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Discovered by von Reichenstein in 1782, tellurium is a brittle metalloid that is relatively rare. It is named from the Latin tellus for "earth". Tellurium can be alloyed with some metals to increase their machinability and is a basic ingredient in the manufacture of blasting caps. Elemental tellurium is occasionally found in nature but is more often recovered from various gold ores, all containing $AuTe_2$.
History
Tellurium was discovered in a gold ore from the mines in Zlatna, near present day Sibiu, Transylvania. The ore was known as "Faczebajer weißes blättriges Golderz" (white leafy gold ore from Faczebaja) or antimonalischer Goldkies (antimonic gold pyrite). In 1782, while serving as the Austrian chief inspector of mines in Transylvania, Franz-Joseph Müller von Reichenstein concluded that a certain ore did not contain antimony, but that it contained bismuth sulfide. However, the following year, he reported that this was erroneous and that the ore contained mostly gold and an unknown metal very similar to antimony. After 3 years of testing, Müller determined the specific gravity of the mineral and noted the radish-like odor of the white smoke, which passed off when the new metal was heated. In 1789, another Hungarian scientist, Pál Kitaibel, also discovered the element independently in an ore from Deutsch-Pilsen which had been regarded as argentiferous molybdenite, but later he gave the credit to Müller. In 1798, it was named by Martin Heinrich Klaproth, who earlier isolated it from the mineral calaverite.
Properties
Tellurium is a semimetallic, lustrous, crystalline, brittle, silver-white element. It is usually available as a dark grey powder and has metal and non-metal properties. Te forms many compounds corresponding to those of sulfur and selenium. When burned in the air, tellurium has a greenish-blue flame and forms tellurium dioxide as a result. Tellurium is unaffected by water or hydrochloric acid, but dissolves in nitric acid. It has an atomic mass of 127.6 g/mol-1 and a density of 6.24 g-cm-3. Its boiling point is 450 degrees Celsius and its melting point is 1390 °C.
Source and Abundance
There are eight naturally occurring isotopes of tellurium, of which three are radioactive. Tellurium is among the rarest stable solid elements in the Earth's crust. At 0.005 ppm, it is comparable to platinum in abundance. However, tellurium is far more abundant in the wider universe. Tellurium was originally and is most commonly found in gold tellurides. However, the largest source for modern production of tellurium is as a byproduct of blister copper refinement. The treatment of 500 tons of copper ore results in 0.45 kg of tellurium. Tellurium can also be found in lead deposits. Other tellurium sources, known as subeconomic deposits because the cost of abstraction outweighs the yield in tellurium, are lower-grade copper and some coal.
Originally, the copper tellurium ore is treated with sodium bicarbonate and elemental oxygen to produce a tellurium oxide salt, copper oxide, and carbon dioxide:
$Cu_2Te + Na_2CO_3 + 2O_2 \rightarrow 2CuO + Na_2TeO_3 + CO_2 \nonumber$
Then, the sodium tellurium oxide is treated with sulfuric acid to precipitate out tellurium dioxide, which can be treated with aqueous sodium hydroxide to reduce to pure tellurium and oxygen gas:
$TeO_2 + 2NaOH \rightarrow Na_2TeO_3 + H_2O \rightarrow Te + 2NaOH + O_2 \nonumber$
Industrial and Commercial Use
Tellurium has many unique industrial and commercial uses that improve product quality and quality-of-life. Many of the technologies that utilize tellurium have important uses for the energy industry, the military, and health industries. Tellurium is used to color glass and ceramics and can improve the machining quality of metal products. When added to copper alloys, tellurium makes the alloy more ductile, whereas it can prevent corrosion in lead products. Tellurium is an important component of infrared detectors used by the military as well as x-ray detectors used by a variety of fields including medicine, science, and security. In addition, tellurium-based catalysts are used to produce higher-quality rubber. CdTe films are one of the highest efficiency photovoltaics, metals that convert sunlight directly into electrical power, at 11-13% efficiency and are, therefore, widely used in solar panels. CdTe is a thin-film semiconductor that absorbs sunlight.
Tellurium can be replaced by other elements in some of its uses. For many metallurgical uses, selenium, bismuth, or lead are effective substitutes. Both selenium and sulfur can replace tellurium in rubber production. Technologies based on tellurium have global impacts. As a photovoltaic, CdTe is the second most utilized solar cell in the world, soon said to surpass crystalline silicon and become the first. According to the US military, tellurium-based infrared detectors are the reason that the military has such an advantage at night, an advantage which, in turn, has an effect on global and domestic politics.
Environmental Impacts
Tellurium extraction, as a byproduct of copper refinement, shares environmental impacts associated with copper mining and extraction. While a generally safe process, the removal of copper from other impurities in the ore can lead to leaching of various hazardous sediments. In addition, the mining of copper tends to lead to reduced water flow and quality, disruption of soils and erosion of riverbanks, and reduction of air quality.
Resource Limitations v. Demand
About 215-220 tons of tellurium are mined across the globe every year. In 2006, the US produced 40% of the global production, Peru produced 30%, Japan produced 20%, and Canada produced 10% of the world's tellurium supply (since the chart can't be any bigger). The leading countries in production are the United States with 50 tons per year, Japan with 40 tons per year, Canada with 16 tons per year, and Peru with 7 tons per year (year 2009). When pure, tellurium costs \$24 per 100 grams. Because tellurium is about as rare as platinum on earth, the United States Department of Energy expects a supply shortfall by the year 2025, despite the always improving extraction methods. As demand increases to provide the tellurium needed for solar panels and other such things, supply will continue to decrease and thus the price will skyrocket. This will cause waves in the sustainable energy movement as well as military practices and modern medicine.
Z084 Chemistry of Polonium (Z84)
Polonium was discovered in 1898 by Marie Curie and named for her native country of Poland. The discovery was made by extraction of the remaining radioactive components of pitchblende following the removal of uranium. There is only about 10-6 g per ton of ore! Current production for research purposes involves the synthesis of the element in the lab rather than its recovery from minerals. This is accomplished by producing Bi-210 from the abundant Bi-209. The new isotope of bismuth is then allowed to decay naturally into Po-210. The sample pictured above is actually a thin film of polonium on stainless steel.
Although radioactive, polonium has a few commercial uses. You can buy your own sample of polonium at a photography store. It is part of the special anti-static brushes for dusting off negatives and prints.
Contributors and Attributions
Stephen R. Marsden
Z116 Chemistry of Livermorium (Z116)
In May of 2012 the IUPAC approved the name "Livermorium" (symbol Lv) for element 116. The new name honors the Lawrence Livermore National Laboratory (1952). A group of researchers of this laboratory with the heavy element research group of the Flerov Laboratory of Nuclear Reactions took part in the work carried out in Dubna on the synthesis of superheavy elements including element 116. Atoms of Lv were formed by the reaction of Cm-248 with Ca-48. The resulting Lv-292 decays by alpha emission to Fl-288.
Contributors and Attributions
Stephen R. Marsden | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Group/Group_16%3A_The_Oxygen_Family/Z052_Chemistry_of_Tellurium_%28Z52%29.txt |
The halogens are located on the left of the noble gases on the periodic table. These five toxic, non-metallic elements make up Group 17 of the periodic table and consist of: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Although astatine is radioactive and only has short-lived isotopes, it behaves similar to iodine and is often included in the halogen group. Because the halogen elements have seven valence electrons, they only require one additional electron to form a full octet. This characteristic makes them more reactive than other non-metal groups.
• Group 17: Physical Properties of the Halogens
It can be seen that there is a regular increase in many of the properties of the halogens proceeding down group 17 from fluorine to iodine. This includes their melting points, boiling points, intensity of their color, the radius of the corresponding halide ion, and the density of the element. On the other hand, there is a regular decrease in the first ionization energy as we go down this group. As a result, there is a regular increase in the ability to form high oxidation states.
• Group 17: Chemical Properties of the Halogens
Covers the halogens in Group 17: fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). Includes trends in atomic and physical properties, the redox properties of the halogens and their ions, the acidity of the hydrogen halides, and the tests for the halide ions.
• Chemistry of Fluorine (Z=9)
Fluorine (F) is the first element in the Halogen group (group 17) in the periodic table. Its atomic number is 9 and its atomic weight is 19, and it's a gas at room temperature. It is the most electronegative element, given that it is the top element in the Halogen Group, and therefore is very reactive. It is a nonmetal, and is one of the few elements that can form diatomic molecules (F2).
• Chemistry of Chlorine (Z=17)
Chlorine is a halogen in group 17 and period 3. It is very reactive and is widely used for many purposes, such as as a disinfectant. Due to its high reactivity, it is commonly found in nature bonded to many different elements.
• Chemistry of Bromine (Z=35)
Bromine is a reddish-brown fuming liquid at room temperature with a very disagreeable chlorine-like smell. In fact its name is derived from the Greek bromos or "stench". It was first isolated in pure form by Balard in 1826. It is the only non-metal that is a liquid at normal room conditions. Bromine on the skin causes painful burns that heal very slowly. It is an element to be treated with the utmost respect in the laboratory.
• Chemistry of Iodine (Z=53)
Elemental iodine is a dark grey solid with a faint metallic luster. When heated at ordinary air pressures it sublimes to a violet gas. The name iodine is taken from the Greek ioeides which means "violet colored". It was discovered in 1811 by Courtois.
• Chemistry of Astatine (Z=85)
Astatine is the last of the known halogens and was synthesized in 1940 by Corson and others at the University of California. It is radioactive and its name, from the Greek astatos, means "unstable". The element can be produced by bombarding targets made of bismuth-209 with high energy alpha particles (helium nuclei). Astatine 211 is the product and has a half-life of 7.2 hours. The most stable isotope of astatine is 210 which has a half-life of 8.1 hours.
Thumbnail: Chlorine gas in an ampoule. (CC-BY-SA; W. Oelen (http://woelen.homescience.net/science/index.html)).
Group 18: The Noble Gases
The noble gases (Group 18) are located in the far right of the periodic table and were previously referred to as the "inert gases" due to the fact that their filled valence shells (octets) make them extremely nonreactive. The noble gases were characterized relatively late compared to other element groups.
Thumbnail: Vial of glowing ultrapure neon. (CC SA; Jurii via http://images-of-elements.com/neon.php).
Z20 The Lanthanides
The Lanthanides consist of the elements in the f-block of period six in the periodic table. While these metals can be considered transition metals, they have properties that set them apart from the rest of the elements.
Z21 The Actinides
The Actinide series contains elements with atomic numbers 89 to 103 and is the sixth period and third group of the periodic table. The series is the row below the Lanthanide series, which is located underneath the main body of the periodic table. Both the Lanthanide and Actinide series are referred to as Rare Earth Metals and have a high diversity in oxidation numbers. All of the Actinides are radioactive. The most famous actinide is Uranium.
• General Properties and Reactions of The Actinides
The Actinide series contains elements with atomic numbers 89 to 103 and is in the sixth period and the third group of the periodic table. The series is the row below the Lanthanide series, which is located underneath the main body of the periodic table. Lanthanide and Actinide Series are both referred to as Rare Earth Metals. These elements all have a high diversity in oxidation numbers and all are radioactive.
• Chemistry of Americium
• Chemistry of Berkelium
• Chemistry of Californium
The fifth element in succession to emerge from the Berkeley, California cyclotron was element 98, californium (named after the State of origin).
• Chemistry of Curium
• Chemistry of Einsteinium
• Chemistry of Fermium
• Chemistry of Lawrencium
• Chemistry of Mendelevium
• Chemistry of Neptunium
• Chemistry of Nobelium
Discovered by Ghiorso, Sikkeland, Walton and Seaborg (see element 106) and named for Alfred Nobel, element 102 was originally named by a Swedish group working at the Nobel Institute of Physics in Stockholm. Details of their work, however, did not yield the expected results and the credit eventually shifted to the American team which decided to retain the original given name. The most stable isotope is No-259 with a half-life of just about 1 minute.
• Chemistry of Plutonium
• Chemistry of Protoactinium
• Chemistry of Thorium
• Chemistry of Uranium
Most of the naturally occurring uranium is the isotope U-238. This form of uranium is not fissionable, i.e., it cannot be used in atomic weapons or power plants. A much smaller percentage of naturally occurring uranium is the isotope U-235, which is fissionable. The process of "enriching" uranium to increase the proportion of U-235 in a sample is expensive and tedious but necessary to produce fuel that is usable in power plants and material for weapons. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Group/Group_17%3A_The_Halogens.txt |
• Period 3 Elements
Covers the elements sodium to argon in Period 3 of the Periodic Table. Includes trends in atomic and physical properties of the elements, some selected reactions of the elements, and some properties of their oxides, chlorides and hydroxides.
• Period 6 Elements: The Lanthanides
The Lanthanides consist of the elements in the f-block of period six in the periodic table. While these metals can be considered transition metals, they have properties that set them apart from the rest of the elements.
• Period 7 Elements: The Actinides
The Actinide series contains elements with atomic numbers 89 to 103 and is the third group in the periodic table. The series is the row below the Lanthanide series, which is located underneath the main body of the periodic table. Lanthanide and Actinide Series are both referred to as Rare Earth Metals. These elements all have a high diversity in oxidation numbers and all are radioactive. The most common and known element is Uranium, which is used as nuclear fuel when its converted into plutonium
Elements Organized by Period
Covers the elements sodium to argon in Period 3 of the Periodic Table. Includes trends in atomic and physical properties of the elements, some selected reactions of the elements, and some properties of their oxides, chlorides and hydroxides.
Period 3 Elements
This page discusses the reactions of the oxides of Period 3 elements (sodium to chlorine) with water, and with acids or bases where relevant (as before, argon is omitted because it does not form an oxide).
A quick summary of the trend
The oxides: The oxides of interest are given below:
Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7
P4O6 SO2 Cl2O
The trend in acid-base behavior can be summarized as follows:
Acidity increases from left to right, ranging from strongly basic oxides on the left to strongly acidic ones on the right, with an amphoteric oxide (aluminum oxide) in the middle. An amphoteric oxide is one which shows both acidic and basic properties.
This trend applies only to the highest oxides of the individual elements (see the top row of the table), in the highest oxidation states for those elements. The pattern is less clear for other oxides. Non-metal oxide acidity is defined in terms of the acidic solutions formed in reactions with water—for example, sulfur trioxide reacts with water to forms sulfuric acid. They will all, however, react with bases such as sodium hydroxide to form salts such as sodium sulfate as explored in detail below.
Sodium Oxide
Sodium oxide is a simple strongly basic oxide. It is basic because it contains the oxide ion, O2-, which is a very strong base with a high tendency to combine with hydrogen ions.
Reaction with water: Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution. A concentrated solution of sodium oxide in water will have pH 14.
$Na_2O + H_2O \rightarrow 2NaOH \nonumber$
Reaction with acids: As a strong base, sodium oxide also reacts with acids. For example, it reacts with dilute hydrochloric acid to produce sodium chloride solution.
$Na_2O + 2HCl \rightarrow 2NaCl + H_2O \nonumber$
Magnesium oxide
Magnesium oxide is another simple basic oxide, which also contains oxide ions. However, it is not as strongly basic as sodium oxide because the oxide ions are not as weakly-bound. In the sodium oxide, the solid is held together by attractions between 1+ and 2- ions. In magnesium oxide, the attractions are between 2+ and 2- ions. Because of the higher charge on the metal, more energy is required to break this association. Even considering other factors (such as the energy released from ion-dipole interactions between the cations and water), the net effect is that reactions involving magnesium oxide will always be less exothermic than those of sodium oxide.
Reaction with water: At first glance, magnesium oxide powder does not appear to react with water. However, the pH of the resulting solution is about 9, indicating that hydroxide ions have been produced. In fact, some magnesium hydroxide is formed in the reaction, but as the species is almost insoluble, few hydroxide ions actually dissolve. The reaction is shown below:
$MgO + H_2O \rightarrow Mg(OH)_2 \nonumber$
Reaction with acids: Magnesium oxide reacts with acids as predicted for a simple metal oxide. For example, it reacts with warm dilute hydrochloric acid to give magnesium chloride solution.
$MgO + 2HCl \rightarrow MgCl_2+H_2O \nonumber$
Aluminum Oxide
Describing the properties of aluminum oxide can be confusing because it exists in a number of different forms. One of those forms is very unreactive (known chemically as alpha-Al2O3) and is produced at high temperatures. The following reactions concern the more reactive forms of the molecule. Aluminium oxide is amphoteric. It has reactions as both a base and an acid.
Reaction with water: Aluminum oxide is insoluble in water and does not react like sodium oxide and magnesium oxide. The oxide ions are held too strongly in the solid lattice to react with the water.
Reaction with acids: Aluminum oxide contains oxide ions, and thus reacts with acids in the same way sodium or magnesium oxides do. Aluminum oxide reacts with hot dilute hydrochloric acid to give aluminum chloride solution.
$Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O \nonumber$
This reaction and others display the amphoteric nature of aluminum oxide.
Reaction with bases: Aluminum oxide also displays acidic properties, as shown in its reactions with bases such as sodium hydroxide. Various aluminates (compounds in which the aluminum is a component in a negative ion) exist, which is possible because aluminum can form covalent bonds with oxygen. This is possible because the electronegativity difference between aluminum and oxygen is small, unlike the difference between sodium and oxygen, for example (electronegativity increases across a period)
Aluminum oxide reacts with hot, concentrated sodium hydroxide solution to produce a colorless solution of sodium tetrahydroxoaluminate:
$Al_2O_3 + 2NaOH +3H_2O \rightarrow 2NaAl(OH)_4 \nonumber$
Silicon dioxide (silicon(IV) oxide)
Silicon is too similar in electronegativity to oxygen to form ionic bonds. Therefore, because silicon dioxide does not contain oxide ions, it has no basic properties. In fact, it is very weakly acidic, reacting with strong bases.
Reaction with water: Silicon dioxide does not react with water, due to the thermodynamic difficulty of breaking up its network covalent structure.
Reaction with bases: Silicon dioxide reacts with hot, concentrated sodium hydroxide solution, forming a colorless solution of sodium silicate:
$SiO_2 + 2NaOH \rightarrow Na_2SiO_3 + H2O \nonumber$
In another example of acidic silicon dioxide reacting with a base, the Blast Furnace extraction of iron, calcium oxide from limestone reacts with silicon dioxide to produce a liquid slag, calcium silicate:
$SiO_2 + CaO \rightarrow CaSiO_3 \nonumber$
Phosphorus Oxides
Two phosphorus oxides, phosphorus(III) oxide, P4O6, and phosphorus(V) oxide, P4O10, are considered here.
Phosphorus(III) oxide: Phosphorus(III) oxide reacts with cold water to produce a solution of the weak acid, H3PO3—known as phosphorous acid, orthophosphorous acid or phosphonic acid:
$P_4O_6 + 6H_2O \rightarrow 4H_3PO_3 \nonumber$
The fully-protonated acid structure is shown below:
The protons remain associated until water is added; even then, because phosphorous acid is a weak acid, few acid molecules are deprotonated. Phosphorous acid has a pKa of 2.00, which is more acidic than common organic acids like ethanoic acid (pKa = 4.76).
Phosphorus(III) oxide is unlikely to be reacted directly with a base. In phosphorous acid, the two hydrogen atoms in the -OH groups are acidic, but the third hydrogen atom is not. Therefore, there are two possible reactions with a base like sodium hydroxide, depending on the amount of base added:
$NaOH + H_3PO_3 \rightarrow NaH_2PO_3 + H_2O \nonumber$
$2NaOH + H_3PO_3 \rightarrow Na_2HPO_3 + 2H_2O \nonumber$
In the first reaction, only one of the protons reacts with the hydroxide ions from the base. In the second case (using twice as much sodium hydroxide), both protons react.
If instead phosphorus(III) oxide is reacted directly with sodium hydroxide solution, the same salts are possible:
$4NaOH + P_4O_6 + 2H_2O \rightarrow 4NaH_2PO_3 \nonumber$
$9NaOH + P_4O_6 \rightarrow 4Na_2HPO_3 + 2H_2O \nonumber$
Phosphorus(V) oxide: Phosphorus(V) oxide reacts violently with water to give a solution containing a mixture of acids, the nature of which depends on the reaction conditions. Only one acid is commonly considered, phosphoric(V) acid, H3PO4 (also known as phosphoric acid or as orthophosphoric acid).
$P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4 \nonumber$
This time the fully protonated acid has the following structure:
Phosphoric(V) acid is another weak acid with a pKa of 2.15, marginally weaker than phosphorous acid. Solutions of each of these acids with concentrations around 1 mol dm-3 have a pH of about 1.
Phosphoric (V) oxide is also unlikely to be reacted directly with a base, but the hypothetical reactions are considered. In its acid form, molecule has three acidic -OH groups, which can cause a three-stage reaction with sodium hydroxide:
$NaOH + H_3PO_4 \rightarrow NaH_2PO_4 + H_2O \nonumber$
$2NaOH + H_3PO_4 \rightarrow Na_2HPO_4 + 2H_2O \nonumber$
$3NaOH + H_3PO_4 \rightarrow Na_3PO_4 + 3H_2O \nonumber$
Similar to phosphorus (III) oxide, if phosphorus(V) oxide reacts directly with sodium hydroxide solution, the same possible salt as in the third step (and only this salt) is formed:
$12NaOH + P_4O_{10} \rightarrow 4Na_3PO_4 + 6H_2O \nonumber$
Sulfur Oxides
Two oxides are considered: sulfur dioxide, SO2, and sulfur trioxide, SO3.
Sulfur dioxide: Sulfur dioxide is fairly soluble in water, reacting to give a solution of sulfurous acid (also known as sulfuric(IV) acid), H2SO3, as shown in the reaction below. This species only exists in solution, and any attempt to isolate it gives off sulfur dioxide.
$SO_2 + H_2O \rightarrow H_2SO_3 \nonumber$
The protonated acid has the following structure:
Sulfurous acid is also a relatively weak acid, with a pKa of around 1.8, but slightly stronger than the two phosphorus-containing acids above. A reasonably concentrated solution of sulfurous acid has a pH of about 1.
Sulfur dioxide also reacts directly with bases such as sodium hydroxide solution. Bubbling sulfur dioxide through sodium hydroxide solution first forms sodium sulfite solution, followed by sodium hydrogen sulfite solution if the sulfur dioxide is in excess.
$SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O \nonumber$
$Na_2SO_3 + H_2O \rightarrow 2NaHSO_3 \nonumber$
Another important reaction of sulfur dioxide is with the base calcium oxide to form calcium sulfite (also known as calcium sulfate(IV)). This is of the important methods of removing sulfur dioxide from flue gases in power stations.
$CaO + SO_2 \rightarrow CaSO_3 \nonumber$
Sulfur trioxide: Sulfur trioxide reacts violently with water to produce a fog of concentrated sulfuric acid droplets.
$SO_3 + H_2O \rightarrow H_2SO_4 \nonumber$
Pure, fully-protonated sulfuric acid has the structure:
Sulfuric acid is a strong acid, and solutions will typically have a pH around 0. The acid reacts with water to give a hydronium ion (a hydrogen ion in solution) and a hydrogen sulfate ion. This reaction runs essentially to completion:
$H_2SO_4 (aq) + H_2O (l) \rightarrow H_3P^+ + HSO_4^- (aq) \nonumber$
The second proton is more difficult to remove. In fact, the hydrogen sulfate ion is a relatively weak acid, similar in strength to the acids discussed above. This reaction is more appropriately described as an equilibrium:
$HSO_4^- (aq) + H_2O \rightleftharpoons H_3O^+ (aq) + SO_4^{2-} (aq) \nonumber$
It is useful if you understand the reason that sulfuric acid is a stronger acid than sulfurous acid. You can apply the same reasoning to other acids that you find on this page as well.
Sulfuric acid is stronger than sulfurous acid because when a hydrogen ion is lost from one of the -OH groups on sulfuric acid, the negative charge left on the oxygen is spread out (delocalized) over the ion by interacting with the doubly-bonded oxygen atoms. It follows that more double bonded oxygen atoms in the ion make more delocalization possible; more delocalization leads to greater stability, making the ion less likely to recombine with a hydrogen ion and revert to the non-ionized acid.
Sulfurous acid only has one double bonded oxygen, whereas sulfuric acid has two; the extra double bond provides much more effective delocalization, a much more stable ion, and a stronger acid. Sulfuric acid displays all the reactions characteristic of a strong acid. For example, a reaction with sodium hydroxide forms sodium sulfate; in this reaction, both of the acidic protons react with hydroxide ions as shown:
$2NaOH +H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \nonumber$
In principle, sodium hydrogen sulfate can be formed by using half as much sodium hydroxide; in this case, only one of the acidic hydrogen atoms is removed.
Sulfur trioxide itself also reacts directly with bases such as calcium oxide, forming calcium sulfate:
$CaO + SO_3 \rightarrow CaSO_4 \nonumber$
This reaction is similar to the reaction with sulfur dioxide discussed above.
Chlorine Oxides
Chlorine forms several oxides, but only two (chlorine(VII) oxide, Cl2O7, and chlorine(I)oxide, Cl2O) are considered here. Chlorine(VII) oxide is also known as dichlorine heptoxide, and chlorine(I) oxide as dichlorine monoxide.
Chlorine(VII) oxide: Chlorine(VII) oxide is the highest oxide of chlorine—the chlorine atom is in its maximum oxidation state of +7. It continues the trend of the highest oxides of the Period 3 elements towards being stronger acids. Chlorine(VII) oxide reacts with water to give the very strong acid, chloric(VII) acid, also known as perchloric acid.
$Cl_2O_7 + H_2O \rightarrow 2HClO_4 \nonumber$
As in sulfuric acid, the pH of typical solutions of perchloric acid are around 0. Neutral chloric(VII) acid has the following structure:
When the chlorate(VII) ion (perchlorate ion) forms by loss of a proton (in a reaction with water, for example), the charge is delocalized over every oxygen atom in the ion. That makes the ion very stable, making chloric(VII) acid very strong.
Chloric(VII) acid reacts with sodium hydroxide solution to form a solution of sodium chlorate(VII):
$NaOH + HClO_4 \rightarrow NaClO_4 + H2O \nonumber$
Chlorine(VII) oxide itself also reacts directly with sodium hydroxide solution to give the same product:
$2NaOH + Cl_2O_7 \rightarrow 2NaClO_4 + H_2O \nonumber$
Chlorine(I) oxide: Chlorine(I) oxide is far less acidic than chlorine(VII) oxide. It reacts with water to some extent to give chloric(I) acid, $HOCl^-$ also known as hypochlorous acid.
$Cl_2O + H_2O \rightleftharpoons 2HOCl \nonumber$
The structure of chloric(I) acid is exactly as shown by its formula, HOCl. It has no doubly-bonded oxygens, and no way of delocalizing the charge over the negative ion formed by loss of the hydrogen. Therefore, the negative ion formed not very stable, and readily reclaims its proton to revert to the acid. Chloric(I) acid is very weak (pKa = 7.43) and reacts with sodium hydroxide solution to give a solution of sodium chlorate(I) (sodium hypochlorite):
$NaOH + HOCl \rightarrow NaOCl + H_2O \nonumber$
Chlorine(I) oxide also reacts directly with sodium hydroxide to give the same product:
$2NaOH + Cl_2O \rightarrow 2NaOCl + H_2O \nonumber$ | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Period/Acid-base_Behavior_of_the_Oxides.txt |
This page discusses the structures of the chlorides of the Period 3 elements (sodium to sulfur), their physical properties and their reactions with water. Chlorine and argon are omitted—chlorine because it is meaningless to talk about "chlorine chloride", and argon because it is inert and does not form a chloride.
A quick summary of the trends
The chlorides of interest are given in the table below:
NaCl MgCl2 AlCl3 SiCl4 PCl5 S2Cl2
PCl3
Sulfur forms three chlorides, but S2Cl2 is most common. Aluminum chloride also exists under some conditions as a dimer, Al2Cl6.
• The structures: Sodium chloride and magnesium chloride are ionic and consist of large ionic lattices at room temperature. Aluminum chloride and phosphorus(V) chloride are more complicated. They change their structures from ionic to covalent as their solids transition to liquids or vapors. This is discussed in greater detail below. The other chlorides are simple covalent molecules.
• Melting and boiling points: Sodium and magnesium chlorides are solids with high melting and boiling points because of the large amount of heat which is needed to break the strong ionic attractions.
• The rest are liquids or low melting point solids. Leaving aside the aluminum chloride and phosphorus(V) chloride cases where the situation is quite complicated, the attractions in the others will be much weaker intermolecular forces such as van der Waals dispersion forces. These vary depending on the size and shape of the molecule, but will always be far weaker than ionic bonds.
• Electrical conductivity: Sodium and magnesium chlorides are ionic and so will undergo electrolysis when they are molten. Electricity is carried by the movement of the ions and their discharge at the electrodes (not electrons). In the aluminum chloride and phosphorus(V) chloride cases, the solid does not conduct electricity because the ions aren't free to move. In the liquid (where it exists - both of these sublime at ordinary pressures), they have converted into a covalent form, and so don't conduct either. The rest of the chlorides do not conduct electricity either solid or molten because they don't have any ions or any mobile electrons.
• Reactions with water: As an approximation, the simple ionic chlorides (sodium and magnesium chloride) just dissolve in water. Although other chlorides all react with water in a variety of ways described below for each individual chloride. The reaction with water is known as hydrolysis.
Sodium chloride (NaCl)
Sodium chloride is an ionic compound consisting of a giant array of sodium and chloride ions. A small representative portion of a sodium chloride lattice looks like this:
This is normally drawn in an exploded form as:
The strong attractions between the positive and negative ions require a large amount of heat energy to break, so sodium chloride has high melting and boiling points. The compound does not conduct electricity in the solid state because it has no mobile electrons, and the ions are constrained by the crystal lattice. However, when it melts it undergoes electrolysis. Sodium chloride dissolves in water to give a neutral solution.
Magnesium chloride (MgCl2)
Like sodium chloride, magnesium chloride also forms an ionic solid, but with a more complicated crystal structure of the ions to accommodate twice as many chloride ions as magnesium ions. As with sodium chloride, large amounts of heat energy are needed to overcome the attractions between the ions (because of the high lattice enthalpy of the compound), so the melting and boiling points are also high. Solid magnesium chloride is a non-conductor of electricity because the ions are constrained. However, upon melting, the compound undergoes electrolysis.
Magnesium chloride dissolves in water to give a slightly acidic solution (with a pH of approximately 6). When magnesium ions are solvated from the solid lattice, there is enough attraction between the 2+ ions and the water molecules to form coordinate (dative covalent) bonds between the magnesium ions and lone pairs on surrounding water molecules. Hexaaquamagnesium complex ions are formed, [Mg(H2O)6]2+, as follows:
$MgCl_{2 (s)} + 6H_2O \rightarrow [Mg(H_2O)_6]^{2+}_{(aq)} + 2Cl^-_{(aq)} \nonumber$
Many complex ions are acidic, the degree of acidity depending on the attraction between the electrons in the water molecules and the metal at the center of the ion. The hydrogen atoms carry less electron density in this state, and are thus more easily removed by a base. For magnesium, the amount of distortion is quite small, and only a small proportion of the hydrogen atoms are removed, in this case by water molecules in the solution:
$[Mg(H_2O)_6]^{2+} + H_2O_{(l)} \rightleftharpoons [Mg(H_2O)_5(OH)^{2+}]^+ +H_3O^+_{(aq)} \nonumber$
The hydronium ions make the solution acidic. Few are formed (the equilibrium lies well to the left) because the solution is only weakly acidic. The previous equation can be simplified as follows:
$[Mg(H_2O)^6]^{2+}_{(aq)} \rightleftharpoons [Mg(H_2O)_5(OH)]^+_{(aq)} + H^+_{(aq)} \nonumber$
It is essential to include the state symbols if the equation is written this way.
Aluminum chloride (AlCl3)
Electronegativity increases across the period; aluminum and chlorine do not differ enough in electronegativity to form a simple ionic bond. The structure of aluminum chloride changes with temperature. At room temperature, the aluminum is 6-coordinated (i.e. each aluminum is surrounded by 6 chlorine atoms). The structure is an ionic lattice, but it has a lot of covalent character.
At atmospheric pressure, aluminum chloride sublimes at about 180°C. If the pressure is increased to just over 2 atmospheres, it melts instead at a temperature of 192°C.
Both of these temperatures are far below the expected range for an ionic compound. They suggest comparatively weak attractions between molecules instead of strong attractions between ions. This is because the coordination of the aluminum changes at these temperatures. It becomes 4-coordinated—each aluminum is surrounded by 4 chlorine atoms rather than 6. The original lattice converts into an Al2Cl6 arrangement of molecules. The structure is shown below:
In the conversion, all ionic character is lost, causing the aluminum chloride to vaporize or melt (depending on the pressure). These dimers and simple AlCl3 molecules exist in equilibrium. As the temperature increases further, the position of equilibrium shifts more and more to the right of the following system:
$Al_2Cl_6 \rightleftharpoons 2AlCl_3 \nonumber$
Summary of AlCl3
• At room temperature, solid aluminum chloride has an ionic lattice with significant of covalent character.
• At temperatures around 180 - 190°C (depending on the pressure), aluminum chloride coverts to its molecular form, Al2Cl6. This causes it to melt or vaporize due to comparatively weak intermolecular attractions.
• As the temperature increases further, more AlCl3 molecules are formed.
Solid aluminum chloride does not conduct electricity at room temperature because the ions are not free to move. Molten aluminum chloride (only possible at increased pressures) is also nonconductive, because it has lost its ionic character.
Aluminum chloride reacts dramatically with water. A drop of water placed onto solid aluminum chloride produces steamy clouds of hydrogen chloride gas. Solid aluminum chloride in an excess of water still splutters, but instead an acidic solution is formed. A solution of aluminum chloride of ordinary concentrations (around 1 mol dm-3, for example) has a pH around 2-3. More concentrated solutions have a lower pH.
The aluminum chloride reacts with the water rather than simply dissolving in it. In the first instance, hexaaquaaluminum complex ions and chloride ions are formed:
$AlCl_3 (s) + 6H_2O (l) \rightarrow [Al(H_2O)_6]^{3+} (aq) + 3Cl^- (aq) \nonumber$
This is very similar to the magnesium chloride equation given above—the only difference is the charge on the ion. The greater charge attracts electrons in the water molecules quite strongly toward the aluminum, making the hydrogen atoms more positive and therefore easier to remove from the ion. Hence, this ion is much more acidic than in the corresponding magnesium case.
The acid-base equilibria for this reaction lie further to the right than those for magnesium, and so the solution formed is more acidic—more hydronium ions are formed, as shown:
$[Al(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H_3O^+ \nonumber$
or, more simply:
$[Al(H_2O)_6]^{3+} (aq) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} (aq) + H^+ \nonumber$
If there is little water present, hydrogen chloride gas is produced. Because of the heat produced in the reaction and the concentration of the solution formed, hydrogen ions and chloride ions in the mixture combine together as hydrogen chloride ($HCl$) molecules and are given off as a gas. In a large excess of water, the temperature is never high enough for this to happen; the ions remain solvated.
Silicon tetrachloride (SiCl4)
Silicon tetrachloride is a simple no-messing-about covalent chloride. There isn't enough electronegativity difference between the silicon and the chlorine for the two to form ionic bonds. Silicon tetrachloride is a colorless liquid at room temperature which fumes in moist air. The only attractions between the molecules are van der Waals dispersion forces. It doesn't conduct electricity because of the lack of ions or mobile electrons.
It fumes in moist air because it reacts with water in the air to produce hydrogen chloride. If you add water to silicon tetrachloride, there is a violent reaction to produce silicon dioxide and fumes of hydrogen chloride. In a large excess of water, the hydrogen chloride will, of course, dissolve to give a strongly acidic solution containing hydrochloric acid.
$SiCl_4 + 2H_2O \rightarrow SiO_2 + 4HCl \nonumber$
The phosphorus chlorides
There are two phosphorus chlorides: phosphorus(III) chloride, PCl3, and phosphorus(V) chloride, PCl5.
Phosphorus(III) chloride (PCl3)
This simple covalent chloride exists as a fuming liquid at room temperature because there are only van der Waals dispersion forces and dipole-dipole attractions between the molecules. The liquid does not conduct electricity because of the lack of ions or mobile electrons.
Phosphorus(III) chloride reacts violently with water to generate phosphorous acid, H3PO3, and hydrogen chloride fumes (or a solution containing hydrochloric acid in excess of water):
$PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl \nonumber$
Phosphorus(V) chloride (PCl5)
Phosphorus(V) chloride is structurally more complicated than phosphorus(III) chloride. At room temperature, it forms a white solid which sublimes at 163°C. Increasing the temperature beyond its sublimation point dissociates (divides reversibly) more the phosphorus(V) chloride into phosphorus(III) chloride and chlorine:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2 \nonumber$
Phosphorus(V) chloride is an ionic solid. The formation of the ions involves two molecules of PCl5. A chloride ion transfers from one of the original molecules to the other, leaving a positive ion, [PCl4]+, and a negative ion, [PCl6]-.
At 163°C, the phosphorus(V) chloride converts to a molecular form containing PCl5 molecules. Because only van der Waals dispersion forces exist between these molecules, the species vaporizes. Solid phosphorus(V) chloride does not conduct electricity.
Phosphorus(V) chloride reacts violently with water, producing hydrogen chloride fumes. As with the other covalent chlorides, if there is enough water present, these dissolve to give a hydrochloric acid solution.
The reaction happens in two stages. The first takes place in cold water; phosphorus oxychloride, POCl3, is produced along with HCl:
$PCl_5 + 4H_2O \rightarrow POCl_3 + 2HCl \nonumber$
As the solution is brought to a boil, the phosphorus(V) chloride reacts further to give phosphoric(V) acid and more HCl. Phosphoric(V) acid is also known as phosphoric acid or as orthophosphoric acid:
$POCl_3 + 3H_2O \rightarrow H_3PO_4 + 3HCl \nonumber$
Combining these equations gives the overall reaction in boiling water:
$PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl \nonumber$
Disulfur Dichloride (S2Cl2)
Disulfur dichloride is one of three sulfur chlorides and is the species formed when chlorine reacts with hot sulfur. Disulfur dichloride is an orange, unpleasant-smelling covalent liquid. Its rather unusual structure is given below:
The molecule's conformation indicates its possible intermolecular interactions:
• There is no plane of symmetry in the molecule; therefore, it has an overall permanent dipole.
• In liquid state, the molecule experiences van der Waals dispersion forces and dipole-dipole attractions.
