically, Δo for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, for [V(H2O)6]2+, Δo = 11,800 cm−1; for [V(H2O)6]3+, Δo = 17,850 cm−1.
Principal Quantum Number of the Metal
For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δo increases with increasing principal quantum number: Δo (3d) < Δo (4d) < Δo (5d). The data for hexaammine complexes of the trivalent group 9 metals illustrate this point:
[Co(NH3)6]3+: Δo = 22,900 cm−1
[Rh(NH3)6]3+: Δo = 34,100 cm−1
[Ir(NH3)6]3+: Δo = 40,000 cm−1
The increase in Δo with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions.
The Nature of the Ligands
Experimentally, it is found that the Δo observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δo decreases as the size of the donor atom increases. For example, Δo values for halide complexes generally decrease in the order F > Cl > Br > I− because smaller, more localized charges, such as we see for F, interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH3, results in significantly larger Δo values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F. The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series, shown here in order of decreasing Δo:
$\mathrm{\underset{\textrm{strong-field ligands}}{CO\approx CN^->}NO_2^->en>NH_3>\underset{\textrm{intermediate-field ligands}}{SCN^->H_2O>oxalate^{2-}}>OH^->F>acetate^->\underset{\textrm{weak-field ligands}}{Cl^->Br^->I^-}}$
The values of Δo listed in Table $1$ illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand.
The largest Δo splittings are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons.
Colors of Transition-Metal Complexes
The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a d–d transition (Figure 24.6.3). For a photon to effect such a transition, its energy must be equal to the difference in energy between the two d orbitals, which depends on the magnitude of Δo.
Recall that the color we observe when we look at an object or a compound is due to light that is transmitted or reflected, not light that is absorbed, and that reflected or transmitted light is complementary in color to the light that is absorbed. Thus a green compound absorbs light in the red portion of the visible spectrum and vice versa, as indicated by the color wheel. Because the energy of a photon of light is inversely proportional to its wavelength, the color of a complex depends on the magnitude of Δo, which depends on the structure of the complex. For example, the complex [Cr(NH3)6]3+ has strong-field ligands and a relatively large Δo. Consequently, it absorbs relatively high-energy photons, corresponding to blue-violet light, which gives it a yellow color. A related complex with weak-field ligands, the [Cr(H2O)6]3+ ion, absorbs lower-energy photons corresponding to the yellow-green portion of the visible spectrum, giving it a deep violet color.
We can now understand why emeralds and rubies have such different colors, even though both contain Cr3+ in an octahedral environment provided by six oxide ions. Although the chemical identity of the six ligands is the same in both cases, the Cr–O distances are different because the compositions of the host lattices are different (Al2O3 in rubies and Be3Al2Si6O18 in emeralds). In ruby, the Cr–O distances are relatively short because of the constraints of the host lattice, which increases the d orbital–ligand interactions and makes Δo relatively large. Consequently, rubies absorb green light and the transmitted or reflected light is red, which gives the gem its characteristic color. In emerald, the Cr–O distances are longer due to relatively large [Si6O18]12− silicate rings; this results in decreased d orbital–ligand interactions and a smaller Δo. Consequently, emeralds absorb light of a longer wavelength (red), which gives the gem its characteristic green color. It is clear that the environment of the transition-metal ion, which is determined by the host lattice, dramatically affects the spectroscopic properties of a metal ion.
Crystal Field Stabilization Energies
Recall that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t2g orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d1 complex such as [Ti(H2O)6]3+ is located in one of the t2g orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δo. The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE). The CFSE of a complex can be calculated by multiplying the number of electrons in t2g orbitals by the energy of those orbitals (−0.4Δo), multiplying the number of electrons in eg orbitals by the energy of those orbitals (+0.6Δo), and summing the two. Table $2$ gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d6 complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d3 configuration.
Table $2$: CFSEs for Octahedral Complexes with Different Electron Configurations (in Units of Δo)
High Spin CFSE (Δo) Low Spin CFSE (Δo)
d 0 0
d 1 0.4
d 2 ↿ ↿ 0.8
d 3 ↿ ↿ ↿ 1.2
d 4 ↿ ↿ ↿ 0.6 ↿⇂ ↿ ↿ 1.6
d 5 ↿ ↿ ↿ ↿ ↿ 0.0 ↿⇂ ↿⇂ ↿ 2.0
d 6 ↿⇂ ↿ ↿ ↿ ↿ 0.4 ↿⇂ ↿⇂ ↿⇂ 2.4
d 7 ↿⇂ ↿⇂ ↿ ↿ ↿ 0.8 ↿⇂ ↿⇂ ↿⇂ 1.8
d 8 ↿⇂ ↿⇂ ↿⇂ ↿ ↿ 1.2
d 9 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ 0.6
d 10 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ 0.0
CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences.
Octahedral d3 and d8 complexes and low-spin d6, d5, d7, and d4 complexes exhibit large CFSEs.
Example $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [CoF6]3−
2. [Rh(CO)2Cl2]
Given: complexes
Asked for: structure, high spin versus low spin, and the number of unpaired electrons
Strategy:
1. From the number of ligands, determine the coordination number of the compound.
2. Classify the ligands as either strong field or weak field and determine the electron configuration of the metal ion.
3. Predict the relative magnitude of Δo and decide whether the compound is high spin or low spin.
4. Place the appropriate number of electrons in the d orbitals and determine the number of unpaired electrons.
Solution
1. A With six ligands, we expect this complex to be octahedral.
B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d6 electron configuration.
C Because of the weak-field ligands, we expect a relatively small Δo, making the compound high spin.
D In a high-spin octahedral d6 complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t2g orbitals, giving four unpaired electrons.
1. A This complex has four ligands, so it is either square planar or tetrahedral.
B C Because rhodium is a second-row transition metal ion with a d8 electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δo, making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order dz2dyz, and dxz (these are degenerate); dxy; and dx2−y2.
D The eight electrons occupy the first four of these orbitals, leaving the dx2−y2. orbital empty. Thus there are no unpaired electrons.
Exercise $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [Mn(H2O)6]2+
2. [PtCl4]2−
Answers
1. octahedral; high spin; five
2. square planar; low spin; no unpaired electrons
Summary
Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δo), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δo is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δo is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δo, whereas weak-field ligands interact more weakly and give a smaller Δo. The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT.
19.7: Molecular Orbital Model
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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_(Zumdahl_and_Decoste)/20%3A_The_Nucleus_A_Chemists_View/20.1%3A_Nuclear_Stability_and_Radioactive_Decay.txt |
Learning Objectives
• To know how to use half-lives to describe the rates of first-order reactions
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-life of the reaction, written as t1/2. Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A]0 to [A]0/2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life.
The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a first-order reaction to produce the following equation:
$\ln\dfrac{[\textrm A]_0}{[\textrm A]}=kt \label{21.4.1}$
Substituting [A]0/2 for [A] and t1/2 for t (to indicate a half-life) into Equation $\ref{21.4.1}$ gives
$\ln\dfrac{[\textrm A]_0}{[\textrm A]_0/2}=\ln 2=kt_{1/2}$
Substituting $\ln{2} \approx 0.693$ into the equation results in the expression for the half-life of a first-order reaction:
$t_{1/2}=\dfrac{0.693}{k} \label{21.4.2}$
Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in Figure $1$, and is independent of [A].
If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion.
Number of Half-Lives Percentage of Reactant Remaining
1 $\dfrac{100\%}{2}=50\%$ $\dfrac{1}{2}(100\%)=50\%$
2 $\dfrac{50\%}{2}=25\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right)(100\%)=25\%$
3 $\dfrac{25\%}{2}=12.5\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right )\left (\dfrac{1}{2}\right)(100\%)=12.5\%$
n $\dfrac{100\%}{2^n}$ $\left(\dfrac{1}{2}\right)^n(100\%)=\left(\dfrac{1}{2}\right)^n\%$
As you can see from this table, the amount of reactant left after n half-lives of a first-order reaction is (1/2)n times the initial concentration.
For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A].
Example $1$
The anticancer drug cis-platin hydrolyzes in water with a rate constant of 1.5 × 10−3 min−1 at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cis-platin has a concentration of 0.053 M, what will be the concentration of cis-platin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives?
Given: rate constant, initial concentration, and number of half-lives
Asked for: half-life, final concentrations, and percent completion
Strategy:
1. Use Equation $\ref{21.4.2}$ to calculate the half-life of the reaction.
2. Multiply the initial concentration by 1/2 to the power corresponding to the number of half-lives to obtain the remaining concentrations after those half-lives.
3. Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion.
Solution
A We can calculate the half-life of the reaction using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5\times10^{-3}\textrm{ min}^{-1}}=4.6\times10^2\textrm{ min}$
Thus it takes almost 8 h for half of the cis-platin to hydrolyze.
B After 5 half-lives (about 38 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^5}=\dfrac{0.053\textrm{ M}}{32}=0.0017\textrm{ M}$
After 10 half-lives (77 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^{10}}=\dfrac{0.053\textrm{ M}}{1024}=5.2\times10^{-5}\textrm{ M}$
C The percent completion after 5 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-0.0017\textrm{ M})(100)}{0.053}=97\%$
The percent completion after 10 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-5.2\times10^{-5}\textrm{ M})(100)}{0.053\textrm{ M}}=100\%$
Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives.
Exercise $1$
Ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10−6 s−1 at 650°C.
1. What is the half-life for the reaction under these conditions?
2. If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 half-lives?
Answer a
4.3 × 105 s = 120 h = 5.0 days;
Answer b
4.8 × 10−3 M
Radioactive Decay Rates
Radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes.
In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using N to represent the number of atoms of the radioactive isotope, we can define the rate of decay of the sample, which is also called its activity (A) as the decrease in the number of the radioisotope’s nuclei per unit time:
$A=-\dfrac{\Delta N}{\Delta t} \label{21.4.3}$
Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm).
The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:
$A = kN \label{21.4.4}$
Here, the symbol k is the radioactive decay constant, which has units of inverse time (e.g., s−1, yr−1) and a characteristic value for each radioactive isotope. If we combine Equation $\ref{21.4.3}$ and Equation $\ref{21.4.4}$, we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:
$-\dfrac{\Delta N}{\Delta t}=kN \label{21.4.5}$
Equation $\ref{21.4.5}$ is the same as the equation for the reaction rate of a first-order reaction, except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a first-order process and can be described in terms of either the differential rate law (Equation $\ref{21.4.5}$) or the integrated rate law:
$N = N_0e^{−kt}$
$\ln \dfrac{N}{N_0}=-kt \label{21.4.6}$
Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the half-life of the isotope. The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in Table 14.6, along with some of their applications.
Table $2$: Half-Lives and Applications of Some Radioactive Isotopes
Radioactive Isotope Half-Life Typical Uses
*The m denotes metastable, where an excited state nucleus decays to the ground state of the same isotope.
hydrogen-3 (tritium) 12.32 yr biochemical tracer
carbon-11 20.33 min positron emission tomography (biomedical imaging)
carbon-14 5.70 × 103 yr dating of artifacts
sodium-24 14.951 h cardiovascular system tracer
phosphorus-32 14.26 days biochemical tracer
potassium-40 1.248 × 109 yr dating of rocks
iron-59 44.495 days red blood cell lifetime tracer
cobalt-60 5.2712 yr radiation therapy for cancer
technetium-99m* 6.006 h biomedical imaging
iodine-131 8.0207 days thyroid studies tracer
radium-226 1.600 × 103 yr radiation therapy for cancer
uranium-238 4.468 × 109 yr dating of rocks and Earth’s crust
americium-241 432.2 yr smoke detectors
Note
Radioactive decay is a first-order process.
Radioisotope Dating Techniques
In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques.
The most common method for measuring the age of ancient objects is carbon-14 dating. The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form 14CO2. As a result, the CO2 that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of 14CO2 molecules as well as nonradioactive 12CO2 and 13CO2. Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as beta decay, which releases low-energy electrons (β particles) that can be detected and measured:
$\ce{^{14}C \rightarrow ^{14}N + \beta^{−}} \label{21.4.7}$
The half-life for this reaction is 5700 ± 30 yr.
The 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g of carbon (Figure $2$). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11.Using this method implicitly assumes that the 14CO2/12CO2 ratio in the atmosphere is constant, which is not strictly correct. Other methods, such as tree-ring dating, have been used to calibrate the dates obtained by radiocarbon dating, and all radiocarbon dates reported are now corrected for minor changes in the 14CO2/12CO2 ratio over time.
Example $2$
In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the 14C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die?
Given: isotope and final activity
Asked for: elapsed time
Strategy:
A Use Equation $\ref{21.4.4}$ to calculate N0/N. Then substitute the value for the half-life of 14C into Equation $\ref{21.4.2}$ to find the rate constant for the reaction.
B Using the values obtained for N0/N and the rate constant, solve Equation $\ref{21.4.6}$ to obtain the elapsed time.
Solution
We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction (Equation $\ref{21.4.6}$) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay).
\begin{align}\ln\dfrac{N}{N_0}&=-kt \ \dfrac{\ln(N/N_0)}{k}&=t\end{align}
A From Equation $\ref{21.4.4}$, we know that A = kN. We can therefore use the initial and final activities (A0 = 15 dpm and A = 8.0 dpm) to calculate N0/N:
$\dfrac{A_0}{A}=\dfrac{kN_0}{kN}=\dfrac{N_0}{N}=\dfrac{15}{8.0}$
Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}$
This equation can be rearranged as follows:
$k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\textrm{ yr}}=1.22\times10^{-4}\textrm{ yr}^{-1}$
B Substituting into the equation for t,
$t=\dfrac{\ln(N_0/N)}{k}=\dfrac{\ln(15/8.0)}{1.22\times10^{-4}\textrm{ yr}^{-1}}=5.2\times10^3\textrm{ yr}$
From our calculations, the man died 5200 yr ago.
Exercise $2$
It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a 14C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample?
Answer
30,000 yr
Summary
• The half-life of a first-order reaction is independent of the concentration of the reactants.
• The half-lives of radioactive isotopes can be used to date objects.
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/20%3A_The_Nucleus_A_Chemists_View/20.2%3A_Radioactive_Kinetics.txt |
Learning Objectives
• Describe the synthesis of transuranium nuclides
After the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War II, combined to usher in the Nuclear Age in the mid-twentieth century. Science learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society’s needs during peace.
Synthesis of Nuclides
Nuclear transmutation is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford’s laboratory in 1919 by a transmutation reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed α particles from a natural radioactive isotope of radium and observed protons resulting from the reaction:
$\ce{^{14}_7N + ^4_2He ⟶ ^{17}_8O + ^1_1H} \nonumber$
The $\ce{^{17}_8O}$ and $\ce{^1_1H}$ nuclei that are produced are stable, so no further (nuclear) changes occur.
To reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors.
CERN Particle Accelerator
Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure $1$). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers.
Figure $1$: A small section of the LHC is shown with workers traveling along it. (credit: Christophe Delaere)
In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2103 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously.
Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are:
\begin{align*} \ce{^{238}_{92}U + ^1_0n} &⟶ \ce{^{239}_{92}U} && \ \ce{^{239}_{92}U} &⟶ \ce{^{239}_{93}Np + ^0_{−1}e} &&\textrm{half-life}=\mathrm{23.5\: min} \ \ce{^{239}_{93}Np } &⟶ \ce{^{239}_{94}Pu + ^0_{−1}e} &&\textrm{half-life}=\mathrm{2.36\: days} \end{align*}
Plutonium is now mostly formed in nuclear reactors as a byproduct during the decay of uranium. Some of the neutrons that are released during U-235 decay combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. It is possible to summarize these equations as:
$\mathrm{\ce{^{238}_{92}U} + {^1_0n}⟶ \ce{^{239}_{92}U} \xrightarrow{β^-} \ce{^{239}_{93}Np} \xrightarrow{β^-} \ce{^{239}_{94}Pu}}$
Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years.
Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses.
The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table $1$.
Table $1$: Preparation of Some of the Transuranium Elements
Name Symbol Atomic Number Reaction
americium Am 95 $\ce{^{239}_{94}Pu + ^1_0n ⟶ ^{240}_{95}Am + ^0_{−1}e}$
curium Cm 96 $\ce{^{239}_{94}Pu + ^4_2He ⟶ ^{242}_{96}Cm + ^1_0n}$
californium Cf 98 $\ce{^{242}_{96}Cm + ^4_2He⟶ ^{243}_{97}Bk + 2^1_0n}$
einsteinium Es 99 $\ce{^{238}_{92}U + 15^1_0n⟶ ^{253}_{99}Es + 7^0_{−1}e}$
mendelevium Md 101 $\ce{^{253}_{99}Es + ^4_2He ⟶ ^{256}_{101}Md + ^1_0n}$
nobelium No 102 $\ce{^{246}_{96}Cm + ^{12}_6C ⟶ ^{254}_{102}No + 4 ^1_0n}$
rutherfordium Rf 104 $\ce{^{249}_{98}Cf + ^{12}_6C⟶ ^{257}_{104}Rf + 4 ^1_0n}$
seaborgium
Sg
106
$\ce{^{206}_{82}Pb + ^{54}_{24}Cr ⟶ ^{257}_{106}Sg + 3 ^1_0n}$
$\ce{^{249}_{98}Cf + ^{18}_8O ⟶ ^{263}_{106}Sg + 4 ^1_0n}$
meitnerium Mt 107 $\ce{^{209}_{83}Bi + ^{58}_{26}Fe ⟶ ^{266}_{109}Mt + ^1_0n}$
20.4: Detections and Applications of Radioactivity
Learning Objectives
• Understand how the Geiger counter can be used to quantify the rate of ionization radiation.
When alpha, beta or gamma particles collide with a target, some of the energy in the particle is transferred to the target, typically resulting in the promotion of an electron to an “excited state”. In many “targets”, especially gasses, this results in ionization. Alpha, beta and gamma radiation are broadly referred to as ionizing radiation. A Geiger counter (or Geiger-Müller counter) takes advantage of this in order to detect these particles. In a Geiger tube, the electron produced by ionization of a captive gas travels to the anode and the change in voltage is detected by the attached circuitry. Most counters of this type are designed to emit an audible “click” in response to the change in voltage, and to also show it on a digital or analog meter. A simple schematic of a Geiger counter is shown in Figure \(1\).
Although scientists were not aware at the time of the Geiger counter's invention, all of us are subjected to a certain amount of radiation every day. This radiation is called background radiation and comes from a variety of natural and artificial radiation sources. Approximately 82% of background radiation comes from natural sources. These natural sources include:
1. Sources in the earth—including naturally occurring radioactive elements—which are incorporated in building materials, and also in the human body.
2. Sources from space in the form of cosmic rays.
3. Sources in the atmosphere, such as radioactive radon gas released from the earth; and radioactive atoms like carbon-14, produced in the atmosphere by bombardment from high-energy cosmic rays. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/20%3A_The_Nucleus_A_Chemists_View/20.3%3A_Nuclear_Transformations.txt |
Learning Objectives
• To understand the factors that affect nuclear stability.
Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus.
The Atomic Nucleus
Each element can be represented by the notation \(^A_Z \textrm X\), where A, the mass number, is the sum of the number of protons and the number of neutrons, and Z, the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called nucleons, and an atom with a particular number of protons and neutrons is called a nuclide. Nuclides with the same number of protons but different numbers of neutrons are called isotopes. Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways:
stable isotopes of oxygen represented in different ways
\(^A_Z \textrm X\) \(\ce{^{16}_8 O}\) \(\ce{^{17}_8 O}\) \(\ce{^{18}_8 O}\)
\(^A \textrm X\) \(\ce{^{16} O}\) \(\ce{^{17} O}\) \(\ce{^{18} O}\)
\(\textrm{element-A:}\) \(\textrm{oxygen-16}\) \(\textrm{oxygen-17}\) \(\textrm{oxygen-18}\)
Because the number of neutrons is equal to AZ, we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s natural abundance. For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17. Any nucleus that is unstable and decays spontaneously is said to be radioactive, emitting subatomic particles and electromagnetic radiation. The emissions are collectively called radioactivity and can be measured. Isotopes that emit radiation are called radioisotopes.
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.
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.
Origin of the Magic Numbers
Multiple models have been formulated to explain the origin of the magic numbers and two popular ones are the Nuclear Shell Model and the Liquid Drop Model. Unfortuneatly, both require advanced quantum mechanics to fully understand and are beyond the scope of this text.
Example \(1\)
Classify each nuclide as stable or radioactive.
1. \(\ce{_{15}^{30} P}\)
2. \(\ce{_{43}^{98} Tc}\)
3. tin-118
4. \(\ce{_{94}^{239} Pu}\)
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. \(\ce{_{90}^{232} Th}\)
2. \(\ce{_{20}^{40} Ca}\)
3. \(\ce{_8^{15} O}\)
4. \(\ce{_{57}^{139} La}\)
Answer a
radioactive
Answer b
stable
Answer c
radioactive
Answer d
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.
20.6: Nuclear Fission and Fusion
https://chem.libretexts.org/Textbook...uclear_Fission
https://chem.libretexts.org/Textbook...Nuclear_Fusion | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/20%3A_The_Nucleus_A_Chemists_View/20.5%3A_Thermodynamic_Stability_of_Nuclei.txt |
Learning Objectives
• To know the differences between ionizing and nonionizing radiation and their effects on matter.
• To identify natural and artificial sources of radiation.
Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an α particle is an ionized helium nucleus (He2+) that can act as a powerful oxidant. In this section, we describe how radiation interacts with matter and the some of the chemical and biological effects of radiation.
Ionizing versus Nonionizing Radiation
The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. Nonionizing radiation is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling.
In contrast, ionizing radiation is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions:
$\mathrm{atom + ionizing\: radiation \rightarrow ion^+ + \, {e^-}\label{Eq1}}$
Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials (Figure $1$). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle:
$\text{1 MeV/particle} = \text{96 billion J/mol}. \nonumber$
The Effects of Ionizing Radiation on Matter
The effects of ionizing radiation depend on four factors:
1. The type of radiation, which dictates how far it can penetrate into matter
2. The energy of the individual particles or photons
3. The number of particles or photons that strike a given area per unit time
4. The chemical nature of the substance exposed to the radiation
The relative abilities of the various forms of ionizing radiation to penetrate biological tissues are illustrated in Figure $2$. Because of its high charge and mass, α radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, γ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop γ rays. Because β particles are intermediate in mass and charge between α particles and γ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal.
Because of their great penetrating ability, γ rays are by far the most dangerous type of radiation when they come from a source outside the body. Alpha particles, however, are the most damaging if their source is inside the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in Table $1$.
Table $1$: Some Properties of Ionizing Radiation
Type Energy Range (MeV) Penetration Distance in Water* Penetration Distance in Air*
*Distance at which half of the radiation has been absorbed.
α particles 3–9 < 0.05 mm < 10 cm
β particles ≤ 3 < 4 mm 1 m
x-rays <10−2 < 1 cm < 3 m
γ rays 10−2–101 < 20 cm > 3 m
There are many different ways to measure radiation exposure, or the dose. The roentgen (R), which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure.Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered x-rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10−4 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose); the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram:
$\mathrm{1\: rad = 0.010\: J/kg \hspace{25 pt} 1\: Gy = 1\: J/kg \label{Eq2}}$
Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10−4 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10−5 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle.
Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the rem (roentgen equivalent in man) was devised to describe the actual amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the RBE (relative biological effectiveness) factor, which is 1 for β particles, γ rays, and x-rays and about 20 for α particles. Because actual radiation doses tend to be very small, most measurements are reported in millirems (1 mrem = 10−3 rem).
Wilhelm Röntgen
Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics.
Natural Sources of Radiation
We are continuously exposed to measurable background radiation from a variety of natural sources, which, on average, is equal to about 150–600 mrem/yr (Figure $3$). One component of background radiation is cosmic rays, high-energy particles and $\gamma$ rays emitted by the sun and other stars, which bombard Earth continuously. Because cosmic rays are partially absorbed by the atmosphere before they reach Earth’s surface, the exposure of people living at sea level (about 30 mrem/yr) is significantly less than the exposure of people living at higher altitudes (about 50 mrem/yr in Denver, Colorado). Every 4 hours spent in an airplane at greater than 30,000 ft adds about 1 mrem to a person’s annual radiation exposure.
A second component of background radiation is cosmogenic radiation, produced by the interaction of cosmic rays with gases in the upper atmosphere. When high-energy cosmic rays collide with oxygen and nitrogen atoms, neutrons and protons are released. These, in turn, react with other atoms to produce radioactive isotopes, such as $\ce{^{14}C}$:
$\ce{^{14}_7 N + ^1_0 n \rightarrow ^{14}_6 C + ^1_1p }\label{Eq3}$
The carbon atoms react with oxygen atoms to form CO2, which is eventually washed to Earth’s surface in rain and taken up by plants. About 1 atom in 1 × 1012 of the carbon atoms in our bodies is radioactive 14C, which decays by beta emission. About 5000 14C nuclei disintegrate in your body during the 15 s or so that it takes you to read this paragraph. Tritium (3H) is also produced in the upper atmosphere and falls to Earth in precipitation. The total radiation dose attributable to 14C is estimated to be 1 mrem/yr, while that due to 3H is about 1000 times less.
The third major component of background radiation is terrestrial radiation, which is due to the remnants of radioactive elements that were present on primordial Earth and their decay products. For example, many rocks and minerals in the soil contain small amounts of radioactive isotopes, such as $\ce{^{232}Th}$ and $\ce{^{238}U}$ as well as radioactive daughter isotopes, such as$\ce{^{226}Ra}$. The amount of background radiation from these sources is about the same as that from cosmic rays (approximately 30 mrem/yr). These isotopes are also found in small amounts in building materials derived from rocks and minerals, which significantly increases the radiation exposure for people who live in brick or concrete-block houses (60–160 mrem/yr) instead of houses made of wood (10–20 mrem/yr). Our tissues also absorb radiation (about 40 mrem/yr) from naturally occurring radioactive elements that are present in our bodies. For example, the average adult contains about 140 g of potassium as the $K^+$ ion. Naturally occurring potassium contains 0.0117% $\ce{^{40}K}$, which decays by emitting both a β particle and a (\gamma\) ray. In the last 20 seconds, about the time it took you to read this paragraph, approximately 40,000 $\ce{^{40}K}$ nuclei disintegrated in your body.
By far the most important source of background radiation is radon, the heaviest of the noble gases (group 18). Radon-222 is produced during the decay of238U, and other isotopes of radon are produced by the decay of other heavy elements. Even though radon is chemically inert, all its isotopes are radioactive. For example, 222Rn undergoes two successive alpha-decay events to give 214Pb:
$\ce{^{222}_{86} Rn \rightarrow ^4_2\alpha + ^{218}_{84} Po + ^4_2\alpha + ^{214}_{82} Pb } \label{Eq4}$
Because radon is a dense gas, it tends to accumulate in enclosed spaces such as basements, especially in locations where the soil contains greater-than-average amounts of naturally occurring uranium minerals. Under most conditions, radioactive decay of radon poses no problems because of the very short range of the emitted α particle. If an atom of radon happens to be in your lungs when it decays, however, the chemically reactive daughter isotope polonium-218 can become irreversibly bound to molecules in the lung tissue. Subsequent decay of $\ce{^{218}Po}$ releases an α particle directly into one of the cells lining the lung, and the resulting damage can eventually cause lung cancer. The $\ce{^{218}Po}$ isotope is also readily absorbed by particles in cigarette smoke, which adhere to the surface of the lungs and can hold the radioactive isotope in place. Recent estimates suggest that radon exposure is a contributing factor in about 15% of the deaths due to lung cancer. Because of the potential health problem radon poses, many states require houses to be tested for radon before they can be sold. By current estimates, radon accounts for more than half of the radiation exposure of a typical adult in the United States.
Artificial Sources of Radiation
In addition to naturally occurring background radiation, humans are exposed to small amounts of radiation from a variety of artificial sources. The most important of these are the x-rays used for diagnostic purposes in medicine and dentistry, which are photons with much lower energy than γ rays. A single chest x-ray provides a radiation dose of about 10 mrem, and a dental x-ray about 2–3 mrem. Other minor sources include television screens and computer monitors with cathode-ray tubes, which also produce x-rays. Luminescent paints for watch dials originally used radium, a highly toxic alpha emitter if ingested by those painting the dials. Radium was replaced by tritium (3H) and promethium (147Pr), which emit low-energy β particles that are absorbed by the watch crystal or the glass covering the instrument. Radiation exposure from television screens, monitors, and luminescent dials totals about 2 mrem/yr. Residual fallout from previous atmospheric nuclear-weapons testing is estimated to account for about twice this amount, and the nuclear power industry accounts for less than 1 mrem/yr (about the same as a single 4 h jet flight).
Example $1$
Calculate the annual radiation dose in rads a typical 70 kg chemistry student receives from the naturally occurring 40K in his or her body, which contains about 140 g of potassium (as the K+ ion). The natural abundance of 40K is 0.0117%. Each 1.00 mol of 40K undergoes 1.05 × 107 decays/s, and each decay event is accompanied by the emission of a 1.32 MeV β particle.
Given: mass of student, mass of isotope, natural abundance, rate of decay, and energy of particle
Asked for: annual radiation dose in rads
Strategy:
1. Calculate the number of moles of 40K present using its mass, molar mass, and natural abundance.
2. Determine the number of decays per year for this amount of 40K.
3. Multiply the number of decays per year by the energy associated with each decay event. To obtain the annual radiation dose, use the mass of the student to convert this value to rads.
Solution
A The number of moles of 40K present in the body is the total number of potassium atoms times the natural abundance of potassium atoms present as 40K divided by the atomic mass of 40K:
$\textrm{moles }^{40}\textrm K= 140\textrm{ g K} \times \dfrac{0.0117\textrm{ mol }^{40}\textrm K}{100\textrm{ mol K}}\times\dfrac{1\textrm{ mol K}}{40.0\textrm{ g K}}=4.10\times10^{-4}\mathrm{\,mol\,^{40}K} \nonumber$
B We are given the number of atoms of 40K that decay per second in 1.00 mol of 40K, so the number of decays per year is as follows:
$\dfrac{\textrm{decays}}{\textrm{year}}=4.10\times10^{-4}\mathrm{\,mol^{40}\,K}\times\dfrac{1.05\times10^7\textrm{ decays/s}}{\mathrm{1.00\,mol\,^{40}K}}\times\dfrac{60\textrm{ s}}{1\textrm{ min}}\times\dfrac{60\textrm{ min}}{1\textrm{ h}}\times\dfrac{24\textrm{ h}}{1\textrm{ day}}\times\dfrac{365\textrm{ days}}{1\textrm{ yr}}$
C The total energy the body receives per year from the decay of 40K is equal to the total number of decays per year multiplied by the energy associated with each decay event:
\begin{align*}\textrm{total energy per year}&=\dfrac{1.36\times10^{11}\textrm{ decays}}{\textrm{yr}}\times\dfrac{1.32\textrm{ MeV}}{\textrm{decays}}\times\dfrac{10^6\textrm{ eV}}{\textrm{MeV}}\times\dfrac{1.602\times10^{-19}\textrm{ J}}{\textrm{eV}}\&=2.87\times10^{-2}\textrm{ J/yr}\end{align*} \nonumber
We use the definition of the rad (1 rad = 10−2 J/kg of tissue) to convert this figure to a radiation dose in rads. If we assume the dose is equally distributed throughout the body, then the radiation dose per year is as follows:
\begin{align*}\textrm{radiation dose per year}&=\dfrac{2.87\times10^{-2}\textrm{ J/yr}}{\textrm{70.0 kg}}\times\dfrac{1\textrm{ rad}}{1\times10^{-2}\textrm{ J/kg}}\&=4.10\times10^{-2}\textrm{ rad/yr}=41\textrm{ mrad/yr}\end{align*} \nonumber
This corresponds to almost half of the normal background radiation most people experience.
Exercise $1$
Because strontium is chemically similar to calcium, small amounts of the Sr2+ ion are taken up by the body and deposited in calcium-rich tissues such as bone, using the same mechanism that is responsible for the absorption of Ca2+. Consequently, the radioactive strontium (90Sr) found in fission waste and released by atmospheric nuclear-weapons testing is a major health concern. A normal 70 kg human body has about 280 mg of strontium, and each mole of 90Sr undergoes 4.55 × 1014 decays/s by the emission of a 0.546 MeV β particle. What would be the annual radiation dose in rads for a 70 kg person if 0.10% of the strontium ingested were 90Sr?
Answer
5.7 × 103 rad/yr (which is 10 times the fatal dose)
Assessing the Impact of Radiation Exposure
One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in Table $2$. Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid.
Table $2$: The Effects of a Single Radiation Dose on a 70 kg Human
Dose (rem) Symptoms/Effects
< 5 no observable effect
5–20 possible chromosomal damage
20–100 temporary reduction in white blood cell count
50–100 temporary sterility in men (up to a year)
100–200 mild radiation sickness, vomiting, diarrhea, fatigue; immune system suppressed; bone growth in children retarded
> 300 permanent sterility in women
> 500 fatal to 50% within 30 days; destruction of bone marrow and intestine
> 3000 fatal within hours
Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess.
The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Both patterns are plotted in Figure $4$, but which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that all exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure.
Summary
Nonionizing radiation is relatively low in energy and can be used as a heat source, whereas ionizing radiation, which is higher in energy, can penetrate biological tissues and is highly reactive. The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas γ rays penetrate more deeply. Common units of radiation exposure, or dose, are the roentgen (R), the amount of energy absorbed by dry air, and the rad (radiation absorbed dose), the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The rem (roentgen equivalent in man) measures the actual amount of tissue damage caused by a given amount of radiation. Natural sources of radiation include cosmic radiation, consisting of high-energy particles and γ rays emitted by the sun and other stars; cosmogenic radiation, which is produced by the interaction of cosmic rays with gases in the upper atmosphere; and terrestrial radiation, from radioactive elements present on primordial Earth and their decay products. The risks of ionizing radiation depend on the intensity of the radiation, the mode of exposure, and the duration of the exposure. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/20%3A_The_Nucleus_A_Chemists_View/20.7%3A_Effects_of_Radiation_on_Matter.txt |
Organic chemistry is the study of the chemistry of carbon compounds. Why focus on carbon? Carbon has properties that give its chemistry unparalleled complexity. It forms four covalent bonds, which give it great flexibility in bonding. It makes fairly strong bonds with itself (a characteristic called catenation), allowing for the formation of large molecules; it also forms fairly strong bonds with other elements, allowing for the possibility of a wide variety of substances. No other element demonstrates the versatility of carbon when it comes to making compounds. So an entire field of chemistry is devoted to the study of the compounds and reactivity of one element.
It was thought that organic compounds could only be manufactured in living organisms, and chemistry was divided into the subfields of inorganic and organic on this basis. This subdivision persists today, but the definition of organic has changed in response to the discovery of numerous ways to make organic compounds from inorganic starting materials. Biochemistry is the study of chemical elements found in living systems, and how these elements combine to form molecules and collections of molecules which carry out the biological functions and behaviors that we associate with life.
21: Organic and Biological Chemistry
alkane
ketone
alkene
aldehyde
alkyne
imine
(Schiff base)
aromatic
carboxylic acid
alkyl halide
ester
alcohol
thioester
thiol
amide
amine
acyl phosphate
ether
acid chloride
thioether
phosphate monoester
phenol
phosphate diester
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
21.5: Polymers
Learning Objectives
• To draw structures for monomers that can undergo addition polymerization and for four-monomer-unit sections of an addition polymer.
The most important commercial reactions of alkenes are polymerizations, reactions in which small molecules, referred to in general as monomers (from the Greek monos, meaning “one,” and meros, meaning “parts”), are assembled into giant molecules referred to as polymers (from the Greek poly, meaning “many,” and meros, meaning “parts”). A polymer is as different from its monomer as a long strand of spaghetti is from a tiny speck of flour. For example, polyethylene, the familiar waxy material used to make plastic bags, is made from the monomer ethylene—a gas.
There are two general types of polymerization reactions: addition polymerization and condensation polymerization. In addition polymerization, the monomers add to one another in such a way that the polymer contains all the atoms of the starting monomers. Ethylene molecules are joined together in long chains. The polymerization can be represented by the reaction of a few monomer units:
The bond lines extending at the ends in the formula of the product indicate that the structure extends for many units in each direction. Notice that all the atoms—two carbon atoms and four hydrogen atoms—of each monomer molecule are incorporated into the polymer structure. Because displays such as the one above are cumbersome, the polymerization is often abbreviated as follows:
nCH2=CH2 [ CH2CH2 ] n
Many natural materials—such as proteins, cellulose and starch, and complex silicate minerals—are polymers. Artificial fibers, films, plastics, semisolid resins, and rubbers are also polymers. More than half the compounds produced by the chemical industry are synthetic polymers.
Some common addition polymers are listed in Table \(1\). Note that all the monomers have carbon-to-carbon double bonds. Many polymers are mundane (e.g., plastic bags, food wrap, toys, and tableware), but there are also polymers that conduct electricity, have amazing adhesive properties, or are stronger than steel but much lighter in weight.
Table \(1\): Some Addition Polymers
Monomer Polymer Polymer Name Some Uses
CH2=CH2 ~CH2CH2CH2CH2CH2CH2~ polyethylene plastic bags, bottles, toys, electrical insulation
CH2=CHCH3 polypropylene carpeting, bottles, luggage, exercise clothing
CH2=CHCl polyvinyl chloride bags for intravenous solutions, pipes, tubing, floor coverings
CF2=CF2 ~CF2CF2CF2CF2CF2CF2~ polytetrafluoroethylene nonstick coatings, electrical insulation
Medical Uses of Polymers
An interesting use of polymers is the replacement of diseased, worn out, or missing parts in the body. For example, about a 250,000 hip joints and 500,000 knees are replaced in US hospitals each year. The artificial ball-and-socket hip joints are made of a special steel (the ball) and plastic (the socket). People crippled by arthritis or injuries gain freedom of movement and relief from pain. Patients with heart and circulatory problems can be helped by replacing worn out heart valves with parts based on synthetic polymers. These are only a few of the many biomedical uses of polymers.
Key Takeaway
• Molecules having carbon-to-carbon double bonds can undergo addition polymerization.
21.6: Natural Polymers
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.
• Addition Polymers
An addition polymer is a polymer which is formed by an addition reaction, where many monomers bond together via rearrangement of bonds without the loss of any atom or molecule under specific conditions of heat, pressure, and/or the presence of a catalyst.
• Condensation Polymers
Condensation polymers are any kind of polymers formed through a condensation reaction—where molecules join together—losing small molecules as byproducts such as water or methanol, as opposed to addition polymers which involve the reaction of unsaturated monomers.
• Introduction to Polymers
Polymers are substances containing a large number of structural units joined by the same type of linkage. These substances often form into a chain-like structure. Polymers in the natural world have been around since the beginning of time. Starch, cellulose, and rubber all possess polymeric properties. Man-made polymers have been studied since 1832. Today, the polymer industry has grown to be larger than the aluminum, copper and steel industries combined.
• Molecular Weights of Polymers
Most polymers are not composed of identical molecules. The HDPE molecules, for example, are all long carbon chains, but the lengths may vary by thousands of monomer units. Because of this, polymer molecular weights are usually given as averages.
• Polyethylene
Polyethylene is the most popular plastic in the world. This is the polymer that makes grocery bags, shampoo bottles, children's toys, and even bullet proof vests. For such a versatile material, it has a very simple structure, the simplest of all commercial polymers. A molecule of polyethylene is nothing more than a long chain of carbon atoms, with two hydrogen atoms attached to each carbon atom.
• Rubber Polymers
Rubber is an example of an elastomer type polymer, where the polymer has the ability to return to its original shape after being stretched or deformed. The rubber polymer is coiled when in the resting state. The elastic properties arise from the its ability to stretch the chains apart, but when the tension is released the chains snap back to the original position. The majority of rubber polymer molecules contain at least some units derived from conjugated diene monomers.
Thumbnail: Space-filling model of a section of the polyethylene terephthalate polymer, also known as PET and PETE, a polyester used in most plastic bottles. Color code: Carbon, C (black), Hydrogen, H (white), and Oxygen, O (red). (Public Domain; Jynto). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/21%3A_Organic_and_Biological_Chemistry/21.4%3A_Hydrocarbon_Derivatives.txt |
When calculating equilibrium constants of solutions, the concentrations of each component in the solution are used to calculate K, which is the equilibrium constant. When working with concentrations, the equilibrium constant is designated as Kc. However, when working with a mixture of gases, concentrations are not used, but instead the gases' partial pressures. By first calculating the equilibrium constant in terms of pressure, designated Kp, Kc can then be calculated by using a simple formula.
How do you begin writing the equilibrium constant for a mixture of gases?
To explain this clearly, let us look at a simple example, the reaction:
2NH3(g) <---> N2(g) + 3H2(g).
1) First, each component of the mixture in the equilibrium constant is written in terms of their activities. The activity of each component is the partial pressure of each component divided by their partial pressure "reference-states." Writing the equilibrium constant in terms of partial pressures requires it to be referred to as Kp.
K = ( (aN2)(aH2)3 / (aNH3)2) eq
aN2 = PN2/Po
aH2 = PH2/Po
aNH3 = PNH3/Po
2) Next, take the activities, showed in terms of pressure, of the products raised to the power of their coefficients, and divided by the activities of the reactants. The activities of the reactants are also raised to the power of their coefficients. To simplify the expression, the partial pressure "reference-state" of each component can be divided out.
Kp = ( (PN2/Po)(PH2/Po)3 / (PNH3/Po)2 ) eq
Po ( (PN2)(PH2)3/(PNH3)2) eq
3) To relate the pressures of this expression to concentrations, the ideal gas law is used. The ideal gas law is PV=nRT, where P is the pressure, V is the volume, n is the number of moles of the substnace, R is the constant 0.08206 L atm/K mol, and T is the temperature in Kelvins. By inverting the ideal gas law equation and solving for the concentration, n/V, or moles per liter, the concentrations of the gases are expressed by their partial pressures divided by RT.
{N2} = n/v = PN2/RT
{H2} = n/V = PH2/RT
{NH3} = n/V = PNH3/RT
4) Now, since we have solved for the concentrations of each gas component, we write them in terms of concentration activities. This is the concentrations of each component divided by the concentration reference state.
aN2 = {N2}/co = (PN2/RT)/co
aH2 = {H2}/co = (PH2/RT)/co
aNH3 = {NH3}/co = (PNH3/RT)/co
The last and final step is entering these activities into the expression we earlier derived in step 2 and dividing out RT.
Kp = Po( ({N2}RT)({H2}RT)3/({NH3}RT)2 ) eq
Kp = (Po/RT)({N2}{H2}3/{NH3}2) eq
This gives the equilibrium constant expression for a mixture of gases (Kp).
Kp and Kc can be related by setting the two equations equal to each other and divided out the reference states. This gives the simple formula:
Kp = Kc(RT)difference in coefficients of gas components only
Please Note:
Po is the partial pressure reference-state and it equals about 1 atm
co is the concentration reference-state and it equals about 1mol/L
eq is the term used to indicate that the components are at equilibrium
{x} is the concentration of the component x where x is a variable used to indicate a substance
Equilibrium constant expression do not include those components in a reaction that are pure solids or liquids (please refer to the corresponding internal link below)
Practice Problems
1) Gas A and Gas B react to form Gas C. The reaction performed can be written as A(g)+B(g)
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C(g). At equilibrium, the partial pressures of A, B, and C are 1 atm, 0.50 atm, and 0.75 atm respectively. Find Kp for the reaction.
Solution:
First, write out the equilibrium constant expression for Kp. Kp= Po(
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)eq
Remember that Po= 1 bar which is essentially equal to 1 atm. Next, plug in the partial pressures for the corresponding gases.
Kp=
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)
=1.5
2) A(g), at 750 mmHg, reacts with B(g), at 760 mmHg, to form C(g). Kp for the reaction is 1.5 X 10-5 and the reaction at equilibrium can be written A(g)+B(g)
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2C(g). Find the partial pressure of C.
Solution
First set up the equilibrium constant expression for Kp. Kp= Po(
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C is raised to the power of 2 since 2 is the coefficient of C. Since we know Kp we can subsitute that into the equation. Also, convert the partial pressure of A and B to atm. PA= 750 mmHg X
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at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
=0.98684 & PB= 1 atm
1.5 X 105 =
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[11]/span/span/span/span/span/span/span, line 1, column 1
at wiki.page()
at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
(1.5 X10-5)(0.98684)= C2
sqrt(1.48026X10-5)=C
C= 3.8474147X10-3 or C= 3.9X10-3
3) For the following reaction at equilibrium 2SO2(g)+ O2
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[15]/span, line 1, column 1
at wiki.page()
at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
2SO3(g) Kp is 3.4 at 1000K. What is Kc for the reaction?
Solution:
We know that Kp=Kc(RT)(Difference in coefficients of gaseous products and reactants)
We also know Kp= 3.4, R is a constant which is 0.08206L atm mol-1K-1, T=1000K
(RT) is raised to the power of (Difference in coefficients of gaseous products and reactants) so since 2SO2(g)+ 1O2
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[19]/span, line 1, column 1
at wiki.page()
at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
2SO3(g) (Difference in coefficients of gaseous products and reactants)= (2)-(2+1)= -1
Rewrite the equation, Kp=Kc(RT)(Difference in coefficients of gaseous products and reactants), to solve for Kc and subsitute in the values. ^Kc=
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[20]/span/span, line 1, column 1
at wiki.page()
at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
Kc= 2.79X102
4) For the following reaction at equilibrium, N2(g)+ 3H2(g)
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[22]/span, line 1, column 1
at wiki.page()
at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
2NH3(g), Kp is 2.25X10-6. If the values of the partial pressures of N2 and NH3 are 3.5X10-3 and 2.0X10-5 respectively, what is the partial pressure of H2?
Solution
Set up the equilibrium constant expression for Kp. Kp= Kp= Po
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[24]/span/span, line 1, column 1
at wiki.page()
at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
eq
Substitute the values in to the correct places.
2.25X10-6=
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[26]/span/span, line 1, column 1
at wiki.page()
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```
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[27]/span/span[1], line 1, column 1
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at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
=
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[27]/span/span[2], line 1, column 1
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```
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[28]/span/span, line 1, column 1
at wiki.page()
at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
=0.0507936508
Take the cubed root of the number to find the Partial pressure of H2
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[30]/span/span/span, line 1, column 1
at wiki.page()
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```
=0.37 atm
5) 1 mol of A and 1 mol of B are placed in a 2.0 L flask. The following reaction at equilibrium is established at 500K. A(g)+B(g)
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[31]/span/span/span/span/span/span/span/span/span, line 1, column 1
at wiki.page()
at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
C(g) Kp= 15.0 For the equilibrium established, find the partial pressure of C.
Solution
First, we must find the partial pressures of A and B. We know that by using P=
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[33]/span/span/span[1]/span/span, line 1, column 1
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```
, we can say PA=PB=
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[33]/span/span/span[3]/span, line 1, column 1
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```
=
20.515
Now create an ICE table.
A(g) + B(g)
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[35]/span/span/span/span/span/span/span, line 1, column 1
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```
C(g)
I 20.515 20.515 0
C -x -x +x
E (20.515-x) (20.515-x) (x)
Now set up the equilibrium constant expression for Kp= 15.0=
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[39]/span/span, line 1, column 1
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```
15(20.515-x)2= x Use FOIL
15(420.865225-41.03x+x2)= x distribute & multiply by 15
15x2-615.45x+6312.978375=x subtract the x to make the equation equal to 0
15x2-616.45x+6312.978375=0 Use the quadratic formula
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[44]/span/span/span[1], line 1, column 2
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```
and
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[44]/span/span/span[2], line 1, column 2
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at (Bookshelves/General_Chemistry/Map:_Chemistry_(Zumdahl_and_Decoste)/6:_Chemical_Equilibrium/6.3:_Equilibrium_Expressions_Involving_Pressures), /content/body/p/span, line 1, column 11
```
a=15 b=-616.45 c= 6312.978375
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[46]/span/span/span[1], line 1, column 2
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```
and
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```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[46]/span/span/span[2], line 1, column 2
at wiki.page()
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```
We find that there are two solutions. PC= 19.38 or 21.7
19.38 is the correct solution since 21.7 is too large.
Contributors and Attributions
• Kathryn Rashel
• Lisa Peterson
• Kyle Catabay (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/6%3A_Chemical_Equilibrium/6.3%3A_Equilibrium_Expressions_Involving_Pressures.txt |
Effective Concentrations in Gases
In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution. By convention, activity is treated as a dimensionless quantity, although its value depends on customary choices of standard state for the species. The activity of pure substances in condensed phases (solid or liquids) is normally taken as unity (the number 1). Activity depends on temperature, pressure and composition of the mixture, among other things. For gases, the activity is the effective partial pressure, and is usually referred to as fugacity.
The difference between activity and other measures of composition arises because molecules in non-ideal gases or solutions interact with each other, either to attract or to repel each other. The activity of an ion is particularly influenced by its surroundings. The use of activities allows chemists to explain various discrepancies between ideal solutions and real solutions. The mathematical description of an ideal solution must be modified to describe a real solution, just as the law for ideal gases (PV = nRT) must be modified to describe real gases.
Example $1$: Ideal vs. Real Gases
Compare the pressures predicted for one mole of ethane at 298.15 K under the following equations of states:
1. ethane is an ideal gas or
2. ethane is a van der Waal gas
What is the deviation of the two?
Solution
a) Ideal gas law equation of state: Calculate the pressure of 1.000 mole of ethane at 298.15 K in a 1.000 L flask using the ideal gas law.
\begin{align} P_{ideal}V &=nRT \label{ideal gas} \[4pt] P_{ideal} &=\dfrac{nRT}{V} \nonumber \[4pt] &= \dfrac{ (1\;mole)(0.0821 L \; atm \; K^{-1} \; mol^{-1}) ( 298.15 \;K)}{1 \;L} \nonumber \[4pt] &= 24.47\; atm \nonumber \end{align}
b) Van der Waal's equation of state: Calculate the pressure of 1.00 mole of ethane at 298 K in a 1.00 L flask using the van der Waals equation. The van der Waals constants for ethane can found in Table A8.
• $a = 5.492 \;atm \; \dfrac{L^2}{M^2}$ and
• $b = 0.06499 \;L/m$
\begin{align} \left(P_{vdW} + \dfrac{an^2}{V^2}\right) (V − nb) &=nRT \label{vdW equation} \[4pt] P_{vdW} &= \dfrac{nRT}{V − nb} - \dfrac{an^2}{V^2} \nonumber \[4pt] &= 20.67\; atm \nonumber \end{align}
Calculate the percent error between the $P_{ideal}$ and the $P_{vdW}$
$Error = \dfrac{P_{vdW} - P_{ideal}}{P_{vdW}} = 18.36\% \nonumber$
An error this large is often too big to ignore when carrying out a gas-phase reaction or designing a vessel in which to carry out such a reaction.
There are two ways to deal with real systems that deviate appreciably from ideal conditions:
1. Use a more accurate phenomenologically (i.e., real) Equation of State like the van der Waal' (Equation $\ref{vdW equation}$) that models the system more accurately. This must explicitly address intermolecular forces and other effects that exist in real system and the concentration used is 1 mol/L (i.e., the real concentration).
2. Use an ideal equation pf state like the ideal gas Equation of State in Equation $\ref{ideal gas}$), but use an "effective concentration" of 0.816 mol/L to generate the observed pressure (that is, the gas behaves as if it has a reduced concentration of 100%-18.36% = 81.6% of the real concentration).
Effective Concentrations in Solutions
In a similar fashion, the difference between the calculated solute concentrations in an ideal solution and in a real solution can lead to wide variations in experimental results. The following three examples compare the results obtained when formal concentrations are used (assuming ideality) and when activities are used (assuming non-ideality). Just like gases, "ideal solutions" have certain predictable physical properties (e.g. colligative properties) that real solutions often deviate from. As with the van der Waal equation in Example 1, this deviation originates from solute-solvent, solvent-solvent and solute-solute interactions. The magnitude of this non-ideality naturally greater with higher with solute concentrations and with greater intermolecular interaction (e.g., ions vs. non-charged species). Lewis introduced idea of 'effective concentration' or 'activity' to deal with this problem by allowing an "ideal solution" description for non-ideal solutions.
Since the Van der Waals equation describes real gases instead of the ideal gases law, activity can be used in place of concentration to describe the behavior of real solutions vs. ideal solutions.
The activity of a substance (abbreviated as a) describes the effective concentration of that substance in the reaction mixture. Activity takes into account the non-ideality of the reaction mixture, including solvent-solvent, solvent-solute, and solute-solute interactions. Thus, activity provides a more accurate description of how all of the particles act in solution. For very dilute solutions, the activities of the substances in the solution closely approach the formal concentration (what the calculated concentration should be based on how much substance was measured out.) As solutions get more concentrated, the activities of all of the species tend to be smaller than the formal concentration. The decrease in activity as concentration increases is much more pronounced for ions than it is for neutral solutes.
Activities are actually unitless ratios that compare an effective pressure or an effective concentration to a standard state pressure or concentration (the correct term for the effective pressure is fugacity). There are several ways to define standard states for the different components of a solution, but a common system is
• the standard state for gas pressure, P°, is 1 bar (often approximated with 1 atm)
• the standard state for solute concentration, C°, is 1 molal (moles solute/kg solvent) for dilute solutions. Often molality is approximated with molarity (moles solute/Liter solution).
• the standard state for a liquid is the pure liquid
• the standard state for a solid is the pure solid
Thus, when we discuss the activity of a gas, we actually are discussing the ratio of the effective pressure to the standard state pressure:
$a_{gas} = \dfrac{P}{P^º} \label{1}$
• $a_{gas}$ is a ratio with no units.
Likewise, the activity of a solute in solution would be:
$a_{solute} = \dfrac{C}{C^º} \label{2}$
• $a_{solute}$ is a ratio with no units.
For all solids, the activity is a ratio of the concentration of a pure solid to the concentration of that same pure solid
$a_{solid} = \dfrac{C_{\text{effective solid}}}{C^º_{\text{effective solid}}} = 1 \label{3}$
• $a_{solid}$ always has a value of 1 with no units.
For all liquids, the activity is a ratio of the concentration of a pure liquid to the concentration of that same pure liquid:
$a_{liquid} =\dfrac{C_{\text{effective liquid}}}{C^º_{\text{effective liquid}}} = 1 \label{4}$
• $a_{liquid}$ always has a value of 1 with no units.
For most experimental situations, solutions are assumed to be dilute with respect to the solvent. This assumption implies the solvent can be approximated with pure liquid. According to Raoult's Law, the vapor pressure of the solvent in a solution is equal to the mole fraction of the solvent in the solution times the vapor pressure of the pure solvent:
$\chi= \dfrac{P}{P^º} \label{5}$
The mole fraction of solvent in a dilute solution is approximately 1, so the vapor pressure of the solution is essentially identical to the vapor pressure of the pure solvent. This means that the activity of a solvent in dilute solution will always has a value of 1, with no units.
Activity indicates how many particles "appear" to be present in the solution, which is different from how many actually are present. Hence, activity is a "fudge factor" to ideal solutions that correct the true concentration.
• $a_{gas}$ is a ratio with no units.
• $a_{solute}$ is a ratio with no units.
• $a_{solid}$ is always 1 with no units.
• $a_{liquid}$ is always 1 with no units.
Estimating Activities
The activity of a substance can be estimated from the nominal concentration of that substance (C) by using an activity coefficient, $\gamma$:
$a = \gamma \cdot [C] \label{6}$
The value of $\gamma$ depends upon the substance, the temperature, and the concentration of all solute particles in the solution. The lower the concentration of all solute particles in the solution, the closer the value of $\gamma$ for each solute approaches 1:
$\lim_{[C] \rightarrow 0} \gamma \rightarrow 1 \label{6a}$
Therefore, as $\gamma$ approaches 1, the value of $a$ for the solute approaches C.
$\lim_{\gamma \rightarrow 1} a \rightarrow [C] \label{6b}$
The activity coefficient for a nonvolatile, neutral solute is often estimated by non-linear curve fitting, taking into account the molality of the solute and the activity of the solvent (usually its vapor pressure). In most situations, it is more practical to look up the values of the activity coefficient for a given solute than it is to carry out the curve fitting.
Table $1$: Activity Coefficients
m/(mol kg-1) HCl LiCl NaCl LiNO3 NaNO3
0.01 0.904 0.903 0.902 0.903 0.900
0.02 0.875 0.873 0.870 0.872 0.866
0.05 0.830 0.825 0.820 0.825 0.811
0.10 0.796 0.790 0.778 0.788 0.762
0.2 0.767 0.757 0.735 0.752 0.703
0.4 0.755 0.740 0.693 0.728 0.638
0.6 0.763 0.743 0.673 0.727 0.599
0.8 0.783 0.755 0.662 0.733 0.570
1.0 0.809 0.774 0.657 0.743 0.548
1.2 0.840 0.796 0.654 0.757 0.530
1.4 0.876 0.823 0.655 0.774 0.514
1.6 0.916 0.853 0.657 0.792 0.501
1.8 0.960 0.885 0.662 0.812 0.489
2.0 1.009 0.921 0.668 0.835 0.478
2.5 1.147 1.026 0.688 0.896 0.455
3.0 1.316 1.156 0.714 0.966 0.437
3.5 1.518 1.317 0.746 1.044 0.422
4.0 1.762 1.510 0.783 1.125 0.408
4.5 2.04 1.741 0.826 1.215 0.396
5.0 2.38 2.02 0.874 1.310 0.386
Estimating the activity coefficient of electrolytes (solutes that dissolve or react with the solvent to form ions) depends upon the number of ions formed by the dissociation of the solute in solution or the reaction of the solute with the solution, because each ion formed is dealt with individually. In a theoretical, infinitely dilute ideal solution, an electrolyte would dissociate or react completely to form an integer number of independent ions. For example, 1 mole of NaCl would dissociate to form 2 moles of ions (1 mole of Na+ ions and 1 mole of Cl- ions). In reality, it is found that electrolytes almost always act as if they contain fewer moles of ions than expected based on the formal concentration. This non-ideality is attributed to the degree of dissociation/reaction of the solute, to the solute-solvent interactions such as complex ion formation, and to the solute-solute interactions such as ion pairing. An activity coefficient incorporates the particle interactions into a single term that modifies the formal concentration to give an estimate of the effective concentration, or activity, of each ion.
At infinite dilution, $gamma$ is solely determined by the Debye-Hückel limiting law (Equation $\ref{Debye Equation}$) and depends only on the number and charges of the cations and anions. This means that the same limiting mean ionic activity coefficient is found for sodium chloride and potassium chloride and that also the values for the 2-1 and 1-2 salts sodium sulfate and calcium chloride are identical. At higher electrolyte concentrations though, these values change very strongly and are usually modeled using empirical parameters regressed to the experimental data.
Electrolytes almost always act as if they contain fewer moles of ions than expected based on the formal concentration.
Single ion activity coefficients are calculated using various forms of the Debye-Hückel equation:
$\log \gamma = \dfrac{-0.51 z^2 \sqrt{\mu}}{1+ \dfrac{\alpha\sqrt{\mu}}{305}} \label{Debye Equation}$
This equation takes into account the solution environment as well as the individual characteristics of the specific ion of interest. It is not difficult to calculate single ion activity coefficients, but tables of these activity coefficients for many common ions in solutions of various concentrations are available (e.g., Table $1$).
Applications of Activities
The law of mass action states that a reaction at a constant temperature will proceed spontaneously and predominantly in one direction until a constant ratio of concentrations of products and reactants is obtained. For the generic reaction
$aA + bB \rightleftharpoons cC + dD \label{8}$
the ratio of concentrations (called the mass action expression or equilibrium constant expression) is
$\dfrac{[C]^c[D]^d}{[A]^a[B]^b} = Q \label{9}$
where [ ] represents concentration in
$\dfrac{\text{moles of solute}}{\text{Liter of solution}} \label{10}$\\\\\\\hhhufd
This ratio can take on any value greater than zero, depending on the reaction conditions. Thus, it is often called the instantaneous reaction quotient, Q. The term “instantaneous” signifies that the reaction will (and is) proceeding spontaneously to reach a constant ratio of products and reactants. When the reaction attains that constant ratio of products and reactants, it has reached a state of dynamic equilibrium, and the ratio of concentrations can be represented by the symbol K, the equilibrium constant:
$\dfrac{[C]_{eq}^c[D]_{eq}^d}{[A]_{eq}^a[B]_{eq}^b} = K \label{11}$
Laws of mass action and equilibrium constants are discussed in most general chemistry textbooks, but they are often discussed as if they were describing ideal systems. For instance, if all of the substances are gases, partial pressures are used in the mass-action expression. If the substances are in solution, molarities are used in the mass-action expression. To be thermodynamically correct, however, the activities of the substances must be compared in the mass-action expression.
Activities are needed for precise work because, unlike concentrations, activities contain information about the effects of the solvent and other surrounding particles on the behavior of the particles of interest. Using any unit of comparison other than activities will give an incorrect value for K, but it is assumed that the approximate value is close enough to the true value for most situations. Many tables list K values to 2-3 significant digits, but this degree of precision is valid only under the exact experimental conditions used to obtain those values.
The Derivation of Mass Action Expressions
Given all of the above information on activities, it is now possible to show how a true mass-action equation involving activities can be approximated by a mass-action equation involving molarites. It should be noted that just as the activities were unitless ratios, the molalities and molarities that appear in the following approximations should also be thought of as unitless ratios of concentrations divided by the standard state concentration.
Example $2$: A Solution of Ammonia in Water
Starting with the mass action equation in terms of activities, show all the approximations needed to obtain the mass action equation in terms of molar concentrations for the reaction:
$NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH^-_{(aq)}$
Solution
$Q = \dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}a_{H_2O}} = \dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}(1)}$
• assume $a_{H_2O} \approx 1$ because water is the solvent in a dilute solution
$Q=\dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}}=\dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{m_{NH_4^+}m_{OH^-}}{m_{NH_3}}$
• assume activity = (activity coefficient)(molality)
$Q =\dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{m_{NH_4^+}m_{OH^-}}{m_{NH_3}} \approx \dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{[NH_4^+][OH^-]}{[NH_3]}$
• assume molarity [ ] ≈ molality in dilute solutions
$Q \approx \dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{[NH_4^+][OH^-]}{[NH_3]}\approx \dfrac{[NH_4^+][OH^-]}{[NH_3]}$
• assume $\dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \approx 1$ because the solution is dilute
Therefore the final mass action equation typically used for this reaction is
$Q \approx \dfrac{[NH_4^+][OH^-]}{[NH_3]}$
Example $3$: A Reaction of Two Solids that Produces a Solution
Starting with the mass action equation in terms of activities, show all the approximations needed to obtain the mass action equation in terms of molar concentrations for the reaction:
$Ba(OH)_2 \cdot 8H_2O(s) + 2NH_4NO_3(s) \rightleftharpoons Ba^{2+}(aq) + 2NO_3^-(aq) + 2NH_3(aq) + 10H_2O(l)$
Solution
$Q = \dfrac{a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2a_{H_2O}^{10}}{a_{Ba(OH)_2·8H_2O}a_{NH_4NO_3}} = \dfrac{a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2·(1)}{(1)·(1)} = a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2$
• a =1 for solids and for solvent
$Q = a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2 = \gamma_{Ba^{2+}} \gamma_{NO_3^-}^2\gamma_{NH_3}^2m_{Ba^{2+}}m_{NO_3^-}^2m_{NH_3}^2$
• activity = (activity coefficient)(molality)
$Q = \gamma_{Ba^{2+}} \gamma_{NO_3^-}^2 \gamma_{NH_3}^2m_{Ba^{2+}} {m_{NO_3^-}}^2 m_{NH_3}^2 \approx \gamma_{Ba^{2+}}\gamma_{NO_3^-}^2 \gamma_{NH_3}^2 [Ba^{2+}] [NO_3^-]^2 [NH_3]^2$
• molarity [ ] ≈ molality in dilute solutions
$Q \approx \gamma_{Ba^{2+}}\gamma_{NO_3^-}^2 \gamma_{NH_3}^2[Ba^{2+}][NO_3^-]^2[NH_3]^2 \approx [Ba^{2+}][NO_3^-]^2[NH_3]^2$
• $\gamma_{Ba^{2+}}\gamma_{NO_3^-}^2\gamma_{NH_3}^2 \approx 1$ because the solution is dilute
$Q \approx [Ba^{2+}][NO_3^-]^2[NH_3]^2$
Salt
Determine the molar solubility $\dfrac {moles}{Liter}$ of the slightly soluble solid, BaSO4, inPure Water and An Aqueous 0.1 M NaCl Solution?
Solution
Pure Water
Barium sulfate is a solid that is slightly soluble in water, with a Ksp value of 1.1 x 10-10:
$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$
This solution is dilute enough that the Ba2+, SO42-, OH-, and H+ ions will not affect each other greatly, thus the activity of the ions closely approaches their formal concentration. In this nearly ideal aqueous solution, the mass action expression would be
$1.1 \times 10^{-10} = a_{Ba} · a_{SO_4} \approx [Ba^{2+}][SO_4^{2-}]$
Remember that BaSO4 is a solid, with an activity equal to 1. At equilibrium, the [Ba2+] = [SO42-] = 1.05 x 10-5 M.
An Aqueous 0.1 M NaCl Solution
A saturated aqueous solution of BaSO4 that also is 0.1 M in NaCl is no longer near to ideality. The Na+ and Cl- ions surround the Ba2+ and SO42- ions and prevent these ions from being able to reform solid BaSO4 readily as they did in pure water. The activities of the Ba2+ and the SO42- ions will be lower than their formal concentrations. However, the product of the activities must still be equal to the true (thermodynamic) equilibrium constant.
$1.1 x 10^{-10} = a_{Ba}·a_{SO_4} = \gamma_{Ba}[Ba^{2+}]·\gamma_{SO_4}[SO_4^{2-}]$
With the total amount of Na+, Cl-, Ba2+, SO42-, OH-, and H+ ions in the solution, $\gamma_{Ba}$ = 0.38 and $\gamma_{SO_4}$ = 0.355.
$1.1 x 10^{-10} = a_{Ba}·a_{SO_4} = (0.38)[Ba^{2+}](0.355)[SO_4^{2-}]$
$8.2 x 10^{-10} =[Ba^{2+}][SO_4^{2-}]$
$[Ba^{2+}] = [SO_4^{2-}] = 2.9 x 10^{-5}$
The net result is that more solid BaSO4 will dissolve in the 0.1M NaCl solution than in water, and the experimental equilibrium constant will seem to be larger than the thermodynamic equilibrium constant.
Colligative Properties
The van 't Hoff factor, i, is a term that often appears in colligative property calculations to account for the fact that electrolytes will form two or more moles of ions per every mole of electrolyte. In most cases, the solutions are treated as if they are ideal, in which case i will equal an integer representing the total number of independent ions per one formula unit of the solute (Table 2).
Table $2$: Integer van 't Hoff factors for Colligative Properties
Compound
i
Sucrose 1
NaCl 2
MgBr2 3
CaCl2 3
Na3PO4 4
Al2(SO4)3 5
The van 't Hoff factor is actually rarely an integer, and was, in fact, developed to take into account the non-ideality of solutes. Tables listing the i values for specific compounds in specific solutions are available, but it is also possible to use activities to estimate to effective concentrations of ions in solution for use in colligative property calculations.
Example $5$
What is the freezing point of a 0.1 m BaCl2 aqueous solution?
Solution
The calculation for an ideal solution would be $\Delta T = mki$, where $m = 0.1 molal$, $k = 1.86\dfrac{ºC}{molal}$, and $i = 3$. The resulting $\Delta T$ is
$\Delta T = (0.1 molal)(1.86\dfrac{ºC}{molal})(3) = 0.558ºC$
if non-ideality is assumed, the calculation becomes
$\Delta T = \gamma_{Ba} \cdot (\gamma_{Cl})^2•(m_{Ba})(m_{Cl})^2$
Substituting in the estimated $\gamma$ values of $\gamma_{Ba} = 0.38$ and$\gamma_{Cl} = 0.755$, the ion activities are
• $a_{Ba} = (0.38)(0.1) = 0.038$
• $a_{Cl} =(0.755)(0.2) = 0.151$
and the $\Delta T$ is
$\Delta T = (0.038 + 0.151)(1.86 ºC) = 0.351 ºC$
The $\Delta T$ obtained using activities is lower than the $\Delta T$ obtained when using an integer value for i because the activity values take into account the fact that the ions in the solution are not able to act as free and independent particles because of their interactions with each other and with the solvent, | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/6%3A_Chemical_Equilibrium/6.4%3A_Activity_is_an_Effective_Concetration.txt |
The equilibrium constant, K, expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.This article explains how to write equilibrium constant expressions, and introduces the calculations involved with both the concentration and the partial pressure equilibrium constant.