• There are no ions in disulfur dichloride and no mobile electrons, making it nonconductive.
Disulfur dichloride reacts slowly with water to produce a complex mixture of hydrochloric acid, sulfur, hydrogen sulfide and various sulfur-containing acids and anions.
Contributors and Attributions
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Period/Chlorides_of_Period_3_Elements.txt |
This page briefly examines how the chemistry of the "hydroxides" of the Period 3 elements vary across the period, from sodium to chlorine.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
Physical Properties of Period 3 Elements
This page describes and explains the trends in atomic and physical properties of the Period 3 elements from sodium to argon. It covers ionization energy, atomic radius, electronegativity, electrical conductivity, melting point and boiling point.
Electronic structures
Across Period 3 of the Periodic Table, the 3s and 3p orbitals fill with electrons. Below are the abbreviated electronic configurations for the eight Period 3 elements:
Na [Ne] 3s1
Mg [Ne] 3s2
Al [Ne] 3s2 3px1
Si [Ne] 3s2 3px1 3py1
P [Ne] 3s2 3px1 3py1 3pz1
S [Ne] 3s2 3px2 3py1 3pz1
Cl [Ne] 3s2 3px2 3py2 3pz1
Ar [Ne] 3s2 3px2 3py2 3pz2
In each case, [Ne] represents the complete electronic configuration of a neon atom.
First Ionization Energy
The first ionization energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of +1.
$X (g) \rightarrow X^+ (g) + e^- \nonumber$
The molar first ionization energy is the energy required to carry out this change per mole of $X$.
The pattern of first ionization energies across Period 3
There is a general upward trend across the period, but this trend is broken by decreases between magnesium and aluminum, and between phosphorus and sulfur.
Explaining the pattern
First ionization energy is dependent on four factors:
• the charge on the nucleus;
• the distance of the outer electron from the nucleus;
• the amount of screening by inner electrons;
• whether the electron is alone in an orbital or one of a pair.
The upward trend: In the whole of period 3, the outer electrons are in 3-level orbitals. These electrons are at approximately the same distance from the nucleus, and are screened by corresponding electrons in orbitals with principal atomic numbers n=1 and n=2. The determining factor in the increase in energy is the increasing number of protons in the nucleus from sodium across to argon. This creates greater attraction between the nucleus and the electrons and thus increases the ionization energies. The increasing nuclear charge also pulls the outer electrons toward the nucleus, further increasing ionization energies across the period.
The decrease at aluminum: The value for aluminum might be expected to be greater than that of magnesium due to the extra proton. However, this effect is offset by the fact that the outer electron of aluminum occupies a 3p orbital rather than a 3s orbital. The 3p electron is slightly farther from the nucleus than the 3s electron, and partially screened by the 3s electrons as well as the inner electrons. Both of these factors offset the effect of the extra proton.
The decrease at sulfur: In this case something other than the transition from a 3s orbital to a 3p orbital must offset the effect of an extra proton. The screening (from the inner electrons and, to some extent, from the 3s electrons) is identical in phosphorus and sulfur , and the electron is removed from an identical orbital. The difference is that in the case of sulfur, the electron being removed is one of the 3px2 pair. The repulsion between the two electrons in the same orbital creates a higher-energy environment, making the electron easier to remove than predicted.
Atomic radius
The diagram below shows how atomic radius changes across Period 3.
The figures used to construct this diagram are based on:
• metallic radii for Na, Mg and Al;
• covalent radii for Si, P, S and Cl;
• the van der Waals radius for Ar (which forms no strong bonds).
It is appropriate to compare metallic and covalent radii because they are both being measured in tightly bonded circumstances. These radii cannot be compared with a van der Waals radius, however, making the diagram deceptive. The general trend towards smaller atoms across the period is not broken at argon. For convenience and clarity, argon is ignored in this discussion.
Explaining the Trend
A metallic or covalent radius is a measure of the distance from the nucleus to the bonding pair of electrons. From sodium to chlorine, the bonding electrons are all in the 3-level, screened by the electrons in the first and second levels. The increasing number of protons in the nucleus across the period attracts the bonding electrons more strongly. The amount of screening is constant across Period 3.
Electronegativity
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values decrease toward cesium and francium which are the least electronegative at 0.7.
The trend
The trend across Period 3 looks like this:
Argon is not included; because it does not form covalent bonds, its electronegativity cannot be assigned.
Explaining the Trend
The explanation is the same as that for the trend in atomic radii. Across the period, the valence electrons for each atom are in the 3-level. They are screened by the same inner electrons. The only difference is the number of protons in the nucleus. From sodium to chlorine, the number of protons steadily increases and so attracts the bonding pair more closely.
Physical Properties
This section discusses electrical conductivity and the melting and boiling points of the Period 3 elements. An understanding of the structure of each element is necessary for this discussion.
Structures of the elements
The structures of the elements vary across the period. The first three are metallic, silicon is network covalent, and the rest are simple molecules.
Three metallic structures
Sodium, magnesium and aluminum all have metallic structures. In sodium, only one electron per atom is involved in the metallic bond, the single 3s electron. In magnesium, both of its outer electrons are involved, and in aluminum all three are involved. One key difference to be aware of is the way the atoms are packed in the metal crystal. Sodium is 8-coordinated with each sodium atom interacting with only 8 other atoms. Magnesium and aluminum are each 12-coordinated, and therefore packed more efficiently, creating less empty space in the metal structures and stronger bonding in the metal.
A network covalent structure
Silicon has a network covalent structure like that of diamond. A representative section of this structure is shown:
The structure is held together by strong covalent bonds in all three dimensions.
Four simple molecular structures
The structures of phosphorus and sulfur vary depending on the type of phosphorus or sulfur in question. In this case, white phosphorus and one of the crystalline forms of sulfur—rhombic or monoclinic—are considered. These structures are shown below:
Aside from argon, the atoms in each of these molecules are held together by covalent bonds. In the liquid or solid state, the molecules are held in close proximity by van der Waals dispersion forces .
Electrical conductivity
• Sodium, magnesium and aluminum are good conductors of electricity. Conductivity increases from sodium to magnesium to aluminum.
• Silicon is a semiconductor.
• Phosphorus, sulfur, chlorine, and argon are nonconductive.
The three metals conduct electricity because the delocalized electrons (as in the "sea of electrons" model) are free to move throughout the solid or the liquid metal. Semiconductor chemistry for substances such as silicon is beyond the scope of most introductory level chemistry courses. The other elements do not conduct electricity because they are simple molecular substances. without free, delocalized electrons..
Melting and boiling points
The chart shows how the melting and boiling points of the elements change as you go across the period. The figures are plotted in kelvin rather than °C to avoid showing negative temperatures.
The metallic structures
Melting and boiling points increase across the three metals because of the increasing strength of their metallic bonds. The number of electrons which each atom can contribute to the delocalized "sea of electrons" increases. The atoms also get smaller and have more protons as you go from sodium to magnesium to aluminum. The attractions and therefore the melting and boiling points increase because:
• The nuclei of the atoms are more positively charged.
• The "sea" is more negatively charged.
• The "sea" is progressively nearer to the nuclei and thus is more strongly attracted.
Silicon
Silicon has high melting and boiling points due to its network covalent structure. Melting or boiling silicon requires the breaking of strong covalent bonds. Because of the two different types of bonding in silicon and aluminum, it makes little sense to directly compare the two melting and boiling points.
The four molecular elements
Phosphorus, sulfur, chlorine and argon are simple molecular substances with only van der Waals attractions between the molecules. Their melting or boiling points are lower than those of the first four members of the period which have complex structures. The magnitudes of the melting and boiling points are governed entirely by the sizes of the molecules, which are shown again for reference:
• Phosphorus: Elemental phosphorus adopts the tetrahedral P4 arrangement. Melting phosphorus breaks no covalent bonds; instead, it disrupts the much weaker van der Waals forces between the molecules.
• Sulfur: Elemental sulfur forms S8 rings of atoms. The molecules are bigger than phosphorus molecules, and thus the van der Waals attractions are stronger, leading to a higher melting and boiling point.
• Chlorine: Chlorine, Cl2, is a much smaller molecule with comparatively weak van der Waals attractions, and thus chlorine will have a lower melting and boiling point than sulfur or phosphorus.
• Argon: Elemental argon is monatomic . The scope for van der Waals attractions between argon atoms is very limited and so the melting and boiling points of argon are lower again.
Contributors and Attributions
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Period/Hydroxides_of_Period_3_Elements.txt |
This page explains the relationship between the physical properties of the oxides of Period 3 elements and their structures (including sodium to chlorine; argon is omitted because it does not form an oxide).
A summary of the trends
The oxides:
Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7
P4O6 SO2 Cl2O
The oxides in the top row are the highest known oxides of the various elements, in which the Period 3 elements are in their highest oxidation states. In these oxides, all the outer electrons in the Period 3 elements are involved in bonding.
• The structures: The metallic oxides on the left adopt giant structures of ions on the left of the period; in the middle, silicon forms a giant covalent oxide (silicon dioxide); the elements on the right form molecular oxides.
• Melting and boiling points: The large structures (the metal oxides and silicon dioxide) have high melting and boiling points because a large amount of energy is needed to break the strong bonds (ionic or covalent) operating in three dimensions. The oxides of phosphorus, sulfur and chlorine consist of individual molecules, simple or polymeric.
The attractive forces between these molecules include van der Waals dispersion and dipole-dipole interactions. These vary in size depending on the size, shape and polarity of the various molecules, but will always be much weaker than the ionic or covalent bonds in a giant structure. These oxides tend to be gases, liquids, or low melting point solids.
• Electrical conductivity: None of the oxides above have any free or mobile electrons, indicating that none of them will conduct electricity when solid. The ionic oxides can, however, undergo electrolysis in a molten state. They can then conduct electricity because of the movement of the ions towards the electrodes and the discharge of the ions when they reach the electrodes.
The metallic oxides
• The structures: Sodium, magnesium and aluminum oxides consist of giant structures containing metal ions and oxide ions. Magnesium oxide is similar in structure to sodium chloride. The other two oxides have more complicated possible arrangements.
• Melting and boiling points: There are strong attractions between the ions in each of the oxides above; breaking these attractions requires much heat energy. These oxides therefore have high melting and boiling points.
• Electrical conductivity: None of the metallic oxides conduct electricity in the solid state, but electrolysis is possible when molten. As discussed above, in a molten state, the oxides conduct electricity because of the movement and discharge of the ions present. The only industrially-important example of this process is the electrolysis of aluminum oxide in the manufacture of aluminum. Whether you can electrolyze molten sodium oxide depends on whether it actually melts (as opposed to subliming or decomposing under ordinary conditions). If it sublimes (experiences a solid-to-vapor transition) , no liquid can electrolyze. Magnesium and aluminum oxides have melting points that are too high for electrolysis under laboratory conditions.
Silicon dioxide (silicon(IV) oxide)
The structure: The electronegativity of the elements increases across the period; silicon and oxygen do not differ enough in electronegativity to form an ionic bond. There are three different crystal forms of silicon dioxide; the most convenient structure to visualize is similar to that of diamond. In silicon dioxide, oxygen atoms fill the empty spaces along the silicon-silicon bond axes, as shown in a representative structure below:
This structure is repeated many times throughout the silicon dioxide substance.
Melting and boiling points: Silicon dioxide has a high melting point that varies depending on the particular structure (the structure given is one of three possible structures), but each is close to 1700°C. Very strong silicon-oxygen covalent bonds must be broken throughout the structure before melting occurs. Silicon dioxide boils at 2230°C. Because two types of bonding are considered, it makes little sense to compare these values directly to those of the metallic oxides; suffice it to say that in both types of structures, the melting and boiling points are very high.
Electrical conductivity: Silicon dioxide has no mobile electrons or ions, and hence does not conduct electricity either as a solid or a liquid.
The molecular oxides
Phosphorus, sulfur and chlorine form molecular oxides. Some of these molecules are fairly simple—others are polymeric. Here the simple structures are considered. Melting and boiling points of these oxides are much lower than those of the metal oxides or silicon dioxide. The intermolecular forces binding one molecule to its neighbors are van der Waals dispersion forces or dipole-dipole interactions. The strength of these vary depending on the size of the molecules. None of these oxides conduct electricity either as solids or as liquids, because none of them contain ions or free electrons.
The phosphorus oxides
Phosphorus has two common oxides, phosphorus(III) oxide, P4O6, and phosphorus(V) oxide, P4O10.
Phosphorus(III) oxide: Phosphorus(III) oxide is a white solid, melting at 24°C and boiling at 173°C. To understand its structure, consider a tetrahedral P4 molecule:
The structure is expanded to display the bonds:
The phosphorus-phosphorus bonds are interrupted with oxygen atoms, in a bent shape similar to water, as shown below:
Only 3 of the valence electrons of phosphorus (the 3 unpaired p electrons) are involved in the phosphorus-oxygen bonds.
Phosphorus(V) oxide: Phosphorus(V) oxide is also a white solid, which sublimes at 300°C. In this case, the phosphorus uses all five of its outer electrons in the bonding. Solid phosphorus(V) oxide exists in several different forms - some of them polymeric. We are going to concentrate on a simple molecular form, and this is also present in the vapor. This is most easily drawn starting from P4O6. The other four oxygens are attached to the four phosphorus atoms via double bonds.
The sulfur oxides
Sulfur has two common oxides, sulfur dioxide (sulfur(IV) oxide), SO2, and sulfur trioxide (sulfur(VI) oxide), SO3.
Sulfur dioxide: S ulfur dioxide is a colorless gas at room temperature with an easily recognized rotten-egg smell. It consists of simple SO2 molecules as shown:
The sulfur uses four of its six valence electrons to form the double bonds with oxygen, leaving the other two as a lone pair on the sulfur. The bent shape of SO2 is due to this lone pair.
Sulfur trioxide: Pure sulfur trioxide is a white solid with a low melting and boiling point. It reacts very rapidly with water vapour in the air to form sulfuric acid. Under laboratory conditions, it forms a white sludge which fumes dramatically in moist air (forming a fog of sulfuric acid droplets). Gaseous sulfur trioxide consists of simple SO3 molecules in which all six of the sulfur's outer electrons are involved in the bonding.
There are various forms of solid sulfur trioxide. The simplest one is a trimer, S3O9, in which three SO3 molecules are joined in a ring.
There are also other polymeric forms in which the SO3 molecules join together in long chains. For example:
The tendency of sulfur trioxide to form polymers determines to its solid nature.
The chlorine oxides
Chlorine forms several oxides. Two are considered here: chlorine(I) oxide, Cl2O, and chlorine(VII) oxide, Cl2O7.
Chlorine(I) oxide: Chlorine(I) oxide is a yellowish-red gas at room temperature. It consists of simple, small molecules.
The physical properties of chlorine (I) oxide are consistent with those predicted for a molecule of its size.
Chlorine(VII) oxide: In chlorine(VII) oxide, the chlorine involves all of its seven valence electrons in bonds with oxygen. This produces a molecule much larger than chlorine (I) oxide, suggesting higher melting and boiling points. Chlorine(VII) oxide is a colorless oily liquid at room temperature. The diagram below shows a simple structural formula, neglecting three-dimensional structure; the geometry is tetrahedral around both chlorine atoms, and V-shaped around the central oxygen. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Period/Physical_Properties_of_Period_3_Oxides.txt |
This page describes the reactions of the Period 3 elements from sodium to argon with water, oxygen and chlorine.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
Structures and Physical Properties of Period 3 Elements
This page describes the structures of the Period 3 elements from sodium to argon, and shows how these structures can be used to explain the physical properties of the elements.
Melting and boiling points
In a moment we shall explain all the ups and downs in this graph.
Electrical conductivity
Sodium, magnesium and aluminum are all good conductors of electricity. Silicon is a semiconductor. None of the rest conduct electricity. These trends are explained below.
Three metallic structures
Sodium, magnesium and aluminum all have metallic structures, which accounts for their electrical conductivity and relatively high melting and boiling points. Melting and boiling points rise across the three metals because of the increasing number of electrons which each atom can contribute to the delocalized "sea of electrons". The atoms also get smaller and have more protons as you go from sodium to magnesium to aluminum.
The attractions and therefore the melting and boiling points increase because:
• The nuclei of the atoms are getting more positively charged.
• The sea is getting more negatively charged.
• The sea is getting progressively nearer to the nuclei and so more strongly attracted.
Silicon - a giant covalent structure
Silicon is a non-metal, and has a giant covalent structure exactly the same as carbon in diamond - hence the high melting point. You have to break strong covalent bonds in order to melt it. There are no obviously free electrons in the structure, and although it conducts electricity, it doesn't do so in the same way as metals. Silicon is a semiconductor.
Four molecular elements
Phosphorus, sulfur, chlorine and argon are simple molecular substances with only van der Waals attractions between the molecules. Their melting or boiling points will be lower than those of the first four members of the period which have giant structures. The presence of individual molecules prevents any possibility of electrons flowing, and so none of them conduct electricity. The sizes of the melting and boiling points are governed entirely by the sizes of the molecules:
Argon molecules consist of single argon atoms.
• Phosphorus: There are several forms of phosphorus. The data in the graph at the top of the page applies to white phosphorus which contains P4 molecules. To melt phosphorus you don't have to break any covalent bonds - just the much weaker van der Waals forces between the molecules.
• Sulfur: sulfur consists of S8 rings of atoms. The molecules are bigger than phosphorus molecules, and so the van der Waals attractions will be stronger, leading to a higher melting and boiling point.
• Chlorine: Chlorine, Cl2, is a much smaller molecule with comparatively weak van der Waals attractions, and so chlorine will have a lower melting and boiling point than sulfur or phosphorus.
• Argon: Argon molecules are just single argon atoms, Ar. The scope for van der Waals attractions between these is very limited and so the melting and boiling points of argon are lower again.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
Period 6 Elements: The Lanthanides
The Lanthanides consist of the elements in the f-block of period six in the periodic table. While these metals can be considered transition metals, they have properties that set them apart from the rest of the elements.
Period 7 Elements: The Actinides
The Actinide series contains elements with atomic numbers 89 to 103 and is the sixth period and third group of the periodic table. The series is the row below the Lanthanide series, which is located underneath the main body of the periodic table. Both the Lanthanide and Actinide series are referred to as Rare Earth Metals and have a high diversity in oxidation numbers. All of the Actinides are radioactive. The most famous actinide is Uranium.
• General Properties and Reactions of The Actinides
The Actinide series contains elements with atomic numbers 89 to 103 and is in the sixth period and the third group of the periodic table. The series is the row below the Lanthanide series, which is located underneath the main body of the periodic table. Lanthanide and Actinide Series are both referred to as Rare Earth Metals. These elements all have a high diversity in oxidation numbers and all are radioactive.
• Chemistry of Americium
• Chemistry of Berkelium
• Chemistry of Californium
The fifth element in succession to emerge from the Berkeley, California cyclotron was element 98, californium (named after the State of origin).
• Chemistry of Curium
• Chemistry of Einsteinium
• Chemistry of Fermium
• Chemistry of Lawrencium
• Chemistry of Mendelevium
• Chemistry of Neptunium
• Chemistry of Nobelium
Discovered by Ghiorso, Sikkeland, Walton and Seaborg (see element 106) and named for Alfred Nobel, element 102 was originally named by a Swedish group working at the Nobel Institute of Physics in Stockholm. Details of their work, however, did not yield the expected results and the credit eventually shifted to the American team which decided to retain the original given name. The most stable isotope is No-259 with a half-life of just about 1 minute.
• Chemistry of Plutonium
• Chemistry of Protoactinium
• Chemistry of Thorium
• Chemistry of Uranium
Most of the naturally occurring uranium is the isotope U-238. This form of uranium is not fissionable, i.e., it cannot be used in atomic weapons or power plants. A much smaller percentage of naturally occurring uranium is the isotope U-235, which is fissionable. The process of "enriching" uranium to increase the proportion of U-235 in a sample is expensive and tedious but necessary to produce fuel that is usable in power plants and material for weapons. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Period/Reactions_of_Period_3_Elements.txt |
• Aluminas
Aluminas are the second most abundant mineral of the earth crust. From the discussion on this page, you will be introduced to various forms of alumina, their structures, and properties so that when you encounter them, you can associate their properties with their chemical identities (compositions) and structures.
• Aluminosilicates
• Silicates
The silicates are the largest, the most interesting and the most complicated class of minerals than any other minerals. Approximately 30% of all minerals are silicates and some geologists estimate that 90% of the Earth's crust is made up of silicate. Thus, oxygen and silicon are the two most abundant elements in the earth's crust.
Aluminosilicates
Discussion Questions
• What are natural occurring aluminum oxides?
• How is bauxite mined and processed?
• What are some of the applications of aluminum oxides?
• How does aluminum oxide protect aluminum from corrosion?
Aside from silicates, aluminas are the most abundant mineral of the earth crust. Thus, it is important for chemical engineers to know some chemistry about aluminas, because they are found and used in so many different places and technologies. Furthermore, aluminum ions often replace silicon ions in silicates forming aluminosilicates, which is discussed in the next page.
From the discussion on this page, you will be introduced to various forms of alumina, their structures, and properties so that when you encounter them, you can associate their properties with their chemical identities (compositions) and structures.
What are natural occurring aluminum oxides?
The most common ore is bauxite, which is aluminum oxide, Al2O3, mixed with oxides of silicon, iron, and other elements and varying small percentages of clay and other silicates. Physically, bauxite can be as hard as rock or as soft as mud, and its color may be red, white, buff, pink, yellow or any combination of these. The picture shows the mining of bauxite at Gove in Australia Bauxite is the product of extreme chemical weathering of aluminum-rich rocks. Australia produces the largest amount of alumina because she has a large body of bauxite. Jamica, South Africa and some other countries also have a good reserve.
Aluminum oxides often coexist with silicates. Natural aluminum oxide minerals include
• Corundum, Al2O3
• Spinel, MgAl2O4
• Hercynite, FeAl2O4
• Galaxite, MnAl2O4
• Gibbsite, Al(OH)3
• Diaspore, AlO(OH)
• Boehmite, AlO(OH)
Bauxites are mainly used for producing pure alumina, which is the feed stock for aluminum metal production, raw material for ceramics, and other applications.
Corundum
There are a few forms of aluminum oxide, and corundum being the most common. The structure of corundum can be viewed as a hexagonal close packed array of oxygen atoms with 2/3 of the octahedral sites occupied by Al3+ ions. Thus, the Al3+ ions are bonded to 6 oxygen in a distorted octahedron. Each such octahedron share a face with one on the upper and one on the lower layers. The distortion is caused by repulsion between Al3+ ions in octahedra sharing the faces.
Corundum is a dense (specific gravity of 3.97), hard (9 on the Mohs' scale, next only to diamond), high melting (melting point 2288 K), and insoluble in water. Crystals of corundum are usually prismatic or barrel-shaped bounded by steep pyramids. A massive grey granular corundum powder is called emery.
Colored corundum are called ruby (deep red due to presence of Cr3+ ions) and sapphire (blue, pink, yellow or green due to various degrees of Fe2+ or 3+, and Ti4+). The color may be modified by heating or irradiation. Some ruby crystals are shown here exposed in a piece of bauxite ore. Grey corundum or emery are used as abrasive, for example, emery paper (sand paper) and ruby and sapphire are for gemstones. They do have technical applications, for example, the first LASER was produced using a ruby crystal.
How is bauxite mined and processed?
Pure aluminas are used for pottery, ceramics, refractories, catalyst supports, and for the production of aluminum by the Hall process. Thus, bauxite and other aluminum containing minerals such as kaolinite ($Al_4Si_4O_{10}(OH)_8$) must be mined and processed to produce pure alumina. Australia produces about \$3 billion worth of alumina a year from six Australian refineries. These refineries use the Bayer Process to extract aluminum hydroxide from the bauxite using hot caustic liquor. Aluminum oxide, Al2O3, is a typical amphoteric oxide, which dissolves in a strong acid and a strong base.
$Al_2O_3 + 6H^+ \rightarrow 2 Al^{3+} + 3 H_2O \tag{1}$
$Al_2O_3 + 6 OH^- + 3 H_2O \rightarrow 2 Al(OH)_6^{3-} \tag{2}$
After separation of the solids residue, the clear liquor is cooled. Depending on the pH of the solution, the aluminate ion Al(OH)63- bears various amounts of charge due to these reactions:
$Al(OH)_6^{3-} + H6+ \rightarrow Al(OH)_5^{2-}(H_2O) \tag{3}$
$Al(OH)_5^{2} \cdot (H_2O) + H^+ = Al(OH)_4 \cdot (H_2O)_2 \tag{4}$
$Al(OH)_4 \cdot (H_2O)_2 + H^+ \rightarrow \underset{\text{a precipitate}}{Al(OH)_3(H_2O)_3} \tag{5}$
In a neutral solution, the compound, Al(OH)3(H2O)3 or Al(OH)3 if water is ignored, forms a gelatinous precipitate. Under controlled manner, the liquor crystallizes to give particles of gibbsite of the desired chemical purity and physical characteristics. A lot of research and development has gone into this crystallization process alone.
The hydroxide ions of gibbsite form two layers similar to layers of closest packed spheres with Al3+ ions filling in some of the octahedral sites. The crystal structure of gibbsite consists of stacked double layers. It is expected that the hydroxide ions form extensive intra- and inter-layer hydrogen bonds.
Further dehydration converts Al(OH)3 into diaspore and boehmite, both of which have the stoichiometry AlO(OH). Gibbsite is is converted to alumina, Al2O3 by calcination. Alumina is marketed as the feed stock to smelters for the production of aluminium metal, ceramics, catalyst supports and other applications.
What are some of the applications of aluminum oxides?
Mineralogists consider a mineral a homogeneous solid body, formed by natural process that has a regular crystal structure with a limit range of atomic compositions. Engineers are mainly interested in properties and their applications. Scientists are interested in correlate the relationship of structures and properties. Engineers deal with natural and synthetic materials alike. Aluminum oxide is a basic material for the ceramic industry. For more details regarding the properties of alumina, consult the data sheet for alumina ceramics.
Aluminas are basic materials for ceramics, and they are useful for lining containers and mass transferring pipes, especially if heat resistance is required. Intricate tools such as the 95 % alumina ceramic rotor for 20 cm rotary valve have been made of these materials. Ceramics are related to many technologies. A list of resources related to ceramics gives many companies, whose main products are made of ceramic materials. For example, aluminas are used for paint, ink, coating and filling paper, adhesives, rubber, pharmaceuticals, tiles, bricks, cooking utilities, table wares, electronic components, porcelain, pottery, dental restoration, and plastics.
Technological changes demand materials with new specific properties. Since changes take place all the time, new materials are also developed all the time. Additions of specific amounts of other oxides to aluminas produce composite materials whose properties differ from both parent materials. This type of blending is a new frontier of material engineering.
Aluminum Oxide protects Aluminum from Corrosion?
Aluminum is a very reactive metal if it is not protected by its aluminum oxide film. It is much more reactive than zinc and iron, but far less reactive than magnesium. Their oxidation reduction potentials are given below for you to compare.
Fe3+ + 3 e- = Fe(s), Eo = -0.037 V.
Fe2+ + 2 e- = Fe(s), Eo = -0.447 V.
Zn2+ + 2 e- = Zn(s), Eo = -0.76 V.
Al3+ + 3 e- = Al(s), Eo = -1.67 V.
Mg2+ + 2 e- = Mg(s), Eo = -2.70 V.
If the oxide film is cracked under aerated water, Al3+ is formed instantly along with OH- ions. Thus, an oxide film is formed immediately, sealing it from further corrosion at the anodic site. Another important fact is that Al2+ ions will not form, and the aluminum oxide is an inert substance.
Example $1$
Calculate the molar volume of aluminum and aluminum oxide.
Solution
The question requires the densities of aluminum and aluminum oxides. The CRC Handbook give their densities as 2.702 and 3.97 g/cc respectively. Thus, molar volumes of Al and aluminum oxides are
Vo of Al = 26.98 / 2.702 = 9.96 cc per mol of Al
Vo of Al2O3 = (26.98 + 24.0) / 3.97 = 12.84 cc
DISCUSSION
By applying the Pilling and Bedworth model, aluminum oxide formed has a larger molar volume per Al than the metal itself. Thus, the Pilling and Bedworth model also apply to aluminum and its oxide.
Furthermore, aluminum oxide and some hydrates have different densities,
Al2O3.H2O, 3.014 g/cc.
Al2O3.3 H2O, 2.42 g/cc.
The formation of Al(OH)3 forms a protective layer. This formula suggest the formation of gibbsite, density 2.44 g/cc leading to a molar volume of 32.0 cc.
Questions
1. What is bauxite?
2. What are the two colored forms of corundum?
3. What is the hardness of corundum?
4. What is the arrangement of the oxygen atoms in corundum?
5. What is an amphoteric oxide?
6. What solution is used to extract aluminum from bauxite in the Bayer process?Is aluminum a more reactive metal than zinc towards oxygen?
7. Why does aluminum not rust?
Solutions
1. Bauxite is an ore of aluminum oxide and other metal oxide.
Skill -
To be able to identify the ore bauxite from the description given in this page.
2. Red corundum is ruby and blue corundum is sapphire.
Skill -
Explain the relationship of various materials, even if they appear very different.
3. 9 on the Mohs' scale, next to diamond.
Skill -
Explain the properties of corundum and relate them to its applications.
4. The oxygen atoms pack as a hexagonal array.
Skill -
Describe the structure of corundum, an important mineral related to ruby and sapphire.
5. An oxide that dissolves in strong acid and strong base.
Skill -
Explain why a concentrated basic solution is used in the Bayer process.
6. A strong base is used to dissolve aluminum oxide.
Skill -
Describe the chemistry of the Bayer process for the extraction of alumina.
7. Yes, it is.
Skill -
Explain the chemical property base on electromotive potentials.
8. A thin layer of aluminum oxide protects it from oxidation.
Discussion -
The thin layer does not cover the luster of aluminum metal. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Aluminosilicates/Aluminas.txt |
While you are admiring this beautiful picture of faujasite, remember that the oxygen atoms have two unshared electron pairs in addition to the (Al,Si)-O-Si(or Al) bonds. Thus the oxygen atoms are sites to interact with positive site of molecules that passes by these structures. At present over 150 synthetic zeolites & zeotypes and 40 natural zeolites are known. Synthesis of zeolite is a very active field of study.
Aluminosilicates have three major minerals: Andalusite, sillimanite, and kyanite. Zeochem has been developing and manufacturing molecular sieve adsorbents since 1977. Simply put, their adsorbents are used to "screen" out impurities from a variety of applications by attracting and trapping the targeted contaminants. For example, in natural gas processing, molecular sieves are used to remove specific molecules from the gas stream to allow for more efficient downstream processing. Faujasite is a typical zeolite.
Applications of Zeolites?
As you have read above that there are many different kinds of zeolites, each with a definite structure and associate with it are unique properties. In terms of applications, we are assuming zeolites as porous aluminosilicates with large tunnels and cages for a fluid (gas and liquid) to pass through. The applications are based on the interactions between the fluid phase and the atoms or ions of the zeolites. In general terms, zeolites have many applications:
1. As selective and strong adsorbers: remove toxic material, selective concentrate a particular chemical, as Molecular Sieve. This link will be a very good to discuss zeolites. Currently, the site is under construction, but it has a very good framework. Even many deorderants are zeolite type.
2. As selective ion exchangers: for example used in water softener.
3. Superb solid acid catalysts, when the cations are protons H+. As catalysts, their environmental advantages include decreased corrosion, improved handling, decreased environmentally toxic waste and minimal undesirable byproducts.
4. As builder: a material that enhance or protecting the cleaning power of a detergent. Sodium aluminosilicate is an ion exchange builder often used in laundry detergent as a builder. A builder inactive the hardness of water by either keeping calcium ions in solution, by precipitation, or by ion exchange.
123 ppm CaCO3 = 123 g per 106 g of water.
``` 1 mol CaCO3 2 mol H+ 1 mol z-A 1926 g z-A 100
123 g CaCO3 ----------- ----------- ---------- ---------- ---
100 g CaCO3 1 mol CaCO3 12 mol H+ 1 mol z-A 80
```
= 494 g zeolite A
That 80 % of protons of the zeolite A is used means that we require a little more zeolite A than stoichiometric quantities.
DISCUSSION
Zeolites are aluminosilicates with open frames strcutures discussed above. Replacement of each Si atom by an Al atom in silicates results in having an extra negative charge on the frame. These charges must be balanced by trapping positive ions: H+, Na K+, Ca2+, Cu2+ or Mg2+. Water molecules are also trapped in the frame work of zeolites.
In this example, we assume that when we soak the zeolite in water containing Ca2+, and Mg2+ ions, these ions are more attrative to the zeolite than the small, singly charged protons. We further assumed that 80 percent of the protons in zeolite are replaced by other ions.
58.5 g/mol.
``` 1 mol 0.8*12 mol NaCl 100 58.5 g NaCl
10 kg z-A -------- --------------- --- -----------
2.190 kg 1 mol z-A 20 1 mol NaCl
= 12822 g NaCl = 12.8 kg NaCl
```
DISCUSSION
How much salt is required if 60% of the sodium ions are effectively used to replace all the divalent ions? | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Aluminosilicates/Aluminosilicates.txt |
Discussion Questions
• How to classify silicates?
• What are asbestoses?
A photograph of quartz, one of the many silicates, is shown here. Quartz alone is a very interesting mineral, used in many technologies ranging from jewry to watch, computers, to scientific instrumentation, because quartz is an important piezoelectric material. A clever application of piezoelectric effect is to use its vibration movement to deliver high viscosity ink in the drop-on-demand ink jet system. Piezoelectric materials are also used in Piezo Electric Transducers and Buzzers Techniques have also been developed to apply piezoelectric material for the conversion of static wave energy into rotation motion. Such a Piezoelectric Motor is developed for the wrist watch application.
The silicates are the largest, the most interesting and the most complicated class of minerals than any other minerals. Approximately 30% of all minerals are silicates and some geologists estimate that 90% of the Earth's crust is made up of silicates, SiO44- based material. Thus, oxygen and silicon are the two most abundant elements in the earth's crust.
Silicates is based on the basic chemical unit SiO44-, tetrahedron shaped anionic group. The central silicon ion has a charge of positive four while each oxygen has a charge of negative two (-2) and thus each silicon-oxygen bond is equal to one half (½ ) the total bond energy of oxygen. This condition leaves the oxygens with the option of bonding to another silicon ion and therefore linking one SiO44- tetrahedron to another.
In the extreme case, the tetrahedra are arranged in a regular, orderly fashion forming a three-dimensional network. Quartz is such a structure (see the diagram), and its formula is SiO2. If silica in the molten state is cooled very slowly it crystallizes at the freezing point. But if molten silica is cooled more rapidly, the resulting solid is a disorderly arrangement which is called glass, often also called quartz.
Classify silicates?
For the study of many silicate based minerals, a classification scheme is required. Otherwise, the huge amount of information becomes unmanageable. We use a simple classification scheme based on the number of \(SiO_4^{4-}\) units that are connected together by sharing the oxygen atoms.
Orthosilicates
Orthosilicates are minerals consisting of only single SiO44- units. The cations are some other metals. For example, the following minerals are orthosilicates:
The Be and Zn ions are tetrahedrally bonded to the oxygen of the silicate in these two minerals: phenacite, Be2SiO4 and willemite, Zn2SiO4. In olivine, (Fe, Mg)2SiO4, the cations are either Fe2+ or Mg2+. This formula suggests that this mineral is a mixed salt of iron and magnesium silicates. These cations are octahedrally coordinated to the oxygen atoms of the silicate. Pure salt Fe2SiO4 is called fayalite, and Mg2SiO4 is called forsterite.
Pyrosilicates
When two SiO44- units are linked together, they form the pyrosilicate group, Si2O76-. For example, thortveitite, Sc2Si2O7 is a pyrosilicate.
Ring and chain silicates
When two oxygen of SiO44- units share with other SiO44- units, the silicates form a ring or an infinite chain. The stoichiometry of the silicates becomes (SiO3)n2n-. Benitoite BaTi(SiO3)3 contain three silica rings, but these are relaxed 6-atom rings
``` SiO2
/ \
O O
| |
O2Si SiO2
\ /
O
```
The precious stone beryl Be3Al2(SiO3)6 contain six-silica rings.
Single chain silica are called pyroxenes. Some synthetic metasilicates Na2(SiO3) have been shown to contain the simple chain silicates (SiO3)n, in which the Si-O bonds of the type Si-O-Si are 168 nm, with the Si-O-Si angles of 137o. The Si=O bonds are shorter, 1.57 nm. The natural pyroxenes include enstatite, MgSiO3, diopside, CaMg(SiO3)2, and jadeite, NaAl(SiO3)2.
Double chain silicates
Double chain silicates are called amphiboles, part of the double chain is shown here, same as the double chain shown in Inorganic Chemistry by Swaddle. These chains have a stoichiometry of (Si4O11)n6n. You can easily identify one such unit in the diagram. Tremolite, Ca2Mg5(Si4O11)2(OH)2, is such a mineral.
The true asbestoses such as crocidolite or blue asbestos consist of double chain silicates. Asbestoses have been identified as carcinogens, and its application has since been limited due to a ban to limit its exposure to the public. Most commercial asbestoses are chrysotile, which contain layers of silicate sheet as we shall below.
Silicates with sheet structures
Sheet slilicates are called phyllosilicates (phyllo means leaflike). These silicates are easy to cleave (as does graphite). Talc is a typical sheet silicate, Mg3(OH)2(Si2O5. Talc is a main ingredient of the soapstone (steatite).
The diagram below shows the arrangement of sheets in brucite, Mg(OH)2, in which the sheets consist of corner sharing octahedrons of Mg(OH)6. In chlorite, there are two types of sheets. Half of the sheets are the same as those of brucite, but half of the brucite-sheets are sandwiched between sheets of silicates. The talc consists of only the sandwiched sheets.
The diagram came from a polysome series which discusses sheet silicates.
Serpentine, Mg3(OH)4Si2O5, has curved sheets.
The comercial asbestos chrysotile is a sheet silicate, but the sheets are rolled up like a tube. These tubes appear as fibers, and they are usually known as asbestos.
Silicates with 3-dimensional framework
As mentioned earlier, the SiO44- units can share every oxygen with other units to form a three dimensional network, and quartz has such a structure. A portion of such a framework is shown here. In this arrangement, the stoichiometry is reduced to SiO2, which is often called silica. A collection of small pieces of quartz is sand.
Please see an animated illustration of the ring structure by Bob Hanson.
Quartz is a group of minerals.
Asbestoses?
Asbestos is the name applied to six naturally occurring minerals that are mined from the earth. The different types of asbestos are:
• Amosite
• Chrysotile
• Tremolite
• ActinoliteUnlink
• Anthophyllite
• Crocidolite
Of these six, three are used more commonly. Chrysotile (white) is the most common, but it is not unusual to encounter. Amosite (brown / off-white), or Crocidolite (blue) as well. Asbestos are noncombustable fibrous material, and they have been used for terminal insulation material, brake linings, construction material, and filters. When mixed with cement, it reinforce the mechanical strength of concrete. It decomposes due to loss of water, and forms forsterite and silica at high temperature.
Example 1
What is the stoichiometry for an infinite chain of silicates by sharing two oxygen atoms with the neighboring units?
Solution
We can draw such a chain first:
``` O O O O O O O O O O O O
Si Si Si Si Si Si
/ \ / \ / \ / \ / \ / \ /
O O O O O O O
```
In this diagram, the basic unit is bolded, and the stoichiometry is apparently SiO3.
DISCUSSION
Is there another type of chain that satisfies the condition given in the question? If you work with 3-dimensional models, you probably realize that this is the only way to construct a single chain. Infinite chains and a ring structures have the same stoichiometry.
Questions
1. What is the basic units for silicate based minerals?
2. What is the chemical composition of quartz?
3. Briefly describe the quartz crystal structure.
4. What are orthosilicates?
5. What is the name given to the silicates that contain the Si2O76- units?
6. Briefly describe the quartz crystal structure.
Solutions
1. SiO4
Skill -
Describe the basic SiO44- unit of silicate and its versatility in connecting to each other or one another.
2. SiO2
Skill -
Identify the units used to build a structure.
3. The tetrahedra form a 3-dimensional network like that of a diamond but with linkages of Si-O-Si in places of C-C bonds of diamond.
Skill -
These diagrams age discussed during the lectures.
4. Minerals consisting of single SiO44- units.
Skill -
Minerals consisting of single Be2+, units are called orthosilicates. These form salts with divalent cations Be2+, Zn2+, Mg2+, Fe2+ etc, but these divalent cations are tightly bonded to the oxygen of the Be2+ units.
5. Pyrosilicates
6. The tetrahedra form a 3-dimensional network like that of diamond but with linkages of Si-O-Si in places of C-C bonds of diamond.
Skill -
These diagrams age discussed during the lectures. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Aluminosilicates/Silicates.txt |
Carbonate is a polyatomic anion with the formula $CO_3^{2-}$ and has a trigonal planar molecular structure which consists of a carbon atom surrounded by three oxygen atoms. The carbonate ion is a moderately strong base, so by definition of a Lewis base, it attracts protons in aqueous solutions. It carries a formal charge of -2. Carbonate bonds to metal cations, generally forming insoluble compounds.
Introduction
The term "carbonate" is usually used to refer to one of its salts or carbonate minerals. The more commonly known carbonates are calcium carbonate ($CaCO_3$) and sodium carbonate ($Na_2CO_3$).
Figure 1. Ball-and-stick model of the carbonate ion: $CO_3^{2−}$
Reaction with Group 1 Elements
All of the alkali metals react with carbonate ions and create thermally stable compounds. The exception to that rule is $Li_2CO_3$. Lithium and magnesium have very similar properties. Their similarities are referred to as a diagonal relationship, possibly due to their comparable size. Therefore, lithium and its compounds do not react the same as other group 1 elements. Some of the examples of alkali metal carbonates are shown below:
• Lithium carbonate, $Li_2CO_3$: can be used to treat patients who are manic depressive. $Li^+ + CO_3^{2-} \longrightarrow Li_2CO_3 \nonumber$
• Sodium carbonate (soda ash), Na2CO3: can be used to manufacture glass.$Na^+ + CO_3^{2-} \longrightarrow Na_2CO_3 \nonumber$
Reaction with Group 2 Elements
The group 2 carbonates are the most important minerals of the alkaline earth metals. Their insolubility in water and their solubility in acidic solution makes them ideal reservoirs for petroleum. One of the most significant group 2 carbonates is calcium carbonate, which is the chief constituent of limestone. Limestones are used primarily for building stones including the manufacturing of glasses, Portland cement, and the formation of limestone caves. Here is the reaction of carbonate calcium:
$Ca^{2+} + CO_3^{2-} \longrightarrow CaCO_3$
To obtain pure CaCO3 from limestone, three steps must be taken:
1. Calcination: decomposing limestone with thermal energy $CaCO_{3 \; (s)} \longrightarrow CaO_{(s)} + CO_{2 \; (g)} \nonumber$
2. Slaking: adding water to $CaO_{(s)}$ $CaO_{(s)} + H_2O_{(l)} \longrightarrow Ca(OH)_{2 \; (s)} \nonumber$
3. Carbonation: Converting $Ca(OH)_2$ in aqueous form to a precipitated $CaCo_3$ $Ca(OH)_{2 \; (s)} + CO_{2 \; (g)} \longrightarrow CaCO_{3 \; (s)} + H_2O_{(l)} \nonumber$
Practical Applications of Carbonates
Permanent hard water contains HCO3-. By adding Na2CO3 (washing soda), the water is softened and hard water precipitates calcium and magnesium. Ammonium sulfide group filtrate, when treated with CO32-, yields precipitate from the fourth group (Mg, Ca, Sr, Ba). Aqueous carbonate anion is the key reagent, earning the name carbonate group. After the series of precipitations, the solution will contain Na, K, NH4 (common water soluble salts). Bicarbonates are used in the lab to prevent injury or damage from use of strong acids; for instance, by laying out bicarbonate powder in areas of potential acid leakage, accidental spills get neutralized.
Interesting Facts about Carbonates
• Carbonate is a moderately strong base.
• Alkali metals can be mined in the form: Na2CO3, sodium carbonate.
• Except for Li2CO3, alkali metal carbonates are thermally stable.
• Lithium Carbonate was used to treat individuals who are manic depressive.
• Sodium Carbonate (soda ash) is used in the production of glass.
• Calcium Carbonate is limestone.
• Sodium Bicarbonate can be isolated and sold or converted to sodium carbonate by heating.
Outside Links
• For more information on carbonate rocks, click here.
• Hawkes, Stephen J. "Glass Doesn't Flow and Doesn't Crystallize and It Isn't a Liquid." J. Chem. Educ. 2000 77 846.
• Lagier, Claudia; Olivieri, Alejandro. "Calculation of solubilities of carbonates and phosphates in water as influenced by competitive acid-base reactions." J. Chem. Educ. 1990, 67, 934.C
Contributors
• Candice Chan (UCD), Vicky Vo (UCD), Margaret Huang(UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Carbonates.txt |
The term hydride is commonly named after binary compounds that hydrogen forms with other elements of the periodic table. Hydride compounds in general form with almost any element, except a few noble gases. The trends and properties vary according to the type of intermolecular force that bonds the elements together, the temperature, its molecular masses, and other components. Hydrides are classified into three major groups, depending on what elements the hydrogen bonds to. The three major groups are covalent, ionic, and metallic hydrides. Formally, hydride is known as the negative ion of a hydrogen, H-, also called a hydride ion. Because of this negative charge, hydrides have reducing, or basic properties. Its special characteristics will be further discussed.
Covalent Hydrides
The first major group is covalent hydrides, which is when a hydrogen atom and one or more non-metals form compounds. This occurs when hydrogen covalently bonds to a more electropositive element by sharing electron pairs. These hydrides can be volatile or non-volatile. Volatile simply means being readily able to be vaporized at low temperatures. One such example of a covalent hydride is when hydrogen bonds with chlorine and forms hydrochloric acid ($HCl$). Examples are listed below:
$\ce{H2(g) + Cl2(g) -> 2HCl(g)} \label{1}$
$\ce{3H2(g) + N2(g) -> 2NH3(g)} \label{2}$
The hydrides of nonmetals on the periodic table become more electronegative as you move from group 13 to 17. This means that they are less capable of donating an electron, and want to keep them because their electron orbital becomes fuller. Instead of donating a $H^-$, they would instead donate a $H^+$ because they are more acidic.
Example $1$: Boron Hydrides
Boron can form many different types of hydrides; one of them is borane ($\ce{BH3}$), which reacts violently with air and is easily oxidized. Borane occurs as a gaseous substance, and can form $\ce{B2H6}$ by two borane molecules combined with each other. Borane is not a stable compound because it does not follow a complete octet rule since it has only six valence electrons.
Example $2$: Nitrogen Hydrides
Ammonia is an important nitrogen hydride that is possible due to the synthesis of nitrogen and water which is called the Haber-Bosch process. The chemical equation for this reaction is:
$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber$
To yield ammonia, there needs to be a catalyst to speed up the reaction, a high temperature and a high pressure. Ammonia is a reagent used in many chemistry experiments and is used as fertilizer. Ammonia can react with sulfuric acid to produce ammonium sulfate, which is also an important fertilizer. In this reaction, ammonia acts as a base since it receives electrons while sulfuric acid gives off electrons.
$\ce{2NH3(aq) + H2SO4(aq) <=> (NH4)2SO4(aq)} \nonumber$
Other hydrides of nitrogen include ammonium chloride, hydrazine and hydroxylamine. Ammonium chloride is widely used in dry-cell batteries and clean metals.
Ionic Hydrides
The second category of hydrides are ionic hydrides (also known as saline hydrides or pseudohalides). These compounds form between hydrogen and the most active metals, especially with the alkali and alkaline-earth metals of group one and two elements. In this group, the hydrogen acts as the hydride ion ($H^-$). They bond with more electropositive metal atoms. Ionic hydrides are usually binary compounds (i.e., only two elements in the compound) and are also insoluble in solutions.
$\ce{A(s) + H2(g) -> 2AH(s)} \label{3}$
with $A$ as any group 1 metal.
$\ce{A(s) + H2(g) -> AH2(s)} \label{4}$
with $A$ as any group 2 metal.
Ionic hydrides combine vigorously with water to produce hydrogen gas.
Example $3$: Alkali Metal Hydrides
As ionic hydrides, alkali metal hydrides contain the hydride ion $H^-$ as well. They are all very reactive and readily react with various compounds. For example, when an alkali metal reacts with hydrogen gas under heat, an ionic hydride is produced. Alkali metal hydrides also react with water to produce hydrogen gas and a hydroxide salt:
$\ce{MH(s) + H2O(l) -> MOH(aq) + H2(g)} \nonumber$
Metallic Hydrides
The third category of hydrides are metallic hydrides, also known as interstitial hydrides. Hydrogen bonds with transition metals. One interesting and unique characteristic of these hydrides are that they can be nonstoichiometric, meaning basically that the fraction of H atoms to the metals are not fixed. Nonstoichiometric compounds have a variable composition. The idea and basis for this is that with metal and hydrogen bonding there is a crystal lattice that H atoms can and may fill in between the lattice while some might, and is not a definite ordered filling. Thus it is not a fixed ratio of H atoms to the metals. Even so, metallic hydrides consist of more basic stoichiometric compounds as well.
Intermolecular Interactions
You may think that hydrides are all intact through hydrogen bonding because of the presence of at least a hydrogen atom, but that is false. Only some hydrides are connected with hydrogen bonding. Hydrogen bonds have energies of the order of 15-40 kJ/mol, which are fairly strong but in comparison with covalent bonds at energies greater than 150 kJ/mol, they are still much weaker. Some hydrogen bonding can be weak if they are mildly encountered with neighboring molecules. Specifically fluorine, oxygen, and nitrogen are more vulnerable to hydrogen bonding.
In hydrides, hydrogen is bonded with a highly electronegative atom so their properties are more distinguished. Such that in the chart below comparing boiling points of groups 14-17 hydrides, the values of ammonia (NH3), water (H2O), and hydrogen fluoride (HF) break the increasing boiling point trend.
Supposedly, as the molecular mass increases, the boiling points increase as well. Due to the hydrogen bonds of the three following hydrides, they distinctly have high boiling points instead of the initial assumption of having the lowest boiling points. What occur in these hydrogen bonds are strong dipole-dipole attractions because of the high ionic character of the compounds.
Contributors
• Tandis Arani (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Hydrides.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
$C + 4HNO_3 \rightarrow CO_2 + 4NO_2 + 2H_2O \nonumber$
Trends in Acid-Base Behavior
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)
Oxides
This page explains the relationship between the physical properties of the oxides of Period 3 elements (sodium to chlorine) and their structures. Argon is obviously omitted because it does not form an oxide.
The oxides
The oxides we'll be looking at are:
Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7
P4O6 SO2 Cl2O
Those oxides in the top row are known as the highest oxides of the various elements. These are the oxides where the Period 3 elements are in their highest oxidation states. In these oxides, all the outer electrons in the Period 3 element are being involved in the bonding - from just the one with sodium, to all seven of chlorine's outer electrons.
The structures
The trend in structure is from the metallic oxides containing giant structures of ions on the left of the period via a giant covalent oxide (silicon dioxide) in the middle to molecular oxides on the right.
Melting and boiling points
The giant structures (the metal oxides and silicon dioxide) will have high melting and boiling points because a lot of energy is needed to break the strong bonds (ionic or covalent) operating in three dimensions. The oxides of phosphorus, sulfur and chlorine consist of individual molecules; some are small and simple and others are polymeric.
The attractive forces between these molecules will be van der Waals dispersion and dipole-dipole interactions. These vary in size depending on the size, shape and polarity of the various molecules - but will always be much weaker than the ionic or covalent bonds you need to break in a giant structure. These oxides tend to be gases, liquids or low melting point solids.
Electrical conductivity
None of these oxides has any free or mobile electrons. That means that none of them will conduct electricity when they are solid. The ionic oxides can, however, undergo electrolysis when they are molten. They can conduct electricity because of the movement of the ions towards the electrodes and the discharge of the ions when they get there.
The metallic oxides
The structures
Sodium, magnesium and aluminium oxides consist of giant structures containing metal ions and oxide ions. Magnesium oxide has a structure just like sodium chloride. The other two have more complicated arrangements.
Melting and boiling points
There are strong attractions between the ions in each of these oxides and these attractions need a lot of heat energy to break. These oxides therefore have high melting and boiling points.
Electrical conductivity
None of these conducts electricity in the solid state, but electrolysis is possible if they are molten. They conduct electricity because of the movement and discharge of the ions present. The only important example of this is in the electrolysis of aluminium oxide in the manufacture of aluminium .
Whether you can electrolyse molten sodium oxide depends, of course, on whether it actually melts instead of subliming or decomposing under ordinary circumstances. If it sublimes, you will not get any liquid to electrolyse! Magnesium and aluminium oxides have melting points far too high to be able to electrolyse them in a simple lab.
Silicon dioxide (silicon(IV) oxide)
The structure
The electronegativity of the elements increases as you go across the period, and by the time you get to silicon, there is not enough electronegativity difference between the silicon and the oxygen to form an ionic bond. Silicon dioxide is a giant covalent structure. There are three different crystal forms of silicon dioxide. The easiest one to remember and draw is based on the diamond structure. Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all you need to do is to modify the silicon structure by including some oxygen atoms.
Notice that each silicon atom is bridged to its neighbours by an oxygen atom. Don't forget that this is just a tiny part of a giant structure extending in all 3 dimensions.
Note
If you want to be fussy, the Si-O-Si bond angles are wrong in this diagram. In reality the "bridge" from one silicon atom to its neighbour is not in a straight line, but via a "V" shape (similar to the shape around the oxygen atom in a water molecule). It's extremely difficult to draw that convincingly and tidily in a diagram involving this number of atoms. The simplification is perfectly acceptable.
Melting and boiling points: Silicon dioxide has a high melting point - varying depending on what the particular structure is (remember that the structure given is only one of three possible structures), but they are all around 1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the structure before melting occurs. Silicon dioxide boils at 2230°C.
Because you are talking about a different form of bonding, it doesn't make sense to try to compare these values directly with the metallic oxides. What you can safely say is that because the metallic oxides and silicon dioxide have giant structures, the melting and boiling points are all high.
Electrical conductivity: Silicon dioxide does not have any mobile electrons or ions and hence does not conduct electricity either as a solid or a liquid.
The molecular oxides
Phosphorus, sulfur and chlorine all form oxides which consist of molecules. Some of these molecules are fairly simple - others are polymeric. We are just going to look at some of the simple ones. Melting and boiling points of these oxides will be much lower than those of the metal oxides or silicon dioxide. The intermolecular forces holding one molecule to its neighbors will be van der Waals dispersion forces or dipole-dipole interactions. The strength of these will vary depending on the size of the molecules. None of these oxides conducts electricity either as solids or as liquids. None of them contains ions or free electrons.
The Phosphorus Oxides
Phosphorus has two common oxides, phosphorus(III) oxide, P4O6, and phosphorus(V) oxide, P4O10.
Phosphorus(III) oxide
Phosphorus(III) oxide is a white solid, melting at 24°C and boiling at 173°C. The structure of its molecule is best worked out starting from a P4 molecule which is a little tetrahedron.
Pull this apart so that you can see the bonds . . .
. . . and then replace the bonds by new bonds linking the phosphorus atoms via oxygen atoms. These will be in a V-shape (rather like in water), but you probably wouldn't be penalised if you drew them on a straight line between the phosphorus atoms in an exam.
The phosphorus is using only three of its outer electrons (the 3 unpaired p electrons) to form bonds with the oxygens.
Phosphorus(V) oxide
Phosphorus(V) oxide is also a white solid, subliming (turning straight from solid to vapour) at 300°C. In this case, the phosphorus uses all five of its outer electrons in the bonding. Solid phosphorus(V) oxide exists in several different forms - some of them polymeric. We are going to concentrate on a simple molecular form, and this is also present in the vapor. This is most easily drawn starting from P4O6. The other four oxygens are attached to the four phosphorus atoms via double bonds.
If you look carefully, the shape of this molecule looks very much like the way we usually draw the repeating unit in the diamond giant structure. Don't confuse the two, though! The \(P_4O_{10}\) molecule stops here. This is not a little bit of a giant structure - it's all there is. In diamond, of course, the structure just continues almost endlessly in three dimensions.
The Sulfur Oxides
Sulfur has two common oxides, sulfur dioxide (sulfur(IV) oxide), SO2, and sulfur trioxide (sulfur(VI) oxide), SO3.
sulfur dioxide
Sulfur dioxide is a colourless gas at room temperature with an easily recognised choking smell. It consists of simple SO2 molecules.
The sulfur uses 4 of its outer electrons to form the double bonds with the oxygen, leaving the other two as a lone pair on the sulfur. The bent shape of SO2 is due to this lone pair.
sulfur trioxide
Pure sulfur trioxide is a white solid with a low melting and boiling point. It reacts very rapidly with water vapour in the air to form sulfuric acid. That means that if you make some in the lab, you tend to see it as a white sludge which fumes dramatically in moist air (forming a fog of sulfuric acid droplets). Gaseous sulfur trioxide consists of simple SO3 molecules in which all six of the sulfur's outer electrons are involved in the bonding.
There are various forms of solid sulfur trioxide. The simplest one is a trimer, S3O9, where three SO3 molecules are joined up and arranged in a ring.
There are also other polymeric forms in which the SO3 molecules join together in long chains. For example:
It is difficult to draw this convincingly. In fact, on each sulfur atom, one of the double bonded oxygens is coming out of the diagram towards you, and the other one is going back in away from you.
The fact that the simple molecules join up in this way to make bigger structures is what makes the sulfur trioxide a solid rather than a gas.
The chlorine oxides
Chlorine forms several oxides. Here we are just looking at two of them : chlorine(I) oxide (Cl2O) and chlorine(VII) oxide (Cl2O7).
Chlorine(I) oxide
Chlorine(I) oxide is a yellowish-red gas at room temperature. It consists of simple small molecules.
There's nothing in the least surprising about this molecule and it's physical properties are just what you would expect for a molecule this size.
Chlorine(VII) oxide
In chlorine(VII) oxide, the chlorine uses all of its seven outer electrons in bonds with oxygen. This produces a much bigger molecule, and so you would expect its melting point and boiling point to be higher than chlorine(I) oxide. Chlorine(VII) oxide is a colourless oily liquid at room temperature. In the diagram, for simplicity I have drawn a standard structural formula. In fact, the shape is tetrahedral around both chlorines, and V-shaped around the central oxygen.
Problems
I intended at this point to quote values for each of the oxides, hoping to show that the melting and boiling points increase as the charges on the positive ion increase from 1+ in sodium to 3+ in aluminium. You would expect that the greater the charge, the greater the attractions. Unfortunately, the oxide with the highest melting and boiling point is magnesium oxide, not aluminium oxide! So that theory bit the dust!
The reason for this probably lies in the increase in electronegativity as you go from sodium to magnesium to aluminium. That would mean that the electronegativity difference between the metal and the oxygen is decreasing. The smaller difference means that the bond won't be so purely ionic. It is also likely that molten aluminium oxide contains complex ions containing both aluminium and oxygen rather than simple aluminium and oxide ions. All this means, of course, that you aren't really comparing like with like - so wouldn't necessarily expect a neat trend.
The other problems I came across lie with sodium oxide. Most sources say that this sublimes (turns straight from solid to vapour) at 1275°C. However, the usually reliable Webelements gives a melting point of 1132°C followed by a decomposition temperature (before boiling) of 1950°C. Other sources talk about it decomposing (to sodium and sodium peroxide) above 400°C. I have no idea what the truth of this is - although I suspect that the Webelements melting point value is probably for a pressure above atmospheric pressure (although it doesn't say so).
Template:Clark | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Oxides/Physical_Properties_of_Period_3_Oxides.txt |
• Compounds
• Hard Water
• Reactions in Aqueous Solutions
• Reactions of Main Group Elements with Carbonates
Carbonates ions are formed by the reaction of carbonic acid with metals or organic compounds. In this case, the reaction of carbonates with the main group elements and its products, salts, is explained.
• Reactions of Main Group Elements with Halogens
This section describes the chemistry of halogens with the main group elements such as the alkali metals, alkaline earth metals, and Groups 13 and 14. The word halogen itself means "salt former" in Greek. Halogens such as chlorine, bromine and iodine have properties that enable them to react with other elements to form important salts such as sodium chloride, also known as table salt.
• Reactions of Main Group Elements with Hydrogen
This module will introduce the basic chemistry of hydrogen and the general reactions between hydrogen and the main group elements. Hydrogen cannot simply be grouped with any other element due to its uniqueness, which is exhibited through its ability of obtaining multiple forms and producing unique compounds in its reactions with other elements.
• Reactions of Main Group Elements with Nitrogen
Within this module will be the primary discussion about the chemistry of nitrogen and its ability and inability of forming reactions with certain main group elements. Although nitrogen is considered a relatively inert element, it is fully capable of creating some very active compounds.
• Reactions of Main Group Elements with Oxygen
Oxygen is a highly reactive element that is very abundant on earth and in the human body. It is found in many compounds that are used to sustain basic life forms and modern civilization. Compounds containing oxygen are of great interest in the field of chemistry.
• Reactions of Main Group Elements with Water
Water is composed of two hydrogen atoms and an oxygen atom. It exhibits polarity and is naturally found in the liquid, solid, and vapor states. Its polarity makes it a good solvent and is commonly known as the universal solvent. Because of its abundance on earth, it is important to note that it is involved in many chemical reactions. Many of these chemical reactions behave in trends that can be categorized using the periodic table.
• The s-Block Elements in Biology
The s-block elements play important roles in biological systems. Covalent hydrides, for example, are the building blocks of organic compounds, and other compounds and ions containing s-block elements are found in tissues and cellular fluids. In this section, we describe some ways in which biology depends on the properties of the group 1 and group 2 elements.
Main Group Reactions
Hard water contains high amounts of minerals in the form of ions, especially the metals calcium and magnesium, which can precipitate out and cause problems in water cconducting or storing vessels like pipes. Hard water can be distinguished from other types of water by its metallic, dry taste and the dry feeling it leaves on skin. It is responsible for the scum rings seen in bathtubs, as well as the inability of soap to lather.
Types of Hard Water
Hard water is water containing high amounts of mineral ions. The most common ions found in hard water are the metal cations calcium (Ca2+) and magnesium (Mg2+), though iron, aluminum, and manganese may also be found in certain areas. These metals are water soluble, meaning they will dissolve in water. The relatively high concentrations of these ions can saturate the solution and consequently cause the equilibrium of these solutes to shift to the left, towards reactants. In other words, the ions can precipitate out of the solution. This displacement of minerals from the solution is responsible for the calcination often seen on water faucets, which is a precipitation of calcium or magnesium carbonate. Hard water may also react with other substances in the solution, such as soap, and form a precipitate called "scum." There are two defined types of hard water, temporary and permanent, which are described below.
Temporary Hard Water
Temporary hard water is hard water that consists primarily of calcium (Ca2+) and bicarbonate (HCO3-) ions. Heating causes the bicarbonate ion in temporary hard water to decompose into carbonate ion (CO32-), carbon dioxide (CO2), and water (H2O). The resultant carbonate ion (CO32-) can then react with other ions in the solution to form insoluble compounds, such as CaCO3 and MgCO3. The interactions of carbonate ion in the solution also cause the well-known mineral build-up seen on the sides of pots used to boil water, a rust known as "boiler scale." Increasing the temperature of temporary hard water, with its resultant decomposition of the bicarbonate ion, signifies a shift in the equilibrium equation (shown below). The high temperature causes the equilibrium to shift to the left, causing precipitation of the initial reactants.
$CaCO_{3 \; (s)} + CO_{2 \; (aq)} + H_2O_{(l)} \rightleftharpoons Ca^{2+}_{(aq)} + 2HCO^-_{3 \; (aq)} \tag{1}$
This shift is responsible for the white scale observed in the boiling containers described above, as well as the mineral deposits that build up inside water pipes, resulting in inefficiency and even explosion due to overheating. The CaCO3 or other scale does not completely dissolve back into the water when it is cooled because it is relatively insoluble, as shown by its small solubility constant. For this reason, this type of hard water is "temporary" because boiling can remove the hardness by displacing the offending ions from solution.
$CaCO_{3 \; (s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2-}_{3 \; (aq)} \tag{2a}$
$K_{(sp)} = 2.8 \times 10^{-9} \tag{2b}$
Permanent Hard Water
Permanent hard water consists of high concentrations of anions, like the sulfate anion (SO42-). This type of hard water is referred to as "permanent" because, unlike temporary hard water, the hardness cannot be removed simply by boiling the water and thereby precipitating out the mineral ions. However, the name is deceiving as "permanent" hard water can be softened by other means. The scale caused by permanent hard water has detrimental effects similar to those seen with temporary hard water, such as obstruction of water flow in pipes. Permanent hard water is also responsible for the bathtub "ring," or soap scum, seen after showering or bathing. As previously mentioned, permanent hard water contains calcium and magnesium cations.These cations react with soap to form insoluble compounds that are then deposited on the sides of the tub. Additionally, the reaction of these cations with soap is the reason it is difficult for soap to foam or lather well in hard water. The equation below gives an example of the reaction of magnesium ion with components of soap, in this case stearate (C18H35O22-), to form the insoluble compound magnesium stearate, which is responsible for the infamous soap scum.
$2(C_{18}H_{35}O_2)^{2-}_{(aq)} + Mg^{2+}_{(aq)} \longrightarrow Mg(C_{18}H_{35}O_2)_{2 \; (s)} \tag{3}$
Effects on the body
Though the taste of hard water may be unpleasant to some, it has many health benefits when compared to soft water. Two of the most prevalent minerals in hard water are calcium and magnesium. Both calcium and magnesium are considered essential nutrients, meaning that they must be provided in the diet in order to maintain healthy body function. Calcium is a critical component of bones, and has many positive effects on the body, such as prevention of serious life-threatening and painful ailments like osteoporosis, kidney stones, hypertension, stroke, obesity, and coronary artery disease. Magnesium also has positive health effects. Inadequate amounts of magnesium in the body increase the risks for some forms of health problems, such as hypertension, cardiac arrhythmia, coronary heart disease, and diabetes mellitus. Studies done on the health effects of hard and soft water have shown that people who drink greater amounts of soft water have much higher incidences of heart disease, as well as higher blood pressure and cholesterol levels, and faster heart rates than those who drink mostly hard water. Furthermore, soft water is corrosive to pipes, which may allow for toxic substances like lead to contaminate drinking water.
How to soften hard water
Some wish to soften hard water to control its irritating, and in many cases damaging, effects. The diminished ability of soap to lather is not only annoying, but can also be potentially harmful economically. Businesses that depend on the foaming of soap, such as car washes and pet groomers, may wish to soften hard water to avoid excessive use of soap due to a decreased ability to lather. Likewise, it is often necessary to soften water that comes into contact with pipes to avoid the destructive and compromising build-up of deposits. Also, many people may find the calcifying effects that hard water has on faucets and other items unfavorable and choose to soften the water to prevent such mineral deposits from forming. Still others may dislike the sticky, dry feeling left by the precipitation of soap scum onto the skin. Whatever the reasons, there are many processes available to soften hard water.
Ion Exchange
One way to soften water is through a process called ion exchange. During ion exchange, the unwanted ions are "exchanged" for more acceptable ions. In many cases, it is desirable to replace the hard water ions, such as Ca2+ and Mg2+, with more agreeable ions, like that of Na+. To do this, the hard water is conducted through a zeolite or resin-containing column, which binds the unwanted ions to its surface and releases the more tolerable ions. In this process, the hard water ions become "fixed" ions because of their attachment to the resin material. These fixed ions displace the desirable ions (Na+), now referred to as counterions, from the column, thus exchanging the ions in the water. This process is illustrated in Figure 1.
Unfortunately, this process has the disadvantage of increasing the sodium content of drinking water, which could be potentially hazardous to the health of people with sodium-restricted diets.
Lime Softening
Another process is called lime softening. In this process, the compound calcium hydroxide, Ca(OH)2, is added to the hard water. The calcium hydroxide, or "slaked lime," raises the pH of the water and causes the calcium and magnesium to precipitate into CaCO3 and Mg(OH)2. These precipitates can then be easily filtered out due to their insolubility in water, shown below by the small solubility constant of magnesium hydroxide (the solubility product constant for calcium carbonate is shown above). After precipitation and removal of the offending ions, acid is added to bring the pH of the water back to normal.
$Mg(OH)_{2 \; (s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2OH^-_{(aq)} \tag{4a}$
$K_{(sp)} = 1.8 \times 10^{-11} \tag{4b}$
Chelation
Chelating agents can also be used to soften hard water. Polydentate ligands, such as the popular hexadentate ligand EDTA, bind the undesirable ions in hard water. These ligands are especially helpful in binding the magnesium and calcium cations, which as already mentioned are highly prevalent in hard water solutions. The chelating agent forms a very stable ring complex with the metal cations, which prevents them from interacting with any other substances that may be introduced to the solution, such as soap. In this way, chelators are able to diminish the negative effects associated with hard water. A simplified equation representing the chelation of the metal calcium cation (Ca2+) with the hexadentate ligand EDTA is shown below. The large value of the formation constant (Kf) reflects the tendency of the reaction to proceed to completion in the forward direction.
$Ca^{2+} + EDTA^{4-} \longrightarrow [Ca(EDTA)]^{2-} \tag{5a}$
$K_f = 4.9 \times 10^{10} \tag{5b}$
Reverse Osmosis
The final process, reverse osmosis, uses high pressures to force the water through a semipermeable membrane. This membrane is generally intended to be impermeable to anything other than water. The membrane serves to filter out the larger ions and molecules responsible for the water's hardness, resulting in softened water. During this process, the water is forced from an area with a high concentration of solute in the form of dissolved metal ions and similar compounds, to an area that is very low in the concentration of these substances. In other words, the water moves from a state of hardness to a softer composition as the ions causing the water's hardness are prevented passage through the membrane. Reverse Osmosis does have a disadvantage of wasting wastewater compared to other water treatment methods. This process is shown in Figure 2 below. Note that this figure describes the desalination of salt water. However, the process for softening hard water is the same.
Practice Problems
1. Name the two main types of hard water. (Highlight blue area for the answers)
Temporary and Permanent.
2. What makes "hard" water hard?
The presence of high concentrations of minerals, typically in the form of metal cations.
3. What are the two most prevalent ions in hard water? How are these important to the proper function of the body?
Calcium (Ca2+) and magnesium (Mg2+) ion. They are important because both are essential nutrients, which means they are necessary for the proper function of the body and are also important for the prevention of many diseases and other ailments.
4. Name the four described processes for softening hard water. It is also important to understand the essential steps of each process.
Ion exchange, chelation, lime softening, and reverse osmosis
5. What is a major disadvantage of ion exchange when Na+ is used as a counterion?
It increases the concentration of sodium in the water, a potential hazard for people with sodium-restricted diets.
Contributors
• Andrea Kubisch, Courtney Korff (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Hard_Water.txt |
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate. Whether or not such a reaction occurs can be determined by using the solubility rules for common ionic solids. Because not all aqueous reactions form precipitates, one must consult the solubility rules before determining the state of the products and writing a net ionic equation. The ability to predict these reactions allows scientists to determine which ions are present in a solution, and allows industries to form chemicals by extracting components from these reactions.
Properties of Precipitates
Precipitates are insoluble ionic solid products of a reaction, formed when certain cations and anions combine in an aqueous solution. The determining factors of the formation of a precipitate can vary. Some reactions depend on temperature, such as solutions used for buffers, whereas others are dependent only on solution concentration. The solids produced in precipitate reactions are crystalline solids, and can be suspended throughout the liquid or fall to the bottom of the solution. The remaining fluid is called supernatant liquid. The two components of the mixture (precipitate and supernate) can be separated by various methods, such as filtration, centrifuging, or decanting.
Precipitation and Double Replacement Reactions
The use of solubility rules require an understanding of the way that ions react. Most precipitation reactions are single replacement reactions or double replacement reactions. A double replacement reaction occurs when two ionic reactants dissociate and bond with the respective anion or cation from the other reactant. The ions replace each other based on their charges as either a cation or an anion. This can be thought of as "switching partners"; that is, the two reactants each "lose" their partner and form a bond with a different partner:
A double replacement reaction is specifically classified as a precipitation reaction when the chemical equation in question occurs in aqueous solution and one of the of the products formed is insoluble. An example of a precipitation reaction is given below:
$CdSO_{4(aq)} + K_2S_{(aq)} \rightarrow CdS_{(s)} + K_2SO_{4(aq)} \nonumber$
Both reactants are aqueous and one product is solid. Because the reactants are ionic and aqueous, they dissociate and are therefore soluble. However, there are six solubility guidelines used to predict which molecules are insoluble in water. These molecules form a solid precipitate in solution.
Solubility Rules
Whether or not a reaction forms a precipitate is dictated by the solubility rules. These rules provide guidelines that tell which ions form solids and which remain in their ionic form in aqueous solution. The rules are to be followed from the top down, meaning that if something is insoluble (or soluble) due to rule 1, it has precedence over a higher-numbered rule.
1. Salts formed with group 1 cations and $NH_4^+$ cations are soluble. There are some exceptions for certain $Li^+$ salts.
2. Acetates ($C_2H_3O_2^-$), nitrates ($NO_3^-$), and perchlorates ($ClO_4^-$) are soluble.
3. Bromides, chlorides, and iodides are soluble.
4. Sulfates ($SO_4^{2-}$) are soluble with the exception of sulfates formed with $Ca^{2+}$, $Sr^{2+}$, and $Ba^{2+}$.
5. Salts containing silver, lead, and mercury (I) are insoluble.
6. Carbonates ($CO_3^{2-}$), phosphates ($PO_4^{3-}$), sulfides, oxides, and hydroxides ($OH^-$) are insoluble. Sulfides formed with group 2 cations and hydroxides formed with calcium, strontium, and barium are exceptions.
If the rules state that an ion is soluble, then it remains in its aqueous ion form. If an ion is insoluble based on the solubility rules, then it forms a solid with an ion from the other reactant. If all the ions in a reaction are shown to be soluble, then no precipitation reaction occurs.
Net Ionic Equations
To understand the definition of a net ionic equation, recall the equation for the double replacement reaction. Because this particular reaction is a precipitation reaction, states of matter can be assigned to each variable pair:
AB(aq) + CD(aq) AD(aq) + CB(s)
The first step to writing a net ionic equation is to separate the soluble (aqueous) reactants and products into their respective cations and anions. Precipitates do not dissociate in water, so the solid should not be separated. The resulting equation looks like that below:
A+(aq) + B-(aq) + C+(aq) + D-(aq) A+(aq) + D-(aq) + CB(s)
In the equation above, A+and D- ions are present on both sides of the equation. These are called spectator ions because they remain unchanged throughout the reaction. Since they go through the equation unchanged, they can be eliminated to show the net ionic equation:
C+ (aq)+ B- (aq) CB (s)
The net ionic equation only shows the precipitation reaction. A net ionic equation must be balanced on both sides not only in terms of atoms of elements, but also in terms of electric charge. Precipitation reactions are usually represented solely by net ionic equations. If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs.
Applications and Examples
Precipitation reactions are useful in determining whether a certain element is present in a solution. If a precipitate is formed when a chemical reacts with lead, for example, the presence of lead in water sources could be tested by adding the chemical and monitoring for precipitate formation. In addition, precipitation reactions can be used to extract elements, such as magnesium from seawater. Precipitation reactions even occur in the human body between antibodies and antigens; however, the environment in which this occurs is still being studied.
Example 1
Complete the double replacement reaction and then reduce it to the net ionic equation.
$NaOH_{(aq)} + MgCl_{2 \;(aq)} \rightarrow \nonumber$
First, predict the products of this reaction using knowledge of double replacement reactions (remember the cations and anions “switch partners”).
$2NaOH_{(aq)} + MgCl_{2\;(aq)} \rightarrow 2NaCl + Mg(OH)_2 \nonumber$
Second, consult the solubility rules to determine if the products are soluble. Group 1 cations ($Na^+$) and chlorides are soluble from rules 1 and 3 respectively, so $NaCl$ will be soluble in water. However, rule 6 states that hydroxides are insoluble, and thus $Mg(OH)_2$ will form a precipitate. The resulting equation is the following:
$2NaOH(aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl_{(aq)} + Mg(OH)_{2\;(s)} \nonumber$
Third, separate the reactants into their ionic forms, as they would exist in an aqueous solution. Be sure to balance both the electrical charge and the number of atoms:
$2Na^+_{(aq)} + 2OH^-_{(aq)} + Mg^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow Mg(OH)_{2\;(s)} + 2Na^+_{(aq)} + 2Cl^-_{(aq)} \nonumber$
Lastly, eliminate the spectator ions (the ions that occur on both sides of the equation unchanged). In this case, they are the sodium and chlorine ions. The final net ionic equation is:
$Mg^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Mg(OH)_{2(s)} \nonumber$
Example 2
Complete the double replacement reaction and then reduce it to the net ionic equation.