Homogeneous Reactions
A homogeneous reaction is one where the states of matter of the products and reactions are all the same (the word "homo" means "same"). In most cases, the solvent determines the state of matter for the overall reaction. For example, the synthesis of methanol from a carbon monoxide-hydrogen mixture is a gaseous homogeneous mixture, which contains two or more substances:
$CO (g)+ 2H_2 (g) \rightleftharpoons CH_3OH (g)$
At equilibrium, the rate of the forward and reverse reaction are equal, which is demonstrated by the arrows. The equilibrium constant, however, gives the ratio of the units (pressure or concentration) of the products to the reactants when the reaction is at equilibrium.
The synthesis of ammonia is another example of a gaseous homogeneous mixture:
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
Heterogeneous Reactions
A heterogeneous reaction is one in which one or more states within the reaction differ (the Greek word "heteros" means "different"). For example, the formation of an aqueous solution of lead(II) iodide creates a heterogeneous mixture dealing with particles in both the solid and aqueous states:
$PbI_{2 (s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2I^-_{(aq)}$
The decomposition of sodium hydrogen carbonate (baking soda) at high elevations is another example of a heterogeneous mixture, this reaction deals with molecules in both the solid and gaseous states:
$2NaHCO_{3 (s)} \rightleftharpoons Na_2CO_{3 (s)} + H_2O_{ (g)} + CO_{2 (g)}$
$C_{(s)} + O_{2 (g)} \rightleftharpoons CO_{2 (g)}$
This difference between homogeneous and heterogeneous reactions is emphasized so that students remember that solids, pure liquids, and solvents are treated differently than gases and solutes when approximating the activities of the substances in equilibrium constant expressions.
Writing Equilibrium Constant Expressions
The numerical value of an equilibrium constant is obtained by letting a single reaction proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. The ratio of the product concentrations to reactant concentrations is calculated. Because the concentrations are measured at equilibrium, the equilibrium constant remains the same for a given reaction independent of initial concentrations. This knowledge allowed scientists to derive a model expression that can serve as a "template" for any reaction. This basic "template" form of an equilibrium constant expression is examined here.
Equilibrium Constant of Activities
The thermodynamically correct equilibrium constant expression relates the activities of all of the species present in the reaction. Although the concept of activity is too advanced for a typical General Chemistry course, it is essential that the explanation of the derivation of the equilibrium constant expression starts with activities so that no misconceptions occur. For the hypothetical reaction:
$bB + cC \rightleftharpoons dD + eE$
the equilibrium constant expression is written as
$K = \dfrac{a_D^d ·a_E^e}{a_B^b · a_C^c}$
*The lower case letters in the balanced equation represent the number of moles of each substance, the upper case letters represent the substance itself.
• If $K > 1$ then equilibrium favors products
• If $K < 1$ then equilibrium favors the reactants
Equilibrium Constant of Concentration
To avoid the use of activities, and to simplify experimental measurements, the equilibrium constant of concentration approximates the activities of solutes and gases in dilute solutions with their respective molarities. However, the activities of solids, pure liquids, and solvents are not approximated with their molarities. Instead these activities are defined to have a value equal to 1 (one).The equilibrium constant expression is written as $K_c$, as in the expression for the reaction:
$HF_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + F^-_{(aq)}$
$K_c = \dfrac{a_{H_3O^+}· a_{F^-}}{a_{HF} · a_{H_2O}} ≈ \dfrac{[H_3O^+][F^-]}{[HF](1)} = \dfrac{[H_3O^+][F^-]}{[HF]}$
Here, the letters inside the brackets represent the concentration (in molarity) of each substance. Notice the mathematical product of the chemical products raised to the powers of their respective coefficients is the numerator of the ratio and the mathematical product of the reactants raised to the powers of their respective coefficients is the denominator. This is the case for every equilibrium constant. A ratio of molarities of products over reactants is usually used when most of the species involved are dissolved in water. A ratio of concentrations can also be used for reactions involving gases if the volume of the container is known. .
Equilibrium Constant of Pressure
Gaseous reaction equilibria are often expressed in terms of partial pressures. The equilibrium constant of pressure gives the ratio of pressure of products over reactants for a reaction that is at equilibrium (again, the pressures of all species are raised to the powers of their respective coefficients). The equilibrium constant is written as $K_p$, as shown for the reaction:
$aA_{(g)} + bB_{(g)} \rightleftharpoons gG_{(g)} + hH_{(g)}$
$K_p= \dfrac{p^g_G \, p^h_H}{ p^a_A \,p^b_B}$
• Where $p$ can have units of pressure (e.g., atm or bar).
Conversion of Kc to Kp
To convert Kc to Kp, the following equation is used:
$K_p = K_c(RT)^{\Delta{n_{gas}}}$
where:
• R=0.0820575 L atm mol-1 K-1 or 8.31447 J mol-1 K-1
• T= Temperature in Kelvin
• Δngas= Moles of gas (product) - Moles of Gas (Reactant)
Reaction Quotient
Another quantity of interest is the reaction quotient, $Q$, which is the numerical value of the ratio of products to reactants at any point in the reaction. The reaction quotient is calculated the same way as is $K$, but is not necessarily equal to $K$. It is used to determine which way the reaction will proceed at any given point in time.
$Q = \dfrac{[G]^g[H]^h}{[A]^a[B]^b}$
• If $Q > K$, then the reactions shifts to the left to reach equilibrium
• If $Q < K$, then the reactions shifts to the right to reach equilibrium
• If $Q = K$ then the reaction is at equilibrium
The same process is employed whether calculating $Q_c$ or $Q_p$.
Heterogeneous Mixture
The most important consideration for a heterogeneous mixture is that solids and pure liquids and solvents have an activity that has a fixed value of 1. From a mathematical perspective, with the activities of solids and liquids and solvents equal one, these substances do not affect the overall K or Q value. This convention is extremely important to remember, especially in dealing with heterogeneous solutions.
Example $1$
In a hypothetical reaction:
$aA_{(s)} + bB_{(l)} \rightleftharpoons gG_{(aq)} + hH_{(aq)}$
The equilibrium constant expression is written as follows:
$K_c = \dfrac{[G]^g[H]^h}{1 \times 1} = [G]^g[H]^h$
In this case, since solids and liquids have a fixed value of 1, the numerical value of the expression is independent of the amounts of A and B. If the product of the reaction is a solvent, the numerator equals one, which is illustrated in the following reaction:
$H^+_{(aq)} + OH^–_{(aq)} \rightarrow H_2O_{ (l)}$
The equilibrium constant expression would be:
$K_c= \dfrac{1}{ [H^+][OH^-]}$
which is the reciprocal of the autoionization constant of water ($K_w$)
$K_c = \dfrac{1}{K_w}=1 \times 10^{14}$
Manipulation of Constants
The equilibrium constant expression must be manipulated if a reaction is reversed or split into elementary steps. When the reaction is reversed, the equilibrium constant expression is inverted. The new expression would be written as:
$K'= \dfrac{1}{\dfrac{[G]^g[H]^h}{[A]^a[B]^b}} = \dfrac{[A]^a[B]^b}{[G]^g[H]^h}$
When there are multiple steps in the reaction, each with its own K (in a scenario similar to Hess's law problems), then the successive K values for each step are multiplied together to calculate the overall K.
Activities
Because the concentration of reactants and products are not dimensionless (i.e. they have units) in a reaction, the actual quantities used in an equilibrium constant expression are activities. Activity is expressed by the dimensionless ratio $\frac{[X]}{c^{\circ}}$ where $[X]$ signifies the molarity of the molecule and c is the chosen reference state:
$a_b=\dfrac{[B]}{c^{\circ}}$
For gases that do not follow the ideal gas laws, using activities will accurately determine the equilibrium constant that changes when concentration or pressure varies. Thus, the units are canceled and $K$ becomes unitless.
Practice Problems
1. Write the equilibrium constant expression for each reaction.
1. $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
2. $N_2O_{ (g)} + \dfrac{1}{2} O_{2(g)} \rightleftharpoons 2NO_{(g)}$
3. $Cu_{(s)} + 2Ag^+_{(aq)} \rightleftharpoons Cu^{+2}_{(aq)} + 2Ag_{(s)}$
4. $CaCO_{3 (g)} \rightleftharpoons CaCO_{(s)} + CO_{2 (g)}$
5. $2NaHCO_{3 (s)} \rightleftharpoons Na_2CO_{3 (s)} + CO_{2 (g)} + H_2O_{ (g) }$
2. What is the $K_c$ of the following reaction? $2SO_{2 (g)} + O_{2 (g)} \rightleftharpoons 2SO_{3 (g)}$ with concentration $SO_{2(g)} = 0.2 M O_{2 (g)} = 0.5 M SO_{3 (g)} = 0.7 \;M$ Also, What is the $K_p$ of this reaction? At room temperature?
3. For the same reaction, the differing concentrations: $SO_{2 (g)} = 0.1\; M O_{2(g)} = 0.3\; M \;SO_{3 (g)} = 0.5\; M$ Would this go towards to product or reactant?
4. Write the Partial Pressure Equilibrium: $C_{(s)} + O_{2 (g)} \rightarrow CO_{2 (g)}$
5. Write the chemicl reaction for the following equilibrium constant: $K_p= \dfrac{P^2_{HI}}{P_{H_2} \times P_{I_2}}$
• For more information on equilibrium constant expressions please visit the Wikipedia site: http://en.Wikipedia.org/wiki/Equilibrium_constant
• The image below can be found here: image.tutorvista.com/content/chemical-equilibrium/reaction-rate-time-graph.gif
Answers to Practice Problems
1. $K_c = \dfrac{ [SO_3]^2}{[O_2][SO_2]^2}$
2. $Kc = \dfrac{ [NO]^2}{[O_2]^{0.5}[N_2O]}$
3. $Kc = \dfrac{ [Cu^{+2}]}{[Ag^+]^2 }$
4. $Kc = \dfrac{ [CO_2]}{[CaCO_3]}$
5. $K_c = [H_2O][CO_2]$
What is $K_c$ for the Reaction
1) Kc: 24.5
Kp: 1.002 Atm
2) Qc= 83.33 > Kc therefore the reaction shifts to the left
1. $K_p= \dfrac{P_{CO_2}} {P_{O_2}}$
2. $H_2 (g)+ I_2 (g) \rightarrow 2HI(g)$
Contributors and Attributions
• Heather Voigt
• Modified by Tom Neils (Grand Rapids Community College)
6.6: Applications of the Equilibrium Constant
The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties. An ideal solution or ideal mixture is a solution in which the enthalpy of solution (\(\Delta{H_{solution}} = 0\)) is zero; with the closer to zero the enthalpy of solution, the more "ideal" the behavior of the solution becomes. Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing (\(\Delta{S_{solution}}\)).
• Raoult's Law
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing.
• Henry's Law
Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/6%3A_Chemical_Equilibrium/6.5%3A_Heterogeneous_Equilibria.txt |
An ICE (Initial, Change, Equilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (e.g., weak acids and weak bases or complex ion formation).
Introduction
ICE tables are composed of the concentrations of molecules in solution in different stages of a reaction, and are usually used to calculate the K, or equilibrium constant expression, of a reaction (in some instances, K may be given, and one or more of the concentrations in the table will be the unknown to be solved for). ICE tables automatically set up and organize the variables and constants needed when calculating the unknown.
ICE is a simple acronym for the titles of the first column of the table.
• I stands for initial concentration. This row contains the initial concentrations of products and reactants.
• C stands for the change in concentration. This is the concentration change required for the reaction to reach equilibrium. It is the difference between the equilibrium and initial rows. The concentrations in this row are, unlike the other rows, expressed with either an appropriate positive (+) or negative (-) sign and a variable; this is because this row represents an increase or decrease (or no change) in concentration.
• E is for the concentration when the reaction is at equilibrium. This is the summation of the initial and change rows. Once this row is completed, its contents can be plugged into the equilibrium constant equation to solve for $K_c$.
The procedure for filling out an ICE table is best illustrated through example.
Example 1
Use an ICE table to determine $K_c$ for the following balanced general reaction:
$\ce{ 2X(g) <=> 3Y(g) + 4Z(g)} \nonumber$
where the capital letters represent the products and reactants.
• This equation will be placed horizontally above the table, with each product and reactant having a separate column.
A sample consisting of 0.500 mol of x is placed into a system with a volume of 0.750 liters.
• This statement implies that there are no initial amounts of Y and Z. For the I row of the Y and Z columns, 0.000 mol will be entered.
• Notice that the initial composition is given in moles. The amounts can either be converted to concentrations before putting them into the ICE table or after the equilibrium amounts have been calculated. This example uses moles for the ICE table, and calculates concentrations later.
At equilibrium, the amount of sample x is known to be 0.350 mol.
• For the equilibrium row of X, 0.350 mol will be entered.
Desired Unknown
$K_c = ? \nonumber$
Solution
The equilibrium constant expression is expressed as products over reactants, each raised to the power of their respective stoichiometric coefficients:
$K_c = \dfrac{[Y]^3[Z]^4}{[X]^2} \nonumber$
The equilibrium concentrations of Y and Z are unknown, but they can be calculated using the ICE table.
STEP 1: Fill in the given amounts
Reaction: 2X 3Y 4Z
Initial amounts 0.500 mol 0.000 mol 0.000 mol
Change in amount ? ? ?
Equilibrium amount 0.350 mol ? ?
This is the first step in setting up the ICE table. As mentioned above, the ICE mnemonic is vertical and the equation heads the table horizontally, giving the rows and columns of the table, respectively. The numerical amounts were given. Any amount not directly given is unknown.
STEP 2: Fill in the amount of change for each compound
Reaction 2X 3Y 4Z
Initial amounts 0.500 mol 0.000 mol 0.000 mol
Change in amount -0.150 mol +0.225 mol +0.300 mol
Equilibrium amounts 0.350 mol ? ?
Notice that the equilibrium in this equation is shifted to the right, meaning that some amount of reactant will be taken away and some amount of product will be added (for the Change row).
The change in amount ($x$) can be calculated using algebra:
$Equilibrium \; Amount = Initial \; Amount + Change \; in \; Amount \nonumber$
Solving for the Change in the amount of $2x$ gives:
$0.350 \; mol - 0.500 \; mol = -0.150 \; mol \nonumber$
The change in reactants and the balanced equation of the reaction is known, so the change in products can be calculated. The stoichiometric coefficients indicate that for every 2 mol of x reacted, 3 mol of Y and 4 mol of Z are produced. The relationship is as follows:
$\begin{eqnarray} Change \; in \; Product &=& -\left(\dfrac{\text{Stoichiometric Coefficient of Product}}{\text{Stoichiometric Coefficient of Reactant}}\right)(\text{Change in Reactant}) \ Change \; in \; Y &=& -\left(\dfrac{3}{2}\right)(-0.150 \; mol) \ &= +0.225 \; mol \end{eqnarray} \nonumber$
Try obtaining the change in Z with this method (the answer is already in the ICE table).
STEP 3: Solve for the equilibrium amounts
Reaction 2X 3Y 4Z
Initial amounts 0.500 mol 0.000 mol 0.000 mol
Change in amounts -0.150 mol +0.225 mol +0.300 mol
Equilibrium amounts 0.350 mol 0.225 mol 0.300 mol
If the initial amounts of Y and/or Z were nonzero, then they would be added together with the change in amounts to determine equilibrium amounts. However, because there was no initial amount for the two products, the equilibrium amount is simply equal to the change:
$\begin{eqnarray} Equilibrium \; Amount &=& Initial \; Amount + Change \; in \; Amount \ Equilibrium \; Amount \; of \; Y &=& 0.000 \; mol\; + 0.225 \; mol \ &=& +0.225 \; mol \end{eqnarray} \nonumber$
Use the same method to find the equilibrium amount of Z. Convert the equilibrium amounts to concentrations. Recall that the volume of the system is 0.750 liters.
$[Equilibrium \; Concentration \; of \; Substance] = \dfrac{Amount \; of \; Substance}{Volume \; of \; System}\nonumber$
$[X] = \dfrac{0.350 \; mol}{0.750 \; L} = 0.467 \; M \nonumber$
$[Y] = \dfrac{0.225 \; mol}{0.750 \; L} = 0.300 \; M \nonumber$
$[Z] = \dfrac{0.300 \; mol}{0.750 \; L} = 0.400 \; M \nonumber$
Use the concentration values to solve the $K_c$ equation:
$\begin{eqnarray} K_c &=& \dfrac{[Y]^3[Z]^4}{[X]^2} \ &=& \dfrac{[0.300]^3[0.400]^4}{[0.467]^2} \ K_c &=& 3.17 \times 10^{-3} \end{eqnarray}\nonumber$
Example 2: Using an ICE Table with Concentrations
n this example an ICE table is used to find the equilibrium concentration of the reactants and products. (This example will be less in depth than the previous example, but the same concepts are applied.) These calculations are often carried out for weak acid titrations.
Find the concentration of A- for the generic acid dissociation reaction:
$\ce{HA(aq) + H_2O(l) <=> A^{-}(aq) + H_3O^{+}(aq)} \nonumber$
with $[HA (aq)]_{initial} = 0.150 M$ and $K_a = 1.6 \times 10^{-2}$
Solution
This equation describes a weak acid reaction in solution with water. The acid (HA) dissociates into its conjugate base ($A^-$) and protons (H3O+). Notice that water is a liquid, so its concentration is not relevant to these calculations.
STEP 1: Fill in the given concentrations
Reaction: HA A- H3O+
I 0.150 M 0.000 M 0.000 M
C ? ? ?
E ? ? ?
• The contents of the leftmost column column are shortened for convenience.
STEP 2: Calculate the change concentrations by using a variable 'x'
Reaction: HA A- H3O+
I 0.150 M 0.000 M 0.000 M
C -x M +x M +x M
E ? ? ?
• The change in concentration is unknown, so the variable x is used to denote the change. x is the same for both products and reactants because equal stoichiometric amounts of A- and H3O+ are generated when HA dissociates in water.
STEP 3: Calculate the concentrations at equilibrium
Reaction: HA A- H3O+
I 0.150 M 0.000 M 0.000 M
C -x M +x M +x M
E 0.150 - x M x M x M
• To find the equilibrium amounts the I row and the C row are added. Use these values and Ka (the equilibrium constant for acids) to find the concentration x.
STEP 4: Use the ICE table to calculate concentrations with $K_a$
The expression for Ka is written by dividing the concentrations of the products by the concentrations of the reactants. Plugging in the values at equilibrium into the equation for Ka gives the following:
$K_a = \dfrac{x^2}{0.150-x} = 1.6 \times 10^{-2} \nonumber$
To find the concentration x, rearrange this equation to its quadratic form, and then use the quadratic formula to find x:
\begin{align*} (1.6 \times 10^{-2})({0.150-x}) &= {x^2} \[4pt] x^2+(1.6 \times 10^{-2})x-(0.150)(1.6 \times 10^{-2}) &= 0 \end{align*}
This is the typical form for a quadratic equation:
$Ax^{2}+Bx+C=0\nonumber$
where, in this case:
• $A = 1$
• $B = 1.6 \times 10^{-2}$
• $C =( -0.150)( 1.6 \times 10^{-2}) = -2.4 \times 10^{-3}$
The quadratic formula gives two solutions (but only one physical solution) for x:
$x = \dfrac{-B+\sqrt{B^2-4AC}}{2A}\nonumber$
and
$x = \dfrac{-B-\sqrt{B^2-4AC}}{2A}\nonumber$
Intuition must be used in determining which solution is correct. If one gives a negative concentration, it can be eliminated, because negative concentrations are unphysical.
The x value can be used to calculate the equilibrium concentrations of each product and reactant by plugging it into the elements in the E row of the ice table.
[Solution: x = 0.0416, -0.0576. x = 0.0416 makes chemical sense and is therefore the correct answer.]
For some problems like example 2, if x is significantly less than the value for Ka, then the x of the reactants (in the denominator) can be omitted and the concentration for x should not be greatly affected. This will make calculations faster by eliminating the necessity of the quadratic formula.
Checklist for ICE tables
• Make sure the reversible equation is balanced at the start of the problem; otherwise, the wrong amounts will be used in the table.
• The given data should be in amounts, concentrations, partial pressures, or somehow able to be converted to such. If it is not, then an ICE table will not help solve the problem.
• If the ICE table has the equilibrium in amounts, make sure to convert equilibrium values to concentrations before plugging in to solve for $K_c$.
• If the given data is in amounts or concentrations, use the ICE table to find $K_c$. If the given data is in partial pressures, use the ICE table to find $K_p$. If you desire to convert from one to the other, remember that $K_p = K_c(RT)^{\Delta n_{gas}}\nonumber$ It is simpler to use the ICE table with the appropriate givens and convert at the end of the problem.
• Enter in known data first, and then calculate the unknown data.
• If there is a negative value in the "initial" or "equilibrium" rows, reexamine the calculation. A negative concentration, amount, or partial pressure is physically impossible. Obviously, the "change" row can contain a negative value.
• Pay attention to the state of each reactant and product. If a compound is a solid or a liquid, its concentrations are not relevant to the calculations. Only concentrations of gaseous and aqueous compounds are used.
• In the "change" row the values will usually be a variable, denoted by x. It must first be understood which direction the equation is going to reach equilibrium (from left to right or from right to left). The value for "change" in the "from" direction of the reaction will be the opposite of x and the "to" direction will be the positive of x (adding concentration to one side and take away an equal amount from the other side).
• Know the direction of the reaction. This knowledge will affect the "change" row of the ICE table (for our example, we knew the reaction would proceed forward, as there was no initial products). Direction of reaction can be calculated using Q, the reaction quotient, which is then compared to a known K value.
• It is easiest to use the same units every time an ICE table is used (molarity is usually preferred). This will minimize confusion when calculating the equilibrium constants. ICE tables are usually used for weak acid or weak base reactions because all of the nature of these solutions. The amount of acid or base that will dissociate is unknown (for strong acids and strong bases it can be assumed that all of the acid or base will dissociate, meaning that the concentration of the strong acid or base is the same as its dissociated particles).
Partial pressures may also be substituted for concentrations in the ICE table, if desired (i.e., if the concentrations are not known, $K_p$ instead of $K_c$ is desired, etc.). "Amount" is also acceptable (the ICE table may be done in amounts until the equilibrium amounts are found, after which they will be converted to concentrations). For simplicity, assume that the word "concentration" can be replaced with "partial pressure" or "amounts" when formulating ICE tables.
Exercise $1$
0.200 M acetic acid is added to water. What is the concentration of H3O+ in solution if $K_c = 1.8 \times 10^{-6}$?
Answer
5.99×10-4
Exercise $2$
If the initial concentration of NH3 is 0.350 M and the concentration at equilibrium is 0.325 M, what is $K_c$ for this reaction?
Answer
1.92×10-3
Exercise $3$
How is $K_c$ derived from $K_p$?
Answer
$K_p = K_c(RT)^{\Delta n}$ then solve for $K_c$
Exercise $4$
Complete this ICE table:
Reaction: [HA] [A-] [H3O+]
I 0.650 mol ? ?
C ? ? ?
E 0.250 mol ? ?
Answer
Reaction: HA A- H3O+
I 0.650 mol 0.000 mol 0.000 mol
C -0.400 mol +0.400 mol +0.400 mol
E 0.250 mol 0.400 mol 0.400 mol
Contributors and Attributions
• Alexander Shei (UCD), Aileen McDuff (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/6%3A_Chemical_Equilibrium/6.7%3A_Solving_Equilibrium_Problems.txt |
Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. This page covers changes to the position of equilibrium due to such changes and discusses briefly why catalysts have no effect on the equilibrium position.
• Case Study: The Manufacture of Ethanol from Ethene
This page describes the manufacture of ethanol by the direct hydration of ethene, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture and the rate of the reaction.
• Effect of Temperature on Equilibrium
A temperature change occurs when temperature is increased or decreased by the flow of heat. This shifts chemical equilibria toward the products or reactants, which can be determined by studying the reaction and deciding whether it is endothermic or exothermic.
• ICE Tables
An ICE (Initial, Change, Equilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (e.g., weak acids and weak bases or complex ion formation).
• Le Chatelier's Principle and Dynamic Equilbria
This page looks at Le Châtelier's Principle and explains how to apply it to reactions in a state of dynamic equilibrium. It covers changes to the position of equilibrium if you change concentration, pressure or temperature. It also explains very briefly why catalysts have no effect on the position of equilibrium.
• Le Chatelier's Principle Fundamentals
Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change.
• The Contact Process
The Contact Process is used in the manufacture of sulfuric acid. This Modules explain the reasons for the conditions used in the process by considering the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process.
• The Effect of Changing Conditions
This page looks at the relationship between equilibrium constants and Le Châtelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as you change the conditions of a reaction, although the equilibrium constant may remain the same.
• The Haber Process
This page describes the Haber Process for the manufacture of ammonia from nitrogen and hydrogen, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/6%3A_Chemical_Equilibrium/6.8%3A_Le_Chatelier%27s_Principle.txt |
Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. This page covers changes to the position of equilibrium due to such changes and discusses briefly why catalysts have no effect on the equilibrium position.
• Case Study: The Manufacture of Ethanol from Ethene
This page describes the manufacture of ethanol by the direct hydration of ethene, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture and the rate of the reaction.
• Effect of Temperature on Equilibrium
A temperature change occurs when temperature is increased or decreased by the flow of heat. This shifts chemical equilibria toward the products or reactants, which can be determined by studying the reaction and deciding whether it is endothermic or exothermic.
• ICE Tables
An ICE (Initial, Change, Equilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (e.g., weak acids and weak bases or complex ion formation).
• Le Chatelier's Principle and Dynamic Equilbria
This page looks at Le Châtelier's Principle and explains how to apply it to reactions in a state of dynamic equilibrium. It covers changes to the position of equilibrium if you change concentration, pressure or temperature. It also explains very briefly why catalysts have no effect on the position of equilibrium.
• Le Chatelier's Principle Fundamentals
Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change.
• The Contact Process
The Contact Process is used in the manufacture of sulfuric acid. This Modules explain the reasons for the conditions used in the process by considering the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process.
• The Effect of Changing Conditions
This page looks at the relationship between equilibrium constants and Le Châtelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as you change the conditions of a reaction, although the equilibrium constant may remain the same.
• The Haber Process
This page describes the Haber Process for the manufacture of ammonia from nitrogen and hydrogen, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/6%3A_Chemical_Equilibrium/6.9%3A_Equilibria_of_Real_Gases.txt |
Thumbnail: Major structural changes accompany binding of the Lewis base to the coordinatively unsaturated, planar Lewis acid BF3. (Public Domain; Ben Mills ).
7.02 Acid Strength
All acids and bases do not ionize or dissociate to the same extent. This leads to the statement that acids and bases are not all of equal strength in producing H+ and OH- ions in solution. The terms "strong" and "weak" give an indication of the strength of an acid or base. The terms strong and weak describe the ability of acid and base solutions to conduct electricity. If the acid or base conducts electricity strongly, it is a strong acid or base. If the acid or base conducts electricity weakly, it is a weak acid or base.
Demonstration of Acid and Base Conductivity
The instructor will test the conductivity of various solutions with a light bulb apparatus. The light bulb circuit is incomplete. If the circuit is completed by a solution containing a large number of ions, the light bulb will glow brightly indicating a strong ability to conduct electricity as shown for HCl. If the circuit is completed by a solution containing large numbers of molecules and either no ions or few ions, the solution does not conduct or conducts very weakly as shown for acetic acid.
An acid or base which strongly conducts electricity contains a large number of ions and is called a strong acid or base and an acid or base which conducts electricity only weakly contains only a few ions and is called a weak acid or base.
Conductivity Behavior of Acids and Bases
Compounds Appearance of light bulb Classification
Weak or Strong
Inference of Ions or Molecules
H2O no light weak molecules
HCl bright strong ions
HC2H3O2 dim weak molecules
H2SO4 bright
H2CO3 dim
NaOH bright
KOH bright
NH4OH dim
Bond Strength
The bond strengths of acids and bases are implied by the relative amounts of molecules and ions present in solution. The bonds are represented as:
acid base
H-A M-OH
where A is a negative ion, and M is a positive ion
• Strong acids have mostly ions in solution, therefore the bonds holding H and A together must be weak. Strong acids easily break apart into ions.
• Weak acids exist mostly as molecules with only a few ions in solution, therefore the bonds holding H and A together must be strong. Weak acids do not readily break apart as ions but remain bonded together as molecules.
Bond Strength Principle
Acids or bases with strong bonds exist predominately as molecules in solutions and are called "weak" acids or bases. Acids or bases with weak bonds easily dissociate into ions and are called "strong" acids or bases.
Table 1: Summary List of Characteristics for Strong and Weak Acids and Bases. All characteristics of acids and bases are related to whether the predominate forms are molecules and ions.
Characteristic Strong Acid or Base Weak Acid or Base
Molecules few large number
Ions large number small number
Conductivity strong weak
Bond Strength weak strong
Acids and bases behave differently in solution based on their strength. Acid or base "strength" is a measure of how readily the molecule ionizes in water.
Introduction Again
Some acids and bases ionize rapidly and almost completely in solution; these are called strong acids and strong bases. For example, hydrochloric acid (HCl) is a strong acid. When placed in water, virtually every HCl molecule splits into a H+ ion and a Cl- ion in the reaction.1
$\ce{HCl(aq) + H2O(l) <=> H3O^{+}(aq) + Cl^{-}(aq)} \nonumber$
For a strong acid like HCl, if you place 1 mole of HCl in a liter of water, you will get roughly 1 mole of H30+ ions and 1 mole of Cl- ions. In a weak acid like hydrofluoric acid (HF), not all of the HF molecules split up, and although there will be some H+ and F- ions released, there will still be HF molecules in solution1. A similar concept applies to bases, except the reaction is different. A strong base like sodium hydroxide (NaOH) will also dissociate completely into water; if you put in 1 mole of NaOH into water, you will get 1 mole of hydroxide ions.1
$\ce{NaOH(aq) + H2O(l) <=> Na^{+}(aq) + OH^{-}(aq) + H2O(l)} \nonumber$
The terms "strong" and "weak" in this context do not relate to how corrosive or caustic the substance is, but only its capability to ionize in water. The ability of a substance to eat through other materials or damage skin is more of a function of the properties of that acid, as well as its concentration. Although, strong acids are more directly dangerous at lower concentrations a strong acid is not necessarily more dangerous than a weak one. For example, hydrofluoric acid is a weak acid1, but it is extremely dangerous and should be handled with great care. Hydrofluoric acid is particularly dangerous because it is capable of eating through glass, as seen in the video in the links sectionV1. The percent dissociation of an acid or base is mathematically indicated by the acid ionization constant (Ka) or the base ionization constant (Kb)1. These terms refer to the ratio of reactants to products in equilibrium when the acid or base reacts with water. For acids the expression will be
Ka = [H3O+][A-]/[HA]
where HA is the concentration of the acid at equilibrium, and A- is the concentration of its conjugate base at equilibrium and for bases the expression will be
$K_b = \dfrac{[\ce{OH^{-}}][\ce{HB^{+}}]}{\ce{B}}$
where B is the concentration of the base at equilibrium and HB+ is the concentration of its conjugate acid at equilibrium
The stronger an acid is, the lower the pH it will produce in solution. pH is calculated by taking the negative logarithm of the concentration of hydronium ions. For strong acids, you can calculate the pH by simply taking the negative logarithm of its molarity as it completely dissociates into its conjugate base and hydronium. The same goes for strong bases, except the negative logarithm gives you the pOH as opposed to the pH. For weak acids and bases, the higher the Ka or Kb, the more acidic or basic the solution. To find the pH for a weak acid or base, you must use the K equation and a RICE table to determine the pH.
All acids have a conjugate base that forms when they react with water, and similarly, all bases have a conjugate acid that reacts when they form with water.1 You can judge the relative strength of a conjugate by the $K_a$ or $K_b$ value of the substance because $K_a \times K_b$ is equal to the ionization constant of water, Kw which is equal to $1 \times 10^{-14}$ at room temperature. The higher the Ka, the stronger the acid is, and the weaker its conjugate base is. Similarly, the higher the Kb, the stronger the substance is as a base, and the more weakly acidic its conjugate acid is.1
Calculation of Ka
For an acid that reacts with water in the reaction
$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}$
$K_a = \dfrac{[H_3O^+][A^-]}{[HA]}$
where each bracketed term represents the concentration of that substance in solution.
Relation of Kw, Kb, Ka
$K_w = K_a \times K_b \nonumber$
Partial List of Strong Acids: Hydrochlroic acid (HCl), Nitric Acid (HNO3), Perchloric Acid (HClO4), Sulfuric Acid (H2SO4)
Partial List of Strong Bases: Sodium Hydroxide (NaOH), Barium Hydroxide (Ba(OH)2), Calcium Hydroxide (Ca(OH)2), Lithium Hydroxide (LiOH) (Hydroxides of Group I and II elements are generally strong bases)
Partial List of Weak Acids: Acetic Acid (CH3COOH), Carbonic Acid (H2CO3), Phosphoric Acid (H3PO4)
Partial List of Weak Bases: Ammonia (NH3), Calcium Carbonate (CaCO3), Sodium Acetate (NaCH3COO)
Example $1$
Find the pH of 0.5 grams of HCl disolved into 100 ml of water:
Solution
First find moles of acid:
grams / molar mass = moles
0.5 grams / (36.5 g/mole) = 0.014 moles HCl
Then find molarity:
moles / volume = molarity
0.014 moles / 0.100 L = 0.14 M
HCl is a strong acid and completely dissociates in water, therefore the pH will be equal to the negative logarithm of the concentration of HCl
pH = -log(H3O+)
pH = -log(0.14) = 0.85
Example $2$
The Ka value for acetic acid is 1.76*10-5, and the Ka value for benzoic acid is 6.46*10-5, if two solutions are made, one from each acid, with equal concentrations, which one will have the lower pH?
Solution
The Ka value is a measure of the ratio between reactants and products at equilibrium. For an acid, the reaction will be HA + H2O --> A- + H3O+ . PH is based on the concentration of the hydronium ion (H3O+) which is a product of the reaction of acid and water. A higher Ka value means a higher ratio of reactants to products, and so the acid with the higher Ka value will be producing more hydronium, and therefore have a lower pH. Therefore the solution of benzoic acid will have a lower pH.
Example $3$
The Ka value of ammonium (NH4+) is 5.6*10-10, the Kb value of ammonia (NH3) 1.8*10-5, is ammonium more strongly acidic than ammonia is basic?
Solution
The relative strength of an acid or base depends on how high its Ka or Kb value is, in this case, the Ka value is far lower than the Kb value so the ammonia is more strongly basic than ammonium is acidic.
Contributors and Attributions
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook,. Lloyd McCarthy (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.01_The_Nature_of_Acids_and_Bases.txt |
Learning Objectives
• To define the pH scale as a measure of acidity of a solution
• Tell the origin and the logic of using the pH scale.
• Apply the same strategy for representing other types of quantities such as pKa, pKb, pKw.
Auto-Ionization of Water
Because of its amphoteric nature (i.e., acts as both an acid or a base), water does not always remain as $H_2O$ molecules. In fact, two water molecules react to form hydronium and hydroxide ions:
$\ce{ 2 H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^{−} (aq)} \label{1}$
This is also called the self-ionization of water. The concentration of $H_3O^+$ and $OH^-$ are equal in pure water because of the 1:1 stoichiometric ratio of Equation $\ref{1}$. The molarity of H3O+ and OH- in water are also both $1.0 \times 10^{-7} \,M$ at 25° C. Therefore, a constant of water ($K_w$) is created to show the equilibrium condition for the self-ionization of water. The product of the molarity of hydronium and hydroxide ion is always $1.0 \times 10^{-14}$ (at room temperature).
$K_w= [H_3O^+][OH^-] = 1.0 \times 10^{-14} \label{2}$
Equation $\ref{2}$ also applies to all aqueous solutions. However, $K_w$ does change at different temperatures, which affects the pH range discussed below.
$H^+$ and $H_3O^+$ is often used interchangeably to represent the hydrated proton, commonly call the hydronium ion.
Equation \ref{1} can also be written as
$H_2O \rightleftharpoons H^+ + OH^- \label{3}$
As expected for any equilibrium, the reaction can be shifted to the reactants or products:
• If an acid ($H^+$) is added to the water, the equilibrium shifts to the left and the $OH^-$ ion concentration decreases
• If base ( $OH^-$) is added to water, the equilibrium shifts to left and the $H^+$ concentration decreases.
pH and pOH
Because the constant of water, Kw is $1.0 \times 10^{-14}$ (at 25° C), the $pK_w$ is 14, the constant of water determines the range of the pH scale. To understand what the pKw is, it is important to understand first what the "p" means in pOH and pH. The addition of the "p" reflects the negative of the logarithm, $-\log$. Therefore, the pH is the negative logarithm of the molarity of H, the pOH is the negative logarithm of the molarity of $\ce{OH^-}$, and the $pK_w$ is the negative logarithm of the constant of water:
\begin{align} pH &= -\log [H^+] \label{4a} \[4pt] pOH &= -\log [OH^-] \label{4b} \[4pt] pK_w &= -\log [K_w] \label{4c} \end{align}
At room temperature,
$K_w =1.0 \times 10^{-14} \label{4d}$
So
\begin{align} pK_w &=-\log [1.0 \times 10^{-14}] \label{4e} \[4pt] &=14 \end{align}
Using the properties of logarithms, Equation $\ref{4e}$ can be rewritten as
$10^{-pK_w}=10^{-14}. \label{4f}$
The equation also shows that each increasing unit on the scale decreases by the factor of ten on the concentration of $\ce{H^{+}}$. Combining Equations \ref{4a} - \ref{4c} and \ref{4e} results in this important relationship:
$pK_w= pH + pOH = 14 \label{5b}$
Equation \ref{5b} is correct only at room temperature since changing the temperature will change $K_w$.
The pH scale is logarithmic, meaning that an increase or decrease of an integer value changes the concentration by a tenfold. For example, a pH of 3 is ten times more acidic than a pH of 4. Likewise, a pH of 3 is one hundred times more acidic than a pH of 5. Similarly a pH of 11 is ten times more basic than a pH of 10.
Properties of the pH Scale
From the simple definition of pH in Equation \ref{4a}, the following properties can be identified:
• This scale is convenient to use, because it converts some odd expressions such as $1.23 \times 10^{-4}$ into a single number of 3.91.
• This scale covers a very large range of $\ce{[H+]}$, from 0.1 to 10-14. When $\ce{[H+]}$ is high, we usually do not use the pH value, but simply the $\ce{[H+]}$. For example, when $\mathrm{[H^+] = 1.0}$, pH = 0. We seldom say the pH is 0, and that is why you consider pH = 0 such an odd expression. A pH = -0.30 is equivalent to a $\ce{[H+]}$ of 2.0 M. Negative pH values are only for academic exercises. Using the concentrations directly conveys a better sense than the pH scales.
• The pH scale expands the division between zero and 1 in a linear scale or a compact scale into a large scale for comparison purposes. In mathematics, you learned that there are infinite values between 0 and 1, or between 0 and 0.1, or between 0 and 0.01 or between 0 and any small value. Using a log scale certainly converts infinite small quantities into infinite large quantities.
• The non-linearity of the pH scale in terms of $\ce{[H+]}$ is easily illustrated by looking at the corresponding values for pH between 0.1 and 0.9 as follows:
pH 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
[H+] 1 0.79 0.63 0.50 0.40 0.32 0.25 0.20 0.16 0.13
• Because the negative log of $\ce{[H+]}$ is used in the pH scale, the pH scale usually has positive values. Furthermore, the larger the pH, the smaller the $\ce{[H+]}$.
The Effective Range of the pH Scale
It is common that the pH scale is argued to range from 0-14 or perhaps 1-14, but neither is correct. The pH range does not have an upper nor lower bound, since as defined above, the pH is an indication of concentration of H+. For example, at a pH of zero the hydronium ion concentration is one molar, while at pH 14 the hydroxide ion concentration is one molar. Typically the concentrations of H+ in water in most solutions fall between a range of 1 M (pH=0) and 10-14 M (pH=14). Hence a range of 0 to 14 provides sensible (but not absolute) "bookends" for the scale. One can go somewhat below zero and somewhat above 14 in water, because the concentrations of hydronium ions or hydroxide ions can exceed one molar. Figure $1$ depicts the pH scale with common solutions and where they are on the scale.
Quick Interpretation
• If pH >7, the solution is basic. The pOH should be looked in the perspective of OH- instead of H+. Whenever the value of pOH is less than 7, then it is considered basic. And therefore there are more OH- than H+ in the solution.
• At pH 7, the substance or solution is at neutral and means that the concentration of H+ and OH- ion is the same.
• If pH < 7, the solution is acidic. There are more H+ than OH- in an acidic solution.
• The pH scale does not have an upper nor lower bound.
Example $1$
If the concentration of $NaOH$ in a solution is $2.5 \times 10^{-4}\; M$, what is the concentration of $H_3O^+$?
Solution
We can assume room temperature, so
$1.0 \times 10^{-14} = [H_3O^+][OH^-] \nonumber$
to find the concentration of H3O+, solve for the [H3O+].
$\dfrac{1.0 \times 10^{-14}}{[OH^-]} = [H_3O^+]$
$\dfrac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}} = [H_3O^+] = 4.0 \times 10^{-11}\; M$
Example $2$
1. Find the pH of a solution of 0.002 M of HCl.
2. Find the pH of a solution of 0.00005 M NaOH.
Solution
1. The equation for pH is -log [H+]
$[H^+]= 2.0 \times 10^{-3}\; M \nonumber$
$pH = -\log [2.0 \times 10^{-3}] = 2.70 \nonumber$
1. The equation for pOH is -log [OH-]
$[OH^-]= 5.0 \times 10^{-5}\; M \nonumber$
$pOH = -\log [5.0 \times 10^{-5}] = 4.30 \nonumber$
$pK_w = pH + pOH \nonumber$
and
$pH = pK_w - pOH \nonumber$
then
$pH = 14 - 4.30 = 9.70 \nonumber$
Example $3$: Soil
If moist soil has a pH of 7.84, what is the H+ concentration of the soil solution?
Solution
$pH = -\log [H^+] \nonumber$
$7.84 = -\log [H^+] \nonumber$
$[H^+] = 1.45 \times 10^{-8} M \nonumber$
Hint
Place -7.84 in your calculator and take the antilog (often inverse log or 10x) = 1.45 x 10-8M
Proper Definition of pH
The pH scale was originally introduced by the Danish biochemist S.P.L. Sørenson in 1909 using the symbol pH. The letter p is derived from the German word potenz meaning power or exponent of, in this case, 10. In 1909, S.P.L. Sørenson published a paper in Biochem Z in which he discussed the effect of H+ ions on the activity of enzymes. In the paper, he invented the term pH (purported to mean pondus hydrogenii in Latin) to describe this effect and defined it as the $-\log[H^+]$. In 1924, Sørenson realized that the pH of a solution is a function of the "activity" of the H+ ion and not the concentration. Thus, he published a second paper on the subject. A better definition would be
$pH = -\log\,a\{\ce{H^{+}}\}$
where $a\{H^+\}$ denotes the activity (an effective concentration) of the H+ ions. The activity of an ion is a function of many variables of which concentration is one.
• Concentration is abbreviated by using square brackets, e.g., $[H_3O^+]$ is the concentration of hydronium ion in solution.
• Activity is abbreviated by using "a" with curly brackets, e.g., $a\{H_3O^+\}$ is the activity of hydronium ions in solution
Because of the difficulty in accurately measuring the activity of the $\ce{H^{+}}$ ion for most solutions the International Union of Pure and Applied Chemistry (IUPAC) and the National Bureau of Standards (NBS) has defined pH as the reading on a pH meter that has been standardized against standard buffers. The following equation is used to calculate the pH of all solutions:
\begin{align} pH &= \dfrac{F(E-E_{standard})}{RT\;\ln 10} + pH_{standard} \label{6a} \[4pt] &= \dfrac{5039.879 (E-E_{standard})}{T} + pH_{standard} \label{6b} \end{align}
with
• $R$ is the ideal gas constant,
• $F$ is the Faraday's constant, and
• $T$ is absolute temperature (in K)
The activity of the H+ ion is determined as accurately as possible for the standard solutions used. The identity of these solutions vary from one authority to another, but all give the same values of pH to ± 0.005 pH unit. The historical definition of pH is correct for those solutions that are so dilute and so pure the H+ ions are not influenced by anything but the solvent molecules (usually water).
When measuring pH, [H+] is in units of moles of H+ per liter of solution. This is a reasonably accurate definition at low concentrations (the dilute limit) of H+. At very high concentrations (10 M hydrochloric acid or sodium hydroxide, for example,) a significant fraction of the ions will be associated into neutral pairs such as H+Cl, thus reducing the concentration of “available” ions to a smaller value which we will call the effective concentration. It is the effective concentration of H+ and OH that determines the pH and pOH. The pH scale as shown above is called sometimes "concentration pH scale" as opposed to the "thermodynamic pH scale". The main difference between both scales is that in thermodynamic pH scale one is interested not in H+concentration, but in H+activity. What a person measures in the solution is just activity, not the concentration. Thus it is thermodynamic pH scale that describes real solutions, not the concentration one.
For solutions in which ion concentrations don't exceed 0.1 M, the formulas pH = –log [H+] and pOH = –log[OH] are generally reliable, but don't expect a 10.0 M solution of a strong acid to have a pH of exactly –1.00! However, this definition is only an approximation (albeit very good under most situations) of the proper definition of pH, which depends on the activity of the hydrogen ion:
$pH= -\log a\{H^+\} \approx -\log [H^+] \label{7}$
The activity is a measure of the "effective concentration" of a substance, is often related to the true concentration via an activity coefficient, $\gamma$:
$a{H^+}=\gamma [H^+] \label{8}$
Calculating the activity coefficient requires detailed theories of how charged species interact in solution at high concentrations (e.g., the Debye-Hückel Theory). In most solutions the pH differs from the -log[H+ ] in the first decimal point. The following table gives experimentally determined pH values for a series of HCl solutions of increasing concentration at 25 °C.
Table $1$: HCl Solutions with corresponding pH values. Data taken from Christopher G. McCarty and Ed Vitz, Journal of Chemical Education, 83(5), 752 (2006) and G.N. Lewis, M. Randall, K. Pitzer, D.F. Brewer, Thermodynamics (McGraw-Hill: New York, 1961; pp. 233-34).
Molar Concentration of $HCl$ pH defined as Concentration Experimentally Determined pH Relative Deviation
0.00050 3.30 3.31 0.3%
0.0100 2 2.04 1.9%
0.100 1 1.10 9%
0.40 0.39 0.52 25%
7.6 -0.88 -1.85 52%
While the pH scale formally measures the activity of hydrogen ions in a substance or solution, it is typically approximated as the concentration of hydrogen ions; this approximation is applicable only under low concentrations.
Living Systems
Molecules that make up or are produced by living organisms usually function within a narrow pH range (near neutral) and a narrow temperature range (body temperature). Many biological solutions, such as blood, have a pH near neutral. pH influences the structure and the function of many enzymes (protein catalysts) in living systems. Many of these enzymes have narrow ranges of pH activity. Cellular pH is so important that death may occur within hours if a person becomes acidotic (having increased acidity in the blood). As one can see pH is critical to life, biochemistry, and important chemical reactions. Common examples of how pH plays a very important role in our daily lives are given below:
• Water in swimming pool is maintained by checking its pH. Acidic or basic chemicals can be added if the water becomes too acidic or too basic.
• Whenever we get a heartburn, more acid build up in the stomach and causes pain. We needs to take antacid tablets (a base) to neutralize excess acid in the stomach.
• The pH of blood is slightly basic. A fluctuation in the pH of the blood can cause in serious harm to vital organs in the body.
• Certain diseases are diagnosed only by checking the pH of blood and urine.
• Certain crops thrive better at certain pH range.
• Enzymes activate at a certain pH in our body.
Table $2$: pH in Living Systems
Compartment pH
Gastric Acid 1
Lysosomes 4.5
Granules of Chromaffin Cells 5.5
Human Skin 5.5
Urine 6
Neutral H2O at 37 °C 6.81
Cytosol 7.2
Cerebrospinal Fluid 7.3
Blood 7.43-7.45
Mitochondrial Matrix 7.5
Pancreas Secretions 8.1
Problems
1. In a solution of $2.4 \times 10^{-3} M$ of HI, find the concentration of $OH^-$.
2. Determine the pH of a solution that is 0.0035 M HCl.
3. Determine the [H3O+] of a solution with a pH = 5.65
4. If the pOH of NH3, ammonia, in water is 4.74. What is the pH?
5. Pepsin, a digestive enzyme in our stomach, has a pH of 1.5. Find the concentration of OH- in the stomach.
Solutions
1. We use the dissociation of water equation to find [OH-].
Kw = [H3O+][OH-] = 1.0 X 10-14
Solve for [OH-]
[OH-] = (1.0 X 10-14)/ [H3O+]
Plug in the molarity of HI and solve for OH-.
[OH-] = (1.0 X 10-14)/ [2.4 X 10-3] = 4.17 X 10-12 M.
2. pH = -log[H3O+]
Plug the molarity of the HCl in and solve for pH.
pH = -log[0.0035] = 2.46
3. pH = -log[H3O+]
Plug in the pH and solve for [H3O+]
5.65 = -log[H3O+]
Move the negative sign to the pH. -5.65 = log[H3O+]
10-5.65=10log[H3O+] = 2.24 X 10-6 M
4. pH + pOH = 14
Solve for pH.
14 - pOH = pH
14 - 4.74 = pH = 9.26
5. There are several ways to do this problem.
Answer 1.
pH + pOH = 14
Solve for pOH.
pOH = 14 - pH
pOH = 14 - 1.5 = 12.5
When the pOH is solved, solve for the concentration by using log.
pOH = -log[OH-]
12.5 = -log[OH-]
-12.5 = log[OH-]
10-12.5 = 10log[OH-] = 3.16 X 10-13 M.
Answer 2.
pH = -log[H+]
Plug in the pH and solve for the molarity of H+ of pepsin.
1.5 = -log[H+]
-1.5 = log[H+]
10-1.5 = 10log[H+] = [H+]= 0.032
Use the concentration of H+ to solve for the concentration of OH-.
[H+][OH-] = 1.0 X 10-14
Plug in the [H+] and solve for [OH-].
[OH-] = (1.0 X 10-14)/[H3O+]
[OH-] = (1.0 X 10-14)/(0.032) = 3.125 X 10-14 M | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.03_The_pH_Scale.txt |
Thus far, we have been discussing problems and answers in equilibria--perhaps the most popular type of problem being how to find the pH of a weak acid solution given a certain concentration of a molecule. However, those problems in particular usually only involve what is called a monoprotic acid. “Mono” in the word “monoprotic” indicates that there is only one ionizeable hydrogen atom in an acid when immersed in water, whereas the concept of allows for two or more ionizeable hydrogen atoms.
Introduction
Consider the following chemical equation as the molecule acetic acid equilibrates in the solution:
$\ce{ CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}} \nonumber$
Although acetic acid carries a four hydrogen atoms, only a single becomes ionized. Not to get into too much detail between monoprotic and polyprotic acids, but if you desire to find the pH given a concentration of a weak acid (in this case, acetic acid), you would create and complete anadjusting for how much acetic acid disassociates. However, if you wish to find the pH of a solution after a polyprotic acid disassociates, there are extra steps that would need to be done. Let's first take a look at a unique example:
Example $1$: Finding the pH of sulfuric Acid
What is the pH of 0.75 M sulfuric acid?
Solution
In sulfuric acid (H2SO4), there are two ionizable hydrogen atoms. What makes this molecule interesting is that its ionization constant for the first hydrogen ($K_{a1}$) ionized is significantly larger than is the second ionization constant ($K_{a2}$). The $K_{a1}$ constant for sulfuric acid is conveniently dubbed “very large” while the $K_{a2}$ constant is 1.1 x 10-2. As such, the sulfuric acid will completely disassociate into HSO4- and H3O+ ions (as a strong acid).
$\ce{H2SO4 (aq) + H2O -> HSO4^{-} (aq) + H_3O^{+}} \nonumber$
Since the sulfuric acid completely disassociates in the solution, we can skip the ICE table process for sulfuric acid, and assert that the concentration $\ce{HSO4^{-}}$ and $\ce{H3O^{+}}$ are the same as that of H2SO4, that is 0.75 M. (This neglects the background concentration of $\ce{H_3O^{+}}$ in water of $1 \times 10^{-7}M$).
Equation:
$\ce{HSO4^{-} (aq) + H2O <=> SO4^{2-} (aq) + H_3O^{+}} \nonumber$
ICE Table:
$HSO_4^-$ $SO_4^{2-}$ $H_3O^+$
Initial 0.75 M 0 M 0.75 M
Change -x M +x M +x M
Equilibrium (0.75 - x) M +x M (0.75 + x) M
$K_{a2} = \dfrac{[SO_4^{2-}][H_3O^+]}{[HSO_4^-]} = 1.1 \times 10^{-2} = 0.011 = \dfrac{x(0.75+x)}{0.75 - x}$
Assume $x$ in the denominator is negligible. Therefore,
$x = 0.011 M = [SO_4^{2-}] \nonumber$
Since we know the value of x, we can use the equation from the ICE table to find the value of [HSO4-].
$[HSO_4^-] = 0.75 \; M - x = 0.75 - 0.011 = 0.74 \; M \nonumber$
We can also find [H3O+] using the equation from the ICE table.
$[H_3O+] = 0.75 \; M + x = 0.75 + 0.011 = 0.76 \; M \nonumber$
We can then find the pH from the calculated [H3O+] value.
$pH = -log[H_3O^+] = -log0.76 = 0.119 \nonumber$
Let's say our task is to find the pH given a polyprotic base which gains protons in water. Thankfully, the process is essentially the same as finding the pH of a polyprotic acid except in this case we deal with the concentration of OH- instead of H3O+.
Example $2$: Finding the pH of a polyprotic base
Let's take a look at how to find the pH of C20H24O2N2, a diprotic base with a concentration of 0.00162 M, and a $K_{b1}$ of 10-6 and a $K_{b2}$ of 10-9.8.
Equation:
$C_{20}H_{24}O_2N \; (aq) + H_2O \rightleftharpoons C_{20}H_{24}O_2N_2H^+ + OH^-$
$C_{20}H_{24}O_2N_2$ $C_{20}H_{24}O_2N_2H^+$ $OH^-$
Initial 0.00162 M 0 M 0 M
Change -x M +x M +x M
Equilibrium 0.00162 M x M x M
$K_{b1} = \dfrac{[C_{20}H_{24}O_2N_2H^+][OH^-]}{[C_{20}H_{24}O_2N_2]} = 10^{-6} = 0.011 = \dfrac{(x)(x)}{0.00162 - x} \nonumber$
Again, assume x in the denominator is negligible. Therefore,
$0.011 \approx \dfrac{x^2}{0.00162}$
Then,
$x \approx 4 \times 10^{-5}$
We can then find the pH.
$pOH = -log(4 \times 10^{-5}) = 4.4$
$pH = 14 - 4.4 = 9.6$
As we determine the pH of the solution, we realize that the OH-gained using the second ionization constant is so insignificant that it does not impact the final pH value. For good measure, the following is the process to determine the pH in case the second use of the ICE table would indeed make a difference.
Equation:
$C_{20}H_{24}O_2N_2H^+ + H_2O \rightleftharpoons C_{20}H_{24}O_2N_2H_2^{2+} + OH^-$
ICE Table:
$C_{20}H_{24}O_2N_2H^+$ $C_{20}H_{24}O_2N_2H_2^{2+}$ $OH^-$
Initial 4 x 10-5 M 0 M 4 x 10-5 M
Change -x M +x M +x M
Equilibrium (4 x 10-5 - x) M x M (4 x 10-5 + x) M
$K_{b2} = \dfrac{[C_{20}H_{24}O_2N_2H_2^{2^+}][OH^-]}{[C_{20}H_{24}O_2N_2H^+]} = 10^{-9.8}$
$10^{-9.8} = \dfrac{(4 \times 10^{-5} + x)(x)}{(4 \times 10^{-5} - x)} \nonumber$
$10^{-9.8} = \dfrac{0.00004 \; x + x^2}{0.00004 - x} \nonumber$
$10^{-9.8}(0.00004 - x) = 0.00004 x + x^2 \nonumber$
$x^2 + (4 \times 10^{-5})x - 6.3 \times 10^{-15} = 0 \nonumber$
\begin{align*} x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \[4pt] &= \dfrac{-4 \times 10^{-5} \pm \sqrt{(4 \times 10^{-5})^2 - 4(1)(6.3 \times 10^{-15})}}{2(1)} \[4pt] & = 0 \end{align*} | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.05_Calculating_the_pH_of_Weak_Acid_Solutions.txt |
Thumbnail: Major structural changes accompany binding of the Lewis base to the coordinatively unsaturated, planar Lewis acid BF3. (Public Domain; Ben Mills ).
7.07 Polyprotic Acids
The name "polyprotic" literally means many protons. Therefore, in this section we will be observing some specific acids and bases which either lose or accept more than one proton. Then, we will be talking about the equations used in finding the degree of dissociation. Finally, with given examples, we will be able to approach problems dealing with polyprotic acids and bases.
Introduction
Polyprotic acids are specific acids that are capable of losing more than a single proton per molecule in acid-base reactions. (In other words, acids that have more than one ionizable H+ atom per molecule). Protons are lost through several stages (one at each stage), with the first proton being the fastest and most easily lost. Contrast with monoprotic acids in section Monoprotic Versus Polyprotic Acids And Bases.
Common Polyprotic Acids Formula Strong/Weak Acid Number of Ionizable Hydrogens Ka1 Ka2 Ka3
Sulfuric acid H2SO4 Strong 2 (diprotic) Very Large 1.1E-2
Sulfurous acid H2SO3 Weak 2 (diprotic) 1.3E-2 6.2E-8
Phosphoric acid H3PO4 Weak 3 (triprotic) 7.1E-3 6.3E-8 4.2E-13
Carbonic acid H2CO3 Weak 2 (diprotic) 4.4E-7 4.7E-11
Hydrosulfuric acid or Hydrogen sulfide H2S Weak 2 (diprotic) 1.0E-7 1E-19
Oxalic acid H2C2O4 Weak 2 (diprotic) 5.4E-2 5.3E-5
Malonic acid H2C3H2O4 Medium Strong 2 (diprotic) 1.5E-3 2.0E-6
From the table above, we see that sulfuric acid is the strongest.
Ionization Constant
It is important to know that K1>K2>K3, where K stands for the acidity constant or acid ionization constant (first, second, and third, respectively). These constants are used to measure the degree of dissociation of hydrogens in the acid. For a more in depth discussion on this, go to Ionization Constants.
Example 1: Hydrosulfuric acid
To find Ka1 of Hydrosulfuric acid (H2S), you must first write the reaction:
$H_2S \rightleftharpoons H^+ + HS^- \nonumber$
Dividing the products by the reactants, we then have:
$K_{a1} = \dfrac{[H^+] [HS^-]}{ [HS-]} \nonumber$
To find Ka2, we start with the reaction:
$HS^- \rightleftharpoons H^+ + S_2^- \nonumber$
Then, like when finding $K_{a1}$, write the products over the reactants:
$K_{a2} = \dfrac{[H^+] [S_2^-]}{[HS^-]} \nonumber$
From these reactions we can observe that it takes two steps to fully remove the H+ ion. This also means that this reaction will produce two equivalence points or stoichiometric points. The equivalence point, by definition, is the point during an acid-base titration in which there has been equal amounts of acid and base reacted. If we were to graph this, we would be able to see exactly just what two equivalence points looks like. Let's check it out:
Note the multiple equivalence points and notice that they are almost straight lines at that point, indicating equal added quantities of acid and base.
Titrations
In strong acid + strong base titrations, the pH changes slowly at first, rapidly through the equivalence point of pH=7, and then slows down again. If it is being titrated in a strong acid, the pH will go up as the base is added to it. Conversely, if it is in a strong base, the pH will fall down as acid is added.
• In strong acid + weak base titrations, the pH changes slowly at the equivalence point and the pH equals the pKa of the acid. The pH is below 7.
• For the weak acid + strong base, the pH is above 7 at the equivalence point.
• If there is strong acid or strong base left over after the equivalence point, this can be used to find the pH of the solution.
Next, let's take a look at sulfuric acid. This unique polyprotic acid is the only one to be completely deprotonated after the first step:
$H_2SO_{4(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + HSO^-_{4(aq)} \nonumber$
Now let's try something a little harder. The ionization of phosphoric acid (three dissociation reactions this time) can be written like this:
Start with H3PO4:
$K_{a1}: H_3PO_{4(aq)} \rightleftharpoons H^+_{(aq)} + H_2PO^-_{4(aq)} \nonumber$
$K_{a2} : H_2PO^-_{4(aq)} \rightleftharpoons HPO_{4(aq)} + H^+_{(aq)} \nonumber$
$K_{a3} : HPO^-_{4(aq)} \rightleftharpoons H^+_{(aq)} + PO^{3-}_{4(aq)} \nonumber$
So from these above reactions we can see that it takes three steps to fully remove the H+ ion. This also means that this reaction will produce three equivalence points. Polyprotic Bases are bases that can accept at least one H+ ion, or proton, in acid-base reactions.
Common Polyprotic Bases Formula Strong/Weak Base Diprotic/Triprotic Base
Phosphate ion PO43- Weak Triprotic
Sulfate ion SO42- Very Weak Diprotic
Carbonate ion CO32- Strong Diprotic
Example 2: Some examples for calculating the constant, Kb
First, start with the reaction A3- + H2O ? HA2- + OH-
Kb1= [OH-][HA2-]/[A3-]=KW/Ka3
Then, we plug in the products over the reactants:
HA2- + H2O ? H2A- + OH-
Kb2 = [OH-][H2A2-]/[HA2-]=KW/Ka2
Finally, we are left with the third dissociation, or Kb3:
H2A- + H2O ? H3A + OH-
Kb3 = [OH-][H3A]/[H2A-]=KW/Ka1
Contributors
• Natalie Kania | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.06_Bases.txt |
Salts, when placed in water, will often react with the water to produce H3O+ or OH-. This is known as a hydrolysis reaction. Based on how strong the ion acts as an acid or base, it will produce varying pH levels. When water and salts react, there are many possibilities due to the varying structures of salts. A salt can be made of either a weak acid and strong base, strong acid and weak base, a strong acid and strong base, or a weak acid and weak base. The reactants are composed of the salt and the water and the products side is composed of the conjugate base (from the acid of the reaction side) or the conjugate acid (from the base of the reaction side). In this section of chemistry, we discuss the pH values of salts based on several conditions.
When is a salt solution basic or acidic?
There are several guiding principles that summarize the outcome:
1. Salts that are from strong bases and strong acids do not hydrolyze. The pH will remain neutral at 7. Halides and alkaline metals dissociate and do not affect the H+ as the cation does not alter the H+ and the anion does not attract the H+ from water. This is why NaCl is a neutral salt. In General: Salts containing halides (except F-) and an alkaline metal (except Be2+) will dissociate into spectator ions.
2. Salts that are from strong bases and weak acids do hydrolyze, which gives it a pH greater than 7. The anion in the salt is derived from a weak acid, most likely organic, and will accept the proton from the water in the reaction. This will have the water act as an acid that will, in this case, leaving a hydroxide ion (OH-). The cation will be from a strong base, meaning from either the alkaline or alkaline earth metals and, like before, it will dissociate into an ion and not affect the H+.
3. Salts of weak bases and strong acids do hydrolyze, which gives it a pH less than 7. This is due to the fact that the anion will become a spectator ion and fail to attract the H+, while the cation from the weak base will donate a proton to the water forming a hydronium ion.
4. Salts from a weak base and weak acid also hydrolyze as the others, but a bit more complex and will require the Ka and Kb to be taken into account. Whichever is the stronger acid will be the dominate factor in determining whether it is acidic or basic. The cation will be the acid, and the anion will be the base and will form either form a hydronium ion or a hydroxide ion depending on which ion reacts more readily with the water.
Salts of Polyprotic Acids
Do not be intimidated by the salts of polyprotic acids. Yes, they're bigger and "badder" then most other salts. But they can be handled the exact same way as other salts, just with a bit more math. First of all, we know a few things:
• It's still just a salt. All of the rules from above still apply. Luckily, since we're dealing with acids, the pH of a salt of polyprotic acid will always be greater than 7.
• The same way that polyprotic acids lose H+ stepwise, salts of polyprotic acids gain H+ in the same manner, but in reverse order of the polyprotic acid.
Take for example dissociation of $\ce{H2CO3}$, carbonic acid.
$\ce{H2CO3(aq) + H2O(l) <=> H3O^{+}(aq) + HCO^{-}3(aq)} \nonumber$
with $K_{a1} = 2.5 \times 10^{-4}$
$\ce{HCO^{-}3(aq) + H2O(l) <=> H3O^{+}(aq) + CO^{2-}3(aq)} \nonumber$
with $K_{a2} = 5.61 \times 10^{-11}$.
This means that when calculating the values for Kb of CO32-, the Kb of the first hydrolysis reaction will be $K_{b1} = \dfrac{K_w}{K_{a2}}$ since it will go in the reverse order.
Summary of Acid Base Properties of Salts
Type of Solution Cations Anions pH
Acidic
From weak bases NH4+, Al3+, Fe3+
From strong acids: Cl-, Br-, I-, NO3-, ClO4-
< 7
Basic From strong bases: Group 1 and Group 2, but not Be2+
From weak acids: F-, NO2-, CN-, CH3COO-
> 7
Neutral
From strong bases: Group 1 and Group 2, but not Be2+.
From strong acids: Cl-, Br-, I-, NO3-, ClO4-
= 7
Questions
1. Predict whether the pH of each of the following salts placed into water is acidic, basic, or neutral.
1. NaOCl(s)
2. KCN(s)
3. NH4NO3(s)
1. Find the pH of a solution of .200 M NH4NO3 where (Ka = 1.8 * 10-5).
2. Find the pH of a solution of .200 M Na3PO4 where (Ka1 = 7.25 * 10-5, Ka2 = 6.31 * 10-8, Ka3 = 3.98 * 10-3).
Answers
1
1. The ions present are Na+ and OCl- as shown by the following reaction:
$NaOCl _{(s)} \rightarrow Na^+_{(aq)} + OCl^-_{(aq)}$
While Na+ will not hydrolyze, OCl- will (remember that it is the conjugate base of HOCl). It acts as a base, accepting a proton from water.
$OCl^-_{(aq)} + H_2O_{(l)} \rightleftharpoons HOCl_{(aq)} + OH^-_{(aq)}$
Na+ is excluded from this reaction since it is a spectator ion.
Therefore, with the production of OH-, it will cause a basic solution and raise the pH above 7.
$pH>7$
1. The KCN(s) will dissociate into K+(aq) and CN_(aq) by the following reaction:
$KCN_{(s)}\rightarrow K^+_{(aq)} + CN^-_{(aq)}$
K+ will not hydrolyze, but the CN- anion will attract an H+away from the water:
$CN^-_{(aq)} + H_2O_{(l)}\rightleftharpoons HCN_{(aq)} + OH^-_{(aq)}$
Because this reaction produces OH-, the resulting solution will be basic and cause a pH>7.
$pH>7$
1. The NH4NO3 (s) will dissociate into NH4+ and NO3- by the following reaction:
$NH_4NO_{3(s)} \rightarrow NH^+_{4(aq)} + NO^-_{3(aq)}$
Now, NO3- won't attract an H+ because it is usually from a strong acid. This means the Kb will be very small. However, NH4+ will lose an electron and act as an acid (NH4+ is the conjugate acid of NH3) by the following reaction:
$NH^+_{4(aq)} + H_2O_{(l)} \rightleftharpoons NH_{3(aq)} + H_3O^+_{(aq)}$
This reaction produces a hydronium ion, making the solution acidic, lowering the pH below 7.