$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow \nonumber$
Solution
The predicted products of this reaction are $CoSO_4$ and $NaCl$. From the solubility rules, $CoSO_4$ is soluble because rule 4 states that sulfates ($SO_4^{2-}$) are soluble. Similarly, we find that $NaCl$ is soluble based on rules 1 and 3. After balancing, the resulting equation is as follows:
$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow CoSO_{4\;(aq)} + 2 NaCl_{(aq)} \nonumber$
Separate the species into their ionic forms, as they would exist in an aqueous solution. Balance the charge and the atoms. Cancel out all spectator ions (those that appear as ions on both sides of the equation.):
Co2- (aq) + 2Cl-(aq) + 2Na+ (aq) + SO42-(aq) Co2- (aq) + SO42-(aq) + 2Na+ (aq) + 2Cl-(aq)
No precipitation reaction
This particular example is important because all of the reactants and the products are aqueous, meaning they cancel out of the net ionic equation. There is no solid precipitate formed; therefore, no precipitation reaction occurs.
Practice Problems
Write the net ionic equation for the potentially double displacement reactions. Make sure to include the states of matter and balance the equations.
1. $Fe(NO_3)_{3\;(aq)} + NaOH_{(aq)} \rightarrow$
2. $Al_2(SO_4)_{3\;(aq)} + BaCl_{2\;(aq)} \rightarrow$
3. $HI_{(aq)} + Zn(NO_3)_{2\;(aq)} \rightarrow$
4. $CaCl_{2\;(aq)} + Na_3PO_{4\;(aq)} \rightarrow$
5. $Pb(NO_3)_{2\;(aq)} + K_2SO_{4 \;(aq)} \rightarrow$
Solutions
1. Regardless of physical state, the products of this reaction are $Fe(OH)_3$ and $NaNO_3$. The solubility rules predict that $NaNO_3$ is soluble because all nitrates are soluble (rule 2). However, $Fe(OH)_3$ is insoluble, because hydroxides are insoluble (rule 6) and $Fe$ is not one of the cations which results in an exception. After dissociation, the ionic equation is as follows:
$Fe^{3+}_{(aq)} + NO^-_{3\;(aq)} + Na^+_{(aq)} + 3OH^-_{(aq)} \rightarrow Fe(OH)_{3\;(s)} + Na^+_{(aq)} + NO^-_{3\;(aq)} \nonumber$
Canceling out spectator ions leaves the net ionic equation:
$Fe^{3+}_{(aq)} + OH^-_{(aq)} \rightarrow Fe(OH)_{\;3(s)} \nonumber$
2. From the double replacement reaction, the products are $AlCl_3$ and $BaSO_4$. $AlCl_3$ is soluble because it contains a chloride (rule 3); however, $BaSO_4$ is insoluble: it contains a sulfate, but the $Ba^{2+}$ ion causes it to be insoluble because it is one of the cations that causes an exception to rule 4. The ionic equation is (after balancing):
$2Al^{3+}_{(aq)} + 6Cl^-_{(aq)} + 3Ba^{2+}_{(aq)} + 3SO^{2-}_{4\;(aq)} \rightarrow 2 Al^{3+}_{(aq)} +6Cl^-_{(aq)} + 3BaSO_{4\;(s)} \nonumber$
Canceling out spectator ions leaves the following net ionic equation:
$Ba^{2+}_{(aq)} + SO^{2-}_{4\;(aq)} \rightarrow BaSO_{4\;(s)} \nonumber$
3. From the double replacement reaction, the products $HNO_3$ and $ZnI_2$ are formed. Looking at the solubility rules, $HNO_3$ is soluble because it contains nitrate (rule 2), and $ZnI_2$ is soluble because iodides are soluble (rule 3). This means that both the products are aqueous (i.e. dissociate in water), and thus no precipitation reaction occurs.
4. The products of this double replacement reaction are $Ca_3(PO_4)_2$ and $NaCl$. Rule 1 states that $NaCl$ is soluble, and according to solubility rule 6, $Ca_3(PO_4)_2$ is insoluble. The ionic equation is:
$Ca^{2+}_{(aq)}+ Cl^-_{(aq)} + Na^+_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} + Na^+_{(aq)} + Cl^-_{(aq)} \nonumber$
After canceling out spectator ions, the net ionic equation is given below:
$Ca^{2+}_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} \nonumber$
5. The first product of this reaction, $PbSO_4$, is soluble according to rule 4 because it is a sulfate. The second product, $KNO_3$, is also soluble because it contains nitrate (rule 2). Therefore, no precipitation reaction occurs.
Contributors and Attributions
• Julie Schaffer (UCD), Corinne Herman (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_in_Aqueous_Solutions/Precipitation_Reactions.txt |
An aqueous solution is one that is occurring in water. What makes water significant is that it can allow for substances to dissolve and/or be dissociated into ions within it.
Electrolytes
Water is generally the solvent found in aqueous solution, where a solvent is the substance that dissolves the solute. The solute is the substance or compound being dissolved in the solvent. A solute has fewer number of particles than a solvent, where it's particles are in random motion. Interestingly, aqueous solutions with ions conduct electricity to some degree. Pure water, having a very low concentration of ions, cannot conduct electricity. When a solute dissociates in water to form ions, it is called an electrolyte, due to the solution being a good electrical conductor. When no ions are produced, or the ion content is low, the solute is a non-electrolyte. Non-electrolytes do not conduct electricity or conduct it to a very small degree. In an aqueous solution a strong electrolyte is considered to be completely ionized, or dissociated, in water, meaning it is soluble. Strong acids and bases are usually strong electrolytes. A weak electrolyte then is considered to be one that is not completely dissociated, therefore still containing whole compounds and ions in the solution. Weak acids and bases are generally weak electrolytes. In other words, strong electrolytes have a better tendency to supply ions to the aqueous solution than weak electrolytes, and therefore strong electrolytes create an aqueous solution that is a better conductor of electricity.
• Most soluble ionic compounds and few molecular compounds are strong electrolytes.
• Most molecular compounds are weak or non electrolytes.
Example $1$
Here's an example of MgCl2 in water:
$MgCl_2 \rightarrow Mg^{2+}_{(aq)} + 2Cl^-_{(aq)} \nonumber$
The ionic compound dissociates completely to form ions in water, therefore, it is a strong electrolyte.
Now let's look at a weak electrolyte:
$HC_2H_3O_2(aq) \rightleftharpoons H^+(aq) + C_2H_3O_2^-(aq) \nonumber$
The ionic compound, $HC_2H_3O_2$ in this situation, only partially dissociates, as expressed by the double arrows in the reaction. This means that the reaction is reversible and never goes to completion.
The $H^+$ cation is a proton that interacts with the $H_2O$ molecules that it is submerged in. The interaction is called hydration. The actual H+ ion does not exist in the aqueous solution. It is the hydronium ion, $H_3O^+$ that interacts with water to create additional species like $H_5O_2^+$, $H_9O_4^+$, and $H_7O_3^+$. $H_3O^+$ can simply be described as the hydration of one H+ and one water molecule. For nonelectrolytes, all that needs to be done is write the molecular formula because no reaction or dissociation occurs. One example of a nonelectrolyte is sugar: written as $C_6H_{12}O_6 (aq)$.
Ion Concentrations
In an aqueous solution the amount of ions of a species is related to the number of moles of that species per concentration of the substance in the aqueous solution. Molarity is the number of moles of a solute ($n$) divided by the total volume ($V$) of the solution:
$M= \dfrac{n}{V} \nonumber$
Molarity, or concentration, can also be represented by placing the solute within brackets (e..g, $[Cl^-]$ for the concentration of chloride ions).
Example $2$
Determine the concentration of $K^+$ in an aqueous solution of 0.238 M $KNO_3$.
Solution
Since there is one mole of potassium in $KNO_3$, multiply the concentration of the species by the number of moles of the atom to obtain:
$[K^+] = (0.238\; M\; KNO_3) \times (1 \;mol\; K^+)= 0.238\; M \nonumber$
Although not asked, there is also one mole of nitrite ions in one mole of $KNO_3$, so its concentration is also 0.238 M:
$[NO_3^-] = (0.238\; M\; KNO_3) \times (1 \;mol\; NO_3^-)= 0.238\; M \nonumber$
FOLLOWUP
The stoichiometry always dictates the concentration, which was a simple 1:1 ratio for $KNO_3$. However, for more complex situations, different ratios will be encountered. For instance, if consider the dissolving of $Al_2(SO_4)_3$:
$Al_2(SO_4)_3 \rightarrow 2 Al^{3+}_{(aq)} + 3 SO^{2-}_{4(aq)} \nonumber$
If the concentration of $Al_2(SO_4)_3$ is 0.019 M, what is the concentration of $Al^{3+}_{(aq)}$? Simply multiply 0.019 M by the stoichiometric factor of $Al^{3+}_{(aq)}$ in $Al_2(SO_4)_3$, which is 2:1. The concentration of $Al^{3+}_{(aq)}$ then becomes 0.038 M:
$[Al^{3+}] = (0.019 \; M\; Al_2(SO_4)_3) \times (2 \;mol\; Al^{3+})= 0.238\; M \nonumber$
Although not asked, the concentration of $SO_4^{2-}$ is 0.057 M via the same argument;
$[SO_4^{2-}]= (0.019 \; M) \times (3 \; mol \; SO_4^{2-}) = 0.057\; M \nonumber$
Precipitation Reactions
Precipitation reactions occur when the resulting product of an aqueous solution is insoluble. This means that there is a solid produced, called the precipitate. The precipitate is a combination of cation and anions forming an ionic bond. Precipitates are used in manufacturing chemicals, so that certain ions can be isolated by forming precipitates with them. We can predict if a precipitate is produced using the solubility rules for common ionic solids:
1. Salts of Group 1 cations (with some exceptions for $Li^+$) and the $NH_4^+$ cation are soluble.
2. Nitrates, acetates, and perchlorates are soluble.
3. Salts of silver, lead, and mercury(I) are insoluble.
4. Chlorides, bromides, and iodides are soluble.
5. Carbonates, phosphates, sulfides, oxides, and hydroxides are insoluble (sulfides of Group 2 cations and hydroxides of $Ca^{2+}$, $Sr^{2+}$, and $Ba^{2+}$ are slightly soluble).
6. Sulfates are soluble except for those of calcium, strontium, and barium.
Example $3$
Here's an example of a precipitation reaction:
$AgNO_{3(aq)} + NaI_{(aq)} \rightarrow AgI_{(s)} + NaNO_{3(aq)} \nonumber$
This is considered the "whole" reaction, because all of the species involved are recognized. However, in an aqueous solution, the particles look more like this:
$Ag^+_{(aq)} + NO^-_{3(aq)} + Na^+_{(aq)} + I^-_{(aq)} \rightarrow AgI_{(s)} + Na^+_{(aq)} + NO^-_{3(aq)} \nonumber$
This is called the "ionic" form, because all the ions in the solution are shown. If the ions that are not involved in creating the solid are removed (the spectator ions), a net ionic equation is generated:
$Ag^+_{(aq)} + I^-_{(aq)} \rightarrow AgI_{(s)} \nonumber$
What are the spectator ions in this reaction?
$NaOH_{(aq)} + MgCl_{2(aq)} \rightarrow NaCl_{(aq)} + Mg(OH)_{2(s)} \nonumber$
As you can see, it is critical to include the symbols for each species to identify what state they're in: either gaseous (g), solid (s), liquid (l) or aqueous (aq). It is also important to remember that these types of ionic equations also must have a balanced charge. You can use stoichiometric coefficients to ensure that both sides of the equation have equal net charges.
Acid Base Reactions
An acid is a substance that gives off $H^+$ ions in an aqueous solution, while a base gives of OH- ions. Strong acids almost completely dissociate to become $H^+$ ions, and strong bases dissociate to become OH-. There are only a few strong acids and bases. Most are weak, meaning they produce few $H^+$ or OH- ions in aqueous solution. Here is a list of common strong acids and bases:
Strong Acids Strong Bases
HCl LiOH
HBr NaOH
HI KOH
HClO4 RbOH
HNO3 CsOH
H2SO4 Ca(OH)2
Sr(OH)2
Ba(OH)2
In a neutralization reaction an acid and a base are combined to produce water and an aqueous salt. The acid and base neutralize/balance each other to get a result that is the salt. For example, determine which is the acid, base, and salt n this neutralization reaction?
$HCl_{(aq)} + NaOH_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)} \nonumber$
Problems
Predict if a reaction is likely to occur, and the resulting product:
1. HI(aq) + Zn(NO3)2(aq)
2. CuSO4(aq) + Na2CO3(aq)
3. AgNO3(aq) + CuCl2(aq)
Determine the concentration:
• [Al3+] in 0.078 M Al2(SO4)3
Contributors and Attributions
• Keerith Singh (UCD), Tenaya Natov (UCD), Zachary Mar (UCD), Mandeep Sohal (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_in_Aqueous_Solutions/Unique_Features_of_Aqueous_Solutions.txt |
Carbonates ions are formed by the reaction of carbonic acid with metals or organic compounds. In this case, the reaction of carbonates with the main group elements and its products, salts, is explained.
Chemical Characteristics
In general, the most common main group elements used as carbonates are the Alkali and Alkaline metals. All main group carbonates, except Na, K, Rb and Cs are unstable to heat and insoluble in water. Usually all carbonates are soluble in acid, because of the formation of bicarbonate ion. Its regular physical appearance is of a white powder. The main uses of carbonates is as raw materials in different industrial processes such as drug development, glass making, pulp and paper industry, sodium chemicals (silicates), soap and detergent production, paper industry, water softener, clay and concrete production, among others. Other carbonates such as Beryllium Carbonate (BeCO3) and Tallium Carbonate (Tl2CO3) are consider toxic and are used in fungicides and poison manufacture.
From main group elements Sodium Carbonate (\(Na_2CO_3\)) and Calcium Carbonate (\(CaCO_3\)) are the most used.
Sodium Carbonate, known as soda ash, is a very important industrial chemical. It is mainly obtained by a method named Solvay process by the chemical reaction of limestone (CaCO3) and sodium chloride (NaCl).
2 NaCl + CaCO3 → Na2CO3 + CaCl
Na2CO3 common uses are in glass making, pulp and paper industry, sodium chemicals (silicates), soap and detergent production, paper industry and water softener.
Calcium Carbonate is the principal constituent of limestone (a sedimentary rock) and its pure state is obtained in three steps by the calcination of limestone and subsequent reaction with water and carbon dioxide.
Ca(OH)2(s) + CO2(aq) → CaCO3 (s) + H2O(l)
Calcium Carbonate common uses are in glass, textile, paint, paper and plastic production, caulks industry, to produce ink and sealant. It is also used as a food additive (non toxic), as a drug development and chalk production.
Carbonate Hardness
Hard water is the term used in relation with high amount of inorganic compounds such as carbonates, bicarbonates, sulfates or chlorides in water. The presence of high levels of carbonates and bicarbonates in water is denominated as temporary hardness of water. It is considered hard water when this inorganic compound exceed the 100 mg/L approximated, and it is usually expressed as the amount of calcium carbonate present in water. This is important when water is used for industrial process or cleaning purposes.
The way precipitation is formed is when inorganic compounds in presence of high molecular organic compounds (such as soaps) produce undesirable insoluble precipitation. These precipitations are the responsible of a dirty or grayish white and low efficiency in cleaners.
Other way to form precipitation is when hard water attains high temperatures; its inorganic compounds such as calcium carbonate precipitates leaving a deposited coating in water pipes or over boilers. Here is an example of what happens:
Ca2+(aq)+2HCO3(aq) → CaCO3(s) + H2O + CO2
One way to reduce temporary hardness such as calcium ions is by boiling or by the addition of calcium hydroxide (lime), but is not often used. The best way of soften water is by the addition of Soda Ash to the water or using an ion exchanged column.
Carbonates in Detergents
Some carbonates such as sodium carbonate or potassium carbonates are used to elaborate some detergents. Soda Ash (other name for sodium carbonate) is used since in one way it is cheap and also it helps soften water by precipitation of calcium and magnesium carbonates. Potassium carbonates are used because of its solubility.
Carbonates in Glass Manufacture
Some inorganic compounds such as calcium carbonate, sodium carbonate and potassium carbonate are used as unprocessed material to elaborate glass since their chemical characteristics, high quality and low prices. Calcium carbonates and sodium carbonates are used as raw material in the production of paper due to their low cost by using them instead of pulp and to improve the white and gloss of the paper.
Group 1: Alkali Metals and Carbonates (X2CO3)
is the reaction between Li, Na, K, Rb and Cs with CO3. All except Lithium are soluble in water and stable to heat.
Lithium Carbonate (Li2CO3)
• Uses: drug development.
• Chemical Characteristics:
• Low solubility in water.
• Unstable to heat.
• Appearance: white powder, fragrance-free
Sodium Carbonate (Na2CO3)
• Uses: glass making, pulp and paper industry, sodium chemicals (silicates), soap and detergent production, paper industry and water softener.
• Known as soda ash
• Very important industrial chemical
• Chemical Characteristics:
• Hygroscopic substance
• Soluble in water
• Stable to heat
• Appearance: crystal white solid
Potassium Carbonate (K2CO3)
• Uses: glass making, soft soap production, textile and photography chemicals.
• Characteristics:
• Hygroscopic.
• Soluble in water.
• Can be heated to high. temperatures.
• Appearance: white solid
Rubidium Carbonate (Rb2CO3)
• Uses: glass making, short-chain alcohol production.
• Characteristics:
• Hygroscopic substance
• Soluble in water.
• Soluble in water.
• Can be heated to high temperatures.
• Appearance: white solid.
Cesium Carbonate (Cs2CO3)
• Uses: production of other cesium salts.
• Characteristics:
• Hygroscopic substance.
• Soluble in water.
• Can be heated to high temperatures.
• Appearance: white powder.
Group 2: Alkaline Earth Metals and Carbonates (XCO3)
Is the reaction between Be, Mg, Ca, Sr and Ba with CO3. All are insoluble in water and unstable to heat.
Beryllium Carbonate (BeCO3)
• Insoluble in water.
Magnesium Carbonate (MgCO3)
• Uses: skin care products, cosmetic, anti-fire products, climbing chalk.
• Characteristics:
• Insoluble in water.
• Hygroscopic substance.
• Appearance: white solid.
Calcium Carbonate (CaCO3)
• Uses: textile, paint, paper, plastic, caulks industry used to produce ink and sealant. It is used as a food additive (non toxic), drug development and chalk production.
• Principal constituent of limestone (sedimentary rock).
• Characteristics:
• Responsible of hard water.
• Low solubility in water.
• Soluble in acids.
• Appearance: white powder.
Strontium Carbonate (SrCO3)
• Uses: fireworks, magnets and ceramic manufacture.
• Characteristics:
• Low solubility in water.
• Reactive with acids.
• Neutralize acids.
• Hygroscopic substance.
• Appearance: white powder, unscented.
Barium Carbonate (BaCO3)
• Uses: glass, cement, ceramic, porcelain, rat poison manufacture.
• Characteristics:
• Low solubility in water.
• Appearance: white crystals.
Group 13: The Boron Family and Carbonates (X2CO3)
Is the reaction between Al and Tl with CO3. Both are insoluble in water and unstable to heat.
Aluminium Carbonate (Al2(CO3)3)
• Uses: drug development.
• Chemical Characteristics:
• Appearance: white powder.
Tallium Carbonate (Tl2CO3)
• Uses: fungicides manufacture.
• Characteristics:
• Toxic.
• Appearance: white crystals, unscented.
Group 14: The Carbon Family and Carbonates (XCO3)
Is the reaction between Pb with CO3. It is insoluble in water and unstable to heat
Lead Carbonate (PbCO3)
• Uses: glass, cement, ceramic, porcelain, rat poison manufacture.
• Characteristics:
• Toxic.
• Low solubility in water.
• Appearance: white powder.
Problems
1) Which are the common carbonates used as raw materials?
Calcium Carbonate and Sodium Carbonate
2) Are carbonates soluble in water?
All main group carbonates, except Na, K, Rb and Cs are insoluble in water.
3) Which carbonate is responsible of hard water?
Most communly Calcium Carbonate
4) Which carbonate can be use as softener of hard water?
Sodium Carbonate ( Soda Ash) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_of_Main_Group_Elements_with_Carbonates.txt |
This section describes the chemistry of halogens with the main group elements such as the alkali metals, alkaline earth metals, and Groups 13 and 14. The word halogen itself means "salt former" in Greek. Halogens such as chlorine, bromine and iodine have properties that enable them to react with other elements to form important salts such as sodium chloride, also known as table salt.
Properties of Halogens
Elements such as fluorine, chlorine, bromine, iodine, and astatine belong to Group 17, the halogen group. At room temperature fluorine is a yellow gas, chlorine is a pale green gas, bromine is a red liquid, and iodine is a purple solid. Astatine is a radioactive element, and exists in nature only in small amounts. All the halogens exist as diatomic molecules. They have high ionization energies and form the most electronegative group of elements. Their electron configuration, ns2np5, allows them to easily react with Group 1 and 2 metals; each halogen tends to pick up one electron, and the Group 1 and Group 2 elements each tend to lose one or two electrons, respectively. Halogens therefore react most vigorously with Group 1 and Group 2 metals of all main group elements.
Reaction with Water
From a standard reduction potential table, it is determined that iodine and bromine cannot oxidize water to oxygen because they have smaller reduction potentials than oxygen. Thus, iodine and bromine do not react with water. However, fluorine and chlorine have larger reduction potentials, and can oxidize water.
Fluorine reacts with water vapor to form oxygen and ozone:
$2F_2(g) + 2H_2O(g) \rightarrow 4HF(g)+ O_2(g) \nonumber$
$3F_2(g) + 2H_2O(g) \rightarrow 6HF(g) + O_3(g) \nonumber$
The reaction of water with chlorine, shown below, proceeds very slowly.
$Cl_2 + H_2O \rightarrow H^+ + Cl^- + HClO \nonumber$
Chlorine and bromine are moderately soluble in water. These solutions form solid hydrates within an ice lattice. These solutions are good oxidizing agents. Chlorine reacts reversibly with water to produce acids as in the following example, in which chloric acid and hydrochloric acid are formed:
$Cl_2 + H_2O \rightleftharpoons HClO + HCl \nonumber$
Iodine is slightly soluble in water. It has the lowest standard reduction potential of the halogens, and is therefore the least powerful oxidizing agent. Air and other reagents can oxidize acidified solution of iodide ions.
Reaction with Hydrogen
All the halogens react directly with hydrogen, forming covalent bonds and—at sufficient levels of purity—colorless gases at room temperature. Hydrogen reacts with fluorine, chlorine, bromine, and iodine, forming HF, HCl, HBr, and HI, respectively. The bond strength of these molecules decreases down the group: $HF > HCl > HBr > HI$.
Iodine and hydrogen react non-spontaneously to produce hydrogen iodide:
$H_2 + I_2 \rightarrow 2HI \nonumber$
All the hydrogen halides are soluble in water, in which they form strong acids (with the exception of $HF$). The general equation of hydrogen halide for the acid reaction is given below:
$HX + H_2O \rightarrow H_3O^+ + X^- \nonumber$
Group 1: The Alkali Metals
All the alkali metals react vigorously with halogens to produce salts, the most industrially important of which are NaCl and KCl.
$2Na(s) + Cl_2(g) \rightarrow 2NaCl(s) \nonumber$
Sodium Chloride is used as a preservative for meat and to melt the ice on the roads (via freezing point depression). KCl is important for plant fertilizers because of the positive impact of potassium on plant growth. These metal halides form white ionic crystalline solids and are all soluble in water except LiF, because of its high lattice enthalpy attributed to strong electrostatic attraction between Li+ and F- ions.
Group 2: The Alkaline Earth Metals
The alkaline earth metals react to form hydrated halides. These halides are ionic except for those involving beryllium (the least metallic of the group). Because alkaline earth metals tend to lose electrons and halogen atoms tend to gain electrons (Table P2), the chemical reaction between these groups is the following:
$M + X_2 \rightarrow MX_2 \nonumber$
where
• $M$ represents any metal from Group 2 and
• $X$ represents fluorine, chlorine, bromine or iodine.
Anhydrous calcium chloride has strong affinity for water, absorbing enough to dissolve its own crystal lattice. It can be produced directly from limestone, or as a by-product by Solvay Process.
Group 13: The Boron Family
All the Group 13 elements react with Halogens to form trihalides.
• All halides of boron are Lewis acids
• Aluminum halides adopt a dimeric structure
Aluminum Halides
Aluminum Fluoride, $AlF_3$, is an ionic compound with a high melting point. However, most of the other aluminum halides form molecules with the formula $Al_2X_6$ ($X$ represents chlorine, bromine, or iodine). When two $AlX_3$ units (or, more generally, any two identical units) join together, the resulting molecule is called a dimer. Aluminum halides are very reactive Lewis acids. They accept electrons and form acid-base compound called adducts, as in the following example:
$AlCl_3 + (C_2H_5)_2O \rightarrow Al(C_2H_5)_2OCl_3 \nonumber$
In this reaction, $AlCl_3$ is the Lewis acid and $(C_2H_5)_2O$ is the Lewis base.
Group 14: The Carbon Family
Group 14 elements form halides with general formula MX4 (CCl4, SiCl4, GeCl4, SnCl4, PbCl4), although some elements such as Ge, Sn, Pb can also form dihalides (MX2). The tetrahalides of carbon, such as CCl4, cannot be hydrolyzed due to non-availability of vacant valence d-orbitals, but other tetrahalides can be hydrolyzed.
Silicon Halides
Silicon reacts with halogens to form compounds of the form SiX4, where X represents any common halogen. At room temperature, SiF4 is a colorless gas, SiCl4 is a colorless liquid, SiBr4 is a colorless liquid, and SiI4 forms colorless crystals. SiF4 and SiCl4 can be completely hydrolyzed, but SiBr4 can be only partially hydrolyzed.
Metal Halides
Lead and tin are metals in Group 14. Tin occurs as both SnO2 and SnO4. SnCl2 is a good reducing agent and is found in tinstone. SnF2 was once used as additive to toothpaste but now is replaced by NaF.
Group 16: The Oxygen Family
Oxygen and Sulfur Halides
Sulfur reacts directly with all the halogens except iodine. It spontaneously combines with fluorine to form sulfur hexafluoride, SF6, a colorless and inert gas. It can also form SF4 which is a powerful fluorinating agent. Sulfur and chlorine form SCl2, a red liquid, which is used in the production of the poisonous mustard gas. This reaction is shown below:
3SF4 + 4BCl3 → 4BF3 + 3SCl2 + 3Cl2
Oxygen combines with fluoride to form the compounds OF2 and O2F2. The structures of these molecules resemble that of hydrogen peroxide, although they are much more reactive.
Halogen Oxides
Common halogen oxides include $Cl_2O$, $ClO_2$, $Cl_2O_4$, and $I_2O_5$. Chlorine monoxide, the anhydride of hypochlorous acid, reacts vigorously with water as shown below, giving off chlorine and oxygen as products.
$Cl_2O + H_2O \rightleftharpoons 2HOCl \nonumber$
Chlorine dioxide and chlorine perchlorate form when sulfuric acid reacts with potassium chlorate. These compounds are similar to the nitrogen compounds $NO_2$ and $N_2O_4$. Iodine pentoxide forms iodic anhydride when reacted with water, as shown:
$I_2O_5 + H_2O \rightarrow 2HIO_3 \nonumber$
Halogen Oxoacids
Compounds that are made up of both oxygen and hydrogen are considered to be oxygen acids, or oxoacids. Common oxoacids are shown in the table below.
Table: Oxoacids of Halogens
HClO
HBrO
HIO
HClO2
HBrO3
HIO3
HClO3
HBrO4
HIO4
HClO4
Halogen Oxoanions
The oxoanions of chlorine are following:
• Hypochlorite, OCl-
• Chlorite, ClO2-
• Chlorate, ClO3-
• Perchlorate, ClO4-
All these compounds have common uses. For example, sodium chlorite is used as bleaching agent for textiles. Chlorate is a very good oxidizing agent and is very important in matches and fireworks.
Group 17: Other Halogens (Interhalogens)
Halogens have the ability to form compounds with other halogens (interhalogens). They are represented with the notation XY, in which the X and Y refer to two different halogens. Examples of this type of molecule include IBr and BrCl. They can also form polyatomic molecules such as XY3, XY5, XY7, corresponding to molecules such as IF3, BrF5, and IF7. Most interhalogen compounds such as CIF3 and BrF3 are very reactive.
Questions
1. Complete the following chemical reaction: $Mg + Br_2 \rightarrow \nonumber$
2. Which halogens cannot oxidize water to oxygen, and why?
3. Name four chlorine oxoanions.
4. Which metal forms a dimer when reacted with halogen?
5. Complete the following acid reaction: $HF + H_2O \rightarrow \nonumber$
6. Why is Aluminium Chloride a covalent compound, while Aluminium Chloride is ionic?
Answers
1. $Mg + Br_2 \rightarrow MgBr_2 \nonumber$
2. Iodine and bromine cannot oxidize water to oxygen because they have low electrode potential.
3. Hypochlorite, chlorite, chlorate, perchlorate
4. Aluminum
5. $HF + H_2O \rightarrow H_3O^+ + F^- \nonumber$
$AlCl_3$ is a molecular compound (molecular formula)
• Apparent charges Al (+3) and Cl (-1)
• Electronegativity of Al is 1.61
• Electronegativity of Cl is 3.16
• The difference in electronegativities is 1.55
• Therefore bonds are covalent
$AlF_3$ is an ionic compound (formula compound)
• Ionic charges Al (+3) and F (-1)
• Electronegativity of Al is 1.61
• Electronegativity of F is 3.98 (The fluorine atom is the most electronegative of all the elements)
• The difference in electronegativities is 2.37
• Therefore bonds are Ionic
Reactions of Main Group Elements with Hydrogen
Hydrogen Reactions
Hydrides are binary compounds of hydrogen. There are three possible hydrides that can be formed: ionic hydrides, covalent hydrides, and metallic hydrides. Metallic hydrides form when hydrogen reacts with transition metals, therefore they will not be introduced in this module.
Hydrogen and the S-Block
Ionic hydrides form when hydrogen reacts with s-block metals, not including Be and Mg. These s-block elements are found in Group 1 and Group 2 of the periodic table and are the most active metals. Group 1 metals are referred to as alkali metals and have a charge of +1 Group 2 metals are referred to as alkaline earth metals and have a charge of +2. Both Group 1 and Group 2 metals have low electronegativity values (less than 1.2).
Problems
1. Write the equation of the reaction between hydrogen and strontium.
2. Write the equation of the reaction between hydrogen and phosphorus.
3. Write the equation of the reaction between hydrogen and potassium.
4. Write the equation of the reaction between hydrogen and bromine.
5. Write 3 possible hydrocarbons that contains 3 carbons atoms.
Answers
1. Sr(s) + H2(s) SrH2(s)
2. P2(g) + 3H2(g) 2PH3(g)
3. 2K(s) + H2(g) 2KH(s)
4. H2(g) +Br2(g) 2HBr(g)
5. C3H8, C3H6, C3H4 ....answers may vary
Reactions of Main Group Elements with Nitrogen
Within this module will be the primary discussion about the chemistry of nitrogen and its ability and inability of forming reactions with certain main group elements. Although nitrogen is considered a relatively inert element, it is fully capable of creating some very active compounds.
Contributors and Attributions
• Kristin K. (University of California, Davis) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_of_Main_Group_Elements_with_Halogens.txt |
Oxygen is a highly reactive element that is very abundant on earth and in the human body. It is found in many compounds that are used to sustain basic life forms and modern civilization. Compounds containing oxygen are of great interest in the field of chemistry.
Background
Oxygen is ubiquitous; it comprises approximately 46% of the crust, 21% of the atmosphere, and 61% of the human body. Because of oxygen's high reactivity, it is most often found in compounds. Oxygen's high reactivity is due to its biradical electron configuration. As shown in a molecular orbital drawing of O2, the two unpaired electrons make the molecule highly susceptible to bond formation.
Oxygen has two allotropes (dioxygen, O2, and ozone, O3), both excellent oxidizing agents (Table P2). Oxygen is typically observed observed in the -2 oxidation state, in the form O2-, but it can also form other ions such as peroxide, O22-, and superoxide, O2-. With different possible oxidation states, many possible molecular compounds can be formed when an element reacts with oxygen. Many reactions involving oxygen occur in biological processes, including cellular respiration and photosynthesis.
Oxides are chemical compounds that contain at least one oxygen atom and at least one atom of another element. There are four principle oxidation states of oxygen: -2, -1, -1/2, and 0. The oxide ion, O2-, has a oxidation state of -2; the peroxide ion, O22-, has a oxidation state of -1; and the superoxide ion, O2-, has a oxidation state of -1/2. With metals, oxygen forms oxides that are largely ionic in character.
There are general trends in the reactions between main group elements and oxygen:
• Most nonmetals form the oxide with the highest possible oxidation state, with the halides excepted. Most metals form oxides with the oxygen in a -2 oxidation state.
• As a general rule, metal oxides are basic and nonmetal oxides are acidic. Basicity of an oxide increases with increasing ionic (metallic) character. Metal oxides, peroxides, and superoxides dissolve in water actually react with water to form basic solutions. Oxygen also forms covalent oxides with non-metals, that react with water to form acidic solutions.
• Oxygen does not react with fluorine or noble gases.
Exceptions to all of these trends are discussed below.
Reactions with Hydrogen
Oxygen reacts with hydrogen to produce two compounds: water ($H_2O$) and hydrogen peroxide ($H_2O_2$). Water is a versatile compound and participates in acid-base equilibrium and oxidation-reduction reactions. It can act as an acid, base, reducing agent, or oxidizing agent. Water's multifaceted abilities make it one of the most important compounds on earth. The reaction between hydrogen and oxygen to form water is given below:
$2H_{2 (g)} + O_{2 (g)} \rightarrow 2H_2O_{(l)} \label{1}$
Hydrogen peroxide's potent oxidizing abilities give it great industrial potential. The following equation shows the reaction of hydrogen and oxygen to form hydrogen peroxide:
$H_2 + O_2 \rightarrow H_2O_2 \label{2}$
The product of this reaction is called a peroxide because oxygen is in the $O_2^{2-}$ form (hydrogen has a +1 oxidation state). This concept is further explained regarding lithium below.
Reactions with Group 1 Elements
Oxygen reacts rapidly with Group 1 elements. All alkali metal oxides form basic solutions when dissolved in water. The principal combustion product is the most stable product with respect to the reactants. For example, with careful control of oxygen, the oxide M2O (where M represents any alkali metal) can be formed with any of the alkali metals. When heated, lithium, sodium, potassium, rubidium, and cesium ignite through combustion reactions with oxygen.
Lithium
Lithium, the first metal in Group 1, reacts with oxygen to form Li2O and burns with a red flame. The oxygen in this compound is an oxide (O2-). The formation of Li2O, the principal combustion product, is illustrated by the equation below:
$4 Li(s) + O_2(g) \rightarrow 2 Li_2O(s)\label{3}$
However, if there is excess oxygen present, it is possible that a small amount of the compound Li2O2 can be formed. Because alkali metals always have a +1 oxidation state, oxygen is in the O22- form. When oxygen is in this state, the compound is called a peroxide. The formation of this peroxide, the less-likely non-principal combustion product, under excess oxygen is illustrated by the equation below:
$2 Li(s) + O_2(g) \rightarrow Li_2O_2(s) \label{4}$
Sodium
Sodium burns in air with often little more than an orange glow. Using larger amounts of sodium or burning it in pure oxygen produces a strong orange flame. A white solid mixture of sodium oxide and sodium peroxide is formed. The equation for the formation of the simple oxide is analogous to that for lithium:
$4Na(s) + O_2 (g) \rightarrow 2Na_2O (s) \label{5}$
Likewise, the reaction for peroxide formation takes the same form for both metals:
$2Na(s) + O_2 (g) \rightarrow Na_2O_2 (s) \label{6}$
Potassium
Small pieces of potassium heated in air tend to melt instantly into a mixture of potassium peroxide and potassium superoxide with no visible flame. Larger pieces of potassium burn with a lilac-colored flame. The equation for the formation of the peroxide is identical to that for sodium (click here for more information):
$2K(s) + O_2 (g) \rightarrow K_2O_2 (s) \label{7}$
The superoxide generating reaction is given below:
$K(s) + O_2 (g) \rightarrow KO_2 (s) \label{8}$
Other alkali metals
The other alkali metals (Rb, Cs, Fr) form superoxide compounds (in which oxygen takes the form O2-) as the principal combustion products. The following equation shows the formation of superoxide, where M represents K, Rb, Cs, or Fr:
$M(s) + O_2(g) \rightarrow MO_2(s) \label{9}$
These compounds tend to be effective oxidizing agents due to the fact that O2- is one electron short of a complete octet and thus has a strong affinity for another electron. It is easily reduced, and therefore act as an effective oxidizing agent.
Reactions with Group 2 Elements
The elements of Group 2 are beryllium, magnesium, calcium, strontium, barium, and radioactive radium. Alkaline earth metals also react with oxygen, though not as rapidly as Group 1 metals; these reactions also require heating. Similarly to Group 1 oxides, most group 2 oxides and hydroxides are only slightly soluble in water and form basic, or alkaline solutions.
All Group 2 metals all react similarly, burning to form oxides (compounds containing the O2- ion) as shown:
$2 M(s) + O_2(g) \rightarrow 2 MO(s) \label{10}$
Once initiated, the reactions with oxygen are vigorous. The only peroxides (compounds containing the O22- ion) that can be formed from alkaline metals are strontium peroxide and barium peroxide. Both reactions require heat and excess oxygen. The general reaction is given below:
$M(s) + O_2(g) \rightarrow MO_2(s)\label{11}$
where M represents Sr or Ba.
Beryllium
Beryllium is unreactive with air and water. The chemical behavior of beryllium is best attributed to its small size and high ionization energy of its atoms.
All other group 2 metals
Except beryllium, the other alkaline earth metals form oxides in air at room temperature.
$2 M(s) + O_2(g) \rightarrow 2 MO(s) \label{12}$
where M represents Be, Mg, Ca, Sr, Ba, or Ra.
Peroxides, of the form MO2, are formed for all these elements except beryllium as shown:
$M(s) + O_2(g) \rightarrow MO_2(s) \label{13}$
Magnesium, calcium, strontium and barium oxides react with water to form hydroxides:
$MO(s) + H_2O(l) \rightarrow M(OH)_2(s) \label{14}$
All the oxides and hydroxides of the group 2 metals, except of those of beryllium, are bases:
$M(OH)_2(s) \rightarrow M^{2+}(aq) + 2OH^-(aq) \label{15}$
Reactions with Group 13 Elements
Group 13 consists of the following elements: boron, aluminum, gallium, indium, and thallium. Boron is the only element in this group that possesses no metallic properties. These elements vary in their reactions with oxygen. Recall that oxides of metals are basic and oxides or nonmetals are acidic; this is true for all elements in Group 13, except Al and Ga.
All other Group 13 elements also produce compounds of the form of M2O3, but adhere to the acid-base rules of metal and nonmetal oxides. Here is the equation of the reaction of oxygen and a Group 13 element:
$4M(s) + 3O_2(g) \rightarrow 2M_2O_3(s) \label{16}$
where M is any Group 13 element. At high temperatures, thallium also reacts with oxygen to produce Tl2O:
$4Tl(s) + O_2(g) \rightarrow 2Tl_2O \label{17}$
The most common oxide form of boron, B2O3 or boron trioxide, is obtained by heating boric acid:
$2B(OH)_3 \xrightarrow{\Delta} B_2O_3 + 3H_2O \label{18}$
Aluminum:
Aluminum occurs almost exclusively in the +3 oxidation state. It rapidly reacts with oxygen in air to give a water-insoluble coating of Al2O3. This oxide layer protects the metal beneath from further corrosion. The reaction is shown below:
$4Al(s) + 3O_2(g) \rightarrow 2Al_2O_3 \label{19}$
Aluminum trioxide, Al2O3, is amphoteric (acts both as an acid and a base):
$Al_2O_3(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2O(l) \label{20}$
$Al_2O_3(s) + 2NaOH(aq) + 3H_2O(l) \rightarrow 2Na[Al(OH)_4](aq) \label{21}$
Except for thallium in which the +1 oxidation state is more stable than the +3 state, aluminum, gallium, and indium favor +3 oxidation states
All of group 13 metal elements are known to form a trivalent oxide.
$4M(s) + 3O_2(g) \rightarrow 2M_2O_3(s) \label{22}$
with M represents Al, Ga, In, or Tl
Thallium is the only element in this group favors the formation of oxide over trioxide.
$2M(s) + O_2(g) \rightarrow 2MO(s) \label{23}$
Reactions with Group 14 Elements
Group 14 is made up of both metals (toward the bottom of the group), metalloids, and nonmetals (at the top of the group). The oxides of the top of Group 4 elements are slightly acidic, and the acidity of the oxides decreases down the group.
• Non-metal: The non-metal carbon of Group 14 (and its compounds) burn to form $CO_{2\;(g)}$ and, in smaller amounts, $CO_{(g)}$; both are acidic under different conditions. Carbon monoxide is only slightly soluble in water and does not react with it. Click here for more Information.
• Metalloid: The metalloid silicon reacts with oxygen to form only one stable compound, $SiO_{2}$, which dissolves slightly in water and is weakly acidic(Figure 2).
• The three metals in this group have many different oxide compounds due to their extended octets. All of these oxides are amphoteric (exhibit both basic and acidic properties). For example:
• Germanium: $GeO$, $GeO_{2}$
• Tin: $SnO$, $SnO_{2}$
• Lead: $PbO$, $PbO_{2}$, $Pb_{3}O_4$
Reactions with Group 15 Elements
The nitrogen family, Group 15, is capable of reacting with oxygen in many different ways. Nitrogen and phosphorus are nonmetallic, arsenic and antimony are metalloids, and bismuth is metallic.
Nitrogen
Nitrogen reacts with oxygen to form many oxides ranging in oxidation states from +1 to +5: All these oxides are gases at room temperature except for N2O5, which is solid. The nitrogen oxides are given below:
NO, N2O, N2O3, NO2, N2O5
All of these reactions are endothermic, requiring energy for oxygen to react directly with N2(g). The oxides of nitrogen are acidic (because they are nonmetal oxides). N2O3 and N2O5 react with water to give acidic solutions of oxoacids. These reactions are shown below:
Nitrous acid: $N_2O_3(s) + H_2O(l) \rightarrow 2HNO_2(aq) \label{24}$
Nitric acid: $N_2O_5(s) + H_2O(l) \rightarrow 2HNO_3(aq) \label{25}$
Phosphorus
There are two forms of allotropes of phosphorus, white phosphorus and red phosphorus. Red phosphorus is less reactive than white phosphorus. Phosphorus reacts with oxygen, usually forming two oxides depending on the amount available oxygen: P4O6 when reacted with a limited supply of oxygen, and P4O10 when reacted with excess oxygen; the latter is shown below.
$P_4O_{10} + 6 H_2O \rightarrow 4 H_3PO \label{26}$
On rare occasions, P4O7, P4O8, and P4O9 are also formed, but in small amounts.
Both P4O4 and P4O10 react with water to generate oxoacids. Reactions are shown below.
Phosphorous acid: $P_4O_6(l) + 6H_2O(l) \rightarrow 4H_3PO_3(aq) \label{27}$
Phosphoric acid: $P_4O_{10}(s) + 6H_2O(l) \rightarrow 4H_3PO_4(aq) \label{28}$
Other Group 15 Elements
Arsenic, antimony and bismuth react with oxygen when burned. The common oxidation states for arsenic, antimony, and bismuth are +3 and +5. There are two main types of oxides for each element:
• Arsenic: As2O3, As2O5
• Antimony: Sb2O3, Sb2O5
• Bismuth: Bi2O3, Bi2O5
There are other oxides, such as Sb4O10, that are not formed directly through reaction with oxygen. Arsenic(III) oxide and antimony(III) oxide are amphoteric, whereas bismuth(III) oxide acts only as a base (this is because it is the most metallic element in the group).
Reactions with Group 16 Elements
The elements in Group 16 include oxygen, sulfur, selenium, tellurium, and polonium. Oxygen reacts with the elements in its own group to form various oxides, mostly in the form of AO2 and AO3.
Oxygen
Although oxygen is located in Group 16, it is unique in its extreme electronegativity; this allows it to readily gain electrons and create hydrogen bonds. Because it is the smallest element in its group, it is capable of forming double bonds. It has no d-orbitals, and cannot expand its valence shell.
Oxygen is capable of reacting with itself, forming allotropes. One of oxygen's allotropes, ozone (O3), is formed when oxygen gas, O2, is subjected to ultraviolet light.
Sulfur
Sulfur dioxide, SO2, and sulfur trioxide, SO3, are the only common sulfur oxides.
$S(s) + O_2(g) \xrightarrow{\Delta} SO_2(g) \label{29}$
Sulfur's reaction with oxygen produces the oxides mentioned above as well as oxoacids. All are powerful oxidizing agents. SO2 is mainly used to make SO3, which reacts with water to produce sulfuric acid (recall that nonmetals form acidic oxides). These sequential reactions are shown below:
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \label{30}$
$2SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq) \label{31}$
Selenium and tellurium
Selenium and tellurium adopt compounds of the forms AO2, AO3, and AO.
Reactions with Group 17 Elements
The elements in Group 17 include fluorine, chlorine, bromine, and iodine. These elements are called halogens, from Greek roots translating to "salt formers." The halogens react with oxygen, but many of the resulting compounds are unstable, lasting for only moments at a time. They range in structure from X2O to X2O7, where X represents a halogen. Their extended octets allow them to bond with many oxygen atoms at a time.
Fluorine:
The most electronegative element adopts the -1 oxidation state. Fluorine and oxygen form OF2, which is known as oxygen fluoride.
$2 F_2 + 2NaOH \rightarrow OF_2 + 2 NaF + H_2O \label{32}$
Other Halogens
The other halogens form oxoacids instead of oxides. For example:
Oxidation state of halogen Chlorine Bromine Iodine
+1 HOCl HOBr HOI
+3 HClO2 –— –—
+5 HClO3 HBrO3 HIO HIO3
+7 HClO4 HBrO4 HIO4; H5IO6
Reactions with Group 18 Elements
The Group 18 noble gases include helium, neon, krypton, xenon, and radon. Noble gases are chemically inert with the exception of xenon, which reacts with oxygen to form XeO3 and XeO4 at low temperatures and high pressures. The ionization energy of xenon is low enough for the electronegative oxygen atom to capture electrons. XeO3 is highly unstable, and is known to spontaneously detonate in a clean, dry environment.
Problems
1. If there are 4.00 g of potassium, how much K2O is created when during combustion (in excess oxygen)?
2. Which noble gas(es), if any, react with oxygen? Why?
3. Complete and balance the following reaction: Al(s) + O2(g) →
4. What is the best environment for nitrogen and oxygen to react in?
5. What are the principal combustion products of each Group 1 element with oxygen?
Solutions
1. 0.00 g K2O. Potassium reacts with oxygen to form K2O2 and KO2 only.
2. Xenon, because the first ionization energy is low enough, allowing oxygen to bond to the xenon atom.
3. 4Al(s) + 3O2(g) → Al2O3(s)
4. High temperatures - it takes energy to cause those reactions.
5. Li generally forms Li2O. Na usually forms Na2O2. The rest of Group 1 elements tend to form comppunds in the form of MO2 where M is K, Rb, Cs, or Fr.
Contributors and Attributions
• Kathleen M. (UCD Spring 2010), Antoinette Mursa (UC Davis Spring 2011)
Reactions of Main Group Elements with Oxygen
From left to right on the periodic table, acid-base character of oxides and hydroxides go from basic to acidic.
• Increasing charge on an anion increases the production of basic solutions.
• As electronegativity increase, production of ionic cations increases because elements are more able to adopt a cation.
• As ionization energy increases, the acidic nature increases.
Metallic Oxides:
- Ionic Bonding: no distribution of electron wave function
- Ionic oxides are usually basic (element act as a base when reacting with H2O)
Na2O(s) + H2O(l) --> 2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
B. Oxide B. Hydroxide
Semimetal Oxides:
- Semimetal are amphoteric (elements acts as an acid and/or base when reacting depending on pH of solution)
Al2O3 --> Al(OH)3 --(3H+)--> [Al(H2O)6]^(3+) (aq)
--(OH-)--> [Al(OH)4]-(aq)
Non-Metal Oxides
- Covalent Bonding: almost complete distribution of electron wave function
- Covalent oxides are usually acidic (elements act as an acid when reacts with H2O)
SO3 + H2O(l) -> H2SO4(aq) -> H+ + HSO4-
A. Oxide A Hydroxide
Ionic Hydrides
- Ionic Bonding: no distribution of electron wave function
- Bronsted Basic because they will react with proton
- Lewis Basic because they can be ligands
CaH2 + 2H2O -> 2H2 + Ca(OH)2
H- H+ H2
-In this case, CaH2 is basic because it reacts with water (an acid in this case) to form many hydrides by reducing a proton.
Covalent Hydrides
- Covalent Bonding: almost complete distribution of electron wave function
HF + H2O -> F- + H3O+ ....can also be written as HF(aq) <--> H+(aq) + F-(aq)
H+ H+ H+
- HF is a weak acid that is bronsted acid because it will loose a proton. Therefore, HF is the weak acid, where the water acts as a silent water, and F- is the weak conjugate base. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_of_Main_Group_Elements_with_Oxygen/Acid-Base_Character_of_Oxides_and_Hydroxides.txt |
Reactions with Group 1 Elements
The elements of Group 1 consist of: Lithium, Sodium, Potassium, Rubidium, Cesium, and Francium. These elements are called the alkali metals because they react strongly with water and create hydroxide ions and hydrogen gas, leaving a basic solution.
2M(s) + 2H2O → 2M+(aq) + 2OH-(aq) + H2(g) M = Group 1 metal
Group 1 metals are very reactive with oxygen and must be kept away from oxygen in order to not get oxidized. These alkali metals rapidly react with oxygen to produce several different ionic oxides.
Oxides: O2- , peroxides: O22- , super oxide: O2- .
The usual oxide, M2O, can be formed with alkali metals generally by limiting the supply of oxygen.
With excess oxygen, the alkali metals can form peroxides, M2O2, or superoxides, MO2.
Lithium: Reacts with oxygen to give 2Li2O, lithium oxide. Reactions are shown below.
4Li(s) + O2(g)→2Li2O(s)
Sodium: Reacts with oxygen to form mostly sodium peroxide, Na2O2 . Na2O2 along with Li2O2 is used in emergency breathing devices in submarines and spacecrafts. Reactions are shown below.
2Na(s) + O2(g) → Na2O2(s)
The rest of the group, K, Rb, Cs, and Fr, forms the superoxides.
M(s) + O2(g) → MO2(s) M= K, Rb, Cs, Fr
Metal oxides, peroxides, and superoxides that dissolve in water react with water to form basic solutions.
Oxide ion with water: O2-(aq) + H2O(l) → 2OH-(aq)
Peroxide ion with water: O22-(aq) + 2H2O(l) → H2O2(aq) + 2OH-(aq)
Superoxide ion with water: 2O2-(aq) + 2H2O(l) → H2O2(aq) + 2OH-(aq) + O2(g)
Tl
Reactions with Group 14 Elements
The elements in Group 14 consist of carbon, silicon, germanium, tin, and lead. Carbon is the only nonmetal element of the group 14. Silicon is mostly nonmetallic. Germanium is a metalloid or semi-metal. Tin and lead have mainly metallic properties.
Carbon: Reacts with oxygen to form oxides. The main form of oxides of carbon are carbon monoxide, CO, and carbon dioxide, CO2.
Carbon dioxide is the primary product of burning organic materials and also a byproduct of respiration. During photosynthesis carbon is combined with water to form carbohydrates.
6CO2+ 6H2O C6H12O6 + 6O2
Carbon is the building block to many organic compounds.
Carbon dioxide is the only oxide formed when carbon is burned in an excess of air.
The reactions are shown below.
C(s) + O2(g) → CO2
2C(s) + O2(g) → 2CO
Silicon: Forms only one stable oxide with the empirical formula SiO2, silica. In silica, each Si atom is bonded to four O atoms and each O atom to two Si atoms forming a network covalent solid with a network of –Si–O–Si– bonds.
Germanium: Forms germanium dioxide which is covalent network solid similar to silicon dioxide.
Tin: Forms two primary oxides, SnO and SnO2. By heating SnO in air, it can be converted to SnO2. SnO2 is used as a jewerly abarasive. The reactions are shown below.
Sn(s) + O2(g) → SnO2(s)
2Sn(s) + O2(g) → 2SnO(s)
Lead: Forms a several forms of oxides. The best known oxides of lead are yellow lead oxide, PbO, red-brown lead dioxide, PbO2,and red lead, Pb3O4. The reactions are shown below.
2Pb(s) + O2(g) → 2PbO(s)
Pb(s) + O2(g) → PbO2(s)
3Pb(s) + 2O2(g) → Pb3O4(s)
Reactions with Group 15 Elements
The elements in Group 15 consist of : nitrogen, phosphorus, arsenic, antimony, and bismuth. Nitrogen and phosphorus are nonmetallic, arsenic and antimony are metalloids, and bismuth is metallic.
Nitrogen: Forms a sires of oxides in which the oxidation state of N can have every value ranging from +1 to +5. All these oxides are gases at room temperature except for N2O5, which is solid.
Preparation of Oxides of Nitrogen: Reactions are shown below.
N2O NH4NO3(s) —∆→ N2O(g) + 2H2O(g)
NO 3Cu(s) + 8H+(aq) + 2NO3-(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)
N2O3 2NO(g) + N2O4(g) —-20°C→ 2N2O3(l)
NO2 2Pb(NO3)2(s) —∆→ 2PbO(s) +4NO2(g) + O2(g)
2NO(g) + O2(g) <=> 2NO2(g)
N2O4 2NO2(g) <=> N2O4(g)
N2O5 4HNO3 (l) + P4O10 ) —-10°C→ 4HPO3(s) + 2N2O5(s) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_of_Main_Group_Elements_with_Oxygen/Main_Group_Oxides_Reactions.txt |
Water is composed of two hydrogen atoms and an oxygen atom. It exhibits polarity and is naturally found in the liquid, solid, and vapor states. Its polarity makes it a good solvent and is commonly known as the universal solvent. Because of its abundance on earth, it is important to note that it is involved in many chemical reactions. Many of these chemical reactions behave in trends that can be categorized using the periodic table.
Group 1: Alkali Metals
A common characteristic of most Alkali Metals is their ability to displace H2(g) from water. This is represented by their large, negative electrode potentials. In this event, the Group 1 metal is oxidized to its metal ion and water is reduced to form hydrogen gas and hydroxide ions. The general reaction of an alkali metal (M) with H2O (l) is given in the following equation:
$\ce{ 2M(s) + 2H2O(l) \longrightarrow 2M^{+}(aq) + 2OH^{-}(aq) + H2 (g)} \nonumber$
From this reaction it is apparent that OH- is produced, creating a basic or alkaline environment. Group 1 elements are called alkali metals because of their ability to displace H2(g) from water and create a basic solution.
Alkali metals are also known to react violently and explosively with water. This is because enough heat is given off during the exothermic reaction to ignite the H2(g).
Alkali Metals Oxides and Water
Oxides of Group 1 elements also react with water to create basic solutions. Alkali metals react with oxygen to form monoxides, peroxides, or superoxides. These species react with water in different ways:
• Monoxides (M2O) produce alkali metal hydroxides:
$\ce{M2O(s) + 2H2O(l) \longrightarrow 2M^{+}(aq) + 2OH^{-} (aq)} \label{1}$
• Peroxides (M2O2) produce metal hydroxides and hydrogen peroxide:
$\ce{M2O2(s) + 2H2O(l) \longrightarrow 2M^{+}(aq) + 2OH^{-} (aq) + H2O2(aq)} \label{2}$
• Superoxides (MO2) produce metal hydroxides, hydrogen peroxide, and oxygen gas:
$\ce{2MO2 (s) + 2H2O(l) \longrightarrow 2M^{+}(aq) + 2OH^{-} (aq) + H2O2 (aq) + O2(g)} \label{3}$
Alkali Metal Hydrides and Water
Similarly to the Group 1 oxides, the hydrides of the Group 1 elements react with water to form a basic solution. In this case, however, hydrogen gas is produced with the metal hydroxide. The general reaction for alkali metal hydrides and water is given below:
$MH_{(s)} + H_2O_{(l)} \longrightarrow M^+_{(aq)}+OH^-_{(aq)}+H_{2(g)} \label{4}$
This reaction can be generalized to all alkali metal hydrides.
Group 2: Alkaline Earth Metals
The majority of Alkaline Earth Metals also produce hydroxides when reacted with water. The hydroxides of calcium, strontium, and barium are only slightly soluble in water; however, enough hydroxide ions are produced to make a basic environment. The general reaction of calcium, strontium, and barium with water is represented below, where M represents calcium, strontium, or barium:
$M_{(s)} + 2H_2O_{(l)} \longrightarrow M(OH)_{2(aq)}+H_{2\;(g)} \label{5}$
Magnesium (Mg) reacts with water vapor to form magnesium hydroxide and hydrogen gas. Beryllium (Be) is the only alkaline earth metal that does not react with water. This is due to its small size and high ionization energy in relation to the other elements in the group.
Alkaline Earth Metal Oxides and Water
Similarly to the alkali metal oxides, alkaline earth metal monoxides combine with water to form metal hydroxide salts (as illustrated in the equation below). The exception to this general assumption is beryllium, whose oxide (BeO) does not react with water.
$MO_{(s)}+H_2O_{(l)} \longrightarrow M(OH)_{2(s)} \label{6}$
One of the most familiar alkaline earth metal oxides is CaO or quicklime. This substance is often used to treat water and to remove harmful $SO_{2(g)}$ from industrial smokestacks.
Alkaline Earth Metal Hydrides and Water
With the exception of beryllium (Be), the alkaline metal hydrides react with water to produce the metal hydroxide and hydrogen gas. The reaction of these metal hydrides can be described below:
$MH_{2(s)}+2H_2O_{(l)} \longrightarrow M(OH)_{2(aq)}+2H_{2(g)} \label{7}$
Hard Water
The two types of hard water include temporary hard water and permanent hard water. Temporary hard water contains bicarbonate (HCO3-) which forms CO3-2(aq), CO2(g), and H2O when heated. The bicarbonate ions react with alkaline earth cations and precipitate out of solution, causing boiler scale and problems in water heaters and plumbing. Common cations in the water include Mg+2 and Ca+2. In order to soften the water, water treatment plants add an alkaline earth metal hydroxide, such as slake lime [Ca(OH)2]. This solid dissolves in the water producing a metal ion (M+2) and hydroxide ions (OH-). The hydroxide ions combine with the bicarbonate ions in the water to produce water and a carbonate ion. The carbonate ion then precipitates out with the metal ion to form MCO3(s). Water treatment plants are able to remove the precipitated metal carbonate and thus soften the water.
The other type of hard water is permanent hard water. Permanent hard water contains bicarbonate ions (HCO3-) as well as other anions such as sulfate ions (SO4-2). The hardening species often cannot be boiled off. To soften permanent water, sodium carbonate (Na2CO3) is added. Sodium carbonate precipitates out the Mg+2 and Ca+2 ions out as the respective metal carbonates and introduces Na+ ions into the solutions.
Group 13: Boron Family
Group 13 elements are not very reactive with water. In fact, boron (B) does not react at with water. One notable reaction within this group is aluminum's (Al) reaction with water. Aluminum does not appear to react with water because an outer layer of aluminum oxide (Al2O3) solid forms and protects the rest of the metal.
Group 14: Carbon Family
For the most part, Group 14 elements do not react with water. One interesting consequence of this is that tin (Sn) is often sprayed as a protective layer on iron cans to prevent the can from corroding.
Group 15: Nitrogen Family
The pure elements in this family do not tend to react with water. Compounds of nitrogen (nitrates and nitrites) as well as nitrogen gas (N2) dissolve in water but do not react.
Group 16: Oxygen Family
As mentioned earlier, many Group 1 and Group 2 oxides react with water to form metal hydroxides. The nonmetal oxides react with water to form oxoacids. Examples include phosphoric acid and sulfuric acid.
Group 17: Halogens
Generally halogens react with water to give their halides and hypohalides. The halogen gases vary in their reactions with water due to their different electronegativities. Because fluorine ($\ce{F2}$) is so electronegative, it can displace oxygen gas from water. The products of this reaction include oxygen gas and hydrogen fluoride. The hydrogen halides react with water to form hydrohalic acids ($\ce{HX}$). With the exception of $\ce{HF}$, the hydrohalic acids are strong acids in water. Hydrochloric acid ($\ce{HCl}$), a strong acid, is an example.
$\ce{Cl2(g) + 2H2O(l) → HCl(aq) + HOCl(aq)} \nonumber$
Hypochlorous ($\ce{HOCl}$) acid is a strong bleaching agent and is not very stable in solution and readily decomcomposes, especially when exposed to sunlight, yielding oxygen.
$\ce{ 2 HClO -> 2 HCl + O2} \nonumber$
Bromine liquid dissolves slowly in water to form a yellowish-brown solution.
$\ce{Br2(g) + 2H2O(l) → HBr(aq) + HOBr(aq)} \nonumber$
Hypobromous ($\ce{HOBr}$) acid is a weak bleaching agent.
$\ce{I2(g) + 2H2O(l) → HI(aq) + HOI(aq)} \nonumber$
Only a little iodine dissolves in water to form a yellowish solution and hypoiodous ($\ce{HOI}$) acid has very weak bleaching characteristic.
Group 18: Noble Gases
The noble gases do not react with water.
General Summary
Reactant #1 Reactant #2 Products
Group 1 metal in period 3 or higher Cold Water Metal hydroxide & molecular hydrogen
Group 2 metal in period 3 or higher Cold Water Metal hydroxide & molecular hydrogen
Nonmetal element (excluding halogens) Cold Water No reaction
Fluorine (F2) Water Hydrogen fluoride (HF) and molecular oxygen (O2)
Halogen Water Hydrohalic acid or hypohalous acid
Metal with E0< -4.14 V for the lowest oxidation state Steam Metal oxide & molecular hydrogen
Nonmetal Halide Water Nonmetal oxide and hydrogen halide
Metal oxide Water Metal hydroxide
Nonmetal oxide Water oxoacid
Problems
1. Predict the products of the following reactions:
1. $Be_{(s)}+2H_{2}O_{(l)} \longrightarrow$
2. $Ne_{(g)}+2H_{2}O_{(l)} \longrightarrow$
3. $Cl_{2\;(g)}+2H_{2}O_{(l)} \longrightarrow$
4. $Li_2O_{(s)}+2H_{2}O_{(l)} \longrightarrow$
2. True/False
1. Metal oxides form basic solutions in water
2. Difluorine does not react with water.
3. Beryllium has a large atomic radius.
4. Sodium is the alkali element that reacts most violently with water.
3. Why are do we called Group 1 and 2 metals "alkali" and "alkaline"?
4. How is aluminum affected by water?
5. Will the following reaction create an acidic or basic solution?
$NaH{(s)}+2H_{2}O_{(l)} \longrightarrow$
Answers
1. Answers
1. No reaction
2. No reaction
3. $Cl_{2\;(g)}+2H_{2}O_{(l)} \longrightarrow HOCl_{(aq)} + H^+_{(aq)} + Cl^-_{(aq)}$
4. $Li_2O_{(s)}+2H_{2}O_{(l)} \longrightarrow 2LiOH_{(aq)}$
2. Answers
1. True
2. False
3. False
4. False
3. These metals react with water to form OH-(aq) ions and create a basic or "alkaline" environment.
4. No reaction occurs.
5. Basic.
$NaH{(s)}+2H_{2}O_{(l)} \longrightarrow Na^+_{(aq)}+OH^-_{(aq)}+H_2 \; {(g)}$
Contributors and Attributions
• Trevor Landas (University of California, Davis) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_of_Main_Group_Elements_with_Water.txt |
Learning Objectives
• To be familiar with the roles of the s-block elements in biology.
The s-block elements play important roles in biological systems. Covalent hydrides, for example, are the building blocks of organic compounds, and other compounds and ions containing s-block elements are found in tissues and cellular fluids. In this section, we describe some ways in which biology depends on the properties of the group 1 and group 2 elements.
Covalent Hydrides
There are three major classes of hydrides—covalent, ionic, and metallic—but only covalent hydrides occur in living cells and have any biochemical significance. Carbon and hydrogen have similar electronegativities, and the C–H bonds in organic molecules are strong and essentially nonpolar. Little acid–base chemistry is involved in the cleavage or formation of these bonds. In contrast, because hydrogen is less electronegative than oxygen and nitrogen (symbolized by Z), the H–Z bond in the hydrides of these elements is polarized (Hδ+–Zδ). Consequently, the hydrogen atoms in these H–Z bonds are relatively acidic. Moreover, S–H bonds are relatively weak due to poor s orbital overlap, so they are readily cleaved to give a proton. Hydrides in which H is bonded to O, N, or S atoms are therefore polar, hydrophilic molecules that form hydrogen bonds. They also undergo acid–base reactions by transferring a proton.
Note
Covalent hydrides in which H is bonded to O, N, or S atoms are polar and hydrophilic, form hydrogen bonds, and transfer a proton in their acid-base reactions.
Hydrogen bonds are crucial in biochemistry, in part because they help hold proteins in their biologically active folded structures. Hydrogen bonds also connect the two intertwining strands of DNA (deoxyribonucleic acid), the substance that contains the genetic code for all organisms. Because hydrogen bonds are easier to break than the covalent bonds that form the individual DNA strands, the two intertwined strands can be separated to give intact single strands, which is essential for the duplication of genetic information.
In addition to the importance of hydrogen bonds in biochemical molecules, the extensive hydrogen-bonding network in water is one of the keys to the existence of life on our planet. Based on its molecular mass, water should be a gas at room temperature (20°C), but the strong intermolecular interactions in liquid water greatly increase its boiling point. Hydrogen bonding also produces the relatively open molecular arrangement found in ice, which causes ice to be less dense than water. Because ice floats on the surface of water, it creates an insulating layer that allows aquatic organisms to survive during cold winter months.
These same strong intermolecular hydrogen-bonding interactions are also responsible for the high heat capacity of water and its high heat of fusion. A great deal of energy must be removed from water for it to freeze. Consequently, large bodies of water act as “thermal buffers” that have a stabilizing effect on the climate of adjacent land areas. Perhaps the most striking example of this effect is the fact that humans can live comfortably at very high latitudes. For example, palm trees grow in southern England at the same latitude (51°N) as the southern end of frigid Hudson Bay and northern Newfoundland in North America, areas known more for their moose populations than for their tropical vegetation. Warm water from the Gulf Stream current in the Atlantic Ocean flows clockwise from the tropical climate at the equator past the eastern coast of the United States and then turns toward England, where heat stored in the water is released. The temperate climate of Europe is largely attributable to the thermal properties of water.
Note
Strong intermolecular hydrogen-bonding interactions are responsible for the high heat capacity of water and its high heat of fusion.
Macrominerals
The members of group 1 and group 2 that are present in the largest amounts in organisms are sodium, potassium, magnesium, and calcium, all of which form monatomic cations with a charge of +1 (group 1, M+) or +2 (group 2, M2+). Biologically, these elements can be classified as macrominerals (Table 1.6).
For example, calcium is found in the form of relatively insoluble calcium salts that are used as structural materials in many organisms. Hydroxyapatite [Ca5(PO4)3OH] is the major component of bones, calcium carbonate (CaCO3) is the major component of the shells of mollusks and the eggs of birds and reptiles, and calcium oxalate (CaO2CCO2) is found in many plants. Because calcium and strontium have similar sizes and charge-to-radius ratios, small quantities of strontium are always found in bone and other calcium-containing structural materials. Normally this is not a problem because the Sr2+ ions occupy sites that would otherwise be occupied by Ca2+ ions. When trace amounts of radioactive 90Sr are released into the atmosphere from nuclear weapons tests or a nuclear accident, however, the radioactive strontium eventually reaches the ground, where it is taken up by plants that are consumed by dairy cattle. The isotope then becomes concentrated in cow’s milk, along with calcium. Because radioactive strontium coprecipitates with calcium in the hydroxyapatite that surrounds the bone marrow (where white blood cells are produced), children, who typically ingest more cow’s milk than adults, are at substantially increased risk for leukemia, a type of cancer characterized by the overproduction of white blood cells.
Ion Transport
The Na+, K+, Mg2+, and Ca2+ ions are important components of intracellular and extracellular fluids. Both Na+ and Ca2+ are found primarily in extracellular fluids, such as blood plasma, whereas K+ and Mg2+ are found primarily in intracellular fluids. Substantial inputs of energy are required to establish and maintain these concentration gradients and prevent the system from reaching equilibrium. Thus energy is needed to transport each ion across the cell membrane toward the side with the higher concentration. The biological machines that are responsible for the selective transport of these metal ions are complex assemblies of proteins called ion pumps. Ion pumps recognize and discriminate between metal ions in the same way that crown ethers and cryptands do, with a high affinity for ions of a certain charge and radius.
Defects in the ion pumps or their control mechanisms can result in major health problems. For example, cystic fibrosis, the most common inherited disease in the United States, is caused by a defect in the transport system (in this case, chloride ions). Similarly, in many cases, hypertension, or high blood pressure, is thought to be due to defective Na+ uptake and/or excretion. If too much Na+ is absorbed from the diet (or if too little is excreted), water diffuses from tissues into the blood to dilute the solution, thereby decreasing the osmotic pressure in the circulatory system. The increased volume increases the blood pressure, and ruptured arteries called aneurysms can result, often in the brain. Because high blood pressure causes other medical problems as well, it is one of the most important biomedical disorders in modern society.
For patients who suffer from hypertension, low-sodium diets that use NaCl substitutes, such as KCl, are often prescribed. Although KCl and NaCl give similar flavors to foods, the K+ is not readily taken up by the highly specific Na+-uptake system. This approach to controlling hypertension is controversial, however, because direct correlations between dietary Na+ content and blood pressure are difficult to demonstrate in the general population. More important, recent observations indicate that high blood pressure may correlate more closely with inadequate intake of calcium in the diet than with excessive sodium levels. This finding is important because the typical “low-sodium” diet is also low in good sources of calcium, such as dairy products.
Some of the most important biological functions of the group 1 and group 2 metals are due to small changes in the cellular concentrations of the metal ion. The transmission of nerve impulses, for example, is accompanied by an increased flux of Na+ ions into a nerve cell. Similarly, the binding of various hormones to specific receptors on the surface of a cell leads to a rapid influx of Ca2+ ions; the resulting sudden rise in the intracellular Ca2+ concentration triggers other events, such as muscle contraction, the release of neurotransmitters, enzyme activation, or the secretion of other hormones.
Within cells, K+ and Mg2+ often activate particular enzymes by binding to specific, negatively charged sites in the enzyme structure. Chlorophyll, the green pigment used by all plants to absorb light and drive the process of photosynthesis, contains magnesium. During photosynthesis, CO2 is reduced to form sugars such as glucose. The structure of the central portion of a chlorophyll molecule resembles a crown ether (part (a) in Figure 13.7) with four five-member nitrogen-containing rings linked together to form a large ring that provides a “hole” the proper size to tightly bind Mg2+.
The structure of the central core of chlorophyll, a magnesium complex present in all photosynthetic tissues.
Ionophores
Because the health of cells depends on maintaining the proper levels of cations in intracellular fluids, any change that affects the normal flux of metal ions across cell membranes could well cause an organism to die. Molecules that facilitate the transport of metal ions across membranes are generally called ionophores (ion plus phore from the Greek phorein, meaning “to carry”). Many ionophores are potent antibiotics that can kill or inhibit the growth of bacteria. An example is valinomycin, a cyclic molecule with a central cavity lined with oxygen atoms (part (a) in Figure 21.14) that is similar to the cavity of a crown ether (part (a) in Figure 13.7). Like a crown ether, valinomycin is highly selective: its affinity for K+ is about 1000 times greater than that for Na+. By increasing the flux of K+ ions into cells, valinomycin disrupts the normal K+ gradient across a cell membrane, thereby killing the cell (part (b) in Figure 21.14).
(a) This model of the structure of the K+–valinomycin complex, determined by x-ray diffraction, shows how the valinomycin molecule wraps itself around the K+ ion, shielding it from the environment, in a manner reminiscent of a crown ether complex. (For more information on the crown ethers, see Chapter 13, Section 13.2.) (b) Valinomycin kills bacteria by facilitating the transport of K+ ions across the cell membrane, thereby disrupting the normal distribution of ions in the bacterium. At the surface of the membrane, valinomycin binds a K+ ion. Because the hydrophobic exterior of the valinomycin molecule forms a “doughnut” that shields the positive charge of the metal ion, the K+–valinomycin complex is highly soluble in the nonpolar interior of the membrane. After the K+–valinomycin complex diffuses across the membrane to the interior of the cell, the K+ ion is released, and the valinomycin is free to diffuse back to the other side of the membrane to bind another K+ ion. Valinomycin thereby destroys the normal K+ gradient across the membrane, killing the cell.
Example 6
A common way to study the function of a metal ion in biology is to replace the naturally occurring metal with one whose reactivity can be traced by spectroscopic methods. The substitute metal ion must bind to the same site as the naturally occurring ion, and it must have a similar (or greater) affinity for that site, as indicated by its charge density. Arrange the following ions in order of increasing effectiveness as a replacement for Ca2+, which has an ionic radius of 100 pm (numbers in parentheses are ionic radii): Na+ (102 pm), Eu2+ (117 pm), Sr2+ (118 pm), F (133 pm), Pb2+ (119 pm), and La3+ (103 pm). Explain your reasoning.
Given: ions and ionic radii
Asked for: suitability as replacement for Ca2+
Strategy:
Use periodic trends to arrange the ions from least effective to most effective as a replacement for Ca2+.
Solution
The most important properties in determining the affinity of a biological molecule for a metal ion are the size and charge-to-radius ratio of the metal ion. Of the possible Ca2+ replacements listed, the F ion has the opposite charge, so it should have no affinity for a Ca2+-binding site. Na+ is approximately the right size, but with a +1 charge it will bind much more weakly than Ca2+. Although Eu2+, Sr2+, and Pb2+ are all a little larger than Ca2+, they are probably similar enough in size and charge to bind. Based on its ionic radius, Eu2+ should bind most tightly of the three. La3+ is nearly the same size as Ca2+ and more highly charged. With a higher charge-to-radius ratio and a similar size, La3+ should bind tightly to a Ca2+ site and be the most effective replacement for Ca2+. The order is F << Na+ << Pb2+ ~ Sr2+ ~ Eu2+ < La3+.
Exercise
The ionic radius of K+ is 138 pm. Arrange the following ions in order of increasing affinity for a K+-binding site in an enzyme (numbers in parentheses are ionic radii): Na+ (102 pm), Rb+ (152 pm), Ba2+ (135 pm), Cl (181 pm), and Tl+ (150 pm).
Answer Cl << Na+ < Tl+ ~ Rb+ < Ba2+
Summary
Covalent hydrides in which hydrogen is bonded to oxygen, nitrogen, or sulfur are polar, hydrophilic molecules that form hydrogen bonds and undergo acid–base reactions. Hydrogen-bonding interactions are crucial in stabilizing the structure of proteins and DNA and allow genetic information to be duplicated. The hydrogen-bonding interactions in water and ice also allow life to exist on our planet. The group 1 and group 2 metals present in organisms are macrominerals, which are important components of intracellular and extracellular fluids. Small changes in the cellular concentration of a metal ion can have a significant impact on biological functions. Metal ions are selectively transported across cell membranes by ion pumps, which bind ions based on their charge and radius. Ionophores, many of which are potent antibiotics, facilitate the transport of metal ions across membranes.
Key Takeaway
• Among their many roles in biology, the s-block elements allow genetic information to be duplicated and are important components of intracellular and extracellular fluids.
Problems
1. Explain the thermochemical properties of water in terms of its intermolecular bonding interactions. How does this affect global climate patterns?