$pH<7$
1. $NH^+ _{4(aq)} + H_2O {(l)} \rightleftharpoons NH_{3(aq)} + H_3O_{(aq)}$
$\dfrac{x^2}{0.2-x} = \dfrac{1*10^{-14}}{1.8 \times 10^{-5}}$
$x = 1.05*10^-5 M = [H_3O^+]$
$pH = 4.98$
1. $PO^3-_{4(aq)} + H_2O_{(l)} \rightleftharpoons HPO^{2-}_{4(aq)} + OH^-_{(aq)}$
The majority of the hydroxide ion will come from this first step. So only the first step will be completed here. To complete the other steps, follow the same manner of this calculation.
$\dfrac{x^2}{0.2-x}=\dfrac{1*10^-14}{3.98 \times 10{-13}}$
$x = 0.0594 = [OH^-]$
$pH = 12.77$
Practice Questions
1. Why does a salt containing a cation from a strong base and an anion from a weak acid form a basic solution?
2. Why does a salt containing a cation from a weak base and an anion from a strong acid form an acidic solution?
3. How do the Ka or Kb values help determine whether a weak acid or weak base will be the dominant driving force of a reaction?
The answers to these questions can be found in the attached files section at the bottom of the page.
Contributors and Attributions
• Christopher Wu (UCD), Christian Dowell (UCD), Nicole Hooper (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.08_Acid-Base_Properties_of_Salts.txt |
Unlike strong acids/bases, weak acids and weak bases do not completely dissociate (separate into ions) at equilibrium in water, so calculating the pH of these solutions requires consideration of a unique ionization constant and equilibrium concentrations. Although this is more difficult than calculating the pH of a strong acid or base solution, most biochemically important acids and bases are considered weak, and so it is very useful to understand how to calculate the pH of these substances. The same basic method can be used to determine the pH of aqueous solutions of many different weak acids and bases.
Introduction
An aqueous solution of a weak acid or base contains both the protonated and unprotonated forms of the compound, so an ICE table can be made and used to plug in concentrations into an equilibrium constant expression. The ionization constant for the acid (Ka) or base (Kb) is a measure of how readily the acid donates protons or how readily a base accepts protons. Because you are calculating pH, you must solve for the unknown concentration of hydronium ions in solution at equilibrium.
The first step in calculating the pH of an aqueous solution of any weak acid or base is to notice whether the initial concentration is high or low relative to 10-7 M (the concentration of hydronium and hydroxide ions in water due to the autoionization of water). If the concentration of the acid or base is very close to or less than 10-7 M, then the solution is considered dilute and additional steps must be taken to calculate pH.
Weak Acids and Bases
You must first be familiar with equilibrium constant expressions and how to write them for a chemical reaction. Then, by making an ICE table, you can find unknown concentration values that can be plugged into this equilibrium expression.
Example $1$: Vinegar as a Weak Acid
What is the pH of 1.5 L of a vinegar that is 3% acetic acid by mass? (Ka = 1.8 x 10-5)
Solution
To start, you must find the initial concentration of acetic acid in the vinegar. Assume that the vinegar is really just a solution of acetic acid in water, and that density = 1 g/mL.
So if the vinegar is 3% acetic acid by mass and the molar mass of HC2H3O2 = 60.05 g/mol, then
$\dfrac{1.5L,vinegar}{} \times \dfrac{1000mL}{1L} \times \dfrac{1g}{1mL} \times \dfrac{3g,acetic acid}{100g,vinegar} \times \dfrac{1mol,acetic acid}{60.05g,acetic acid} = 0.75 \;mol\; HC_2H_3O_2 \nonumber$
Divide 0.75 mol by 1.5 L to get an initial concentration of 0.50 M.
Now make an ICE table, considering the ionization of acetic acid in water into acetate ion and hydronium ion. Because only solutes and gases are incorporated into the equilibrium expression, you can ignore the concentration of water (a pure liquid) in our calculations.
$HC_2H_3O_{2(aq)} + H_2O_{(l)} \rightleftharpoons C_2H_3O^-_{2(aq)} + H_3O^+_{(aq)} \nonumber$
HC2H3O2 H2O C2H3O2- H3O+
Initial 0.5 --- 0 0
Change -x --- +x +x
Equilibrium 0.5 - x --- x x
For every acetic acid molecule that dissociates, one acetate ion and one hydronium ion is produced. This can be represented by subtracting "x" from the original acetic acid concentration, and adding "x" to the original concentrations of the dissociated ions.
You can create a modified equilibrium constant expression
$K_a = \dfrac{[C_2H_3O_2^-][H_3O^+]}{[HC_2H_3O_2]} \nonumber$
and then plug in the concentration values you found in the ICE table
$1.8 \times 10^{-5} = \dfrac{x^2}{0.5 - x} \nonumber$
so
$x^2 + (1.8 \times 10^{-5})x - (9 \times 10^{-6}) = 0 \nonumber$
then use the quadratic formula to calculate
$x = 0.0030\; M = [H_3O^+] \nonumber$
which can be plugged into the formula
$pH = -\log[H_3O^+] \nonumber$
$-\log(0.0030) = pH = 2.5 \nonumber$
The same thing can be done for calculating the pH of a weak base.
Example $2$: Ammonia as a Weak Base
What is the pH of a $7.0 \times 10^{-3}$ M NH3 solution? (pKb = 4.74)
Solution
$NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH^-_{(aq)} \nonumber$
NH3 H2O NH4+ OH-
I 7.0 x 10-3 --- 0 0
C -x --- +x +x
E 7.0x10-3 - x --- x x
Instead of Kb, you were given pKb. So to get Kb
pKb = -log(Kb) = 4.74
Kb = 10-4.74 = 1.8 x 10-5
Plug these values into the equilibrium expression to get
$K_b = \dfrac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5} = \dfrac{x^2}{7 \times 10^{-3} - x} \nonumber$
and use the quadratic formula to find that
$x = 3.46 \times 10^{-4}\; M = [OH^-] \nonumber$
so
$pOH = -\log(3.46 \times 10^{-4}) = 3.46 \nonumber$
and in water at 25 degrees
$pH + pOH = 14 \nonumber$
Hence:
$14 - 3.46 = pH = 10.54 \nonumber$
Weak Polyprotic Acids and Bases
Polyprotic acids have more than one proton to donate to water, and so they have more than one ionization constant (Ka1, Ka2, etc) that can be considered. Polyprotic bases take more than one proton from water, and also have more than one ionization constant (Kb1, Kb2, etc).
Most often the first proton exchange is the only one that considerably affects pH. This is discussed more at the end of the first example.
Example $3$: Citric Acid as a Polyprotic Acid
What is the pH of a grapefruit that contains 0.007 M citric acid solution (C6H8O7)?
(Ka1 = 7.5 x 10-4, Ka2 = 1.7 x 10-5, Ka3 = 4.0 x 10-7)
Solution
Make an ICE table for the first dissociation
$C_6H_8O_{7(aq)} + H_2O_{(l)} \rightleftharpoons C_6H_7O_7^- + H_3O^+_{(aq)} \nonumber$
C6H8O7 H2O C6H7O7- H3O+
I 0.007 --- 0 0
C -x --- +x +x
E 0.007 - x --- x x
$K_{a1} = \dfrac{[C_6H_7O_7^-][H_3O^+]}{[C_6H_8O_7]} = 7.5 \times 10^{-4} = \dfrac{x^2}{0.007 - x} \nonumber$
and use the quadratic formula to find that
$x = 0.00195 \;M = [H_3O^+] \nonumber$
Then a second ICE table can be made for the second dissociation
$C_6H_7O^-_{7(aq)} + H_2O_{(l)} \rightleftharpoons C_6H_6O^{2-}_7 + H_3O^+_{(aq)} \nonumber$
C6H7O7- H2O C6H6O72- H3O+
I 0.00195 --- 0 0.00195
C -x --- +x +x
E 0.00195 - x --- x 0.00195 + x
Remember that, for the first dissociation, x = [H3O+] = [C6H7O7-], so you can plug in the first value of x in for the initial concentrations of C6H7O7- and H3O+.
$K_{a2} = \dfrac{[C_6H_6O_7^{2-}][H_3O^+]}{[C_6H_7O_7^-]} = 1.7 \times 10^{-5} = \dfrac{(x)(0.00195 + x)}{0.00195- x} \nonumber$
and use the quadratic formula to find that
$x = 1.67 \times 10^{-5} \nonumber$
$[H_3O^+] = 0.00195 + 1.67 \times 10^{-5} = 0.00197 \;M \nonumber$
$-\log(0.00197) = pH = 2.71 \nonumber$
Note that if you ignored the addition of hydronium from the second dissociation, then [H3O+] = 0.00195 M, and using this value to calculate pH still gives you the answer of 2.71.
So even though you made two ICE tables (you could even make a third table for Ka3), the protons donated in the second dissociation were negligible compared to the first dissociation. So you can see that it is really only the first dissociation that affects pH. Most often this is the case, and only one ICE table is necessary. It is up to you how certain you want to be and how many ICE tables you want to make when you calculate these problems.
Example $4$: Soda Ash as a Polyprotic Base
What is the pH of a saturated solution of sodium carbonate (Na2CO3)?
(solubility in water is 21.6 g/100mL at room temperature and for carbonic acid, H2CO3, Ka1 = 4.5 x 10-7, Ka2 = 4.7 x 10-11
Solution
First, you have to find the find the initial concentration of CO32- which can be found from
$\dfrac{21.6g,Na_2CO_3}{} \times \dfrac{1\;mol}{105.99\;g} = 0.204\; mol\; Na_2CO_3 = 0.204\; mol\; CO_3^{2-} \nonumber$
then divide 0.204 mol by 0.100 L to get 2.04 M CO32-
Plug into an ICE table
$CO^{2-}_{3(aq)} + H_2O_{(l)} \rightleftharpoons HCO^-_{3(aq)} + OH^-_{(aq)} \nonumber$
CO32- H2O HCO3- OH-
I 2.04 --- 0 0
C -x --- +x +x
E 2.04 - x --- x x
But notice that the equilibrium constants are for carbonic acid. If you were considering the dissociation of carbonic acid, you would write the following expressions
$K_{a1} = \dfrac{[HCO_3^-][H_3O^+]}{[H_2CO_3]} = 4.5 \times 10^{-7} \nonumber$
for
$H_2CO_3 + H_2O \rightleftharpoons HCO_3^- + H_3O^+ \nonumber$
$K_{a2} = \dfrac{[CO_3^{2-}][H_3O^+]}{[HCO_3^-]} = 4.7 \times 10^{-11} \nonumber$
for
$HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+ \nonumber$
The second acid ionization constant corresponds to the first base ionization constant (because the base reactions go backwards). To convert the second acid ionization constant to the first base ionization constant, you use the equation
$K_a \times K_b = K_w = 10^{-14} \nonumber$
so that
$K_{a2} \times K_{b1} = 10^{-14} \nonumber$
$K_{b1} = \dfrac{10^{-14}}{4.7 \times 10^{-11}} = 2.13 \times 10^{-4} \nonumber$
Use the same equation to convert the first acid ionization constant to the second base ionization constant
$K_{a1} \times K_{b2} = 10^{-14} \nonumber$
$K_{b2} = \dfrac{10^{-14}}{4.5 \times 10^{-7}} = 2.22 \times 10^{-8} \nonumber$
The expressions for the protonation of carbonate are now known to be
$K_{b1} = \dfrac{[HCO_3^-][OH^-]}{[CO_3^{2-}]} = 2.13 \times 10^{-4} \nonumber$
for
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^- \nonumber$
$K_{b2} = \dfrac{[H_2CO_3][OH^-]}{[HCO_3^-]} = 2.22 \times 10^{-8} \nonumber$
for
$HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^- \nonumber$
Plug the ICE tables values into the first equilibrium expression
$K_{b1} = \dfrac{x^2}{2.04 - x} = 2.13 \times 10^{-4} \nonumber$
and use the quadratic formula to solve
$x = 0.0207\; M = [OH^-] \nonumber$
You can ignore the second base ionization constant because it removes a negligible amount of protons from water. If you want to test this by making an ICE table, you should get that the total hydroxide concentration is 0.0207000222 M \approx 0.0207 M
so pOH = -log(0.0207) = 1.68
pH + pOH = 14
14 - 1.68 = pH = 12.32
Dilute Weak Acids and Bases
'Dilute' refers to the concentration of the acid or base in water. If the concentration is close to or below 10-7 M, then you must consider the donation of hydronium ions from water as well as from your acid or base. This is done by making an ICE table to find the protonation of the acid or base, while also incorporating the ion product of water.
Example $5$: Dilute Weak Acid
An average bee sting contains 5 micrograms of formic acid ($HCO_2H$). What is the pH of a 500 mL solution of formic acid? (pKa = 3.75)
Solution
First calculate the number of moles of formic acid was excreted.
$\dfrac{5.00 \mu g}{} \times \dfrac{1 g}{10^6 \mu g} \times \dfrac{1\;mol,HCO_2H}{46.03g,HCO_2H} = 1.09 \times 10^{-7} mol,HCO_2H \nonumber$
$\dfrac{1.09 \times 10^{-7} mol}{0.500 L} = 2.17 \times 10^{-7}\; M \nonumber$
$HCO_2H_{(aq)} + H_2O_{(l)} \rightleftharpoons CO_2H^-_{(aq)} + H_3O^+_{(aq)} \nonumber$
HCO2H H2O CO2H- H3O+
I $2.17 \times 10^{-7}$ --- 0 0
C $-x$ --- +x +x
E $2.17 \times 10^{-7} - x$ --- x x
A second ICE table can be made for the autoionization of water
$2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)} \nonumber$
2H2O H3O+ OH-
I --- 0 0
C --- +y +y
E --- y y
Notice that total [H3O+] = x + y
$pK_a = 3.75 \nonumber$
$K_a = 10^{-3.75} = 1.78 \times 10^{-4} \nonumber$
So you can simultaneously solve both equations
$K_a = \dfrac{[CO_2H^-][H_3O^+]}{[HCO_2H]} = \dfrac{x(x + y)}{2.17 \times 10^{-7} - x} = 1.78 \times 10^{-4} \nonumber$
and
$K_w = [H_3O^+][OH^-] = (x + y)(y) = 10^{-14} \nonumber$
These calculations can be tricky, and it is very easy to make mistakes. It is usually easier to use variables to solve these problems instead of handling awkward numbers.
For this problem use
• a = Ka = 1.78 x 10-4
• c = initial [HCO2H] = 2.17 x 10-7
• w = Kw = 10-14
$a = \dfrac{x(x+y)}{c-x} \nonumber$
and
$(x+y)(y) = w \nonumber$
$ac-ax = x^2 +xy \nonumber$
and
$x=\dfrac{w}{y}-y \nonumber$
$ac-a\left(\dfrac{w}{y}-y\right)=\left(\dfrac{w}{y}-y\right)^2+\left(\dfrac{w}{y}-y\right)y \nonumber$
$ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-\dfrac{2wy}{y}+y^2+\dfrac{wy}{y}-y^2 \nonumber$
$ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-2w+y^2+w-y^2 \nonumber$
$ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-w \nonumber$
$ac-\dfrac{aw}{y}+ay+w=\dfrac{w^2}{y^2} \nonumber$
$acy^2-awy+ay^3+wy^2=w^2 \nonumber$
$ay^3+(ac+w)y^2-awy-w^2=0 \nonumber$
so when we plug back in the values
$(1.78 \times 10^{-4})y^3+((1.78 \times 10^{-4})(2.17 \times 10^{-7})+(10^{-14})) y^2-(1.78 \times 10^{-4})(10^{-14})y-(10^{-14})^2=0 \nonumber$
$(1.78 \times 10^{-4})y^3+(3.86 \times 10^{-11})y^2-(1.78 \times 10^{-18})y-10^{-28}=0 \nonumber$
and use a graphing calculator to find that
$y = 3.91 \times 10^{-8} \nonumber$
and
$x=\dfrac{w}{y}-y = \dfrac{10^{-14}}{3.91 \times 10^{-8}}-3.91 \times 10^{-8} = 2.17 \times 10^{-7} \nonumber$
and total
$[H_3O^+] = (x + y) = (2.17 \times 10^{-7} + 3.91 \times 10^{-8}) = 2.56 \times 10^{-7} \;M \nonumber$
so
$-\log[H_3O^+] = -\log(2.56 \times 10^{-7}) = pH = 6.59 \nonumber$
Compare this value to pH = 6.66, which is what would have been calculated if the autoprotonization of water was not considered.
Buffer Solutions
Buffer solutions resist pH change when more acid or base is added. They are made from a weak acid and its conjugate base or a weak base and its conjugate acid. The Henderson-Hasselbalch approximation can be used to find the pH of a buffer solution, and is derived from the acid equilibrium expression.
$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons A^-_{(aq)} + H_3O^+_{(aq)} \nonumber$
$K_a = \dfrac{[A^-][H_3O^+]}{[HA]} \nonumber$
$\log(K_a) = log\left( \dfrac{[A^-][H_3O^+]}{[HA]} \right) \nonumber$
$\log(K_a) = \log[H_3O^+] + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber$
$-pK_a = -pH + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber$
$pH = pK_a + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber$
A similar equation can be used for bases
$pOH = pK_b + \log \left( \dfrac{[HB^+]}{[B]} \right) \nonumber$
Example $6$: Blood
The pH of blood plasma is 7.40, and is maintained by a carbonic acid/hydrogen carbonate buffer system. What mass of sodium bicarbonate (NaHCO3) should be added to a one liter solution of 0.250 M H2CO3 to maintain the solution at pH of 7.40? (pKa = 6.35)
Solution
$pH = pK_a + \log \left( \dfrac{[HCO_3^-]}{[H_2CO_3]} \right) \nonumber$
$7.40 = 6.35 + \log \left( \dfrac{[HCO_3^-]}{[0.250]} \right) \nonumber$
$1.05 = \log[HCO_3^-] -\log(0.250) \nonumber$
$0.448 = \log[HCO_3^-] \nonumber$
$[HCO_3^-] = 0.356\; M \nonumber$
$(0.356\; M) \times (1\; L\; solution) = 0.356\; mol\; HCO_3^- \nonumber$
$[0.356 mol\,HCO_3^{-} \times \dfrac{1 mol\,NaHCO_3}{1 mol\,HCO_3^-} \times \dfrac{84.01g}{1 mol,NaHCO_3} = 29.9 g \text{sodium bicarbonate} \nonumber$
It is also possible to use the Henderson-Hasselbalch equation to find pKa, pH, or [HA] if the other variables are given or calculated. Also notice that because $\dfrac{[A^-]}{[HA]}$ will cancel out the unit of volume, moles of $HA$ and $A^-$ can be used instead of molarity. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.09%3A_Acid_Solutions_that_Water_Contributes_p.txt |
Oxides are chemical compounds with one or more oxygen atoms combined with another element (e.g. Li2O). Oxides are binary compounds of oxygen with another element, e.g., CO2, SO2, CaO, CO, ZnO, BaO2, H2O, etc. These are termed as oxides because here, oxygen is in combination with only one element. Based on their acid-base characteristics oxides are classified as acidic, basic, amphoteric or neutral:
1. An oxide that combines with water to give an acid is termed as an acidic oxide.
2. The oxide that gives a base in water is known as a basic oxide.
3. An amphoteric solution is a substance that can chemically react as either acid or base.
4. However, it is also possible for an oxide to be neither acidic nor basic, but is a neutral oxide.
There are different properties which help distinguish between the three types of oxides. The term anhydride ("without water") refers to compounds that assimilate H2O to form either an acid or a base upon the addition of water.
Acidic Oxides
Acidic oxides are the oxides of non-metals (Groups 14-17) and these acid anhydrides form acids with water:
• Sulfurous Acid
$\ce{SO_2 + H_2O \rightarrow H_2SO_3} \label{1}$
• Sulfuric Acid
$\ce{ SO_3 + H_2O \rightarrow H_2SO_4} \label{2}$
• Carbonic Acid
$\ce{CO_2 + H_2O \rightarrow H_2CO_3} \label{3}$
Acidic oxides are known as acid anhydrides (e.g., sulfur dioxide is sulfurous anhydride and sulfur trioxide is sulfuric anhydride) and when combined with bases, they produce salts, e.g.,
$\ce{ SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O} \label{4}$
Basic Oxides
Generally Group 1 and Group 2 elements form bases called base anhydrides or basic oxides e.g.,
$\ce{K_2O \; (s) + H_2O \; (l) \rightarrow 2KOH \; (aq) } \label{5}$
Basic oxides are the oxides of metals. If soluble in water, they react with water to produce hydroxides (alkalies) e.g.,
$\ce{ CaO + H_2O \rightarrow Ca(OH)_2} \label{6}$
$\ce{ MgO + H_2O \rightarrow Mg(OH)_2} \label{7}$
$\ce{ Na_2O + H_2O \rightarrow 2NaOH } \label{8}$
These metallic oxides are known as basic anhydrides. They react with acids to produce salts, e.g.,
$\ce{ MgO + 2HCl \rightarrow MgCl_2 + H_2O } \label{9}$
$\ce{ Na_2O + H_2SO_4 \rightarrow Na_2SO_4 + H_2O} \label{10}$
Amphoteric Oxides
An amphoteric solution is a substance that can chemically react as either acid or base. For example, when HSO4- reacts with water it will make both hydroxide and hydronium ions:
$HSO_4^- + H_2O \rightarrow SO_4^{2^-} + H_3O^+ \label{11}$
$HSO_4^- + H_2O \rightarrow H_2SO_4 + OH^- \label{12}$
Amphoteric oxides exhibit both basic as well as acidic properties. When they react with an acid, they produce salt and water, showing basic properties. While reacting with alkalies they form salt and water showing acidic properties.
• For example $ZnO$ exhibits basic behavior with $HCl$
$ZnO + 2HCl \rightarrow \underset{\large{zinc\:chloride}}{ZnCl_2}+H_2O\,(basic\: nature) \label{13}$
• and acidic behavior with $NaOH$
$ZnO + 2NaOH \rightarrow \underset{\large{sodium\:zincate}}{Na_2ZnO_2}+H_2O\,(acidic\: nature) \label{14}$
• Similarly, $Al_2O_3$ exhibits basic behavior with $H_2SO_4$
$Al_2O_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2O\,(basic\: nature) \label{15}$
• and acidic behavior with $NaOH$
$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2+H_2O\,(acidic\: nature) \label{16}$
Neutral Oxides
Neutral oxides show neither basic nor acidic properties and hence do not form salts when reacted with acids or bases, e.g., carbon monoxide (CO); nitrous oxide (N2O); nitric oxide (NO), etc., are neutral oxides.
Peroxides and Dioxides
Oxides: Group 1 metals react rapidly with oxygen to produce several different ionic oxides, usually in the form of $M_2O$. With the oyxgen exhibiting an oxidation number of -2.
$4 Li + O_2 \rightarrow 2Li_2O \label{19}$
Peroxides: Often Lithium and Sodium reacts with excess oxygen to produce the peroxide, $M_2O_2$. with the oxidation number of the oxygen equal to -1.
$H_2 + O_2 \rightarrow H_2O_2 \label{20}$
Superoxides: Often Potassium, Rubidium, and Cesium react with excess oxygen to produce the superoxide, $MO_2$. with the oxidation number of the oxygen equal to -1/2.
$Cs + O_2 \rightarrow CsO_2 \label{21}$
A peroxide is a metallic oxide which gives hydrogen peroxide by the action of dilute acids. They contain more oxygen than the corresponding basic oxide, e.g., sodium, calcium and barium peroxides.
$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 \label{22}$
$Na_2O_2 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2 \label{23}$
Dioxides like PbO2 and MnO2 also contain higher percentage of oxygen like peroxides and have similar molecular formulae. These oxides, however, do not give hydrogen peroxide by action with dilute acids. Dioxides on reaction with concentrated HCl yield Cl2 and on reacting with concentrated H2SO4 yield O2.
$PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O \label{24}$
$2PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O + O_2 \label{25}$
Compound Oxides
Compound oxides are metallic oxides that behave as if they are made up of two oxides, one that has a lower oxidation and one with a higher oxidation of the same metal, e.g.,
$\textrm{Red lead: } Pb_3O_4 = PbO_2 + 2PbO \label{26}$
$\textrm{Ferro-ferric oxide: } Fe_3O_4 = Fe_2O_3 + FeO \label{27}$
On treatment with an acid, compound oxides give a mixture of salts.
$\underset{\text{Ferro-ferric oxide}}{Fe_3O_4} + 8HCl \rightarrow \underset{\text{ferric chloride}}{2FeCl_3} + \underset{\text{ferrous chloride}}{FeCl_2} + 4H_2O \label{28}$
Preparation of Oxides
Oxides can be generated via multiple reactions. Below are a few.
By direct heating of an element with oxygen: Many metals and non-metals burn rapidly when heated in oxygen or air, producing their oxides, e.g.,
$2Mg + O_2 \xrightarrow{Heat} 2MgO \nonumber$
$2Ca + O_2 \xrightarrow{Heat} 2CaO \nonumber$
$S + O_2 \xrightarrow{Heat} SO_2 \nonumber$
$P_4 + 5O_2 \xrightarrow{Heat} 2P_2O_5 \nonumber$
By reaction of oxygen with compounds at higher temperatures: At higher temperatures, oxygen also reacts with many compounds forming oxides, e.g.,
• sulfides are usually oxidized when heated with oxygen.
$2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2 \nonumber$
$2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2 \nonumber$
• When heated with oxygen, compounds containing carbon and hydrogen are oxidized.
$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \nonumber$
• By thermal decomposition of certain compounds like hydroxides, carbonates, and nitrates
$CaCO_3 \xrightarrow{\Delta} CaO + CO_2 \nonumber$
$2Cu(NO_3)_2 \xrightarrow{\Delta} 2CuO + 4NO_2 + O_2 \nonumber$
$Cu(OH)_2 \xrightarrow{\Delta} CuO + H_2O \nonumber$
By oxidation of some metals with nitric acid
$2Cu + 8HNO_3 \xrightarrow{Heat} 2CuO + 8NO_2 + 4H_2O + O_2 \nonumber$
$Sn + 4HNO_3 \xrightarrow{Heat} SnO_2 + 4NO_2 + 2H_2O \nonumber$
By oxidation of some non-metals with nitric acid
$C + 4HNO_3 \rightarrow CO_2 + 4NO_2 + 2H_2O \nonumber$
The oxides of elements in a period become progressively more acidic as one goes from left to right in a period of the periodic table. For example, in third period, the behavior of oxides changes as follows:
$\underset{\large{Basic}}{\underbrace{Na_2O,\: MgO}}\hspace{20px} \underset{\large{Amphoteric}}{\underbrace{Al_2O_3,\: SiO_2}}\hspace{20px} \underset{\large{Acidic}}{\underbrace{P_4O_{10},\: SO_3,\:Cl_2O_7}}\hspace{20px}$
If we take a closer look at a specific period, we may better understand the acid-base properties of oxides. It may also help to examine the physical properties of oxides, but it is not necessary. Metal oxides on the left side of the periodic table produce basic solutions in water (e.g. Na2O and MgO). Non-metal oxides on the right side of the periodic table produce acidic solutions (e.g. Cl2O, SO2, P4O10). There is a trend within acid-base behavior: basic oxides are present on the left side of the period and acidic oxides are found on the right side.
Aluminum oxide shows acid and basic properties of an oxide, it is amphoteric. Thus Al2O3 entails the marking point at which a change over from a basic oxide to acidic oxide occurs. It is important to remember that the trend only applies for oxides in their highest oxidation states. The individual element must be in its highest possible oxidation state because the trend does not follow if all oxidation states are included. Notice how the amphoteric oxides (shown in blue) of each period signify the change from basic to acidic oxides,
Groups
1 2 3 14 15 16 17
Li
Be
B
C
N
O
F
Na
Mg
Al
Si
P
S
Cl
K
Ca
Ga
Ge
As
Se
Br
Rb
Sr
In
Sn
Sb
Te
I
Cs
Ba
Tl
Pb
Bi
Po
At
The figure above show oxides of the s- and p-block elements.
purple: basic oxides blue: amphoteric oxides pink: acidic oxides
Problems
1. Can an oxide be neither acidic nor basic?
2. $Rb + O_2\: (excess) \rightarrow \:?$
3. $Na + O_2 \rightarrow \:?$
4. BaO2 is which of the following: hydroxide, peroxide, or superoxide?
5. What is an amphoteric solution?
6. Why is it difficult to obtain oxygen directly from water?
Solutions
1. Yes, an example is carbon monoxide (CO). CO doesn’t produce a salt when reacted with an acid or a base.
2. $Rb + O_2 \; (excess) \rightarrow RbO_2$
With the presence of excess oxygen, Rubidium forms a superoxide. Please review section regarding basic oxides above for more detail.
3. $2 Na + O_2 \rightarrow Na_2O$
Note: The problem does not specify that the oxygen was in excess, so it cannot be a peroxide. Please review section regarding basic oxides for more detail.
4. BaO2 is a peroxide. Barium has an oxidation state of +2 so the oxygen atoms have oxidation state of -1. As a result, the compound is a peroxide, but more specifically referred to as barium peroxide.
5. An amphoteric solution is a substance that can chemically react as either acid or base. See section above on Properties of Amphoteric Oxides for more detail.
6. Water as such is a neutral stable molecule. It is difficult to break the covalent O-H bonds easily. Hence, electrical energy through the electrolysis process is applied to separate dioxygen from water. When a small amount of acid is added to water ionization is initiated which helps in electrochemical reactions as follows.
$[H_2O\:(acidulated)\rightleftharpoons H^+\,(aq)+OH]^-\times4 \nonumber$
At cathode:
$[H^+\,(aq)+e^-\rightarrow\dfrac{1}{2}H_2(g)]\times4 \nonumber$
At anode:
$4OH^-\,(aq)\rightarrow O_2+2H_2O + 4e^- \nonumber$
Net reaction:
$2H_2O \xrightarrow{\large{electrolysis}} 2H_2\,(g) + O_2\,(g) \nonumber$
Oxygen can thus be obtained from acidified water by its electrolysis.
Contributors and Attributions
Binod Shrestha (University of Lorraine)
7.11: Approaches to Solve Acid-Base Problems
Thumbnail: Major structural changes accompany binding of the Lewis base to the coordinatively unsaturated, planar Lewis acid BF3. (Public Domain; Ben Mills ). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.10%3A_Strong_Acid_Solutions_that_Water_Contributes_pH.txt |
Chemistry is the study of matter and the changes that material substances undergo. Of all the scientific disciplines, it is perhaps the most extensively connected to other fields of study. As you begin your study of college chemistry, those of you who do not intend to become professional chemists may well wonder why you need to study chemistry. You will soon discover that a basic understanding of chemistry is useful in a wide range of disciplines and career paths. You will also discover that an understanding of chemistry helps you make informed decisions about many issues that affect you, your community, and your world. A major goal of this text is to demonstrate the importance of chemistry in your daily life and in our collective understanding of both the physical world we occupy and the biological realm of which we are a part.
• 1.1: Studying Chemistry
An understanding of chemistry is essential for understanding much of the natural world and is central to many other disciplines. Chemistry is the study of matter and the changes material substances undergo. It is essential for understanding much of the natural world and central to many other scientific disciplines, including astronomy, geology, paleontology, biology, and medicine.
• 1.2: Classification of Matter
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 & properties.
• 1.3: Properties of Matter
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.
• 1.4: Units of Measurement
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.
• 1.5: Uncertainty in Measurement
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.
• 1.6: Dimensional Analysis
Dimensional analysis is used in numerical calculations, and in converting units. It can help us identify whether an equation is set up correctly (i.e. the resulting units should be as expected). Units are treated similarly to the associated numerical values, i.e., if a variable in an equation is supposed to be squared, then the associated dimensions are squared, etc.
• 1.E: Matter and Measurement (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 1.S: Matter and Measurement (Summary)
This is the summary Module for the chapter "Matter and Measurement" in the Brown et al. General Chemistry Textmap.
Thumbnail: Chemicals in flasks (including Ammonium hydroxide and Nitric acid) lit in different colours (CC BY 2.0 Generic; Joe Sullivan via Flickr)
01: Introduction - Matter and Measurement
Learning Objectives
• To recognize the breadth, depth, and scope of chemistry.
Chemistry is the study of matter and the changes that material substances undergo. Of all the scientific disciplines, it is perhaps the most extensively connected to other fields of study. Geologists who want to locate new mineral or oil deposits use chemical techniques to analyze and identify rock samples. Oceanographers use chemistry to track ocean currents, determine the flux of nutrients into the sea, and measure the rate of exchange of nutrients between ocean layers. Engineers consider the relationships between the structures and the properties of substances when they specify materials for various uses. Physicists take advantage of the properties of substances to detect new subatomic particles. Astronomers use chemical signatures to determine the age and distance of stars and thus answer questions about how stars form and how old the universe is. The entire subject of environmental science depends on chemistry to explain the origin and impacts of phenomena such as air pollution, ozone layer depletion, and global warming.
The disciplines that focus on living organisms and their interactions with the physical world rely heavily on biochemistry, the application of chemistry to the study of biological processes. A living cell contains a large collection of complex molecules that carry out thousands of chemical reactions, including those that are necessary for the cell to reproduce. Biological phenomena such as vision, taste, smell, and movement result from numerous chemical reactions. Fields such as medicine, pharmacology, nutrition, and toxicology focus specifically on how the chemical substances that enter our bodies interact with the chemical components of the body to maintain our health and well-being. For example, in the specialized area of sports medicine, a knowledge of chemistry is needed to understand why muscles get sore after exercise as well as how prolonged exercise produces the euphoric feeling known as “runner’s high.”
Examples of the practical applications of chemistry are everywhere (Figure \(1\)). Engineers need to understand the chemical properties of the substances when designing biologically compatible implants for joint replacements or designing roads, bridges, buildings, and nuclear reactors that do not collapse because of weakened structural materials such as steel and cement. Archaeology and paleontology rely on chemical techniques to date bones and artifacts and identify their origins. Although law is not normally considered a field related to chemistry, forensic scientists use chemical methods to analyze blood, fibers, and other evidence as they investigate crimes. In particular, DNA matching—comparing biological samples of genetic material to see whether they could have come from the same person—has been used to solve many high-profile criminal cases as well as clear innocent people who have been wrongly accused or convicted. Forensics is a rapidly growing area of applied chemistry. In addition, the proliferation of chemical and biochemical innovations in industry is producing rapid growth in the area of patent law. Ultimately, the dispersal of information in all the fields in which chemistry plays a part requires experts who are able to explain complex chemical issues to the public through television, print journalism, the Internet, and popular books.
By this point, it shouldn’t surprise you to learn that chemistry was essential in explaining a pivotal event in the history of Earth: the disappearance of the dinosaurs. Although dinosaurs ruled Earth for more than 150 million years, fossil evidence suggests that they became extinct rather abruptly approximately 66 million years ago. Proposed explanations for their extinction have ranged from an epidemic caused by some deadly microbe or virus to more gradual phenomena such as massive climate changes. In 1978 Luis Alvarez (a Nobel Prize–winning physicist), the geologist Walter Alvarez (Luis’s son), and their coworkers discovered a thin layer of sedimentary rock formed 66 million years ago that contained unusually high concentrations of iridium, a rather rare metal (part (a) in Figure \(2\)). This layer was deposited at about the time dinosaurs disappeared from the fossil record. Although iridium is very rare in most rocks, accounting for only 0.0000001% of Earth’s crust, it is much more abundant in comets and asteroids. Because corresponding samples of rocks at sites in Italy and Denmark contained high iridium concentrations, the Alvarezes suggested that the impact of a large asteroid with Earth led to the extinction of the dinosaurs. When chemists analyzed additional samples of 66-million-year-old sediments from sites around the world, all were found to contain high levels of iridium. In addition, small grains of quartz in most of the iridium-containing layers exhibit microscopic cracks characteristic of high-intensity shock waves (part (b) in Figure \(2\)). These grains apparently originated from terrestrial rocks at the impact site, which were pulverized on impact and blasted into the upper atmosphere before they settled out all over the world.
Scientists calculate that a collision of Earth with a stony asteroid about 10 kilometers (6 miles) in diameter, traveling at 25 kilometers per second (about 56,000 miles per hour), would almost instantaneously release energy equivalent to the explosion of about 100 million megatons of TNT (trinitrotoluene). This is more energy than that stored in the entire nuclear arsenal of the world. The energy released by such an impact would set fire to vast areas of forest, and the smoke from the fires and the dust created by the impact would block the sunlight for months or years, eventually killing virtually all green plants and most organisms that depend on them. This could explain why about 70% of all species—not just dinosaurs—disappeared at the same time. Scientists also calculate that this impact would form a crater at least 125 kilometers (78 miles) in diameter. Recently, satellite images from a Space Shuttle mission confirmed that a huge asteroid or comet crashed into Earth’s surface across the Yucatan’s northern tip in the Gulf of Mexico 65 million years ago, leaving a partially submerged crater 180 kilometers (112 miles) in diameter (Figure \(3\)). Thus simple chemical measurements of the abundance of one element in rocks led to a new and dramatic explanation for the extinction of the dinosaurs. Though still controversial, this explanation is supported by additional evidence, much of it chemical.
This is only one example of how chemistry has been applied to an important scientific problem. Other chemical applications and explanations that we will discuss in this text include how astronomers determine the distance of galaxies and how fish can survive in subfreezing water under polar ice sheets. We will also consider ways in which chemistry affects our daily lives: the addition of iodine to table salt; the development of more effective drugs to treat diseases such as cancer, AIDS (acquired immunodeficiency syndrome), and arthritis; the retooling of industry to use nonchlorine-containing refrigerants, propellants, and other chemicals to preserve Earth’s ozone layer; the use of modern materials in engineering; current efforts to control the problems of acid rain and global warming; and the awareness that our bodies require small amounts of some chemical substances that are toxic when ingested in larger doses. By the time you finish this text, you will be able to discuss these kinds of topics knowledgeably, either as a beginning scientist who intends to spend your career studying such problems or as an informed observer who is able to participate in public debates that will certainly arise as society grapples with scientific issues.
Summary
An understanding of chemistry is essential for understanding much of the natural world and is central to many other disciplines. Chemistry is the study of matter and the changes material substances undergo. It is essential for understanding much of the natural world and central to many other scientific disciplines, including astronomy, geology, paleontology, biology, and medicine. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.01%3A_Studying_Chemistry.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 118 elements are known, but millions of chemical compounds have been prepared from these 118 elements. The known elements are listed in the periodic table.
Different Definitions of Matter: Different Definitions of Matter, YouTube (opens in new window) [youtu.be]
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 called electrolysis. 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 A
solution
Answer B
element
Answer C
heterogeneous mixture
Answer D
compound
Different Definitions of Changes: Different Definitions of Changes, YouTube(opens in new window) [youtu.be] (opens in new window)
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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.02%3A_Classification_of_Matter.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:
\begin{align*} \text{density} &={\text{mass} \over \text{volume}} \[4pt] \rho &={m \over V} \label{Eq1} \end{align*}
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” (Figure $2$).
Table $1$: Densities of Common Substances
Substance Density at 25 °C (g/cm3) Substance Density at 25 °C (g/cm3)
blood 1.035 corn oil 0.922
body fat 0.918 mayonnaise 0.910
whole milk 1.030 honey 1.420
Physical Property and Change
Physical changes are changes in which no chemical bonds are broken or formed. This means that the same types of compounds or elements that were there at the beginning of the change are there at the end of the change. Because the ending materials are the same as the beginning materials, the properties (such as color, boiling point, etc) will also be the same. Physical changes involve moving molecules around, but not changing them. Some types of physical changes include:
• Changes of state (changes from a solid to a liquid or a gas and vice versa)
• Separation of a mixture
• Physical deformation (cutting, denting, stretching)
• Making solutions (special kinds of mixtures) .
As an ice cube melts, its shape changes as it acquires the ability to flow. However, its composition does not change. Melting is an example of a physical change (Figure $3$), since some properties of the material change, but the identity of the matter does not. Physical changes can further be classified as reversible or irreversible. The melted ice cube may be refrozen, so melting is a reversible physical change. Physical changes that involve a change of state are all reversible. Other changes of state include vaporization (liquid to gas), freezing (liquid to solid), and condensation (gas to liquid). Dissolving is also a reversible physical change. When salt is dissolved into water, the salt is said to have entered the aqueous state. The salt may be regained by boiling off the water, leaving the salt behind.
Figure $3$: Ice Melting is a physical change. When solid water ($\ce{H_2O}$) as ice melts into a liquid (water), 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 changes occur when bonds are broken and/or formed between molecules or atoms. This means that one substance with a certain set of properties (such as melting point, color, taste, etc) is turned into a different substance with different properties. Chemical changes are frequently harder to reverse than physical changes.
One good example of a chemical change is burning paper. In contrast to the act of ripping paper, the act of burning paper actually results in the formation of new chemicals (carbon dioxide and water, to be exact). Another example of chemical change occurs when water is formed. Each molecule contains two atoms of hydrogen and one atom of oxygen chemically bonded.
Another example of a chemical change is what occurs when natural gas is burned in your furnace. This time, before the reaction we have a molecule of methane, $\ce{CH_4}$, and two molecules of oxygen, $\ce{O_2}$, while after the reaction we have two molecules of water, $\ce{H_2O}$, and one molecule of carbon dioxide, $\ce{CO_2}$. In this case, not only has the appearance changed, but the structure of the molecules has also changed. The new substances do not have the same chemical properties as the original ones. Therefore, this is a chemical change.
The combustion of magnesium metal is also chemical change (Magnesium + Oxygen → Magnesium Oxide):
$\ce{2 Mg + O_2 \rightarrow 2 MgO } \nonumber$
as is the rusting of iron (Iron + Oxygen → Iron Oxide/ Rust):
$\ce{4 Fe + 3O_2 \rightarrow 2 Fe_2O_3} \nonumber$
Using the components of composition and properties, we have the ability to distinguish one sample of matter from the others.
Different Definitions of Changes: Different Definitions of Changes, YouTube(opens in new window) [youtu.be]
Different Definitions of Properties: Different Definitions of Properties, YouTube(opens in new window) [youtu.be]
Contributors and Attributions
• Samantha Ma (UC Davis) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.03%3A_Properties_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 1015 deci d 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.
Pseudo-Si Units
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:
$V = x · y · z \nonumber$
We say, therefore, that volume has the dimensions of length-cubed:
$dim\{V\} = L^3 \nonumber$
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,
$dim\{V\} = M^0L^3T^0Q^0 Θ^0 \nonumber$
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.
Units of Temperature: Units of Temperature, YouTube(opens in new window) [youtu.be] (opens in new window)
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:
$1\; Pa = 1\; N \;m^{–2} \nonumber$
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
$f = ma = mg = (10340 \;kg)(9.81\; m\; s^{–2}) = 1.013 \times 10^5 \;kg \;m \;s^{–2} = 1.013 \times 10^5\; N \nonumber$
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.
International System of Units (SI Units): International System of Units (SI Units), YouTube(opens in new window) [youtu.be]
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_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.04%3A_Units_of_Measurement.txt |
Learning Objectives
• To introduce the fundamental mathematical skills you will need to complete basic chemistry questions and problems
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.
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:
Masses Obtained from the three balances
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:
$\text{average} = \dfrac{\text{sum of measurements} }{\text{number of measurements}} \label{Eq1}$
2. Calculate the deviation of each measurement, which is the absolute value of the difference between each measurement and the average value:
$\text{deviation} = |\text{measurement − average}| \label{Eq2}$
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:
$\text{average} = \dfrac{\text{sum of deviations}}{\text{number of measurements}} \label{Eq3}$
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
${1.125 \;g + 1.158 \;g + 1.067\; g \over 3} = 1.117 \;g \nonumber$
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
${0.008 \:g + 0.041 \;g + 0.050 \;g \over 3} = 0.033\; g \nonumber$
The precision of this set of measurements is therefore
${0.033\;g \over 1.117\;g} \times 100 = 3.0 \% \nonumber$
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?
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.
\begin{align*} \text {precision (Zn)} &= \dfrac {0.2 \%}{93.2 \% } \times 100 = 0.2 \% \[4pt] \text {precision (Cu)} &= \dfrac {0.2 \%}{2.8 \% } \times 100 = 7 \% \end{align*} \nonumber
Significant Figures
No measurement is free from error. Error is introduced by the limitations of instruments and measuring devices (such as the size of the divisions on a graduated cylinder) and the imperfection of human senses (i.e., detection). 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.
Significant Figure Rules
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: $1 \, \text{ft} = 1, \text{in} \nonumber$
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?
Answer
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:
$3240.7 + 21.236 = 3261.9|36 \nonumber$
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,
$42.9 × 8.323 = 357.057 = 357. \nonumber$
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.
Example $5$
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
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: Significant Figures, YouTube(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.05%3A_Uncertainty_in_Measurement.txt |
Learning Objectives
• To be introduced to the dimensional analysis and how it can be used to aid basic chemistry problem solving.
• To use dimensional analysis to identify whether an equation is set up correctly in a numerical calculation
• To use dimensional analysis to facilitate the conversion of units.
Dimensional analysis is amongst the most valuable tools physical scientists use. Simply put, it is the conversion between an amount in one unit to the corresponding amount in a desired unit using various conversion factors. This is valuable because certain measurements are more accurate or easier to find than others.
A Macroscopic Example: Party Planning
If you have every planned a party, you have used dimensional analysis. The amount of beer and munchies you will need depends on the number of people you expect. For example, if you are planning a Friday night party and expect 30 people you might estimate you need to go out and buy 120 bottles of sodas and 10 large pizza's. How did you arrive at these numbers? The following indicates the type of dimensional analysis solution to party problem:
$(30 \; \cancel{humans}) \times \left( \dfrac{\text{4 sodas}}{1 \; \cancel{human}} \right) = 120 \; \text{sodas} \label{Eq1}$
$(30 \; \cancel{humans}) \times \left( \dfrac{\text{0.333 pizzas}}{1 \; \cancel{human}} \right) = 10 \; \text{pizzas} \label{Eq2}$
Notice that the units that canceled out are lined out and only the desired units are left (discussed more below). Finally, in going to buy the soda, you perform another dimensional analysis: should you buy the sodas in six-packs or in cases?
$(120\; { sodas}) \times \left( \dfrac{\text{1 six pack}}{6\; {sodas}} \right) = 20 \; \text{six packs} \label{Eq3}$
$(120\; {sodas}) \times \left( \dfrac{\text{1 case }}{24\; {sodas}} \right) = 5 \; \text{cases} \label{Eq4}$
Realizing that carrying around 20 six packs is a real headache, you get 5 cases of soda instead.
In this party problem, we have used dimensional analysis in two different ways:
• In the first application (Equations $\ref{Eq1}$ and Equation $\ref{Eq2}$), dimensional analysis was used to calculate how much soda is needed need. This is based on knowing: (1) how much soda we need for one person and (2) how many people we expect; likewise for the pizza.
• In the second application (Equations $\ref{Eq3}$ and $\ref{Eq4}$), dimensional analysis was used to convert units (i.e. from individual sodas to the equivalent amount of six packs or cases)
Using Dimensional Analysis to Convert Units
Consider the conversion in Equation $\ref{Eq3}$:
$(120\; {sodas}) \times \left( \dfrac{\text{1 six pack}}{6\; {sodas}} \right) = 20 \; \text{six packs} \label{Eq3a}$
If we ignore the numbers for a moment, and just look at the units (i.e. dimensions), we have:
$\text{soda} \times \left(\dfrac{\text{six pack}}{\text{sodas}}\right) \nonumber$
We can treat the dimensions in a similar fashion as other numerical analyses (i.e. any number divided by itself is 1). Therefore:
$\text{soda} \times \left(\dfrac{\text{six pack}}{\text{sodas}}\right) = \cancel{\text{soda}} \times \left(\dfrac{\text{six pack}}{\cancel{\text{sodas}}}\right) \nonumber$
So, the dimensions of the numerical answer will be "six packs".
How can we use dimensional analysis to be sure we have set up our equation correctly? Consider the following alternative way to set up the above unit conversion analysis:
$120 \cancel{\text{soda}} \times \left(\dfrac{\text{6 sodas}}{\cancel{\text{six pack}}}\right) = 720 \; \dfrac{\text{sodas}^2}{\text{1 six pack}} \nonumber$
• While it is correct that there are 6 sodas in one six pack, the above equation yields a value of 720 with units of sodas2/six pack.
• These rather bizarre units indicate that the equation has been setup incorrectly (and as a consequence you will have a ton of extra soda at the party).
Using Dimensional Analysis in Calculations
In the above case it was relatively straightforward keeping track of units during the calculation. What if the calculation involves powers, etc? For example, the equation relating kinetic energy to mass and velocity is:
$E_{kinetics} = \dfrac{1}{2} \text{mass} \times \text{velocity}^2 \label{KE}$
An example of units of mass is kilograms (kg) and velocity might be in meters/second (m/s). What are the dimensions of $E_{kinetic}$?
$(kg) \times \left( \dfrac{m}{s} \right)^2 = \dfrac{kg \; m^2}{s^2} \nonumber$
The $\frac{1}{2}$ factor in Equation \ref{KE} is neglected since pure numbers have no units. Since the velocity is squared in Equation \ref{KE}, the dimensions associated with the numerical value of the velocity are also squared. We can double check this by knowing the the Joule ($J$) is a measure of energy, and as a composite unit can be decomposed thusly:
$1\; J = kg \dfrac{m^2}{s^2} \nonumber$
Units of Pressure
Pressure (P) is a measure of the Force (F) per unit area (A):
$P =\dfrac{F}{A} \nonumber$
Force, in turn, is a measure of the acceleration ($a$) on a mass ($m$):
$F= m \times a \nonumber$
Thus, pressure ($P$) can be written as:
$P= \dfrac{m \times a}{A} \nonumber$
What are the units of pressure from this relationship? (Note: acceleration is the change in velocity per unit time)
$P =\dfrac{kg \times \frac{\cancel{m}}{s^2}}{m^{\cancel{2}}} \nonumber$
We can simplify this description of the units of Pressure by dividing numerator and denominator by $m$:
$P =\dfrac{\frac{kg}{s^2}}{m}=\dfrac{kg}{m\; s^2} \nonumber$
In fact, these are the units of a the composite Pascal (Pa) unit and is the SI measure of pressure.
Performing Dimensional Analysis
The use of units in a calculation to ensure that we obtain the final proper units is called dimensional analysis. For example, if we observe experimentally that an object’s potential energy is related to its mass, its height from the ground, and to a gravitational force, then when multiplied, the units of mass, height, and the force of gravity must give us units corresponding to those of energy.
Energy is typically measured in joules, calories, or electron volts (eV), defined by the following expressions:
• 1 J = 1 (kg·m2)/s2 = 1 coulomb·volt
• 1 cal = 4.184 J
• 1 eV = 1.602 × 10−19 J
Performing dimensional analysis begins with finding the appropriate conversion factors. Then, you simply multiply the values together such that the units cancel by having equal units in the numerator and the denominator. To understand this process, let us walk through a few examples.
Example $1$
Imagine that a chemist wants to measure out 0.214 mL of benzene, but lacks the equipment to accurately measure such a small volume. The chemist, however, is equipped with an analytical balance capable of measuring to $\pm 0.0001 \;g$. Looking in a reference table, the chemist learns the density of benzene ($\rho=0.8765 \;g/mL$). How many grams of benzene should the chemist use?
Solution
$0.214 \; \cancel{mL} \left( \dfrac{0.8765\; g}{1\;\cancel{mL}}\right)= 0.187571\; g \nonumber$
Notice that the mL are being divided by mL, an equivalent unit. We can cancel these our, which results with the 0.187571 g. However, this is not our final answer, since this result has too many significant figures and must be rounded down to three significant digits. This is because 0.214 mL has three significant digits and the conversion factor had four significant digits. Since 5 is greater than or equal to 5, we must round the preceding 7 up to 8.
Hence, the chemist should weigh out 0.188 g of benzene to have 0.214 mL of benzene.
Example $2$
To illustrate the use of dimensional analysis to solve energy problems, let us calculate the kinetic energy in joules of a 320 g object traveling at 123 cm/s.
Solution
To obtain an answer in joules, we must convert grams to kilograms and centimeters to meters. Using Equation \ref{KE}, the calculation may be set up as follows:
\begin{align*} KE&=\dfrac{1}{2}mv^2=\dfrac{1}{2}(g) \left(\dfrac{kg}{g}\right) \left[\left(\dfrac{cm}{s}\right)\left(\dfrac{m}{cm}\right) \right]^2 \[4pt] &= (\cancel{g})\left(\dfrac{kg}{\cancel{g}}\right) \left(\dfrac{\cancel{m^2}}{s^2}\right) \left(\dfrac{m^2}{\cancel{cm^2}}\right) = \dfrac{kg⋅m^2}{s^2} \[4pt] &=\dfrac{1}{2}320\; \cancel{g} \left( \dfrac{1\; kg}{1000\;\cancel{g}}\right) \left[\left(\dfrac{123\;\cancel{cm}}{1 \;s}\right) \left(\dfrac{1 \;m}{100\; \cancel{cm}}\right) \right]^2=\dfrac{0.320\; kg}{2}\left[\dfrac{123 m}{s(100)}\right]^2 \[4pt] &=\dfrac{1}{2} 0.320\; kg \left[ \dfrac{(123)^2 m^2}{s^2(100)^2} \right]= 0.242 \dfrac{kg⋅m^2}{s^2} = 0.242\; J \end{align*} \nonumber
Alternatively, the conversions may be carried out in a stepwise manner:
Step 1: convert $g$ to $kg$
$320\; \cancel{g} \left( \dfrac{1\; kg}{1000\;\cancel{g}}\right) = 0.320 \; kg \nonumber$
Step 2: convert $cm$ to $m$
$123\;\cancel{cm} \left(\dfrac{1 \;m}{100\; \cancel{cm}}\right) = 1.23\ m \nonumber$
Now the natural units for calculating joules is used to get final results
\begin{align*} KE &=\dfrac{1}{2} 0.320\; kg \left(1.23 \;ms\right)^2 \[4pt] &=\dfrac{1}{2} 0.320\; kg \left(1.513 \dfrac{m^2}{s^2}\right)= 0.242\; \dfrac{kg⋅m^2}{s^2}= 0.242\; J \end{align*} \nonumber
Of course, steps 1 and 2 can be done in the opposite order with no effect on the final results. However, this second method involves an additional step.
Example $3$
Now suppose you wish to report the number of kilocalories of energy contained in a 7.00 oz piece of chocolate in units of kilojoules per gram.
Solution
To obtain an answer in kilojoules, we must convert 7.00 oz to grams and kilocalories to kilojoules. Food reported to contain a value in Calories actually contains that same value in kilocalories. If the chocolate wrapper lists the caloric content as 120 Calories, the chocolate contains 120 kcal of energy. If we choose to use multiple steps to obtain our answer, we can begin with the conversion of kilocalories to kilojoules:
$120 \cancel{kcal} \left(\dfrac{1000 \;\cancel{cal}}{\cancel{kcal}}\right)\left(\dfrac{4.184 \;\cancel{J}}{1 \cancel{cal}}\right)\left(\dfrac{1 \;kJ}{1000 \cancel{J}}\right)= 502\; kJ \nonumber$
We next convert the 7.00 oz of chocolate to grams:
$7.00\;\cancel{oz} \left(\dfrac{28.35\; g}{1\; \cancel{oz}}\right)= 199\; g \nonumber$
The number of kilojoules per gram is therefore
$\dfrac{ 502 \;kJ}{199\; g}= 2.52\; kJ/g \nonumber$
Alternatively, we could solve the problem in one step with all the conversions included:
$\left(\dfrac{120\; \cancel{kcal}}{7.00\; \cancel{oz}}\right)\left(\dfrac{1000 \;\cancel{cal}}{1 \;\cancel{kcal}}\right)\left(\dfrac{4.184 \;\cancel{J}}{1 \; \cancel{cal}}\right)\left(\dfrac{1 \;kJ}{1000 \;\cancel{J}}\right)\left(\dfrac{1 \;\cancel{oz}}{28.35\; g}\right)= 2.53 \; kJ/g \nonumber$
The discrepancy between the two answers is attributable to rounding to the correct number of significant figures for each step when carrying out the calculation in a stepwise manner. Recall that all digits in the calculator should be carried forward when carrying out a calculation using multiple steps. In this problem, we first converted kilocalories to kilojoules and then converted ounces to grams.
Converting Between Units: Converting Between Units, YouTube(opens in new window) [youtu.be]
Summary
Dimensional analysis is used in numerical calculations, and in converting units. It can help us identify whether an equation is set up correctly (i.e. the resulting units should be as expected). Units are treated similarly to the associated numerical values, i.e., if a variable in an equation is supposed to be squared, then the associated dimensions are squared, etc.
Contributors and Attributions
• Mark Tye (Diablo Valley College) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.06%3A_Dimensional_Analysis.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.
1.2
CONCEPTUAL PROBLEMS
Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9) before proceeding to the Conceptual Problems.
1. What is the difference between mass and weight? Is the mass of an object on Earth the same as the mass of the same object on Jupiter? Why or why not?
2. Is it accurate to say that a substance with a mass of 1 kg weighs 2.2 lb? Why or why not?
3. What factor must be considered when reporting the weight of an object as opposed to its mass?
4. Construct a table with the headings “Solid,” “Liquid,” and “Gas.” For any given substance, state what you expect for each of the following:
1. the relative densities of the three phases
2. the physical shapes of the three phases
3. the volumes for the same mass of compound
4. the sensitivity of the volume of each phase to changes in temperature
5. the sensitivity of the volume to changes in pressure
5. Classify each substance as homogeneous or heterogeneous and explain your reasoning.
1. platinum
2. a carbonated beverage
3. bronze
4. wood
5. natural gas
6. Styrofoam
6. Classify each substance as homogeneous or heterogeneous and explain your reasoning.
1. snowflakes
2. gasoline
3. black tea
4. plastic wrap
5. blood
6. water containing ice cubes
7. Classify each substance as a pure substance or a mixture and explain your reasoning.
1. seawater
2. coffee
3. 14-karat gold
4. diamond
5. distilled water
8. Classify each substance as a pure substance or a mixture.
1. cardboard
2. caffeine
3. tin
4. a vitamin tablet
5. helium gas
9. Classify each substance as an element or a compound.
1. sugar
2. silver
3. rust
4. rubbing alcohol
5. copper
10. Classify each substance as an element or a compound.
1. water
2. iron
3. hydrogen gas
4. glass
5. nylon
11. What techniques could be used to separate each of the following?
1. sugar and water from an aqueous solution of sugar
2. a mixture of sugar and sand
3. a heterogeneous mixture of solids with different solubilities
12. What techniques could be used to separate each of the following?
1. solid calcium chloride from a solution of calcium chloride in water
2. the components of a solution of vinegar in water
3. particulates from water in a fish tank
13. Match each separation technique in (a) with the physical/chemical property that each takes advantage of in (b).
1. crystallization, distillation, filtration
2. volatility, physical state, solubility
14. The following figures illustrate the arrangement of atoms in some samples of matter. Which figures are related by a physical change? By a chemical change?
15. Classify each statement as an extensive property or an intensive property.
1. Carbon, in the form of diamond, is one of the hardest known materials.
2. A sample of crystalline silicon, a grayish solid, has a mass of 14.3 g.
3. Germanium has a density of 5.32 g/cm3.
4. Gray tin converts to white tin at 13.2°C.
5. Lead is a bluish-white metal.
16. Classify each statement as a physical property or a chemical property.
1. Fluorine etches glass.
2. Chlorine interacts with moisture in the lungs to produce a respiratory irritant.
3. Bromine is a reddish-brown liquid.
4. Iodine has a density of 11.27 g/L at 0°C.
NUMERICAL 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. If a person weighs 176 lb on Earth, what is his or her mass on Mars, where the force of gravity is 37% of that on Earth?
2. If a person weighs 135 lb on Earth, what is his or her mass on Jupiter, where the force of gravity is 236% of that on Earth?
3. Calculate the volume of 10.00 g of each element and then arrange the elements in order of decreasing volume. The numbers in parentheses are densities.
1. copper (8.92 g/cm3)
2. calcium (1.54 g/cm3)
3. titanium (4.51 g/cm3)
4. iridium (22.85 g/cm3)
4. Given 15.00 g of each element, calculate the volume of each and then arrange the elements in order of increasing volume. The numbers in parentheses are densities.
1. gold (19.32 g/cm3)
2. lead (11.34 g/cm3)
3. iron (7.87 g/cm3)
4. sulfur (2.07 g/cm3)
5. A silver bar has dimensions of 10.00 cm × 4.00 cm × 1.50 cm, and the density of silver is 10.49 g/cm3. What is the mass of the bar?
6. Platinum has a density of 21.45 g/cm3. What is the mass of a platinum bar measuring 3.00 cm × 1.50 cm × 0.500 cm?
7. Complete the following table.
Density (g/cm3) Mass (g) Volume (cm3) Element
3.14 79.904 Br
3.51 3.42 C
39.1 45.5 K
11.34 207.2 Pb
107.868 10.28 Ag
6.51 14.0 Zr
8. Gold has a density of 19.30 g/cm3. If a person who weighs 85.00 kg (1 kg = 1000 g) were given his or her weight in gold, what volume (in cm3) would the gold occupy? Are we justified in using the SI unit of mass for the person’s weight in this case?
9. An irregularly shaped piece of magnesium with a mass of 11.81 g was dropped into a graduated cylinder partially filled with water. The magnesium displaced 6.80 mL of water. What is the density of magnesium?
10. The density of copper is 8.92 g/cm3. If a 10.00 g sample is placed in a graduated cylinder that contains 15.0 mL of water, what is the total volume that would be occupied?
11. At 20°C, the density of fresh water is 0.9982 kg/m3, and the density of seawater is 1.025 kg/m3. Will a ship float higher in fresh water or in seawater? Explain your reasoning.
Numerical Answers
1. Unlike weight, mass does not depend on location. The mass of the person is therefore the same on Earth and Mars: 176 lb ÷ 2.2 lb/kg = 80 kg.
3.
a. Cu: 1.12 cm3
b. Ca: 6.49 cm3
c. Ti: 2.22 cm3
d. Ir: 0.4376 cm3
Volume decreases: Ca > Ti > Cu > Ir
5. 629 g
9. 1.74 g/cm3
1.3 Problems
1. Milk turns sour. This is a ________________
• Chemical Change
• Physical Change
• Chemical Property
• Physical Property
• None of the above
2. HCl being a strong acid is a __________, Wood sawed in two is ___________
• Chemical Change, Physical Change
• Physical Change, Chemical Change
• Chemical Property, Physical Change
• Physical Property, Chemical Change
• None of the above
3. CuSO4 is dissolved in water
• Chemical Change
• Physical Change
• Chemical Property
• Physical Property
• None of the above
4. Aluminum Phosphate has a density of 2.566 g/cm3
• Chemical Change
• Physical Change
• Chemical Property
• Physical Property
• None of the above
5. Which of the following are examples of matter?
• A Dog
• Carbon Dioxide
• Ice Cubes
• copper (II) nitrate
• A Moving Car
6. The formation of gas bubbles is a sign of what type of change?
7. True or False: Bread rising is a physical property. 8. True or False: Dicing potatoes is a physical change. 9. Is sunlight matter? 10. The mass of lead is a _____________property.
1.3 Solutions
1)chemical change 2) chemical property, physical change 3) physical change 4) physical property 5) All of the above 6) chemical 7) False 8) True 9) No 10) physical property
1.6
1. Write a single equation to show how to convert
1. $cm/min$ to $km/h$;
2. $cal/oz$ to $J/g$
3. $lb/in^2$ to $kg/m^2$ and
4. $°C/s$ to $K/h$.
2. How many Calories are contained in an 8.0 oz serving of green beans if their fuel value is 1.5 kJ/g?
3. Gasoline has a fuel value of 48 kJ/g. How much energy in joules can be obtained by filling an automobile’s 16.3 gal tank with gasoline, assuming gasoline has a density of 0.70 g/mL?
Solutions
1. Converting from one compound unit to another
1. $\left(\dfrac{\cancel{cm}}{\cancel{min}}\right)\left(\dfrac{1\;\cancel{m}}{100\; \cancel{cm}}\right)\left(\dfrac{1\; km}{1000\;\cancel{m}}\right)\left(\dfrac{60\;\cancel{min}}{1\; h}\right)= km/h \nonumber$
2. $\left(\dfrac{\cancel{cal}}{\cancel{oz}}\right) \left(\dfrac{4.184 \;J}{1\; \cancel{cal}}\right) \left( \dfrac{16\; \cancel{oz}}{1\; \cancel{lb}}\right) \left(\dfrac{1\; \cancel{lb}}{453.59 \;g}\right)= J/g \nonumber$
3. $\left(\dfrac{\cancel{lb}}{\cancel{in^2}}\right)\left(\dfrac{16 \;\cancel{oz}}{\cancel{lb}}\right) \left(\dfrac{28.35\; \cancel{g}}{\cancel{oz}}\right)\left(\dfrac{1\; kg}{1000\; \cancel{g}}\right) \left[\dfrac{(36\; \cancel{in.})^2}{(1\;\cancel{yd})^2}\right] \left[\dfrac{1.09 \;\cancel{yd^2}}{1 \;m^2}\right] =kg/m^2 \nonumber$
4. $\left(\dfrac{°C}{\cancel{s}}\right)\left(\dfrac{60\;\cancel{s}}{1\;\cancel{min}}\right)\left(\dfrac{60\;\cancel{min}}{h}\right)+273.15 K = K/h \nonumber$
2. Our goal is to convert 1.5 kJ/g to Calories in 8 oz:$\left(\dfrac{1.5 \; \cancel{kJ}}{1\; \cancel{g}}\right)\left(\dfrac{1000\; \cancel{J}}{1\; \cancel{kJ}}\right)\left(\dfrac{1\; \cancel{cal}}{4.184\; \cancel{J}}\right)\left(\dfrac{1\; Cal}{1000\; \cancel{cal}}\right)\left(\dfrac{28.35 \;\cancel{g}}{1\; \cancel{oz}}\right)\left(8.0\; \cancel{oz}\right)= 81\; Cal \nonumber$
3. Our goal is to use the energy content, 48 kJ/g, and the density, 0.70 g/mL, to obtain the number of joules in 16.3 gal of gasoline: $\left(\dfrac{48\; \cancel{kJ}}{g}\right)\left(\dfrac{1000\; J}{\cancel{kJ}}\right)\left(\dfrac{0.70\; \cancel{g}}{\cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{\cancel{L}}\right)\left(\dfrac{3.79 \;\cancel{L}}{\cancel{gal}}\right)\left(16.3 \;\cancel{gal}\right)= 2.1 \times 10^9 J \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.E%3A_Matter_and_Measurement_%28Exercises%29.txt |
1.1: The Study of Chemistry
An understanding of chemistry is essential for understanding much of the natural world and is central to many other disciplines. Chemistry is the study of matter and the changes material substances undergo. It is essential for understanding much of the natural world and central to many other scientific disciplines, including astronomy, geology, paleontology, biology, and medicine.
The Molecular Perspective of Chemistry
• Matter – physical material of the universe, has mass and takes up space
• Atoms – the building blocks of matter
• Molecules – groups of combined atoms
Why Study Chemistry?
• important in understanding our world
• chemistry is the central science
• many various subjects have some kind of relation to chemistry
1.2: Classification of Matter
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 & properties.
States of Matter
• states of matter: liquid, solid, and gas
Pure Substances and Mixtures
• pure substance – matter that has a fixed composition and distinct properties
• substances can be classified as elements or compounds
• elements – composed of only one atom
• compounds – two ore more elements
• mixtures – combination of two or more substances
• heterogeneous – mixtures that are not uniform throughout
• homogeneous – mixtures that are uniform throughout; also called solutions
Separation of Mixtures
• components in a mixture retain their own properties
• mixtures can be separated by using the different properties of each substance
• types of separation: filtration, distillation, chromatography
Elements
• over 90% of earth’s crust consists of oxygen, silicon, aluminum, iron, and calcium
• human body consists of 90% of oxygen, carbon, and hydrogen
Compounds
• law of constant composition (law of definite proportions) – elemental composition of a pure compound is always the same
1.3: Properties of Matter
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.
• physical properties – properties measured by not changing the identity and composition of the substance
• chemical properties – the way a substance may change or react to form other substances
• intensive properties – identify substances
• extensive properties – amount of substance
Physical and Chemical Changes
• physical change results in a change in appearance but not composition
• changes of state are physical changes
• chemical changes (chemical reactions) results in a chemically different substance
1.4: Units of Measurement
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.