2. Of the three classes of hydrides, which is (are) biochemically significant? How do you account for this?
3. Many proteins are degraded and become nonfunctional when heated higher than a certain temperature, even though the individual protein molecules do not undergo a distinct chemical change. Propose an explanation for this observation.
4. Los Angeles has moderate weather throughout the year, with average temperatures between 57°F and 70°F. In contrast, Palm Springs, which is just 100 miles inland, has average temperatures between 55°F and 95°F. Explain the difference in the average temperature ranges between the two cities.
5. Although all group 1 ions have the same charge (+1), Na+ and K+ ions are selectively transported across cell membranes. What strategy do organisms employ to discriminate between these two cations?
Structure and Reactivity
1. A 0.156 g sample of a chloride salt of an alkaline earth metal is dissolved in enough water to make 20.5 mL of solution. If this solution has an osmotic pressure of 2.68 atm at 25°C, what is the identity of the alkaline earth metal?
2. The thermal buffering capacity of water is one of the reasons the human body is capable of withstanding a wide range of temperatures. How much heat (in kilojoules) is required to raise the temperature of a 70.0 kg human from 37.0°C to 38.0°C? Assume that 70% of the mass of the body is water and that body fluids have the same specific heat as water.
3. During illness, body temperature can increase by more than 2°C. One piece of folklore is that you should “feed a fever.” Using the data in Table 5.5, how many fried chicken drumsticks would a 70.0 kg person need to eat to generate a 2.0°C change in body temperature? Assume the following: there is complete conversion of the caloric content of the chicken to thermal energy, 70% of the mass of the body is solely due to water, and body fluids have the same specific heat as water.
4. Hydrogen bonding is partly responsible for the high enthalpy of vaporization of water (ΔHvap = 43.99 kJ/mol at 25°C), which contributes to cooling the body during exercise. Assume that a 50.0 kg runner produces 20.0 g of perspiration during a race, and all the perspiration is converted to water vapor at 37.0°C. What amount of heat (in joules) is removed from the runner’s skin if the perspiration consists of only water?
Answer
1. Ba
• Anonymous | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/The_s-Block_Elements_in_Biology.txt |
Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements.
• Periodic Properties of the Elements
The elements in the periodic table are arranged in order of increasing atomic number. All of these elements display several other trends and we can use the periodic law and table formation to predict their chemical, physical, and atomic properties. Understanding these trends is done by analyzing the elements electron configuration; all elements prefer an octet formation and will gain or lose electrons to form that stable configuration.
• Periodic Trends
Page notifications Off Share Table of contents Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool.
• Periodic Trends in Ionic Radii
An understanding of periodic trends is necessary when analyzing and predicting molecular properties and interactions. Common periodic trends include those in ionization energy, atomic radius, and electron affinity. One such trend is closely linked to atomic radii -- ionic radii. Neutral atoms tend to increase in size down a group and decrease across a period. When a neutral atom gains or loses an electron, creating an anion or cation, the atom's radius increases or decreases, respectively.
• Table Basics
The periodic table of elements is one of the ways that scientists keep track of all the known elements.
• The Periodic Law
The periodic law was developed independently by Dmitri Mendeleev and Lothar Meyer in 1869. Mendeleev created the first periodic table and was shortly followed by Meyer. They both arranged the elements by their mass and proposed that certain properties periodically reoccur. Meyer formed his periodic law based on the atomic volume or molar volume, which is the atomic mass divided by the density in solid form.
Periodic Trends of Elemental Properties
The elements in the periodic table are arranged in order of increasing atomic number. All of these elements display several other trends and we can use the periodic law and table formation to predict their chemical, physical, and atomic properties. Understanding these trends is done by analyzing the elements electron configuration; all elements prefer an octet formation and will gain or lose electrons to form that stable configuration.
Atomic Radius
We can never determine the atomic radius of an atom because there is never a zero probability of finding an electron, and thus never a distinct boundary to the atom. All that we can measure is the distance between two nuclei (internuclear distance). A covalent radius is one-half the distance between the nuclei of two identical atoms. An ionic radius is one-half the distance between the nuclei of two ions in an ionic bond. The distance must be apportioned for the smaller cation and larger anion. A metallic radius is one-half the distance between the nuclei of two adjacent atoms in a crystalline structure. The noble gases are left out of the trends in atomic radii because there is great debate over the experimental values of their atomic radii. The SI units for measuring atomic radii are the nanometer (nm) and the picometer (pm). $1 \, nm = 1 \times 10^{-9}\, m$ and $1\, pm = 1 \times 10^{-12}\, m$.
To explain this trend, the concept of screening and penetration must be understood. Penetration is commonly known as the distance that an electron is from the nucleus. Screening is defined as the concept of the inner electrons blocking the outer electrons from the nuclear charge. Within this concept we assume that there is no screening between the outer electrons and that the inner electrons shield the outer electrons from the total positive charge of the nucleus. In order to comprehend the extent of screening and penetration within an atom, scientists came up with the effective nuclear charge, $Z_{eff}$. The equation for calculating the effective nuclear charge is shown below.
$Z_{eff}= Z - S \nonumber$
In the equation S represents the number of inner electrons that screen the outer electrons. Students can easily find S by using the atomic number of the noble gas that is one period above the element. For example, the S we would use for Chlorine would be 10 (the atomic number of Neon). Z is the total number of electrons in the atom. Since we know that a neutral atom has an identical number of protons and electrons, we can use the atomic number to define Z. For example, Chlorine would have a Z value of 17 (the atomic number of Chlorine). Continuing to use Chlorine as an example, the 10 inner electrons (S) would screen out the positive charge of ten protons. Therefore there would be and effective nuclear charge of 17-10 or +7. The effective nuclear charge shows that the nucleus is pulling the outer electrons with a +7 charge and therefore the outer electrons are pulled closer to the nucleus and the atomic radii is smaller. In summary, the greater the nuclear charge, the greater pull the nucleus has on the outer electrons and the smaller the atomic radii. In contrast, the smaller nuclear charge, the lesser pull the nucleus has on the outer electrons, and the larger atomic radii. Additionally, as the atomic number increases, the effective nuclear charge also increases. Figure 3 depicts the effect that the effective nuclear charge has on atomic radii.
Now we are ready to describe the atomic radius trend in the periodic table. The atomic number increases moving left to right across a period and subsequently so does the effective nuclear charge. Therefore, moving left to right across a period the nucleus has a greater pull on the outer electrons and the atomic radii decreases. Moving down a group in the periodic table, the number of filled electron shells increases. In a group, the valence electrons keep the same effective nuclear charge, but now the orbitals are farther from the nucleus. Therefore, the nucleus has less of a pull on the outer electrons and the atomic radii are larger.
We can now use these concept to explain the atomic radius differences of cations and anions. A cation is an atom that has lost one of its outer electrons. Cations have a smaller radius than the atom that they were formed from. With the loss of an electron, the positive nuclear charge out powers the negative charge that the electrons exert. Therefore, the positive nucleus pulls the electrons tighter and the radius is smaller. An anion is an atom that has gained an outer electron. Anions have a greater radius than the atom that they were formed from. The gain of an electron does not alter the nuclear charge, but the addition of an electron causes a decrease in the effective nuclear charge. Therefore, the electrons are held more loosely and the atomic radius is increased.
Ionization Energy (ionization potential)
Expelling an electron from an atom requires enough energy to overcome the magnetic pull of the positive charge of the nucleus. Therefore, ionization energy (I.E. or I) is the energy required to completely remove an electron from a gaseous atom or ion. The Ionization Energy is always positive.
The energy required to remove one valence electron is the first ionization energy, the second ionization energy is the energy required to remove a second valence electron, and so on.
• 1st ionization energy
$\ce{Na(g) -> Na^{+}(g) + e^{-}(g)} \nonumber$
• 2nd ionization energy
$\ce{ Na^{+}(g) -> Na^{2+}(g) + e^{-}} \nonumber$
Ionization energies increase relative to high effective charge. The highest ionization energies are the noble gases because they all have high effective charge due to their octet formation and require a high amount of energy to destroy that stable configuration. The highest amount of energy required occurs with the elements in the upper right hand corner. Additionally, elements in the left corner have a low ionization energy because losing an electron allows them to have the noble gas configuration. Therefore, it requires less energy to remove one of their valence electrons
Table 1: Ionization Energies of certain elements (1st IE, 2nd IE, etc)
Element 1st 2nd 3rd 4th 5th 6th 7th
Na 496 4562
Mg 738 1451 7733
Al 577 1817 2745 11580
Si 786 1577 3232 4356 16090
P 1060 1903 2912 4957 6274 21270
S 999.6 2251 3361 4564 7013 8496 27110
Cl 1256 2297 3822 5158 6542 9362 11020
Ar 1520 2666 3931 5771 7238 8781 12000
These are the ionization energies for the period three elements. Notice how Na after in the second I.E, Mg in the third I.E., Al in the fourth I.E., and so on, all have a huge increase in energy compared to the proceeding one. This occurs because the proceeding configuration was in a stable octet formation; therefore it requires a much larger amount of energy to ionize.
Ionization Energies increase going left to right across a period and increase going up a group. As you go up a group, the ionization energy increases, because there are less electron shielding the outer electrons from the pull of the nucleus. Therefore, it requires more energy to out power the nucleus and remove an electron. As we move across the periodic table from left to right, the ionization energy increases , due to the effective nuclear charge increasing. This is because the larger the effective nuclear charge, the stronger the nucleus is holding onto the electron and the more energy it takes to release an electron.
The ionization energy is only a general rule. There are some instances when this trend does not prove to be correct. These can typically be explained by their electron configuration. For example, Magnesium has a higher ionization energy than Aluminum. Magnesium has an electron configuration of [Ne]3s2. Magnesium has a high ionization energy because it has a filled 3s orbital and it requires a higher amount of energy to take an electron from the filled orbital.
Electron Affinity
Electron affinity (E.A.) is the energy change that occurs when an electron is added to a gaseous atom. Electron affinity can further be defined as the enthalpy change that results from the addition of an electron to a gaseous atom. It can be either positive or negative value. The greater the negative value, the more stable the anion is.
• (Exothermic) The electron affinity is positive
$\ce{X(g) + e^{-} -> X^{-} + Energy} \nonumber$
• (Endothermic) The electron affinity is negative
$\ce{X(g) + e^{-} + Energy -> X^{-}} \nonumber$
It is more difficult to come up with trends that describe the electron affinity. Generally, the elements on the right side of the periodic table will have large negative electron affinity. The electron affinities will become less negative as you go from the top to the bottom of the periodic table. However, Nitrogen, Oxygen, and Fluorine do not follow this trend. The noble gas electron configuration will be close to zero because they will not easily gain electrons.
Electronegativity
Electronegativity is the measurement of an atom to compete for electrons in a bond. The higher the electronegativity, the greater its ability to gain electrons in a bond. Electronegativity will be important when we later determine polar and nonpolar molecules. Electronegativity is related with ionization energy and electron affinity. Electrons with low ionization energies have low electronegativities because their nuclei do not exert a strong attractive force on electrons. Elements with high ionization energies have high electronegativities due to the strong pull exerted by the positive nucleus on the negative electrons. Therefore the electronegativity increases from bottom to top and from left to right.
Metallic Character
The metallic character is used to define the chemical properties that metallic elements present. Generally, metals tend to lose electrons to form cations. Nonmetals tend to gain electrons to form anions. They also have a high oxidation potential therefore they are easily oxidized and are strong reducing agents. Metals also form basic oxides; the more basic the oxide, the higher the metallic character.
As you move across the table from left to right, the metallic character decreases, because the elements easily accept electrons to fill their valance shells. Therefore, these elements take on the nonmetallic character of forming anions. As you move up the table, the metallic character decreases, due to the greater pull that the nucleus has on the outer electrons. This greater pull makes it harder for the atoms to lose electrons and form cations.
Other Trends
Melting Points: Trends in melting points and molecular mass of binary carbon-halogen compounds and hydrogen halides are due to intermolecular forces. Melting destroys the arrangement of atoms in a solid, therefore the amount of heat necessary for melting to occur depends on the strength of attraction between the atoms. This strength of attraction increases as the number of electrons increase. Increase in electrons increases bonding.
Example: Melting point of HF should be approximately -145 °C based off melting points of HCl, HBr, and HI, but the observed value is -83.6°C.
Heat and electricity conductibility vary regularly across a period. Melting points may increase gradually or reach a peak within a group then reverse direction.
Example: Third period elements Na, Mg, and Al are good conductors of heat and electricity while Si is only a fair conductor and the nonmetals P, S, Cl and Ar are poor conductors.
Redox Potentials
Oxidation Potential
Oxidation is a reaction that results in the loss of an electron. Oxidation potential follows the same trends as the ionization energy. That is because the smaller the ionization energy, the easier it is to remove an electron. (e.g)
$K_{(s)} \rightarrow K^+ + e^- \nonumber$
Reduction Potential
Reduction is a reaction that results in the gaining of an electron. Reduction potentials follow the same trend as the electron affinity. That is because the larger, negative electron affinity, the easier it is to give an electron. Example of Reduction:
$F_{(s)} + e^- \rightarrow F^- \nonumber$
Uses in knowing the Periodic Properties of Elements
1. Predicting greater or smaller atomic size and radial distribution in neutral atoms and ions
2. Measuring and comparing ionization energies
3. Comparing electron affinities and electronegativities
• Predicting redox potential
• Comparing metallic character with other elements; its ability to form cations
• Predicting what reaction may or may not occur due to the trends
• Determining greater cell potential (sum of oxidation and reduction potential) between reactions
• Completing chemical reactions according to trends
Summary of Periodic Trends
The Periodic Table of Elements categorizes like elements together. Dmitri Mendeleev, a Russian scientist, was the first to create a widely accepted arrangement of the elements in 1869. Mendeleev believed that when the elements are arranged in order of increasing atomic mass, certain sets of properties recur periodically. Although most modern periodic tables are arranged in eighteen groups (columns) of elements, Mendeleev's original periodic table had the elements organized into eight groups and twelve periods (rows).
On the periodic table, elements that have similar properties are in the same groups (vertical). From left to right, the atomic number (z) of the elements increases from one period to the next (horizontal). The groups are numbered at the top of each column and the periods on the left next to each row. The main group elements are groups 1,2 and 13 through 18. These groups contain the most naturally abundant elements, and are the most important for life. The elements shaded in light pink in the table above are known as transition metals. The two rows of elements starting at z=58, are sometimes called inner transition metals and have that have been extracted and placed at the bottom of the table, because they would make the table too wide if kept continuous. The 14 elements following lanthanum (z=57) are called lanthanides, and the 14 following actinium (z=89) are called actinides.
Elements in the periodic table can be placed into two broad categories, metals and nonmetals. Most metals are good conductors of heat and electricity, are malleable and ductile, and are moderate to high melting points. In general, nonmetals are nonconductors of heat and electricity, are nonmalleable solids, and many are gases at room temperature. Just as shown in the table above, metals and nonmetals on the periodic table are often separated by a stairstep diagonal line, and several elements near this line are often called metalloids (Si, Ge, As, Sb, Te, and At). Metalloids are elements that look like metals and in some ways behave like metals but also have some nonmetallic properties. The group to the farthest right of the table, shaded orange, is known as the noble gases. Noble gases are treated as a special group of nonmetals.
Alkali Metals/Alkali Earth Metals
The Alkali metals are comprised of group 1 of the periodic table and consist of Lithium, Sodium, Rubidium, Cesium, and Francium. These metals are highly reactive and form ionic compounds (when a nonmetal and a metal come together) as well as many other compounds. Alkali metals all have a charge of +1 and have the largest atom sizes than any of the other elements on each of their respective periods.
Alkali Earth Metals are located in group 2 and consist of Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium. Unlike the Alkali metals, the earth metals have a smaller atom size and are not as reactive. These metals may also form ionic and other compounds and have a charge of +2.
Transition Metals
The transition metals range from groups IIIB to XIIB on the periodic table. These metals form positively charged ions, are very hard, and have very high melting and boiling points. Transition metals are also good conductors of electricity and are malleable.
Lanthanides and Actinides
Lanthanides (shown in row ** in chart above) and Actinides (shown in row * in chart above), form the block of two rows that are placed at the bottom of the periodic table for space issues. These are also considered to be transition metals. Lanthanides are form the top row of this block and are very soft metals with high boiling and melting points. Actinides form the bottom row and are radioactive. They also form compounds with most nonmetals. To find out why these elements have their own section, check out the electron configurations page.
Metalloids
As mentioned in the introduction, metalloids are located along the staircase separating the metals from the nonmetals on the periodic table. Boron, silicon, germanium, arsenic, antimony, and tellurium all have metal and nonmetal properties. For example, Silicon has a metallic luster but is brittle and is an inefficient conductor of electricity like a nonmetal. As the metalloids have a combination of both metallic and nonmetal characteristics, they are intermediate conductors of electricity or "semiconductors".
Halogens
Halogens are comprised of the five nonmetal elements Flourine, Chlorine, Bromine, Iodine, and Astatine. They are located on group 17 of the periodic table and have a charge of -1. The term "halogen" means "salt-former" and compounds that contain one of the halogens are salts. The physical properties of halogens vary significantly as they can exist as solids, liquids, and gases at room temperature. However in general, halogens are very reactive, especially with the alkali metals and earth metals of groups 1 and 2 with which they form ionic compounds.
Noble Gases
The noble gases consist of group 18 (sometimes reffered to as group O) of the periodic table of elements. The noble gases have very low boiling and melting points and are all gases at room temperature. They are also very nonreactive as they already have a full valence shell with 8 electrons. Therefore, the noble gases have little tendency to lose or gain electrons.
Useful Relationships from the Periodic Table
The periodic table of elements is useful in determining the charges on simple monoatomic ions. For main-group elements, those categorized in groups 1, 2, and 13-18, form ions they lose the same number of electrons as the corresponding group number to which they fall under. For example, K atoms (group 1) lose one electron to become K+ and Mg atoms (group 2) lose two electrons to form Mg2+. The other main-group elements found in group 13 and higher form more than one possible ion.
The elements in groups 3-12 are called transition elements, or transition metals. Similar to the main-group elements described above, the transition metals form positive ions but due to their capability of forming more than two or more ions of differing charge, a relation between the group number and the charge is non-existent.
Exercise $1$
Arrange these elements according to decreasing atomic size: Na, C, Sr, Cu, Fr
Answer
Fr, Sr, Cu, Na, C
Exercise $1$
Arrange these elements according to increasing negative E. A.: Ba, F, Si, Ca, O
Answer
Ba, Ca, Si, O, F
Exercise $1$
Arrange these elements according to increasing metallic character: Li, S, Ag, Cs, Ge
Answer
Li, S, Ge, Ag, Cs
Exercise $1$
Which reaction do you expect to have the greater cell potential?
1. $\ce{2Na(s) + Cl2(g)→ 2NaCl(s) or 2Cs(s) +Cl2(g) → 2RbCl(s)$
2. $\ce{2Na(s) + Cl2(g)→ 2NaCl(s) or Be(s) + Cl2(g) → BeCl2(s)}$
Answer
Second equation
First equation
Exercise $1$
Which equation do you expect to occur?
1. $\ce{I2(s) + 2Br (aq) → Br2(l) + 2I(aq)}$
2. $\ce{Cl2(g) + 2I (aq) → I2 (s) + 2Cl(aq)}$
Answer
5. A) Yes
B) No
Problems
*Highlight Answer:_____ to view answers.
1. An element that is an example of a metalloid is (a) S; (b) Zn; (c) Ge; (d) Re; (e) none of these
Answer: (c) Ge
2. In the periodic table, the vertical (up and down) columns are called (a) periods; (b) transitions; (c) families/groups; (d) metalloids; (e) none of these.
Answer: (c) families/groups
3. Why are noble gases inert (nonreactive)?
Answer: Noble gases are inert because they already have a full valence electron shell and have little tendency to gain or lose electrons.
4. What are compounds that contain a halogen called?
Answer: Salts
5. Lanthanides and Actinides are: (a) alkali earth metals; (b) transition metals; (c) metalloids; (d) alkali metals; (e) none of these
Answer: (b) transition metals | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Periodic_Properties_of_the_Elements.txt |
Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements.
Electronegativity Trends
Electronegativity can be understood as a chemical property describing an atom's ability to attract and bind with electrons. Because electronegativity is a qualitative property, there is no standardized method for calculating electronegativity. However, the most common scale for quantifying electronegativity is the Pauling scale (Table A2), named after the chemist Linus Pauling. The numbers assigned by the Pauling scale are dimensionless due to the qualitative nature of electronegativity. Electronegativity values for each element can be found on certain periodic tables. An example is provided below.
Electronegativity measures an atom's tendency to attract and form bonds with electrons. This property exists due to the electronic configuration of atoms. Most atoms follow the octet rule (having the valence, or outer, shell comprise of 8 electrons). Because elements on the left side of the periodic table have less than a half-full valence shell, the energy required to gain electrons is significantly higher compared with the energy required to lose electrons. As a result, the elements on the left side of the periodic table generally lose electrons when forming bonds. Conversely, elements on the right side of the periodic table are more energy-efficient in gaining electrons to create a complete valence shell of 8 electrons. The nature of electronegativity is effectively described thus: the more inclined an atom is to gain electrons, the more likely that atom will pull electrons toward itself.
• From left to right across a period of elements, electronegativity increases. If the valence shell of an atom is less than half full, it requires less energy to lose an electron than to gain one. Conversely, if the valence shell is more than half full, it is easier to pull an electron into the valence shell than to donate one.
• From top to bottom down a group, electronegativity decreases. This is because atomic number increases down a group, and thus there is an increased distance between the valence electrons and nucleus, or a greater atomic radius.
• Important exceptions of the above rules include the noble gases, lanthanides, and actinides. The noble gases possess a complete valence shell and do not usually attract electrons. The lanthanides and actinides possess more complicated chemistry that does not generally follow any trends. Therefore, noble gases, lanthanides, and actinides do not have electronegativity values.
• As for the transition metals, although they have electronegativity values, there is little variance among them across the period and up and down a group. This is because their metallic properties affect their ability to attract electrons as easily as the other elements.
According to these two general trends, the most electronegative element is fluorine, with 3.98 Pauling units.
Ionization Energy Trends
Ionization energy is the energy required to remove an electron from a neutral atom in its gaseous phase. Conceptually, ionization energy is the opposite of electronegativity. The lower this energy is, the more readily the atom becomes a cation. Therefore, the higher this energy is, the more unlikely it is the atom becomes a cation. Generally, elements on the right side of the periodic table have a higher ionization energy because their valence shell is nearly filled. Elements on the left side of the periodic table have low ionization energies because of their willingness to lose electrons and become cations. Thus, ionization energy increases from left to right on the periodic table.
Another factor that affects ionization energy is electron shielding. Electron shielding describes the ability of an atom's inner electrons to shield its positively-charged nucleus from its valence electrons. When moving to the right of a period, the number of electrons increases and the strength of shielding increases. As a result, it is easier for valence shell electrons to ionize, and thus the ionization energy decreases down a group. Electron shielding is also known as screening.
Trends
• The ionization energy of the elements within a period generally increases from left to right. This is due to valence shell stability.
• The ionization energy of the elements within a group generally decreases from top to bottom. This is due to electron shielding.
• The noble gases possess very high ionization energies because of their full valence shells as indicated in the graph. Note that helium has the highest ionization energy of all the elements.
Some elements have several ionization energies; these varying energies are referred to as the first ionization energy, the second ionization energy, third ionization energy, etc. The first ionization energy is the energy requiredto remove the outermost, or highest, energy electron, the second ionization energy is the energy required to remove any subsequent high-energy electron from a gaseous cation, etc. Below are the chemical equations describing the first and second ionization energies:
First Ionization Energy:
$X_{(g)} \rightarrow X^+_{(g)} + e^- \nonumber$
Second Ionization Energy:
$X^+_{(g)} \rightarrow X^{2+}_{(g)} + e^- \nonumber$
Generally, any subsequent ionization energies (2nd, 3rd, etc.) follow the same periodic trend as the first ionization energy.
Ionization energies decrease as atomic radii increase. This observation is affected by $n$ (the principal quantum number) and $Z_{eff}$ (based on the atomic number and shows how many protons are seen in the atom) on the ionization energy (I). The relationship is given by the following equation:
$I = \dfrac{R_H Z^2_{eff}}{n^2} \nonumber$
• Across a period, $Z_{eff}$ increases and n (principal quantum number) remains the same, so the ionization energy increases.
• Down a group, $n$ increases and $Z_{eff}$ increases slightly; the ionization energy decreases.
Electron Affinity Trends
As the name suggests, electron affinity is the ability of an atom to accept an electron. Unlike electronegativity, electron affinity is a quantitative measurement of the energy change that occurs when an electron is added to a neutral gas atom. The more negative the electron affinity value, the higher an atom's affinity for electrons.
Electron affinity generally decreases down a group of elements because each atom is larger than the atom above it (this is the atomic radius trend, discussed below). This means that an added electron is further away from the atom's nucleus compared with its position in the smaller atom. With a larger distance between the negatively-charged electron and the positively-charged nucleus, the force of attraction is relatively weaker. Therefore, electron affinity decreases. Moving from left to right across a period, atoms become smaller as the forces of attraction become stronger. This causes the electron to move closer to the nucleus, thus increasing the electron affinity from left to right across a period.
• Electron affinity increases from left to right within a period. This is caused by the decrease in atomic radius.
• Electron affinity decreases from top to bottom within a group. This is caused by the increase in atomic radius.
Atomic Radius Trends
The atomic radius is one-half the distance between the nuclei of two atoms (just like a radius is half the diameter of a circle). However, this idea is complicated by the fact that not all atoms are normally bound together in the same way. Some are bound by covalent bonds in molecules, some are attracted to each other in ionic crystals, and others are held in metallic crystals. Nevertheless, it is possible for a vast majority of elements to form covalent molecules in which two like atoms are held together by a single covalent bond. The covalent radii of these molecules are often referred to as atomic radii. This distance is measured in picometers. Atomic radius patterns are observed throughout the periodic table.
Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are added to the same shell. However, at the same time, protons are being added to the nucleus, making it more positively charged. The effect of increasing proton number is greater than that of the increasing electron number; therefore, there is a greater nuclear attraction. This means that the nucleus attracts the electrons more strongly, pulling the atom's shell closer to the nucleus. The valence electrons are held closer towards the nucleus of the atom. As a result, the atomic radius decreases.
Down a group, atomic radius increases. The valence electrons occupy higher levels due to the increasing quantum number (n). As a result, the valence electrons are further away from the nucleus as ‘n’ increases. Electron shielding prevents these outer electrons from being attracted to the nucleus; thus, they are loosely held, and the resulting atomic radius is large.
• Atomic radius decreases from left to right within a period. This is caused by the increase in the number of protons and electrons across a period. One proton has a greater effect than one electron; thus, electrons are pulled towards the nucleus, resulting in a smaller radius.
• Atomic radius increases from top to bottom within a group. This is caused by electron shielding.
Melting Point Trends
The melting points is the amount of energy required to break a bond(s) to change the solid phase of a substance to a liquid. Generally, the stronger the bond between the atoms of an element, the more energy required to break that bond. Because temperature is directly proportional to energy, a high bond dissociation energy correlates to a high temperature. Melting points are varied and do not generally form a distinguishable trend across the periodic table. However, certain conclusions can be drawn from Figure $7$.
• Metals generally possess a high melting point.
• Most non-metals possess low melting points.
• The non-metal carbon possesses the highest melting point of all the elements. The semi-metal boron also possesses a high melting point.
Metallic Character Trends
The metallic character of an element can be defined as how readily an atom can lose an electron. From right to left across a period, metallic character increases because the attraction between valence electron and the nucleus is weaker, enabling an easier loss of electrons. Metallic character increases as you move down a group because the atomic size is increasing. When the atomic size increases, the outer shells are farther away. The principal quantum number increases and average electron density moves farther from nucleus. The electrons of the valence shell have less attraction to the nucleus and, as a result, can lose electrons more readily. This causes an increase in metallic character.
• Metallic characteristics decrease from left to right across a period. This is caused by the decrease in radius (caused by Zeff, as stated above) of the atom that allows the outer electrons to ionize more readily.
• Metallic characteristics increase down a group. Electron shielding causes the atomic radius to increase thus the outer electrons ionizes more readily than electrons in smaller atoms.
• Metallic character relates to the ability to lose electrons, and nonmetallic character relates to the ability to gain electrons.
Another easier way to remember the trend of metallic character is that moving left and down toward the bottom-left corner of the periodic table, metallic character increases toward Groups 1 and 2, or the alkali and alkaline earth metal groups. Likewise, moving up and to the right to the upper-right corner of the periodic table, metallic character decreases because you are passing by to the right side of the staircase, which indicate the nonmetals. These include the Group 8, the noble gases, and other common gases such as oxygen and nitrogen.
• In other words:
• Move left across period and down the group: increase metallic character (heading towards alkali and alkaline metals)
• Move right across period and up the group: decrease metallic character (heading towards nonmetals like noble gases)
Problems
The following series of problems reviews general understanding of the aforementioned material.
1. Based on the periodic trends for ionization energy, which element has the highest ionization energy?
1. Fluorine (F)
2. Nitrogen (N)
3. Helium (He)
2.) Nitrogen has a larger atomic radius than oxygen.
1. A.) True
2. B.) False
3.) Which has more metallic character, Lead (Pb) or Tin (Sn)?
4.) Which element has a higher melting point: chlorine (Cl) or bromine (Br)?
5.) Which element is more electronegative, sulfur (S) or selenium (Se)?
6) Why is the electronegativity value of most noble gases zero?
7) Arrange these atoms in order of decreasing effective nuclear charge by the valence electrons: Si, Al, Mg, S
8) Rewrite the following list in order of decreasing electron affinity: fluorine (F), phosphorous (P), sulfur (S), boron (B).
9) An atom with an atomic radius smaller than that of sulfur (S) is __________.
1. A.) Oxygen (O)
2. B.) Chlorine (Cl)
3. C.) Calcium (Ca)
4. D.) Lithium (Li)
5. E.) None of the above
10) A nonmetal has a smaller ionic radius compared with a metal of the same period.
1. A.) True B.) False
Solutions
1. Answer: C.) Helium (He)
Explanation: Helium (He) has the highest ionization energy because, like other noble gases, helium's valence shell is full. Therefore, helium is stable and does not readily lose or gain electrons.
2. Answer: A.) True
Explanation: Atomic radius increases from right to left on the periodic table. Therefore, nitrogen is larger than oxygen.
3. Answer: Lead (Pb)
Explanation: Lead and tin share the same column. Metallic character increases down a column. Lead is under tin, so lead has more metallic character.
4. Answer: Bromine (Br)
Explanation: In non-metals, melting point increases down a column. Because chlorine and bromine share the same column, bromine possesses the higher melting point.
5. Answer: Sulfur (S)
Explanation: Note that sulfur and selenium share the same column. Electronegativity increases up a column. This indicates that sulfur is more electronegative than selenium.
6. Answer: Most noble gases have full valence shells.
Explanation: Because of their full valence electron shell, the noble gases are extremely stable and do not readily lose or gain electrons.
7. Answer: S > Si > Al > Mg.
Explanation: The electrons above a closed shell are shielded by the closed shell. S has 6 electrons above a closed shell, so each one feels the pull of 6 protons in the nucleus.
8. Answer: Fluorine (F)>Sulfur (S)>Phosphorous (P)>Boron (B)
Explanation: Electron affinity generally increases from left to right and from bottom to top.
9. Answer: C.) Oxygen (O)
Explanation: Periodic trends indicate that atomic radius increases up a group and from left to right across a period. Therefore, oxygen has a smaller atomic radius sulfur.
10. Answer: B.) False
Explanation: The reasoning behind this lies in the fact that a metal usually loses an electron in becoming an ion while a non-metal gains an electron. This results in a smaller ionic radius for the metal ion and a larger ionic radius for the non-metal ion. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Periodic_Trends.txt |
An understanding of periodic trends is necessary when analyzing and predicting molecular properties and interactions. Common periodic trends include those in ionization energy, atomic radius, and electron affinity. One such trend is closely linked to atomic radii -- ionic radii. Neutral atoms tend to increase in size down a group and decrease across a period. When a neutral atom gains or loses an electron, creating an anion or cation, the atom's radius increases or decreases, respectively. This module explains how this occurs and how this trend differs from that of atomic radii.
Shielding and Penetration
Electromagnetic interactions between electrons in an atom modify the effective nuclear charge ($Z_{eff}$) on each electron. Penetration refers to the presence of an electron inside the shell of an inner electron, and shielding is the process by which an inner electron masks an outer electron from the full attractive force of the nucleus, decreasing $Z_{eff}$. Differences in orbital characteristics dictate differences in shielding and penetration. Within the same energy level (indicated by the principle quantum number, n), due to their relative proximity to the nucleus, s-orbital electrons both penetrate and shield more effectively than p-orbital electrons, and p electrons penetrate and shield more effectively than d-orbital electrons. Shielding and penetration along with the effective nuclear charge determine the size of an ion. An overly-simplistic but useful conceptualization of effective nuclear charge is given by the following equation:
$Z_{eff} = Z - S \nonumber$
where
• $Z$ is the number of protons in the nucleus of an atom or ion (the atomic number), and
• $S$ is the number of core electrons.
Figure $1$ illustrates how this equation can be used to estimate the effective nuclear charge of sodium:
Cations and Anions
Neutral atoms that have lost an electron exhibit a positive charge and are called cations. These cations are smaller than their respective atoms; this is because when an electron is lost, electron-electron repulsion (and therefore, shielding) decreases and the protons are better able to pull the remaining electrons towards the nucleus (in other words, $Z_{eff}$increases). A second lost electron further reduces the radius of the ion. For instance, the ionic radius of Fe2+ is 76 pm, while that of Fe3+ is 65 pm. If creation of an ion involves completely emptying an outer shell, then the decrease in radius is especially great.
Neutral atoms that have gained an electron are called anions, and they are much larger than their respective atoms. As an additional electron occupies an outer orbital, there is increased electron-electron repulsion (and hence, increased shielding) which pushes the electrons further apart. Because the electrons now outnumber the protons in the ion, the protons can not pull the extra electrons as tightly toward the nucleus; this results in decreased $Z_{eff}$.
Figure 2 shows an isoelectric series of atoms and ions (each has the same number of electrons, and thus the same degree of electron-electron repulsion and shielding) with differing numbers of protons (and thus different nuclear attraction), giving the relative ionic sizes of each atom or ion.
The Periodic Trend
Due to each atom’s unique ability to lose or gain an electron, periodic trends in ionic radii are not as ubiquitous as trends in atomic radii across the periodic table. Therefore, trends must be isolated to specific groups and considered for either cations or anions.
Consider the s- and d-block elements. All metals can lose electrons and form cations. The alkali and alkali earth metals (groups 1 and 2) form cations which increase in size down each group; atomic radii behave the same way. Beginning in the d-block of the periodic table, the ionic radii of the cations do not significantly change across a period. However, the ionic radii do slightly decrease until group 12, after which the trend continues (Shannon 1976). It is important to note that metals, not including groups 1 and 2, can have different ionic states, or oxidation states, (e.g. Fe2+ or Fe3+ for iron) so caution must be employed when generalizing about trends in ionic radii across the periodic table.
All non-metals (except for the noble gases which do not form ions) form anions which become larger down a group. For non-metals, a subtle trend of decreasing ionic radii is found across a pegroup theoryriod (Shannon 1976). Anions are almost always larger than cations, although there are some exceptions (i.e. fluorides of some alkali metals).
Measurement and Factors Affecting Ionic Radii
The ionic radius of an atom is measured by calculating its spatial proportions in an ionic bond with another ion within a crystal lattice. However, it is to consistently and accurately determine the proportions of the ionic bonds. After comparing many compounds, chemist Linus Pauling assign a radius of 140 pm to O2- and use this as a reference point to determine the sizes of other Ionic Radii (Jensen 2010).
Ionic radius is not a permanent trait of an ion, but changes depending on coordination number, spin state, and other variables (Shannon 1976). For a given ion, the ionic radius increases with increasing coordination number and is larger in a high-spin state than in a low-spin state.
According to group theory, the idea of ionic radii as a measurement of spherical shapes only applies to ions that form highly-symmetric crystal lattices like Na+ and Cl -. The point group symmetry of a lattice determines whether or not the ionic radii in that lattice can be accurately measured (Johnson 1973). For instance, lattices with Oh and Td symmetries are considered to have high symmetry; thus the electron densities of the component ions occupy relatively-spherical regions and ionic radii can be measured fairly accurately. However, for less symmetrical and more polar lattices such as those with Cn, Cnh, and Cnv symmetries, significant changes in the electron density can occur, causing deviations from spherical shape; these deviations make ionic radii more difficult to measure.
Problems
1. Why are cations smaller and anions larger than their respective atoms?
1. Cations are smaller than their respective atoms because of increased electron-electron repulsion
2. Anions are larger than their respective atoms because of increased effective nuclear charge
3. Cations are smaller than their respective atoms because of decreased electron-electron repulsion
4. Anions are larger than their respective atoms because of decreased effective nuclear charge
2. Which of the following are isoelectronic: $F^-$, $Cl^-$, $Ca^{2+}$, $Ar$
3. List the ions from smallest to largest: $Se^{2-}$,$Zr^{4+}$, $Na^+$,$Mg$, $Rb^+$, $Br^-$, $K^+$
4. How are ionic radii measured?
5. What are some of the problems with generalizing ionic trends?
Solution
1. C & D: Cations are formed when an electron is lost. When this occurs there are less electron-electron repulsions and there is a greater net nuclear attraction per electron. So, the newly formed ion becomes a more condensed version of its neutral atom. Anions are formed when an electron is gained. When this occurs there are more electron-electron repulsions and there is a lower net nuclear attraction per electron. This will cause the electrons push each other away and spread out, causing the atom to become larger.
2. Cl-, Ca2+, Ar all have 18 electrons; therefore, they are isoelectronic (F- has 10 electrons). An isoelectronic series is useful in understanding the effects of gained or loss electrons on atom size.
3. Zr4+<K+<Rb+<Mg<Br-<Se2-: Ionic radii shorten with increasing positive charge and lengthen with increasing negative charge, and thus, anions are almost always larger than cations.
4. Ionic radii are measured by proportioning ionic bond lengths between two ions within a crystal lattice. After studying many compounds, Linus Pauling found that the approximate ionic radii of O2- was 140pm. With this reference point, Pauling was able to calculate the ionic radii of other ions.
5. Ionic radii are not fixed properties of ions. For the same ion, the radii can differ in different crystal lattices due certain variables such as coordination number and electron spin. Group Theory suggests that only ions in high-symmetric non-polar crystal lattices can accurately be measured for their radii.
Contributors and Attributions
• Michael H. Nguyen. University of California, Davis | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Periodic_Trends_in_Ionic_Radii.txt |
The Periodic table characterizes the known elements in increasing order of atomic number. It starts on the top right hand corner with Hydrogen and continue from left to right which then repeats in the horizontal row below the last element. This is not just a list of elements it is organized in very different ways like properties and atomic mass. At first glance the periodic table may seem disorganized with only a couple elements on the top row and a block on the last row but it is very specific in the way that they are organized.
The Periodic Table is organized by Properties
Elements themselves can be one of the following metals, nonmetals and metalloids.
Metals are:
• Solid at room temperature (except for mercury which will be liquid)
• Malleable (the ability to be flattened into a sheet)
• Ductile (the ability to be made into wire wires)
• good conductors of heat and electricity
• shiny in appearance
Non-metals are the complete opposite so they are:
• poor conductors of heat
• At different phases at room temperature
• Nitrogen, Oxygen, and Chloride are gases
• Silicon and Sulfur are brittle solids
• Bromine is liquid
Metalloids have the properties from both metals and non-metals.
Position
One way that the elements are organized is vertically in groups or families. An example of this is group 8, the noble gases with include Helium, Neon and Argon which has a specific name called alkali metals. Group 1 isn’t the only family with a special name there is group 17 which includes Fluorine and Chlorine, that is called the halogens. When looking at groups, elements at the top are the beginning of the group and the ones at the bottom are the end of the group. The main group elements are the ones in groups 1, 2 and the ones ranging from 13 – 18 and the metals in between 3 and 12 are the transition metals.
Elements are organized in horizontally in periods. There are a total of 7 groups in the periodic table and each vary in how many elements they contain. Period 1 only has two elements, Hydrogen and Helium. Period 2 has eight elements whereas group 6 has the most with the addition of the top section of the block of elements on the bottom of the table called lanthanides. Below them are the actinides which follow the same rules as the lanthanide but in group 7.
The individual elements are represented in their own blocks by stating the atomic number (Z) on the top of the box, in the middle is the elements symbol and on the bottom is the average atomic mass of the element. This information can differ in the type of periodic table you are looking at. Some can include information about the actual chemical name or omit information described above.
Practice problems
Determine whether the following elements are metals, non-metals or metalloids,
1. calcium,
2. phosphorus,
3. silicon
4. krypton
Identify the group and period that the following elements are in:
1. hydrogen,
2. aluminum
3. silver
Classify which elements are considered as the main group or transition metals. If they are transition metals, state if they are lanthanides or actinides. The elements are:
1. Magnesium,
2. Lanthanide
3. Uranium
4. Holmium
5. Selenium
Arrange the elements from the lowest to highest group number: nitrogen, fluorine, boron, oxygen and carbon.
Arrange the following elements from the lowest to highest period number: aluminum, polonium, germanium, and antimony.
From looking at the periodic table, information about the following elements:
1. cobalt,
2. barium
3. chromium.
Answers
1. Determine whether the following elements are metals, non-metals or metalloids:
2. Calcium is a metal,
3. phosphorus is a non-metal
4. silicon is a metalloid
5. krypton is a nonmetal
1. Identify the group and period that the following elements are in:
2. Hydrogen is in group 1and period 1.
3. Aluminum is in the group 13 but group 3 for the main group elements, the period is 3.
4. Silver is Ag which is in group 11 and period 5.
1. Identify which elements are considered as the main group or transition metals. If they are transition metals, state if they are lanthanides or actinides. The elements are:
2. Magnesium, Selenium, and Lanthanide are all main group elements.
3. Uranium is a transition metal which is part of the Actinide series.
4. Holmium is a transition metal as well but is part of the Lanthanides.
1. Arrange the elements from the lowest to highest group number: nitrogen, fluorine, boron, oxygen and carbon.
Boron, carbon, nitrogen, oxygen, fluorine
1. Arrange the following elements from the lowest to highest period number: aluminum, polonium, germanium, and antimony.
aluminum, germanium, antimony, and polonium
6. From looking at the periodic table, information about the following elements:
This depends on what period table you use!
1. Cobalt has an atomic number (Z) of 27. Elemental symbol of Co and weighs 58.693
2. Barium has an atomic number (Z) of 56. Elemental symbol of Ba and weighs 137.327.
3. Chromium has an atomic number (Z) of 24. Elemental symbol of Cr and weighs 51.9961.
Contributors and Attributions
• Jayne Aclan (UCD) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Table_Basics.txt |
The periodic law was developed independently by Dmitri Mendeleev and Lothar Meyer in 1869. Mendeleev created the first periodic table and was shortly followed by Meyer. They both arranged the elements by their mass and proposed that certain properties periodically reoccur. Meyer formed his periodic law based on the atomic volume or molar volume, which is the atomic mass divided by the density in solid form. Mendeleev's table is noteworthy because it exhibits mostly accurate values for atomic mass and it also contains blank spaces for unknown elements.
Introduction
In 1804 physicist John Dalton advanced the atomic theory of matter, helping scientists determine the mass of the known elements. Around the same time, two chemists Sir Humphry Davy and Michael Faraday developed electrochemistry which aided in the discovery of new elements. By 1829, chemist Johann Wolfgang Doberiner observed that certain elements with similar properties occur in group of three such as; chlorine, bromine, iodine; calcium, strontium, and barium; sulfur, selenium, tellurium; iron, cobalt, manganese. However, at the time of this discovery too few elements had been discovered and there was confusion between molecular weight and atomic weights; therefore, chemists never really understood the significance of Doberiner's triad.
In 1859 two physicists Robert Willhem Bunsen and Gustav Robert Kirchoff discovered spectroscopy which allowed for discovery of many new elements. This gave scientists the tools to reveal the relationships between elements. Thus in 1864, chemist John A. R Newland arranged the elements in increasing of atomic weights. Explaining that a given set of properties reoccurs every eight place, he named it the law of Octaves.
The Periodic Law
In 1869, Dmitri Mendeleev and Lothar Meyer individually came up with their own periodic law "when the elements are arranged in order of increasing atomic mass, certain sets of properties recur periodically." Meyer based his laws on the atomic volume (the atomic mass of an element divided by the density of its solid form), this property is called Molar volume.
$\text{Atomic (molar) volume (cm}^3\text{/mol)} = \dfrac{\text{ molar mass (g/ mol)}}{\rho \text{ (cm}^3\text{/g)}} \nonumber$
Mendeleev's Periodic Table
Mendeleev's periodic table is an arrangement of the elements that group similar elements together. He left blank spaces for the undiscovered elements (atomic masses, element: 44, scandium; 68, gallium; 72, germanium; & 100, technetium) so that certain elements can be grouped together. However, Mendeleev had not predicted the noble gases, so no spots were left for them.
In Mendeleev's table, elements with similar characteristics fall in vertical columns, called groups. Molar volume increases from top to bottom of a group3
Example
The alkali metals (Mendeleev's group I) have high molar volumes and they also have low melting points which decrease in the order:
Li (174 oC) > Na (97.8 oC) > K (63.7 oC) > Rb (38.9 oC) > Cs (28.5 oC)
Atomic Number as the Basis for the Periodic Law
Assuming there were errors in atomic masses, Mendeleev placed certain elements not in order of increasing atomic mass so that they could fit into the proper groups (similar elements have similar properties) of his periodic table. An example of this was with argon (atomic mass 39.9), which was put in front of potassium (atomic mass 39.1). Elements were placed into groups that expressed similar chemical behavior.
In 1913 Henry G.J. Moseley did researched the X-Ray spectra of the elements and suggested that the energies of electron orbitals depend on the nuclear charge and the nuclear charges of atoms in the target, which is also known as anode, dictate the frequencies of emitted X-Rays. Moseley was able to tie the X-Ray frequencies to numbers equal to the nuclear charges, therefore showing the placement of the elements in Mendeleev's periodic table. The equation he used:
$\nu = A(Z-b)^2 \nonumber$
with
• $\nu$: X-Ray frequency
• $Z$: Atomic Number
• $A$ and $b$: constants
With Moseley's contribution the Periodic Law can be restated:
Similar properties recur periodically when elements are arranged according to increasing atomic number."
Atomic numbers, not weights, determine the factor of chemical properties. As mentioned before, argon weights more than potassium (39.9 vs. 39.1, respectively), yet argon is in front of potassium. Thus, we can see that elements are arranged based on their atomic number. The periodic law is found to help determine many patterns of many different properties of elements; melting and boiling points, densities, electrical conductivity, reactivity, acidic, basic, valance, polarity, and solubility.
The table below shows that elements increase from left to right accordingly to their atomic number. The vertical columns have similar properties within their group for example Lithium is similar to sodium, beryllium is similar to magnesium, and so on.
Group 1 2 13 14 15 16 17 18
Element Li Be B C N O F Ne
Atomic Number 3 4 5 6 7 8 9 10
Atomic Mass 6.94 9.01 10.81 12.01 14.01 15.99 18.99 20.18
Element Na Mg Al Si P S Cl Ar
Atomic Number 11 12 13 14 15 16 17 18
Atomic Mass 22.99 24.31 26.98 20.09 30.97 32.07 35.45 39.95
Elements in Group 1 (periodic table) have similar chemical properties and are called alkali metals. Elements in Group 2 have similar chemical properties, they are called the alkaline earth metals.
Short form periodic table
The short form periodic table is a table where elements are arranged in 7 rows, periods, with increasing atomic numbers from left to right. There are 18 vertical columns known as groups. This table is based on Mendeleev's periodic table and the periodic law.
Long form Periodic Table
In the long form, each period correlates to the building up of electronic shell; the first two groups (1-2) (s-block) and the last 6 groups (13-18) (p-block) make up the main-group elements and the groups (3-12) in between the s and p blocks are called the transition metals. Group 18 elements are called noble gases, and group 17 are called halogens. The f-block elements, called inner transition metals, which are at the bottom of the periodic table (periods 8 and 9); the 15 elements after barium (atomic number 56) are called lanthanides and the 14 elements after radium (atomic number 88) are called actinides.
Problems
1) The periodic law states that
1. similar properties recur periodically when elements are arranged according to increasing atomic number
2. similar properties recur periodically when elements are arranged according to increasing atomic weight
3. similar properties are everywhere on the periodic table
4. elements in the same period have same characteristics
2) Which element is most similar to Sodium
1. Potassium
2. Aluminum
3. Oxygen
4. Calcium
3) According to the periodic law, would argon be in front of potassium or after? Explain why.
4) Which element is most similar to Calcium?
1. Carbon
2. Oxygen
3. Strontium
4. Iodine
5) Who were the two chemists that came up with the periodic law?
1. John Dalton and Michael Faraday
2. Dmitri Mendeleev and Lothar Meyer
3. Michael Faraday and Lothar Meyer
4. John Dalton and Dmitri Mendeleev
Answers
1. A
2. A
3. Argon would in front of potassium because the periodic law states that the periodic table increases from left to right based on atomic number not atomic weights
4. C
5. B | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/The_Periodic_Law.txt |
Ligand field theory (LFT) describes the bonding, orbital arrangement, and other characteristics of It represents an application of molecular orbital theory to transition metal complexes. A transition metal ion has nine valence atomic orbitals: five nd, one (n+1)s, and three (n+1)p orbitals. These orbitals are of appropriate energy to form bonding interaction with ligands. The LFT analysis is highly dependent on the geometry of the complex, but most explanations begin by describing octahedral complexes, where six ligands coordinate to the metal.
• Ligand Field Theory Fundamentals
Ligand Field Theory can be considered an extension of Crystal Field Theory such that all levels of covalent interactions can be incorporated into the model. Treatment of the bonding in LFT is generally done using Molecular Orbital Theory.
Ligand Field Theory
Ligand Field Theory can be considered an extension of Crystal Field Theory such that all levels of covalent interactions can be incorporated into the model. Treatment of the bonding in LFT is generally done using Molecular Orbital Theory. A qualitative approach that can be used for octahedral metal complexes is given in the following 3 diagrams.
In the first diagram, the 3d, 4s and 4p metal ion atomic orbitals are shown together with the ligand group orbitals that would have the correct symmetry to be able to overlap with them. The symmetry adapted linear comination of ligand orbitals are generated by taking 6 sigma orbitals from the ligands, designated as σx, σ-x, σy, σ-y, σz, σ-z and then combining them to make 6 ligand group orbitals. (labelled eg, a1g, t1u)
In the second diagram only sigma bonding is considered and it shows the combination of the metal 3d, 4s and 4p orbitals with OCCUPIED ligand group orbitals (using 1 orbital from each ligand). The result is that that the metal electrons would be fed into t2g and eg* molecular orbitals which is similar to the CFT model except that the eg orbital is now eg*.
For example: B - [M(II)I6]4- A - [M(II)(H2O)6]2+ C - [M(II)(CN)6]4-
In the third diagram, π (pi) bonding is considered. In general π bonds are weaker than σ (sigma) bonds and so the effect is to modify rather than dramatically alter the description. 2 orbitals from each ligand are combined to give a total of 12 which are subdivided into four sets with three ligand group orbitals in each set. These are labelled t1g, t1u, t2g and t2u. The metal t2g orbital is the most suitable for interaction and this is shown in the 2 cases above.
Case A is the same as above, ignoring π interactions.
• For case B, the ligand π orbitals are full and at lower energy than the metal t2g. This causes a decrease in the size of Δ.
• For case C, the ligand π orbitals are empty and at higher energy than the metal t2g. This causes an increase in the size of Δ.
Returning to the problem of correctly placing ligands in the Spectrochemical series, the halides are examples of case A and groups like CN- and CO are examples of case B. It is possible then to explain the Spectrochemical series once covalent effects are considered.
Some convincing arguments for covalency and effects on Δ come from a study of the IR spectra recorded for simple carbonyl compounds e.g. M(CO)6.
• The CO molecule has a strong triple bond which in the IR gives rise to a strong absorption at ~2140 cm-1. For the series [Mn(CO)6]+, [Cr(CO)6] and [V(CO)6]-, which are isoelectronic, the IR bands for the CO have shifted to 2090, 2000 and 1860 cm-1 respectively. Despite the fact that the metals have the same number of electrons (isoelectronic) the frequency of force constant of the CO bond is seen to vary Mn+ > Cr > V-.
• This can not be explained on an ionic basis but is consistent with the π bonding scheme since the greater the positive charge on the metal, the less readily the metal can delocalize electrons back into the π* orbitals of the CO group.
Note that the IR values we are dealing with relate to the CO bond and not the M-C so when the CO frequency gets less then it is losing triple bond character and becoming more like a double bond. This is expected if electrons are pushed back from the metal into what were empty π* antibonding orbitals.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Ligand_Field_Theory/Ligand_Field_Theory_Fundamentals.txt |
• Catenation
Catenation of chemical bonds leads to the formation of inorganic polymers. However, inorganic polymers are mostly solids in the form of crystals. Typical inorganic polymers are diamond, graphite, silicates, and other solids in which all atoms are connected by covalent bonds.
• Macromolecules
A macromolecule is a very large molecule having a polymeric chain structure. Proteins, polysaccharides, genes, ruber, and synthetic polymers consist of macromolecules. There are only a few known inorganic macromolecules. Inorganic macromolecules can be divided into several categories: solids formed mainly due to covalent bonds, organosilanes, siloxanes and organosiloxanes.
Macromolecules
Discussion Questions
• See Assignment 3 on Macromolecules
Catenation of chemical bonds leads to the formation of inorganic polymers. However, inorganic polymers are mostly solids in the form of crystals. Typical inorganic polymers are diamond, graphite, silicates, and other solids in which all atoms are connected by covalent bonds.
Atoms and Their Nuclei
During the 20th century, the investigation of the material world turned to the very heart of material world - the structure of atoms. The discovery of electrons in 1897 by J.J. Thomson showed that there were more fundamental particles present in the atoms. Fourteen years later, Rutherford discovered that most of the mass of an atom resides in a tiny nucleus whose radius is 100,000 times smaller than that of an atom. In the mean time, light beams were discovered to be made of photons which are equivalent to particles of wave motion. These discoveries created new concepts. When these concepts and discoveries are integrated, new ideas emerge. The result is quantum theory. This theory gives good interpretations of the phenomena of the atomic and subatomic world. In this microscopic world, distances are measured in nanometers (10-9 or 1e-9 meter) and fantometers (1e-15 meter, also called fermi, in honour of Fermi who built the first nuclear reactor).
The electrons in an atom are confined by the electromagnetic force of the atomic nuclei. At this level, we need a quantum mechanical approach to understand the energy states of the electrons in the atom. However, we do not have the time to discuss this in details.
Quantum Number and Atomic Orbitals
Quantum mechanics on atomic structures is a mathematical approach to describe the behavior of electrons in atoms. The electrons are represented by wavefunctions, and each of them are characterized by a set of numbers. Each set of numbers represent a state, which is often called an orbital. Quantum Numbers and Atomic Orbitals are pages that give a bit more details on this subject, but a summary of the atomic orbitals is given below:
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f 5g
6s 6p 6d 6f 6g 7h
7s 7p 7d 7f 7g 7h 8i
A film has shown how these are related to the Period Tables of Elements, and it also shows the shape and concepts of atomic orbitals. These concepts are vital for the understanding of bonding, such as the bonds formed between carbon atoms of diamond, silicon, graphite etc.
In the assignment, you have been asked to apply these concept to describe the bonding for carbon. The same argument also applies to the bonding of silicon.
Electronic Configurations
The electronic configuration of an element or atom shows the energy states of electrons in it. Pauli exclusion principle and Hund's rules are some of the theories involved in assigning electronic configurations. For the discussion of bondings in some light elements, please note the following:
H: 1s1
He: 1s2
Li: 1s2*2s1 after * are valence electrons
Be: 1s2*2s2
B: 1s2*2s22p1
C: 1s2*2s22p2
N: 1s2*2s22p3
O: 1s2*2s22p4
F: 1s2*2s22p5
Ne: 1s2*2s22p6
...
Si: 1s22s22p6*3s23p2
P: 1s22s22p6*3s23p3
S: 1s22s22p6*3s23p4
Cl: 1s22s22p6*3s23p5
Ar: 1s22s22p63s23p6
Na: {Ar}*4s1 ...
Hybrid Orbitals
Electrons in an atom may have properties of several orbitals, and they share each other's characters. In other words, atomic orbitals may be combined to form hybrid orbitals. These hybrid orbitals are particularly useful in the discussion of chemical bonding. For carbon, the hybrid orbitals are made up of
2s, 2po, 2p+ and 2p-
orbitals. Since 1 s and 3 p orbitals are used, the 4 orbitals sharing s and p characters are called sp3 hybrid orbitals. The shapes and directions of these orbitals should have been demonstrated in lectures, and diagrams are needed here. The bonding of diamond is beautifully described using the sp3 hybrid orbitals.
Bonding in Benzene and Graphite
The bonding of benzene should have been fully discussed in the organic chemistry course you have taken. Simply, the orbitals used to form the sigma bond are sp2 hybrid orbitals resulting from combining
2s, 2p+ and 2p-
orbitals. Furthermore, the overlap of the 2po orbitals leads to the formation of the pi bond.
Resonance, Benzene and Graphite
Again, the structure of benzene serves an excellent example for the concept called resonance. If one insisted on the fact that the 3 double bonds and 3 single bonds in benzene alternate along the ring, one can start with a single or a double bond. Nether structure represent the structure of benzene, beccause all 6 bond lengths are about the same. Thus a combination of the two structures is used to represent the structure of benzene, and such an approach is called resonance. In other words, electrons in the double bonds delocalize over the entire ring.
The bonding description for benzene can be applied to that of a sheet of graphite. The electrons are delocalized on two planes in graphite. Thus, it no surprize that graphite is a good conductor along the sheet.
Buckminsterfullerenes
The graphite structure is the result of expanding the pi electrons into planes. Since all rings in graphite consist of 6 carbon atoms, the sheets are flat. If the hybrid orbitals are somewhat flexible, it is easy to understand that the 5-member rings are also possibilities. However, formation of 5-member rings reults in buckling of a flat structure, and we usually do not think this will happen.
The discoverer of the buckminsterfullerenes spent a long time figuring out the structure of a cluster of carbon consisting of 60 carbon atoms, which is represented by C60. However, once they have deducted the structure, its shape is very common. The carbon atoms are at the junction of lines on a soccer ball (other parts of the world call it foot ball). A geometric description is a truncated icosahedron. The fullerenes, or buckyballs have become the talk of the news media since the award of Nobel Prize to its discoverers. The fullerenes actually are common, and their discovery adds a nice touch to theories of bonding and electronic structures. The electrons on the surface of the ball perhaps think they are on a very large atom. The discoverers are still very active in the study of fullerenes.
Boron Nitride, BN
A compound with equal number of boron and nitrogen atoms, BN, has on average 4 valence electrons per atom, same as that of diamond or graphite in carbon. Thus, we anticipate BN to form solids with simlar bonding and structures as diamond and graphite.
In fact, the bonding of boron and boron compounds is also very interesting. The following items are mentioned here for future development.
• Geometry of significant cages. Comparison of organic and inorganic polymers. Survey of catenation in the Periodic Table. Bonding by bridging groups.
• Boron: borides (mainly but not exclusively MB2, MB6, MB12), borates, B-N systems.
• Silane
Structural chemistry of silicates, including clays, and zeolites.
• Phosphorus, phosphates, P-N systems. Sulphur, polycations, S-N compounds, (SN)x and other inorganic conductors.
Questions
1. How many valence electrons do carbon, silicon, and germanium have?
Skill -
Give the electronic configuration of an element.
Solutions
1. They all have 4. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Macromolecules/Catenation.txt |
Discussion Questions
• What are macromolecules?
• What are some examples of inorganic macromolecules?
• How is the properties of sulfur related to its structure?
A macromolecule is a very large molecule having a polymeric chain structure. Proteins, polysaccharides, genes, ruber, and synthetic polymers consist of macromolecules. For synthetic polymers, here are the abbreviations for some common polymers:
• HDPE: high density polyethylene
• LDPE: low density polyethylene
• PET: polyethylene terephthalate
• PP: polypropylene
• PS: polystyrene
• PVA: polyvinyl alcohol
• PVC: polyvinyl chloride
There are only a few known inorganic macromolecules. For example, when liquid sulfur is poured into cold water, long chains of ...-S-S-S-S-S-... are formed. These molecules are present in a phase known as elastic sulfur. Inorganic macromolecules can be divided into several categories: solids formed mainly due to covalent bonds, organosilanes, siloxanes and organosiloxanes.
Solids formed mainly due to covalent bonds
We have mentioned diamond, graphite, silicon, germanium, etc as large molecules. In this section, some more examples are given: Zinc sulfide has two forms (or phases): the wurtzite and the zinc blende. These are typical structures, because many other common compounds have the same structure. Knowing the bonding, geometry, symmetry of these structures is important, because scientific discussion and applications are based on their structures. The difference between wurtzite and zinc blende illustrates some fundamental geometry and symmetry difference.
Wurtzite is a typical mineral, often involving iron and zinc sulfide, and formulated as (Fe,Zn)S For example: ZnO, SiC, AlN, CaSe, BN, C(Hexagonal Diamond) all have the same crystal structure as wurtzite in terms of bonding, symmetry, packing sequence etc.
The zinc blende is cubic. It can be perceived as a f.c.c. lattice of S with half of the tetrahedral sites occupied by Zn. A unit cell of the crystal structure is shown in Zinc blende, and you can see this structure from various perspective. The diagram on the page can be manipulated by moving the mouse. The same group has also got a wurtzite structure that you can manipulate. By the way, the zinc blende structure has the same bonding skeleton as the diamond structure. Thus, the wurtzite and zinc blende structures are two typical structural types for inorganic macromolecules.
Silanes and Organosilanes
Since Si-Si is much weaker than C-C bonds, silanes SinH2n+2, are not as stable as alkanes. Methane is a stable gas, but silane SiH4 explode as soon as it comes in contact with air. Silane is made by reacting Mg2Si with dilute HCl, but the silane produced burns as soon as it comes in contact with air at the surface of the solution:
$Mg_2Si + 4 HCl \rightarrow 2 MgCl_2 + SiH_4 \nonumber$
$SiH_4 + 2 O_2 \rightarrow SiO_2 + 2 H_2O \nonumber$
When an organosilane reacts with water, the Si-Si bonds break:
$R_3Si-SiR_3 + 2 H_2O \rightarrow 2 R_3SiOH + H_2 \nonumber$
where $R$ is an alkyl or aryl group. Organosilanes have some unique applications, and silane coupling agents, RnSiX4-n (X being a halogen) are useful.
Siloxanes and Organosiloxanes
Siloxanes and organosiloxanes have Si-O-Si-O-Si linkage, and these are stable polymers due to the strong Si-O-Si bonds. Since Si atoms tend to form four bonds, these polymers form two- and three-dimensional networks, making them excellent sealants.
R R
| |
- Si-O-Si-O-Si-R
| | |
R O O
| |
R-Si-O-Si-O
| |
Silicates are based on Si-O-Si linkages. Quartz for example is based on three-dimensional frame work of these linkages.
How is the properties of sulfur related to its structure?
Sulfur is in the same group as oxygen, and its valence electrons have the electron configuration:
3s23p4.
These six electrons usually occupy the four sp3 hybrid orbitals, two of which have a pair of electrons each, and the other two have only one electron each. Thus, sulfur usually form two bonds such as H2S, its structure similar to H2O. Sulfur atoms bond to each other forming the cyclic molecules such as S6 and S8. The two bonds and the two lone pairs of electrons around the sulfur point to the direction of a tetrahedron, making the -S- angle approximately 100 degrees.
Sulfur has three allotropes: rhombic, monoclinic, and plastic sulfur. At room temperature, monoclinic sulfur is the the stable form. When heated, monoclinic sulfur melts to form a viscous liquid at 119 degree C at the atmosphere pressure. At higher pressure, the monoclinic sulfur transforms into the rhombic sulfur. Both crystalline forms have the S8 crown shaped molecule and the plastic sulfur has a chain structure of unspecified number of atoms Sn (n is a very large unspecified number).
Since we are talking about the element sulfur, we might include some information on it. Sulfur occur as minerals: pyrite, FeS2 (also known as fool's gold); gypsum, CaSO4ºH2O, (when dehydrated it is called paster of paris). Sulfur also occur as an element in nature because some bacterial converts sulfur oxides and other compounds to elemental sulfur. Elemental sulfur is extrated from under ground by hot water and air in a process called the Frasch process.
Sulfur is easily oxidized to sulfur dioxide
$S_8 + 8 O_2 \rightarrow 8 SO_2 \nonumber$
The SO2 is a bent molecule with an O-S-O angle of 120 degrees. It reacts further with water and with oxygen as examplified by the following reaction:
SO2 + H2O = H2SO3 (sulfurous acid)
2 SO2 + O2 = 2 SO3 (catalyst required)
SO3 + H2O = H2SO4 (sulfuric acid)
Na2SO3 + S = Na2S(=S)O3 (known as hypo)
What are the properties and structures of phosphorus?
Phosphorus has only 5 valence elections, one less electron than sulfur. Thus it tends to form three bonds around a P atom. The analogous of ammonia is phosphine, PH3, which is air stable with a melting point of -88o C. White phosphorus can be obtained by reducing phosphorus oxide with carbon:
$P_4O_{10} + 10 C \rightarrow P_4 + 10 CO \nonumber$
The crystals consist of P4 molecules as shown here, and it undergoes natural combustion if not stored under water. The combustion leads to the formation of P4O10. White phosphorus converts to stable red, black, violet and scarlet phosphorus, which have complicated network of macromolecules. The bonding in P4 can be explained in the same manner as that described for sulfur, but that is left as an exercise. As most other non-metallic elements, phosphorus also form a complicated covalent solid. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Macromolecules/Macromolecules.txt |
Molecular geometry is the 3-dimensional shape that a molecule occupies in space. It is determined by the central atom and the surrounding atoms and electron pairs. The shape of most molecules can be predicted using the Valence Shell Electron Pair Repulsion (VSEPR) method. This method states a few rules to help one determine the shape of a substance without using high technology methods such as X-ray crystallography, NMR Spectroscopy, or electron microscopy. Some of the most common shapes that can be taken are linear, trigonal planar, tetrahedral, pyramidal, and angular (or bent).
Molecular Geometry
An example of bent molecular geometry that results from tetrahedral electron pair geometry is H2O. The water molecule is so common that it is wise to just memorize that water is a BENT molecule.
Bent Molecular Geometry
AX2E2
Shape: bent
Steric Number: 4
Lone Pair2: 2
Polar/NonPolar: Polar
Hybridization: sp3
Examples: H2O, OF2
GIF File: ax2e2.gif (52K)
NOTES: This molecule is made up of 4 equally spaced sp3 hybrid orbitals forming bond angles of approximately 109.5o. The shape of the orbitals is tetrahedral. Two of the orbitals contain lone pairs of electrons. The two atoms connected to the central atom form a molecule with a bent shape.
Nonlinear
AX2E
Shape: nonlinear
Steric Number: 3
Lone Pair: 1
Polar/NonPolar: Polar
Hybridization: sp2
Examples: SO2
GIF File: ax2e.gif (49K)
NOTES: This molecule is made up of 3 equally spaced sp2 hybrid orbitals arranged at approximately 120o angles. The shape of the orbitals is planar triangular. One orbital contains a lone pair of electrons.The two atoms attached to the central atom form a molecule that is nonlinear.
Limitations of VSEPR
Athough the VSEPR model is useful in predicting molecular geometry, it fails to predict the shapes of isoelectronic species and transition metal compounds. This model does not take relative sizes of substituents and stereochemically inactive lone pairs into account. As a result, VSEPR cannot be applied to heavy d-block species that experience the stereochemical inert pair effect.
Introduction
The VSEPR model is a powerful tool used by chemists to predict the shapes of molecules; yet like many other theories, it has exceptions and limitations. This article discusses in detail the various limitations of VSEPR and gives explanations for these exceptions.
VSEPR fails for isoelectronic species
Isoelectronic species are elements, ions and molecules that share the same number of electrons. According to the VSEPR model, chemists determine the shape of the molecules based on numbers of valence electrons (i.e. bond pairs and lone pairs). However, two isoelectronic species, can differ in geometry despite the fact that they have the same numbers of valence electrons. For example, both IF7and [TeF7]- have 56 valence electrons; VSEPR theory predicts both to be pentagonal bipyramidal. However, electron diffraction and X-ray diffraction data indicate that the equatorial F atoms of [TeF7]- are not coplanar. Moreover, the bond lengths of equatorial I-F and Te-F bonds are different. The VSEPR model fails to accommodate the correct shape for [TeF7]- and other species because it does not account for such irregularities.
VSEPR fails for transition metal compounds
The VSEPR model also fails to predict the structure of certain compounds because it does not take relative sizes of the substituents and stereochemically inactive lone pairs into account. Elements in the d-block have relatively high atomic masses and they tend to have stereochemically inactive electron pairs. In other words, valence shell s-electrons in these elements tend to adopt a non-bonding role. This is known as the inert pair effect. As a result, VSEPR does not give correct geometries for these transitional metal complex compounds.. For example, [SeCl6]2- , [TeCl6]2- ,and [BrF6]- are predicted to adopt pentagonal bipyramidal geometries according to VSEPR since the central atom can have seven electron pairs. However, due to the stereochemical inert pair effect, these molecules are found to be regular octahedral because one of the electron pairs is stereochemically inactive.
Re
Problems
1. What is the shape of [XeF8]2- ?(Hint: Xenon contains one pair of stereochemically inactive lone pair)
2. Give some examples of isoelectronic species.
3. Why are d-block elements exceptions to VSEPR?
Answers
1. Antiprismatic
2. IF7 and [TeF7]-
3. Because of stereochemical inert pair effect
Linear Molecular Geometry
AX2E3
Shape: Linear
Steric Number: 5
Lone Pairs: 3
Polar/NonPolar: NonPolar
Hybridization: sp3d
Example: I3-
NOTES: This molecule is made up of 5 sp3d hybrid orbitals. Three orbitals are arranged around the equator of the molecule with bond angles of 120o. Two orbitals are arranged along the vertical axis at 90o from the equatorial orbitals. The shape of the orbitals is trigonal bipyramidal. All three equatorial orbitals contain lone pairs of electrons. The three atom molecule has a linear shape.
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Molecular_Geometry/Bent_Molecular_Geometry/Bent.txt |
The specific three dimensional arrangement of atoms in molecules is referred to as molecular geometry. We also define molecular geometry as the positions of the atomic nuclei in a molecule. There are various instrumental techniques such as X-Ray crystallography and other experimental techniques which can be used to tell us where the atoms are located in a molecule. Using advanced techniques, very complicated structures for proteins, enzymes, DNA, and RNA have been determined. Molecular geometry is associated with the chemistry of vision, smell and odors, taste, drug reactions and enzyme controlled reactions to name a few.
Note: The AXE Method
It is common practice to represent bonding patterns by "generic" formulas such as \(AX_4\), \(AX_2E_2\), etc., in which "X" stands for bonding pairs and "E" denotes lone pairs. This convention is known as the "AXE Method."
Molecular geometry is associated with the specific orientation of bonding atoms. A careful analysis of electron distributions in orbitals will usually result in correct molecular geometry determinations. In addition, the simple writing of Lewis diagrams can also provide important clues for the determination of molecular geometry. Click on a picture to link to a page with the GIF file and a short discussion of the molecule.
Steric Number (# bonded atoms + # electron pairs)
6 5 4 3 2
AX6
octahedral
AX5
trigonal bipyramidal
AX4
tetrahedral
AX3
trigonal planar
AX2
linear
1 lone pair of electrons
AX5E
square pyramidal
AX4E
distorted tetrahedron
AX3E
pyramidal
AX2E
nonlinear
AXE
linear
2 lone pairs of electrons
AX4E2
square planar
AX3E2
T-shaped
AX2E2
bent
AXE2
linear
3 lone pairs of electrons
AX3E3
T-shaped
AX2E3
linear
AXE3
linear
4 lone pairs
AX2E4
linear
AXE4
linear
5 lone pairs
AXE5
linear
Valence Shell Electron Pair Repulsion (VSEPR) theory
Electron pairs around a central atom arrange themselves so that they can be as far apart as possible from each other. The valence shell is the outermost electron-occupied shell of an atom that holds the electrons involved in bonding. In a covalent bond, a pair of electrons is shared between two atoms. In a polyatomic molecule, several atoms are bonded to a central atom using two or more electron pairs. The repulsion between negatively charged electron pairs in bonds or as lone pairs causes them to spread apart as much as possible.
The idea of "electron pair repulsion can be demonstrated by tying several inflated balloons together at their necks. Each balloon represents an electron pair. The balloons will try to minimize the crowding and will spread as far apart as possible. According to VSEPR theory, molecular geometry can be predicted by starting with the electron pair geometry about the central atom and adding atoms to some or all of the electron pairs. This model produces good agreement with experimental determinations for simple molecules. With this model in mind, the molecular geometry can be determined in a systematic way.
• Lewis diagrams provide information about what atoms are bonded to each other and the total electron pairs involved.
• Electron pair geometry is determined from the total electron pairs.
Molecules can then be divided into two groups:
• Group 1: Molecules with NO lone electron pairs. In this case the molecular geometry is identical to the electron pair geometry.
• Group 2: Molecules with one or more lone electron pairs. In this case an extra step is needed to to translate from electron pair geometry to the final molecular geometry, since only the positions of bonded atoms are considered in molecular geometry.
Outside Links
• Pfennig, Brian W.; Frock, Richard L. "The Use of Molecular Modeling and VSEPR Theory in the Undergraduate Curriculum to Predict the Three-Dimensional Structure of Molecules." J. Chem. Educ.1999 76 1018.
• Covalent Radii - Wikipedia: http://en.Wikipedia.org/wiki/Covalent_radius
Contributors and Attributions
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
• Robyn Rindge (Class of '98) who now works for PDI Dreamworks (look for his name in the credits of Shrek2.). Robyn drew these rotating molecules using Infini-D (MetaCreations).
• Paul Groves, chemistry teacher at South Pasadena High School and Chemmy Bear
Octahedral
AX6
Shape: octahedral
Steric Number: 6
Lone Pairs: 0
Polar/NonPolar: NonPolar
Hybridization: sp3d2
Example: SF6
NOTES: This molecule is made up of 6 equally spaced sp3d2 hybrid orbitals arranged at 90o angles. The shape of the orbitals is octahedral. Since there is an atom at the end of each orbital, the shape of the molecule is also octahedral. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Molecular_Geometry/Molecular_Geometry_Overview.txt |
This page explains how to work out the shapes of molecules and ions containing only single bonds. The examples on this page are all simple in the sense that they only contain two sorts of atoms joined by single bonds - for example, ammonia only contains a nitrogen atom joined to three hydrogen atoms by single bonds. If you are given a more complicated example, look carefully at the arrangement of the atoms before you start to make sure that there are only single bonds present.
For example, if you had a molecule such as COCl2, you would need to work out its structure, based on the fact that you know that carbon forms 4 covalent bonds, oxygen 2, and chlorine (normally) 1. If you did that, you would find that the carbon is joined to the oxygen by a double bond, and to the two chlorines by single bonds. That means that you couldn't use the techniques on this page, because this page only considers single bonds.
The electron pair repulsion theory
The shape of a molecule or ion is governed by the arrangement of the electron pairs around the central atom. All you need to do is to work out how many electron pairs there are at the bonding level, and then arrange them to produce the minimum amount of repulsion between them. You have to include both bonding pairs and lone pairs.
First you need to work out how many electrons there are around the central atom:
• Write down the number of electrons in the outer level of the central atom. That will be the same as the Periodic Table group number, except in the case of the noble gases which form compounds, when it will be 8.
• Add one electron for each bond being formed. (This allows for the electrons coming from the other atoms.)
• Allow for any ion charge. For example, if the ion has a 1- charge, add one more electron. For a 1+ charge, deduct an electron.
Now work out how many bonding pairs and lone pairs of electrons there are:
• Divide by 2 to find the total number of electron pairs around the central atom.
• Work out how many of these are bonding pairs, and how many are lone pairs. You know how many bonding pairs there are because you know how many other atoms are joined to the central atom (assuming that only single bonds are formed).
For example, if you have 4 pairs of electrons but only 3 bonds, there must be 1 lone pair as well as the 3 bonding pairs.