SI Units
• seven base units
SI Units
Prefix Abbreviation Meaning
Mega- M 106
Kilo- K 103
Deci- D 10-1
Centi- C 10-2
Milli- m 10-3
Micro- µa 10-6
Nano- n 10-9
Pico- p 10-12
Femto- f 10-15
Length and Mass
• SI base unit for length is the meter (m)
• Mass is a measure of the amount of material in an object
Temperature
• Celsius and Kelvin scales are used commonly in science
• Both have equal sized units
• K = °C + 273.15
• °C = (5/9)(°F –32)
• °F = (9/5)(°C) + 32
Derived SI Units
Volume
• SI unit is cubic meter
• Equipment to measure volume accurately : syringes, burets, and pipets
Density
• density = mass/volume
1.5: Uncertainty in Measurement
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.
1.5.1 Precision and Accuracy
• precision – the closeness of individual measurements to one another
• accuracy – the correctness of individual measurements
Significant Figures
• Nonzero digits are always significant
• Zeros between nonzero digits are always significant
• Zeros at the beginning of a number are never significant
• Zeros that fall both at the end of a number and after the decimal point are always significant
• When a number ends in zeros but contains no decimal point, the zeros may or may not be significant
• Scientific notation can be used to get the correct significant numbers
Significant Figures in Calculations
• Addition/Subtraction: the number of decimal places is determined by the number that has the fewest decimal places in the calculation
• Multiplication/Division: the number of significant figures is determined by the number that has the fewest significant figures in the calculation
1.6: Dimensional Analysis
Dimensional analysis is used in numerical calculations, and in converting units. It can help us identify whether an equation is set up correctly (i.e. the resulting units should be as expected). Units are treated similarly to the associated numerical values, i.e., if a variable in an equation is supposed to be squared, then the associated dimensions are squared, etc.
• aid in problem solving
• conversion factor – fraction where the numerator and denominator are the same quantity | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.S%3A_Matter_and_Measurement_%28Summary%29.txt |
In this chapter, you will learn how to describe the composition of chemical compounds. We introduce chemical nomenclature—the language of chemistry—that will enable you to recognize and name the most common kinds of compounds. An understanding of chemical nomenclature not only is essential for your study of chemistry but also has other benefits—for example, it helps you understand the labels on products found in the supermarket and the pharmacy. You will also be better equipped to understand many of the important environmental and medical issues that face society. By the end of this chapter, you will be able to describe what happens chemically when a doctor prepares a cast to stabilize a broken bone, and you will know the composition of common substances such as laundry bleach, the active ingredient in baking powder, and the foul-smelling compound responsible for the odor of spoiled fish. Finally, you will be able to explain the chemical differences among different grades of gasoline.
• 2.1: The Atomic Theory of Matter
This article explains the theories that Dalton used as a basis for his theory: (1) the Law of Conservation of Mass, (2) the Law of Constant Composition, (3) the Law of Multiple Proportions.
• 2.2: The Discovery of Atomic Structure
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.
• 2.3: The Modern View of Atomic Structure
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).
• 2.4: Atomic Mass
The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, and 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 different mass number.
• 2.5: The Periodic Table
The periodic table is used as a predictive tool that arranges 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. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals.
• 2.6: Molecules and Molecular Compounds
There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties. The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. The molecular formula of a covalent compound gives the types and numbers of atoms present. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two.
• 2.7: Ions and Ionic Compounds
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. 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).
• 2.8: Naming Inorganic Compounds
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.
• 2.9: Some Simple Organic Compounds
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.
• 2.E: Atoms, Molecules, and Ions (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 2.S: Atoms, Molecules, and Ions (Summary)
This is the summary Module for the chapter "Atoms, Molecules, and Ions" in the Brown et al. General Chemistry Textmap.
Thumbnail: Spinning Buckminsterfullerene (\(\ce{C60}\)). (CC BY-SA 3.0; unported; Sponk).
02: Atoms Molecules and Ions
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 section explains the theories that Dalton used as a basis for his theory: (1) Law of Conservation of Mass, (2) Law of Definite Proportions, and (3) Law of Multiple Proportions
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.
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.
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 in action.
Law of Definite Proportions states that in a given type of chemical substance, the elements are always combined in the same proportions by mass.
The Law of Definite Proportions applies when elements are reacted together to form the same product. Therefore, while the Law of Definite Proportions can be used to compare two experiments in which hydrogen and oxygen react to form water, the Law of Definite Proportions can not be used to compare one experiment in which hydrogen and oxygen react to form water, and another experiment in which hydrogen and oxygen react to form hydrogen peroxide (peroxide is another material that can be made from 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) = 14:32. (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 consistent 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:
$\dfrac{0.571\; \text{g oxygen}}{0.429 \;\text{g carbon}} = 1.33\; \dfrac{\text{g oxygen}}{\text{g carbon}}\nonumber$
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
$\dfrac{0.727\; \text{g oxygen}}{0.273 \;\text{g carbon}} = 2.66\; \dfrac{\text{g oxygen}}{\text{g carbon}}\nonumber$
Dividing the mass of oxygen per g of carbon of the second compound:
$\dfrac{2.66}{1.33} = 2\nonumber$
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: Fundamental Experiments in Chemistry, YouTube(opens in new window) [youtu.be]
Summary
This article explains the theories that Dalton used as a basis for his theory: (1) the Law of Conservation of Mass, (2) the Law of Constant Composition, (3) the Law of Multiple Proportions. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.01%3A_The_Atomic_Theory_of_Matter.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:
$\dfrac {mass}{charge} \times {charge} ={mass} \nonumber$
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
Measuring e/m For an Electron. Video from Davidson College demonstrating Thompson's e/m experiment.
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 the 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.
The Atomic Model
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 $6$).
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 (Figure $\PageIndex{7a}$) 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 (Figure $\PageIndex{7b}$). 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 (Figure $\PageIndex{7c}$). 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.”
The Nuclear Atom: The Nuclear Atom, YouTube(opens in new window) [youtu.be]
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 $8$. 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 $9$ shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today.
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_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.02%3A_The_Discovery_of_Atomic_Structure.txt |
Learning Objectives
• To know the meaning of isotopes and atomic masses.
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.
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.
Elements
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. Atoms consist of electrons, protons, and neutrons. Although this is an oversimplification that ignores the other subatomic particles that have been discovered, 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
electron $9.109 \times 10^{-28}$ 0.0005486 −1.602 × 10−19 −1
proton $1.673 \times 10^{-24}$ 1.007276 +1.602 × 10−19 +1
neutron $1.675 \times 10^{-24}$ 1.008665 0 0
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 $2$.
Table $2$: 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
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.
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.
Figure $2$: Formalism used for identifying specific nuclide (any particular kind of nucleus)
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 archaeology. 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 $3$.
Table $3$: 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: How Elements Are Represented on the Periodic Table, YouTube(opens in new window) [youtu.be]
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. 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_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.03%3A_The_Modern_View_of_Atomic_Structure.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
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_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.04%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 \(1\): Classifying Elements
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
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.
The noble gases are unreactive at room temperature and pressure.
Summary
The periodic table is used as a predictive tool. It arranges 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_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.05%3A_The_Periodic_Table.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.
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 in Section 2.8.
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:
1. Identify the symbol for each element in the molecule. Then identify the substance as either an organic compound or an inorganic compound.
2. 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.
3. From the information given, add a subscript for each kind of atom to write the molecular formula.
Solution
a
1. The molecule has 4 phosphorus atoms and 3 sulfur atoms. Because the compound does not contain mostly carbon and hydrogen, it is inorganic.
2. Phosphorus is in group 15, and sulfur is in group 16. Because phosphorus is to the left of sulfur, it is written first.
3. Writing the number of each kind of atom as a right-hand subscript gives P4S3 as the molecular formula.
b.
1. Ethyl alcohol contains predominantly carbon and hydrogen, so it is an organic compound.
2. 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.
3. Adding subscripts gives the molecular formula ]\(C_2H_6O\).
c
1. 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.
2. 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.
3. 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 a
N2O
Answer b
C12H22O11
Answer c
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 (see 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:
Figures \(1\), \(2\), and \(3\) 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 (Figure \(\PageIndex{4a}\)) gives only the number of each kind of atom; writing methanol as CH4O tells nothing about its structure. In contrast, the structural formula (Figure \(\PageIndex{4b}\)) 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 (Figure \(\PageIndex{4d}\)) 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 \(2\): Molecular Formulas
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 \(2\): Molecular Formulas
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 \(\ce{CHCl3}\).
2. Hydrazine is used as a propellant in the attitude jets of the space shuttle. Its condensed structural formula is \(\ce{H2NNH2}\).
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 \(\ce{H2N(CH2)4NH2}\) to indicate that there are four CH2 fragments linked together.
Answer a
CHCl3
Answer b
N2H4
Answer c
C4H12N2
Summary
There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties. The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. 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. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.06%3A_Molecules_and_Molecular_Compounds.txt |
Learning Objectives
• Explain the bonding nature of ionic compounds.
• Relating microscopic bonding properties to macroscopic solid properties.
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.
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 $1$). As shown in Equation $\ref{Eq1}$, 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:
$\text {electrostatic energy} \propto {Q_1Q_2 \over r} \label{Eq1}$
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.
Ionic compounds contain both cations and anions in a ratio that results in zero electrical charge.
As shown in Figure $2$, 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.
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; Figure $3$), 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 $1$. 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. Some common monatomic ions are listed in Table $1$.
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 $1$: 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 $1$
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:
1. 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.
2. 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 $1$
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 a
Ca2+
Answer b
I
Answer c
Zr4+
Ions of Atoms: Ions of Atoms, YouTube(opens in new window) [youtu.be]
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 Figure $\PageIndex{4a}$ which shows the three-dimensional array of alternating positive and negative ions held together by strong electrostatic attractions. In contrast, as shown in Figure $\PageIndex{4b}$ 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 $\ce{CH4}$ molecules. A comparison of the different physical properties of ionic compounds and covalent molecular substances is given in Table $2$.
Table $2$: 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
Summary
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. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.07%3A_Ions_and_Ionic_Compounds.txt |
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 \(\ce{C4H10}\). 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 \(\ce{C2H5}\). 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 \(\ce{C2H5}\), but it contains two \(\ce{C2H5}\) formula units, giving a molecular formula of \(\ce{C4H10}\).
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 \(\ce{Na^{+}}\) and \(\ce{Cl^{−}}\), then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be \(\ce{NaCl}\). 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 \(\ce{Mg^{2+}}\) and \(\ce{Cl^{−}}\), for example, two Cl ions are needed to balance the two positive charges on each Mg2+ ion, giving an empirical formula of \(\ce{MgCl2}\). Similarly, the formula for the ionic compound that contains Na+ and O2− ions is Na2O.
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 \(\ce{MgCl_2}\), 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:
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 \(\ce{Mg2O2}\):
This simplifies to its correct empirical formula MgO. The empirical formula has one Mg2+ ion and one O2− ion.
Example \(1\): Binary Ionic Compounds
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 \(\ce{Ga3As3}\):
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 a
Li3N
Answer b
Al2O3
Answer c
Li2O
Nomenclature of Metals: Nomenclature of Metals(opens in new window) [youtu.be]
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 a
Ca(H2PO4)2: calcium dihydrogen phosphate is one of the ingredients in baking powder.
Answer b
NaHCO3: sodium bicarbonate is found in antacids and baking powder; in pure form, it is sold as baking soda.
Answer c
(NH4)2SO4: ammonium sulfate is a common source of nitrogen in fertilizers.
Polyatomics: Polyatomics, YouTube(opens in new window) [youtu.be] (opens in new window)
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 \(\ce{Ca(H2PO4)2}\) 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 \(\ce{Ca(H2PO4)2·H2O}\). In contrast, copper sulfate usually forms a blue solid that contains five waters of hydration per formula unit, with the empirical formula \(\ce{CuSO4·5H2O}\). When heated, all five water molecules are lost, giving a white solid with the empirical formula \(\ce{CuSO4}\).
Compounds that differ only in the numbers of waters of hydration can have very different properties. For example, \(\ce{CaSO4·½H2O}\) is plaster of Paris, which was often used to make sturdy casts for broken arms or legs, whereas \(\ce{CaSO4·2H2O}\) 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 \(\ce{CaSO4·2H2O}\). Similar processes are used in the setting of cement and concrete.
Binary Acids
Some compounds containing hydrogen are members of an important class of substances known as acids. The chemistry of these compounds is explored in more detail in later chapters of this text, but for now, it will suffice to note that many acids release hydrogen ions, H+, when dissolved in water. To denote this distinct chemical property, a mixture of water with an acid is given a name derived from the compound’s name. If the compound is a binary acid (comprised of hydrogen and one other nonmetallic element):
1. The word “hydrogen” is changed to the prefix hydro-
2. The other nonmetallic element name is modified by adding the suffix -ic
3. The word “acid” is added as a second word
For example, when the gas \(\ce{HCl}\) (hydrogen chloride) is dissolved in water, the solution is called hydrochloric acid. Several other examples of this nomenclature are shown in Table \(2\).
Table \(2\): Names of Some Simple Acids
Name of Gas Name of Acid
HF(g), hydrogen fluoride HF(aq), hydrofluoric acid
HCl(g), hydrogen chloride HCl(aq), hydrochloric acid
HBr(g), hydrogen bromide HBr(aq), hydrobromic acid
HI(g), hydrogen iodide HI(aq), hydroiodic acid
H2S(g), hydrogen sulfide H2S(aq), hydrosulfuric acid
Oxyacids
Many compounds containing three or more elements (such as organic compounds or coordination compounds) are subject to specialized nomenclature rules that you will learn later. However, we will briefly discuss the important compounds known as oxyacids, compounds that contain hydrogen, oxygen, and at least one other element, and are bonded in such a way as to impart acidic properties to the compound (you will learn the details of this in a later chapter). Typical oxyacids consist of hydrogen combined with a polyatomic, oxygen-containing ion. To name oxyacids:
1. Omit “hydrogen”
2. Start with the root name of the anion
3. Replace –ate with –ic, or –ite with –ous
4. Add “acid”
For example, consider H2CO3 (which you might be tempted to call “hydrogen carbonate”). To name this correctly, “hydrogen” is omitted; the –ate of carbonate is replace with –ic; and acid is added—so its name is carbonic acid. Other examples are given in Table \(3\). There are some exceptions to the general naming method (e.g., H2SO4 is called sulfuric acid, not sulfic acid, and H2SO3 is sulfurous, not sulfous, acid).
Table \(3\): Names of Common Oxyacids
Formula Anion Name Acid Name
HC2H3O2 acetate acetic acid
HNO3 nitrate nitric acid
HNO2 nitrite nitrous acid
HClO4 perchlorate perchloric acid
H2CO3 carbonate carbonic acid
H2SO4 sulfate sulfuric acid
H2SO3 sulfite sulfurous acid
H3PO4 phosphate phosphoric acid
Nomenclature of Acids: Nomenclature of Acids, YouTube(opens in new window) [youtu.be]
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.
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.
• 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.
• 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 \(3\)). 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 \(3\): 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.
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 \(3\): Binary Covalent Compounds
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 a
iodine heptafluoride
Answer b
dinitrogen pentoxide
Answer c
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 a
SiCl4
Answer b
S2F10
The structures of some of the compounds in Examples \(3\) and \(4\) 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: Nomenclature of Nonmetals, YouTube(opens in new window) [youtu.be] (opens in new window)
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.
2.09: Some Simple Organic Compounds
Learning Objectives
• To be familiar with the classes of basic organic chemistry compounds
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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.08%3A_Naming_Inorganic_Compounds.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.
2.1: The Atomic Theory of Matter
Conceptual Problems
1. Which of the following elements exist as diatomic molecules?
1. helium
2. hydrogen
3. iodine
4. gold
2. Which of the following elements exist as diatomic molecules?
1. chlorine
2. potassium
3. silver
4. oxygen
3. Why is it proper to represent the elemental form of helium as He but improper to represent the elemental form of hydrogen as H?
4. Why is it proper to represent the elemental form of chlorine as Cl2 but improper to represent the elemental form of calcium as Ca2?
Conceptual Solutions
1. Hydrogen and Iodine.
2. Chlorine and Oxygen.
3. Hydrogen exists as a diatomic molecule in its elemental form; helium does not exist as a diatomic molecule.
4. Chlorine exists naturally in its diatomic form, while Calcium does not.
Exercises
1. What is the modern atomic theory?
2. What are atoms?
Conceptual Answers
1. The modern atomic theory states that all matter is composed of atoms, which are unique between elements. They cannot be created or destroyed, but can combine with one another in whole number ratios to form compounds.
2. Atoms are the smallest parts of an element that maintain the identity of that element.
Numerical 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.
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?
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 sodium chloride was produced in the second experiment. Show that these results are consistent with the law of constant 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.
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?
Numerical Solutions
1. 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. 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. 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.
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) = 0.0793664144 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.
1. 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.
2.3: The Modern View of Atomic Structure
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?
Conceptual Solutions
1. Ernest Rutherford's 1911 Alpha-ray scattering experiment proved that the nucleus of an atom is positively charged. In the experiment, he fired high-energy streams of α particles at a thin gold sheet. Many passed through unimpeded, but some were deflected slightly and few were reflected back, proving that the positive charge of an atom is contained in the nucleus.
2. Rutherford proposed the existence of the neutron when discovering that electrons and protons alone did not fully account for the mass of the atom, leading him to work with James Chadwick to prove the existence of a neutral particle.
3. While Rutherford's model established the positively charged nucleus surrounded by negatively charged electrons, it stated that the electrons were evenly distributed throughout, rather than in "specific energy levels" of electrons, which are a significant part of the modern model.
4. In the absence of any other fields affecting the cathode ray, a lack of deflection implies an absence of a magnetic field in a direction that would affect their path.
5. If the nucleus were negatively charged, one could expect α particles to be deflected towards the nucleus rather than away. If it were neutral, it would have no effect on the particles.
6. α particles are helium nuclei (effectively He ions lacking electrons), β particle are fast-moving electrons, and γ rays are very short electromagnetic rays. Out of the three, γ rays have the greatest penetrative ability.
Numerical Problems
Please be sure you are familiar with the topics discussed in Section 1.6 before proceeding to the Numerical Problems.
1. Using the data in Table 1.3 and the periodic table, calculate the percentage of the mass of a silicon atom that is due to
1. electrons.
2. protons.
1. Using the data in Table 1.3 and the periodic table, calculate the percentage of the mass of a helium atom that is due to
1. electrons.
2. 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?
Numerical Solutions
1. The total atomic mass of a silicon atom is 28.085amu. Using the data from the table (as well as silicon's atomic number of 14), we can calculate that the mass of electrons in the atom is 0.0076804amu, while the mass of protons is 14.101864amu. From this we can calculate that electrons take up 0.02735% of the mass of the atom, while protons take up 50.2114% of the atom's mass.
2. The total atomic mass of a helium atom is 4.0026amu. Using the data from the table we see that the mass of electrons in the atom is 0.0010972amu, while the mass of protons is 2.015amu. Therefore, electrons take up 0.0274% of the mass of the atom, while protons take up 50.342%.
3. If the nucleus were to have a radius of 1.0cm, the atom would have a radius of 10000cm, or 0.06 miles.
4. The electrical charge of a single electron is equal to -1.602 × 10-19. Thus, we can divide the total charge of the oil drop by this number to see that there are 24 electrons contained within it.
2.4: Atomic Mass
Conceptual Problems
1. Complete the following table for the missing elements, symbols, and numbers of electrons.
Element Symbol Number of Electrons
molybdenum
19
titanium
B
53
Sm
helium
14
2. Complete the following table for the missing elements, symbols, and numbers of electrons.
Element Symbol Number of Electrons
lanthanum
Ir
aluminum
80
sodium
Si
9
Be
3. Is the mass of an ion the same as the mass of its parent atom? Explain your answer.
4. What isotopic standard is used for determining the mass of an atom?
5. Give the symbol \(^A_Z X\) for these elements, all of which exist as a single isotope.
1. beryllium
2. ruthenium
3. phosphorus
4. aluminum
5. cesium
6. praseodymium
7. cobalt
8. yttrium
9. arsenic
6. Give the symbol \(_Z^AX\) for these elements, all of which exist as a single isotope.
1. fluorine
2. helium
3. terbium
4. iodine
5. gold
6. scandium
7. sodium
8. niobium
9. manganese
7. Identify each element, represented by X, that has the given symbols.
1. \(_{26}^{55}X\)
2. \(_{33}^{74}X\)
3. \(_{12}^{24}X\)
4. \(_{53}^{127}X\)
5. \(_{18}^{40}X\)
6. \(_{63}^{152}X\)
Conceptual Solutions
1. Element Symbol Number of Electrons
molybdenum Mo 42
potassium K 19
titanium Ti 22
boron B 5
iodine I 53
samarium Sm 62
helium He 2
silicon Si 14
2. Element Symbol Number of Electrons
lanthanum La 57
iridium Ir 77
aluminum Al 13
mercury Hg 80
sodium Na 11
silicon Si 14
flourine F 9
beryllium Be 4
3. Although the mass of an ion is not exactly the same as its parent atom, electrons weigh so little that removing them will not change the mass in any significant way. Thus, we can say that the mass of an ion is the same.
4. The reference isotope which is used for determining the mass of all other atoms is the carbon-12 nuclide, which is set to weigh exactly 12 amu. Therefore other elements are massed in relation to this value.
5. Give the symbol \(^A_Z X\) for these elements, all of which exist as a single isotope.
1. beryllium
1. \(^9_4 Be\)
2. ruthenium
1. \(^{102}_{44} Ru\)
3. phosphorus
1. \(^{31}_{15} P\)
4. aluminum
1. \(^{27}_{13} Al\)
5. cesium
1. \(^{133}_{55} Cs\)
6. praseodymium
1. \(^{141}_{59} Pr\)
7. cobalt
1. \(^{59}_{27} Co\)
8. yttrium
1. \(^{89}_{39} Y\)
9. arsenic
1. \(^{75}_{33} As\)
6. Give the symbol \(_Z^AX\) for these elements.
1. fluorine
1. \(^{19}_{9} F\)
2. helium
1. \(^{4}_{2} He\)
3. terbium
1. \(^{159}_{65} Tb\)
4. iodine
1. \(^{127}_{53} I\)
5. gold
1. \(^{197}_{79} Au\)
6. scandium
1. \(^{45}_{21} Sc\)
7. sodium
1. \(^{23}_{11} Na\)
8. niobium
1. \(^{93}_{41} Nb\)
9. manganese
1. \(^{55}_{25} Mn\)
7. Identify each element, represented by X, that has the given symbols.
1. \(_{26}^{55}X\)
1. Iron (Fe)
2. \(_{33}^{74}X\)
1. Arsenic (As)
3. \(_{12}^{24}X\)
1. Magnesium (Mg)
4. \(_{53}^{127}X\)
1. Iodine (I)
5. \(_{18}^{40}X\)
1. Argon (Ar)
6. \(_{63}^{152}X\)
1. Europium (Eu)
Numerical Problems
Please be sure you are familiar with the topics discussed in Section 1.6 before proceeding to the Numerical Problems.
1. The isotopes 131I and 60Co are commonly used in medicine. Determine the number of neutrons, protons, and electrons in a neutral atom of each.
2. Determine the number of protons, neutrons, and electrons in a neutral atom of each isotope:
1. \(^{97}Tc\)
2. \(^{113}In\)
3. \(^{63}Ni\)
4. \(^{55}Fe\)
3. Both technetium-97 and americium-240 are produced in nuclear reactors. Determine the number of protons, neutrons, and electrons in the neutral atoms of each.
4. The following isotopes are important in archeological research. How many protons, neutrons, and electrons does a neutral atom of each contain?
1. \(^{207}Pb\)
2. \(^{16}O\)
3. \(^{40}K\)
4. \(^{137}Cs\)
5. \(^{40}Ar\)
5. Copper, an excellent conductor of heat, has two isotopes: 63Cu and 65Cu. Use the following information to calculate the average atomic mass of copper:
Isotope Percent Abundance (%) Atomic Mass (amu)
63Cu 69.09 62.9298
65Cu 30.92 64.9278
6. Silicon consists of three isotopes with the following percent abundance:
Isotope Percent Abundance (%) Atomic Mass (amu)
28Si 92.18 27.976926
29Si 4.71 28.976495
30Si 3.12 29.973770
Calculate the average atomic mass of silicon.
7. Complete the following table for neon. The average atomic mass of neon is 20.1797 amu.
Isotope Percent Abundance (%) Atomic Mass (amu)
20Ne 90.92 19.99244
21Ne 0.257 20.99395
22Ne
8. Are \(_{28}^{63} X \) and \( _{29}^{62} X \) isotopes of the same element? Explain your answer.
9.Complete the following table:
Isotope Number of Protons Number of Neutrons Number of Electrons
238X 95
238U
75 112
10.Complete the following table:
Isotope Number of Protons Number of Neutrons Number of Electrons
57Fe
40X 20
36S
11. Using a mass spectrometer, a scientist determined the percent abundances of the isotopes of sulfur to be 95.27% for 32S, 0.51% for 33S, and 4.22% for 34S. Use the atomic mass of sulfur from the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") and the following atomic masses to determine whether these data are accurate, assuming that these are the only isotopes of sulfur: 31.972071 amu for 32S, 32.971459 amu for 33S, and 33.967867 amu for 34S.
12. The percent abundances of two of the three isotopes of oxygen are 99.76% for 16O, and 0.204% for 18O. Use the atomic mass of oxygen given in the periodic table and the following data to determine the mass of 17O: 15.994915 amu for 16O and 17.999160 amu for 18O.
13. Which element has the higher proportion by mass in NaI?
14. Which element has the higher proportion by mass in KBr?
Numerical Solutions
1. Determine the number of neutrons, protons, and electrons in a neutral atom of:
1. 131I
1. 53 Protons, 53 Electrons, 78 Neutrons. In a neutral atom, the amount of protons is equal to the amount of electrons. Additionally, the amount of neutrons can be found by taking [Mass Number] - [Amount of Protons].
2. 60Co
1. 27 Protons, 27 Electrons, 33 Neutrons.
2. Determine the number of protons, neutrons, and electrons in a neutral atom of each isotope:
1. \(^{97}Tc\)
1. 43 Protons, 43 Electrons, 54 Neutrons
2. \(^{113}In\)
1. 49 Protons, 49 Electrons, 64 Neutrons
3. \(^{63}Ni\)
1. 28 Protons, 28 Electrons, 35 Neutrons
4. \(^{55}Fe\)
1. 26 Protons, 26 Electrons, 29 Neutrons
3. Determine the number of protons, neutrons, and electrons in the neutral atoms of each.
1. Technetium-97
1. 43 Protons, 43 Electrons, 54 Neutrons
2. Americium-240
1. 95 Protons, 95 Electrons, 145 Neutrons
4. How many protons, neutrons, and electrons does a neutral atom of each contain?
1. \(^{207}Pb\)
1. 82 Protons, 82 Electrons, 125 Neutrons
2. \(^{16}O\)
1. 8 Protons, 8 Electrons, 8 Neutrons
3. \(^{40}K\)
1. 19 Protons, 19 Electrons, 21 Neutrons
4. \(^{137}Cs\)
1. 55 Protons, 55 Electrons, 82 Neutrons
5. \(^{40}Ar\)
1. 18 Protons, 18 Electrons, 22 Neutrons
5. The average atomic mass of an element from its isotopes can be found via the equation: Average A. Mass = f1M1 + f2M2 +… + fnMn, where f is the abundance of an isotope and M is that isotope's mass. For example, for the given table of copper, we can plug in (0.6909 × 62.9298) + (0.3092 × 64.9278) = 63.5539amu.
6. Average Atomic Mass of Silicon: (0.9218 × 27.796926) + (0.0471 × 28.976495) + (0.0312 × 29.97377) = 27.9232amu.
7. Isotope Percent Abundance (%) Atomic Mass (amu)
20Ne 90.92 19.99244
21Ne 0.257 20.99395
22Ne 8.823 22.08568
8. They are not isotopes of the same element, because the number of protons (and therefore the atomic number) are different. For two isotopes to be of the same element, they must share an atomic number.
9. Isotope Number of Protons Number of Neutrons Number of Electrons
238X 95 143 95
238U 92 146 92
187Re 75 112 75
10. Isotope Number of Protons Number of Neutrons Number of Electrons
57Fe 26 31 26
40X 20 20 20
36S 16 20 16
11. To solve this problem, we will use the same method we did for problem 6. With the values given we can plug them into the formula to obtain (0.9572 × 31.9720) + (0.0051 × 32.9715) + (0.0422 × 33.9679) = 32.2052amu. The periodic table states that the mass of sulfur is 32.06amu, meaning that the measurements are not entirely accurate.
12. From the given percent abundances, we can calculate that the abundance of 17O is 0.036%. Plugging this into the formula we can solve for the mass of 17O, which is 15.985amu. Obviously, it does not make much sense for an isotope of oxygen containing more neutrons to be lighter, meaning the data is not entirely correct.
13. Iodine has a higher proportion by mass in NaI. For every 22.99amu of Na, there are 126.9amu of I.
14. Bromine has a higher proportion by mass in KBr. For every 39.098amu of K, there are 79.904amu of Br.
2.5: The Periodic Table
Conceptual Problems
1. Classify each element in Conceptual Problem 1 of section 2.4 as a metal, a nonmetal, or a semimetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal.
2. Classify each element in Conceptual Problem 2 of section 2.4 as a metal, a nonmetal, or a semimetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal.
3. Classify each element as a metal, a semimetal, or a nonmetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal.
1. iron
2. tantalum
3. sulfur
4. silicon
5. chlorine
6. nickel
7. potassium
8. radon
9. zirconium
4. Which of these sets of elements are all in the same period?
1. potassium, vanadium, and ruthenium
2. lithium, carbon, and chlorine
3. sodium, magnesium, and sulfur
4. chromium, nickel, and krypton
5. Which of these sets of elements are all in the same period?
1. barium, tungsten, and argon
2. yttrium, zirconium, and selenium
3. potassium, calcium, and zinc
4. scandium, bromine, and manganese
6. Which of these sets of elements are all in the same group?
1. sodium, rubidium, and barium
2. nitrogen, phosphorus, and bismuth
3. copper, silver, and gold
4. magnesium, strontium, and samarium
7. Which of these sets of elements are all in the same group?
1. iron, ruthenium, and osmium
2. nickel, palladium, and lead
3. iodine, fluorine, and oxygen
4. boron, aluminum, and gallium
8. Indicate whether each element is a transition metal, a halogen, or a noble gas.
1. manganese
2. iridium
3. fluorine
4. xenon
5. lithium
6. carbon
7. zinc
8. sodium
9. tantalum
10. hafnium
11. antimony
12. cadmium
9. Which of the elements indicated in color in the periodic table shown below is most likely to exist as a monoatomic gas? As a diatomic gas? Which is most likely to be a semimetal? A reactive metal?
10. Based on their locations in the periodic table, would you expect these elements to be malleable? Why or why not?
1. phosphorus
2. chromium
3. rubidium
4. copper
5. aluminum
6. bismuth
7. neodymium
11. Based on their locations in the periodic table, would you expect these elements to be lustrous? Why or why not?
1. sulfur
2. vanadium
3. nickel
4. arsenic
5. strontium
6. cerium
7. sodium
Conceptual Solution
1. Element Symbol Type
molybdenum Mo metal; transition metal
potassium K metal; alkali metal
titanium Ti metal; transition metal
boron B semimetal
iodine I nonmetal
samarium Sm metal; lanthanide
helium He nonmetal
silicon Si semimetal
2. Element Symbol Type
lanthanum La metal; lanthanide
iridium Ir metal; transition metal
aluminum Al metal; post-transition metal
mercury Hg metal; transition metal
sodium Na metal; alkali metal
silicon Si semimetal
flourine F nonmetal
beryllium Be metal; alkaline earth metal
3. Symbol Type
Fe metal: transition metal
Ta metal: transition metal
S nonmetal
Si semimetal
Cl nonmetal (halogen)
Ni metal: transition metal
K metal: alkali metal
Rn nonmetal (noble gas)
Zr metal: transition metal
4. C and D. All elements in set C are in the 3rd period, and all elements in set D are in the 4th period.
5. C and D. All elements in both of these sets are in the 4th period.
6. B and C. All elements in set B are in group 15, while all elements in set C are in group 11.
7. A and D. All elements in set A are in group 8, and all elements in set D are in group 13.
8. Element Symbol Type
manganese Mn transition metal
iridium Ir transition metal
flourine F halogen
xenon Xe noble gas
lithium Li alkali metal
carbon C nonmetal
zinc Zn transition metal
sodium Na alkali metal
tantalum Ta transition metal
hafnium Hf transition metal
antimony Sb semimetal
cadmium Cd transition metal
9. The red element is most likely to exist as a monoatomic gas, as it is a noble gas meaning that all electrons are paired, and it has no need to bond with itself. The yellow element is most likely to exist as a diatomic gas, as it is in the halogen group meaning it has one unpaired electron. The orange element is most likely to be a semimetal, being within both the group and period ranges that include semimetals. The green element is most likely to be a reactive metal, as it is in the first group making it an alkali metal.
10. Note that though some elements are expected to be malleable based on their position, that is not always the case. For example, Bismuth would be expected to be malleable due to its status as a post-transition metal, but it is in fact brittle.
1. Element Expected to be Malleable? Why?
phosphorus No Reactive Nonmetal
chromium Yes Transition Metal
rubidium Yes Alkali Metal
copper Yes Transition Metal
aluminium Yes Post-Transition Metal
bismuth Yes Post-Transition Metal
neodymium Yes Lanthanoid (Metal)
11. Element Lustrous? Why?
sulfur No Reactive Nonmetal
vanadium Yes Transition Metal
nickel Yes Transition Metal
arsenic Yes Metalloid
strontium Yes Alkaline Earth Metal
cerium Yes Lanthanoid (Metal)
sodium Yes Alkali Metal
2.6: Molecules and Molecular Compounds
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
1. a. the forces that hold the atoms together.
2. b. melting points.
3. c. physical states at room temperature and pressure.
7. Why do covalent compounds generally tend to have lower melting points than ionic compounds?
Conceptual Answer
1. Ionic compounds are held together by powerful electromagnetic forces, usually caused by a more electronegative atom "taking" an electron from another to become an ion. Covalent compounds, however, share electrons between them in a steady balance between attractive and repulsive charges.
2. Cations, being positively charged, possess fewer electrons than their neutral atoms.
3. Organic compounds involve Carbon and Hydrogen in nearly all cases, while inorganic compounds consist of the other elements.
4. Condensed formulas can be written out more quickly and easily, and are helpful when showing that multiple atoms are connected to a single one in a compound.
5. Group 17, Halogens
6. Covalent Ionic
Forces Electrostatic Attraction between Nuclei and Electrons Electrostatic Attraction between Cations and Anions
Melting Points Very Low Very High
Physical States at STP Liquid/Gas Solid
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 2.6.2. Based on this information, draw the structural formula of dichloromethane (CH2Cl2).
2. What is the total number of electrons present in each ion?
1. F
2. Rb+
3. Ce3+
4. Zr4+
5. Zn2+
6. Kr2+
7. B3+
3. What is the total number of electrons present in each ion?
1. Ca2+
2. Se2
3. In3+
4. Sr2+
5. As3+
6. N3−
7. Tl+
4. Predict how many electrons are in each ion.
1. an oxygen ion with a −2 charge
2. a beryllium ion with a +2 charge
3. a silver ion with a +1 charge
4. a selenium ion with a +4 charge
5. an iron ion with a +2 charge
6. a chlorine ion with a −1 charge
5. Predict how many electrons are in each ion.
1. copper ion with a +2 charge
2. a molybdenum ion with a +4 charge
3. an iodine ion with a −1 charge
4. a gallium ion with a +3 charge
5. an ytterbium ion with a +3 charge
6. a scandium ion with a +3 charge
6. Predict the charge on the most common monatomic ion formed by each element.
1. chlorine
2. phosphorus
3. scandium
4. magnesium
5. arsenic
6. oxygen
7. Predict the charge on the most common monatomic ion formed by each element.
1. sodium
2. selenium
3. barium
4. rubidium
5. nitrogen
6. 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.
1. \(_4^9X^{2+} \)
2. \(_1^1X^-\)
3. \(_8^{16}X^{2-} \)
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.
1. \(_3^7X^+ \)
2. \(_9^{19}X^-\)
3. \(_{13}^{27}X^{3+}\)
Numerical Answers
1. Total number of electrons present in each ion:
1. F
1. 10
2. Rb+
1. 36
3. Ce3+
1. 55
4. Zr4+
1. 36
5. Zn2+
1. 28
6. Kr2+
1. 34
7. B3+
1. 2
2. Total number of electrons present in each ion:
1. Ca2+
1. 18
2. Se2
1. 36
3. In3+
1. 46
4. Sr2+
1. 36
5. As3+
1. 30
6. N3−
1. 10
7. Tl+
1. 80
3. How many electrons are in each ion:
1. an oxygen ion with a −2 charge
1. 10
2. a beryllium ion with a +2 charge
1. 2
3. a silver ion with a +1 charge
1. 46
4. a selenium ion with a +4 charge
1. 30
5. an iron ion with a +2 charge
1. 24
6. a chlorine ion with a −1 charge
1. 18
4. How many electrons are in each ion:
1. copper ion with a +2 charge
1. 27
2. a molybdenum ion with a +4 charge
1. 38
3. an iodine ion with a −1 charge
1. 54
4. a gallium ion with a +3 charge
1. 28
5. an ytterbium ion with a +3 charge
1. 67
6. a scandium ion with a +3 charge
1. 18
5. The charge on the most common monatomic ion formed by each element:
1. chlorine
1. -1
2. phosphorus
1. -3
3. scandium
1. +3
4. magnesium
1. +2
5. arsenic
1. -3
6. oxygen
1. -2
6. The charge on the most common monatomic ion formed by each element:
1. sodium
1. +1
2. selenium
1. -2
3. barium
1. +2
4. rubidium
1. +1
5. nitrogen
1. +3
6. aluminum
1. +3
7. 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.
1. \(_4^9X^{2+} \)
1. Parent Atom
1. Be
2. Formula of Ion
1. Be2+
3. Period and Group
1. Period 2, Group 2
2. \(_1^1X^-\)
1. Parent Atom
1. H
2. Formula of Ion
1. H-
3. Period and Group
1. Period 1, Group 1
3. \(_8^{16}X^{2-} \)
1. Parent Atom
1. O
2. Formula of Ion
1. O2-
3. Period and Group
1. Period 2, Group 16
8. 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.
1. \(_3^7X^+ \)
1. Parent Atom
1. Li
2. Formula of Ion
1. Li+
3. Period and Group
1. Period 2, Group 1
2. \(_9^{19}X^-\)
1. Parent Atom
1. F
2. Formula of Ion
1. F-
3. Period and Group
1. Period 2, Group 17
3. \(_{13}^{27}X^{3+}\)
1. Parent Atom
1. Al
2. Formula of Ion
1. Al3+
3. Period and Group
1. Period 3, Group 13
2.8: Naming Inorganic Compounds
Conceptual Problems
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?
Conceptual Solutions
1. Polyatomic ions and molecules are similar in that both are groups of multiple atoms held together by covalent bonds. However, polyatomic ions have net electrical charges, either positive or negative, while molecules do not.
2. Classify each compound as ionic or covalent.
1. Zn3(PO4)2
1. Ionic
2. C6H5CO2H
1. Covalent
3. K2Cr2O7
1. Ionic
4. CH3CH2SH
1. Covalent
5. NH4Br
1. Covalent
6. CCl2F2
1. Covalent
3. Classify each compound as ionic or covalent, organic or inorganic.
1. CH3CH2CO2H
1. Covalent, organic
2. CaCl2
1. Ionic, inorganic
3. Y(NO3)3
1. Ionic, inorganic
4. H2S
1. Covalent, inorganic
5. NaC2H3O2
1. Ionic, organic
4. In order to find the molecular formula from the empirical formula, the molar mass of the compound is also needed.
5. Structural Formulas allow us to see the types of bonds connecting atoms within a molecule or ion (single, double, or triple bonds), as well as a general shape in which the atoms are arranged.
6. OH is the functional group of the alcohol family, which confers upon it its characteristic properties. By using the ROH notation, one can tell with a single glance that a compound is an alcohol.
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.
8. The correct formula is Mg(OH)2. Being a neutral molecule, the charges on the components must even out to 0 — Mg has a charge of +2, O has a charge of -2, while H has a charge of +1. MgOH2 would result in a polyatomic ion with charge +2, since there would only be one O. Additionally, Hydroxide refers specifically to the diatomic anion OH-, so Magnesium Hydroxide would need to be composed of OH- due to the name.
9. The functional group present in cyanides is the cyano group, which is CN. The cyano group has a charge of -1. In order for the molecule to be a Cyanide, it must contain this functional group, and since this is a neutral molecule the charges must add up to 0. Given that Mg has a charge of +2, there must be two CN groups in order to even out the charges.
10. Yes. Each specific hydrate has a certain ratio of waters of hydration bound to it, differing between compounds (but not within the same hydrate).
Numerical Problems
1. Write the formula for each compound.
1. magnesium sulfate, which has 1 magnesium atom, 4 oxygen atoms, and 1 sulfur atom
2. ethylene glycol (antifreeze), which has 6 hydrogen atoms, 2 carbon atoms, and 2 oxygen atoms
3. acetic acid, which has 2 oxygen atoms, 2 carbon atoms, and 4 hydrogen atoms
4. potassium chlorate, which has 1 chlorine atom, 1 potassium atom, and 3 oxygen atoms
5. 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.
1. cadmium acetate, which has 1 cadmium atom, 4 oxygen atoms, 4 carbon atoms, and 6 hydrogen atoms
2. barium cyanide, which has 1 barium atom, 2 carbon atoms, and 2 nitrogen atoms
3. iron(III) phosphate dihydrate, which has 1 iron atom, 1 phosphorus atom, 6 oxygen atoms, and 4 hydrogen atoms
4. manganese(II) nitrate hexahydrate, which has 1 manganese atom, 12 hydrogen atoms, 12 oxygen atoms, and 2 nitrogen atoms
5. 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.
1. zinc and sulfur
2. barium and iodine
3. magnesium and chlorine
4. silicon and oxygen
5. 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.
1. lithium and nitrogen
2. cesium and chlorine
3. germanium and oxygen
4. rubidium and sulfur
5. arsenic and sodium
6. Write the empirical formula for each compound.
1. Na2S2O4
2. B2H6
3. C6H12O6
4. P4O10
5. KMnO4
7. Write the empirical formula for each compound.
1. Al2Cl6
2. K2Cr2O7
3. C2H4
4. (NH2)2CNH
5. CH3COOH
Numerical Answers
1. Formula of each compound:
1. magnesium sulfate
1. MgSO4
2. ethylene glycol (antifreeze)
1. (CH2OH)2
3. acetic acid
1. CH₃COOH
4. potassium chlorate
1. KClO3
5. sodium hypochlorite pentahydrate
1. NaOCl·5H2O
2. Formula of each compound:
1. cadmium acetate, which has 1 cadmium atom, 4 oxygen atoms, 4 carbon atoms, and 6 hydrogen atoms
1. Cd(O2CCH3)2
2. barium cyanide, which has 1 barium atom, 2 carbon atoms, and 2 nitrogen atoms
1. Ba(CN)2
3. iron(III) phosphate dihydrate, which has 1 iron atom, 1 phosphorus atom, 6 oxygen atoms, and 4 hydrogen atoms
1. FePO4 · 2H2O
4. manganese(II) nitrate hexahydrate, which has 1 manganese atom, 12 hydrogen atoms, 12 oxygen atoms, and 2 nitrogen atoms
1. Mn(NO3)2 · 6H2O
5. silver phosphate, which has 1 phosphorus atom, 3 silver atoms, and 4 oxygen atoms
1. Ag3PO4
3. Ion K+ Fe3+ NH4+ Ba2+
Cl KCl FeCl3 NH4Cl BaCl2
SO42 K2SO4 Fe2(SO4)3 (NH4)2SO4 BaSO4
PO4 3 K3PO4 FePO4 (NH4)3PO4 Ba3(PO4)2
NO3 KNO3 Fe(NO3)3 NH4NO3 Ba(NO3)2
OH KOH Fe(OH)3 NH4OH Ba(OH)2
4. The empirical formula for the binary compound:
1. zinc and sulfur
1. ZnS
2. barium and iodine
1. BaI2
3. magnesium and chlorine
1. MgCl2
4. silicon and oxygen
1. SiO2
5. sodium and sulfur
1. Na2S
5. The empirical formula for the binary compound:
1. lithium and nitrogen
1. Li3N
2. cesium and chlorine
1. CsCl
3. germanium and oxygen
1. GeO2
4. rubidium and sulfur
1. Rb2S
5. arsenic and sodium
1. Na3As
6. The empirical formula for each compound:
1. Na2S2O4
1. NaSO2
2. B2H6
1. BH3
3. C6H12O6
1. CH2O
4. P4O10
1. P2O5
5. KMnO4
1. KMnO4
7. The empirical formula for each compound:
1. Al2Cl6
1. AlCl3
2. K2Cr2O7
1. K2Cr2O7
3. C2H4
1. CH2
4. (NH2)2CNH
1. CH5N3
5. CH3COOH
1. CH2O
2.9: Some Simple Organic Compounds
Conceptual Problems
1. Benzene (C6H6) is an organic compound, and KCl is an ionic compound. The sum of the masses of the atoms in each empirical formula is approximately the same. How would you expect the two to compare with regard to each of the following? What species are present in benzene vapor?
1. melting point
2. type of bonding
3. rate of evaporation
4. structure
2. Can an inorganic compound be classified as a hydrocarbon? Why or why not?
3. Is the compound NaHCO3 a hydrocarbon? Why or why not?
4. Name each compound.
1. NiO
2. TiO2
3. N2O
4. CS2
5. SO3
6. NF3
7. SF6
5. Name each compound.
1. HgCl2
2. IF5
3. N2O5
4. Cl2O
5. HgS
6. PCl5
6. For each structural formula, write the condensed formula and the name of the compound.
a.
b.
c.
d.
e.
7. For each structural formula, write the condensed formula and the name of the compound.
a.
b.
c.
d.
8. Would you expect PCl3 to be an ionic compound or a covalent compound? Explain your reasoning.
9. What distinguishes an aromatic hydrocarbon from an aliphatic hydrocarbon?
10. The following general formulas represent specific classes of hydrocarbons. Refer to Table 2.7 "The First 10 Straight-Chain Alkanes" and Table 2.8 "Some Common Acids That Do Not Contain Oxygen" and Figure 2.16 and identify the classes.
1. CnH2n + 2
1.
2. CnH2n
1.
3. CnH2n − 2
1.
11. Using R to represent an alkyl or aryl group, show the general structure of an
1. alcohol.
2. phenol.
Conceptual Answer
1. Benzene KCl
Melting Point Lower Higher
Type of Bonding Covalent Ionic
Rate of Evaporation Higher Lower
Structure Planar Molecule Ionic Crystal
2. An inorganic compound cannot be classified as a hydrocarbon, as hydrocarbons are by definition organic.
3. NaHCO3 is not a hydrocarbon. Hydrocarbons consist entirely of carbon and hydrogen atoms, while NaHCO3 also contains nitrogen and oxygen.
4. Name each compound.
1. NiO
1. Nickel Oxide
2. TiO2
1. Titanium Dioxide
3. N2O
1. Nitrous Oxide
4. CS2
1. Carbon Disulfide
5. SO3
1. Sulfur Trioxide
6. NF3
1. Nitrogen Trifluoride
7. SF6
1. Sulfur Hexafluoride
5. Name each compound.
1. HgCl2
1. Mercury(II) Chloride
2. IF5
1. Iodine Pentafluoride
3. N2O5
1. Dinitrogen Pentoxide
4. Cl2O
1. Dichlorine Monoxide
5. HgS
1. Mercury(II) Sulfide
6. PCl5
1. Phosphorus Pentachloride
6.
7.
8. PCl3 would be expected to be a covalent compound. The electronegativity difference between P and Cl is not great enough to create an ionic bond.
9. An aromatic hydrocarbon contains a conjugated ring system (in other words, a ring system composed of alternating single and double bonds). An aliphatic hydrocarbon, on the other hand, can be linear, or it can contain a non-conjugated ring system.
10. Identify the classes.
1. CnH2n + 2
1. Alkane
2. CnH2n
1. Alkene
3. CnH2n − 2
1. Alkyne
11.
1. ROH, where R is an alkyl group, in other words an alkane where one hydrogen is removed. The OH is attached to a saturated carbon atom.
2. ROH, where R is an aryl group, in other words an aromatic ring where one hydrogen is removed. The OH is attached directly to the ring.
Numerical Problems
1. Write the formula for each compound.
1. dinitrogen monoxide
2. silicon tetrafluoride
3. boron trichloride
4. nitrogen trifluoride
5. phosphorus tribromide
2. Write the formula for each compound.
1. dinitrogen trioxide
2. iodine pentafluoride
3. boron tribromide
4. oxygen difluoride
5. arsenic trichloride
3. Write the formula for each compound.
1. thallium(I) selenide
2. neptunium(IV) oxide
3. iron(II) sulfide
4. copper(I) cyanide
5. nitrogen trichloride
4. Name each compound.
1. RuO4
2. PbO2
3. MoF6
4. Hg2(NO3)2·2H2O
5. WCl4
5. Name each compound.
1. NbO2
2. MoS2
3. P4S10
4. Cu2O
5. ReF5
6. Draw the structure of each compound.
1. propyne
2. ethanol
3. n-hexane
4. cyclopropane
5. benzene
7. Draw the structure of each compound.
1. 1-butene
2. 2-pentyne
3. cycloheptane
4. toluene
5. phenol
Numerical Answers
1.
1. N2O
2. SiF4
3. BCl3
4. NF3
5. PBr3
2.
1. N2O3
2. IF5
3. BBr3
4. OF2
5. AsCl3
3.
1. TlSe
2. NpO2
3. FeS
4. CuCN
5. NCl3
4.
1. Ruthenium Tetroxide
2. Lead Dioxide (or Lead(IV) Oxide)
3. Molybdenum Hexafluoride (or Molybdenum(VI) Fluoride)
4. Mercury(I) Nitrate Dihydrate
5. Tungsten Tetrachloride (or Tungsten(IV) Chloride)
5.
1. niobium (IV) oxide
2. molybdenum (IV) sulfide
3. tetraphosphorus decasulfide
4. copper(I) oxide
5. rhenium(V) fluoride
6.
7.
a.
b.
c.
d.
e. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.E%3A_Atoms_Molecules_and_Ions_%28Exercises%29.txt |
This chapter introduced some of the fundamental concepts of chemistry, with particular attention to the basic properties of atoms and elements. These entities are the building blocks of all substances we encounter, yet most common substances do not consist of only pure elements or individual atoms. Instead, nearly all substances are chemical compounds or mixtures of chemical compounds. Although there are only about 115 elements (of which about 86 occur naturally), millions of chemical compounds are known, with a tremendous range of physical and chemical properties. Consequently, a major emphasis of modern chemistry focuses on understanding the relationship between the structures and properties of chemical compounds.
2.1: The Atomic Theory of Matter
• Dalton’s atomic theory of matter:
1. Each element is composed of extremely small particles called atoms
2. All atoms of a given element are identical; the atoms of different elements are different and have different properties
3. Atoms of an element are not changed into different types of atoms by chemical reactions
4. Compounds are formed when atoms of more than one element combine
• explains the law of constant composition, law of conservation of mass
2.2: The Discovery of Atomic Structure
• subatomic particles – what atoms are composed of
• Like charges repel each other; unlike charges attract
Cathode Rays and Electrons
• cathode rays – radiation resulting from a high voltage
• cause certain materials to give off light (fluoresce)
• mass of an electron 9.10939*10-28g
• 2000 times smaller than hydrogen
1.2.2 Radioactivity
• radioactivity – spontaneous emission of radiation
• three types of radiation: alpha (a ), beta (b ), gamma (g )
• alpha and beta radiation are affected by an electric field
• beta particles have a charge of 1-
• alpha particles have a charge of 2+
• gamma radiation has no particles and no charge
1.2.3 The Nuclear Atom
• Rutherford determined that there was a nucleus in every atom
• Protons discovered by Rutherford in 1919
• Neutrons discovered by James Chadwich in 1932
2.3: The Modern View of Atomic Structure
• charge of an electron is –1.602*10-19
• charge of a proton is +1.602*10-19
• 1.602*10-19 is called to electronic charge
Particle, Charge and Mass (amu)
Particle Charge Mass (amu)
Proton Positive 1.0073
Neutron None 1.0087
Electron Negative 5.486*10-4
• atomic mass unit (amu) – equals 1.66054*10-24 grams
• angstrom (Å) – unit of length to measure atomic dimensions
• 1 angstrom = 10-10m
• atoms have diameters of 1-5 Å
• nucleus – diameter of 10-4 Å
2.4: Atomic Mass
• all atoms of an element have the same number of protons in the nucleus
• isotopes - atoms of the same element that have a different number of neutrons
• atomic number – the number of protons in an atom
• mass number – number of protons + number of neutrons
• nuclide – atom of a specific isotope
2.5: The Periodic Table
• periodic table – the arrangement of all the elements by atomic number and similarities into a table
• columns = groups
• metallic elements – all elements on the left side and in the middle of the periodic table
• nonmetallic elements – elements on the periodic table that are divided by a diagonal steplike line from boron to astatine
• metalloids – properties of metals and nonmetals
2.6: Molecules and Molecular Compounds
• molecule – two or more atoms bonded together
1. Molecules and Chemical Formulas
• chemical formula – way of representing molecules
• diatomic molecule – any molecule made up of two atoms
• molecular compounds – contains more than one type of atom
1. Molecular and Empirical Formulas
• molecular formulas – chemical formulas that indication the actual number of atoms
• empirical formula – chemical formulas that only give the relative number of atoms
1. Picturing Molecules
• structural formulas – shows which atoms are attached to other atoms
• perspective drawing – gives an idea of the three-dimensional shape of a molecule
• ball-and-stick models – shows atoms as balls bonded by sticks
• space-filling model – accurate representation of what atoms would look like
2.7: Ions and Ionic Compounds
• ion – charged particle formed by the removal or addition of an electron
• cation – ion with a positive charge
• anion – ion with a negative charge
• metal atoms tend to lose electrons
• nonmetal atoms tend to gain electrons
• polyatomic ions – joined atoms that have a net positive or negative charge
1. Predicting Ionic Charges
• alkalie metals form 1+ ions
• alkaline earth from 2+ ions
• halogens form 1- ions
• group 6A from 2- ions
1. Ionic Compounds
• ionic compound – a compound that contains positively and negatively charged ions
• ionic compounds are generally combinations of metals and nonmetals
• molecular compounds are generally nonmetals only
2.8: Naming Inorganic Compounds
• chemical nomenclature – the naming of substances
• over 10 million known chemical substances
• organic compounds – contain carbon
• inorganic compounds – everything else
• positive ions
1. cations formed from atoms have the same name as the metal
2. if a metal can form cations of differing charges, the positive charge is given by a roman numeral in parentheses following the name of the metal
3. cations formed from nonmetal atoms have names that end in –ium
• Negative Ions
• monatomic anions have names formed by dropping the ending of the name of the element and adding the ending –ide
• polyatomic anions containing oxygen have names ending in –ate or –ite
• called oxyanions
1. anions derived by adding H+ to an oxyanion are named by adding as a prefix the word hydrogen or dihydrogen, as appropriate
1. Ionic compounds
1. names of ionic compounds are the cation name followed by the anion name
1. Names and formulas of Acids
1. acids bases on anions whose names end in –ide have associated acids that have the hydro- prefix and an –ic ending
2. acids based on anions whose names end in –ate or -ite
Anion and Acid Fill in the Blanks
Anion Acid
____ide Hydro____ic acid
____ate _____ic acid
____ite _____ous acid
1. Names and Formulas of Binary Molecular Compounds
• the name of the element farthest to the left in the periodic table is usually written first
• if elements in same group lower one written first
• name of second element is given an –ide ending
• greek prefixes used to indicate number of atoms of each element
• if prefix ends in a or o and the name of the anion begins with a vowel, the a or o is dropped
Prefixs and Meanings
Prefix Meaning
Mono- 1
Di- 2
Tri- 3
Tetra- 4
Penta- 5
Hexa- 6
Hepta- 7
Octa- 8
Nona- 9
Deca- 10 | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.S%3A_Atoms_Molecules_and_Ions_%28Summary%29.txt |
Stoichiometry is the calculation of relative quantities of reactants and products in chemical reactions. Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of product can be empirically determined, then the amount of the other reactants can also be calculated.
We begin this chapter by describing the relationship between the mass of a sample of a substance and its composition. We then develop methods for determining the quantities of compounds produced or consumed in chemical reactions, and we describe some fundamental types of chemical reactions. By applying the concepts and skills introduced in this chapter, you will be able to explain what happens to the sugar in a candy bar you eat, what reaction occurs in a battery when you start your car, what may be causing the “ozone hole” over Antarctica, and how we might prevent the hole’s growth.
• 3.1: Chemical Equations
A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products. In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right.
• 3.2: Some Simple Patterns of Chemical Reactivity
By recognizing general patterns of chemical reactivity, you will be able to successfully predict the products formed by a given combination of reactants We can often predict a reaction if we have seen a similar reaction before.
• 3.3: Formula Masses
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.
• 3.4: Avogadro's Number and 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 defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12, Avogadro’s number ($6.022 \times 10^{23}$) of atoms of carbon-12.
• 3.5: Empirical Formulas from Analysis
Chemical formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's molecular formula cannot be reduced any more, then the empirical formula is the same as the chemical formula. Combustion analysis can determine the empirical formula of a compound, but cannot determine the chemical formula (other techniques can though).
• 3.6: Quantitative Information from Balanced Equations
Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. 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.
• 3.7: Limiting Reactants
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 reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first is the limiting reagent.
• 3.E: Stoichiometry (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 3.S: Stoichiometry (Summary)
This is the summary Module for the chapter "Stoichiometry" in the Brown et al. General Chemistry Textmap.
Thumbnail: The combustion of methane, a hydrocarbon. (CC BY-SA 4.0 International; Yulo1985 (cropper) via Wikipedia)
03: Stoichiometry- Chemical Formulas and Equations
Learning Objectives
• To describe a chemical reaction.
• To calculate the quantities of compounds produced or consumed in a chemical reaction
What happens to matter when it undergoes chemical changes? The Law of conservation of mass says that "Atoms are neither created, nor destroyed, during any chemical reaction." Thus, the same collection of atoms is present after a reaction as before the reaction. The changes that occur during a reaction just involve the rearrangement of atoms. In this section we will discuss stoichiometry (the "measurement of elements").
Chemical Equations
As shown in Figure $1$, applying a small amount of heat to a pile of orange ammonium dichromate powder results in a vigorous reaction known as the ammonium dichromate volcano. Heat, light, and gas are produced as a large pile of fluffy green chromium(III) oxide forms. This reaction is described with a chemical equation, an expression that gives the identities and quantities of the substances in a chemical reaction.
Chemical reactions are represented on paper by chemical equations. For example, hydrogen gas (H2) can react (burn) with oxygen gas (O2) to form water (H2O). The chemical equation for this reaction is written as:
$\ce{2H_2 + O_2 \rightarrow 2H_2O} \nonumber$
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
$\underbrace{\ce{(NH_4)_2Cr_2O_7}}_{ reactant } \rightarrow \underbrace{\ce{Cr_2O_3 + N_2 + 4H_2O}}_{products }\label{3.1.1}$
The arrow is read as “yields” or “reacts to form.” Equation $\ref{3.1.1}$ 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:
$\ce{ (NH4)2Cr2O7(s) \rightarrow Cr2O3(s) + N2(g) + 4H2O(g)} \label{3.1.2}$
Equation $\ref{3.1.2}$ is identical to Equation $\ref{3.1.1}$ 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 $\ref{3.1.1}$ and $\ref{3.1.2}$. 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 $\ref{3.1.1}$ and $\ref{3.1.2}$ 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.
A chemical reaction changes only the distribution of atoms, not the number of atoms.
Introduction to Chemical Reaction Equations: Introduction to Chemical Reaction Equations, YouTube(opens in new window) [youtu.be]
Balancing Simple Chemical Equations
When a chemist encounters a new reaction, it does not usually come with a label that shows the balanced chemical equation. Instead, the chemist must identify the reactants and products and then write them in the form of a chemical equation that may or may not be balanced as first written. Consider, for example, the combustion of n-heptane ($C_7H_{16}$), an important component of gasoline:
$\ce{C_7H_{16} (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) } \label{3.1.3}$
The complete combustion of any hydrocarbon with sufficient oxygen always yields carbon dioxide and water.
Equation $\ref{3.1.3}$ is not balanced: the numbers of each type of atom on the reactant side of the equation (7 carbon atoms, 16 hydrogen atoms, and 2 oxygen atoms) is not the same as the numbers of each type of atom on the product side (1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms). Consequently, the coefficients of the reactants and products must be adjusted to give the same numbers of atoms of each type on both sides of the equation. Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure $3$.
Balancing Combustion Reactions: Balancing Combustions Reactions, YouTube(opens in new window) [youtu.be]
The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method.
Steps in Balancing a Chemical Equation
1. Identify the most complex substance.
2. Beginning with that substance, choose an element that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element on both sides.
3. Balance polyatomic ions (if present) as a unit.
4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
5. Check your work by counting the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.
Example $\PageIndex{1A}$: Combustion of Heptane
To demonstrate this approach, let’s use the combustion of n-heptane (Equation $\ref{3.1.3}$) as an example.
1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $\ce{C_7H_{16}}$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.
2. Adjust the coefficients. Try to adjust the coefficients of the molecules on the other side of the equation to obtain the same numbers of atoms on both sides. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:
$\ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + H_2O } \label{3.1.4}$
1. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.
2. Balance the remaining atoms. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side: $\ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + 8H_2O} \label{3.1.5}$ The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:$\ce{C_7H_{16} (l) + 11O_2 (g) \rightarrow 7CO_2 (g) + 8H_2O (g)} \label{3.1.6}$
3. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start.
Example $\PageIndex{1B}$: Combustion of Isooctane
Consider, for example, a similar reaction, the combustion of isooctane ($\ce{C8H18}$). Because the combustion of any hydrocarbon with oxygen produces carbon dioxide and water, the unbalanced chemical equation is as follows:
$\ce{C_8H_{18} (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) } \label{3.1.7}$
1. Identify the most complex substance. Begin the balancing process by assuming that the final balanced chemical equation contains a single molecule of isooctane.
2. Adjust the coefficients. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products:$\ce{C_8H_{18} + O_2 \rightarrow 8CO_2 + H_2O} \label{3.1.8}$
3. Balance polyatomic ions as a unit. This step does not apply to this equation.
4. Balance the remaining atoms. Eighteen hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products:$\ce{C_8H_{18} + O_2 \rightarrow 8CO_2 + 9H_2O } \label{3.1.9}$The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient (25/2) to balance the oxygen atoms: $\ce{C_8H_{18} + 25/2 O_2 \rightarrow 8CO_2 + 9H_2O} \label{3.1.10}$Equation $\ref{3.1.10}$ is now balanced, but we usually write equations with whole-number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2: $\ce{2C_8H_{18} (l) + 25O_2 (g) \rightarrow 16CO_2 (g) + 18H_2O (g) }\label{3.11}$
5. Check your work. The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side.
Balancing Complex Chemical Equations: Balancing Complex Chemical Equations, YouTube(opens in new window) [youtu.be]
Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.
Example $\PageIndex{1C}$: Hydroxyapatite
The reaction of the mineral hydroxyapatite ($\ce{Ca5(PO4)3(OH)}$) with phosphoric acid and water gives $\ce{Ca(H2PO4)2•H2O}$ (calcium dihydrogen phosphate monohydrate). Write and balance the equation for this reaction.
Given: reactants and product
Asked for: balanced chemical equation
Strategy:
1. Identify the product and the reactants and then write the unbalanced chemical equation.
2. Follow the steps for balancing a chemical equation.
Solution:
A We must first identify the product and reactants and write an equation for the reaction. The formulas for hydroxyapatite and calcium dihydrogen phosphate monohydrate are given in the problem (recall that phosphoric acid is H3PO4). The initial (unbalanced) equation is as follows:
$\ce{ Ca5(PO4)3(OH)(s) + H_3PO4 (aq) + H_2O_{(l)} \rightarrow Ca(H_2PO_4)_2 \cdot H_2O_{(s)} } \nonumber$
1. B Identify the most complex substance. We start by assuming that only one molecule or formula unit of the most complex substance, $\ce{Ca5(PO4)3(OH)}$, appears in the balanced chemical equation.
2. Adjust the coefficients. Because calcium is present in only one reactant and one product, we begin with it. One formula unit of $\ce{Ca5(PO4)3(OH)}$ contains 5 calcium atoms, so we need 5 Ca(H2PO4)2•H2O on the right side:
$\ce{Ca5(PO4)3(OH) + H3PO4 + H2O \rightarrow 5Ca(H2PO4)2 \cdot H2O} \nonumber$
3. Balance polyatomic ions as a unit. It is usually easier to balance an equation if we recognize that certain combinations of atoms occur on both sides. In this equation, the polyatomic phosphate ion (PO43), shows up in three places. In H3PO4, the phosphate ion is combined with three H+ ions to make phosphoric acid (H3PO4), whereas in Ca(H2PO4)2 • H2O it is combined with two H+ ions to give the dihydrogen phosphate ion. Thus it is easier to balance PO4 as a unit rather than counting individual phosphorus and oxygen atoms. There are 10 PO4 units on the right side but only 4 on the left. The simplest way to balance the PO4 units is to place a coefficient of 7 in front of H3PO4:
$\ce{Ca_5(PO_4)_3(OH) + 7H_3PO_4 + H_2O \rightarrow 5Ca(H_2PO_4)_2 \cdot H_2O } \nonumber$
Although OH is also a polyatomic ion, it does not appear on both sides of the equation. So oxygen and hydrogen must be balanced separately.
4. Balance the remaining atoms. We now have 30 hydrogen atoms on the right side but only 24 on the left. We can balance the hydrogen atoms using the least complex substance, H2O, by placing a coefficient of 4 in front of H2O on the left side, giving a total of 4 H2O molecules:
$\ce{Ca_5(PO_4)_3(OH) (s) + 7H_3PO_4 (aq) + 4H_2O (l) \rightarrow 5Ca(H_2PO_4)_2 \cdot H_2O (s) } \nonumber$
The equation is now balanced. Even though we have not explicitly balanced the oxygen atoms, there are 41 oxygen atoms on each side.
5. Check your work. Both sides of the equation contain 5 calcium atoms, 10 phosphorus atoms, 30 hydrogen atoms, and 41 oxygen atoms.
Exercise $1$: Fermentation
Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Humans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose are converted to ethanol ($CH_3CH_2OH$ and carbon dioxide $CO_2$. Write a balanced chemical reaction for the fermentation of glucose.
Commercial use of fermentation. (a) Microbrewery vats are used to prepare beer. (b) The fermentation of glucose by yeast cells is the reaction that makes beer production possible.
Answer
$C_6H_{12}O_6(s) \rightarrow 2C_2H_5OH(l) + 2CO_2(g) \nonumber$
Balancing Reactions Which Contain Polyatomics: Balancing Reactions Which Contain Polyatomics, YouTube(opens in new window) [youtu.be]
Interpreting Chemical Equations
In addition to providing qualitative information about the identities and physical states of the reactants and products, a balanced chemical equation provides quantitative information. Specifically, it gives the relative amounts of reactants and products consumed or produced in a reaction. The number of atoms, molecules, or formula units of a reactant or a product in a balanced chemical equation is the coefficient of that species (e.g., the 4 preceding H2O in Equation $\ref{3.1.1}$). When no coefficient is written in front of a species, the coefficient is assumed to be 1. As illustrated in Figure $4$, the coefficients allow Equation $\ref{3.1.1}$ to be interpreted in any of the following ways:
• Two NH4+ ions and one Cr2O72 ion yield 1 formula unit of Cr2O3, 1 N2 molecule, and 4 H2O molecules.
• One mole of (NH4)2Cr2O7 yields 1 mol of Cr2O3, 1 mol of N2, and 4 mol of H2O.
• A mass of 252 g of (NH4)2Cr2O7 yields 152 g of Cr2O3, 28 g of N2, and 72 g of H2O.
• A total of 6.022 × 1023 formula units of (NH4)2Cr2O7 yields 6.022 × 1023 formula units of Cr2O3, 6.022 × 1023 molecules of N2, and 24.09 × 1023 molecules of H2O.
These are all chemically equivalent ways of stating the information given in the balanced chemical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratio. For example, the mole ratio of $H_2O$ to $N_2$ in Equation $\ref{3.1.1}$ is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass:
$252 \;g \;\text{of}\; \ce{(NH_4)_2Cr_2O_7} \nonumber$
yield
$152 + 28 + 72 = 252 \; g \; \text{of products.} \nonumber$
The chemical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. These issues are considered in more detail in later chapters.
An important chemical reaction was analyzed by Antoine Lavoisier, an 18th-century French chemist, who was interested in the chemistry of living organisms as well as simple chemical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process chemists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose ($C_6H_{12}O_6$), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of chemical reactions.
Example $2$: Combustion of Glucose
The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows:
$\ce{C_6H_{12}O6(s) + 6O2(g) \rightarrow 6CO2(g) + 6H2O(l)} \nonumber$
Construct a table showing how to interpret the information in this equation in terms of
1. a single molecule of glucose.
2. moles of reactants and products.
3. grams of reactants and products represented by 1 mol of glucose.
4. numbers of molecules of reactants and products represented by 1 mol of glucose.
Given: balanced chemical equation
Asked for: molecule, mole, and mass relationships
Strategy:
1. Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios.
2. Use the molar masses of the reactants and products to convert from moles to grams.
3. Use Avogadro’s number to convert from moles to the number of molecules.
Solution:
This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information.
1. One molecule of glucose reacts with 6 molecules of O2 to yield 6 molecules of CO2 and 6 molecules of H2O.
2. One mole of glucose reacts with 6 mol of O2 to yield 6 mol of CO2 and 6 mol of H2O.
3. To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O2, 31.9988; CO2, 44.010; and H2O, 18.015.
\begin{align*} \text{mass of reactants} &= \text{mass of products} \[4pt] g \, glucose + g \, O_2 &= g \, CO_2 + g \, H_2O \end{align*} \nonumber
$1\,mol\,glucose \left ( {180.16 \, g \over 1 \, mol \, glucose } \right ) + 6 \, mol \, O_2 \left ( { 31.9988 \, g \over 1 \, mol \, O_2} \right ) \nonumber$
$= 6 \, mol \, CO_2 \left ( {44.010 \, g \over 1 \, mol \, CO_2} \right ) + 6 \, mol \, H_2O \left ( {18.015 \, g \over 1 \, mol \, H_2O} \right ) \nonumber$
$372.15 \, g = 372.15 \, g \nonumber$
C One mole of glucose contains Avogadro’s number (6.022 × 1023) of glucose molecules. Thus 6.022 × 1023 glucose molecules react with (6 × 6.022 × 1023) = 3.613 × 1024 oxygen molecules to yield (6 × 6.022 × 1023) = 3.613 × 1024 molecules each of CO2 and H2O.
In tabular form:
Solution to Example 3.1.2
$C_6H_{12}O_{6\;(s)}$ + $6O_{2\;(g)}$ $6CO_{2\;(g)}$ $6H_2O_{(l)}$
a. 1 molecule 6 molecules 6 molecules 6 molecules
b. 1 mol 6 mol 6 mol 6 mol
c. 180.16 g 191.9928 g 264.06 g 108.09 g
d. 6.022 × 1023 molecules 3.613 × 1024 molecules 3.613 × 1024 molecules 3.613 × 1024 molecule
Exercise $2$: Ammonium Nitrate Explosion
Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas.
The explosion resulted from the following reaction:
$2NH_4NO_{3\;(s)} \rightarrow 2N_{2\;(g)} + 4H_2O_{(g)} + O_{2\;(g)} \nonumber$
Construct a table showing how to interpret the information in the equation in terms of
1. individual molecules and ions.
2. moles of reactants and products.
3. grams of reactants and products given 2 mol of ammonium nitrate.
4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.
Answer:
Answer to Exercise 3.1.2
$2NH_4NO_{3\;(s)}$ $2N_{2\;(g)}$ + $4H_2O_{(g)}$ + $O_{2\;(g)}$
a. 2NH4+ ions and 2NO3 ions 2 molecules 4 molecules 1 molecule
b. 2 mol 2 mol 4 mol 1 mol
c. 160.0864 g 56.0268 g 72.0608 g 31.9988 g
d. 1.204 × 1024 formula units 1.204 × 1024 molecules 2.409 × 1024 molecules 6.022 × 1023 molecules
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be]
Summary
A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products. In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right, separated by an arrow. In a balanced chemical equation, the numbers of atoms of each element and the total charge are the same on both sides of the equation. The number of atoms, molecules, or formula units of a reactant or product in a balanced chemical equation is the coefficient of that species. The mole ratio of two substances in a chemical reaction is the ratio of their coefficients in the balanced chemical equation.
3.02: Some Simple Patterns of Chemical Reactivity
Learning Objectives
• To introduce the basic patterns of chemical reaction.
There is an underlying explanation of why a reaction takes place from an underlying chemical theory based on chemical reactivity. In this sections, we discussed various reactions to identify key chemical trends that allows chemists to predict the outcome of chemical reactions. By recognizing general patterns of chemical reactivity, you will be able to successfully predict the products formed by a given combination of reactants We can often predict a reaction if we have seen a similar reaction before. For example, sodium ($\ce{Na}$) reacts with water ($\ce{H_2O}$) to form sodium hydroxide ($\ce{NaOH}$) and $\ce{H_2}$ gas:
$\ce{ 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)} \nonumber$
with $\ce{(aq)}$ indicating aqueous liquid. As discussed later, potassium ($\ce{K}$) is in the same family (column) of elements as sodium and exhibits similar chemistry. Therefore, one might predict that the reaction of $\ce{K}$ with $\ce{H2O}$ would be similar to that of $\ce{Na}$:
$\ce{2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g)} \nonumber$
In fact, all alkali metals react with water to form their hydroxide compounds and hydrogen.
Combination and Decomposition Reactions
In combination reactions, two or more compounds react to form a single, more complex compound. Many elements react with one another in this fashion to form compounds. The general chemical equation for combination reaction is:
$\ce{ A + B -> C} \nonumber$
and an example includes the generation of ammonia ($\ce{NH_3}$) from nitrogen ($\ce{N_2}$), and hydrogen ($\ce{H_2}$):
$\ce{ N2(g) + 3H2(g) -> 2NH3(g)} \nonumber$
In decomposition reactions one substance undergoes a reaction to form two or more simpler products. Such reactions often occur when compounds are heated or electricity is added. The general chemical equation for decomposition reaction is:
$\ce{A -> B + C } \nonumber$
For example, the thermal decomposition of limestone ($\ce{CaCO3(s)}$) generates quicklime ($\ce{CaO(s)}$) and carbon dioxide $\ce{CO2(g)}$
$\ce{CaCO3(s) -> CaO(s) + CO2(g)} \nonumber$
Combustion in Air
Combustion reactions are rapid reactions that produce a flame. Most common combustion reactions involve oxygen (O2) from the air as a reactant. For example, wood will combust in the presence of oxygen in that air (with an ignition event) to generate a fire.
A common class of compounds which can participate in combustion reactions are hydrocarbons (compounds that contain only carbon and hydrogen).
Table $1$: Examples of common hydrocarbons
Hydrocarbon Molecular formula
methane CH4
propane C3H8
butane C4H10
octane C8H18
When hydrocarbons are combusted they react with oxygen ($\ce{O_2}$) to form carbon dioxide ($\ce{CO_2}$) and water ($\ce{H_2O}$). For example, when propane is burned the reaction is:
$\ce{ C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)} \label{3.2.6}$
Other compounds which contain carbon, hydrogen and oxygen (e.g. the alcohol methanol CH3OH, and the sugar glucose C6H12O6) also combust in the presence of oxygen ($\ce{O_2}$) to produce $\ce{CO_2}$ and $\ce{H_2O}$. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/03%3A_Stoichiometry-_Chemical_Formulas_and_Equations/3.01%3A_Chemical_Equations.txt |
Learning Objectives
• To determine the empirical formula of a compound from its composition by mass.
• To derive the molecular formula of a compound from its empirical formula.
When a new chemical compound, such as a potential new pharmaceutical, is synthesized in the laboratory or isolated from a natural source, chemists determine its elemental composition, its empirical formula, and its structure to understand its properties. This section focuses on how to determine the empirical formula of a compound and then use it to determine the molecular formula if the molar mass of the compound is known.
Formula and Molecular Weights
The formula weight of a substance is the sum of the atomic weights of each atom in its chemical formula. For example, water (H2O) has a formula weight of:
$2\times(1.0079\;amu) + 1 \times (15.9994 \;amu) = 18.01528 \;amu \nonumber$
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:
$6\times(12\; amu) + 12\times(1.00794\; amu) + 6\times(15.9994\; amu) = 180.0 \;amu \nonumber$
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 ($\ce{NaCl}$), for example, has a formula weight of:
$23.0\; amu + 35.5 \;amu = 58.5 \;amu \nonumber$
Molar Masses of Compounds, YouTube: Molar Masses of Compounds(opens in new window) [youtu.be]
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:
$1\times (12.011 \;amu) + 4 \times (1.008) = 16.043 \;amu \nonumber$
the relative (mass) percentages of carbon and hydrogen are
$\%C = \dfrac{1 \times (12.011\; amu)}{16.043 amu} = 0.749 = 74.9\% \nonumber$
$\%H = \dfrac{4 \times (1.008 \;amu)}{16.043\; amu} = 0.251 = 25.1\% \nonumber$
A more complex example is sucrose (table sugar), which 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:
$\text {mass of C/mol of sucrose} = 12 \, mol \, C \times {12.011 \, g \, C \over 1 \, mol \, C} = 144.132 \, g \, C \label{3.1.1a}$
$\text {mass of H/mol of sucrose} = 22 \, mol \, H \times {1.008 \, g \, H \over 1 \, mol \, H} = 22.176 \, g \, H \label{3.1.1b}$
$\text {mass of O/mol of sucrose} = 11 \, mol \, O \times {15.999 \, g \, O \over 1 \, mol \, O} = 175.989 \, g \, O \label{3.1.1c}$
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:
$\text {mass % C in Sucrose} = {\text {mass of C/mol sucrose} \over \text {molar mass of sucrose} } \times 100 = {144.132 \, g \, C \over 342.297 \, g/mol } \times 100 = 42.11 \% \nonumber$
$\text {mass % H in Sucrose} = {\text {mass of H/mol sucrose} \over \text {molar mass of sucrose} } \times 100 = {22.176 \, g \, H \over 342.297 \, g/mol } \times 100 = 6.48 \% \nonumber$
$\text {mass % O in Sucrose} = {\text {mass of O/mol sucrose} \over \text {molar mass of sucrose} } \times 100 = {175.989 \, g \, O \over 342.297 \, g/mol } \times 100 = 51.41 \% \nonumber$
This can be checked by verifying that the sum of the percentages of all the elements in the compound is 100%:
$42.11\% + 6.48\% + 51.41\% = 100.00\% \nonumber$
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 42.11 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 42.11% 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 $1$: NutraSweet
Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is $\ce{C14H18N2O5}$.
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:
$14 \,C (14 \, mol \, C)(12.011 \, g/mol \, C) = 168.154 \, g \nonumber$
$18 \,H (18 \, mol \, H)(1.008 \, g/mol \, H) = 18.114 \, g\nonumber$
$2 \,N (2 \, mol \, N)(14.007 \, g/mol \, N) = 28.014 \, g\nonumber$
$+5 \,O (5 \, mol \, O)(15.999 \, g/mol \, O) = 79.995 \, g\nonumber$
$C_{14}H_{18}N_2O_5 \text {molar mass of aspartame} = 294.277 \, g/mol \nonumber$
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:
$mass \% \, C = {168.154 \, g \, C \over 294.277 \, g \, aspartame } \times 100 = 57.14 \% C\nonumber$
$mass \% \, H = {18.114 \, g \, H \over 294.277 \, g \, aspartame } \times 100 = 6.16 \% H\nonumber$
$mass \% \, N = {28.014 \, g \, N \over 294.277 \, g \, aspartame } \times 100 = 9.52 \% \nonumber$
$mass \% \, O = {79.995 \, g \, O \over 294.277 \, g \, aspartame } \times 100 = 27.18 \%\nonumber$
As a check, we can add the percentages together:
$57.14\% + 6.16\% + 9.52\% + 27.18\% = 100.00\% \nonumber$
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:
$\text {mass of C} = 1.00 \, g \, aspartame \times {57.14 \, g \, C \over 100 \, g \, aspartame } = 0.571 \, g \, C \nonumber$
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: Percent Composition, YouTube(opens in new window) [youtu.be]HNS6lItns10 (opens in new window)
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:
$\text {moles Ca} = 38.77 \, g \, Ca \times {1 \, mol \, Ca \over 40.078 \, g \, Ca} = 0.9674 \, mol \, Ca\nonumber$
$\text {moles P} = 19.97 \, g \, P \times {1 \, mol \, P \over 30.9738 \, g \, P} = 0.6447 \, mol \, Ca\nonumber$
$\text {moles O} = 41.27 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O} = 2.5800 \, mol \, O\nonumber$
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:
$P: {0.6447 \, mol\, P \over 0.6447 \, mol\, P} = 1.000 \, \, \, \, Ca: {0.9674 \over 0.6447} = 1.501\, \, \, \, O: {2.5800\over 0.6447}= 4.002\nonumber$
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 $2$: Ammonium Nitrate
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.
Answer
N2H4O3 is NH4+NO3, written as NH4NO3
Determining Empirical and Molecular Formulas from % Composition: Determining Empirical and Molecular Formulas from % Composition, YouTube(opens in new window) [youtu.be]
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:
$C: {3.304 \over 3.304} = 1.000 \, \, \, \, H: {6.53 \over 3.304} = 1.98 \, \, \, \, O: {3.362 \over 3.304} = 1.018 \nonumber$
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.
$moles \, C = 49.18 \, g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 4.095 \, mol \, C \nonumber$
$moles \, H = 5.39 \, g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 5.35 \, mol \, H\nonumber$
$moles \, N = 28.65 \, g \, N \times {1 \, mol \, N \over 14.0067 \, g \, N} = 2.045 \, mol \, N \nonumber$
$moles \, O = 16.68 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O} = 1.043 \, mol \, O \nonumber$
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:
$O: {1.043 \over 1.043} = 1.000 \, \, \, \, C: {4.095 \over 1.043} = 3.926 \, \, \, \, H: {5.35 \over 1.043} = 5.13 \, \, \, \, N: {2.045 \over 1.043} = 1.960 \nonumber$
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:
$4C \, \, \, ( 4 \, atoms \, C) (12.011 \, g/ atom \, C) = 48.044 \, g\nonumber$
$5H \, \, \, ( 5 \, atoms \, H ) (1.0079 \, g/ atom \, H) = 5.0395 \, g \nonumber$
$2N \, \, \, (2 \, atoms \, N) (14.0067 \, g/ atom \, N) = 28.0134 \, g \nonumber$
$+1O \, \, \, (1 \, atom \, O) (15.9994 \, g/ atom \, O) = 15.9994 \, g \nonumber$
$C_4H_5N_2O \, \, \, \, \text {formula mass of caffeine} = 97.096 \, g\nonumber$
Dividing the measured molar mass of caffeine (196 g/mol) by the calculated formula mass gives
${196 g/mol \over 97.096 g/C_4H_5N_2O } = 2.02 \approx 2\, C_4H_5N_2O \, \text {empirical formula units}\nonumber$
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
$C_2Cl_2F_4 \nonumber$
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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/03%3A_Stoichiometry-_Chemical_Formulas_and_Equations/3.03%3A_Formula_Masses.txt |
Learning Objectives
• To calculate the molecular mass of a covalent compound and the formula mass of an ionic compound, and to calculate the number of atoms, molecules, or formula units in a sample of a substances.
As discussed previosuly, the mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass. Although the mass number is unitless, it is assigned units called atomic mass units (amu). Because a molecule or a polyatomic ion is an assembly of atoms whose identities are given in its molecular or ionic formula, the average atomic mass of any molecule or polyatomic ion can be calculated from its composition by adding together the masses of the constituent atoms. The average mass of a monatomic ion is the same as the average mass of an atom of the element because the mass of electrons is so small that it is insignificant in most calculations.
Molecular and Formula Masses
The molecular mass of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example $1$.
Example $1$: Molecular Mass of Ethanol
Calculate the molecular mass of ethanol, whose condensed structural formula is $\ce{CH3CH2OH}$. Among its many uses, ethanol is a fuel for internal combustion engines.
Given: molecule
Asked for: molecular mass
Strategy:
1. Determine the number of atoms of each element in the molecule.
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 molecular mass.
Solution:
A The molecular formula of ethanol may be written in three different ways: $\ce{CH3CH2OH}$ (which illustrates the presence of an ethyl group, CH3CH2, and an −OH group), $\ce{C2H5OH}$, and $\ce{C2H6O}$; all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom.
B Taking the atomic masses from the periodic table, we obtain
\begin{align*} 2 \times \text { atomic mass of carbon} &= 2 \, atoms \left ( {12.011 \, amu \over atoms } \right ) \[4pt] &= 24.022 \,amu \end{align*}\nonumber
\begin{align*} 6 \times \text { atomic mass of hydrogen} &= 2 \, atoms \left ( {1.0079 \, amu \over atoms } \right ) \[4pt] &= 6.0474 \,amu \end{align*}\nonumber
\begin{align*} 1 \times \text { atomic mass of oxygen} &= 1 \, atoms \left ( {15.9994 \, amu \over atoms } \right ) \[4pt] &= 15.994 \,amu \end{align*}\nonumber
C Adding together the masses gives the molecular mass:
$24.022 \,amu + 6.0474 \,amu + 15.9994 \,amu = 46.069 \,amu \nonumber \nonumber$
Alternatively, we could have used unit conversions to reach the result in one step:
$\left [ 2 \, atoms\, C \left ( {12.011 \, amu \over 1 \, atom C} \right ) \right ] + \left [ 6 \, atoms\, H \left ( {1.0079 \, amu \over 1 \, atom H} \right ) \right ] + \left [ 1 \, atoms\, C \left ( {15.9994 \, amu \over 1 \, atom 0} \right ) \right ] = 46.069 \, amu \nonumber$
The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:
$2 \times C \, \, \, (2\, atoms) (12.011 \, amu/atom ) = 24.022 \, amu \nonumber$
$6 \times H \, \, \, (6\, atoms) (1.0079 \, amu/atom ) = 6.0474 \, amu \nonumber$
$1 \times \,O \, \, \, (1\, atoms) (15.9994 \, amu/atom ) = 15.9994 \, amu \nonumber$
$C_2H_6O \, \, \, \, \, \text {molecular mass of ethanol} = 46.069 \, amu \nonumber$
Exercise $1$: Molecular Mass of Freon
Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is $\ce{CCl_3F}$. 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.
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 $\ce{Ca3(PO4)2}$, 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
$3 \times \text {atomic mass of calcium} = 3 \, atoms \left ( {40.078 \, amu \over atom } \right ) = 120.234 \, amu \nonumber \nonumber$
$2 \times \text {atomic mass of phosphorus} = 2 \, atoms \left ( {30.973761 \, amu \over atom } \right ) = 61.947522 \, amu \nonumber \nonumber$
$8 \times \text {atomic mass of oxygen} = 8 \, atoms \left ( {15.9994 \, amu \over atom } \right ) = 127.9952 \, amu \nonumber \nonumber$
C Adding together the masses gives the formula mass of $\ce{Ca3(PO4)2}$:
$120.234 \,amu + 61.947522 \, amu + 127.9952 \, amu = 310.177 \, amu \nonumber \nonumber$
We could also find the formula mass of $\ce{Ca3(PO4)2}$ in one step by using unit conversions or a tabular format:
$\left [ 3 \, atoms Ca \left ({40.078 \, amu \over 1 \, atom Ca } \right ) \right ] + \left [ 2 \, atoms P \left ({30.973761 \, amu \over 1 \, atom P } \right ) \right ] + \left [ 8 \, atoms O \left ({15.9994 \, amu \over 1 \, atom O } \right ) \right ] = 310.177 \,amu \nonumber$
$3Ca \, \, \, \, (3\, atoms)(40.078 \, amu/atom) = 120.234 \, amu \nonumber$
$2P \, \, \, \, (2\, atoms)(30.973761 \, amu/atom) = 61.947522 \, amu \nonumber$
$+ 8O \, \, \, \, (8\, atoms)(15.9994 \, amu/atom) = 127.9952 \, amu \nonumber$
$Ca_3P_2O_8 \, \, \, \, \text {formula mass of Ca}_3(PO_4)_2 = 310.177 \, amu \nonumber$
Exercise $2$: Formula Mass of Silicon Nitride
Calculate the formula mass of $\ce{Si3N4}$, commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.
Answer
$140.29 \,amu$
Molar Masses of Compounds: Molar Masses of Compounds, YouTube(opens in new window) [youtu.be]
The Mole
Dalton’s theory that each chemical compound has a particular combination of atoms and that the ratios of the numbers of atoms of the elements present are usually small whole numbers. It also describes the law of multiple proportions, which states that the ratios of the masses of elements that form a series of compounds are small whole numbers. The problem for Dalton and other early chemists was to discover the quantitative relationship between the number of atoms in a chemical substance and its mass. Because the masses of individual atoms are so minuscule (on the order of 10−23 g/atom), chemists do not measure the mass of individual atoms or molecules. In the laboratory, for example, the masses of compounds and elements used by chemists typically range from milligrams to grams, while in industry, chemicals are bought and sold in kilograms and tons. To analyze the transformations that occur between individual atoms or molecules in a chemical reaction, it is therefore essential for chemists to know how many atoms or molecules are contained in a measurable quantity in the laboratory—a given mass of sample. The unit that provides this link is the mole (mol), from the Latin moles, meaning “pile” or “heap.”
Many familiar items are sold in numerical quantities with distinct names. For example, cans of soda come in a six-pack, eggs are sold by the dozen (12), and pencils often come in a gross (12 dozen, or 144). Sheets of printer paper are packaged in reams of 500, a seemingly large number. Atoms are so small, however, that even 500 atoms are too small to see or measure by most common techniques. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extremely large numerical unit is needed to count them. The mole is used for this purpose.
A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142 × 1023 atoms, but for most purposes 6.022 × 1023 provides an adequate number of significant figures. Just as 1 mole of atoms contains 6.022 × 1023 atoms, 1 mole of eggs contains 6.022 × 1023 eggs. This number is called Avogadro’s number, after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain.
It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mole of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 1023.
One mole always has the same number of objects: 6.022 × 1023.
To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 1017 mi high, or almost six times the diameter of the Milky Way galaxy. If a mole of pennies were distributed equally among the entire population on Earth, each person would have more than one trillion dollars. The mole is so large that it is useful only for measuring very small objects, such as atoms.
The concept of the mole allows scientists to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon-12 atoms, one weighs out 12 g of isotopically pure carbon-12. Because each element has a different atomic mass, however, a mole of each element has a different mass, even though it contains the same number of atoms (6.022 × 1023). This is analogous to the fact that a dozen extra large eggs weighs more than a dozen small eggs, or that the total weight of 50 adult humans is greater than the total weight of 50 children. Because of the way the mole is defined, for every element the number of grams in a mole is the same as the number of atomic mass units in the atomic mass of the element. For example, the mass of 1 mol of magnesium (atomic mass = 24.305 amu) is 24.305 g. Because the atomic mass of magnesium (24.305 amu) is slightly more than twice that of a carbon-12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon-12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about one-third that of 1 mol of carbon-12. Using the concept of the mole, Dalton’s theory can be restated: 1 mol of a compound is formed by combining elements in amounts whose mole ratios are small whole numbers. For example, 1 mol of water (H2O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms.
Molar Mass
The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.
The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole.
The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol:
Atomic mass of carbon as 12.011 amu; the average molar mass of carbon
Substance (formula) Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol)
carbon (C) 12.011 (atomic mass) 12.011
ethanol (C2H5OH) 46.069 (molecular mass) 46.069
calcium phosphate [Ca3(PO4)2] 310.177 (formula mass) 310.177
Determining the Molar Mass of a Molecule, YouTube: Determining the Molar Mass of a Molecule(opens in new window) [youtu.be]
The molar mass of naturally-occurring carbon is different from that of carbon-12, and is not an integer because carbon occurs as a mixture of carbon-12, carbon-13, and carbon-14. One mole of carbon still has 6.022 × 1023 carbon atoms, but 98.89% of those atoms are carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 1012) are carbon-14. (For more information, see Section 1.6.) Similarly, the molar mass of uranium is 238.03 g/mol, and the molar mass of iodine is 126.90 g/mol. When dealing with elements such as iodine and sulfur, which occur as a diatomic molecule (I2) and a polyatomic molecule (S8), respectively, molar mass usually refers to the mass of 1 mol of atoms of the element—in this case I and S, not to the mass of 1 mol of molecules of the element (I2 and S8).
The molar mass of ethanol is the mass of ethanol (C2H5OH) that contains 6.022 × 1023 ethanol molecules. As in Example $1$, the molecular mass of ethanol is 46.069 amu. Because 1 mol of ethanol contains 2 mol of carbon atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 × 1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994 g), its molar mass is 46.069 g/mol. Similarly, the formula mass of calcium phosphate [Ca3(PO4)2] is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that contains 6.022 × 1023 formula units.
The mole is the basis of quantitative chemistry. It provides chemists with a way to convert easily between the mass of a substance and the number of individual atoms, molecules, or formula units of that substance. Conversely, it enables chemists to calculate the mass of a substance needed to obtain a desired number of atoms, molecules, or formula units. For example, to convert moles of a substance to mass, the following relationship is used:
$(\text{moles}) \times (\text{molar mass}) \rightarrow \text{mass} \label{3.4.1}$
or, more specifically,
$\cancel{\text{moles}} \times \left ( {\text{grams} \over \cancel{\text{mole}} } \right ) = \text{grams} \nonumber$
Conversely, to convert the mass of a substance to moles:
$\left ( {\text{grams} \over \text{molar mass} } \right ) \rightarrow \text{moles} \label{3.4.2A}$
$\left ( { \text{grams} \over \text{grams/mole}} \right ) = \cancel{\text{grams}} \left ( {\text{mole} \over \cancel{\text{grams}} } \right ) = \text{moles} \label{3.4.2B}$
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:
$\ce{2H2(g) + O2(g) \rightarrow 2H2O(\ell) } \nonumber$
the production of two moles of water would require the consumption of 2 moles of $\ce{H_2}$ and one mole of $\ce{O_2}$. Therefore, when considering this particular reaction
• 2 moles of $\ce{H2}$
• 1 mole of $\ce{O2}$ and
• 2 moles of $\ce{H2O}$
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 $O_2$?
$(1.57\; mol\; \ce{O_2}) \left( \dfrac{2\; mol\, \ce{H_2O}}{1\;mol\; \ce{O_2}} \right) = 3.14\; mol\; \ce{H_2O}\nonumber$
The ratio $\left( \dfrac{2\; mol\; \ce{H_2O}}{1\;mol\;\ce{O_2}} \right)$ is the stoichiometric relationship between $\ce{H_2O}$ and $\ce{O_2}$ from the balanced equation for this reaction.
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 Examples $3$ and $4$.
Example $3$: Combustion of Butane
For the combustion of butane ($\ce{C_4H_{10}}$) the balanced equation is:
$\ce{2C4H_{10}(l) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l) }\nonumber$
Calculate the mass of $\ce{CO_2}$ that is produced in burning 1.00 gram of $\ce{C_4H_{10}}$.
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:
$(1.00\; g\; \ce{C_4H_{10}}) \left(\dfrac{1\; mol\; \ce{C_4H_{10}}}{58.0\;g\; \ce{C_4H_{10}}}\right) = 1.72 \times 10^{-2} \; mol\; \ce{C_4H_{10}}\nonumber$
Now, the stoichiometric relationship between $\ce{C_4H_{10}}$ and $\ce{CO_2}$ is:
$\left( \dfrac{8\; mol\; \ce{CO_2}}{2\; mol\; \ce{C_4H_{10}}}\right)\nonumber$
Therefore:
$\left(\dfrac{8\; mol\; \ce{CO_2}}{2\; mol\; \ce{C_4H_{10}}} \right) \times 1.72 \times 10^{-2} \; mol\; \ce{C_4H_{10}} = 6.88 \times 10^{-2} \; mol\; \ce{CO_2} \nonumber$
The question called for the determination of the mass of $\ce{CO_2}$ produced, thus we have to convert moles of $\ce{CO_2}$ into grams (by using the molecular weight of $\ce{CO_2}$):
$6.88 \times 10^{-2} \; mol\; \ce{CO_2} \left( \dfrac{44.0\; g\; \ce{CO_2}}{1\; mol\; \ce{CO_2}} \right) = 3.03\;g \; \ce{CO_2}\nonumber$
Example $4$: Ethylene Glycol
For 35.00 g of ethylene glycol (\ce{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 The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example $1$:
$2C (2 \,atoms )(12.011 \, amu/atom) = 24.022 \, amu \nonumber$
$6H (6 \,atoms )(1.0079 \, amu/atom) = 6.0474 \, amu \nonumber$
$2O (2 \,atoms )(15.9994 \, amu/atom) = 31.9988 \, amu \nonumber$
$C_2H_6O_2 \text {molecular mass of ethylene glycol} = 62.068 \, amu \nonumber$
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):
${ \text {mass of ethylene glycol (g)} \over \text {molar mass (g/mol)} } = \text {moles ethylene glycol (mol) }\nonumber$
So
$35.00 \, g \text {ethylene glycol} \left ( {1 \, mole\, \text {ethylene glycol} \over 62.068 \, g \, \text {ethylene glycol } } \right ) = 0.5639 \,mol \text {ethylene glycol} \nonumber$
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.
C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:
\begin{align*} \text {molecules of ethylene glycol} &= 0.5639 \, \cancel{mol} \left ( {6.022 \times 10^{23} \, molecules \over 1 \, \cancel{mol} } \right ) \[4pt] &= 3.396 \times 10^{23} \, molecules \end{align*} \nonumber
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.
Exercise $4$: Freon-11
For 75.0 g of CCl3F (Freon-11), calculate the number of
1. moles.
2. molecules.
Answer a
0.546 mol
Answer b
3.29 × 1023 molecules
Example $5$
Calculate the mass of 1.75 mol of each compound.
1. S2Cl2 (common name: sulfur monochloride; systematic name: disulfur dichloride)
2. Ca(ClO)2 (calcium hypochlorite)
Given: number of moles and molecular or empirical formula
Asked for: mass
Strategy:
A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic).
B Convert from moles to mass by multiplying the moles of the compound given by its molar mass.
Solution:
We begin by calculating the molecular mass of S2Cl2 and the formula mass of Ca(ClO)2.
A The molar mass of S2Cl2 is obtained from its molecular mass as follows:
$2S (2 \, atoms)(32.065 \, amu/atom ) = 64.130 \, amu \nonumber$
$+ 2Cl (2 \, atoms )(35.453 \, amu/atom ) = 70.906 \, amu \nonumber$
$S_2 Cl_2 \text {molecular mass of } S_2Cl_2 = 135.036 \, amu \nonumber$
The molar mass of S2Cl2 is 135.036 g/mol.
B The mass of 1.75 mol of S2Cl2 is calculated as follows:
$moles S_2Cl_2 \left [\text {molar mass}\left ({ g \over mol} \right )\right ] \rightarrow mass of S_2Cl_2 \, (g) \nonumber$
$1.75 \, mol S_2Cl_2\left ({135.036 \, g S_2Cl_2 \over 1 \, mol S_2Cl_2 } \right ) = 236 \, g S_2Cl_2 \nonumber$
A The formula mass of Ca(ClO)2 is obtained as follows:
$1Ca (1 \, atom)(40.078 \, amu/atom) = 40.078 \, amu \nonumber$
$2Cl (2 \, atoms)(35.453 \, amu/atom) = 70.906 \, amu \nonumber$
$+ 2O (2 \, atoms)(15.9994 \, amu/atom) = 31.9988 \, amu \nonumber$
$Ca (ClO)_2 \text { formula mass of } Ca (ClO)_2 = 142.983 \, amu\nonumber$
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:
$moles Ca(ClO)_2 \left [{\text {molar mass} Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2} \right ]=mass Ca(ClO)_2\nonumber$
$1.75 \, mol Ca(ClO)_2 \left [ {142.983 \, g Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2 } \right ] = 250 \, g Ca(ClO)_2 \nonumber$
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 a
1.71 g
Answer b
0.721 g
Conversions Between Grams, Mol, & Atoms: Conversions Between Grams, Mol & Atoms, YouTube(opens in new window) [youtu.be]
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:
$\ce{2H2(g) + O2(g) \rightarrow 2H_2O(l)} \nonumber \nonumber$
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 $O_2$?
$(1.57\; mol\; O_2) \left( \dfrac{2\; mol H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O \nonumber \nonumber$
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:
$\ce{2C4H_{10} (l) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l)} \nonumber \nonumber$
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:
$(1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10} \nonumber \nonumber$
Now, the stoichiometric relationship between $C_4H_{10}$ and $CO_2$ is:
$\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right) \nonumber \nonumber$
Therefore:
$\left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2 \nonumber \nonumber$
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 $CO_2$):
$6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2 \nonumber \nonumber$
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): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios), YouTube(opens in new window) [youtu.be]
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_Chemistry_-_The_Central_Science_(Brown_et_al.)/03%3A_Stoichiometry-_Chemical_Formulas_and_Equations/3.04%3A_Avogadro%27s_Number_and_the_Mole.txt |
Learning Objectives
• To understand the definition and difference between empirical formulas and chemical formulas
• To understand how combustion analysis can be used to identify chemical formulas
Chemical formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's chemical formula cannot be reduced any more, then the empirical formula is the same as the chemical formula. Combustion analysis can determine the empirical formula of a compound, but cannot determine the chemical formula (other techniques can though). Once known, the chemical formula can be calculated from the empirical formula.
Empirical Formulas
An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. We can also work backwards from molar ratios since if we know the molar amounts of each element in a compound we can determine the empirical formula.
Example $1$: Mercury Chloride
Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula?
Solution
Let's say we had a 100 gram sample of this compound. The sample would therefore contain 73.9 grams of mercury and 26.1 grams of chlorine. How many moles of each atom do the individual masses represent?
For Mercury:
$(73.9 \;g) \times \left(\dfrac{1\; mol}{200.59\; g}\right) = 0.368 \;moles \nonumber$
For Chlorine:
$(26.1\; g) \times \left(\dfrac{1\; mol}{35.45\; g}\right) = 0.736\; mol \nonumber$
What is the molar ratio between the two elements?
$\dfrac{0.736 \;mol \;Cl}{0.368\; mol\; Hg} = 2.0 \nonumber$
Thus, we have twice as many moles (i.e. atoms) of $\ce{Cl}$ as $\ce{Hg}$. The empirical formula would thus be (remember to list cation first, anion last):
$\ce{HgCl2} \nonumber$
Chemical Formula from Empirical Formula
The chemical formula for a compound obtained by composition analysis is always the empirical formula. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound. The chemical formula will always be some integer multiple of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula). The general flow for this approach is shown in Figure $1$ and demonstrated in Example $2$.
Example $2$: Ascorbic Acid
Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What is the empirical and chemical formula for ascorbic acid?
Solution
Consider an arbitrary amount of 100 grams of ascorbic acid, so we would have:
• 40.92 grams C
• 4.58 grams H
• 54.50 grams O
This would give us how many moles of each element?
• Carbon
$(40.92\; \cancel{g\; C}) \times \left( \dfrac{1\; mol\; C}{12.011\; \cancel{g\; C}} \right) = 3.407\; mol \; C \nonumber$
• Hydrogen
$(4.58\; \cancel{g\; H}) \times \left( \dfrac{1\; mol\; H}{1.008\; \cancel{g\; H}} \right) = 4.544\; mol \;H \nonumber$
• Oxygen
$(54.50\; \cancel{g\; O}) \times \left( \dfrac{1\; mol\; O}{15.999\; \cancel{g\; O}} \right) = 3.406\; mol \; O \nonumber$
Determine the simplest whole number ratio by dividing by the smallest molar amount (3.406 moles in this case - see oxygen):
• Carbon
$C= \dfrac{3.407\; mol}{3.406\; mol} \approx 1.0 \nonumber$
• Hydrogen
$C= \dfrac{4.5.44\; mol}{3.406\; mol} = 1.0 \nonumber$
• Oxygen
$C= \dfrac{3.406\; mol}{3.406\; mol} = 1.0 \nonumber$
The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of hydrogen to be equal to an integer. 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom.
C = (1.0)*3 = 3
H = (1.333)*3 = 4
O = (1.0)*3 = 3
or
$\ce{C3H4O3} \nonumber$
This is our empirical formula for ascorbic acid.
What about the chemical formula? We are told that the experimentally determined molecular mass is 176 amu. What is the molecular mass of our empirical formula?
(3*12.011) + (4*1.008) + (3*15.999) = 88.062 amu
The molecular mass from our empirical formula is significantly lower than the experimentally determined value. What is the ratio between the two values?
(176 amu/88.062 amu) = 2.0
Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. If we multiplied our empirical formula by '2', then the molecular mass would be correct. Thus, the actual chemical formula is:
2* C3H4O3 = C6H8O6
Empirical Formulas: Empirical Formulas, YouTube(opens in new window) [youtu.be]
Combustion Analysis
When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O (Figure $2$). The amount of carbon produced can be determined by measuring the amount of CO2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate.
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 $3$ and a typical combustion analysis is illustrated in Examples $3$ and $4$.
Example $3$: Combustion of Isopropyl Alcohol
What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O?
Solution
From this information quantitate the amount of C and H in the sample.
$(0.561\; \cancel{g\; CO_2}) \left( \dfrac{1 \;mol\; CO_2}{44.0\; \cancel{g\;CO_2}}\right)=0.0128\; mol \; CO_2 \nonumber$
Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this?
$(0.0128 \; \cancel{mol\; C}) \left( \dfrac{12.011\; g \; C}{1\; \cancel{mol\;C}}\right)=0.154\; g \; C \nonumber$
How about the hydrogen?
$(0.306 \; \cancel{g\; H_2O}) \left( \dfrac{1\; mol \; H_2O}{18.0\; \cancel{g \;H_2O}}\right)=0.017\; mol \; H_2O \nonumber$
Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample.
When we add our carbon and hydrogen together we get:
0.154 grams (C) + 0.034 grams (H) = 0.188 grams
But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol:
0.255 grams - 0.188 grams = 0.067 grams oxygen
This much oxygen is how many moles?
$(0.067 \; \cancel{g\; O}) \left( \dfrac{1\; mol \; O}{15.994\; \cancel{g \;O}}\right)=0.0042\; mol \; O \nonumber$
Overall therefore, we have:
• 0.0128 moles Carbon
• 0.0340 moles Hydrogen
• 0.0042 moles Oxygen
Divide by the smallest molar amount to normalize:
• C = 3.05 atoms
• H = 8.1 atoms
• O = 1 atom
Within experimental error, the most likely empirical formula for propanol would be $C_3H_8O$
Example $4$: Combustion of Naphalene
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 $\ce{CO2}$ 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:
$mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C} \nonumber$
$= 1.883 \times 10^{-2} \, g \, C \nonumber$
$mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H} \nonumber$
$= 1.264 \times 10^{-3} \, g \, H \nonumber$
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:
$moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C \nonumber$
$moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H \nonumber$
Dividing each number by the number of moles of the element present in the smaller amount gives
$H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250 \nonumber$
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 chemical 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 chemical formula is C6H6). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO2 and H2O will be produced?
Answer a
The empirical formula is C4H5. (The chemical formula of xylene is actually C8H10.)
Answer b
33.81 mg of CO2; 6.92 mg of H2O
Combustion Analysis: Combustion Analysis, YouTube(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/03%3A_Stoichiometry-_Chemical_Formulas_and_Equations/3.05%3A_Empirical_Formulas_from_Analysis.txt |
Learning Objectives
• To balance equations that describe reactions in solution.
• To calculate the quantities of compounds produced or consumed in a chemical reaction.
• To solve quantitative problems involving the stoichiometry of reactions in solution.
A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced. Stoichiometry is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. A stoichiometric quantity is the amount of product or reactant specified by the coefficients in a balanced chemical equation. This section describes how to use the stoichiometry of a reaction to answer questions like the following: How much oxygen is needed to ensure complete combustion of a given amount of isooctane? (This information is crucial to the design of nonpolluting and efficient automobile engines.) How many grams of pure gold can be obtained from a ton of low-grade gold ore? (The answer determines whether the ore deposit is worth mining.) If an industrial plant must produce a certain number of tons of sulfuric acid per week, how much elemental sulfur must arrive by rail each week?
All these questions can be answered using the concepts of the mole, molar and formula masses, and solution concentrations, along with the coefficients in the appropriate balanced chemical equation.
Stoichiometry Problems
When carrying out a reaction in either an industrial setting or a laboratory, it is easier to work with masses of substances than with the numbers of molecules or moles. The general method for converting from the mass of any reactant or product to the mass of any other reactant or product using a balanced chemical equation is outlined in and described in the following text.
Steps in Converting between Masses of Reactant and Product
1. Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass.
2. From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients).
3. Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are present in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess.
Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons).
To illustrate this procedure, consider the combustion of glucose. Glucose reacts with oxygen to produce carbon dioxide and water:
$C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \label{3.6.1}$
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 above 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:
$moles \, glucose = 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose } = 0.251 \, mol \, glucose \nonumber$
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
$moles \, CO_2 = mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber$
$= 0.251 \, mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber$
$= 1.51 \, mol \, CO_2 \nonumber$
3. Use the molar mass of CO2 (44.010 g/mol) to calculate the mass of CO2 corresponding to 1.51 mol of CO2:
$mass\, of\, CO_2 = 1.51 \, mol \, CO_2 \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.5 \, g \, CO_2 \nonumber$
These operations can be summarized as follows:
$45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose} \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose} \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.4 \, g \, CO_2 \nonumber$
Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (Remember that you should generally carry extra significant digits through a multistep calculation to the end to avoid this!) 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 $1$, the mass of one reactant that is required to consume a given mass of another reactant.
Example $1$: The US Space Shuttle
The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb).
The US space shuttle Discovery during liftoff. The large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine.
Given: reactants, products, and mass of one reactant
Asked for: mass of other reactant
Strategy:
1. Write the balanced chemical equation for the reaction.
2. Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons.
Solution:
We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons.
A We first use the information given to write a balanced chemical equation. Because we know the identity of both the reactants and the product, we can write the reaction as follows:
$H_2 (g) + O_2 (g) \rightarrow H_2O (g) \nonumber$
This equation is not balanced because there are two oxygen atoms on the left side and only one on the right. Assigning a coefficient of 2 to both H2O and H2 gives the balanced chemical equation:
$2 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g) \nonumber$
Thus 2 mol of H2 react with 1 mol of O2 to produce 2 mol of H2O.
1. B To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors:
$mass \, of \, O_2 = 1.00 \, tn \times { 2000 \, lb \over tn} \times {453.6 \, g \over lb} = 9.07 \times 10^5 \, g \, O_2 \nonumber$
Using the molar mass of O2 (32.00 g/mol, to four significant figures), we can calculate the number of moles of O2 contained in this mass of O2:
$mol \, O_2 = 9.07 \times 10^5 \, g \, O_2 \times {1 \, mol \, O_2 \over 32.00 \, g \, O_2} = 2.83 \times 10^4 \, mol \, O_2 \nonumber$
2. Now use the coefficients in the balanced chemical equation to obtain the number of moles of H2 needed to react with this number of moles of O2:
$mol \, H_2 = mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} \nonumber$
$= 2.83 \times 10^4 \, mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} = 5.66 \times 10^4 \, mol \, H_2 \nonumber$
3. The molar mass of H2 (2.016 g/mol) allows us to calculate the corresponding mass of H2:
$mass \, of \, H_2 = 5.66 \times 10^4 \, mol \, H_2 \times {2.016 \, g \, H_2 \over mol \, H_2} = 1.14 \times 10^5 \, g \, H_2 \nonumber$
Finally, convert the mass of H2 to the desired units (tons) by using the appropriate conversion factors:
$tons \, H_2 = 1.14 \times 10^5 \, g \, H_2 \times {1 \, lb \over 453.6 \, g} \times {1 \, tn \over 2000 \, lb} = 0.126 \, tn \, H_2 \nonumber$
The space shuttle had to be designed to carry 0.126 tn of H2 for each 1.00 tn of O2. Even though 2 mol of H2 are needed to react with each mole of O2, the molar mass of H2 is so much smaller than that of O2 that only a relatively small mass of H2 is needed compared to the mass of O2.
Exercise $1$: Roasting Cinnabar
Cinnabar, (or Cinnabarite) $HgS$ is the common ore of mercury. Because of its mercury content, cinnabar can be toxic to human beings; however, because of its red color, it has also been used since ancient times as a pigment.
Alchemists produced elemental mercury by roasting cinnabar ore in air:
$HgS (s) + O_2 (g) \rightarrow Hg (l) + SO_2 (g) \nonumber$
The volatility and toxicity of mercury make this a hazardous procedure, which likely shortened the life span of many alchemists. Given 100 g of cinnabar, how much elemental mercury can be produced from this reaction?
Answer
86.2 g
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. An expanded version of the flowchart for stoichiometric calculations is shown in Figure $2$. 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, as illustrated in the following examples.
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 $2$ : 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:
$Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s) \nonumber$
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 $2$ 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 $2$ : 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
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be]
Summary
Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/03%3A_Stoichiometry-_Chemical_Formulas_and_Equations/3.06%3A_Quantitative_Information_from_Balanced_Equations.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
$1 \,\text{box mix} + 2 \,\text{eggs} \rightarrow 1 \, \text{batch brownies} \label{3.7.1}$
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.
Introduction to Limiting Reactant Problems: Introduction to Limiting Reactant Problems, YouTube(opens in new window) [youtu.be]
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 ($\ce{TiCl4}$) and carbon dioxide.
$\ce{TiO2 (s) + Cl2 (g) \rightarrow TiCl4 (g) + CO2 (g)} \nonumber$
Titanium tetrachloride is then converted to metallic titanium by reaction with molten magnesium metal at high temperature:
$\ce{ TiCl4 (g) + 2 \, Mg (l) \rightarrow Ti (s) + 2 \, MgCl2 (l)} \label{3.7.2}$
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 Equation \ref{3.7.2}?
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.
Step 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: \begin{align} \text{moles} \; \ce{TiCl4} &= \dfrac{\text{mass} \, \ce{TiCl4}}{\text{molar mass} \, \ce{TiCl4}}\nonumber \[4pt] &= 1000 \, \cancel{g} \; \ce{TiCl4} \times {1 \, mol \; TiCl_4 \over 189.679 \, \cancel{g} \; \ce{TiCl4}}\nonumber \[4pt] &= 5.272 \, mol \; \ce{TiCl4} \[4pt] \text{moles }\, \ce{Mg} &= {\text{mass }\, \ce{Mg} \over \text{molar mass }\, \ce{Mg}}\nonumber \[4pt] &= 200 \, \cancel{g} \; \ce{Mg} \times {1 \; mol \, \ce{Mg} \over 24.305 \, \cancel{g} \; \ce{Mg} }\nonumber \[4pt] &= 8.23 \, \text{mol} \; \ce{Mg} \end{align}\nonumber
Step 2: There are more moles of magnesium than of titanium tetrachloride, but the ratio is only the following: ${mol \, \ce{Mg} \over mol \, \ce{TiCl4}} = {8.23 \, mol \over 5.272 \, mol } = 1.56 \nonumber$ Because the ratio of the coefficients in the balanced chemical equation is, ${ 2 \, mol \, \ce{Mg} \over 1 \, mol \, \ce{TiCl4}} = 2 \nonumber$ 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 $\ce{Mg}$, so (8.23 ÷ 2) = 4.12 mol of $\ce{TiCl4}$ are required for complete reaction. Because there are 5.272 mol of $\ce{TiCl4}$, titanium tetrachloride is present in excess. Conversely, 5.272 mol of $\ce{TiCl4}$ requires 2 × 5.272 = 10.54 mol of Mg, but there are only 8.23 mol. Therefore, magnesium is the limiting reactant.
Step 3: Because magnesium is the limiting reactant, the number of moles of magnesium determines the number of moles of titanium that can be formed: $mol \; \ce{Ti} = 8.23 \, mol \; \ce{Mg} = {1 \, mol \; \ce{Ti} \over 2 \, mol \; \ce{Mg}} = 4.12 \, mol \; \ce{Ti} \nonumber$ Thus only 4.12 mol of Ti can be formed.
Step 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): \begin{align} \text{moles }\, \ce{Ti} &= \text{mass }\, \ce{Ti} \times \text{molar mass } \, \ce{Ti}\nonumber \[6pt] &= 4.12 \, mol \; \ce{Ti} \times {47.867 \, g \; \ce{Ti} \over 1 \, mol \; \ce{Ti}}\nonumber\[6pt] &= 197 \, g \; \ce{Ti}\nonumber \end{align} \nonumber
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 $\ce{TiCl4}$ 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\nonumber \[6pt] Mg: {8.23 \, mol \, (actual) \over 2 \, mol \, (stoich)} = 4.12\nonumber$
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.
Example $1$: Fingernail Polish Remover
Ethyl acetate ($\ce{CH3CO2C2H5}$) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethanol ($\ce{C2H5OH}$) with acetic acid ($\ce{CH3CO2H}$); 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:
$\ce{ C2H5OH (l) + CH3CO2H (aq) \rightarrow CH3CO2C2H5 (aq) + H2O (l)}\nonumber$
B We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. Recall that the density of a substance is the mass divided by the volume:
$\text{density} = {\text{mass} \over \text{volume} }\nonumber$
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):
\begin{align*} \text{moles} \; \ce{C2H5OH} & = { \text{mass} \; \ce{C2H5OH} \over \text{molar mass } \; \ce{C2H5OH} }\nonumber \[6pt] & = {( \text{volume} \; \ce{C2H5OH} ) \times (\text{density} \, \ce{C2H5OH}) \over \text{molar mass } \; \ce{C2H5OH}}\nonumber \[6pt] &= 10.0 \, \cancel{ml} \; \ce{C2H5OH} \times {0.7893 \, \cancel{g} \; \ce{C2H5OH} \over 1 \, \cancel{ml} \, \ce{C2H5OH} } \times {1 \, mol \; \ce{C2H5OH} \over 46.07 \, \cancel{g}\; \ce{C2H5OH}}\nonumber \[6pt] &= 0.171 \, mol \; \ce{C2H5OH} \[6pt] \text{moles} \; \ce{CH3CO2H} &= {\text{mass} \; \ce{CH3CO2H} \over \text{molar mass} \, \ce{CH3CO2H}}\nonumber \[6pt] &= { (\text{volume} \; \ce{CH3CO2H} )\times (\text{density} \; \ce{CH3CO2H}) \over \text{molar mass} \, \ce{CH3CO2H}}\nonumber \[6pt] &= 10.0 \, \cancel{ml} \; \ce{CH3CO2H} \times {1.0492 \, \cancel{g} \; \ce{CH3CO2H} \over 1 \, \cancel{ml} \; \ce{CH3CO2H}} \times {1 \, mol \; \ce{CH3CO2H} \over 60.05 \, \cancel{g} \; \ce{CH3CO2H} } \[6pt] &= 0.175 \, mol \; \ce{CH3CO2H}\nonumber \end{align*} \nonumber
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:
\begin{align*} moles \; \text{ethyl acetate} &= mol \, \text{ethanol} \times {1 \, mol \; \text{ethyl acetate} \over 1 \, mol \; \text{ethanol}}\nonumber \[6pt] &= 0.171 \, mol \; \ce{C2H5OH} \times {1 \, mol \, \ce{CH3CO2C2H5} \over 1 \, mol \; \ce{C2H5OH}} \[6pt] &= 0.171 \, mol \; \ce{CH3CO2C2H5}\nonumber \end{align*} \nonumber
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:
\begin{align*} \text{ mass of ethyl acetate} &= mol \; \text{ethyl acetate} \times \text{molar mass}\; \text{ethyl acetate}\nonumber \[6pt] &= 0.171 \, mol \, \ce{CH3CO2C2H5} \times {88.11 \, g \, \ce{CH3CO2C2H5} \over 1 \, mol \, \ce{CH3CO2C2H5}}\nonumber \[6pt] &= 15.1 \, g \, \ce{CH3CO2C2H5}\nonumber \end{align*} \nonumber
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:
\begin{align*} \text{volume of ethyl acetate} & = 15.1 \, g \, \ce{CH3CO2C2H5} \times { 1 \, ml \; \ce{CH3CO2C2H5} \over 0.9003 \, g\; \ce{CH3CO2C2H5}} \[6pt] &= 16.8 \, ml \, \ce{CH3CO2C2H5} \end{align*} \nonumber
Exercise $1$
Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound $P_4S_{10}$. How much $P_4S_{10}$ can be prepared starting with 10.0 g of $\ce{P4}$ and 30.0 g of $S_8$?
Answer
35.9 g
Determining the Limiting Reactant and Theoretical Yield for a Reaction: Determining the Limiting Reactant and Theoretical Yield for a Reaction, YouTube(opens in new window) [youtu.be]
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 $2$: Breathalyzer reaction
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:
$\ce{3CH_3 CH_2 OH(aq)} + \underset{yellow-orange}{\ce{2Cr_2 O_7^{2 -}}}(aq) + \ce{16H^+ (aq)} \underset{\ce{H2SO4 (aq)}}{\xrightarrow{\hspace{10px} \ce{Ag^{+}}\hspace{10px}} } \ce{3CH3CO2H(aq)} + \underset{green}{\ce{4Cr^{3+}}}(aq) + \ce{11H2O(l)}\nonumber$
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 $\ce{Cr2O7^{2−}}$ 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 $\ce{Cr2O7^{2−}}$ 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 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 $\ce{Cr2O7^{2−}}$ 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 $\ce{Cr2O7^{2−}}$ 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 $\ce{Cr2O7^{2−}}$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:
$\dfrac{moles\: K_2 Cr_2 O_7} {1\: mL} = \dfrac{(0 .25\: \cancel{mg}\: K_2 Cr_2 O_7 )} {mL} \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg}} \right) \left( \dfrac{1\: mol} {294 .18\: \cancel{g}\: K_2 Cr_2 O_7} \right) = 8.5 \times 10 ^{-7}\: moles\nonumber$
B Because 1 mol of K2Cr2O7 produces 1 mol of $\ce{Cr2O7^{2−}}$ 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
$moles\: Cr_2 O_7^{2-} = \left( \dfrac{8 .5 \times 10^{-7}\: mol} {1\: \cancel{mL}} \right) ( 3 .0\: \cancel{mL} ) = 2 .6 \times 10^{-6}\: mol\: Cr_2 O_7^{2–}\nonumber$
C The balanced chemical equation tells us that 3 mol of C2H5OH is needed to consume 2 mol of $\ce{Cr2O7^{2−}}$ ion, so the total number of moles of C2H5OH required for complete reaction is
$moles\: of\: \ce{C2H5OH} = ( 2.6 \times 10 ^{-6}\: \cancel{mol\: \ce{Cr2O7^{2-}}} ) \left( \dfrac{3\: mol\: \ce{C2H5OH}} {2\: \cancel{mol\: \ce{Cr2O7^{2 -}}}} \right) = 3 .9 \times 10 ^{-6}\: mol\: \ce{C2H5OH}\nonumber$
As indicated in the strategy, this number can be converted to the mass of C2H5OH using its molar mass:
$mass\: \ce{C2H5OH} = ( 3 .9 \times 10 ^{-6}\: \cancel{mol\: \ce{C2H5OH}} ) \left( \dfrac{46 .07\: g} {\cancel{mol\: \ce{C2H5OH}}} \right) = 1 .8 \times 10 ^{-4}\: g\: \ce{C2H5OH}\nonumber$
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 $\ce{NaOH}$ 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 $1$ and $2$, 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:
$\ce{2AgNO3(aq) + K2Cr2O7(aq) \rightarrow Ag2Cr2O7(s) + 2KNO3(aq) }\nonumber$
What mass of Ag2Cr2O7 is formed when 500 mL of 0.17 M $\ce{K2Cr2O7}$ 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 8.3.2). The first step is to calculate the number of moles of each reactant in the specified volumes:
$moles\: K_2 Cr_2 O_7 = 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .17\: mol\: K_2 Cr_2 O_7} {1\: \cancel{L}} \right) = 0 .085\: mol\: K_2 Cr_2 O_7\nonumber$
$moles\: AgNO_3 = 250\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .57\: mol\: AgNO_3} {1\: \cancel{L}} \right) = 0 .14\: mol\: AgNO_3\nonumber$
B Now determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient:
\begin{align*} \ce{K2Cr2O7}: \: \dfrac{0 .085\: mol} {1\: mol} &= 0.085 \[4pt] \ce{AgNO3}: \: \dfrac{0 .14\: mol} {2\: mol} &= 0 .070 \end{align*} \nonumber
Because 0.070 < 0.085, we know that $\ce{AgNO3}$ is the limiting reactant.
C Each mole of $\ce{Ag2Cr2O7}$ formed requires 2 mol of the limiting reactant ($\ce{AgNO3}$), so we can obtain only 0.14/2 = 0.070 mol of $\ce{Ag2Cr2O7}$. Finally, convert the number of moles of $\ce{Ag2Cr2O7}$ to the corresponding mass:
$mass\: of\: Ag_2 Cr_2 O_7 = 0 .070\: \cancel{mol} \left( \dfrac{431 .72\: g} {1 \: \cancel{mol}} \right) = 30\: g \: Ag_2 Cr_2 O_7\nonumber$
The Ag+ and Cr2O72 ions form a red precipitate of solid $\ce{Ag2Cr2O7}$, while the $\ce{K^{+}}$ and $\ce{NO3^{−}}$ 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:
$\ce{2NaHCO3(aq) + H2SO4(aq) \rightarrow 2CO2(g) + Na2SO4(aq) + 2H2O(l)}\nonumber$
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: Limiting Reactant Problems Using Molarities, YouTube(opens in new window) [youtu.be]eOXTliL-gNw (opens in new window)
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:
$\text{percent yield} = {\text{actual yield } \; (g) \over \text{theoretical yield} \; (g) } \times 100\% \label{3.7.3}$
The method used to calculate the percent yield of a reaction is illustrated in Example $4$.
Example $4$: 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
$\underset {\text{p-amino benzoic acid}}{\ce{C7H7NO2}} + \underset {\text{2-diethylaminoethanol}}{\ce{C6H15NO}} \ce{->[\ce{H2SO4}]} \underset {\text{procaine}}{\ce{C13H20N2O2}} + \ce{H2O}\nonumber$
If this reaction were carried out with 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:
$mol \; \text{p-aminobenzoic acid} = 10.0 \, g \, \times \, {1 \, mol \over 137.14 \, g } = 0.0729 \, mol \; \text{p-aminbenzoic acid}\nonumber$
$mol \; \text{2-diethylaminoethanol} = 10.0 \, g \times {1 \, mol \over 117.19 \, g} = 0.0853 \, mol \; \text{2-diethylaminoethanol}\nonumber$
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.
$\text{theoretical yield of procaine} = 0.0729 \, mol \times {236.31 \, g \over 1 \, mol } = 17.2 \, g\nonumber$
C The actual yield was only 15.7 g of procaine, so the percent yield (via Equation \ref{3.7.3}) is
$\text{percent yield} = {15.7 \, g \over 17.2 \, g } \times 100 = 91.3 \%\nonumber$
(If the product were pure and dry, this yield would indicate very good lab technique!)
Exercise $4$: 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 ($\ce{PbS}$), which is easily converted to lead oxide ($\ce{PbO}$) in 100% yield by roasting in air via the following reaction:
$\ce{ 2PbS (s) + 3O2 \rightarrow 2PbO (s) + 2SO2 (g)}\nonumber$
The resulting $\ce{PbO}$ 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:
$\ce{PbO (s) + C(s) \rightarrow Pb (l) + CO (g)}\nonumber$
If 93.3 kg of $\ce{PbO}$ 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.
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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/03%3A_Stoichiometry-_Chemical_Formulas_and_Equations/3.07%3A_Limiting_Reactants.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.
3.1: Chemical Equations
Conceptual Problems
1. How does a balanced chemical equation agree with the law of definite proportions?
2. What is the difference between S8 and 8S? Use this example to explain why subscripts in a formula must not be changed.
3. What factors determine whether a chemical equation is balanced?
4. What information can be obtained from a balanced chemical equation? Does a balanced chemical equation give information about the rate of a reaction?
Numerical Problems
1. Balance each chemical equation.
a. KI(aq) + Br2(l) → KBr(aq) + I2(s)
b. MnO2(s) + HCl(aq) → MnCl2(aq) + Cl2(g) + H2O(l)
c. Na2O(s) + H2O(l) → NaOH(aq)
d. Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s)
e. SO2(g) + H2O(l) → H2SO3(aq)
f. S2Cl2(l) + NH3(l) → S4N4(s) + S8(s) + NH4Cl(s)
2. Balance each chemical equation.
a. Be(s) + O2(g) → BeO(s)
b. N2O3(g) + H2O(l) → HNO2(aq)
c. Na(s) + H2O(l) → NaOH(aq) + H2(g)
d. CaO(s) + HCl(aq) → CaCl2(aq) + H2O(l)
e. CH3NH2(g) + O2(g) → H2O(g) + CO2(g) + N2(g)
f. Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
3. Balance each chemical equation.
a. N2O5(g) → NO2(g) + O2(g)
b. NaNO3(s) → NaNO2(s) + O2(g)
c. Al(s) + NH4NO3(s) → N2(g) + H2O(l) + Al2O3(s)
d. C3H5N3O9(l) → CO2(g) + N2(g) + H2O(g) + O2(g)
e. reaction of butane with excess oxygen
f. IO2F(s) + BrF3(l) → IF5(l) + Br2(l) + O2(g)
4. Balance each chemical equation.
a. H2S(g) + O2(g) → H2O(l) + S8(s)
b. KCl(aq) + HNO3(aq) + O2(g) → KNO3(aq) + Cl2(g) + H2O(l)
c. NH3(g) + O2(g) → NO(g) + H2O(g)
d. CH4(g) + O2(g) → CO(g) + H2(g)
e. NaF(aq) + Th(NO3)4(aq) → NaNO3(aq) + ThF4(s)
f. Ca5(PO4)3F(s) + H2SO4(aq) + H2O(l) → H3PO4(aq) + CaSO4•2H2O(s) + HF(aq)
5. Balance each chemical equation.
a. NaCl(aq) + H2SO4(aq) → Na2SO4(aq) + HCl(g)
b. K(s) + H2O(l) → KOH(aq) + H2(g)
c. reaction of octane with excess oxygen
d. S8(s) + Cl2(g) → S2Cl2(l)
e. CH3OH(l) + I2(s) + P4(s) → CH3I(l) + H3PO4(aq) + H2O(l)
f. (CH3)3Al(s) + H2O(l) → CH4(g) + Al(OH)3(s)
6. Write a balanced chemical equation for each reaction.
a. Aluminum reacts with bromine.
b. Sodium reacts with chlorine.
c. Aluminum hydroxide and acetic acid react to produce aluminum acetate and water.
d. Ammonia and oxygen react to produce nitrogen monoxide and water.
e. Nitrogen and hydrogen react at elevated temperature and pressure to produce ammonia.
f. An aqueous solution of barium chloride reacts with a solution of sodium sulfate.
7. Write a balanced chemical equation for each reaction.
a. Magnesium burns in oxygen.
b. Carbon dioxide and sodium oxide react to produce sodium carbonate.
c. Aluminum reacts with hydrochloric acid.
d. An aqueous solution of silver nitrate reacts with a solution of potassium chloride.
e. Methane burns in oxygen.
f. Sodium nitrate and sulfuric acid react to produce sodium sulfate and nitric acid.
3.3: Formula Masses
Conceptual Problems
1. What is the relationship between an empirical formula and a molecular formula
2. Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition.
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems.
a. What is the formula mass of each species?
a. ammonium chloride
b. sodium cyanide
c. magnesium hydroxide
d. calcium phosphate
e. lithium carbonate
f. hydrogen sulfite ion
b. What is the molecular or formula mass of each compound?
a. potassium permanganate
b. sodium sulfate
c. hydrogen cyanide
d. potassium thiocyanate
e. ammonium oxalate
f. lithium acetate
1. What is the mass percentage of water in each hydrate?
a. H3AsO4·5H2O
b. NH4NiCl3·6H2O
c. Al(NO3)3·9H2O
2. What is the mass percentage of water in each hydrate?
a. CaSO4·2H2O
b. Fe(NO3)3·9H2O
c. (NH4)3ZrOH(CO3)3·2H2O
3. Which of the following has the greatest mass percentage of oxygen—KMnO4, K2Cr2O7, or Fe2O3?
4. Which of the following has the greatest mass percentage of oxygen—ThOCl2, MgCO3, or NO2Cl?
5. Calculate the percent composition of the element shown in bold in each compound.
a. SbBr3
b. As2I4
c. AlPO4
d. C6H10O
6. Calculate the percent composition of the element shown in bold in each compound.
a. HBrO3
b CsReO4
c. C3H8O
d. FeSO4
7. A sample of a chromium compound has a molar mass of 151.99 g/mol. Elemental analysis of the compound shows that it contains 68.43% chromium and 31.57% oxygen. What is the identity of the compound?
8. The percentages of iron and oxygen in the three most common binary compounds of iron and oxygen are given in the following table. Write the empirical formulas of these three compounds.
Compound % Iron % Oxygen Empirical Formula
1 69.9 30.1
2 77.7 22.3
3 72.4 27.6
9. What is the mass percentage of water in each hydrate?
a. LiCl·H2O
b. MgSO4·7H2O
c. Sr(NO3)2·4H2O
10. What is the mass percentage of water in each hydrate?
a. CaHPO4·2H2O
b. FeCl2·4H2O
c. Mg(NO3)2·4H2O
11. Two hydrates were weighed, heated to drive off the waters of hydration, and then cooled. The residues were then reweighed. Based on the following results, what are the formulas of the hydrates?
Compound Initial Mass (g) Mass after Cooling (g)
NiSO4·xH2O 2.08 1.22
CoCl2·xH2O 1.62 0.88
12. Which contains the greatest mass percentage of sulfur—FeS2, Na2S2O4, or Na2S?
13. Given equal masses of each, which contains the greatest mass percentage of sulfur—NaHSO4 or K2SO4?
14. Calculate the mass percentage of oxygen in each polyatomic ion.
a. bicarbonate
b. chromate
c. acetate
d. sulfite
15. Calculate the mass percentage of oxygen in each polyatomic ion.
a. oxalate
b. nitrite
c. dihydrogen phosphate
d. thiocyanate
16. The empirical formula of garnet, a gemstone, is Fe3Al2Si3O12. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula?
17. A compound has the empirical formula C2H4O, and its formula mass is 88 g. What is its molecular formula?
18. Mirex is an insecticide that contains 22.01% carbon and 77.99% chlorine. It has a molecular mass of 545.59 g. What is its empirical formula? What is its molecular formula?
19. How many moles of CO2 and H2O will be produced by combustion analysis of 0.010 mol of styrene?
20. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0080 mol of aniline?
21. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0074 mol of aspartame?
22. How many moles of CO2, H2O, N2, and SO2 will be produced by combustion analysis of 0.0060 mol of penicillin G?
23. Combustion of a 34.8 mg sample of benzaldehyde, which contains only carbon, hydrogen, and oxygen, produced 101 mg of CO2 and 17.7 mg of H2O.
a. What was the mass of carbon and hydrogen in the sample?
b. Assuming that the original sample contained only carbon, hydrogen, and oxygen, what was the mass of oxygen in the sample?
c. What was the mass percentage of oxygen in the sample?
d. What is the empirical formula of benzaldehyde?
e. The molar mass of benzaldehyde is 106.12 g/mol. What is its molecular formula?
24. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg of H2O.
a. What is the mass of oxygen in the sample?
b. What is the mass percentage of oxygen in the sample?
c. What is the empirical formula of salicylic acid?
d. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula?
25. Given equal masses of the following acids, which contains the greatest amount of hydrogen that can dissociate to form H+—nitric acid, hydroiodic acid, hydrocyanic acid, or chloric acid?
26. Calculate the formula mass or the molecular mass of each compound.
a. heptanoic acid (a seven-carbon carboxylic acid)
b. 2-propanol (a three-carbon alcohol)
c. KMnO4
d. tetraethyllead
e. sulfurous acid
f. ethylbenzene (an eight-carbon aromatic hydrocarbon)
27. Calculate the formula mass or the molecular mass of each compound.
a. MoCl5
b. B2O3
c. bromobenzene
d. cyclohexene
e. phosphoric acid
f. ethylamine
28. Given equal masses of butane, cyclobutane, and propene, which contains the greatest mass of carbon?
29. Given equal masses of urea [(NH2)2CO] and ammonium sulfate, which contains the most nitrogen for use as a fertilizer?
Conceptual Answers
1) What is the relationship between an empirical formula and a molecular formula
• An empirical formula refers to the simplest ratio of elements that is obtained from a chemical formula while a molecular formula is calculated to show the actual formula of a molecular compound.
2) Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition.
Numerical Answers
a. What is the formula mass of each species?
a. 53.49146 amu
b. 49.0072 amu
c. 58.3197 amu
d. 310.177 amu
e. 73.891 amu
f. 81.07 amu
b. What is the molecular or formula mass of each compound?
a.158.034 amu
b. 142.04 amu
c. 27.0253 amu
d. 97.181 amu
e. 124.1 amu
f. 65.99 amu
1. To two decimal places, the percentages are:
a. 5.97%
b. 37.12%
c. 43.22%
2. To two decimal places, the percentages of water are:
a. 20.93%
b. 40.13%
c. 9.52%
3. % oxygen: KMnO4, 40.50%; K2Cr2O7, 38.07%; Fe2O3, 30.06%
4. % oxygen: ThOCl2, 5.02%; MgCO3, 56.93%; NO2Cl, 39.28%
5. To two decimal places, the percentages are:
a. 66.32% Br
b. 22.79% As
c. 25.40% P
d. 73.43% C
6.
a. 61.98% Br
b. 34.69% Cs
c. 59.96% C
d. 21.11% S
7. Cr2O3.
8. Empirical Formulas
1. Fe2O3
2. Fe4O4
3. Fe6O8
9. To two decimal places, the percentages are:
a. 29.82%
b. 51.16%
c. 25.40%
10. What is the mass percentage of water in each hydrate?
a. 20.94%
b. 36.25%
c. 32.70%
11. NiSO4 · 6H2O and CoCl2 · 6H2O
12. FeS2
13. NaHSO4
14. Calculate the mass percentage of oxygen in each polyatomic ion.
a. 78.66%
b. 55.17%
c. 54.19%
d. 59.95%
15.
a. 72.71%
b. 69.55%
c. 65.99%
d. 0%
16. The empirical formula of garnet, a gemstone, is Fe3Al2Si3O12. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula?
No, the calculated mass percentage of silicon in garnet is 16.93%
17. C4H8O2
18.
Empirical Formula: C10Cl12
Molecular Formula: C10Cl12
19. How many moles of CO2 and H2O will be produced by combustion analysis of 0.010 mol of styrene?
Moles of CO2: 0.08 mol CO2
Moles of H2O: 0.04 mol H2O
20. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0080 mol of aniline?
Mole of CO2: 0.048 mol CO2
Mole of H2O: 0.028 mol H2O
Mole of N2: 0.004 mol N2
21. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0074 mol of aspartame?
Mole of CO2: 0.104 mol CO2
Mole of H2O: 0.666 mol H2O
Mole of N2: 0.0074 mol N2
22. How many moles of CO2, H2O, N2, and SO2 will be produced by combustion analysis of 0.0060 mol of penicillin G?
Mole of CO2: 0.096 mol CO2
Mole of H2O: 0.054 mol H2O
Mole of N2: 0.060 mol N2
Mole of SO2: 0.060 mol SO2
23.
a. 27.6 mg C and 1.98 mg H
b. 5.2 mg O
c. 15%
d. C7H6O
e. C7H6O
24. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg of H2O.
a. What is the mass of oxygen in the sample?
70.4mg
b. What is the mass percentage of oxygen in the sample?
61.70%
c. What is the empirical formula of salicylic acid?
C7H6O3
d. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula?
C7H6O3
25. hydrocyanic acid, HCN
26. Calculate the formula mass or the molecular mass of each compound.
a. 130.1849 amu
b. 60.1 amu
c. 158.034 amu
d. 323.4 amu
e. 82.07 amu
f. 106.17 amu
27. To two decimal places, the values are:
a. 273.23 amu
b. 69.62 amu
c. 157.01 amu
d. 82.14 amu
e. 98.00 amu
f. 45.08 amu
28. Cyclobutene
29. Urea
3.4: Avogadro's Number and the Mole
Conceptual Problems
Please be sure you are familiar with the topics discussed in Essential Skills 2 before proceeding to the Conceptual Problems.
1. Describe the relationship between an atomic mass unit and a gram.
2. Is it correct to say that ethanol has a formula mass of 46? Why or why not?
3. If 2 mol of sodium reacts completely with 1 mol of chlorine to produce sodium chloride, does this mean that 2 g of sodium reacts completely with 1 g of chlorine to give the same product? Explain your answer.
4. Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi3O8), a mineral used in the manufacture of porcelain.
5. Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass.
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 2 before proceeding to the Numerical Problems.
1. Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass.
2. Calculate the molecular mass or formula mass of each compound.
1. KCl (potassium chloride)
2. NaCN (sodium cyanide)
3. H2S (hydrogen sulfide)
4. NaN3 (sodium azide)
5. H2CO3 (carbonic acid)
6. K2O (potassium oxide)
7. Al(NO3)3 (aluminum nitrate)
8. Cu(ClO4)2 [copper(II) perchlorate]
3. Calculate the molecular mass or formula mass of each compound.
1. V2O4 (vanadium(IV) oxide)
2. CaSiO3 (calcium silicate)
3. BiOCl (bismuth oxychloride)
4. CH3COOH (acetic acid)
5. Ag2SO4 (silver sulfate)
6. Na2CO3 (sodium carbonate)
7. (CH3)2CHOH (isopropyl alcohol)
4. Calculate the molar mass of each compound.
a.
b.
c.
d.
e.
5. Calculate the molar mass of each compound.
a.
b.
c.
d.
6. For each compound, write the condensed formula, name the compound, and give its molar mass.
a.
b.
7. For each compound, write the condensed formula, name the compound, and give its molar mass.
a.
b.
8. Calculate the number of moles in 5.00 × 102 g of each substance. How many molecules or formula units are present in each sample?
a. CaO (lime)
b. CaCO3(chalk)
c. C12H22O11 [sucrose (cane sugar)]
d. NaOCl (bleach)
e. CO2 (dry ice)
9. Calculate the mass in grams of each sample.
a. 0.520 mol of N2O4
b. 1.63 mol of C6H4Br2
c. 4.62 mol of (NH4)2SO3
10. Give the number of molecules or formula units in each sample.
a. 1.30 × 10−2 mol of SCl2
b. 1.03 mol of N2O5
c. 0.265 mol of Ag2Cr2O7
11. Give the number of moles in each sample.
a. 9.58 × 1026 molecules of Cl2
b. 3.62 × 1027 formula units of KCl
c. 6.94 × 1028 formula units of Fe(OH)2
12. Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I2)?
13. What is the total number of atoms in each sample?
a. 0.431 mol of Li
b. 2.783 mol of methanol (CH3OH)
c. 0.0361 mol of CoCO3
d. 1.002 mol of SeBr2O
14. What is the total number of atoms in each sample?
a. 0.980 mol of Na
b. 2.35 mol of O2
c. 1.83 mol of Ag2S
d. 1.23 mol of propane (C3H8)
15. What is the total number of atoms in each sample?
a. 2.48 g of HBr
b. 4.77 g of CS2
c. 1.89 g of NaOH
d. 1.46 g of SrC2O4
16. Decide whether each statement is true or false and explain your reasoning.
1. There are more molecules in 0.5 mol of Cl2than in 0.5 mol of H2.
2. One mole of H2 has 6.022 × 1023 hydrogen atoms.
3. The molecular mass of H2O is 18.0 amu.
4. The formula mass of benzene is 78 amu.
17. Complete the following table.
Substance Mass (g) Number of Moles Number of Molecules or Formula Units Number of Atoms or Ions
MgCl2 37.62 a. b. c.
AgNO3 d. 2.84 e. f.
BH4Cl g. h. 8.93 × 1025 i.
K2S j k. l. 7.69 × 1026
H2SO4 m. 1.29 n. o.
C6H14 11.84 p. q. r.
HClO3 s. t. 2.45 × 1026 u.
18. Give the formula mass or the molecular mass of each substance.
1. $PbClF$
2. $Cu_2P_2O_7$
3. $BiONO_3$
4. $Tl_2SeO_4$
19. Give the formula mass or the molecular mass of each substance.
1. $MoCl_5$
2. $B_2O_3$
3. $UO_2CO_3$
4. $NH_4UO_2AsO_4$
Conceptual Answers
1. While both are units of mass, a gram is Avogadro’s number of atomic mass units so you would multiply the number of amu by 6.022x10^23 to find total number of grams
2. The correct way to state formula mass of ethanol is to show the units of mass which is amu.
3. No because moles and weight operate on different set of standards meaning that they’re not equal to each other. This means that moles of different compounds contain different weights. For example, 2 moles of Na = 2 x 22.989 g = 45.98g while 1 mole of Cl = 1 x 35.453 g = 35.453 g Cl. This makes the sodium react completely with chlorine. 2g of sodium would react with = (35.453/45.978) x 2 = 1.542 g Cl
4. Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi3O8), a mineral used in the manufacture of porcelain
1. .
5. Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass.
6. The information required to determine the mass of the solute would be the molarity of the solution because once that is achieved, volume of the solution and molar mass of the solute can be used to calculate the total mass. A derivatization that achieves this goes as: Molarity = moles of solute / volume of solution in liter -> Moles = molarity x volume in liter -> Mass= moles x molar mass.
Numerical Answers
1. Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass.
2. Calculate the molecular mass or formula mass of each compound.
1. 74.55 amu
2. 49.01 amu
3. 34.08 amu
4. 65.01 amu
5. 62.02 amu
6. 94.20 amu
7. 213.00 amu
8. 262.45 amu
3. Calculate the molecular mass or formula mass of each compound.
1. 165.88 amu
2. 116.16 amu
3. 260.43 amu
4. 60.05 amu
5. 311.80 amu
6. 105.99 amu
7. 60.10 amu
4. Calculate the molar mass of each compound.
a. 153.82 g/mol
b. 80.06 g/mol
c. 92.01 g/mol
d. 70.13 g/mol
e. 74.12 g/mol
5. Calculate the molar mass of each compound.
a. 92.45 g/mol
b. 135.04 g/mol
c. 44.01 g/mol
d. 40.06 g/mol
6. For each compound, write the condensed formula, name the compound, and give its molar mass.
a. C5H10O2, Valeric Acid, 102.13 g/mol
b. H3PO3, Phosphorous acid, 82 g/mol
7. For each compound, write the condensed formula, name the compound, and give its molar mass.
a. C2H5NH2, Ethylamine, 45.08 g/mol
b. HIO3, Iodic acid, 175.91 g/mol
8. Calculate the number of moles in 5.00 × 102 g of each substance. How many molecules or formula units are present in each sample?
a. 5.37 × 1024 mol
b. 3.01 × 1024 mol
c. 8.80 × 1023 mol
d. 4.04 × 1024 mol
e. 6.84 × 1024 mol
9. Calculate the mass in grams of each sample.
a. 47.85 grams
b. 384.52 grams
c. 536.57 grams
10. Give the number of molecules or formula units in each sample.
a. 7.83x1021 molecules
b. 6.20x1023 molecules
c. 1.60x1023 molecules
11. Give the number of moles in each sample.
a. 1590.8 moles
b. 6011.3 moles
c. 115244.1 moles
12. Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I2)?
2.61 x1022 molecules
13. What is the total number of atoms in each sample?
a. 2.60x1023 atoms
b. 1.01x1025 atoms
c. 1.09x1023 atoms
d. 2.41x1024 atoms
14. What is the total number of atoms in each sample?
a. 5.9x1023 atoms
b. 2.8x1024 atoms
c. 3.31x1024 atoms
d. 8.15x1024 atoms
15. What is the total number of atoms in each sample?
a. 3.69x1022 atoms
b. 1.13x1023 atoms
c. 8.54x1022 atoms
d. 3.50x1023 atoms
16.
a. False, the number of molecules in 0.5 mol Cl2 are the same amount of molecules in H2
b. False, the number of molecules in H2 is 2 x (6.022 x10^23) H atoms
c. True, 2 H (1.01 amu) + 1 O (16.01) = 18.0 amu
d. True, C6H6 -> 12(6) + 1(6) = 78 amu
17. Complete the following table
a. 0.39
b. 2.36x10^23
c. 7.08x10^23
d. 482.8
e. 1.71x10^24
f. 8.55x10^24
g. 7932.7
h. 148.3
i. 5.36x10^26
j. 46938.5
k. 425.7
l. 1276.98
m. 126.5
n. 7.77x10^23
o. 5.44x10^24
p. 0.14
q. 8.27x10^22
r. 1.65x10^24
s. 34358
t. 406.8
u. 1.23x10^2
18. Give the formula mass or the molecular mass of each substance.
1. 261.67 amu
2. 301.04 amu
3. 286.98 amu
4. 551.73 amu
19. Give the formula mass or the molecular mass of each substance.
1. 273.21 amu
2. 69.62 amu
3. 330.04 amu
4. 426.99 amu
3.6: Quantitative Information from Balanced Equations
Conceptual Problems
1. What information is required to determine the mass of solute in a solution if you know the molar concentration of the solution?
Numerical Problems
1. Refer to the Breathalyzer test described in Example 8. How much ethanol must be present in 89.5 mL of a person’s breath to consume all the potassium dichromate in a Breathalyzer ampul containing 3.0 mL of a 0.40 mg/mL solution of potassium dichromate?
2. Phosphoric acid and magnesium hydroxide react to produce magnesium phosphate and water. If 45.00 mL of 1.50 M phosphoric acid are used in the reaction, how many grams of magnesium hydroxide are needed for the reaction to go to completion?
3. Barium chloride and sodium sulfate react to produce sodium chloride and barium sulfate. If 50.00 mL of 2.55 M barium chloride is used in the reaction, how many grams of sodium sulfate are needed for the reaction to go to completion?
4. How many grams of sodium phosphate are obtained in solution from the reaction of 75.00 mL of 2.80 M sodium carbonate with a stoichiometric amount of phosphoric acid? The second product is water; what is the third product? How many grams of the third product are obtained?
5. How many grams of ammonium bromide are produced from the reaction of 50.00 mL of 2.08 M iron(II) bromide with a stoichiometric amount of ammonium sulfide? What is the second product? How many grams of the second product are produced?
Conceptual Answers
1. The information required to determine the mass of the solution with a known molar concentration are the molar mass of the solute and the volume of the solute.
Numerical Answers
1. Refer to the Breathalyzer test described in Example 8. How much ethanol must be present in 89.5 mL of a person’s breath to consume all the potassium dichromate in a Breathalyzer ampul containing 3.0 mL of a 0.40 mg/mL solution of potassium dichromate?
2. 5.905 g Mg(OH)2
3. 18.105 g Na2SO4
4. Third Product: CO2, Grams of Third Product: 9.24 g, Grams of Sodium Phosphate: 22.26g
5. Grams Produced: 20.37 grams NH4Br2, Second Product: Iron (II) Sulfide, 9.15 grams FeS
3.7: LIMITING REACTANTS
Conceptual Problems
1. Engineers use conservation of mass, called a “mass balance,” to determine the amount of product that can be obtained from a chemical reaction. Mass balance assumes that the total mass of reactants is equal to the total mass of products. Is this a chemically valid practice? Explain your answer.
2. Given the equation $2H_{2\;(g)} + O_{2\;(g)} \rightarrow 2H+2O_{(g)},$ is it correct to say that 10 g of hydrogen will react with 10 g of oxygen to produce 20 g of water vapor?
3. What does it mean to say that a reaction is stoichiometric?
4. When sulfur is burned in air to produce sulfur dioxide, what is the limiting reactant? Explain your answer.
5. Is it possible for the percent yield to be greater than the theoretical yield? Justify your answer.
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems.
12. Write a balanced chemical equation for each reaction and then determine which reactant is in excess.
a. 2.46 g barium(s) plus 3.89 g bromine(l) in water to give barium bromide
b. 1.44 g bromine(l) plus 2.42 g potassium iodide(s) in water to give potassium bromide and iodine
c. 1.852 g of Zn metal plus 3.62 g of sulfuric acid in water to give zinc sulfate and hydrogen gas
d. 0.147 g of iron metal reacts with 0.924 g of silver acetate in water to give iron(II) acetate and silver metal
e. 3.142 g of ammonium phosphate reacts with 1.648 g of barium hydroxide in water to give ammonium hydroxide and barium phosphate
13. Under the proper conditions, ammonia and oxygen will react to form dinitrogen monoxide (nitrous oxide, also called laughing gas) and water. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants.
a. 24.6 g of ammonia and 21.4 g of oxygen
b. 3.8 mol of ammonia and 84.2 g of oxygen
c. 3.6 × 1024 molecules of ammonia and 318 g of oxygen
d. 2.1 mol of ammonia and 36.4 g of oxygen
14. When a piece of zinc metal is placed in aqueous hydrochloric acid, zinc chloride is produced, and hydrogen gas is evolved. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants.
a. 12.5 g of HCl and 7.3 g of Zn
b. 6.2 mol of HCl and 100 g of Zn
c. 2.1 × 1023 molecules of Zn and 26.0 g of HCl
d. 3.1 mol of Zn and 97.4 g of HCl
15. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced.
a. NaI(aq) + Cl2(g) → NaCl(aq) + I2(s); 1.0 mol of NaCl
b. NaCl(aq) + H2SO4(aq) → HCl(g) + Na2SO4(aq); 0.50 mol of HCl
c. NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq); 1.5 mol of HNO3
16. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced.
a. AgNO3(aq) + CaCl2(s) → AgCl(s) + Ca(NO3)2(aq); 1.25 mol of AgCl
b. Pb(s) + PbO2(s) + H2SO4(aq) → PbSO4(s) + H2O(l); 3.8 g of PbSO4
c. H3PO4(aq) + MgCO3(s) → Mg3(PO4)2(s) + CO2(g) + H2O(l); 6.41 g of Mg3(PO4)2
17. Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.)
a. KClO3(s) $\underrightarrow {\Delta}$ KCl(s)+O2(g);2.14 g of KClO3 produces 0.67 g of O2
b. Cu(s) + H2SO4(aq) → CuSO4(aq) + SO2(g) + H2O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide
c. AgC2H3O2(aq) + Na3PO4(aq) → Ag3PO4(s) + NaC2H3O2(aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate
18. Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction?
19. A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction?
20. Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting.
A + 3B → 2C
21. Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol.
a. Write a balanced chemical equation for this reaction.
b. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide.
22. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction:
NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq)
a. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO2?
b. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide.
23. The reaction of propane gas (CH3CH2CH3) with chlorine gas (Cl2) produces two monochloride products: CH3CH2CH2Cl and CH3CHClCH3. The first is obtained in a 43% yield and the second in a 57% yield.
a. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion?
b. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane?
c. Use the actual percent yield to calculate how many grams of each product would actually be obtained.
24. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions:
$\ce{2PaI_5 (s) ->[\Delta] 2Pa (s) + 5I_2 (s)} \nonumber$
a. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized?
b. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction?
c. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation?
25. Aniline (C6H5NH2) can be produced from chlorobenzene (C6H5Cl) via the following reaction:
C6H5Cl(l) + 2NH3(g) → C6H5NH2(l) + NH4Cl(s)
Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia.
a. Which is the limiting reactant?
b. Which reactant is present in excess?
c. What is the theoretical yield of ammonium chloride in grams?
d. If 4.78 g of NH4Cl was recovered, what was the percent yield?
e. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess.
26. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following:
a. 9.36 × 1024 formula units of NaCl
b. 8.5 × 104 mol of Br2
c. 3.7 × 108 g of NaCl
Conceptual Answers
1. It is not a chemically valid practice because the law of conservation of mass applies to closed/isolated systems only, so in this case of an open system converting mass into energy and allowing it to escape the system wouldn’t be valid.
2. No it is not correct because in order to examine the molecular weight you need to take log of both molecules to see what there true weight is. In this case, log of O2 = 10g/(32 g/mol) = 0.3125 g/mol while log of H2 = 10g/(2 g/mol) = 5 g/mol. With that when hydrogen and oxygen react against each other ((2 x (0.3125 g/L)) x (18 g/mol H2O)) = 11.25 g of water vapor
3. When a reaction is stoichiometric it means that the none of the reactants remain after being introduced leaving no excess or total leaving quantity.
4. From a reaction of S + O2 -> SO2 we’re able to determine that the limiting reactant would be sulfur because the amount of O2 in the air is higher than the amount of sulfur that is burned in the reaction.
5. Yes it is possible, given the equation of: percent yield = (actual yield)/(Theoretical Yield) its entirely possible to have the percent yield become greater than the theoretical yield when the product of the reaction has impurities that increase the mass compared if the product was pure.
Numerical Answers
12. Write a balanced chemical equation for each reaction and then determine which reactant is in excess.
a. Ba(s) + Br2(l) = BaBr2(aq)
Reactant in excess: Br2
b. Br2(l) + 2KI(s) = 2KBr + I2
Reactant in Excess: Br2
c. Zn + H2SO4 = ZnSO4 + H2
Reactant in excess: H2SO4
d. Fe + 2 AgC2H3O2 = Fe(C2H3O2)2 + 2 Ag
Reactant in excess: AgC2H3O2
e. 2 (NH4)3PO4 + 3 Ba(OH)2 = 6 NH4OH + Ba3(PO4)2
Reactant in excess: Ba(OH)2
13. The balanced chemical equation for this reaction is
2NH3 + 2O2 → N2O + 3H2O
a. NH3
b. NH3
c. O2
d. NH3
14. a. Excess: HCl
b. Excess: HCl
c. Excess: HCl
d. Excess: Zn
15.
a. 150 g NaI and 35 g Cl2
b. 29 g NaCl and 25 g H2SO4
c. 140 g NO2 and 27 g H2O
16.
a. 2 AgNO3(aq) + CaCl2(s) = 2 AgCl(s) + Ca(NO3)2(aq); Mass of Reactant: 212.34g AgNO3 and 69.4 g CaCl2
b. Pb(s) + PbO2(s) + 2 H2SO4(aq) = 2 PbSO4(s) + 2 H2O(l); Mass of Reactant: 1.3g Pb, 1.5g PbO2, 1.2g H2SO4
c. 2 H3PO4(aq) + 3 MgCO3(s) = Mg3(PO4)2(s) + 3 CO2(g) + 3 H2O(l); Mass of Reactant: 6.17 g of MgCO3, 4.78g of H3PO4
17.
a. 80%
b. 30%
c. 35.7%
18. 81%
19. 45%.
20. (x/molecular mass of B) x (2/3) x (molecular mass of C)
21.
a. CO + 2H2 → CH3OH
b. 58.28%
22. 2 NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq)
a. 11.9 g Cl2
b. (0.7458g= Grams of ClO2 = (x/(90.44 g/mol)) x (2/2) x (67.45 g ClO2/ 1 mol ClO2
23.
a. 2.24 g Cl2
b. 4.95 g
c. 2.13 g CH3CH2CH2Cl plus 2.82 g CH3CHClCH3
24. a. 4.14mg Pa
b. 82.1%
c. 0.0162 g PaI5
25.
a. chlorobenzene
b. ammonia
c. 8.74 g ammonium chloride.
d. 55%
e. $Theoretical \, yield \, (NH_4Cl) = {mass \, of \, chlorobenzene \, (g) \times 0.92 \times \times 53.49 \, g/mol \over 112.55 \, g/mol}$
26.
a. 6619.19g Cl2
b. 6.77 x 106 g Cl2
c. 2.52 x 108 g Cl2
Conceptual Problems II
1. Engineers use conservation of mass, called a “mass balance,” to determine the amount of product that can be obtained from a chemical reaction. Mass balance assumes that the total mass of reactants is equal to the total mass of products. Is this a chemically valid practice? Explain your answer.
2. Given the equation $2H_{2\;(g)} + O_{2\;(g)} \rightarrow 2H+2O_{(g)},$ is it correct to say that 10 g of hydrogen will react with 10 g of oxygen to produce 20 g of water vapor?
3. What does it mean to say that a reaction is stoichiometric?
4. When sulfur is burned in air to produce sulfur dioxide, what is the limiting reactant? Explain your answer.
5. Is it possible for the percent yield to be greater than the theoretical yield? Justify your answer.
Numerical Problems II
Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems.
17. Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.)
a. KClO3(s) $\underrightarrow {\Delta}$ KCl(s)+O2(g);2.14 g of KClO3 produces 0.67 g of O2
b. Cu(s) + H2SO4(aq) → CuSO4(aq) + SO2(g) + H2O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide
c. AgC2H3O2(aq) + Na3PO4(aq) → Ag3PO4(s) + NaC2H3O2(aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate
18. Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction?
19. A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction?
20. Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting.
A + 3B → 2C
21. Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol.
a. Write a balanced chemical equation for this reaction.
b. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide.
22. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction:
NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq)
a. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO2?
b. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide.
23. The reaction of propane gas (CH3CH2CH3) with chlorine gas (Cl2) produces two monochloride products: CH3CH2CH2Cl and CH3CHClCH3. The first is obtained in a 43% yield and the second in a 57% yield.
a. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion?
b. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane?
c. Use the actual percent yield to calculate how many grams of each product would actually be obtained.
24. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions:
$2PaI_5 (s) \underrightarrow {\Delta} 2Pa (s) + 5I_2 (s)$
a. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized?
b. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction?
c. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation?
25. Aniline (C6H5NH2) can be produced from chlorobenzene (C6H5Cl) via the following reaction:
C6H5Cl(l) + 2NH3(g) → C6H5NH2(l) + NH4Cl(s)
Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia.
a. Which is the limiting reactant?
b. Which reactant is present in excess?
c. What is the theoretical yield of ammonium chloride in grams?
d. If 4.78 g of NH4Cl was recovered, what was the percent yield?
e. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess.
26. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following:
a. 9.36 × 1024 formula units of NaCl
b. 8.5 × 104 mol of Br2
c. 3.7 × 108 g of NaCl
Numerical Answers II
17.
a. 80%
b. 30%
c. 35.7%
19. 45%.
21.
a. CO + 2H2 → CH3OH
b. 58.28%
23.
a. 2.24 g Cl2
b. 4.95 g
c. 2.13 g CH3CH2CH2Cl plus 2.82 g CH3CHClCH3
25.
a. chlorobenzene
b. ammonia
c. 8.74 g ammonium chloride.
d. 55%
e. $Theoretical \, yield \, (NH_4Cl) = {mass \, of \, chlorobenzene \, (g) \times 0.92 \times \times 53.49 \, g/mol \over 112.55 \, g/mol}$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/03%3A_Stoichiometry-_Chemical_Formulas_and_Equations/3.E%3A_Stoichiometry_%28Exercises%29.txt |
• atoms are neither created nor destroyed during any chemical reaction
• stoichiometry – quantitative nature of chemical formulas and chemical reactions
3.1: Chemical Equations
A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products. In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right.
• chemical equations – the way chemical reactions are represented
• reactants – starting substances
• products – substances produced from a reaction
• balanced equation – equation with equal atoms on both sides of the equation
• subscripts should never be changed in balancing an equation
• coefficients changes only the amount and not identity of the substance
3.2: Some Simple Patterns of Chemical Reactivity
By recognizing general patterns of chemical reactivity, you will be able to successfully predict the products formed by a given combination of reactants We can often predict a reaction if we have seen a similar reaction before.
3.2.1 Using the Periodic Table
• periodic table can be used to determine reactivity of substances
• all alkali metals react with water to form their hydroxide compounds and hydrogen
3.2.2 Combustion in Air
• rapid reaction that produces a flame
• most combustion reactions in air involve oxygen
• hydrocarbons and related compounds produce CO2 and H2O during combustion
3.2.3 Combination and Decomposition Reactions
• combination reactions two or more substances react to form one product
• decomposition reaction one substance produces two or more substances
3.3: Formula Masses
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.
3.3.1 The Atomic Mass Scale
• atomic mass unit (amu) – unit in measuring mass of atoms
• 1 amu = 1.66054*10-24g and 1 amu = 6.02214*1024amu
3.3.2 Average Atomic Masses
• atomic weight – average atomic mass
3.3.3 Formula and Molecular Weights
• formula weight – sum of the atomic weights of each atom in its chemical formula
• molecular weight – same as formula weight
3.3.4 Percentage composition from Formulas
• ((atoms of element)(AW)/(FW of compound) * 100
3.3.5 The Mole
• avogadro’s number – 6.02*1023 atoms
• molar mass – numerically equal to its formula weight
• grams <use molar mass> moles <use avogadro’s number> molecules
3.5: Empirical Formulas from Analysis
Molecular formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's molecular formula cannot be reduced any more, then the empirical formula is the same as the molecular formula. Combustion analysis can determine the empirical formula of a compound, but cannot determine the molecular formula (other techniques can though).
• empirical formula gives relative number of atoms in each element
• mass % elements >>> assume 100g sample >>> grams of each element >>> use atomic weights >>> moles of each element >>> calculate mole ratio >>> empirical formula
• "percent to mass, mass to mol, divide by small, multiply ‘til whole/"
3.5.1 Molecular Formula from Empirical Formula
• the subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula
3.6: Quantitative Information from Balanced Equations
Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. 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.
• 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 numbers of moles
• stoichiometrically equivalent quantities
• grams reactant >> moles reactant >> moles product >> grams product
• grams of substance A >> use molar mass of A >> moles of substance A >> use coefficients of A and B from balanced equation >> moles of substance B >> use molar mass of B >> grams of substance B
3.7: Limiting Reactants
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 reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first is the limiting reagent.
• limiting reactant – limits the amount of product formed
3.7.1 Theoretical Yields
• theoretical yield – the amount of product that is calculated to form
• actual yield – the amount of product actually formed
$\text{percent yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/03%3A_Stoichiometry-_Chemical_Formulas_and_Equations/3.S%3A_Stoichiometry_%28Summary%29.txt |
A solution is a homogeneous mixture in which substances present in lesser amounts, called solutes, are dispersed uniformly throughout the substance in the greater amount, the solvent. An aqueous solution is a solution in which the solvent is water, whereas in a nonaqueous solution, the solvent is a substance other than water. Familiar examples of nonaqueous solvents are ethyl acetate, used in nail polish removers, and turpentine, used to clean paint brushes. In this chapter, we focus on reactions that occur in aqueous solution.
There are many reasons for carrying out reactions in solution. For a chemical reaction to occur, individual atoms, molecules, or ions must collide, and collisions between two solids, which are not dispersed at the atomic, molecular, or ionic level, do not occur at a significant rate. In addition, when the amount of a substance required for a reaction is so small that it cannot be weighed accurately, using a solution of that substance, in which the solute is dispersed in a much larger mass of solvent, enables chemists to measure its quantity with great precision. Chemists can also more effectively control the amount of heat consumed or produced in a reaction when the reaction occurs in solution, and sometimes the nature of the reaction itself can be controlled by the choice of solvent.
This chapter introduces techniques for preparing and analyzing aqueous solutions, for balancing equations that describe reactions in solution, and for solving problems using solution stoichiometry. By the time you complete this chapter, you will know enough about aqueous solutions to explain what causes acid rain, why acid rain is harmful, and how a Breathalyzer measures alcohol levels. You will also understand the chemistry of photographic development, be able to explain why rhubarb leaves are toxic, and learn about a possible chemical reason for the decline and fall of the Roman Empire.
• 4.1: General Properties of Aqueous Solutions
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.
• 4.2: Precipitation Reactions
A complete ionic equation consists of the net ionic equation and spectator ions. Predicting the solubility of ionic compounds gives insight into feasibility of reactions occuring. The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances in their undissociated forms; the complete ionic equation shows substances in the form in which they actually exist in solution; and the net ionic equation omits all spectator ions.
• 4.3: Acid-Base Reactions
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 and a base is a substance that can accept a proton. Acids also differ in their tendency to donate a proton, a measure of their acid strength. The acidity or basicity of an aqueous solution is described quantitatively using the pH scale.
• 4.4: Oxidation-Reduction Reactions
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, in which the overall reaction is separated into an oxidation equation and a reduction equation. The outcome of these reactions can be predicted using the activity series.
• 4.5: Concentration of Solutions
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. 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.
• 4.6: Solution Stoichiometry and Chemical Analysis
The topic solution stoichiometry deals with quantities in chemical reactions taking place in solutions. Quantitative analysis of an unknown solution can be achieved using titration methods. In a titration, 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.
• 4.E: Reactions in Aqueous Solution (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 4.S: Reactions in Aqueous Solution (Summary)
This is the summary Module for the chapter "Reactions in Aqueous Solution" in the Brown et al. General Chemistry Textmap.
04: Reactions in Aqueous Solution
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
$\ce{NaCl(s) ->[\ce{H_2O(l)}] Na^{+} (aq) + Cl^{-} (aq)} \label{4.1.1}$
where (aq) indicates that Na+ and Cl are hydrated ions.
Rule of Thumb
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.
Rule of Thumb
Solutions that contain ions conduct electricity.
An electrolyte is any compound that can form ions when dissolved in water. Electrolytes may be strong or weak. 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 ($\ce{BaCl_2}$) and sodium hydroxide ($\ce{NaOH}$), which are both strong electrolytes and dissociate as follows:
$\ce{BaCl_2(s) ->[\ce{H_2O(l)}] Ba^{2+} (aq) + 2Cl^{-} (aq)} \label{4.1.2}$
$\ce{ NaOH(s) ->[\ce{H_2O(l)}] Na^{+} (aq) + OH^{-} (aq)} \label{4.1.3}$
The single arrows from reactant to products in Equations $\ref{4.1.2}$ and $\ref{4.1.3}$ indicate that dissociation is complete.
When weak electrolytes dissolve, they produce relatively few ions in solution. This does not necessarily 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 ($\ce{CH3CO2H}$), which contains the $\ce{–CO2H}$ unit. Although it is soluble in water, it is a weak acid and therefore also a weak electrolyte. Similarly, ammonia ($\ce{NH3}$) is a weak base and therefore a weak electrolyte. The behavior of weak acids and weak bases will be described in more detail later.
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 $5$.
Rule of Thumb
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
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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.01%3A_General_Properties_of_Aqueous_Solutions.txt |
Learning Objectives
• To identify a precipitation reaction and predict solubilities.
Exchange (Double-Displacement) Reactions
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:
$\ce{AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + KNO_3(aq)} \label{4.2.1}$
This unbalanced equation has the general form of an exchange reaction:
$\overbrace{\ce{AC}}^{\text{soluble}} + \overbrace{\ce{BD}}^{\text{soluble}} \rightarrow \underbrace{\ce{AD}}_{\text{insoluble}} + \overbrace{\ce{BC}}^{\text{soluble}} \label{4.2.2}$
The solubility and insoluble annotations are specific to the reaction in Equation \ref{4.2.1} and not characteristic of all exchange reactions (e.g., both products can be soluble or insoluble). 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 $1$: Mixing Potassium Chromate and Silver Nitrate together to initiate a precipitation reaction (Equation $\ref{4.2.1}$).
While full chemical equations show the identities of the reactants and the products and give the stoichiometries of the reactions, they are less effective at describing what is actually 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 above. 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:
$\ce{2AgNO_3(aq)} + \ce{K_2Cr_2O_7(aq)} \rightarrow \ce{Ag_2Cr_2O_7(s) }+ \ce{2KNO_3(aq)} \label{4.2.1a}$
Although Equation $\ref{4.2.1a}$ 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 $\ce{AgNO3}$ and $\ce{K2Cr2O7}$ are strong electrolytes (i.e., they dissociate completely in aqueous solution to form ions). In contrast, because $\ce{Ag2Cr2O7}$ 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:
$\ce{2Ag^{+}(aq) + 2NO_3^{-} (aq) + 2K^{+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2K^{+}(aq) + 2NO_3^{-}(aq)}\label{4.2.2a}$
Note that $\ce{K^+ (aq)}$ and $\ce{NO3^{−} (aq)}$ ions are present on both sides of Equation $\ref{4.2.2a}$ 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:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{4.2.3}$
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 $\ref{4.2.3}$, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral $\ce{Ag2Cr2O7}$ formula unit on the right side.
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:
$2AgF(aq) + (NH_4)_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4F(aq)\label{4.2.4}$
The complete ionic equation for this reaction is as follows:
$\ce{2Ag^{+}(aq)} + \cancel{\ce{2F^{-}(aq)}} + \cancel{\ce{2NH_4^{+}(aq)}} + \ce{Cr_2O_7^{2-}(aq)} \rightarrow \ce{Ag_2Cr_2O_7(s)} + \cancel{\ce{2NH_4^{+}(aq)}} + \cancel{\ce{2F^{-}(aq)}} \label{4.2.5}$
Because two $\ce{NH4^{+}(aq)}$ and two $\ce{F^{−} (aq)}$ ions appear on both sides of Equation $\ref{4.2.5}$, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation $\ref{4.2.6}$), which is identical to Equation $\ref{4.2.3}$:
$\ce{2Ag^{+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)} \label{4.2.6}$
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.
Determining the Products for Precipitation Reactions: Determining the Products for Precipitation Reactions, YouTube(opens in new window) [youtu.be]
Example $1$: Balancing Precipitation Equations
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:
$\ce{Ba(NO_3)_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaNO_3(aq)} \nonumber$
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:
$\ce{3Ba(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaNO_3(aq)} \nonumber$
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:
$\ce{3Ba^{2+}(aq)} + \cancel{\ce{6NO_3^{-}(aq)}} + \cancel{\ce{6Na^{+} (aq)}} + \ce{2PO_4^{3-} (aq)} \rightarrow \ce{Ba_3(PO_4)_2(s)} + \cancel{\ce{6Na^+(aq)}} + \cancel{\ce{6NO_3^{-}(aq)}} \nonumber$
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:
$\ce{3Ba^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s)} \nonumber$
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:
$\ce{3AgF(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaF(aq) } \nonumber$
complete ionic equation:
$\ce{3Ag^+(aq) + 3F^{-}(aq) + 3Na^{+}(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^{+}(aq) + 3F^{-}(aq) } \nonumber$
net ionic equation:
$\ce{3Ag^{+}(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s)} \nonumber$
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.
Predicting Solubilities
Table $1$ 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 $1$ 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 $1$: 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 $1$, 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 $1$ 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 $1$ 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
$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) \nonumber$
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 $1$, 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 $1$, both AlBr3 (rule 4) and Sr(NO3)2 (rule 2) are soluble. Thus no net reaction will occur.
1. A According to Table $1$, 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 $1$, 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 $1$, 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 a
$Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s)$
Answer b
$2PO_4^{3-}(aq) + 3Hg^{2+}(aq) \rightarrow Hg_3(PO_4)_2(s)$
Answer c
$NaF(s)$ dissolves; no net reaction
Answer d
$Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s)$
Predicting the Solubility of Ionic Compounds: Predicting the Solubility of Ionic Compounds, YouTube(opens in new window) [youtu.be] (opens in new window) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.02%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 can define acids as substances that dissolve in water to produce H+ ions, whereas bases are 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.
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 $\ref{4.3.1}$), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH) ions (Equation $\ref{4.3.2}$):
$\underset{an\: Arrhenius\: acid}{HCl_{(g)}} \xrightarrow {H_2 O_{(l)}} H^+_{(aq)} + Cl^-_{(aq)} \label{4.3.1}$
$\underset{an\: Arrhenius\: base}{NaOH_{(s)}} \xrightarrow {H_2O_{(l)}} Na^+_{(aq)} + OH^-_{(aq)} \label{4.3.2}$
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 $\ref{4.3.3}$) is not an acid–base reaction because it does not involve $H^+$ and $OH^−$:
$NH_{3\;(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \label{4.3.3}$
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 $\ref{4.3.3}$, $NH_3$ 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.
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 is known as a polyprotic acid. 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), (Equation $\ref{4.3.4}$, Equation $\ref{4.3.5}$, and Equation $\ref{4.3.6}$):
$H_3 PO_4 (l) \overset{H_2 O(l)}{\rightleftharpoons} H ^+ ( a q ) + H_2 PO_4 ^- (aq) \label{4.3.4}$
$H_2 PO_4 ^- (aq) \rightleftharpoons H ^+ (aq) + HPO_4^{2-} (aq) \label{4.3.5}$
$HPO_4^{2-} (aq) \rightleftharpoons H^+ (aq) + PO_4^{3-} (aq) \label{4.3.6}$
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 react essentially completely with water to give $H^+$ and the corresponding anion. Similarly, strong bases 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 and weak bases 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:
$CH_3 CO_2 H(l) \overset{H_2 O(l)}{\rightleftharpoons} H^+ (aq) + CH_3 CO_2^- (aq) \label{4.3.7}$
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). Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation $\ref{4.3.8}$) but a weak acid when it donates its second proton (Equation $\ref{4.3.9}$) as indicated by the single and double arrows, respectively:
$\underset{strong\: acid}{H_2 SO_4 (l)} \xrightarrow {H_2 O(l)} H ^+ (aq) + HSO_4 ^- (aq) \label{4.3.8}$
$\underset{weak\: acid}{HSO_4^- (aq)} \rightleftharpoons H^+ (aq) + SO_4^{2-} (aq) \label{4.3.9}$
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. All other polyprotic acids, such as H3PO4, are weak acids.
The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion:
$NH_3 (g) + H_2 O(l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq) \label{4.3.10}$
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.
Definition of Strong/Weak Acids & Bases: Definition of Strong/Weak Acids & Bases, YouTube (opens in new window) [Definition of Strong] [Definition of Strong] [youtu.be] (opens in new window)
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 Heavy Group 2 Elements
HCl HNO3 LiOH Ca(OH)2
HBr H2SO4 NaOH Sr(OH)2
HI HClO4 KOH Ba(OH)2
RbOH
CsOH
Example $1$: Acid Strength
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 $\ref{4.3.7}$), 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 $1$: Acid Strength
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 a
strong base
Answer b
strong acid
Answer c
weak acid
Answer d
weak base
Answer e
none of these; formaldehyde is a neutral molecule | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.03%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, 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. Therefore, these reactions are known as oxidation-reduction reactions, or "redox" reactions for short.
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 vapor). The reaction is as follows:
$\ce{Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g)} \label{4.4.1}$
Oxidation-reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants by a transfer of electrons, 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
$\ce{ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s)} \label{4.4.2}$
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 small overset number emphasizes the oxidation state of the elements):
$4 \overset{0}{\ce{Al}} + 3 \overset{0}{\ce{O2}} \rightarrow \ce{4 Al^{3+} + 6 O^{2-} }\label{4.4.3}$
Equation $\ref{4.4.1}$ and Equation $\ref{4.4.2}$ 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 $\ref{4.4.3}$, for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:
\begin{align*} \text{electrons lost} &= \ce{4 Al} \, \text{atoms} \times {3 \, e^- \, \text{lost} \over \ce{Al} \, \text{atom} } \[4pt] &= 12 \, e^- \, \text{lost} \label{4.4.4a} \end{align*}
\begin{align*} \text{electrons gained} &= \ce{6 O} \, \text{atoms} \times {2 \, e^- \, \text{gained} \over \ce{O} \, \text{atom}} \[4pt] &= 12 \, e^- \, \text{gained} \label{4.4.4b}\end{align*}
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 chlorine is illustrated in Figure $1$.
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. Previously, 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. However, we can still assign oxidation states to the elements involved by treating them as if they were ionic (that is, as if all the bonding electrons were transferred to the more attractive element). 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 (fractional) 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 $\ref{4.4.5}$ 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:
$\overset {\color{ref}{+1}}{\ce{Cu_2}} \overset {\color{ref}-2}{\ce{O}} (s) + \overset {\color{ref}0}{\ce{H_2}} (g) \rightarrow 2 \overset {\color{ref}0}{\ce{Cu}} (s) + \overset {\color{ref}+1}{\ce{H}}_2 \overset {\color{ref}-2}{\ce{O}} (g) \label{4.4.5}$
Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). Thus, 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:
$\text{electrons lost} = 2 \, H \, \text{atoms} \times {1 \, e^- \, \text{lost} \over H \, \text{atom} } = 2 \, e^- \, \text{lost} \label{4.4.6a}$
$\text{electrons gained} = 2 \, Cu \, \text{atoms} \times {1 \, e^- \, \text{gained} \over Cu \, \text{atom}} = 2 \, e^- \, \text{gained} \label{4.4.6b}$
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 $1$: Oxidation States
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
$\underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \nonumber$
Thus the sum of the oxidation states of the two carbon atoms is indeed zero.
Exercise $1$: Oxidation States
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 a
Ba, +2; F, −1
Answer b
C, 0; H, +1; O, −2
Answer c
K, +1; Cr, +6; O, −2
Answer d
Cs, +1; O, −½
Answer e
C, −3; H, +1; C, −1; H, +1; O, −2; H, +1
Types of Redox Reactions
Many types of chemical reactions are classified as redox reactions, and it would be impossible to memorize all of them. However, there are a few important types of redox reactions that you are likely to encounter and should be familiar with. These include:
• Synthesis reactions: The formation of any compound directly from the elements is a redox reaction, for example, the formation of water from hydrogen and oxygen: $\ce{ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \nonumber$
• Decomposition reactions: Conversely, the decomposition of a compound to its elements is also a redox reaction, as in the electrolysis of water: $\ce{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)} \nonumber$
• Combustion reactions: Many chemicals combust (burn) with oxygen. In particular, organic chemicals such as hydrocarbons burn in the presence of oxygen to produce carbon dioxide and water as the products: $\ce{ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)} \nonumber$
The following sections describe another important class of redox reactions: single-displacement reactions of metals in solution.
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 $2$). 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:
$\ce{ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)} \label{4.4.81}$
In subsequent steps, $\ce{FeCl2}$ undergoes oxidation to form a reddish-brown precipitate of $\ce{Fe(OH)3}$.
Many metals dissolve through reactions of this type, which have the general form
$\text{metal} + \text{acid} \rightarrow \text{salt} + \text{hydrogen} \label{4.4.82}$
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:
$\ce{Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) } \label{4.4.83}$
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 $\ref{4.4.84}$) and the reduction of silver salts by copper (Equation $\ref{4.4.85}$ and Figure $3$):
$\ce{Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)} \label{4.4.84}$
$\ce{ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \label{4.4.85}$
The reaction in Equation $\ref{4.4.84}$ 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 $\ce{Zn2+}$. 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:
$\ce{Zn(s) + Mg^{2+}(aq) \xcancel{\rightarrow} Zn^{2+}(aq) + Mg(s)} \label{4.4.10}$
$\ce{Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)} \label{4.4.11}$
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 $2$: Activity
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 $4$ 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:
$\ce{ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)} \nonumber$
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:
$\ce{Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) } \nonumber$
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 a
$no\: reaction$
Answer b
$3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)$
Answer c | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.04%3A_Oxidation-Reduction_Reactions.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 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:
$molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \label{4.5.1}$
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,
$[\rm{sucrose}] = 1.00\: M \nonumber$
is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either
$V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \label{4.5.2}$
or
$V_{mL} M_{mmol/mL} = \cancel{mL} \left( \dfrac{mmol} {\cancel{mL}} \right) = mmoles \label{4.5.3}$
Figure $1$ illustrates the use of Equations $\ref{4.5.2}$ and $\ref{4.5.3}$.
Example $1$: Calculating Moles from Concentration of NaOH
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 \ref{4.5.2} or Equation \ref{4.5.3}, 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 \ref{4.5.2} 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 $1$: Calculating Moles from Concentration of Alanine
Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.
Answer
41.6 mmol
Calculations Involving Molarity (M): Calculations Involving Molarity (M), YouTube(opens in new window) [youtu.be]
Concentrations are also 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. These concentrations and their units are summarized in Table $1$.
Table $1$: 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
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 $1$ 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 $2$, for some substances this effect can be significant, especially for concentrated solutions.
Example $2$
The solution 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 $\ce{CoCl2•2H2O}$?
Given: mass of solute and volume of solution
Asked for: concentration (M)
Strategy:
To find the number of moles of $\ce{CoCl2•2H2O}$, 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,
$moles\: CoCl_2 \cdot 2H_2O = \left( \dfrac{10.0 \: \cancel{g}} {165 .87\: \cancel{g} /mol} \right) = 0 .0603\: mol \nonumber$
The volume of the solution in liters is
$volume = 500\: \cancel{mL} \left( \dfrac{1\: L} {1000\: \cancel{mL}} \right) = 0 .500\: L \nonumber$
Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is
$molarity = \dfrac{0.0603\: mol} {0.500\: L} = 0.121\: M = CoCl_2 \cdot H_2O \nonumber$
Exercise $2$
The solution shown in Figure $2$ 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
$(NH_4)_2Cr_2O_7 = 1.43\: M \nonumber$
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 $\ref{4.5.2}$. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example $3$.
Example $3$: D5W Solution
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 $3$
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 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 $3$. 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
$(V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)\label{4.5.4}$
where the subscripts s and d indicate the stock and dilute solutions, respectively. Example $4$ demonstrates the calculations involved in diluting a concentrated stock solution.
Example $4$
What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example $3$?
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 4.5.3 was 0.310 M glucose. We begin by using Equation 4.5.4 to calculate the number of moles of glucose contained in 2500 mL of the solution:
$moles\: glucose = 2500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .775\: mol\: glucose \nonumber$
B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:
$volume\: of\: stock\: soln = 0 .775\: \cancel{mol\: glucose} \left( \dfrac{1\: L} {3 .00\: \cancel{mol\: glucose}} \right) = 0 .258\: L\: or\: 258\: mL \nonumber$
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 4.5.4 for Vs and substituting the appropriate values:
$V_s = \dfrac{( V_d )(M_d )}{M_s} = \dfrac{(2 .500\: L)(0 .310\: \cancel{M} )} {3 .00\: \cancel{M}} = 0 .258\: L \nonumber$
As we have noted, there is often more than one correct way to solve a problem.
Exercise $4$
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
Ion Concentrations in Solution
In Example $2$, 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:
$(NH_4 )_2 Cr_2 O_7 (s) \xrightarrow {H_2 O(l)} 2NH_4^+ (aq) + Cr_2 O_7^{2-} (aq)\label{4.5.5}$
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 $4$).
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.
Concentration of Ions in Solution from a Soluble Salt: Concentration of Ions in Solution from a Soluble Salt, YouTube(opens in new window) [youtu.be]
Example $5$
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)2CHOH
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 $5$
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 a
$[Ba^{2+}] = 0.0012\: M; \: [OH^-] = 0.0024\: M$
Answer b
$[Na^+] = 0.34\: M; \: [SO_4^{2-}] = 0.17\: M$
Answer c
$[(CH_3)_2CO] = 0.50\: M$
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: $molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \nonumber$
• relationship among volume, molarity, and moles: $V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \nonumber$
• relationship between volume and concentration of stock and dilute solutions: $(V_s)(M_s) = moles\: of\: solute = (V_d)(M_d) \nonumber$
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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.05%3A_Concentration_of_Solutions.txt |
Learning Objectives
• 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. 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 titration, a carefully measured volume of a solution of known concentration, called the titrant, 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 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:
Figure $1$: Dimensional Analysis flowchart for titration problems.
Example $1$: Potassium Permanganate
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.
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. 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 permaganate added equals the number of moles of oxalate in the solution, thus determining how many moles of oxalate we started with.
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
$2MnO_4^-(aq) + 5HO_2CCO_2H(aq) + 6H^+(aq) \rightarrow 2Mn^{2+}(aq) + 10CO_2(g) + 8H_2O(l) \nonumber$
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 $MnO_4^−$
$15.4\; \cancel{mL}\left ( \dfrac{1 \: \cancel{L}}{1000\; \cancel{mL}} \right )\left ( \dfrac{0.0247\; mol\; MnO_{4}^{-}}{1\; \cancel{L}} \right )=3.80\times 10^{4\;} mol\; MnO_{4}^{-} \nonumber$
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} \nonumber
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:
$Ca(O_2CCO_2)(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + HO_2CCO_2H(aq) \nonumber$
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
$\% CaC_2 O_4 = \dfrac{0 .122\: g} {10 .0\: g} \times 100 = 1 .22\% \nonumber$
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 $1$: 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:
$\ce{2GSH(aq) + I_2(aq) \rightarrow GSSG(aq) + 2HI(aq)} \nonumber$
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
Limiting Reactant Problems Using Molarities: Limiting Reactant Problems Using Molarities, YouTube(opens in new window) [youtu.be]
Standard Solutions
In Example $1$, 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 solution, 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.06%3A_Solution_Stoichiometry_and_Chemical_Analysis.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.
4.1: General Properties of Aqueous Solutions
Conceptual Problems
1. What are the advantages to carrying out a reaction in solution rather than simply mixing the pure reactants?
2. What types of compounds dissolve in polar solvents?
3. Describe the charge distribution in liquid water. How does this distribution affect its physical properties?
4. Must a molecule have an asymmetric charge distribution to be polar? Explain your answer.
5. Why are many ionic substances soluble in water?
6. Explain the phrase like dissolves like.
7. What kinds of covalent compounds are soluble in water?
8. Why do most aromatic hydrocarbons have only limited solubility in water? Would you expect their solubility to be higher, lower, or the same in ethanol compared with water? Why?
9. Predict whether each compound will dissolve in water and explain why.
1. toluene
2. acetic acid
3. sodium acetate
4. butanol
5. pentanoic acid
10. Predict whether each compound will dissolve in water and explain why.
1. ammonium chloride
2. 2-propanol
3. heptane
4. potassium dichromate
5. 2-octanol
11. Given water and toluene, predict which is the better solvent for each compound and explain your reasoning.
1. sodium cyanide
2. benzene
3. acetic acid
4. sodium ethoxide (CH3CH2ONa)
12. Of water and toluene, predict which is the better solvent for each compound and explain your reasoning.
1. t-butanol
2. calcium chloride
3. sucrose
4. cyclohexene
13. Compound A is divided into three equal samples. The first sample does not dissolve in water, the second sample dissolves only slightly in ethanol, and the third sample dissolves completely in toluene. What does this suggest about the polarity of A?
14. You are given a mixture of three solid compounds—A, B, and C—and are told that A is a polar compound, B is slightly polar, and C is nonpolar. Suggest a method for separating these three compounds.
15. A laboratory technician is given a sample that contains only sodium chloride, sucrose, and cyclodecanone (a ketone). You must tell the technician how to separate these three compounds from the mixture. What would you suggest?
16. Many over-the-counter drugs are sold as ethanol/water solutions rather than as purely aqueous solutions. Give a plausible reason for this practice.
17. What distinguishes a weak electrolyte from a strong electrolyte?
18. Which organic groups result in aqueous solutions that conduct electricity?
19. It is considered highly dangerous to splash barefoot in puddles during a lightning storm. Why?
20. Which solution(s) would you expect to conduct electricity well? Explain your reasoning.
1. an aqueous solution of sodium chloride
2. a solution of ethanol in water
3. a solution of calcium chloride in water
4. a solution of sucrose in water
21. Which solution(s) would you expect to conduct electricity well? Explain your reasoning.
1. an aqueous solution of acetic acid
2. an aqueous solution of potassium hydroxide
3. a solution of ethylene glycol in water
4. a solution of ammonium chloride in water
22. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning.
1. potassium hydroxide
2. ammonia
3. calcium chloride
4. butanoic acid
23. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning.
1. magnesium hydroxide
2. butanol
3. ammonium bromide
4. pentanoic acid
24. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in aqueous solution? Explain your reasoning.
1. H2SO4
2. diethylamine
3. 2-propanol
4. ammonium chloride
5. propanoic acid
Conceptual Answers
1.
2.
3.
4.
5. Ionic compounds such as NaCl are held together by electrostatic interactions between oppositely charged ions in the highly ordered solid. When an ionic compound dissolves in water, the partially negatively charged oxygen atoms of the H2O molecules surround the cations, and the partially positively charged hydrogen atoms in H2O surround the anions. The favorable electrostatic interactions between water and the ions compensate for the loss of the electrostatic interactions between ions in the solid.
6.
7.
8.
9.
1. Because toluene is an aromatic hydrocarbon that lacks polar groups, it is unlikely to form a homogenous solution in water.
2. Acetic acid contains a carboxylic acid group attached to a small alkyl group (a methyl group). Consequently, the polar characteristics of the carboxylic acid group will be dominant, and acetic acid will form a homogenous solution with water.
3. Because most sodium salts are soluble, sodium acetate should form a homogenous solution with water.
4. Like all alcohols, butanol contains an −OH group that can interact well with water. The alkyl group is rather large, consisting of a 4-carbon chain. In this case, the nonpolar character of the alkyl group is likely to be as important as the polar character of the –OH, decreasing the likelihood that butanol will form a homogeneous solution with water.
5. Like acetic acid, pentanoic acid is a carboxylic acid. Unlike acetic acid, however, the alkyl group is rather large, consisting of a 4-carbon chain as in butanol. As with butanol, the nonpolar character of the alkyl group is likely to be as important as the polar character of the carboxylic acid group, making it unlikely that pentanoic acid will form a homogeneous solution with water. (In fact, the solubility of both butanol and pentanoic acid in water is quite low, only about 3 g per 100 g water at 25°C.)
10.
11.
12.
13.
14.
15.
16.
17. An electrolyte is any compound that can form ions when it dissolves in water. When a strong electrolyte dissolves in water, it dissociates completely to give the constituent ions. In contrast, when a weak electrolyte dissolves in water, it produces relatively few ions in solution.
4.2: Precipitation Reactions
Conceptual Problems
1. What information can be obtained from a complete ionic equation that cannot be obtained from the overall chemical equation?
1. Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate.
1. FeCl2(aq) + Na2S(aq)
2. NaOH(aq) + H3PO4(aq)
3. ZnCl2(aq) + (NH4)2S(aq)
2. Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate.
1. KOH(aq) + H3PO4(aq)
2. K2CO3(aq) + BaCl2(aq)
3. Ba(NO3)2(aq) + Na2SO4(aq)
3. Which representation best corresponds to an aqueous solution originally containing each of the following?
1. 1 M NH4Cl
2. 1 M NaO2CCH3
3. 1 M NaOH + 1 M HCl
4. 1 M Ba(OH)2 + 1 M H2SO4
4. Which representation in Problem 3 best corresponds to an aqueous solution originally containing each of the following?
1. 1 M CH3CO2H + 1 M NaOH
2. 1 M NH3 + 1 M HCl
3. 1 M Na2CO3 + 1 M H2SO4
4. 1 M CaCl2 + 1 M H3PO4
3
1. 1
2. 1
3. 1
4. 2
Numerical Problems
1. What mass of precipitate would you expect to obtain by mixing 250 mL of a solution containing 4.88 g of Na2CrO4 with 200 mL of a solution containing 3.84 g of AgNO3? What is the final nitrate ion concentration?
2. Adding 10.0 mL of a dilute solution of zinc nitrate to 246 mL of 2.00 M sodium sulfide produced 0.279 g of a precipitate. How many grams of zinc(II) nitrate and sodium sulfide were consumed to produce this quantity of product? What was the concentration of each ion in the original solutions? What is the concentration of the sulfide ion in solution after the precipitation reaction, assuming no further reaction?
Numerical Answer
1. 3.75 g Ag2CrO4; 5.02 × 10−2 M nitrate
4.3: Acid-Base Reactions
Conceptual Problems
1. Why was it necessary to expand on the Arrhenius definition of an acid and a base? What specific point does the Brønsted–Lowry definition address?
2. State whether each compound is an acid, a base, or a salt.
1. CaCO3
2. NaHCO3
3. H2SO4
4. CaCl2
5. Ba(OH)2
3. State whether each compound is an acid, a base, or a salt.
1. NH3
2. NH4Cl
3. H2CO3
4. CH3COOH
5. NaOH
4. Classify each compound as a strong acid, a weak acid, a strong base, or a weak base in aqueous solution.
1. sodium hydroxide
2. acetic acid
3. magnesium hydroxide
4. tartaric acid
5. sulfuric acid
6. ammonia
7. hydroxylamine (NH2OH)
8. hydrocyanic acid
5. Decide whether each compound forms an aqueous solution that is strongly acidic, weakly acidic, strongly basic, or weakly basic.
1. propanoic acid
2. hydrobromic acid
3. methylamine
4. lithium hydroxide
5. citric acid
6. sodium acetate
7. ammonium chloride
8. barium hydroxide
6. What is the relationship between the strength of an acid and the strength of the conjugate base derived from that acid? Would you expect the CH3CO2 ion to be a strong base or a weak base? Why? Is the hydronium ion a strong acid or a weak acid? Explain your answer.
7. What are the products of an acid–base reaction? Under what circumstances is one of the products a gas?
8. Explain how an aqueous solution that is strongly basic can have a pH, which is a measure of the acidity of a solution.
Answer
1.
2.
3.
4.
5.
1. weakly acidic
2. strongly acidic
3. weakly basic
4. strongly basic
5. weakly acidic
6. weakly basic
7. weakly acidic
8. strongly basic
6.
7.
8.
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 3 (section 4.1 "Aqueous Solutions"0) before proceeding to the Numerical Problems.
1. Derive an equation to relate the hydrogen ion concentration to the molarity of a solution of a strong monoprotic acid.
2. Derive an equation to relate the hydroxide ion concentration to the molarity of a solution of
1. a group I hydroxide.
2. a group II hydroxide.
3. Given the following salts, identify the acid and the base in the neutralization reactions and then write the complete ionic equation:
1. barium sulfate
2. lithium nitrate
3. sodium bromide
4. calcium perchlorate
4. What is the pH of each solution?
1. 5.8 × 10−3 mol of HNO3 in 257 mL of water
2. 0.0079 mol of HI in 750 mL of water
3. 0.011 mol of HClO4 in 500 mL of water
4. 0.257 mol of HBr in 5.00 L of water
5. What is the hydrogen ion concentration of each substance in the indicated pH range?
1. black coffee (pH 5.10)
2. milk (pH 6.30–7.60)
3. tomatoes (pH 4.00–4.40)
6. What is the hydrogen ion concentration of each substance in the indicated pH range?
1. orange juice (pH 3–4)
2. fresh egg white (pH 7.60–7.80)
3. lemon juice (pH 2.20–2.40)
7. What is the pH of a solution prepared by diluting 25.00 mL of 0.879 M HCl to a volume of 555 mL?
8. Vinegar is primarily an aqueous solution of acetic acid. Commercial vinegar typically contains 5.0 g of acetic acid in 95.0 g of water. What is the concentration of commercial vinegar? If only 3.1% of the acetic acid dissociates to CH3CO2 and H+, what is the pH of the solution? (Assume the density of the solution is 1.00 g/mL.)
9. If a typical household cleanser is 0.50 M in strong base, what volume of 0.998 M strong monoprotic acid is needed to neutralize 50.0 mL of the cleanser?
10. A 25.00 mL sample of a 0.9005 M solution of HCl is diluted to 500.0 mL. What is the molarity of the final solution? How many milliliters of 0.223 M NaOH are needed to neutralize 25.00 mL of this final solution?
11. If 20.0 mL of 0.10 M NaOH are needed to neutralize 15.0 mL of gastric fluid, what is the molarity of HCl in the fluid? (Assume all the acidity is due to the presence of HCl.) What other base might be used instead of NaOH?
12. Malonic acid (C3H4O4) is a diprotic acid used in the manufacture of barbiturates. How many grams of malonic acid are in a 25.00 mL sample that requires 32.68 mL of 1.124 M KOH for complete neutralization to occur? Malonic acid is a dicarboxylic acid; propose a structure for malonic acid.
13. Describe how you would prepare 500 mL of a 1.00 M stock solution of HCl from an HCl solution that is 12.11 M. Using your stock solution, how would you prepare 500 mL of a solution that is 0.012 M in HCl?
14. Given a stock solution that is 8.52 M in HBr, describe how you would prepare a 500 mL solution with each concentration.
1. 2.50 M
2. 4.00 × 10−3 M
3. 0.989 M
15. How many moles of solute are contained in each?
1. 25.00 mL of 1.86 M NaOH
2. 50.00 mL of 0.0898 M HCl
3. 13.89 mL of 0.102 M HBr
16. A chemist needed a solution that was approximately 0.5 M in HCl but could measure only 10.00 mL samples into a 50.00 mL volumetric flask. Propose a method for preparing the solution. (Assume that concentrated HCl is 12.0 M.)
17. Write the balanced chemical equation for each reaction.
1. perchloric acid with potassium hydroxide
2. nitric acid with calcium hydroxide
18. Write the balanced chemical equation for each reaction.
1. solid strontium hydroxide with hydrobromic acid
2. aqueous sulfuric acid with solid sodium hydroxide
19. A neutralization reaction gives calcium nitrate as one of the two products. Identify the acid and the base in this reaction. What is the second product? If the product had been cesium iodide, what would have been the acid and the base? What is the complete ionic equation for each reaction?
Answers
1. [H3O+] = [HA] M
2.
3.
1. H2SO4 and Ba(OH)2; 2H+ + SO42 + Ba2+ + 2OH → 2H2O + Ba2+ + SO42
2. HNO3 and LiOH; H+ + NO3 + Li+ + OH → H2O + Li+ + NO3
3. HBr and NaOH; H+ + Br + Na+ + OH → H2O + Na+ + Br
4. HClO4 and Ca(OH)2; 2H+ + 2ClO4 + Ca2+ + 2OH → 2H2O + Ca2+ + 2ClO4
4.
5.
1. 7.9 × 10−6 M H+
2. 5.0 × 10−7 to 2.5 × 10−8 M H+
3. 1.0 × 10−4 to 4.0 × 10−5 M H+
6.
7. pH = 1.402
8.
9. 25 mL
10.
11. 0.13 M HCl; magnesium carbonate, MgCO3, or aluminum hydroxide, Al(OH)3
12.
13. 1.00 M solution: dilute 41.20 mL of the concentrated solution to a final volume of 500 mL. 0.012 M solution: dilute 12.0 mL of the 1.00 M stock solution to a final volume of 500 mL.
14.
15.
1. 4.65 × 10−2 mol NaOH
2. 4.49 × 10−3 mol HCl
3. 1.42 × 10−3 mol HBr
16.
17.
1. HClO4 + KOH → KClO4 + H2O
2. 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O
18.
19. The acid is nitric acid, and the base is calcium hydroxide. The other product is water.
$2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O$
The acid is hydroiodic acid, and the base is cesium hydroxide. The other product is water.
$HI + CsOH \rightarrow CsI + H_2O$
The complete ionic equations are
$2H^+ + 2NO_3^- + Ca^{2+} + 2OH^- \rightarrow Ca^{2+} + 2NO_3^- + H_2O$
$H^+ + I^- + Cs^+ + OH^- \rightarrow Cs^+ + I^- + H_2O$
4.4: Oxidation-Reduction Reactions
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.
Conceptual Solutions
1. In the periodic table, the trend for oxidation increases towards the top-right of the table, while the trend for reduction increases towards the bottom-left. That is to say, the nonmetals and metalloids (with the exception of Noble Gases) tend to be better oxidants, and alkali and transition metals tend to be better reductants.
2. If the compounds mixed are both poor oxidants and poor reductants, no redox reaction will occur. The poor oxidizing agent has a weak ability to gain electrons, while the reducing agent has a weak ability to donate electrons. Due to this, no electrons (or very few electrons) will be transferred, and no reaction will occur. If one compound is a strong oxidant and the other is a weak reductant, the reaction will progress, but not as far as it would have with a strong reductant. Some reductants, such as gold, are so weak that only the strongest of oxidants can oxidize them.
3. Which is Oxidized and Reduced:
1. Zn is oxidized, and H is reduced.
2. Cu is oxidized, and N is reduced.
3. Mn is oxidized, and Br is reduced.
4. A single-displacement reaction occurs when one "free" element replaces another similar element in a compound. What is oxidized and what is reduced depends on the elements in question. For example, in the single-displacement reaction Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g), the free element Zinc is oxidized while Hydrogen is reduced. In Cl2(g)+2NaBr(aq)→2NaCl(aq)+Br2(l), on the other hand, the free element Chlorine is reduced while Bromine is oxidized.
1. An example of a non-single-displacement reaction is a double-displacement reaction, such as: 2KI(aq)+Pb(NO3)2(aq)→2KNO3(aq)+PbI2(s)
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)Conceptual
Numerical Solutions
1. Balanced Redox Reactions:
1. 3CuS(s) + 8NO3(aq) + 8H+(aq) → 3Cu2+(aq) + 3SO42(aq) + 8NO(g) + 4H2O(l)
2. 6Ag(s) + 2CrO42-(aq) + 3HS-(aq) + 5H2O(l) → 3Ag2S(s) + 2Cr(OH)3(s) + 7OH-(aq)
3. Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
4. 2Sb(s) + 3O2(g) + 2H2O(l) + 2OH-(aq) → 2SbO2-(aq) + 3H2O2(aq)
5. Te(s) + 3UO22+(aq) + 4H+(aq) → TeO42-(aq) + 3U4+(aq) + 2H2O(l)
Conceptual Problems
1. Which of the representations best corresponds to a 1 M aqueous solution of each compound? Justify your answers.
1. NH3
2. HF
3. CH3CH2CH2OH
4. Na2SO4
2. Which of the representations shown in Problem 1 best corresponds to a 1 M aqueous solution of each compound? Justify your answers.
1. CH3CO2H
2. NaCl
3. Na2S
4. Na3PO4
5. acetaldehyde
3. Would you expect a 1.0 M solution of CaCl2 to be a better conductor of electricity than a 1.0 M solution of NaCl? Why or why not?
4. An alternative way to define the concentration of a solution is molality, abbreviated m. Molality is defined as the number of moles of solute in 1 kg of solvent. How is this different from molarity? Would you expect a 1 M solution of sucrose to be more or less concentrated than a 1 m solution of sucrose? Explain your answer.
5. What are the advantages of using solutions for quantitative calculations?
Conceptual Answer
1. Compound Corresponding Beaker Reasoning
NH3 2 Weak base dissociates into both ions and molecules.
HF 2 Weak acid dissociates into both ions and molecules.
CH3CH2CH2OH 3 Covalent compound
Na2SO4 1 Soluble ionic compound
2. CH3CO2H 2 Weak acid dissociates into both ions and molecules.
NaCl 1 Soluble ionic compound
Na2S 1 Soluble ionic compound
Na3PO4 1 Soluble ionic compound
Acetaldehyde 1 Aldehydes with short chain lengths are miscible
3. CaCl2 would be expected to be a better conductor than NaCl due to the fact that it contains more charges. The conductivity of a solution depends on the amount of mobile charges within it. While NaCl only has two charges (Na being + and Cl being -), CaCl2 contains three (Ca being 2+ and the two Cls being -). With more free charges when dissolved in the solution, CaCl2 would be a better conductor.
4. Assuming that the solvent is water, the solution with a Molarity (M) of 1 would be more concentrated than the solution with a Molality (m) of 1. Water is approximately 1 kg per liter, meaning that the solute is dissolved in 1 liter of water. This increases the amount of solution past 1 liter. Molarity, on the other hand, means that there is 1 liter of solution (that is, including both solute and solvent). Since in both cases there will be one mole, the 1M solution will be more concentrated than the 1m solution because there is less volume, but the same amount of solute.
5. If the amount of a substance required for a reaction is too small to be weighed accurately, the use of a solution of the substance, in which the solute is dispersed in a much larger mass of solvent, allows chemists to measure the quantity of the substance more accurately.
Numerical Problems
1. Calculate the number of grams of solute in 1.000 L of each solution.
1. 0.2593 M NaBrO3
2. 1.592 M KNO3
3. 1.559 M acetic acid
4. 0.943 M potassium iodate
2. Calculate the number of grams of solute in 1.000 L of each solution.
1. 0.1065 M BaI2
2. 1.135 M Na2SO4
3. 1.428 M NH4Br
4. 0.889 M sodium acetate
3. If all solutions contain the same solute, which solution contains the greater mass of solute?
1. 1.40 L of a 0.334 M solution or 1.10 L of a 0.420 M solution
2. 25.0 mL of a 0.134 M solution or 10.0 mL of a 0.295 M solution
3. 250 mL of a 0.489 M solution or 150 mL of a 0.769 M solution
4. Complete the following table for 500 mL of solution.
Compound Mass (g) Moles Concentration (M)
calcium sulfate 4.86
acetic acid 3.62
hydrogen iodide dihydrate 1.273
barium bromide 3.92
glucose 0.983
sodium acetate 2.42
5. What is the concentration of each species present in the following aqueous solutions?
1. 0.489 mol of NiSO4 in 600 mL of solution
2. 1.045 mol of magnesium bromide in 500 mL of solution
3. 0.146 mol of glucose in 800 mL of solution
4. 0.479 mol of CeCl3 in 700 mL of solution
6. What is the concentration of each species present in the following aqueous solutions?
1. 0.324 mol of K2MoO4 in 250 mL of solution
2. 0.528 mol of potassium formate in 300 mL of solution
3. 0.477 mol of KClO3 in 900 mL of solution
4. 0.378 mol of potassium iodide in 750 mL of solution
7. What is the molar concentration of each solution?
1. 8.7 g of calcium bromide in 250 mL of solution
2. 9.8 g of lithium sulfate in 300 mL of solution
3. 12.4 g of sucrose (C12H22O11) in 750 mL of solution
4. 14.2 g of iron(III) nitrate hexahydrate in 300 mL of solution
8. What is the molar concentration of each solution?
1. 12.8 g of sodium hydrogen sulfate in 400 mL of solution
2. 7.5 g of potassium hydrogen phosphate in 250 mL of solution
3. 11.4 g of barium chloride in 350 mL of solution
4. 4.3 g of tartaric acid (C4H6O6) in 250 mL of solution
9. Give the concentration of each reactant in the following equations, assuming 20.0 g of each and a solution volume of 250 mL for each reactant.
1. BaCl2(aq) + Na2SO4(aq) →
2. Ca(OH)2(aq) + H3PO4(aq) →
3. Al(NO3)3(aq) + H2SO4(aq) →
4. Pb(NO3)2(aq) + CuSO4(aq) →
5. Al(CH3CO2)3(aq) + NaOH(aq) →
10. An experiment required 200.0 mL of a 0.330 M solution of Na2CrO4. A stock solution of Na2CrO4 containing 20.0% solute by mass with a density of 1.19 g/cm3 was used to prepare this solution. Describe how to prepare 200.0 mL of a 0.330 M solution of Na2CrO4 using the stock solution.
11. Calcium hypochlorite [Ca(OCl)2] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl)2 concentration of 3.4 g per 100 mL of solution, what is the molarity of hypochlorite?
12. Phenol (C6H5OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol?
13. If a tablet containing 100 mg of caffeine (C8H10N4O2) is dissolved in water to give 10.0 oz of solution, what is the molar concentration of caffeine in the solution?
14. A certain drug label carries instructions to add 10.0 mL of sterile water, stating that each milliliter of the resulting solution will contain 0.500 g of medication. If a patient has a prescribed dose of 900.0 mg, how many milliliters of the solution should be administered?
Numerical Answers
1. Grams of solute in 1.000 L of each solution:
1. 0.2593 M NaBrO3
1.
2. 1.592 M KNO3
1.
3. 1.559 M acetic acid
1.
4. 0.943 M potassium iodate
1.
2.
3.
4. Compound Mass (g) Moles Concentration (M)
calcium sulfate 4.86 0.0357 0.07
acetic acid 217.39 3.62 7.24
hydrogen iodide dihydrate 103.79 0.637 1.273
barium bromide 3.92 0.013 0.026
glucose 88.55 0.491 0.983
sodium acetate 198.52 2.42 4.84
5. Molar Concentration in each solution:
1. 0.815M NiSO4
2. 2.09M Magnesium Bromide
3. 0.183M Glucose
4. 0.684M CeCl3
6. Molar Concentration in each solution:
1. 1.296M K2MoO4
2. 1.760M Potassium Formate
3. 0.530M KClO3
4. 0.503M Potassium Iodide
7. Molar Concentration in each solution:
1. 0.174M Calcium Bromide
2. 0.297M Lithium Sulfate
3. 0.048M Sucrose
4. 0.135M Iron(III) Nitrate Hexahydrate
8. Molar Concentration in each solution:
1. 0.267M Sodium Hydrogen Sulfate
2. 0.172M Potassium Hydrogen Phosphate
3. 0.156M Barium Chloride
4. 0.115M Tartaric Acid
9. Concentration of Reactants in each solution:
1. 0.192M of BaCl2, 0.282M of Na2SO4
2. 0.541M of Ca(OH)2, 0.408M of H3PO4
3. 0.188M of Al(NO3)3, 0.408M H2SO4(aq)
4. 0.121M of Pb(NO3)2, 0.251M CuSO4
5. 0.196M of Al(CH3CO2)3, 1M NaOH
10.
11. 0.4756 M ClO
12. 0.159M Phenol
13. 1.74 × 10−3 M caffeine
14. 1.8mL of solution
4.6: Solution Stoichiometry and Chemical Analysis
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.
Conceptual Answers
1. In a titration procedure, the analyte is considered to be the limiting reactant, as the amount of moles of it remain constant while more and more titrant can be added (making the titrant the excess reactant). By using the fact that the analyte is limited, we can continually add the excess reactant to find out just how many moles are in the analyte.
2. To determine the concentration of a solution, one must titrate it with an acid (if the solution is a base) or a base (if the solution is an acid). By titrating the unknown solution with a solution of known concentration, we can determine how many moles of the titrant were needed to fully neutralize the analyte. This point can be determined by finding the equivalence point during the titration, and from this knowledge as well as the mole ratio from the balanced neutralization equation, we can determine how many moles are in the analyte. With this, we can simply divide by the original volume to find its concentration.
3. Graph (a) corresponds to the titration of NaOH with acetic acid, while graph (b) corresponds to the titration of acetic acid with NaOH. This can be determined due to our knowledge that acids have a pH below 7 while bases have a pH above 7. Therefore, titrating a base with an acid would decrease its pH while titrating an acid with a base would increase its pH.
4. Graph (a) corresponds to the titration of ammonia with HCl, while graph (b) corresponds to the titration of NH4Cl with NaOH. This can be determined due to our knowledge that acids have a pH below 7 while bases have a pH above 7. Therefore, titrating a base with an acid would decrease its pH while titrating an acid with a base would increase its pH.
5.
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?
Numerical Answers
1. Final Molarity = 0.0428M, and there are 0.76g of KHP in the initial sample.
2. 15mL of 0.978 M NaOH, or 0.0147 moles.
3. Concentration of basic solution = 0.536 M. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.E%3A_Reactions_in_Aqueous_Solution_%28Exercises%29.txt |
aqueous solutions – solutions in which water is the dissolving medium
4.1: General Properties of Aqueous Solutions
• electrolyte – substance whose aqueous solution contains ions
• nonelectrolyte – substance that does not form ions in solution
4.2.1 Ionic Compounds in Water
• dissociate – when ions separate from a solid being dissolved
4.2.2 Molecular Compounds in Water
• the molecular structure is maintained
4.2.3 Strong and Weak Electrolytes
• strong electrolytes – ionic compounds that exists entirely of ions in solution
• weak electryolytes – molecular compounds that produce a small amound of ions
• chemical equilibrium – equilibrium of forming ions and recrystalizing ions
4.2: Precipitation Reactions
$AX + BY \rightarrow AY + BX \nonumber$
• for methathesis to occur:
1. the formation of an insoluble product
2. the formation of either a weak electrolyte or a nonelectrolyte
3. the formation of a gas that escapes from solution
4.5.1 Precipitation Reactions
• precipitate – insoluble solid formed by a reaction in solution
• solubility – amount of substance that can be dissolved in a given quantity
4.5.2 Solubility Guidelines for Ionic Compounds
• all common ionic compounds of the alkali metal ions and of the ammonium ion are soluble in water
4.5.3 Reactions in Which a Weak Electrolyte or Nonelectrolyte Forms
• hydrogen and hydroxide react to form water
• insoluble metal oxides react with acids
4.3: Acid-Base Reactions
4.3.1 Acids
• substances that ionize to form hydrogen ions
• proton donors
4.3.2 Bases
• substances that ionize to form hydroxide ions
4.3.3 Strong and Weak Acids and Bases
• strong acid, strong base – strong electrolyte
• weak acid, weak base – weak electrolyte
4.3.4 Neutralization Reactions and Salts
• neutralization reaction – when an acid and base are mixed
• produces water and a salt
4.4 Ionic Equations
• molecular formula – and equation written to show the complete chemical formulas of reactants and products
• spectator ions – ions that do not play a role in a reaction
• net ionic equation – equation where the spectator ions are removed
• only soluble strong electrolytes are written in ionic form
4.4: Oxidation-Reduction Reactions
4.6.1 Reactions in Which a Gas Forms
• carbonates and bicarbonates
4.6.2 Oxidation and Reduction
• oxidation – loss of electrons
• reduction – gain of electrons
4.6.3 Oxidation of Metals by Acids and Salts
• whenever one substance is oxidized, some other substance must be reduced
• metals react with acids to form salts and hydrogen gas
4.6.4 The Activity Series
• activity series – list of metals arranged in order of decreasing ease of oxidation
• active metals – alkali metals and alkaline earth metals
• any metal on the list can be oxidized by ions of elements below it
4.5: Concentration of Solutions
• solution – homogeneous mixture of two or more substances
• solvent – component that is present in greatest quantity
• solutes – substances dissolved in the solvent
4.1.1 Molarity
• concentration – the amount of solute dissolved in a given quantity of solvent or solution
• molarity – number of moles of solute in a liter of solution
4.1.2 Dilution
• dilution - obtaining a lower concentration of a solution by adding water
• moles solute before dilution = moles solute after dilution
4.6: Solution Stoichiometry and Chemical Analysis
4.7.1 Titrations
• statndard solution – solution of known concentration
• titration – a known solution that undergoes a specific chemical reaction of known stoichiometry with the solution of unknown concentration
• equivalence point – stoichiometrically equivalent quantities of reactants are brought together
• indicator – used to show the endpoint of the titration | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.S%3A_Reactions_in_Aqueous_Solution_%28Summary%29.txt |
This chapter introduces you to thermochemistry, a branch of chemistry that describes the energy changes that occur during chemical reactions. In some situations, the energy produced by chemical reactions is actually of greater interest to chemists than the material products of the reaction. For example, the controlled combustion of organic molecules, primarily sugars and fats, within our cells provides the energy for physical activity, thought, and other complex chemical transformations that occur in our bodies. Similarly, our energy-intensive society extracts energy from the combustion of fossil fuels, such as coal, petroleum, and natural gas, to manufacture clothing and furniture, heat your home in winter and cool it in summer, and power the car or bus that gets you to class and to the movies. By the end of this chapter, you will know enough about thermochemistry to explain why ice cubes cool a glass of soda, how instant cold packs and hot packs work, and why swimming pools and waterbeds are heated. You will also understand what factors determine the caloric content of your diet and why even “nonpolluting” uses of fossil fuels may be affecting the environment.
• 5.1: The Nature of Energy
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.
• 5.2: 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.
• 5.3: Enthalpy
At constant pressure, heat flow (q) and internal energy (U) are related to the system’s enthalpy (H).
• 5.4: Enthalpy of Reaction
For a chemical reaction, the enthalpy of reaction (\(ΔH_{rxn}\)) is the difference in enthalpy between products and reactants; the units of \(ΔH_{rxn}\) are kilojoules per mole. Reversing a chemical reaction reverses the sign of \(ΔH_{rxn}\).
• 5.5: Calorimetry
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. 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.
• 5.6: Hess's Law
Hess's law argues that 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. Reversing a chemical reaction reverses the sign of ΔHrxn. 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 (ΔHfus) and the enthalpy of vaporiz
• 5.7: Enthalpies of Formation
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 (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 ga
• 5.8: Foods and Fuels
Thermochemical concepts can be applied to determine the actual energy available in food. The nutritional Calorie is equivalent to 1 kcal (4.184 kJ). The caloric content of a food is its \(ΔH_{comb}\) per gram. The typical caloric contents for food are 9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals.
• 5.E: Thermochemistry (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 5.S: Thermochemistry (Summary)
Thumbnail: Dancing Flames of burning charcoal in the dark. (CC BY-SA 3.0 Unported; Oscar(opens in new window) via Wikipedia(opens in new window))
05: 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.
Energy, Heat, and Work
One definition of energy is the capacity to do work. The easiest form of work to visualize is mechanical work (Figure $3$), which is the energy required to move an object a distance d when opposed by a force F, such as gravity:
work = force x distance
$w=F\,d \label{5.1.1}$
Because the force (F) that opposes the action is equal to the mass (m) of the object times its acceleration (a), we can also write Equation $\ref{5.1.1}$ as follows:
work= mass x acceleration x distance
$w = m\,a\,d \label{5.1.2}$
Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth.
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 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.
In contrast, heat (q) is thermal energy that can be transferred from an object at one temperature to an object at another temperature. The net transfer of thermal energy stops when the two objects reach the same temperature.
Energy is an extensive property of matter—for example, the amount of thermal energy in an object is proportional to both its mass and its temperature. A water heater that holds 150 L of water at 50°C contains much more thermal energy than does a 1 L pan of water at 50°C. Similarly, a bomb contains much more chemical energy than does a firecracker. We now present a more detailed description of kinetic and potential energy.
Kinetic and Potential Energy
The kinetic energy of an object is related to its mass $m$ and velocity $v$:
$KE=\dfrac{1}{2}mv^2 \label{5.1.4}$
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
$KE=\dfrac{1}{2}(1360 kg)(26.8 ms)^2= 4.88 \times 10^5 g \cdot m^2 \label{5.1.5}$
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 $\ref{5.1.2}$, 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:
$PE= F\;d = m\,a\;d = m\,g\,h \label{5.1.6a}$
$PE=(1360, Kg)\left(\dfrac{9.81\, m}{s^2}\right)(36.6\;m) = 4.88 \times 10^5\; \frac{Kg \cdot m}{s^2} \label{5.1.6b}$
$=4.88 \times 10^5 J = 488\; kJ \label{5.1.6c}$
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:
$1 \;cal = 4.184 \;J \;\text{exactly} \label{5.1.7a}$
$1 \;J = 0.2390\; cal \label{5.1.7b}$
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 5.1.4 to calculate the kinetic energy and Equation 5.1.6 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: $v= \left(\dfrac{100\; \cancel{mi}}{1\;\cancel{h}} \right) \left(\dfrac{1 \;\cancel{h}}{60 \;\cancel{min}} \right) \left(\dfrac{1 \; \cancel{min}}{60 \;s} \right)\left(\dfrac{1.61\; \cancel{km}}{1 \;\cancel{mi}} \right) (\dfrac{1000\; m}{1\; \cancel{km}})= 44.7 \;m/s \nonumber$
The kinetic energy of the baseball is therefore $KE= 1492 \;\cancel{g} \left(\dfrac{1\; kg}{1000 \;\cancel{g}} \right) \left(\dfrac{44.7 \;m}{s} \right)^2= 1.49 \times 10^2 \dfrac{kg⋅m^2}{s^2}= 1.49 \times10^2\; J \nonumber$
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 $PE= 149\;\cancel{g} \left(\dfrac{1\; kg}{1000\; \cancel{g}} \right)\left(\dfrac{9.81\; m}{s^2} \right) \left(247\; \cancel{ft} \right) \left(\dfrac{0.3048\; m}{1 \;\cancel{ft}} \right)= 1.10 \times 10^2 \dfrac{kg⋅m^2}{s^2}= 1.10 \times 10^2\; J \nonumber$
Exercise $1$
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 a
3.10 × 102 J
Answer b
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. 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_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.01%3A_The_Nature_of_Energy.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 (Figure $3$). The balanced chemical equation for the reaction is as follows:
$\ce{ 2Al(s) + Fe_2O_3(s) -> 2Fe(s) + Al_2O_3(s)} \label{5.2.1}$
We can also write this chemical equation as
$\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s)} + \text{heat} \label{5.2.2}$
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:
$\ce{heat + H_2O(s) \rightarrow H_2O(l)} \label{5.2.3}$
When heat is transferred to a system from its surroundings, the process is endothermic. By convention, $q > 0$ for an endothermic reaction.
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 $\ref{5.2.2}$ and $\ref{5.2.3}$ 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:
$U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \label{5.2.4a}$
$\Delta{U_{sys}}=−ΔU_{surr} \label{5.2.4b}$
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.
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):
$ΔU_{sys} = q + w \label{5.2.5}$
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.
Although $q$ and $w$ are not state functions, their sum ($ΔU_{sys}$) is independent of the path taken and therefore is a state function.
Thus, because of the first law, we can determine $ΔU$ for any process if we can measure both $q$ and $w$. Heat, $q$, may be calculated by measuring a change in temperature of the surroundings. Work, $w$, may come in different forms, but it too can be measured. One important form of work for chemistry is pressure-volume work done by an expanding gas. At a constant external pressure (for example, atmospheric pressure)
$w = −PΔV \label{5.2.6}$
The negative sign associated with $PV$ work done indicates that the system loses energy when the volume increases. That is, an expanding gas does work on its surroundings, while a gas that is compressed has work done on it by the surroundings.
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 $\ref{5.2.5}$.
2. From Equation $\ref{5.2.6}$ calculate $w$ from the values given. Substitute this value into Equation $\ref{5.2.5}$ to calculate $ΔU$.
Solution
A From Equation $\ref{5.2.5}$, 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 $\ref{5.2.5}$,
$w=-P_{\textrm{ext}}\Delta V=-8.00\textrm{ atm}(\textrm{0.0500 L} - \textrm{0.400 L})\left(\dfrac{\textrm{101.3 J}}{\mathrm{L\cdot atm}} \right)=284\textrm{ J} \nonumber$
Thus
\begin{align*} ΔU &= q + w \[4pt] &= −140 \,J + 284\, J \[4pt] &= 144\, J\end{align*} \nonumber
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
By convention (to chemists), both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa.
Summary
In chemistry, the small part of the universe that we are studying is the system, and the rest of the universe is the surroundings. Open systems can exchange both matter and energy with their surroundings, closed systems can exchange energy but not matter with their surroundings, and isolated systems can exchange neither matter nor energy with their surroundings. A state function is a property of a system that depends on only its present state, not its history. A reaction or process in which heat is transferred from a system to its surroundings is exothermic. A reaction or process in which heat is transferred to a system from its surroundings is endothermic. 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. 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_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.02%3A_The_First_Law_of_Thermodynamics.txt |
Learning Objectives
• To know the key features of a state function.
• To use Hess’s law and thermochemical cycles to calculate enthalpy changes of chemical reactions.
Both heat and work represent energy transfer mechanisms. As discussed previously, the term "heat" refers to a process in which a body (the contents of a tea kettle, for example) acquires or loses energy as a direct consequence of its having a different temperature than its surroundings. Similarly, work refers to the transfer of energy that does not involve temperature differences. Hence, work, like energy, can take various forms, the most familiar being mechanical and electrical.
• Mechanical work arises when an object moves a distance $Δx$ against an opposing force $f$: $w_{mechanical} = f Δx \nonumber$
• Electrical work is done when a body having a charge $q$ moves through a potential difference $ΔV$. $w_{electrical}=q\Delta V \nonumber$
The unit of work is Joules. Work, like heat, exists only when energy is being transferred. When two bodies are placed in thermal contact and energy flows from the warmer body to the cooler one, this process is called “heat”. A transfer of energy to or from a system by any means other than heat is called “work”.
Enthalpy as a Composite Function
To further understand the relationship between heat flow ($q$) and the resulting change in internal energy ($ΔU$), we can look at two sets of limiting conditions: reactions that occur at constant volume and reactions that occur at constant pressure. We will assume that PV work is the only kind of work possible for the system, so we can substitute its definition $w=-P\Delta V$ into the first Law of Thermodynamics ($\Delta U = q + w$) to obtain the following:
$ΔU = q − PΔV \label{5.2.5}$
where the subscripts have been deleted.
If the reaction occurs in a closed vessel, the volume of the system is fixed, and ΔV is zero. Under these conditions, the heat flow (often given the symbol qv to indicate constant volume) must equal ΔU:
$\underset{\textrm{constant volume}}{q_{\textrm v}=\Delta U} \label{5.3.3}$
At constant volume, no $PV$ work can be done, and the change in the internal energy of the system is equal to the amount of heat transferred from the system to the surroundings or vice versa.
Many chemical reactions are not, however, carried out in sealed containers at constant volume but in open containers at a more or less constant pressure of about 1 atm. The heat flow under these conditions is given the symbol $q_p$ to indicate constant pressure. Replacing q in Equation $\ref{5.3.3}$ by $q_p$ and rearranging to solve for qp,
$\underset{\textrm{constant pressure}}{q_{\textrm p}=\Delta U+P\Delta V} \label{5.3.4}$
Thus, at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done.
Because conditions of constant pressure are so important in chemistry, a new state function called enthalpy (H) is defined as
$H =U + PV \nonumber$
At constant pressure, the change in the enthalpy of a system is as follows:
$ΔH = ΔU + Δ(PV) = ΔU + PΔV \label{5.3.5}$
Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: $ΔH = q_p$. This expression is consistent with our definition of enthalpy, where we stated that enthalpy is the heat absorbed or produced during any process that occurs at constant pressure.
At constant pressure, the change in the enthalpy of a system is equal to the heat flow: $ΔH = q_p$.
Example $1$
The molar enthalpy of fusion for ice at 0.0°C and a pressure of 1.00 atm is 6.01 kJ, and the molar volumes of ice and water at 0°C are 0.0197 L and 0.0180 L, respectively. Calculate ΔH and ΔU for the melting of ice at 0.0°C.
Given: enthalpy of fusion for ice, pressure, and molar volumes of ice and water
Asked for: ΔH and ΔU for ice melting at 0.0°C
Strategy:
1. Determine the sign of q and set this value equal to ΔH.
2. Calculate Δ(PV) from the information given.
3. Determine ΔU by substituting the calculated values into Equation $\ref{5.3.5}$.
Solution
A Because 6.01 kJ of heat is absorbed from the surroundings when 1 mol of ice melts, q = +6.01 kJ. When the process is carried out at constant pressure, q = qp = ΔH = 6.01 kJ.
B To find ΔU using Equation $\ref{18.11}$, we need to calculate Δ(PV). The process is carried out at a constant pressure of 1.00 atm, so
\begin{align}\Delta(PV)&=P\Delta V=P(V_{\textrm f}-V)=(1.00\textrm{ atm})(\textrm{0.0180 L}-\textrm{0.0197 L}) \ &=(-1.7\times10^{-3}\;\mathrm{L\cdot atm})(101.3\;\mathrm{J/L\cdot atm})=-0.0017\textrm{ J}\end{align} \nonumber
C Substituting the calculated values of ΔH and PΔV into Equation 18.11,
ΔU = ΔH − PΔV = 6010 J − (−0.0017 J) = 6010 J = 6.01 kJ
Exercise $1$
At 298 K and 1 atm, the conversion of graphite to diamond requires the input of 1.850 kJ of heat per mole of carbon. The molar volumes of graphite and diamond are 0.00534 L and 0.00342 L, respectively. Calculate ΔH and ΔU for the conversion of C (graphite) to C (diamond) under these conditions.
Answer a
ΔH = 1.85 kJ/mol
Answer b
ΔU = 1.85 kJ/mol
The Relationship between ΔH and ΔU
If ΔH for a reaction is known, we can use the change in the enthalpy of the system (Equation $\ref{5.3.5}$) to calculate its change in internal energy. When a reaction involves only solids, liquids, liquid solutions, or any combination of these, the volume does not change appreciably (ΔV = 0). Under these conditions, we can simplify Equation $\ref{5.3.5}$ to ΔH = ΔU. If gases are involved, however, ΔH and ΔU can differ significantly. We can calculate ΔU from the measured value of ΔH by using the right side of Equation $\ref{5.3.5}$ together with the ideal gas law, PV = nRT. Recognizing that Δ(PV) = Δ(nRT), we can rewrite Equation $\ref{5.3.5}$ as follows:
$ΔH = ΔU + Δ(PV) = ΔU + Δ(nRT) \label{5.3.6}$
At constant temperature, Δ(nRT) = RTΔn, where Δn is the difference between the final and initial numbers of moles of gas. Thus
$ΔU = ΔH − RTΔn \label{5.3.7}$
For reactions that result in a net production of gas, Δn > 0, so ΔU < ΔH. Conversely, endothermic reactions (ΔH > 0) that result in a net consumption of gas have Δn < 0 and ΔU > ΔH. The relationship between ΔH and ΔU for systems involving gases is illustrated in Example $2$.
For reactions that result in a net production of gas, ΔU < ΔH. For endothermic reactions that result in a net consumption of gas, ΔU > ΔH.
Example $2$: The Relationship Between ΔH and ΔU in Gas Evolving Reactions
The combustion of graphite to produce carbon dioxide is described by the equation
$C_{(graphite, s)} + O_{2(g)} → CO_{2(g)} \nonumber$
At 298 K and 1.0 atm, ΔH = −393.5 kJ/mol of graphite for this reaction, and the molar volume of graphite is 0.0053 L. What is ΔU for the reaction?
Given: balanced chemical equation, temperature, pressure, ΔH, and molar volume of reactant
Asked for: ΔU
Strategy:
1. Use the balanced chemical equation to calculate the change in the number of moles of gas during the reaction.
2. Substitute this value and the data given into Equation $\ref{5.3.7}$ to obtain ΔU.
Solution
A In this reaction, 1 mol of gas (CO2) is produced, and 1 mol of gas (O2) is consumed. Thus Δn = 1 − 1 = 0.
B Substituting this calculated value and the given values into Equation $\ref{5.3.7}$,
ΔU=ΔH−RTΔn=(−393.5 kJ/mol)−[8.314 J/(mol⋅K)](298 K)(0)
=(−393.5 kJ/mol)−(0 J/mol)=−393.5 kJ/mol
To understand why only the change in the volume of the gases needs to be considered, notice that the molar volume of graphite is only 0.0053 L. A change in the number of moles of gas corresponds to a volume change of 22.4 L/mol of gas at standard temperature and pressure (STP), so the volume of gas consumed or produced in this case is (1)(22.4 L) = 22.4 L, which is much, much greater than the volume of 1 mol of a solid such as graphite.
Exercise $2$
Calculate $ΔU$ for the conversion of oxygen gas to ozone at 298 K:
$3O_2(g) \rightarrow 2O_3(g). \nonumber$
The value of $ΔH$ for the reaction is 285.4 kJ.
Answer
288 kJ
As Example $2$ illustrates, the magnitudes of ΔH and ΔU for reactions that involve gases are generally rather similar, even when there is a net production or consumption of gases.
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. Enthalpy is a state function whose change indicates the amount of heat transferred from a system to its surroundings or vice versa, at constant pressure. 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_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.03%3A_Enthalpy.txt |
Learning Objectives
• To understand how enthalpy pertains to 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:
$\ce{Cu(s) + 4HNO3(aq) \rightarrow Cu(NO3)2(aq) + 2H_2O(l) + 2NO2(g)} \nonumber$
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 $1$). 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),
$w = −PΔV \label{5.4.2}$
The negative sign associated with $PV$ work done indicates that the system loses energy when the volume increases. 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 internal energy $U$ of a system 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 $V$:
$H =U + PV \label{5.4.3}$
Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. So we can define a change in enthalpy ($\Delta H$) accordingly
$ΔH = H_{final} − H_{initial} \nonumber$
If a chemical change occurs at constant pressure (i.e., for a given $P$, $ΔP = 0$), the change in enthalpy ($ΔH$) is
\begin{align} ΔH &= Δ(U + PV) \[4pt] &= ΔU + ΔPV \[4pt] &= ΔU + PΔV \label{5.4.4} \end{align}
Substituting $q + w$ for $ΔU$ (First Law of Thermodynamics) and $−w$ for $PΔV$ (Equation $\ref{5.4.2}$) into Equation $\ref{5.4.4}$, we obtain
\begin{align} ΔH &= ΔU + PΔV \[4pt] &= q_p + \cancel{w} −\cancel{w} \[4pt] &= q_p \label{5.4.5} \end{align}
The subscript $p$ is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation $\ref{5.4.5}$ we see that at constant pressure the change in enthalpy, $ΔH$ of the system, is equal to the heat gained or lost.
\begin{align} ΔH &= H_{final} − H_{initial} \[4pt] &= q_p \label{5.4.6} \end{align}
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.
To find $ΔH$ for a reaction, measure $q_p$.
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:
sign conventions for heat flow and enthalpy changes
Reaction Type q ΔHrxn
exothermic < 0 < 0 (heat flows from a system to its surroundings)
endothermic > 0 > 0 (heat flows from the surroundings to a system)
If ΔHrxn 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 $2a$). Conversely, if ΔHrxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (Figure $\PageIndex{2b}$). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.
Bond breaking ALWAYS requires an input of energy; bond making ALWAYS releases energy.y.
• 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):
$\begin{matrix} heat+ H_{2}O(s) \rightarrow H_{2}O(l) & \Delta H > 0 \end{matrix} \label{5.4.7}$
$\begin{matrix} H_{2}O(l) \rightarrow H_{2}O(s) + heat & \Delta H < 0 \end{matrix} \label{5.4.8}$
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 $\ref{5.4.9}$, 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:
$\ce{2Al(s) + Fe2O3(s) -> 2Fe (s) + Al2O3 (s) } + 815.5 \; kJ \label{5.4.9}$
Thus ΔH = −851.5 kJ/mol of 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 $\ref{5.4.10}$, it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:
$\ce{ 2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3 (s)} \quad \Delta H_{rxn}= - 851.5 \; kJ \label{5.4.10}$
If 4 mol of Al and 2 mol of $\ce{Fe2O3}$ 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:
$- \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{5.4.6a}$
The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example $1$.
Example $1$: Melting Icebergs
Certain parts of the world, such as southern California and Saudi Arabia, are short of freshwater 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 at 0°C and constant pressure:
$\ce{H2O(s) → H_2O(l)} \nonumber$
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{\text{metric ton }} \ce{H2O} \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{\text{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 \,\ce{H2O} \end{align*} \nonumber
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 below.
Possible sources of the approximately $3.34 \times 10^{11}\, kJ$ needed to melt a $1.00 \times 10^6$ 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
• $1.1 \times 10^8$ kilowatt-hours of electricity
Alternatively, we can rely on ambient temperatures to slowly melt the iceberg. The main issue with this idea is the cost of dragging the iceberg to the desired place.
Exercise $1$: Thermite Reaction
If 17.3 g of powdered aluminum are allowed to react with excess $\ce{Fe2O3}$, how much heat is produced?
Answer
273 kJ
Enthalpies of Reaction
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, since enthalpy is a state function, all we have to know is the initial and final states of the reaction. This 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 (ΔHcomb) 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 (ΔHfus) The enthalpy change that acompanies 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 (ΔHvap) 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 (ΔHsoln) The change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent. The enthalpy change 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: Enthalpy of Reaction, YouTube(opens in new window) [youtu.be]
Summary
For a chemical reaction, the enthalpy of reaction ($ΔH_{rxn}$) is the difference in enthalpy between products and reactants; the units of $ΔH_{rxn}$ are kilojoules per mole. Reversing a chemical reaction reverses the sign of $ΔH_{rxn}$. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.04%3A_Enthalpy_of_Reaction.txt |
Learning Objectives
• To use calorimetric data to calculate enthalpy changes.
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.
Heat Capacity
We have seen that the temperature of an object changes when it absorbs or loses thermal energy. The magnitude of the temperature change depends on both the amount of thermal energy transferred (q) and the heat capacity of the object. Its heat capacity (C) is the amount of energy needed to raise the temperature of the object exactly 1°C; the units of C are joules per degree Celsius (J/°C). Note that a degree Celsius is exactly the same as a Kelvin, so the heat capacities can be expresses equally well, and perhaps a bit more correctly in SI, as joules per Kelvin, J/K
The change in temperature ($ΔT$) is
$\Delta T = \dfrac{q}{C} \label{5.5.1}$
where q is the amount of heat (in joules), C is the heat capacity (in joules per degree Celsius), and $ΔT$ is $T_{final} − T_{initial}$ (in degrees Celsius). Note that $ΔT$ is always written as the final temperature minus the initial temperature.
Which Temperature units to use
Since the scaling for Kelvin (K) and degrees Celsius (°C) are exactly the same, the DIFFERENCE
$ΔT = T_{final} − T_{initial} \nonumber$
is the same is either is used for temperature calculations, but make sure not to mix these two temperature units for BOTH $T_{final}$ and $T_{initial}$.
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.
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 ($C_p$) is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of $C_p$ 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:
$q = nC_pΔT \label{5.5.2}$
where $n$ is the number of moles of substance and $C_p$ is the molar heat capacity or via mass:
$q = mC_sΔT \label{5.5.3}$
where $m$ is the mass of substance in grams and $C_s$ is the specific heat.
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.
Table $1$: Specific Heats of Selected Substances at 25°C
Compound Specific Heat [J/(g•°C)] Compound Specific Heat [J/(g•°C)]
H2O (s) 2.108 Al(s) 0.897
H2O (l) 4.184 Fe(s) 0.449
H2O (g) 2.062 Cu(s) 0.385
CH3OH (methanol) 2.531 Au(s) 0.129
CH3CH2OH (ethanol) 2.438 Hg(l) 0.140
n-C6H14 (n-hexane) 2.270 NaCl(s) 0.864
C6H6 (benzene) 1.745 MgO(s) 0.921
C(s) (graphite) 0.709 SiO2(s) (quartz) 0.742
C(s) (diamond) 0.509 CaCO3(s) (calcite) 0.915
The high specific heat of liquid water has important implications for life on Earth. A given mass of water releases more than five times as much heat for a 1°C temperature change as does the same mass of limestone or granite. Consequently, coastal regions of our planet tend to have less variable climates than regions in the center of a continent. After absorbing large amounts of thermal energy from the sun in summer, the water slowly releases the energy during the winter, thus keeping coastal areas warmer than otherwise would be expected (Figure $1$). Water’s capacity to absorb large amounts of energy without undergoing a large increase in temperature also explains why swimming pools and waterbeds are usually heated. Heat must be applied to raise the temperature of the water to a comfortable level for swimming or sleeping and to maintain that level as heat is exchanged with the surroundings. Moreover, because the human body is about 70% water by mass, a great deal of energy is required to change its temperature by even 1 °C. Consequently, the mechanism for maintaining our body temperature at about 37 °C does not have to be as finely tuned as would be necessary if our bodies were primarily composed of a substance with a lower specific heat.
Example $1$: 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 $m$, $C_s$, and $ΔT$ into Equation $\ref{5.5.1}$.
Solution:
A The mass of water is
\begin{align*} \text{mass of } \ce{H2O} &= 400 \; \cancel{L}\left ( \dfrac{1000 \; \cancel{mL}}{1 \; \cancel{L}} \right ) \left ( \dfrac{0.998 \; g}{1 \; \cancel{mL}} \right ) \[4pt] &= 3.99\times 10^{5}g\; \ce{H2O} \end{align*} \nonumber
The temperature change ($ΔT$) is
$38.0 ^oC − 22.0 ^oC = +16.0^oC. \nonumber$
B From Table $1$, the specific heat of water is 4.184 J/(g•°C). From Equation $\ref{5.5.3}$, the heat absorbed by the water is thus
\begin{align*} q &=mC_{s}\Delta T \[4pt] &= \left ( 3.99 \times 10^{5} \; \cancel{g} \right )\left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right ) \left ( 16.0 \; \cancel{^{o}C} \right ) \[4pt] &= 2.67 \times 10^{7}\,J = 2.67 \times 10^{4}\,kJ \end{align*} \nonumber
Both $q$ and $ΔT$ are positive, consistent with the fact that the water has absorbed energy.
Exercise $1$: 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 ($\ce{SiO2}$) in Table $1$.
Answer
$2.7 \times 10^4 \,kJ$
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:
$q_{cold} + q_{hot} = 0 \label{5.5.4}$
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:
$q_{cold} = −q_{hot} \label{5.5.5}$
Thus heat is conserved in any such process, consistent with the law of conservation of energy.
Equation $\ref{5.5.5}$ argues that the amount of heat lost by a warmer object equals the amount of heat gained by a cooler object.
Substituting for q from Equation $\ref{5.5.2}$ into Equation $\ref{5.5.5}$ gives
$\left [ mC_{s} \Delta T \right ] _{cold} + \left [ mC_{s} \Delta T \right ] _{hot}=0 \label{5.5.6}$
which can be rearranged to give
$\left [ mC_{s} \Delta T \right ] _{cold} = - \left [ mC_{s} \Delta T \right ] _{hot} \label{5.5.7}$
When two objects initially at different temperatures are placed in contact, we can use Equation $\ref{5.5.7}$ 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 $\ref{5.5.6}$ and writing $ΔT$ as $T_{final} − T_{initial}$ for both the copper and the water, substitute the appropriate values of $m$, $C_s$, and $T_{initial}$ into the equation and solve for $T_{final}$.
Solution
We can adapt Equation $\ref{5.5.6}$ to solve this problem, remembering that $ΔT$ is defined as $T_{final} − T_{initial}$:
$\left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0 \nonumber$
Substituting the data provided in the problem and Table $1$ gives
\newcommand{\celsius}{\,^\circ\text{C}} \begin{align*} \left [ \left (30 \; g \right ) \left (0.385 \; J \right ) \left (T_{final} - 80.0 \celsius \right ) \right ] _{Cu} + \left [ (100\,g) (4.18\, J/\celsius) \left (T_{final} - 27.0 \celsius \right ) \right ] _{\ce{H2O}} &=0 \[4pt] \left[T_{final}\left ( 11.6 \; J/ \celsius \right ) -924 \; J \right] + \left[ T_{final}\left ( 418.4 \; J/\celsius \right ) -11,300 \; J \right]&= 0 \[4pt] T_{final}\left ( 430 \; J/\left ( g \cdot \celsius \right ) \right ) &= 12,224 \; J \[4pt] T_{final} &= 28.4 \; \celsius \end{align*} \nonumber
It is expected that the water will increase in temperature since the added copper pipe was hotter. However, the amount of the increase is not great and that is because there is more mass of water (three-fold), but mostly because the greater specific heat of water (order of magnitude) vs. copper.
Exercise $\PageIndex{2A}$: 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 $\PageIndex{2B}$: 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
In Example $1$, radiant energy from the sun was used to raise the temperature of water. A calorimetric experiment uses essentially the same procedure, except that 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 ($q_{rxn} < 0$), then heat is absorbed by the calorimeter ($q_{calorimeter} > 0$) and its temperature increases. Conversely, if the reaction absorbs heat ($q_{rxn} > 0$), then heat is transferred from the calorimeter to the system ($q_{calorimeter} < 0$) 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.
Conservation of Energy: The Movement of Heat between Substances:
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). Because the heat released or absorbed at constant pressure is equal to $ΔH$, the relationship between heat and $ΔH_{rxn}$ is
$\Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mC_{s} \Delta T \label{5.5.8}$
The use of a constant-pressure calorimeter is illustrated in Example $3$.
Example $3$: Heat of solution for Potassium Hydroxide
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 $ΔH_{soln}$ (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 \ref{5.5.8}.
3. Use the molar mass of $\ce{KOH}$ to calculate $ΔH_{soln}$.
Solution:
A To calculate ΔHsoln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is
$\left (100.0 \; \cancel{mL}\; \ce{H2O} \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; \ce{KOH}=104.72 \; g \nonumber$
The temperature change is therefor
$(34.7\, ^oC − 23.0 \,^oC) = +11.7\, ^oC. \nonumber$
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
\begin{align*} q_{calorimater}&=mC_{s} \Delta T \[4pt] &=\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right ) \[4pt] &=5130 \; J = 5.13 \; kJ \end{align*} \nonumber
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 $\ce{KOH}$ dissolving in water. From Equation $\ref{5.5.1}$, we see that
$ΔH_{rxn} = −q_{calorimeter} = −5.13\, kJ \nonumber$
This experiment tells us that dissolving 5.03 g of $\ce{KOH}$ in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of $\ce{KOH}$ in water must be exothermic.
C The last step is to use the molar mass of $\ce{KOH}$ to calculate $ΔH_{soln}$, i.e., the heat released when dissolving 1 mol of $\ce{KOH}$:
\begin{align*} \Delta H_{soln} &= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right ) \[4pt] &=-57.2 \; kJ/mol \end{align*} \nonumber
Exercise $3$: Heat of solution for ammonium bromide
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 $ΔH_{soln}$ for $\ce{NH4Br }$ (in kilojoules per mole).
Answer
16.6 kJ/mol
Conservation of Energy: Coffee Cup Calorimetry: Conservation of Energy: Coffee Cup Calorimetry, YouTube(opens in new window) [youtu.be] (opens in new window)
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 $ΔH_{comb}$ is given in Equation $\ref{5.5.9}$, where Cbomb is the total heat capacity of the steel bomb and the water surrounding it:
$\Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{5.5.9}$
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 ($\ce{C6H5CO2H}$) 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 $ΔH_{comb} = −26.38\, kJ/g$). This value and the measured increase in temperature of the calorimeter can be used in Equation $\ref{5.5.9}$ to determine $C_{bomb}$. The use of a bomb calorimeter to measure the $ΔH_{comb}$ of a substance is illustrated in Example $4$.
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 $ΔH_{comb}$ of glucose?
Given: mass and $ΔT$ for combustion of standard and sample
Asked for: $ΔH_{comb}$ of glucose
Strategy:
1. Calculate the value of $q_{rxn}$ for benzoic acid by multiplying the mass of benzoic acid by its $ΔH_{comb}$. Then use Equation $ref{5.5.9}$ to determine the heat capacity of the calorimeter ($C_{bomb}$) from $q_{comb}$ 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 $\ref{5.5.9}$ 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:
\begin{align*} q_{comb} &= \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) \[4pt] &= - 15.3 \; kJ \end{align*} \nonumber
From Equation $\ref{5.5.9}$,
\begin{align*} -C_{bomb} &= \dfrac{q_{comb}}{\Delta T} \[4pt] &= \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} \[4pt] &=- 7.34 \; kJ/^{o}C \end{align*} \nonumber
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:
\begin{align*} q_{comb} &=-C_{bomb}\Delta T \[4pt] &= \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right ) \[4pt] &=- 26.7 \; kJ \end{align*} \nonumber
Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the ΔHcomb of glucose is
\begin{align*} \Delta H_{comb} &=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right ) \[4pt] &= -2780 \; kJ/mol \[4pt] &=2.78 \times 10^{3} \; J/mol \end{align*} \nonumber
This result is in good agreement (< 1% error) with the value of $ΔH_{comb} = −2803\, kJ/mol$ 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: Conservation of Energy: Bomb Calorimetry, YouTube(opens in new window) [youtu.be]
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. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.05%3A_Calorimetry.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.
Hess's Law argues that ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This is nothing more than arguing that ΔH is a state function.
We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation $\ref{5.6.1}$ can be viewed as occurring in three distinct steps with known ΔH values. As shown in Figure 5.6.1, the first reaction produces 1 mol of solid aluminum oxide (Al2O3) and 2 mol of liquid iron at its melting point of 1758°C (part (a) in Equation $\ref{5.6.1}$); the enthalpy change for this reaction is −732.5 kJ/mol of Fe2O3. The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (part (b) in Equation 5.6.1); the enthalpy change for this reaction is −13.8 kJ/mol of Fe (−27.6 kJ per 2 mol Fe). In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (part (c) in Equation $\ref{5.6.1}$); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol Fe). As you can see in Figure 5.6.1, 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 parts (a), (b), and (c) in Equation 5.6.1 gives the overall reaction, shown in part (d):
\small \newcommand{\Celsius}{^{\circ}\text{C}} \begin{align*} \ce{2 Al (s, 25 \Celsius) + Fe2O3 (s, 25 \Celsius) &-> 2 Fe (l, 1758 \Celsius) + Al2O3 (s, 1758 \Celsius)} & \Delta H = - 732.5\,\text{kJ} && \text{(a)} \ \ce{2 Fe (l, 1758 \Celsius) &-> 2 Fe (s, 1758 \Celsius)} & \Delta H = -\phantom{0}27.6\,\text{kJ} && \text{(b)} \ \ce{2 Fe (s, 1758 \Celsius) + Al2O3 (s, 1758 \Celsius) &-> 2 Fe (s, 25 \Celsius) + Al2O3 (s, 25 \Celsius) } & \Delta H = -\phantom{0}91.0\,\text{kJ} && \text{(c)} \[2ex] \hline \ce{2 Al (s, 25 \Celsius) + Fe2O3 (s, 25 \Celsius) &-> Al2O3 (s, 25 \Celsius) + 2 Fe (s, 25 \Celsius) } & \Delta H = -851.1\,\text{kJ} && \text{(d)} \ \end{align*} \label{5.6.1} \tag{5.6.1}
The net reaction in part (d) in Equation $\ref{5.6.1}$ is identical to the equation for the thermite reaction that we saw in a previous section. 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 $\ref{5.6.1}$ 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 $\ref{5.6.1}$), 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 $\ref{5.6.1}$). 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 (O2), carbon monoxide (CO) is the main product:
$\left ( 1 \right ) \; \ce{2C (s) + O2 (g) -> 2 CO (g)} \quad \Delta H=-221.0 \; \text{kJ} \nonumber$
When carbon is burned in excess O2, carbon dioxide (CO2) is produced:
$\left ( 2 \right ) \; \ce{C (s) + O2 (g) -> CO2 (g)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber$
Use this information to calculate the enthalpy change per mole of CO for the reaction of CO with O2 to give 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:
$\left ( 3 \right ) \; \ce{CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H_{rxn}=? \nonumber$
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 CO, a reactant in Equation 3, is a product in Equation 1, we can reverse Equation (1) to give
$\ce{2 CO (g) -> 2 C (s) + O2 (g)} \quad \Delta H=+221.0 \; \text{kJ} \nonumber$
Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use Equation 2 as written because its product, CO2, is the product we want in Equation 3:
$\ce{C (s) + O2 (g) -> CO2 (s)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber$
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:
$\ce{2 C (s) + 2 O2 (g) -> 2 CO2 (s)} \quad \Delta H=-787.0 \; \text{kJ} \nonumber$
Writing the resulting equations as a sum, along with the enthalpy change for each, gives
\begin{align*} \ce{2 CO (g) &-> \cancel{2 C(s)} + \cancel{O_2 (g)} } & \Delta H & = -\Delta H_1 = +221.0 \; \text{kJ} \ \ce{\cancel{2 C (s)} + \cancel{2} O2 (g) &-> 2 CO2 (g)} & \Delta H & = -2\Delta H_2 =-787.0 \; \text{kJ} \[2ex] \hline \ce{2 CO (g) + O2 (g) &-> 2 CO2 (g)} & \Delta H &=-566.0 \; \text{kJ} \end{align*} \nonumber
Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of CO with O2, and the problem asks for the amount per mole of CO. Consequently, we must divide both sides of the final equation and the magnitude of ΔH by 2:
$\ce{ CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H = -283.0 \; \text{kJ} \nonumber$
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:
\begin{alignat*}{3} \text{(A)} \quad && \ce{ 2 C (s) + O2 (g) &-> \cancel{2 CO (g)}} \qquad & \Delta H_A &= \Delta H_1 &&= + 221.0 \; \text{kJ} \ \text{(B)} \quad && \ce{ \cancel{2 CO (g)} + O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H_B && &= ? \ \text{(C)} \quad && \ce{2 C (s) + 2 O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H &= 2 \Delta H_2 &= 2 \times \left ( -393.5 \; \text{kJ} \right ) &= -787.0 \; \text{kJ} \end{alignat*}
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
$\begin{matrix} -221.0 \; kJ + \Delta H_{B} = -787.0 \; kJ \ \Delta H_{B} = -566.0 \end{matrix} \nonumber$
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
Hess’s Law: Hess's Law, YouTube(opens in new window) [youtu.be]
Summary
Hess's law is arguing 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 (ΔHrxn) 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 ΔHrxn. 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 (ΔHfus) and the enthalpy of vaporization (ΔHvap), 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 (ΔHcomb) is the enthalpy change that occurs when a substance is burned in excess oxygen. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.06%3A_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_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.07%3A_Enthalpies_of_Formation.txt |
Learning Objectives
• To understand the relationship between thermochemistry and nutrition.
The thermochemical quantities that you probably encounter most often are the caloric values of food. Food supplies the raw materials that your body needs to replace cells and the energy that keeps those cells functioning. About 80% of this energy is released as heat to maintain your body temperature at a sustainable level to keep you alive. The nutritional Calorie (with a capital C) that you see on food labels is equal to 1 kcal (kilocalorie). The caloric content of food is determined from its enthalpy of combustion (ΔHcomb) per gram, as measured in a bomb calorimeter, using the general reaction
$\ce{food + excess\ O2 (g) -> CO2 (g) + H2O (l) + N2 (g)} \label{5.8.1}$
There are two important differences, however, between the caloric values reported for foods and the ΔHcomb of the same foods burned in a calorimeter. First, the ΔHcomb described in joules (or kilojoules) are negative for all substances that can be burned. In contrast, the caloric content of a food is always expressed as a positive number because it is stored energy. Therefore,
$\mathrm{caloric\ content} = - \Delta H_\mathrm{comb} \label{5.8.2}$
Second, when foods are burned in a calorimeter, any nitrogen they contain (largely from proteins, which are rich in nitrogen) is transformed to N2. In the body, however, nitrogen from foods is converted to urea [(H2N)2C=O], rather than N2 before it is excreted. The ΔHcomb of urea measured by bomb calorimetry is −632.0 kJ/mol. Consequently, the enthalpy change measured by calorimetry for any nitrogen-containing food is greater than the amount of energy the body would obtain from it. The difference in the values is equal to the ΔHcomb of urea multiplied by the number of moles of urea formed when the food is broken down. This point is illustrated schematically in the following equations:
$\ce{food + excess\ O2 (g) ->[\Delta H_{1} < 0] CO2 (g) + H2O (l) + \cancel{(H2N)2C=O (s) }} \nonumber$
$\ce{ \cancel{(H2N)2C=O (s)} +3/2 O2 (g) ->[\Delta H_{2} = 632.0 \mathrm{kJ/mol}] CO2 (g) + 2H2O (l) + N2 (g)} \nonumber$
which adds up to
$\ce{food + excess\ O2 (g) ->[\Delta H_{3}=\Delta H_{1}+\Delta H_{2} < 0] 2 CO2 (g) + 3 H2O (l) + N2 (g)} \nonumber$
All three ΔH values are negative, and, by Hess’s law, ΔH3 = ΔH1 + ΔH2. The magnitude of ΔH1 must be less than ΔH3, the calorimetrically measured ΔHcomb for a food. By producing urea rather than N2, therefore, humans are excreting some of the energy that was stored in their food.
Because of their different chemical compositions, foods vary widely in caloric content. As we saw previously, for instance, a fatty acid such as palmitic acid produces about 39 kJ/g during combustion, while a sugar such as glucose produces 15.6 kJ/g. Fatty acids and sugars are the building blocks of fats and carbohydrates, respectively, two of the major sources of energy in the diet. Nutritionists typically assign average values of 38 kJ/g (about 9 Cal/g) and 17 kJ/g (about 4 Cal/g) for fats and carbohydrates, respectively, although the actual values for specific foods vary because of differences in composition. Proteins, the third major source of calories in the diet, vary as well. Proteins are composed of amino acids, which have the following general structure:
In addition to their amine and carboxylic acid components, amino acids may contain a wide range of other functional groups: R can be hydrogen (–H); an alkyl group (e.g., –CH3); an aryl group (e.g., –CH2C6H5); or a substituted alkyl group that contains an amine, an alcohol, or a carboxylic acid (Figure $1$). Of the 20 naturally occurring amino acids, 10 are required in the human diet; these 10 are called essential amino acids because our bodies are unable to synthesize them from other compounds. Because R can be any of several different groups, each amino acid has a different value of ΔHcomb. Proteins are usually estimated to have an average ΔHcomb of 17 kJ/g (about 4 Cal/g).
Example $1$
Calculate the amount of available energy obtained from the biological oxidation of 1.000 g of alanine (an amino acid). Remember that the nitrogen-containing product is urea, not N2, so biological oxidation of alanine will yield less energy than will combustion. The value of ΔHcomb for alanine is −1577 kJ/mol.
Given: amino acid and ΔHcomb per mole
Asked for: caloric content per gram
Strategy:
1. Write balanced chemical equations for the oxidation of alanine to CO2, H2O, and urea; the combustion of urea; and the combustion of alanine. Multiply both sides of the equations by appropriate factors and then rearrange them to cancel urea from both sides when the equations are added.
2. Use Hess’s law to obtain an expression for ΔH for the oxidation of alanine to urea in terms of the ΔHcomb of alanine and urea. Substitute the appropriate values of ΔHcomb into the equation and solve for ΔH for the oxidation of alanine to CO2, H2O, and urea.
3. Calculate the amount of energy released per gram by dividing the value of ΔH by the molar mass of alanine.
Solution:
The actual energy available biologically from alanine is less than its ΔHcomb because of the production of urea rather than N2. We know the ΔHcomb values for alanine and urea, so we can use Hess’s law to calculate ΔH for the oxidation of alanine to CO2, H2O, and urea.
A We begin by writing balanced chemical equations for (1) the oxidation of alanine to CO2, H2O, and urea; (2) the combustion of urea; and (3) the combustion of alanine. Because alanine contains only a single nitrogen atom, whereas urea and N2 each contain two nitrogen atoms, it is easier to balance Equations 1 and 3 if we write them for the oxidation of 2 mol of alanine:
$\left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+6O_{2}\left ( g \right ) \rightarrow 5CO_{2}\left ( g \right )+5H_{2}O\left ( l \right )+\left ( H_{2}N \right )_{2}C=O\left ( s \right ) \nonumber$
$\left ( 2 \right ) \; \; \left( H_{2}N \right )_{2}C=O\left ( s \right ) + \dfrac{3}{2}O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )+ N_{2}\left ( g \right ) \nonumber$
$\left ( 3 \right ) \; \; \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+\dfrac{15}{2}O_{2}\left ( g \right )\rightarrow 6CO_{2}\left ( g \right )+7H_{2}O\left ( l \right )+N_{2}\left ( g \right ) \nonumber$
Adding Equations 1 and 2 and canceling urea from both sides give the overall chemical equation directly:
$\left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+6O_{2}\left ( g \right ) \rightarrow 5CO_{2}\left ( g \right )+5H_{2}O\left ( l \right )+\cancel{\left ( H_{2}N \right )_{2}C=O\left ( s \right )} \nonumber$
$\cancel{ \left ( 2 \right ) \; \; \left( H_{2}N \right )_{2}C=O\left ( s \right )} + \dfrac{3}{2}O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )+ N_{2}\left ( g \right ) \nonumber$
$\left ( 3 \right ) \; \; \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+\dfrac{15}{2}O_{2}\left ( g \right )\rightarrow 6CO_{2}\left ( g \right )+7H_{2}O\left ( l \right )+N_{2}\left ( g \right ) \nonumber$
B By Hess’s law, ΔH3 = ΔH1 + ΔH2. We know that ΔH3 = 2ΔHcomb (alanine), ΔH2 = ΔHcomb (urea), and ΔH1 = 2ΔH (alanine → urea). Rearranging and substituting the appropriate values gives
$\Delta {H_{1}} = \Delta {H_{3}} - \Delta {H_{2}} \nonumber$
$=2\left ( -1577 \; kJ/mol \right )-\left ( -632.0 \; kJ/mol \right ) \nonumber$= -2522 \; kJ/\left ( 2 \;mol\; analine \right ) \nonumber \]
Thus ΔH (alanine → urea) = −2522 kJ/(2 mol of alanine) = −1261 kJ/mol of alanine. Oxidation of alanine to urea rather than to nitrogen therefore results in about a 20% decrease in the amount of energy released (−1261 kJ/mol versus −1577 kJ/mol).
C The energy released per gram by the biological oxidation of alanine is
$\left (\dfrac{-1261 \; kJ}{1 \; \cancel{mol}} \right )\left (\dfrac{1 \; \cancel{mol}}{89.094 \; g} \right )= -14.15 \; kJ/g \nonumber$
This is equal to −3.382 Cal/g.
Exercise $1$
Calculate the energy released per gram from the oxidation of valine (an amino acid) to CO2, H2O, and urea. Report your answer to three significant figures. The value of ΔHcomb for valine is −2922 kJ/mol.
Answer
−22.2 kJ/g (−5.31 Cal/g)
The reported caloric content of foods does not include ΔHcomb for those components that are not digested, such as fiber. Moreover, meats and fruits are 50%−70% water, which cannot be oxidized by O2 to obtain energy. So water contains no calories. Some foods contain large amounts of fiber, which is primarily composed of sugars. Although fiber can be burned in a calorimeter just like glucose to give carbon dioxide, water, and heat, humans lack the enzymes needed to break fiber down into smaller molecules that can be oxidized. Hence fiber also does not contribute to the caloric content of food.
Table $1$: Approximate Composition and Fuel Value of an 8 oz Slice of Roast Beef
Composition Calories
97.5 g of water × 0 Cal/g = 0
58.7 g of protein × 4 Cal/g = 235
69.3 g of fat × 9 Cal/g = 624
0 g of carbohydrates × 4 Cal/g = 0
1.5 g of minerals × 0 Cal/g = 0
Total mass: 227.0 g Total calories: about 900 Cal
We can determine the caloric content of foods in two ways. The most precise method is to dry a carefully weighed sample and carry out a combustion reaction in a bomb calorimeter. The more typical approach, however, is to analyze the food for protein, carbohydrate, fat, water, and “minerals” (everything that doesn’t burn) and then calculate the caloric content using the average values for each component that produces energy (9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals). An example of this approach is shown in Table $1$ for a slice of roast beef. The compositions and caloric contents of some common foods are given in Table $2$.
Table $2$: Approximate Compositions and Fuel Values of Some Common Foods
Food (quantity) Approximate Composition (%) Food Value (Cal/g) Calories
Water Carbohydrate Protein Fat
beer (12 oz) 92 3.6 0.3 0 0.4 150
coffee (6 oz) 100 ~0 ~0 ~0 ~0 ~0
milk (1 cup) 88 4.5 3.3 3.3 0.6 150
egg (1 large) 75 2 12 12 1.6 80
butter (1 tbsp) 16 ~0 ~0 79 7.1 100
apple (8 oz) 84 15 ~0 0.5 0.6 125
bread, white (2 slices) 37 48 8 4 2.6 130
brownie (40 g) 10 55 5 30 4.8 190
hamburger (4 oz) 54 0 24 21 2.9 326
fried chicken (1 drumstick) 53 8.3 22 15 2.7 195
carrots (1 cup) 87 10 1.3 ~0 0.4 70
Because the Calorie represents such a large amount of energy, a few of them go a long way. An average 73 kg (160 lb) person needs about 67 Cal/h (1600 Cal/day) to fuel the basic biochemical processes that keep that person alive. This energy is required to maintain body temperature, keep the heart beating, power the muscles used for breathing, carry out chemical reactions in cells, and send the nerve impulses that control those automatic functions. Physical activity increases the amount of energy required but not by as much as many of us hope (Table $2$). A moderately active individual requires about 2500−3000 Cal/day; athletes or others engaged in strenuous activity can burn 4000 Cal/day. Any excess caloric intake is stored by the body for future use, usually in the form of fat, which is the most compact way to store energy. When more energy is needed than the diet supplies, stored fuels are mobilized and oxidized. We usually exhaust the supply of stored carbohydrates before turning to fats, which accounts in part for the popularity of low-carbohydrate diets.
Table $3$: Approximate Energy Expenditure by a 160 lb Person Engaged in Various Activities
Activity Cal/h
sleeping 80
driving a car 120
standing 140
eating 150
walking 2.5 mph 210
mowing lawn 250
swimming 0.25 mph 300
roller skating 350
tennis 420
bicycling 13 mph 660
running 10 mph 900
Example $2$
What is the minimum number of Calories expended by a 72.6 kg person who climbs a 30-story building? (Assume each flight of stairs is 14 ft high.) How many grams of glucose are required to supply this amount of energy? (The energy released during the combustion of glucose was calculated in Example 5.5.4).
Given: mass, height, and energy released by combustion of glucose
Asked for: calories expended and mass of glucose needed
Strategy:
1. Convert mass and height to SI units and then substitute these values into Equation 5.6 to calculate the change in potential energy (in kilojoules). Divide the calculated energy by 4.184 Cal/kJ to convert the potential energy change to Calories.
2. Use the value obtained in Example 5.5.4 for the combustion of glucose to calculate the mass of glucose needed to supply this amount of energy.
Solution:
The energy needed to climb the stairs equals the difference between the person’s potential energy (PE) at the top of the building and at ground level.
A Recall that PE = mgh. Because m and h are given in non-SI units, we must convert them to kilograms and meters, respectively
$PE = \left ( 72.6 \; kg \right )\left ( 9.81 \; m/s^{2} \right )\left ( 128 m \right ) = 8.55 × 10^{4} \left ( kg \cdot m^{2}/s^{2} \right ) = 91.2 kJ \nonumber$
To convert to Calories, we divide by 4.184 kJ/kcal:
$PE = \left ( 91.2 \; \cancel{kJ} \right ) \left ( \dfrac{1 \; kcal}{4.184 \; \cancel{kJ}} \right )=21.8 \;kcal = 21.8 \; Cal \nonumber$
B Because the combustion of glucose produces 15.6 kJ/g (Example 5), the mass of glucose needed to supply 85.5 kJ of energy is
$PE = \left ( 91.2 \; \cancel{kJ} \right )\left ( \dfrac{1 \; g \; glucose}{15.6 \; \cancel{kJ}} \right )=5.85\; g \; glucose \nonumber$
This mass corresponds to only about a teaspoonful of sugar! Because the body is only about 30% efficient in using the energy in glucose, the actual amount of glucose required would be higher: (100%/30%) × 5.85 g = 19.5 g. Nonetheless, this calculation illustrates the difficulty many people have in trying to lose weight by exercise alone.
Exercise $2$
Calculate how many times a 160 lb person would have to climb the tallest building in the United States, the 110-story Willis Tower in Chicago, to burn off 1.0 lb of stored fat. Assume that each story of the building is 14 ft high and use a calorie content of 9.0 kcal/g of fat.
Answer
About 55 times
The calculations in Example 5.8.2 ignore various factors, such as how fast the person is climbing. Although the rate is irrelevant in calculating the change in potential energy, it is very relevant to the amount of energy actually required to ascend the stairs. The calculations also ignore the fact that the body’s conversion of chemical energy to mechanical work is significantly less than 100% efficient. According to the average energy expended for various activities listed in Table 5.8.3, a person must run more than 4.5 h at 10 mph or bicycle for 6 h at 13 mph to burn off 1 lb of fat (1.0 lb × 454 g/lb × 9.0 Cal/g = 4100 Cal). But if a person rides a bicycle at 13 mph for only 1 h per day 6 days a week, that person will burn off 50 lb of fat in the course of a year (assuming, of course, the cyclist doesn’t increase his or her intake of calories to compensate for the exercise).
Summary
Thermochemical concepts can be applied to determine the actual energy available in food. The nutritional Calorie is equivalent to 1 kcal (4.184 kJ). The caloric content of a food is its ΔHcomb per gram. The combustion of nitrogen-containing substances produces N2(g), but the biological oxidation of such substances produces urea. Hence the actual energy available from nitrogen-containing substances, such as proteins, is less than the ΔHcomb of urea multiplied by the number of moles of urea produced. The typical caloric contents for food are 9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.08%3A_Foods_and_Fuels.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.
5.1: The Nature of Energy
Conceptual Problems
1. What is the relationship between mechanical work and energy?
2. Does a person with a mass of 50 kg climbing a height of 15 m do work? Explain your answer. Does that same person do work while descending a mountain?
3. If a person exerts a force on an immovable object, does that person do work? Explain your answer.
4. Explain the differences between electrical energy, nuclear energy, and chemical energy.
5. The chapter describes thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy. Which form(s) of energy are represented by each of the following?
1. sunlight
2. the energy produced by a cathode ray tube, such as that found in a television
3. the energy emitted from radioactivity
4. the energy emitted from a burning candle
5. the energy associated with a steam engine
6. the energy emitted by a cellular phone
7. the energy associated with a stick of dynamite
6. Describe the various forms of energy that are interconverted when a flashlight is switched on.
7. Describe the forms of energy that are interconverted when the space shuttle lifts off.
8. Categorize each of the following as representing kinetic energy or potential energy.
1. the energy associated with a laptop computer sitting on the edge of a desk
2. shoveling snow
3. water pouring out of a fire hydrant
4. the energy released by an earthquake
5. the energy in a volcano about to erupt
6. the energy associated with a coiled spring
9. Are the units for potential energy the same as the units for kinetic energy? Can an absolute value for potential energy be obtained? Explain your answer.
10. Categorize each of the following as representing kinetic energy or potential energy.
1. water cascading over Niagara Falls
2. a beaker balanced on the edge of a sink
3. the energy released during a mudslide
4. rollerblading
5. the energy in a block of ice on a rooftop before a thaw
11. Why does hammering a piece of sheet metal cause the metal to heat up?
Conceptual Answers
3. Technically, the person is not doing any work, since the object does not move.
11. The kinetic energy of the hammer is transferred to the metal.
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 4 before proceeding to the Numerical Problems.
1. Describe the mathematical relationship between (a) the thermal energy stored in an object and that object’s mass and (b) the thermal energy stored in an object and that object’s temperature.
2. How much energy (in kilojoules) is released or stored when each of the following occurs?
1. A 230 lb football player is lifted to a height of 4.00 ft.
2. An 11.8 lb cat jumps from a height of 6.50 ft.
3. A 3.75 lb book falls off of a shelf that is 5.50 ft high.
3. Calculate how much energy (in kilojoules) is released or stored when each of the following occurs:
1. A 130 lb ice skater is lifted 7.50 ft off the ice.
2. A 48 lb child jumps from a height of 4.0 ft.
3. An 18.5 lb light fixture falls from a 10.0 ft ceiling.
4. A car weighing 1438 kg falls off a bridge that is 211 ft high. Ignoring air resistance, how much energy is released when the car hits the water?
5. A 1 tn roller coaster filled with passengers reaches a height of 28 m before accelerating downhill. How much energy is released when the roller coaster reaches the bottom of the hill? Assume no energy is lost due to friction.
Numerical Answers
1.
1. The thermal energy content of an object is directly proportional to its mass.
2. The thermal energy content of an object is directly proportional to its temperature.
3.
1. 1.3 kJ stored
2. 0.26 kJ released
3. 0.251 kJ released
5.
1. 250 kJ released
5.2: The First Law of Thermodynamics
Conceptual Problems
1. Describe how a swinging pendulum that slows with time illustrates the first law of thermodynamics.
2. When air is pumped into a bicycle tire, the air is compressed. Assuming that the volume is constant, express the change in internal energy in terms of q and w.
3. What is the relationship between enthalpy and internal energy for a reaction that occurs at constant pressure?
4. An intrepid scientist placed an unknown salt in a small amount of water. All the salt dissolved in the water, and the temperature of the solution dropped several degrees.
1. What is the sign of the enthalpy change for this reaction?
2. Assuming the heat capacity of the solution is the same as that of pure water, how would the scientist calculate the molar enthalpy change?
3. Propose an explanation for the decrease in temperature.
1. For years, chemists and physicists focused on enthalpy changes as a way to measure the spontaneity of a reaction. What arguments would you use to convince them not to use this method?
1. What is the relationship between enthalpy and internal energy for a reaction that occurs at constant volume?
1. The enthalpy of combustion (ΔHcomb) is defined thermodynamically as the enthalpy change for complete oxidation. The complete oxidation of hydrocarbons is represented by the following general equation: hydrocarbon + O2(g) → CO2(g) + H2O(g). Enthalpies of combustion from reactions like this one can be measured experimentally with a high degree of precision. It has been found that the less stable the reactant, the more heat is evolved, so the more negative the value of ΔHcomb. In each pair of hydrocarbons, which member do you expect to have the greater (more negative) heat of combustion? Justify your answers.
1. cyclopropane or cyclopentane
2. butane or 2-methylpropane
3. hexane or cyclohexane
1. Using a structural argument, explain why the trans isomer of 2-butene is more stable than the cis isomer. The enthalpies of formation of cis- and trans-2-butene are −7.1 kJ/mol and −11.4 kJ/mol, respectively.
1. Using structural arguments, explain why cyclopropane has a positive ΔHf° (12.7 kJ/mol), whereas cyclopentane has a negative ΔHf° (−18.4 kJ/mol). (Hint: consider bond angles.)
Conceptual Answers
1. At constant pressure, ΔH = ΔU + PΔV.
2.
3. In some simple case, when a reaction only involves solids, liquids or liquid solution, we can say that $ΔH=ΔU$. When a reaction involves gases, we can say that $ΔU = ΔH − RTΔn$. However, if we have a complicated case, it is hard to use $ΔH$ to measure the spontaneity of a reaction. For example, when a reaction involved gases, liquid and solid, it is really hard for us to defined $Δ(PV)$.
1. With bond angles of 60°, cyclopropane is highly strained, causing it to be less stable than cyclopentane, which has nearly ideal tetrahedral geometry at each carbon atom.
Numerical Problems
1. A block of CO2 weighing 15 g evaporates in a 5.0 L container at 25°C. How much work has been done if the gas is allowed to expand against an external pressure of 0.98 atm under isothermal conditions? The enthalpy of sublimation of CO2 is 25.1 kJ/mol. What is the change in internal energy (kJ/mol) for the sublimation of CO2 under these conditions?
2. Zinc and HCl react according to the following equation:
$Zn_{(s)} + 2HCl_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Cl^−_{(aq)} + H_2(g)$
When 3.00 g of zinc metal is added to a dilute HCl solution at 1.00 atm and 25°C, and this reaction is allowed to go to completion at constant pressure, 6.99 kJ of heat must be removed to return the final solution to its original temperature. What are the values of q and w, and what is the change in internal energy?
1. Acetylene torches, used industrially to cut and weld metals, reach flame temperatures as high as 3000°C. The combustion reaction is as follows:
$2C_2H_{2\; (g)}+5O_{2\; (g)} \rightarrow 4CO_{2\; (g)} + 2H_2O_{(l)} \tag{ΔH=−2599 kJ}$
Calculate the amount of work done against a pressure of 1.0 atm when 4.0 mol of acetylene are allowed to react with 10 mol of O2 at 1.0 atm at 20°C. What is the change in internal energy for the reaction?
1. When iron dissolves in 1.00 M aqueous HCl, the products are FeCl2(aq) and hydrogen gas. Calculate the work done if 30 g of Fe react with excess hydrochloric acid in a closed vessel at 20°C. How much work is done if the reaction takes place in an open vessel with an external pressure of 1.0 atm?
Numerical Answers
1. −350 J; 8.2 kJ
5.4: Enthalpy of Reaction
Conceptual Problems
Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Conceptual Problems.
1. Heat implies the flow of energy from one object to another. Describe the energy flow in an
a. exothermic reaction.
b. endothermic reaction.
2. When a thermometer is suspended in an insulated thermos that contains a block of ice, the temperature recorded on the thermometer drops. Describe the direction of heat flow.
3. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. State whether each process is endothermic or exothermic.
1. Water is added to sodium hydroxide pellets, and the flask becomes hot.
2. The body metabolizes glucose, producing carbon dioxide and water.
3. Ammonium nitrate crystals are dissolved in water, causing the solution to become cool.
4. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. Determine whether each process is endothermic or exothermic.
1. Concentrated acid is added to water in a flask, and the flask becomes warm.
2. Water evaporates from your skin, causing you to shiver.
3. A container of ammonium nitrate detonates.
5. Is Earth’s environment an isolated system, an open system, or a closed system? Explain your answer.
6. Why is it impossible to measure the absolute magnitude of the enthalpy of an object or a compound?
7. Determine whether energy is consumed or released in each scenario. Explain your reasoning.
1. A leaf falls from a tree.
2. A motorboat maneuvers against a current.
3. A child jumps rope.
4. Dynamite detonates.
5. A jogger sprints down a hill.
8. The chapter states that enthalpy is an extensive property. Why? Describe a situation that illustrates this fact.
9. The enthalpy of a system is affected by the physical states of the reactants and the products. Explain why.
10. Is the distance a person travels on a trip a state function? Why or why not?
5.5: Calorimetry
Conceptual Problems
1. Can an object have a negative heat capacity? Why or why not?
2. What two factors determine the heat capacity of an object? Does the specific heat also depend on these two factors? Explain your answer.
3. Explain why regions along seacoasts have a more moderate climate than inland regions do.
4. Although soapstone is more expensive than brick, soapstone is frequently the building material of choice for fireplaces, particularly in northern climates with harsh winters. Propose an explanation for this.
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 9.9) before proceeding to the Numerical Problems.
1. Using Equation 9.6.2 and Equation 9.6.3, derive a mathematical relationship between Cs and Cp.
2. Complete the following table for 28.0 g of each element at an initial temperature of 22.0°C.
Element q (J) Cp [J/(mol·K)] Final T (°C)
nickel 137 26.07
silicon 19.789 3.0
zinc 603 77.5
mercury 137 57
3. Using Table 9.6.1, how much heat is needed to raise the temperature of a 2.5 g piece of copper wire from 20°C to 80°C? How much heat is needed to increase the temperature of an equivalent mass of aluminum by the same amount? If you were using one of these metals to channel heat away from electrical components, which metal would you use? Once heated, which metal will cool faster? Give the specific heat for each metal.
4. Gold has a molar heat capacity of 25.418 J/(mol·K), and silver has a molar heat capacity of 23.350 J/(mol·K).
1. If you put silver and gold spoons of equal mass into a cup of hot liquid and wait until the temperature of the liquid is constant, which spoon will take longer to cool down when removed from the hot liquid?
2. If 8.00 g spoons of each metal at 20.0°C are placed in an insulated mug with 50.0 g of water at 97.0°C, what will be the final temperature of the water after the system has equilibrated? (Assume that no heat is transferred to the surroundings.)
5. In an exothermic reaction, how much heat would need to be evolved to raise the temperature of 150 mL of water 7.5°C? Explain how this process illustrates the law of conservation of energy.
6. How much heat must be evolved by a reaction to raise the temperature of 8.0 oz of water 5.0°C? What mass of lithium iodide would need to be dissolved in this volume of water to produce this temperature change?
7. A solution is made by dissolving 3.35 g of an unknown salt in 150 mL of water, and the temperature of the water rises 3.0°C. The addition of a silver nitrate solution results in a precipitate. Assuming that the heat capacity of the solution is the same as that of pure water, use the information in Table 9.5.1 and solubility rules to identify the salt.
8. Using the data in Table 9.8.2, calculate the change in temperature of a calorimeter with a heat capacity of 1.78 kJ/°C when 3.0 g of charcoal is burned in the calorimeter. If the calorimeter is in a 2 L bath of water at an initial temperature of 21.5°C, what will be the final temperature of the water after the combustion reaction (assuming no heat is lost to the surroundings)?
9. A 3.00 g sample of TNT (trinitrotoluene, C7H5N3O6) is placed in a bomb calorimeter with a heat capacity of 1.93 kJ/°C; the ΔHcomb of TNT is −3403.5 kJ/mol. If the initial temperature of the calorimeter is 19.8°C, what will be the final temperature of the calorimeter after the combustion reaction (assuming no heat is lost to the surroundings)? What is the ΔHf of TNT?
Answers
1. Cp = Cs × (molar mass)
2.
3. For Cu: q = 58 J; For Al: q = 130 J; Even though the values of the molar heat capacities are very similar for the two metals, the specific heat of Cu is only about half as large as that of Al, due to the greater molar mass of Cu versus Al: Cs = 0.385 and 0.897 J/(g•K) for Cu and Al, respectively. Thus loss of one joule of heat will cause almost twice as large a decrease in temperature of Cu versus Al.
4.
5. 4.7 kJ
6.
7. ΔHsoln = −0.56 kJ/g; based on reaction with AgNO3, salt contains halide; dividing ΔHsoln values in Table 5.2 by molar mass of salts gives lithium bromide as best match, with −0.56 kJ/g.
8.
9. Tfinal = 43.1°C; the combustion reaction is $4C_7H_5N_3O_{6(s)} + 21O_{2(g)} \rightarrow 28CO_{2(g)} + 10H_2O_{(g)} + 6N_{2(g)}$ with $Δ_f^οH (TNT) = −65.5\; kJ/mol$
5.6: Hess's Law
Conceptual Problems
Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Conceptual Problems.
1. Based on the following energy diagram,
a. write an equation showing how the value of ΔH2 could be determined if the values of ΔH1 and ΔH3 are known.
b. identify each step as being exothermic or endothermic.
2. Based on the following energy diagram,
a. write an equation showing how the value of ΔH3 could be determined if the values of ΔH1 and ΔH2 are known.
b. identify each step as being exothermic or endothermic.
3. Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly.
4. When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state?
1. the melting of a solid
2. the conversion of a gas to a liquid
3. the solidification of a liquid
4. the dissolution of a solid into water
5. In their elemental form, A2 and B2 exist as diatomic molecules. Given the following reactions, each with an associated ΔH°, describe how you would calculate ΔHof for the compound AB2.
$\begin{matrix} 2AB & \rightarrow & A_{2} + B _{2} & \Delta H_{1}^{o}\ 3AB & \rightarrow & AB_{2} + A _{2}B & \Delta H_{2}^{o} \ 2A_{2}B &\rightarrow & 2A_{2} + B _{2} & \Delta H_{3}^{o} \end{matrix}$
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 4 before proceeding to the Numerical Problems.
1. Methanol is used as a fuel in Indianapolis 500 race cars. Use the following table to determine whether methanol or 2,2,4-trimethylpentane (isooctane) releases more energy per liter during combustion.
Fuel ΔHocombustion(kJ/mol) Density (g/mL)
methanol −726.1 0.791
2,2,4-trimethylpentane −5461.4 0.692
2.
a. Use the enthalpies of combustion given in the following table to determine which organic compound releases the greatest amount of energy per gram during combustion.
Fuel ΔHocombustion(kJ/mol)
methanol −726.1
1-ethyl-2-methylbenzene −5210.2
n-octane −5470.5
b. Calculate the standard enthalpy of formation of 1-ethyl-2-methylbenzene.
3. Given the enthalpies of combustion, which organic compound is the best fuel per gram?
Fuel ΔHof(kJ/mol)
ethanol −1366.8
benzene −3267.6
cyclooctane −5434.7
Numerical Answers
1.
2.
a. To one decimal place
methanol: ΔH/g = −22.6 kJ
C9H12: ΔH/g = −43.3 kJ
octane: ΔH/g = −47.9 kJ
Octane provides the largest amount of heat per gram upon combustion.
b, ΔHf(C9H17) = −46.1 kJ/mol
5.7: Enthalpies of Formation
Conceptual Problems
Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Conceptual Problems.
1. Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly.
2. When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state?
3. What is the difference between ΔHof and ΔHf?
4. How can ΔHof of a compound be determined if the compound cannot be prepared by the reactions used to define its standard enthalpy of formation?
5. For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
1. HBr
2. CH3OH
3. NaHCO3
6. Describe the distinction between ΔHsoln and ΔHf.
7. The following table lists ΔHosoln values for some ionic compounds. If 1 mol of each solute is dissolved in 500 mL of water, rank the resulting solutions from warmest to coldest.
Compound ΔHosoln(kJ/mol)
KOH −57.61
LiNO3 −2.51
KMnO4 43.56
NaC2H3O2 −17.32
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Numerical Problems.
1. Using "Appendix A, calculate ΔHorxn for each chemical reaction.
a. 2Mg(s) + O2(g) → 2MgO(s)
b. CaCO3(s, calcite) → CaO(s) + CO2(g)
c. AgNO3(s) + NaCl(s) → AgCl(s) + NaNO3(s)
2. Using "Appendix A, determine ΔHorxn for each chemical reaction.
a. 2Na(s) + Pb(NO3)2(s) → 2NaNO3(s) + Pb(s)
b. Na2CO3(s) + H2SO4(l) → Na2SO4(s) + CO2(g) + H2O(l)
c. 2KClO3(s) → 2KCl(s) + 3O2(g)
3. Calculate ΔHorxn for each chemical equation. If necessary, balance the chemical equations.
a. Fe(s) + CuCl2(s) → FeCl2(s) + Cu(s)
b. (NH4)2SO4(s) + Ca(OH)2(s) → CaSO4(s) + NH3(g) + H2O(l)
c. Pb(s) + PbO2(s) + H2SO4(l) → PbSO4(s) + H2O(l)
4. Calculate ΔHorxn for each reaction. If necessary, balance the chemical equations.
a. 4HBr(g) + O2(g) → 2H2O(l) + 2Br2(l)
b. 2KBr(s) + H2SO4(l) → K2SO4(s) + 2HBr(g)
c. 4Zn(s) + 9HNO3(l) → 4Zn(NO3)2(s) + NH3(g) + 3H2O(l)
5. Use the data in "Appendix A to calculate ΔHof for the reaction Sn(s, white) + 4HNO3(l) → SnO2(s) + 4NO2(g) + 2H2O(l).
6. Use the data in "Appendix A to calculate ΔHof for the reaction P4O10(s) + 6H2O(l) → 4H3PO4(l).
7. How much heat is released or required in the reaction of 0.50 mol of HBr(g) with 1.0 mol of chlorine gas to produce bromine gas?
8. How much energy is released or consumed if 10.0 g of N2O5 is completely decomposed to produce gaseous nitrogen dioxide and oxygen?
9. In the mid-1700s, a method was devised for preparing chlorine gas from the following reaction:
NaCl(s) + H2SO4(l) + MnO2(s) → Na2SO4(s) + MnCl2(s) + H2O(l) + Cl2(g)
Calculate ΔHorxn for this reaction. Is the reaction exothermic or endothermic?
10. Would you expect heat to be evolved during each reaction?
1. solid sodium oxide with gaseous sulfur dioxide to give solid sodium sulfite
2. solid aluminum chloride reacting with water to give solid aluminum oxide and hydrogen chloride gas
11. How much heat is released in preparing an aqueous solution containing 6.3 g of calcium chloride, an aqueous solution containing 2.9 g of potassium carbonate, and then when the two solutions are mixed together to produce potassium chloride and calcium carbonate.
Numerical Answers
1.
a. −1203 kJ/mol O2
b. 179.2 kJ
c. −59.3 kJ
2.
3.
4.
5. −174.1 kJ/mol
6.
7. −20.3 kJ
8.
9. −34.3 kJ/mol Cl2; exothermic
10.
11. ΔH = −2.86 kJ CaCl2: −4.6 kJ; K2CO3, −0.65 kJ; mixing, −0.28 kJ
5.8: Foods and Fuels
Conceptual Problems
1. Can water be considered a food? Explain your answer.
2. Describe how you would determine the caloric content of a bag of popcorn using a calorimeter.
3. Why do some people initially feel cold after eating a meal and then begin to feel warm?
4. In humans, one of the biochemical products of the combustion/digestion of amino acids is urea. What effect does this have on the energy available from these reactions? Speculate why conversion to urea is preferable to the generation of N2.
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 9.9 ) before proceeding to the Numerical Problems.
1. Determine the amount of energy available from the biological oxidation of 1.50 g of leucine (an amino acid, ΔHcomb = −3581.7 kJ/mol).
2. Calculate the energy released (in kilojoules) from the metabolism of 1.5 oz of vodka that is 62% water and 38% ethanol by volume, assuming that the total volume is equal to the sum of the volume of the two components. The density of ethanol is 0.824 g/mL. What is this enthalpy change in nutritional Calories?
3. While exercising, a person lifts an 80 lb barbell 7 ft off the ground. Assuming that the transformation of chemical energy to mechanical energy is only 35% efficient, how many Calories would the person use to accomplish this task? From Figure 9.4.2, how many grams of glucose would be needed to provide the energy to accomplish this task?
4. A 30 g sample of potato chips is placed in a bomb calorimeter with a heat capacity of 1.80 kJ/°C, and the bomb calorimeter is immersed in 1.5 L of water. Calculate the energy contained in the food per gram if, after combustion of the chips, the temperature of the calorimeter increases to 58.6°C from an initial temperature of 22.1°C. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.E%3A_Thermochemistry_%28Exercises%29.txt |
• thermodynamics – study of energy and its transformations
• thermochemistry – relationships between chemical reactions and energy changes
5.1: The Nature of Energy
• Force: symbolized by F, any kind of push or pull exerted on an object
• Work: symbolized by w, movement of an object against some force
$w= F \times d \nonumber$
• where w is work, F is force, and d is distance that the object is being moved
• Heat: energy that is transferred from one object to another because of difference in temperature
• Energy: the capacity to do work or to transfer heat
Kinetic and Potential Energy
Kinetic energy: energy of motion
Ek = ½ mv2 where Ek is the kinetic energy, m is the mass, and v is the velocity.
Potential energy: energy that an object possesses as a result of its composition or its position with respect to another object
Energy Units
• Joule: SI unit for energy, 1J = kgm2/s2
• Calorie: abbreviated cal, originally defined as the amount of energy required to raise the temperature of 1 g of water by 1°C. 1 cal = 4.184 J
Systems and Surroundings
• System: portion singled out for study
• Surroundings: everything beyond the system
Systems exchanges energy with surroundings in the form of heat and work, but not matter
Lowering the Energy of the System
Systems tend to attain as low an energy as possible. Systems with a high potential energy are less stable and more likely to undergo change than systems with a low potential energy.
5.2: The First Law of Thermodynamics
First law of thermodynamics: aka The Law of Conservation of Energy; a statement of our experience that energy is conserved in any process. We can express the first law in many ways. One of the more useful expressions is that the change in internal energy, ∆E, of a system in any process is equal to the heat, q, added to the system, plus the work, w, done on the system by its surroundings: ∆E = q + w.
Internal Energy
The total energy of a system is the sum of all the kinetic and potential energies of its components parts.
Internal energy: total energy of the system (see above)
Because there are so many types of motion and interactions, we cannot determine the exact energy of any system. However, we can measure the changes in internal energy that accompany chemical and physical processes.
?E = EfinalEinitial
Efinal > Einitial = system has gained energy
Efinal < Einitial = system has lost energy
Initial state refers to reactants, final state to products.
Relating ∆E to Heat and Work
The internal energy of a system changes in magnitude as heat is added to or removed from the system, or as work is done on it or by it.
?E = q + w where ∆E is internal energy, q is heat, and w is work
Heat added to the system is assigned a positive sign. Likewise, work on the system is positive. On the other hand, both the heat lost by the system and the work done by the system on its surroundings are negative; they reduce the internal energy.
Ex. A system absorbs 50 J of heat and does 10 J of work on its surroundings, q = 50 J and w = 10 J. Thus, ∆E = 50 J + (10 J) = 40 J.
State Functions
The total energy of a system is proportional to the total quantity of matter in the system; energy is an extensive property.
State function: a property of a system that is determined by the state or condition of the system and not by how it got to that state; its value is fixed when temperature, pressure, composition, and physical form are specified. The internal energy of a system is a state function.
The value of a state function does not depend on the particular history of the sample, only on its present condition. Because E is a state function, ∆E depends only on the initial and final states of the system and not on how the change occurs.
However, the work done by a system in a given process is not a state function!!!
5.3: Enthalpy
Heat flows between system and surroundings until both are at the same temperature.
Endothermic process: a process in which a system absorbs heat from its surroundings
Exothermic process: a process in which a system releases heat to its surroundings
Enthalpy
Enthalpy: represented by H; deals with the amount of heat absorbed or released during a chemical reaction under constant pressure
The change in enthalpy, ∆H, equals the heat, qp, added to or lost by the system when the process occurs under constant pressure:
?H = qp
(The subscript P on q is a reminder that we are considering a special case where pressures is constant)
∆H = HfinalHinitial
Hfinal > Hinitial = positive ∆H; system has gained heat from surroundings (endothermic)
Hfinal < Hinitial = negative ∆H; system has lost heat to the surroundings (exothermic)
5.4: Enthalpy of Reaction
∆H = H(products) – H(reactants)
Thermochemical equations: balanced chemical equations that show the associated enthalpy change
1. Enthalpy is an extensive property (depends on amount of matter present). This fact means that the magnitude of ∆H is directly proportional to the amount of reactant consumed in the process
2. The enthalpy change for a reaction is equal in magnitude but opposite in sign to ∆H for the reverse reaction. When we reverse a reaction, the reactants become the products, and vice versa. Reversing the products and reactants leads to the same magnitude but a change in sign for ∆H.
3. The enthalpy change for a reaction depends on the state of the reactants and products.
5.5: Calorimetry
Calorimetry: measurement of heat flow
Calorimeter: apparatus that measures heat flow
Heat Capacity and Specific Heat
Heat capacity: amount of energy required to raise the temperature of a given object by 1°C. The greater the capacity of a body, the more heat it requires to raise its temperature.
Molar heat capacity: when referring to pure substances; heat capacity of 1 mol of substance
Specific heat: heat capacity of 1 g of a substance (as opposed to a mole)
Specific heat= quantity of heat transferred / (grams of substance) X (temperature change)
= q / m X ∆T
q = (specific heat) X (grams of substance) X ∆T
ConstantPressure Calorimetry
Calorimeters are simple instruments to control pressure. Because the calorimeter prevents the gain or loss of heat from its surroundings, the heat released by the reaction, qrxn, equals that gained by the solution, qsoln. Thus,
qrxn = qsoln = (specific heat) X (grams of solution) X ∆T
Bomb Calorimetry
Bomb calorimeter: device for measuring the heat evolved in the combustion of a substance under constant volume conditions. Most used for combustion reactions.
qevolved = Ccalorimeter X ∆T , where Ccalorimeter is the heat capacity of the calorimeter
5.6: Hess's Law
Because enthalpy is a state function, the enthalpy change, ∆H, associated with any chemical process depends only on the amount of matter that undergoes change, and on the nature of the initial state of the reactants and the final state of the products.
CH4 (g) + 2O2 (g) ? CO2 (g) + 2H2O (g)H = 802 kJ
(Add) 2H2O (g) ? 2H2O (l) ?H = 88 kJ
__________________________________________________________
CH4 (g) + 2O2 (g) + 2H2O (g) → CO2 (g) + 2H2O (l) + 2H2O (g)
Net equation:
CH4 (g) + 2O2 (g) ? CO2 (g) + 2H2O (l)
To obtain the net equation, the sum of the reactants of the two equations is placed on one side of the arrow, and the sum of the products on the other side. Because 2H2O (g) occurs on both sides of the equation, it can be cancelled.
Hess’s law: states that if a reaction is carried out in a series of steps, ∆H for the reaction will be equal to the sum of the enthalpy changes for individual steps. The overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.
The first law of thermodynamics, in the form of Hess’s law, teaches us that we can never expect to obtain more (or less) energy from a chemical reaction by changing the method of carrying out the reaction.
5.7: Enthalpies of Formation
• Enthalpies of vaporization (∆H for converting liquids to gas)
• Enthalpies of fusion (∆H for melting solids)
• Enthalpies of combustion (∆H for combusting a substance in oxygen)
Enthalpy of formation: aka Heat of formation; enthalpy change that accompanies the formation of a substance from the most stable forms of its component elements. Labeled ∆Hf, where the subscript f indicates that the substance has been formed from its elements.
The magnitude of any enthalpy change depends on the conditions of temperature, pressure, and state (gas, liquid, solid) of the reactants and product.
Standard state: for a substance is the form most stable at the particular temperature of interest and at standard atmospheric pressure.
Standard enthalpy of formation: ∆H°f. change in enthalpy that accompanies the formation of 1 mol of that substance from its elements, with all substances in their standard states.
Using Enthalpies of Formation to Calculate Enthalpies of Reaction
∆H°rxn = ∑ n∆H°f (products) ∑ m∆H°f (reactants)
∑ = the sum of
m and n = stoichiometric coefficients of the chemical reaction
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
∆H°rxn = [3∆H°f (CO2) + 4∆H°f (H2O)] – [∆H°f (C3H8) + 5∆H°f (O2)]
∆H°rxn = [(3 mol CO2) (393.5 kJ/mol) + (4 mol H2O) (285.5 kJ/mol)]
= (2324 kJ) – (103.85 kJ)
=2220 kJ
5.8: Foods and Fuels
Fuel value: energy released when 1g of a material is combusted
Foods
• Main sources of energy for the body are from carbohydrates and fat
• Breakdown of carbohydrates is rapid, so energy is quickly supplied to the body
• Fats are well suited for the body because (1) they are insoluble in water, which permits their storage in the human body. (2) they produce more energy per gram thatneither proteins or carbohydrates.
Fuels
• Fossil fuels: coal, petroleum, and natural gas
• Natural gas: consists of gaseous hydrocarbons, compounds of hydrogen and carbon
• Petroleum: liquid composed of hundreds of compounds
• Coal: solid; contains hydrocarbons of high molecular weight as well as compounds containing sulfur, oxygen, and nitrogen. Most abundant fossil fuel
• Syngas: synthetic gas abbreviation | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.S%3A_Thermochemistry_%28Summary%29.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.
• 6.1: The Wave Nature of Light
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.
• 6.2: Quantized Energy and Photons
Blackbody radiation is the radiation emitted by hot objects and could not be explained with classical physics. Max Planck postulated that energy was quantized and may 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 the quantization of energy to explain the photoelectric effect
• 6.3: Line Spectra and the Bohr Model
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 e
• 6.4: The Wave Behavior of Matter
An electron possesses both particle and wave properties. 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. The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. 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.
• 6.5: Quantum Mechanics and Atomic Orbitals
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 wavefunctions to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies.
• 6.6: 3D Representation of Orbitals
Orbitals with l = 0 are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. Orbitals with values of n > 1 and l = 0 contain one or more nodes. Orbitals with l = 1 are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with l = 2 are d orbitals and have more complex shapes with at least two nodal surfaces. l = 3 orbitals are f orbitals, which are still more complex.
• 6.7: Many-Electron Atoms
In addition to the three quantum numbers (n, l, ml) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number (ms), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. This is important for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin).
• 6.8: Electron Configurations
Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with parallel spins.
• 6.9: Electron Configurations and the Periodic Table
The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively.
• 6.E: Electronic Structure of Atoms (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 6.S: Electronic Structure of Atoms (Summary)
This is the summary Module for the chapter "Electronic Structure of Atoms" in the Brown et al. General Chemistry Textmap.
Thumbnail: Electron shell diagram for Sodium, the 19th element in the periodic table of elements. (CC BY-SA; 2.5; Pumbaa)
06: Electronic Structure 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 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. 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 passes a particular point on the surface of the water, anything floating there moves up and down. —between the midpoints of two peaks, for example, or two troughs—is the ($λ$, lowercase Greek lambda). Wavelengths are described by a unit of distance, typically meters. The ($u$, lowercase Greek nu) of a is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations per second (1/s = s), which in the SI system is called the hertz (Hz). is defined as half the peak-to-trough height; as the amplitude of a 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 per unit time is its speed ($v$), which is typically measured in meters per second (m/s). The speed of a is equal to the product of its wavelength and frequency: }} \right )\left ( \dfrac{\cancel{\text{}}}{\text{second}} \right ) &=\dfrac{\text{meters}}{\text{second}} \label{6.1.1b} \end{align} \] (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 . (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 (), which turns out to be a physical constant with a value of 2.99792458 × 10 m/s (about 3.00 ×10 m/s or 1.86 × 10 mi/s). This is about a million times faster than the speed of sound. , with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the of the atmosphere absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation. requires an understanding of the properties of waves and electromagnetic radiation. A is a periodic oscillation by which energy is transmitted through space. All waves are , repeating regularly in both space and time. Waves are characterized by several interrelated properties: ($λ$), the distance between successive waves; ($u$), the number of waves that pass a fixed point per unit time; ($v$), the rate at which the propagates through space; and , the magnitude of the oscillation about the mean position. The speed of a is equal to the product of its wavelength and frequency. consists of two perpendicular waves, one electric and one magnetic, propagating at the ($c$). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.01%3A_The_Wave_Nature_of_Light.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 given off by a hot object. The wavelength (i.e. color) of radiant energy emitted by a blackbody 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 (in what he called "an act of despair") 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.
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 by the equation
$E=h u \label{6.2.1}$
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:
$h = 6.626 \times 10^{−34}\, J•s\, (\text{joule-segundos}) \nonumber$
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.
Energy of a Photon:
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 $\ref{6.2.1}$ 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 ($E_o= u_o$). 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:
\begin{align} \text{ kinetic energy of ejected electron} &=E-E_{o} \nonumber \[4pt] &=h u -h u _{o} \nonumber \[4pt] &=h\left ( u - u _{o} \right ) \label{6.2.2} \end{align}
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. Moreover, the idea that light could behave not only as a wave but as a particle in the form of photons suggested that matter and energy might not be such unrelated phenomena after all.
The Photoelectric Effect:
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:
Use Equation $\ref{6.2.1}$ and the relationship between wavelength and frequency to calculate the energy in joules.
Solution:
The energy of a single photon is given by Equaton \ref{6.2.1}$E = hu = \dfrac{hc}{λ}. \nonumber$
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 a
$1.287 \times 10^{-15}\; J/photon$
Answer b
4497 times
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, $E = h u$. Both energy and matter have fundamental building blocks: quanta and atoms, respectively. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.02%3A_Quantized_Energy_and_Photons.txt |
Learning Objectives
• To know the relationship between atomic spectra and the electronic structure of atoms.
The concept of the photon emerged from experimentation with thermal radiation, electromagnetic radiation emitted as the result of a source’s temperature, which produces a continuous spectrum of energies.The photoelectric effect provided indisputable evidence for the existence of the photon and thus the particle-like behavior of electromagnetic radiation. However, more direct evidence was needed to verify the quantized nature of energy in all matter. In this section, we describe how observation of the interaction of atoms 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 of particular wavelengths, called a line spectrum, are observed rather than a continuous range of wavelengths (Figure $1$). 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 Planck’s equation: the observation of only a few values of λ (or $u$) 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:
$u=constant\; \left ( \dfrac{1}{2^{2}}-\dfrac{1}{n^{^{2}}} \right ) \label{6.3.1}$
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:
$\dfrac{1}{\lambda }=\Re\; \left ( \dfrac{1}{n^{2}_{1}}-\dfrac{1}{n^{2}_{2}} \right ) \label{6.3.2}$
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 predicted 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.
Bohr's Model
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
$E_{n}=\dfrac{-\Re hc}{n^{2}} \label{6.3.3}$
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 $\ref{6.3.3}$ indicates that the electron-nucleus pair is more tightly bound (i.e. at a lower potential energy) 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) 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 $\PageIndex{2a}$). A hydrogen atom with an electron in an orbit with n > 1 is therefore in an excited state, defined as any arrangement of electrons that is higher in energy than 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 \ref{6.3.3}) for each energy value gives
\begin{align*} \Delta E &=E_{final}-E_{initial} \[4pt] &=-\dfrac{\Re hc}{n_{2}^{2}}-\left ( -\dfrac{\Re hc}{n_{1}^{2}} \right ) \[4pt] &=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.4} \end{align*}
If $n_2 > n_1$, 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
$\Delta E = \dfrac{hc}{\lambda }=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.5}$
Canceling $hc$ on both sides gives
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.6}$
Except for the negative sign, this is the same equation that Rydberg obtained experimentally. The negative sign in Equations $\ref{6.3.5}$ and $\ref{6.3.6}$ indicates that energy is released as the electron moves from orbit $n_2$ to orbit $n_1$ because orbit $n_2$ is at a higher energy than orbit $n_1$. 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 ($\PageIndex{3b}$); 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 (Figure $\PageIndex{3a}$). 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 (Figure $\PageIndex{1a}$). 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 $4$
The Bohr Atom:
Using Atoms to Time
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 \ref{6.3.2} (the Rydberg equation) and solve for $\lambda$.
2. Use Figure 2.2.1 to locate the region of the electromagnetic spectrum corresponding to the calculated wavelength.
Solution:
We can use the Rydberg equation to calculate the wavelength:
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \nonumber$
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.
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right )=1.097\times m^{-1}\left ( \dfrac{1}{1}-\dfrac{1}{4} \right )=8.228 \times 10^{6}\; m^{-1} \nonumber$
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.
$\widetilde{ u} =\dfrac{1}{\lambda }=8.228\times 10^{6}\cancel{m^{-1}}\left (\dfrac{\cancel{m}}{100\;cm} \right )=82,280\: cm^{-1} \nonumber$
and
$\lambda = 1.215 \times 10^{−7}\; m = 122\; nm \nonumber$
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:
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.
The Energy States of the Hydrogen Atom
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).
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.
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.
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. 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. Bohr’s model revolutionized the understanding of the atom but could not explain the spectra of atoms heavier than hydrogen.
Key Concepts
• 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.03%3A_Line_Spectra_and_the_Bohr_Model.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 famous equation:
$E=h u=h\dfrac{c}{\lambda }=mc^{2} \label{6.4.1}$
According to this theory, a photon of wavelength $λ$ and frequency $u$ has a nonzero mass, which is given as follows:
$m=\dfrac{E}{c^{2}}=\dfrac{h u }{c^{2}}=\dfrac{h}{\lambda c} \label{6.4.2}$
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
$\lambda =\dfrac{h}{mv} \label{6.4.3}$
where
• $h$ 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.
The de Broglie Equation:
Example $1$: Wavelength of a Baseball in Motion
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 $\ref{6.4.3}$ 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
$\lambda =\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )}= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )}=9.95\times 10^{-35}\; m \nonumber$
(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 $1$: Wavelength of a Neutron in Motion
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, however, 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 And Interference
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 $\PageIndex{2A}$ shows an arbitrary phase difference between two wave and Figure $\PageIndex{2B}$ shows what happens when the two waves are 180 degrees out of phase. The green line is their sum. Figure $\PageIndex{2C}$ 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. A pattern of constructive and destructive interference is obtained when two (or more) diffracting waves interact with each other. This principle of diffraction and interference was used to prove the wave properties of electrons and is the basis for how electron microscopes work.
Photograph of an interference pattern produced by circular water waves in a ripple tank.
For a mathematical analysis of phase aspects in sinusoids, check the math Libretexts library.
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) has wavelength
\begin{align} \dfrac{\lambda }{2} & =L \nonumber \ \lambda &= 2L \nonumber \end{align} \label{6.4.4}
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
$\lambda=\dfrac{2L}{n} \label{6.4.5}$
where n has any integral value. When plucked, all other frequencies die out immediately. Only the resonant frequencies survive and are heard. Thus, we can think of the resonant frequencies of the string as being quantized. Notice in Figure $3$ that all overtones have one or more nodes, 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
$2πr = nλ \label{6.4.6}$
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 nonresonant 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 $\ref{6.4.6}$, one complete wavelength would close the circle. Higher energy levels would have successively higher values of n with a corresponding number of nodes.
Like 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 now 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 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. 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. An "uncertainty principle" for light is, if you will, merely a conclusion about the nature of electromagnetic waves and nothing new. De Broglie's idea of wave-particle duality means that particles such as electrons which exhibit wavelike characteristics will also undergo diffraction from slits whose size is on 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 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:
$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\ge \dfrac{h}{4\pi } \label{6.4.7}$
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:
$\left ( \Delta \left [ mv \right ] \right )= \dfrac{h}{4\pi \left ( \Delta x \right ) }=\dfrac{h}{4\pi \left ( 0 \right ) }=\infty \label{6.4.8}$
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 $\ref{6.4.8}$, 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 $1$: Quantum Nature of Baseballs
Calculate the minimum uncertainty in the position of the pitched baseball from Example $\ref{6.4.1}$ 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 $\ref{6.4.7}$) 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 (Equation \ref{6.4.7}) tells us that $(Δx)(Δ(mv)) = h/4π \nonumber$. 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
$\Delta u =\left ( \dfrac{1\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1\; \cancel{min}}{60\; s} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )=0.4469\; m/s \nonumber$
C Therefore,
$\Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; J\cdot s}{4\left ( 3.1416 \right )} \right ) \left ( \dfrac{1}{\left ( 0.149\; kg \right )\left ( 0.4469\; m\cdot s^{-1} \right )} \right ) \nonumber$
Inserting the definition of a joule (1 J = 1 kg•m2/s2) gives
$\Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; \cancel{kg} \cdot m^{\cancel{2}} \cdot s}{4\left ( 3.1416 \right )\left ( \cancel{s^{2}} \right )} \right ) \left ( \dfrac{1\; \cancel{s}}{\left ( 0.149\; \cancel{kg} \right )\left ( 0.4469\; \cancel{m} \right )} \right ) \nonumber$
$\Delta x \ge 7.92 \pm \times 10^{-34}\; m \nonumber$
This is equal to $3.12 \times 10^{−32}$ 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)
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.
$\lambda =\dfrac{h}{mv} \nonumber$
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.
$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.04%3A_The_Wave_Behavior_of_Matter.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.
Erwin 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 wavefunctions that are the solutions of Schrödinger’s equations.
Wavefunctions
A wavefunction (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 wavefunction is associated with a particular energy E. The properties of wavefunctions derived from quantum mechanics are summarized here:
• A wavefunction 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 were the captain of a ship trying 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.
• The magnitude of the wavefunction at a particular point in space is proportional to the amplitude of the wave at that point. Many wavefunctions 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 wavefunction (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 wavefunction should not be confused with a positive or negative electrical charge.
• The square of the wavefunction 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 wavefunction ($\Psi^2$) is always a real quantity [recall that that $\sqrt{-1}^2=-1$] and is proportional to the probability of finding an electron at a given point. More accurately, the probability is given by the product of the wavefunction 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 wavefunction. From the patterns of one- and two-dimensional standing waves shown previously, 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 wavefunctions. 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 (wavefunctions), each of which is specified by a set of numbers called quantum numbers.
• Each wavefunction 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 wavefunction. 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.
The Principal Quantum Number
The principal quantum number (n) tells the average relative distance of an electron from the nucleus:
$n = 1, 2, 3, 4,… \label{6.5.1}$
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 wavefunctions 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:
$l = 0, 1, 2,…, n − 1 \label{6.5.2}$
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 wavefunctions 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:
The Magnetic Quantum Number
The third quantum number is the magnetic quantum number ($m_l$). 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:
$m_l = −l, −l + 1,…, 0,…, l − 1, l \label{6.5.3}$
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 wavefunction 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 = 16
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:
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 historical terms for 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 6.5.1):
• 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.
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 $m_l$ 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): Magnetic Quantum Number (ml) & Spin Quantum Number (ms) YouTube(opens in new window) [youtu.be] (opens in new window)
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 wavefunctions (Ψ) to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies. Wavefunctions have five important properties:
1. the wavefunction uses three variables (Cartesian axes x, y, and z) to describe the position of an electron;
2. the magnitude of the wavefunction 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 wavefunction 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 wavefunction; and
5. each spatial distribution of the electron described by a wavefunction 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 wavefunctions 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. Wavefunctions 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 wavefunction 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_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.05%3A_Quantum_Mechanics_and_Atomic_Orbitals.txt |
Learning Objectives
• To understand the 3D representation of electronic orbitals
An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of containing an electron.
One way of representing electron probability distributions was illustrated previously for the 1s orbital of hydrogen. Because Ψ2 gives the probability of finding an electron in a given volume of space (such as a cubic picometer), a plot of Ψ2 versus distance from the nucleus (r) is a plot of the probability density. The 1s orbital is spherically symmetrical, so the probability of finding a 1s electron at any given point depends only on its distance from the nucleus. The probability density is greatest at $r = 0$ (at the nucleus) and decreases steadily with increasing distance. At very large values of r, the electron probability density is very small but not zero.
In contrast, we can calculate the radial probability (the probability of finding a 1s electron at a distance r from the nucleus) by adding together the probabilities of an electron being at all points on a series of x spherical shells of radius r1, r2, r3,…, rx − 1, rx. In effect, we are dividing the atom into very thin concentric shells, much like the layers of an onion (Figure $\PageIndex{1a}$), and calculating the probability of finding an electron on each spherical shell. Recall that the electron probability density is greatest at r = 0 (Figure $\PageIndex{1b}$), so the density of dots is greatest for the smallest spherical shells in part (a) in Figure $1$. In contrast, the surface area of each spherical shell is equal to $4πr^2$, which increases very rapidly with increasing r (Figure $\PageIndex{1c}$). Because the surface area of the spherical shells increases more rapidly with increasing r than the electron probability density decreases, the plot of radial probability has a maximum at a particular distance (Figure $\PageIndex{1d}$). Most important, when r is very small, the surface area of a spherical shell is so small that the total probability of finding an electron close to the nucleus is very low; at the nucleus, the electron probability vanishes (Figure $\PageIndex{1d}$).
For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle.
Figure $2$ compares the electron probability densities for the hydrogen 1s, 2s, and 3s orbitals. Note that all three are spherically symmetrical. For the 2s and 3s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r. Instead, a series of minima and maxima are observed in the radial probability plots (Figure $\PageIndex{2c}$). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability. The existence of these nodes is a consequence of changes of wave phase in the wavefunction Ψ.
s Orbitals (l=0)
Three things happen to s orbitals as n increases (Figure $2$):
1. They become larger, extending farther from the nucleus.
2. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude.
3. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus.
Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density, as was shown for the hydrogen 1s, 2s, and 3s orbitals in part (b) in Figure $2$. Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2s and 3s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding.
p Orbitals (l=1)
Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2p subshell has l = 1, with three values of ml (−1, 0, and +1), there are three 2p orbitals.
The electron probability distribution for one of the hydrogen 2p orbitals is shown in Figure $3$. Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a $2p_z$ orbital. As shown in Figure $4$, the other two 2p orbitals have identical shapes, but they lie along the x axis ($2p_x$) and y axis ($2p_y$), respectively. Note that each p orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges.
The surfaces shown enclose 90% of the total electron probability for the 2px, 2py, and 2pz orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2p orbital. The phase of the wave function is positive (orange) in the region of space where x, y, or z is positive and negative (blue) where x, y, or z is negative. Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3p, 4p, and higher-energy p orbitals are, however, essentially the same as those shown in Figure $4$.
d Orbitals (l=2)
Subshells with l = 2 have five d orbitals; the first principal shell to have a d subshell corresponds to n = 3. The five d orbitals have ml values of −2, −1, 0, +1, and +2.
The hydrogen 3d orbitals, shown in Figure $5$, have more complex shapes than the 2p orbitals. All five 3d orbitals contain two nodal surfaces, as compared to one for each p orbital and zero for each s orbital. In three of the d orbitals, the lobes of electron density are oriented between the x and y, x and z, and y and z planes; these orbitals are referred to as the $3d_{xy}$, \)3d_{xz}\), and $3d_{yz}$ orbitals, respectively. A fourth d orbital has lobes lying along the x and y axes; this is the $3d_{x^2−y^2}$ orbital. The fifth 3d orbital, called the $3d_{z^2}$ orbital, has a unique shape: it looks like a $2p_z$ orbital combined with an additional doughnut of electron probability lying in the xy plane. Despite its peculiar shape, the $3d_{z^2}$ orbital is mathematically equivalent to the other four and has the same energy. In contrast to p orbitals, the phase of the wave function for d orbitals is the same for opposite pairs of lobes. As shown in Figure $5$, the phase of the wave function is positive for the two lobes of the $dz^2$ orbital that lie along the z axis, whereas the phase of the wave function is negative for the doughnut of electron density in the xy plane. Like the s and p orbitals, as n increases, the size of the d orbitals increases, but the overall shapes remain similar to those depicted in Figure $5$.
f Orbitals (l=3)
Principal shells with n = 4 can have subshells with l = 3 and ml values of −3, −2, −1, 0, +1, +2, and +3. These subshells consist of seven f orbitals. Each f orbital has three nodal surfaces, so their shapes are complex. Because f orbitals are not particularly important for our purposes, we do not discuss them further, and orbitals with higher values of l are not discussed at all.
Orbital Energies
Although we have discussed the shapes of orbitals, we have said little about their comparative energies. We begin our discussion of orbital energies by considering atoms or ions with only a single electron (such as H or He+).
The relative energies of the atomic orbitals with n ≤ 4 for a hydrogen atom are plotted in Figure $6$; note that the orbital energies depend on only the principal quantum number n. Consequently, the energies of the 2s and 2p orbitals of hydrogen are the same; the energies of the 3s, 3p, and 3d orbitals are the same; and so forth. Quantum mechanics predicts that in the hydrogen atom, all orbitals with the same value of n (e.g., the three 2p orbitals) are degenerate, meaning that they have the same energy. The orbital energies obtained for hydrogen using quantum mechanics are exactly the same as the allowed energies calculated by Bohr. In contrast to Bohr’s model, however, which allowed only one orbit for each energy level, quantum mechanics predicts that there are 4 orbitals with different electron density distributions in the n = 2 principal shell (one 2s and three 2p orbitals), 9 in the n = 3 principal shell, and 16 in the n = 4 principal shell.The different values of l and ml for the individual orbitals within a given principal shell are not important for understanding the emission or absorption spectra of the hydrogen atom under most conditions, but they do explain the splittings of the main lines that are observed when hydrogen atoms are placed in a magnetic field. Figure $6$ shows that the energy levels become closer and closer together as the value of n increases, as expected because of the 1/n2 dependence of orbital energies.
The energies of the orbitals in any species with only one electron can be calculated by a minor variation of Bohr’s equation, which can be extended to other single-electron species by incorporating the nuclear charge $Z$ (the number of protons in the nucleus):
$E=−\dfrac{Z^2}{n^2}Rhc \label{6.6.1}$
In general, both energy and radius decrease as the nuclear charge increases. Thus the most stable orbitals (those with the lowest energy) are those closest to the nucleus. For example, in the ground state of the hydrogen atom, the single electron is in the 1s orbital, whereas in the first excited state, the atom has absorbed energy and the electron has been promoted to one of the n = 2 orbitals. In ions with only a single electron, the energy of a given orbital depends on only n, and all subshells within a principal shell, such as the $p_x$, $p_y$, and $p_z$ orbitals, are degenerate.
Summary
The four chemically important types of atomic orbital correspond to values of $\ell = 0$, $1$, $2$, and $3$. Orbitals with $\ell = 0$ are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. All orbitals with values of $n > 1$ and $ell = 0$ contain one or more nodes. Orbitals with $\ell = 1$ are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with $\ell = 2$ are d orbitals and have more complex shapes with at least two nodal surfaces. Orbitals with $\ell = 3$ are f orbitals, which are still more complex.
Because its average distance from the nucleus determines the energy of an electron, each atomic orbital with a given set of quantum numbers has a particular energy associated with it, the orbital energy.
$E=−\dfrac{Z^2}{n^2}Rhc \nonumber$
In atoms or ions with only a single electron, all orbitals with the same value of $n$ have the same energy (they are degenerate), and the energies of the principal shells increase smoothly as $n$ increases. An atom or ion with the electron(s) in the lowest-energy orbital(s) is said to be in its ground state, whereas an atom or ion in which one or more electrons occupy higher-energy orbitals is said to be in an excited state. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.06%3A_3D_Representation_of_Orbitals.txt |
Learning Objectives
• To write the electron configuration of any element and relate its electron configuration to its position in the periodic table.
The quantum mechanical model allowed us to determine the energies of the hydrogen atomic orbitals; now we would like to extend this to describe the electronic structure of every element in the Periodic Table. The process of describing each atom’s electronic structure consists, essentially, of beginning with hydrogen and adding one proton and one electron at a time to create the next heavier element in the table; however, interactions between electrons make this process a bit more complicated than it sounds. All stable nuclei other than hydrogen also contain one or more neutrons. Because neutrons have no electrical charge, however, they can be ignored in the following discussion. Before demonstrating how to do this, however, we must introduce the concept of electron spin and the Pauli principle.
Orbitals and their Energies
Unlike in hydrogen-like atoms with only one electron, in multielectron atoms the values of quantum numbers n and l determine the energies of an orbital. The energies of the different orbitals for a typical multielectron atom are shown in Figure \(1\). Within a given principal shell of a multielectron atom, the orbital energies increase with increasing l. An ns orbital always lies below the corresponding np orbital, which in turn lies below the nd orbital.
These energy differences are caused by the effects of shielding and penetration, the extent to which a given orbital lies inside other filled orbitals. For example, an electron in the 2s orbital penetrates inside a filled 1s orbital more than an electron in a 2p orbital does. Since electrons, all being negatively charged, repel each other, an electron closer to the nucleus partially shields an electron farther from the nucleus from the attractive effect of the positively charged nucleus. Hence in an atom with a filled 1s orbital, the effective nuclear charge (Zeff) experienced by a 2s electron is greater than the Zeff experienced by a 2p electron. Consequently, the 2s electron is more tightly bound to the nucleus and has a lower energy, consistent with the order of energies shown in Figure \(1\).
Due to electron shielding, \(Z_{eff}\) increases more rapidly going across a row of the periodic table than going down a column.
Notice in Figure \(1\) that the difference in energies between subshells can be so large that the energies of orbitals from different principal shells can become approximately equal. For example, the energy of the 3d orbitals in most atoms is actually between the energies of the 4s and the 4p orbitals.
Electron Spin: The Fourth Quantum Number
When scientists analyzed the emission and absorption spectra of the elements more closely, they saw that for elements having more than one electron, nearly all the lines in the spectra were actually pairs of very closely spaced lines. Because each line represents an energy level available to electrons in the atom, there are twice as many energy levels available as would be predicted solely based on the quantum numbers \(n\), \(l\), and \(m_l\). Scientists also discovered that applying a magnetic field caused the lines in the pairs to split farther apart. In 1925, two graduate students in physics in the Netherlands, George Uhlenbeck (1900–1988) and Samuel Goudsmit (1902–1978), proposed that the splittings were caused by an electron spinning about its axis, much as Earth spins about its axis. When an electrically charged object spins, it produces a magnetic moment parallel to the axis of rotation, making it behave like a magnet. Although the electron cannot be viewed solely as a particle, spinning or otherwise, it is indisputable that it does have a magnetic moment. This magnetic moment is called electron spin.
In an external magnetic field, the electron has two possible orientations (Figure Figure \(2\)). These are described by a fourth quantum number (ms), which for any electron can have only two possible values, designated +½ (up) and −½ (down) to indicate that the two orientations are opposites; the subscript s is for spin. An electron behaves like a magnet that has one of two possible orientations, aligned either with the magnetic field or against it.
The Pauli Exclusion Principle
The implications of electron spin for chemistry were recognized almost immediately by an Austrian physicist, Wolfgang Pauli (1900–1958; Nobel Prize in Physics, 1945), who determined that each orbital can contain no more than two electrons. He developed the Pauli exclusion principle: No two electrons in an atom can have the same values of all four quantum numbers (n, l, ml, ms).
By giving the values of n, l, and ml, we also specify a particular orbital (e.g., 1s with n = 1, l = 0, ml = 0). Because ms has only two possible values (+½ or −½), two electrons, and only two electrons, can occupy any given orbital, one with spin up and one with spin down. With this information, we can proceed to construct the entire periodic table, which was originally based on the physical and chemical properties of the known elements.
Example \(1\)
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in a 2p orbital and predict the maximum number of electrons the 2p subshell can accommodate.
Given: orbital
Asked for: allowed quantum numbers and maximum number of electrons in orbital
Strategy:
1. List the quantum numbers (n, l, ml) that correspond to an n = 2p orbital. List all allowed combinations of (n, l, ml).
2. Build on these combinations to list all the allowed combinations of (n, l, ml, ms).
3. Add together the number of combinations to predict the maximum number of electrons the 2p subshell can accommodate.
Solution:
A For a 2p orbital, we know that n = 2, l = n − 1 = 1, and ml = −l, (−l +1),…, (l − 1), l. There are only three possible combinations of (n, l, ml): (2, 1, 1), (2, 1, 0), and (2, 1, −1).
B Because ms is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of (n, l, ml, ms): (2, 1, 1, +½), (2, 1, 1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, −1, +½), and (2, 1, −1, −½).
C Hence the 2p subshell, which consists of three 2p orbitals (2px, 2py, and 2pz), can contain a total of six electrons, two in each orbital.
Exercise \(1\)
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for a 6s orbital, and predict the total number of electrons it can contain.
Answer
(6, 0, 0, +½), (6, 0, 0, −½); two electrons
Magnetic Quantum Number (ml) & Spin Quantum Number (ms): Magnetic Quantum Number (ml) & Spin Quantum Number (ms), YouTube(opens in new window) [youtu.be] (opens in new window)
Summary
The arrangement of atoms in the periodic table arises from the lowest energy arrangement of electrons in the valence shell. In addition to the three quantum numbers (n, l, ml) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number (ms), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. The concept of electron spin has important consequences for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.07%3A_Many-Electron_Atoms.txt |
Learning Objectives
• To understand the basics of adding electrons to atomic orbitals
• To understand the basics of the Aufbau principle
The electron configuration of an element is the arrangement of its electrons in its atomic orbitals. By knowing the electron configuration of an element, we can predict and explain a great deal of its chemistry.
The Aufbau Principle
We construct the periodic table by following the aufbau principle (from German, meaning “building up”). First we determine the number of electrons in the atom; then we add electrons one at a time to the lowest-energy orbital available without violating the Pauli principle. We use the orbital energy diagram of Figure \(1\), recognizing that each orbital can hold two electrons, one with spin up ↑, corresponding to ms = +½, which is arbitrarily written first, and one with spin down ↓, corresponding to ms = −½. A filled orbital is indicated by ↑↓, in which the electron spins are said to be paired. Here is a schematic orbital diagram for a hydrogen atom in its ground state:
From the orbital diagram, we can write the electron configuration in an abbreviated form in which the occupied orbitals are identified by their principal quantum number n and their value of l (s, p, d, or f), with the number of electrons in the subshell indicated by a superscript. For hydrogen, therefore, the single electron is placed in the 1s orbital, which is the orbital lowest in energy (Figure \(1\)), and the electron configuration is written as 1s1 and read as “one-s-one.”
A neutral helium atom, with an atomic number of 2 (Z = 2), has two electrons. We place one electron in the orbital that is lowest in energy, the 1s orbital. From the Pauli exclusion principle, we know that an orbital can contain two electrons with opposite spin, so we place the second electron in the same orbital as the first but pointing down, so that the electrons are paired. The orbital diagram for the helium atom is therefore
written as 1s2, where the superscript 2 implies the pairing of spins. Otherwise, our configuration would violate the Pauli principle.
The next element is lithium, with Z = 3 and three electrons in the neutral atom. We know that the 1s orbital can hold two of the electrons with their spins paired; the third electron must enter a higher energy orbital. Figure 6.29 tells us that the next lowest energy orbital is 2s, so the orbital diagram for lithium is
This electron configuration is written as 1s22s1.
The next element is beryllium, with Z = 4 and four electrons. We fill both the 1s and 2s orbitals to achieve a 1s22s2 electron configuration:
When we reach boron, with Z = 5 and five electrons, we must place the fifth electron in one of the 2p orbitals. Because all three 2p orbitals are degenerate, it doesn’t matter which one we select. The electron configuration of boron is 1s22s22p1:
At carbon, with Z = 6 and six electrons, we are faced with a choice. Should the sixth electron be placed in the same 2p orbital that already has an electron, or should it go in one of the empty 2p orbitals? If it goes in an empty 2p orbital, will the sixth electron have its spin aligned with or be opposite to the spin of the fifth? In short, which of the following three orbital diagrams is correct for carbon, remembering that the 2p orbitals are degenerate?
Because of electron-electron interactions, it is more favorable energetically for an electron to be in an unoccupied orbital than in one that is already occupied; hence we can eliminate choice a. Similarly, experiments have shown that choice b is slightly higher in energy (less stable) than choice c because electrons in degenerate orbitals prefer to line up with their spins parallel; thus, we can eliminate choice b. Choice c illustrates Hund’s rule (named after the German physicist Friedrich H. Hund, 1896–1997), which today says that the lowest-energy electron configuration for an atom is the one that has the maximum number of electrons with parallel spins in degenerate orbitals. By Hund’s rule, the electron configuration of carbon, which is 1s22s22p2, is understood to correspond to the orbital diagram shown in c. Experimentally, it is found that the ground state of a neutral carbon atom does indeed contain two unpaired electrons.
When we get to nitrogen (Z = 7, with seven electrons), Hund’s rule tells us that the lowest-energy arrangement is
with three unpaired electrons. The electron configuration of nitrogen is thus 1s22s22p3.
At oxygen, with Z = 8 and eight electrons, we have no choice. One electron must be paired with another in one of the 2p orbitals, which gives us two unpaired electrons and a 1s22s22p4 electron configuration. Because all the 2p orbitals are degenerate, it doesn’t matter which one has the pair of electrons.
Similarly, fluorine has the electron configuration 1s22s22p5:
When we reach neon, with Z = 10, we have filled the 2p subshell, giving a 1s22s22p6 electron configuration:
Notice that for neon, as for helium, all the orbitals through the 2p level are completely filled. This fact is very important in dictating both the chemical reactivity and the bonding of helium and neon, as you will see.
Electron Configuration of Atoms:
Valence Electrons
As we continue through the periodic table in this way, writing the electron configurations of larger and larger atoms, it becomes tedious to keep copying the configurations of the filled inner subshells. In practice, chemists simplify the notation by using a bracketed noble gas symbol to represent the configuration of the noble gas from the preceding row because all the orbitals in a noble gas are filled. For example, [Ne] represents the 1s22s22p6 electron configuration of neon (Z = 10), so the electron configuration of sodium, with Z = 11, which is 1s22s22p63s1, is written as [Ne]3s1:
Electron Configuration of Neon and Sodium
Neon Z = 10 1s22s22p6
Sodium Z = 11 1s22s22p63s1 = [Ne]3s1
Because electrons in filled inner orbitals are closer to the nucleus and more tightly bound to it, they are rarely involved in chemical reactions. This means that the chemistry of an atom depends mostly on the electrons in its outermost shell, which are called the valence electrons. The simplified notation allows us to see the valence-electron configuration more easily. Using this notation to compare the electron configurations of sodium and lithium, we have:
Electron Configuration of Sodium and Lithium
Sodium 1s22s22p63s1 = [Ne]3s1
Lithium 1s22s1 = [He]2s1
It is readily apparent that both sodium and lithium have one s electron in their valence shell. We would therefore predict that sodium and lithium have very similar chemistry, which is indeed the case.
As we continue to build the eight elements of period 3, the 3s and 3p orbitals are filled, one electron at a time. This row concludes with the noble gas argon, which has the electron configuration [Ne]3s23p6, corresponding to a filled valence shell.
Example \(1\): Electronic Configuration of Phosphorus
Draw an orbital diagram and use it to derive the electron configuration of phosphorus, Z = 15. What is its valence electron configuration?
Given: atomic number
Asked for: orbital diagram and valence electron configuration for phosphorus
Strategy:
1. Locate the nearest noble gas preceding phosphorus in the periodic table. Then subtract its number of electrons from those in phosphorus to obtain the number of valence electrons in phosphorus.
2. Referring to Figure \(1\), draw an orbital diagram to represent those valence orbitals. Following Hund’s rule, place the valence electrons in the available orbitals, beginning with the orbital that is lowest in energy. Write the electron configuration from your orbital diagram.
3. Ignore the inner orbitals (those that correspond to the electron configuration of the nearest noble gas) and write the valence electron configuration for phosphorus.
Solution:
A Because phosphorus is in the third row of the periodic table, we know that it has a [Ne] closed shell with 10 electrons. We begin by subtracting 10 electrons from the 15 in phosphorus.
B The additional five electrons are placed in the next available orbitals, which Figure \(1\) tells us are the 3s and 3p orbitals:
Because the 3s orbital is lower in energy than the 3p orbitals, we fill it first:
Hund’s rule tells us that the remaining three electrons will occupy the degenerate 3p orbitals separately but with their spins aligned:
The electron configuration is [Ne]3s23p3.
C We obtain the valence electron configuration by ignoring the inner orbitals, which for phosphorus means that we ignore the [Ne] closed shell. This gives a valence-electron configuration of 3s23p3.
Exercise \(1\)
Draw an orbital diagram and use it to derive the electron configuration of chlorine, Z = 17. What is its valence electron configuration?
Answer
[Ne]3s23p5; 3s23p5
Definition of Valence Electrons:
The general order in which orbitals are filled is depicted in Figure \(2\). Subshells corresponding to each value of n are written from left to right on successive horizontal lines, where each row represents a row in the periodic table. The order in which the orbitals are filled is indicated by the diagonal lines running from the upper right to the lower left. Accordingly, the 4s orbital is filled prior to the 3d orbital because of shielding and penetration effects. Consequently, the electron configuration of potassium, which begins the fourth period, is [Ar]4s1, and the configuration of calcium is [Ar]4s2. Five 3d orbitals are filled by the next 10 elements, the transition metals, followed by three 4p orbitals. Notice that the last member of this row is the noble gas krypton (Z = 36), [Ar]4s23d104p6 = [Kr], which has filled 4s, 3d, and 4p orbitals. The fifth row of the periodic table is essentially the same as the fourth, except that the 5s, 4d, and 5p orbitals are filled sequentially.
The sixth row of the periodic table will be different from the preceding two because the 4f orbitals, which can hold 14 electrons, are filled between the 6s and the 5d orbitals. The elements that contain 4f orbitals in their valence shell are the lanthanides. When the 6p orbitals are finally filled, we have reached the next noble gas, radon (Z = 86), [Xe]6s24f145d106p6 = [Rn]. In the last row, the 5f orbitals are filled between the 7s and the 6d orbitals, which gives the 14 actinide elements. Because the large number of protons makes their nuclei unstable, all the actinides are radioactive.
Example \(2\): Electron Configuration of Mercury
Write the electron configuration of mercury (Z = 80), showing all the inner orbitals.
Given: atomic number
Asked for: complete electron configuration
Strategy:
Using the orbital diagram in Figure \(1\) and the periodic table as a guide, fill the orbitals until all 80 electrons have been placed.
Solution:
By placing the electrons in orbitals following the order shown in Figure \(2\) and using the periodic table as a guide, we obtain
Solution to Example 6.8.2
1s2 row 1 2 electrons
2s22p6 row 2 8 electrons
3s23p6 row 3 8 electrons
4s23d104p6 row 4 18 electrons
5s24d105p6 row 5 18 electrons
row 1–5 54 electrons
After filling the first five rows, we still have 80 − 54 = 26 more electrons to accommodate. According to Figure \(2\), we need to fill the 6s (2 electrons), 4f (14 electrons), and 5d (10 electrons) orbitals. The result is mercury’s electron configuration:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 = Hg = [Xe]6s24f145d10
with a filled 5d subshell, a 6s24f145d10 valence shell configuration, and a total of 80 electrons. (You should always check to be sure that the total number of electrons equals the atomic number.)
Exercise \(2\): Electron Configuration of Flerovium
Although element 114 is not stable enough to occur in nature, atoms of element 114 were created for the first time in a nuclear reactor in 1998 by a team of Russian and American scientists. The element is named after the Flerov Laboratory of Nuclear Reactions of the Joint Institute for Nuclear Research in Dubna, Russia, where the element was discovered. The name of the laboratory, in turn, honors the Russian physicist Georgy Flyorov. Write the complete electron configuration for element 114.
Answer
s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p2
The electron configurations of the elements are presented in Figure \(2\), which lists the orbitals in the order in which they are filled. In several cases, the ground state electron configurations are different from those predicted by Figure \(1\). Some of these anomalies occur as the 3d orbitals are filled. For example, the observed ground state electron configuration of chromium is [Ar]4s13d5 rather than the predicted [Ar]4s23d4. Similarly, the observed electron configuration of copper is [Ar]4s13d10 instead of [Ar]s23d9. The actual electron configuration may be rationalized in terms of an added stability associated with a half-filled (ns1, np3, nd5, nf7) or filled (ns2, np6, nd10, nf14) subshell. (In fact, this "special stability" is really another consequence of the instability caused by pairing an electron with another in the same orbital, as illustrated by Hund's rule.) Given the small differences between higher energy levels, this added stability is enough to shift an electron from one orbital to another. In heavier elements, other more complex effects can also be important, leading to many additional anomalies. For example, cerium has an electron configuration of [Xe]6s24f15d1, which is impossible to rationalize in simple terms. In most cases, however, these apparent anomalies do not have important chemical consequences.
Additional stability is associated with half-filled or filled subshells.
Electron Configuration of Transition Metals: Electron Configuration of Transition Metals, YouTube(opens in new window) [youtu.be]
Summary
Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with their spins parallel. For chemical purposes, the most important electrons are those in the outermost principal shell, the valence electrons. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/06%3A_Electronic_Structure_of_Atoms/6.08%3A_Electron_Configurations.txt |
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