Finally, you have to use this information to work out the shape:
• Arrange these electron pairs in space to minimize repulsions. How this is done will become clear in the examples which follow.
Two electron pairs around the central atom
The only simple case of this is beryllium chloride, BeCl2. The electronegativity difference between beryllium and chlorine is not enough to allow the formation of ions.
Beryllium has 2 outer electrons because it is in group 2. It forms bonds to two chlorines, each of which adds another electron to the outer level of the beryllium. There is no ionic charge to worry about, so there are 4 electrons altogether - 2 pairs. It is forming 2 bonds so there are no lone pairs. The two bonding pairs arrange themselves at 180° to each other, because that's as far apart as they can get. The molecule is described as being linear.
Three electron pairs around the central atom
The simple cases of this would be BF3 or BCl3.
Boron is in group 3, so starts off with 3 electrons. It is forming 3 bonds, adding another 3 electrons. There is no charge, so the total is 6 electrons - in 3 pairs.
Because it is forming 3 bonds there can be no lone pairs. The 3 pairs arrange themselves as far apart as possible. They all lie in one plane at 120° to each other. The arrangement is called trigonal planar.
In the diagram, the other electrons on the fluorines have been left out because they are irrelevant.
Four electron pairs around the central atom
There are lots of examples of this. The simplest is methane, CH4. Carbon is in group 4, and so has 4 outer electrons. It is forming 4 bonds to hydrogens, adding another 4 electrons - 8 altogether, in 4 pairs. Because it is forming 4 bonds, these must all be bonding pairs.
Four electron pairs arrange themselves in space in what is called a tetrahedral arrangement. A tetrahedron is a regular triangularly-based pyramid. The carbon atom would be at the centre and the hydrogens at the four corners. All the bond angles are 109.5°.
Note
It is important that you understand the use of various sorts of line to show the 3-dimensional arrangement of the bonds. In diagrams of this sort, an ordinary line represents a bond in the plane of the screen or paper. A dotted line shows a bond going away from you into the screen or paper. A wedge shows a bond coming out towards you.
Other examples with four electron pairs around the central atom
Ammonia, NH3
Nitrogen is in group 5 and so has 5 outer electrons. Each of the 3 hydrogens is adding another electron to the nitrogen's outer level, making a total of 8 electrons in 4 pairs. Because the nitrogen is only forming 3 bonds, one of the pairs must be a lone pair. The electron pairs arrange themselves in a tetrahedral fashion as in methane.
In this case, an additional factor comes into play. Lone pairs are in orbitals that are shorter and rounder than the orbitals that the bonding pairs occupy. Because of this, there is more repulsion between a lone pair and a bonding pair than there is between two bonding pairs. That forces the bonding pairs together slightly - reducing the bond angle from 109.5° to 107°.
Greatest repulsion lone pair - lone pair
lone pair - bond pair
Least repulsion bond pair - bond pair
Be very careful when you describe the shape of ammonia. Although the electron pair arrangement is tetrahedral, when you describe the shape, you only take notice of the atoms. Ammonia is pyramidal - like a pyramid with the three hydrogens at the base and the nitrogen at the top.
Water, H2O
Following the same logic as before, you will find that the oxygen has four pairs of electrons, two of which are lone pairs. These will again take up a tetrahedral arrangement. This time the bond angle closes slightly more to 104°, because of the repulsion of the two lone pairs. The shape is not described as tetrahedral, because we only "see" the oxygen and the hydrogens - not the lone pairs. Water is described as bent or V-shaped.
The ammonium ion, NH4+
The nitrogen has 5 outer electrons, plus another 4 from the four hydrogens - making a total of 9. But take care! This is a positive ion. It has a 1+ charge because it has lost 1 electron. That leaves a total of 8 electrons in the outer level of the nitrogen. There are therefore 4 pairs, all of which are bonding because of the four hydrogens.
The ammonium ion has exactly the same shape as methane, because it has exactly the same electronic arrangement. NH4+ is tetrahedral.
Methane and the ammonium ion are said to be isoelectronic. Two species (atoms, molecules or ions) are isoelectronic if they have exactly the same number and arrangement of electrons (including the distinction between bonding pairs and lone pairs).
The hydroxonium ion, H3O+
Oxygen is in group 6 - so has 6 outer electrons. Add 1 for each hydrogen, giving 9. Take one off for the +1 ion, leaving 8. This gives 4 pairs, 3 of which are bond pairs. The hydroxonium ion is isoelectronic with ammonia, and has an identical shape - pyramidal.
Five electron pairs around the central atom
Phosphorus(V) fluoride, PF5
Phosphorus (in group 5) contributes 5 electrons, and the five fluorines 5 more, giving 10 electrons in 5 pairs around the central atom. Since the phosphorus is forming five bonds, there can't be any lone pairs. (The argument for phosphorus(V) chloride, PCl5, would be identical.)
The 5 electron pairs take up a shape described as a trigonal bipyramid - three of the fluorines are in a plane at 120° to each other; the other two are at right angles to this plane. The trigonal bipyramid therefore has two different bond angles - 120° and 90°.
A tricky example, ClF3
Chlorine is in group 7 and so has 7 outer electrons. The three fluorines contribute one electron each, making a total of 10 - in 5 pairs. The chlorine is forming three bonds - leaving you with 3 bonding pairs and 2 lone pairs, which will arrange themselves into a trigonal bipyramid. There are actually three different ways in which you could arrange 3 bonding pairs and 2 lone pairs into a trigonal bipyramid. The right arrangement will be the one with the minimum amount of repulsion - and you can't decide that without first drawing all the possibilities.
These are the only possible arrangements. Anything else you might think of is simply one of these rotated in space. We need to work out which of these arrangements has the minimum amount of repulsion between the various electron pairs. A new rule applies in cases like this:
Note
If you have more than four electron pairs arranged around the central atom, you can ignore repulsions at angles of greater than 90°.
One of these structures has a fairly obvious large amount of repulsion.
In this diagram, two lone pairs are at 90° to each other, whereas in the other two cases they are at more than 90°, and so their repulsions can be ignored. ClF3 certainly won't take up this shape because of the strong lone pair-lone pair repulsion.
To choose between the other two, you need to count up each sort of repulsion.
In the next structure, each lone pair is at 90° to 3 bond pairs, and so each lone pair is responsible for 3 lone pair-bond pair repulsions.
Because of the two lone pairs there are therefore 6 lone pair-bond pair repulsions. And that's all. The bond pairs are at an angle of 120° to each other, and their repulsions can be ignored.
Now consider the final structure.
Each lone pair is at 90° to 2 bond pairs - the ones above and below the plane. That makes a total of 4 lone pair-bond pair repulsions - compared with 6 of these relatively strong repulsions in the last structure. The other fluorine (the one in the plane) is 120° away, and feels negligible repulsion from the lone pairs.
The bond to the fluorine in the plane is at 90° to the bonds above and below the plane, so there are a total of 2 bond pair-bond pair repulsions. The structure with the minimum amount of repulsion is therefore this last one, because bond pair-bond pair repulsion is less than lone pair-bond pair repulsion. ClF3 is described as T-shaped.
Six electron pairs around the central atom
A simple example: SF6
6 electrons in the outer level of the sulphur, plus 1 each from the six fluorines, makes a total of 12 - in 6 pairs. Because the sulfur is forming 6 bonds, these are all bond pairs. They arrange themselves entirely at 90°, in a shape described as octahedral.
Two slightly more difficult examples
XeF4
Xenon forms a range of compounds, mainly with fluorine or oxygen, and this is a typical one. Xenon has 8 outer electrons, plus 1 from each fluorine - making 12 altogether, in 6 pairs. There will be 4 bonding pairs (because of the four fluorines) and 2 lone pairs.
There are two possible structures, but in one of them the lone pairs would be at 90°. Instead, they go opposite each other. XeF4 is described as square planar.
ClF4-
Chlorine is in group 7 and so has 7 outer electrons. Plus the 4 from the four fluorines. Plus one because it has a 1- charge. That gives a total of 12 electrons in 6 pairs - 4 bond pairs and 2 lone pairs. The shape will be identical with that of XeF4. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Molecular_Geometry/Shapes_of_Molecules_and_Ions.txt |
AX4E2
Shape: square planar
Steric Number: 6
Lone Pairs: 2
Polar/NonPolar: NonPolar
Hybridization: sp3d2
Example: XeF4
NOTES: This molecule is made up of 6 equally spaced sp3d2 hybrid orbitals arranged at 90o angles. The shape of the orbitals is octahedral. Two orbitals contain lone pairs of electrons on opposite sides of the central atom. The remaining four atoms connected to the central atom gives the molecule a square planar shape.
Square Pyramidal
AX5E
Shape: square pyramidal
Steric Number: 6
Lone Pair: 1
Polar/NonPolar: Polar
Hybridization: sp3d2
Example: IF5
NOTES: This molecule is made up of 6 equally spaced sp3d2 hybrid orbitals arranged at 90o angles. The shape of the orbitals is octahedral. One orbital contains a lone pair of electrons so the remaining five atoms connected to the central atom gives the molecule a square pyramidal shape.
T-shaped
AX3E2
Shape: T-shaped
Steric Number: 5
Lone Pairs: 2
Polar/NonPolar: Polar
Hybridization: sp3d
Example: ClF3
NOTES: This molecule is made up of 5 sp3d hybrid orbitals. Three orbitals are arranged around the equator of the molecule with bond angles of 120o. Two orbitals are arranged along the vertical axis at 90o from the equatorial orbitals. The shape of the orbitals is trigonal bipyramidal. Two of the equatorial orbitals contain lone pairs of electrons. The three atoms are arranged around the central atom to form a T-shaped molecule.
Tetrahedral Molecular Geometry
This molecule is made up of 4 equally spaced \(sp^3\) hybrid orbitals forming bond angles of 109.5o. The shape of the orbitals is tetrahedral. Since there is an atom at the end of each orbital, the shape of the molecule is also tetrahedral.
AX4
Shape: tetrahedral
Steric Number: 4
Lone Pairs: 0
Polar/NonPolar: NonPolar
Hybridization: sp3
Examples: CH4, SiH4
Trigonal Bipyramidal Molecular Geometry
AX5
Shape: trigonal bipyramidal
Steric Number: 5
Lone Pairs: 0
Polar/NonPolar: NonPolar
Hybridization: sp3d
Examples: PCl5, SbF5
NOTES: This molecule is made up of 5 sp3d hybrid orbitals. Three orbitals are arranged around the equator of the molecule with bond angles of 120o. Two orbitals are arranged along the vertical axis at 90o from the equatorial orbitals. The shape of the orbitals is trigonal bipyramidal. Since there is an atom at the end of each orbital, the shape of the molecule is also trigonal bipyramidal.
Trigonal Planar Molecular Geometry
AX3
Shape: trigonal planar
Steric Number: 3
Lone Pairs: 0
Polar/NonPolar: NonPolar
Hybridization: sp2
Examples: BF3, CO32-
NOTES: This molecule is made up of 3 equally spaced sp2 hybrid orbitals arranged at 120o angles. The shape of the orbitals is planar triangular. Since there is an atom at the end of each orbital, the shape of the molecule is also planar triangular.
Trigonal Pyramidal Molecular Geometry
An example of trigonal pyramid molecular geometry that results from tetrahedral electron pair geometry is NH3. The nitrogen has 5 valence electrons and thus needs 3 more electrons from 3 hydrogen atoms to complete its octet. This then leaves a lone electron pair that is not bonded to any other atom. The three hydrogen atoms and the lone electron pair are as far apart as possible at nearly 109o bond angle. This is tetrahedral electron pair geometry.
The lone electron pairs exerts a little extra repulsion on the three bonding hydrogen atoms to create a slight compression to a 107o bond angle.The molecule is trigonal pyramid molecular geometry because the lone electron pair, although still exerting its influence, is invisible when looking at molecular geometry. The molecule is three dimensional as opposed to the boron hydride case which was a flat trigonal planar molecular geometry because it did not have a lone electron pair.
Hydronium Ion
In this example, H3O+, the Lewis diagram shows O at the center with one lone electron pair and three hydrogen atoms attached. Compare this with ammonia, NH3, which also has a lone pair. Compare it to the water molecule which has 2 hydrogen atoms and 2 lone electron pairs.. The third hydrogen bonds to the water molecule as a hydrogen ion (no electrons) bonding to the lone pair on the oxygen.
This shows tetrahedral geometry for the electron pair geometry and and trigonal pyramid the molecular geometry. Hydronium ion is a more accurate method to depict the hydrogen ion associated with acid properties of some molecules in water solution.
Sulfite Ion
In this example, SO32-, the Lewis diagram shows sulfur at the center with one lone electron pair. The sulfur and and one oxygen are bonded through a double bond which counts as "one electron pair". Hence the molecule has four electron pairs and is tetrahedral.
The Lewis diagram is as follows:
S = 6 e-
O = 6e- x 3 = 18e-
2- charge = 2e-
Total electrons = 26
Sulfur atoms and all oxygen atoms have an octet of electrons. Sulfite and bisulfite ions are used as a preservative in wines. It is also found as a component of acid rain, formed by the interaction of sulfur dioxide and water molecules.
VSEPR
Valence Shell Electron Pair Repulsion (VSPER) theory is used to predict the geometric shape of the molecules based on the electron repulsive force. There are some limitation to VSEPR.
Introduction
The shapes of the molecules is determined mainly by the electrons surrounding the central atom. Therefore, VSEPR theory gives simple directions on how to predict the shape of the molecules. The VSEPR model combines the original ideas of Sidwick and Powell and further development of Nyholm and Gillespie.
How VSEPR works
In a molecule EXn, the valence shell electron pair around the central atom E and the E-X single bonds are very important due to the repulsion in which determine the shape of the molecule. The repulsions decrease in order of: lone pair-lone pair, lone pair-bonding pair, bonding pair-bonding pair. At the same time, the repulsion would decrease in order of: triple bond-single bond, double bond-single bond, and single bond-single bond if the central atom E has multiple bonds. The difference between the electronegativities of E and X also determine the repulsive force between the bonding pairs. If electron-electron repulsive force is less, then more electron density is drawn away from the central atom E.
Shape determination:
VSEPR model works better for simple halides of the p-block elements but can also be used with other substituents. It does not take steric factors, size of the substituents into account. Therefore, the shape of the molecules are arranged so that the energy is minimized. For example:
• BeCl2 has minimum energy when it is a linear molecule.
• BCl3 takes the shape of trigonal planar.
Lone pair electrons are also taken into account. When lone pair electrons are present, the "parent structure" are used as a guideline for determining the shape..
Problems
1. What is VSEPR used in chemistry?
It is used to predict the molecular shape of molecules
2. How to predict a molecule structure using VSEPR theory?
First step is to count the total number of valence electrons. After the total number of electrons is determined, this number is divided by two to give the total number of electron pairs. With the electron pairs of the molecule, the shape of the molecule is determined based on the table shown above.
3. What is the shape of PF5 ?
It is trigonal bipyramidal because it has total of 20 electron pairs. Each Fluorine atom give 1 electron to the Phosphorus central atom which creates total of 5 pairs. Also, each Fluorine atom has 3 electron pairs. With the presence of 5 Fluorine atom, there are 15 more electron pairs so there are 20 electron pairs total. | textbooks/chem/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Molecular_Geometry/Square_Planar.txt |
The study of chemistry will open your eyes to a fascinating world. Chemical processes are continuously at work all around us. They happen as you cook and eat food, strike a match, shampoo your hair, and even read this page. Chemistry is called the central science because a knowledge of chemical principles is essential for other sciences. You might be surprised at the extent to which chemistry pervades your life.
• 1.0: Prelude to Chemistry, Matter, and Measurement
Quantities and measurements are as important in our everyday lives as they are in medicine. The posted speed limits on roads and highways, such as 55 miles per hour (mph), are quantities we might encounter all the time. Both parts of a quantity, the amount (55) and the unit (mph), must be properly communicated to prevent potential problems. In chemistry, as in any technical endeavor, the proper expression of quantities is a necessary fundamental skill.
• 1.1: What is Chemistry?
Chemistry is the study of matter—what it consists of, what its properties are, and how it changes. Being able to describe the ingredients in a cake and how they change when the cake is baked is called chemistry. Matter is anything that has mass and takes up space—that is, anything that is physically real.
• 1.2: The Classification of Matter
Matter can be described with both physical properties and chemical properties. Matter can be identified as an element, a compound, or a mixture.
• 1.3: Measurements
Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5.
• 1.4: Expressing Numbers - Scientific Notation
Scientific notation is a system for expressing very large or very small numbers in a compact manner. It uses the idea that such numbers can be rewritten as a simple number multiplied by 10 raised to a certain exponent, or power. Scientific notation expressed numbers using powers of 10.
• 1.5: Expressing Numbers - Significant Figures
Significant figures properly report the number of measured and estimated digits in a measurement. There are rules for applying significant figures in calculations.
• 1.6: The International System of Units
Recognize the SI base units. Combining prefixes with base units creates new units of larger or smaller sizes.
• 1.7: Converting Units
The ability to convert from one unit to another is an important skill. A unit can be converted to another unit of the same type with a conversion factor.
• 1.8: Dosage Calculations
Conversion factors are important in calculating dosages.
• 1.E: Chemistry, Matter, and Measurement (Exercises)
These are homework exercises to accompany Chapter 1 of the Ball et al. "The Basics of GOB Chemistry" Textmap.
• 1.S: Chemistry, Matter, and Measurement (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Thumbnail: Two small test tubes held in spring clamps. (CC BY-SA 3.0; mitchell125).
01: Chemistry Matter and Measurement
In April 2003, the US Pharmacopeia, a national organization that establishes quality standards for medications, reported a case in which a physician ordered “morphine [a powerful painkiller] 2–3 mg IV [intravenously] every 2–3 hours for pain.” A nurse misread the dose as “23 mg” and thus administered approximately 10 times the proper amount to an 8-year-old boy with a broken leg. The boy stopped breathing but was successfully resuscitated and left the hospital three days later.
Quantities and measurements are as important in our everyday lives as they are in medicine. The posted speed limits on roads and highways, such as 55 miles per hour (mph), are quantities we might encounter all the time. Both parts of a quantity, the amount (55) and the unit (mph), must be properly communicated to prevent potential problems. In chemistry, as in any technical endeavor, the proper expression of quantities is a necessary fundamental skill. As we begin our journey into chemistry, we will learn this skill so that errors—from homework mistakes to traffic tickets to more serious consequences—can be avoided. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.00%3A_Prelude_to_Chemistry_Matter_and_Measurement.txt |
Learning Objectives
1. Define chemistry in relation to other sciences.
2. Identify the general steps in the scientific method.
Chemistry is the study of matter—what it consists of, what its properties are, and how it changes. Being able to describe the ingredients in a cake and how they change when the cake is baked is called chemistry. Matter is anything that has mass and takes up space—that is, anything that is physically real. Some things are easily identified as matter—this book, for example. Others are not so obvious. Because we move so easily through air, we sometimes forget that it, too, is matter.
Chemistry is one branch of science. Science is the process by which we learn about the natural universe by observing, testing, and then generating models that explain our observations. Because the physical universe is so vast, there are many different branches of science (Figure \(1\)). Thus, chemistry is the study of matter, biology is the study of living things, and geology is the study of rocks and the earth. Mathematics is the language of science, and we will use it to communicate some of the ideas of chemistry.
Although we divide science into different fields, there is much overlap among them. For example, some biologists and chemists work in both fields so much that their work is called biochemistry. Similarly, geology and chemistry overlap in the field called geochemistry. Figure \(1\) shows how many of the individual fields of science are related.
There are many other fields of science in addition to the ones listed here.
Alchemy
As our understanding of the universe has changed over time, so has the practice of science. Chemistry in its modern form, based on principles that we consider valid today, was developed in the 1600s and 1700s. Before that, the study of matter was known as alchemy and was practiced mainly in China, Arabia, Egypt, and Europe.
Alchemy was a somewhat mystical and secretive approach to learning how to manipulate matter. Practitioners, called alchemists, thought that all matter was composed of different proportions of the four basic elements—fire, water, earth, and air—and believed that if you changed the relative proportions of these elements in a substance, you could change the substance. The long-standing attempts to “transmute” common metals into gold represented one goal of alchemy. Alchemy’s other major goal was to synthesize the philosopher’s stone, a material that could impart long life—even immortality. Alchemists used symbols to represent substances, some of which are shown in the accompanying figure. This was not done to better communicate ideas, as chemists do today, but to maintain the secrecy of alchemical knowledge, keeping others from sharing in it.
In spite of this secrecy, in its time alchemy was respected as a serious, scholarly endeavor. Isaac Newton, the great mathematician and physicist, was also an alchemist.
Exercise \(1\)
Which fields of study are branches of science? Explain.
1. sculpture
2. astronomy
Answer a
Sculpture is not considered a science because it is not a study of some aspect of the natural universe.
Answer b
Astronomy is the study of stars and planets, which are part of the natural universe. Astronomy is therefore a field of science.
Exercise \(2\)
Which fields of study are branches of science?
1. physiology (the study of the function of an animal’s or a plant’s body)
2. geophysics
3. agriculture
4. politics
Answer
a, b and c only
How do scientists work? Generally, they follow a process called the scientific method. The scientific method is an organized procedure for learning answers to questions and making explanations for observations. To find the answer to a question (for example, “Why do birds fly toward Earth’s equator during the cold months?”), a scientist goes through the following steps, which are also illustrated in Figure \(2\):
The steps may not be as clear-cut in real life as described here, but most scientific work follows this general outline.
1. Propose a hypothesis. A scientist generates a testable idea, or hypothesis, to try to answer a question or explain an observation about how the natural universe works. Some people use the word theory in place of hypothesis, but the word hypothesis is the proper word in science. For scientific applications, the word theory is a general statement that describes a large set of observations and data. A theory represents the highest level of scientific understanding.
2. Test the hypothesis. A scientist evaluates the hypothesis by devising and carrying out experiments to test it. If the hypothesis passes the test, it may be a proper answer to the question. If the hypothesis does not pass the test, it may not be a good answer.
3. Refine the hypothesis if necessary. Depending on the results of experiments, a scientist may want to modify the hypothesis and then test it again. Sometimes the results show the original hypothesis to be completely wrong, in which case a scientist will have to devise a new hypothesis.
Not all scientific investigations are simple enough to be separated into these three discrete steps. But these steps represent the general method by which scientists learn about our natural universe.
Exercise \(3\)
Define science and chemistry.
Answer
Science is a process by which we learn about the natural universe by observing, testing, and then generating models that explain our observations. Chemistry is the study of matter.
Exercise \(4\)
Name the steps of the scientific method.
Answer
After identifying the problem or making an observation, propose a hypothesis, test the hypothesis, and refine the hypothesis, if necessary
Key Takeaways
• Chemistry is the study of matter and how it behaves.
• The scientific method is the general process by which we learn about the natural universe.
1. Based on what you know, which fields are branches of science?
1. meteorology (the study of weather)
2. astrophysics (the physics of planets and stars)
3. economics (the study of money and monetary systems)
4. astrology (the prediction of human events based on planetary and star positions)
5. political science (the study of politics)
2. Based on what you know, which fields are a branches of science?
1. history (the study of past events)
2. ornithology (the study of birds)
3. paleontology (the study of fossils)
4. zoology (the study of animals)
5. phrenology (using the shape of the head to determine personal characteristics)
3. Which of the following are examples of matter?
1. a baby
2. an idea
3. the Empire State Building
4. an emotion
5. the air
6. Alpha Centauri, the closest known star (excluding the sun) to our solar system
4. Which of the following are examples of matter?
1. your textbook
2. brain cells
3. love
4. a can of soda
5. breakfast cereal
5. Suggest a name for the science that studies the physics of rocks and the earth.
6. Suggest a name for the study of the physics of living organisms.
7. Engineering is the practical application of scientific principles and discoveries to develop things that make our lives easier. Is medicine science or engineering? Justify your answer.
8. Based on the definition of engineering in Exercise 7, would building a bridge over a river or road be considered science or engineering? Justify your answer.
9. When someone says, “I have a theory that excess salt causes high blood pressure,” does that person really have a theory? If it is not a theory, what is it?
10. When a person says, “My hypothesis is that excess calcium in the diet causes kidney stones,” what does the person need to do to determine if the hypothesis is correct?
11. Some people argue that many scientists accept many scientific principles on faith. Using what you know about the scientific method, how might you argue against that assertion?
12. Most students take multiple English classes in school. Does the study of English use the scientific method?
Answers
1.
1. science
2. science
3. not science
4. not science
5. not science
2.
1. not science
2. science
3. science
4. science
5. not science
1.
1. matter
2. not matter
3. matter
4. not matter
5. matter
6. matter
4.
• matter
• matter
• not matter
• matter
• matter
1. geophysics
6. biophysics
1. Medicine is probably closer to a field of engineering than a field of science, but this may be arguable. Ask your doctor.
8. Engineering
1. In scientific terms, this person has a hypothesis.
2. Conduct experiments to determine if kidney stones contain calcium.
3. Science is based on reproducible facts, not blind belief.
4. No. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.01%3A_What_is_Chemistry.txt |
Learning Objectives
• Use physical and chemical properties, including phase, to describe matter.
• Identify a sample of matter as an element, a compound, or a mixture.
Part of understanding matter is being able to describe it. One way chemists describe matter is to assign different kinds of properties to different categories.
Physical and Chemical Properties
The properties that chemists use to describe matter fall into two general categories. Physical properties are characteristics that describe matter. They include characteristics such as size, shape, color, and mass. These characteristics can be observed or measured without changing the identity of the matter in question. Chemical properties are characteristics that describe how matter changes its chemical structure or composition. An example of a chemical property is flammability—a material’s ability to burn—because burning (also known as combustion) changes the chemical composition of a material. The observation of chemical properties involves a chemical change of the matter in question, resulting in matter with a different identity and different physical and chemical properties.
Elements and Compounds
Any sample of matter that has the same physical and chemical properties throughout the sample is called a substance. There are two types of substances. A substance that cannot be broken down into chemically simpler components is called an element. Aluminum, which is used in soda cans and is represented by the symbol Al, is an element. A substance that can be broken down into chemically simpler components (because it consists of more than one element) is called a compound. Water is a compound composed of the elements hydrogen and oxygen and is described by the chemical formula, H2O. Today, there are about 118 elements in the known universe. In contrast, scientists have identified tens of millions of different compounds to date.
Sometimes the word pure is used to describe a substance, but this is not absolutely necessary. By definition, any single substance, element or compound is pure.
The smallest part of an element that maintains the identity of that element is called an atom. Atoms are extremely tiny; to make a line of iron atoms that is 1 inch long, you would need approximately 217 million iron atoms. The smallest part of a compound that maintains the identity of that compound is called a molecule. Molecules are composed of two or more different atoms that are attached together and behave as a unit. Scientists usually work with millions and millions of atoms and molecules at a time. When a scientist is working with large numbers of atoms or molecules at a time, the scientist is studying the macroscopic viewpoint of the universe. However, scientists can also describe chemical events on the level of individual atoms or molecules, which is referred to as the microscopic viewpoint. We will see examples of both macroscopic and microscopic viewpoints throughout this book (Figure \(2\)).
Mixtures
A material composed of two or more substances is a mixture. In a mixture, the individual substances maintain their chemical identities. Many mixtures are obvious combinations of two or more substances, such as a mixture of sand and water. Such mixtures are called heterogeneous mixtures. In some mixtures, the components are so intimately combined that they act like a single substance (even though they are not). Mixtures with a consistent or uniform composition throughout are called homogeneous mixtures (or solutions). For example, when sugar is dissolved in water to form a liquid solution, the individual properties of the components cannot be distinguished. Other examples or homogenous mixtures include solid solutions, like the metal alloy steel, and gaseous solutions, like air which is a mixture of mainly nitrogen and oxygen.
Example \(1\)
How would a chemist categorize each example of matter?
1. saltwater
2. soil
3. water
4. oxygen
Answer a
Saltwater acts as if it were a single substance even though it contains two substances—salt and water. Saltwater is a homogeneous mixture, or a solution.
Answer b
Soil is composed of small pieces of a variety of materials, so it is a heterogeneous mixture.
Answer c
Water is a substance; more specifically, because water is composed of hydrogen and oxygen, it is a compound.
Answer d
Oxygen, a substance, is an element.
Exercise \(2\)
How would a chemist categorize each example of matter?
1. breakfast coffee
2. hydrogen
3. an egg
Answer a
homogeneous mixture or solution
Answer b
element
Answer c
heterogeneous mixture
Phases or Physical States of Matter
All matter can be further classified by one of three physical states or phases, solid, liquid or gas. These three descriptions each imply that the matter has certain physical properties when in these states. A solid has a definite shape and a definite volume. Liquids ordinarily have a definite volume but not a definite shape; they take the shape of their containers. Gases have neither a definite shape nor a definite volume, and they expand to fill their containers.
We encounter matter in each phase every day; in fact, we regularly encounter water in all three phases: ice (solid), water (liquid), and steam (gas) (Figure \(2\)).
We know from our experience with water that substances can change from one phase to another if the conditions are right. Typically, varying the temperature of a substance (and, less commonly, the pressure exerted on it) can cause a phase change, a physical process in which a substance changes from one phase to another (Figure \(5\)). Phase changes are identified by particular names depending on what phases are involved, as summarized in Table \(1\).
Table \(1\): Phase Changes
Change Name
solid to liquid melting, fusion
solid to gas sublimation
liquid to gas boiling, evaporation
liquid to solid solidification, freezing
gas to liquid condensation
gas to solid deposition
Figure \(3\) illustrates the relationships between the different ways matter can be classified.
Concept Review Exercises
1. Explain the differences between the physical properties of matter and the chemical properties of matter.
2. What is the difference between a heterogeneous mixture and a homogeneous mixture? Give an example of each.
3. Give at least two examples of a phase change and state the phases involved in each.
Answers
1. Physical properties describe the existence of matter, and chemical properties describe how substances change into other substances.
2. A heterogeneous mixture is obviously a mixture, such as dirt; a homogeneous mixture behaves like a single substance, such as saltwater.
3. solid to liquid (melting) and liquid to gas (boiling) (answers will vary)
Key Takeaways
• Matter can be described with both physical properties and chemical properties.
• Matter can be identified as an element, a compound, or a mixture
Exercise \(3\)
Does each statement refer to a chemical property or a physical property?
1. Balsa is a very light wood.
2. If held in a flame, magnesium metal burns in air.
3. Mercury has a density of 13.6 g/mL.
4. Human blood is red.
Answer
1. physical property
2. chemical property
3. physical property
4. physical property
Exercise \(4\)
Does each statement refer to a chemical property or a physical property?
1. The elements sodium and chlorine can combine to make table salt.
2. The metal tungsten does not melt until its temperature exceeds 3,000°C.
3. The ingestion of ethyl alcohol can lead to disorientation and confusion.
4. The boiling point of isopropyl alcohol, which is used to sterilize cuts and scrapes, is lower than the boiling point of water
Answer
1. chemical property
2. physical property
3. chemical property
4. physical property
Exercise \(5\)
Define element. How does it differ from a compound?
Answer
An element is a substance that cannot be broken down into chemically simpler components. Compounds can be broken down into simpler substances.
Exercise \(6\)
Define compound. How does it differ from an element?
Answer
A compound is composed of two or more elements combined in a fixed ratio. An element is the simplest chemical substance.
Exercise \(7\)
Give two examples of a heterogeneous mixture.
Answer
a salt and pepper mix and a bowl of cereal (answers will vary)
Exercise \(8\)
Give two examples of a homogeneous mixture.
Answer
vinegar and rubbing alcohol (answers will vary)
Exercise \(9\)
Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. xenon, a substance that cannot be broken down into chemically simpler components
2. blood, a substance composed of several types of cells suspended in a salty solution called plasma
3. water, a substance composed of hydrogen and oxygen
Answer
1. element
2. heterogeneous mixture
3. compound
Exercise \(10\)
Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. sugar, a substance composed of carbon, hydrogen, and oxygen
2. hydrogen, the simplest chemical substance
3. dirt, a combination of rocks and decaying plant matter
Answer
1. compound
2. element
3. heterogeneous mixture
Exercise \(11\)
Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. air, primarily a mixture of nitrogen and oxygen
2. ringer’s lactate, a standard fluid used in medicine that contains salt, potassium, and lactate compounds all dissolved in sterile water
3. tartaric acid, a substance composed of carbon, hydrogen, and oxygen
Answer
1. heterogeneous mixture
2. solution
3. compound
Exercise \(12\)
What word describes each phase change?
1. solid to liquid
2. liquid to gas
3. solid to gas
Answer
1. melting or fusion
2. boiling or evaporation
3. sublimation
Exercise \(13\)
1. What word describes each phase change?
1. liquid to solid
2. gas to liquid
3. gas to solid
Answer
1. freezing
2. condensation
3. deposition | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.02%3A_The_Classification_of_Matter.txt |
Learning Objectives
• Express quantities properly using a number and a unit.
A coffee maker’s instructions tell you to fill the coffeepot with 4 cups of water and use 3 scoops of coffee. When you follow these instructions, you are measuring. When you visit a doctor’s office, a nurse checks your temperature, height, weight, and perhaps blood pressure (Figure $1$); the nurse is also measuring.
Chemists measure the properties of matter using a variety of devices or measuring tools, many of which are similar to those used in everyday life. Rulers are used to measure length, balances (scales) are used to measure mass (weight), and graduated cylinders or pipettes are used to measure volume. Measurements made using these devices are expressed as quantities. A quantity is an amount of something and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5.2 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5.2.
$\color{red} \underbrace{5.2}_{\text{number}} \color{blue} \underbrace{\text{kilometers}}_{\text{unit}} \nonumber$
If you ask a friend how far he or she walks from home to school, and the friend answers “12” without specifying a unit, you do not know whether your friend walks—for example, 12 miles, 12 kilometers, 12 furlongs, or 12 yards.
Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient’s seizures and states a dosage of “100” without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount.
Both a number and a unit must be included to express a quantity properly.
To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers. The next two sections examine the rules for expressing numbers.
Exercise $1$
Identify the number and the unit in each quantity.
1. one dozen eggs
2. 2.54 centimeters
3. a box of pencils
4. 88 meters per second
Answer a
The number is one, and the unit is dozen.
Answer b
The number is 2.54, and the unit is centimeter.
Answer c
The number 1 is implied because the quantity is only a box. The unit is box of pencils.
Answer d
The number is 88, and the unit is meters per second. Note that in this case the unit is actually a combination of two units: meters and seconds.
Exercise $2$
Identify the number and the unit in each quantity.
1. 99 bottles of soda
2. 60 miles per hour
3. 32 fluid ounces
4. 98.6 degrees Fahrenheit
Answer a
The number is 99, and the unit is bottles of soda.
Answer b
The number is 60, and the unit is miles per hour.
Answer c
The number 32, and the unit is fluid ounces
Answer d
The number is 98.6, and the unit is degrees Fahrenheit
Exercise $2$
What are the two necessary parts of a quantity?
Answer
The two necessary parts are the number and the unit.
Key Takeaway
• Identify a quantity properly with a number and a unit. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.03%3A_Measurements.txt |
Learning Objectives
• Express a large number or a small number in scientific notation.
• Convert a number in scientific notation to standard conventional form.
The instructions for making a pot of coffee specified 3 scoops (rather than 12,000 grounds) because any measurement is expressed more efficiently with units that are appropriate in size. In science, however, we often must deal with quantities that are extremely small or incredibly large. For example, you may have 5,000,000,000,000 red blood cells in a liter of blood, and the diameter of an iron atom is 0.000000014 inches. Numbers with many zeros can be cumbersome to work with, so scientists use scientific notation.
Scientific notation is a system for expressing very large or very small numbers in a compact manner. It uses the idea that such numbers can be rewritten as a simple number multiplied by 10 raised to a certain exponent, or power.
Let us look first at very large numbers. Suppose a spacecraft is 1,500,000 miles from Mars. The number 1,500,000 can be thought of as follows:
That is, 1,500,000 is the same as 1.5 times 1 million, and 1 million is 10 × 10 × 10 × 10 × 10 × 10, or 106 (which is read as “ten to the sixth power”). Therefore, 1,500,000 can be rewritten as 1.5 times 106, or 1.5 × 106. The distance of the spacecraft from Mars can therefore be expressed as 1.5 × 106 miles.
Recall that:
• 100 = 1
• 101 = 10
• 102 = 100
• 103 = 1,000
• 104 = 10,000
• and so forth
The standard convention for expressing numbers in scientific notation is to write a single nonzero first digit, a decimal point, and the rest of the digits, excluding any trailing zeros (see rules for significant figures in the next section for more details on what to exclude). This number is followed by a multiplication sign and then by 10 raised to the power necessary to reproduce the original number. For example, although 1,500,000 can also be written as 15. × 105 (which would be 15. × 100,000), the convention is to have only one digit before the decimal point. How do we know to what power 10 is raised? The power is the number of places you have to move the decimal point to the left to place it after the first digit, so that the number being multiplied is between 1 and 10:
Example $1$: Scientific Notation
Express each number in scientific notation.
1. 67,000,000,000
2. 1,689
3. 12.6
Answer a
Moving the decimal point 10 places to the left gives 6.7 × 1010.
Answer b
The decimal point is assumed to be at the end of the number, so moving it three places to the left gives 1.689 × 103.
Answer c
In this case, we need to move the decimal point only one place to the left, which yields 1.26 × 101.
Exercise $1$
Express each number in scientific notation.
1. 1,492
2. 102,000,000
3. 101,325
Answer a
Moving the decimal point 3 places to the left gives 1.492 × 103.
Answer b
The decimal point is assumed to be at the end of the number, so moving it 8 places to the left gives 1.02 × 108.
Answer c
Moving the decimal point 5 places to the left yields 1.01325 × 105.
To change a number in scientific notation to standard form, we reverse the process, moving the decimal point to the right. Add zeros to the end of the number being converted, if necessary, to produce a number of the proper magnitude. Lastly, we drop the number 10 and its power.
Example $2$
Express each number in standard, or conventional notation.
1. 5.27 × 104
2. 1.0008 × 106
Answer a
Moving the decimal four places to the right and adding zeros give 52,700.
Answer b
Moving the decimal six places to the right and adding zeros give 1,000,800.
Exercise $2$
Express each number in standard, or conventional notation.
1. $6.98 \times 10^8$
2. $1.005 \times 10^2$
Answer a
Moving the decimal point eight places to the right and adding zeros give 698,000,000.
Answer b
Moving the decimal point two places to the right gives 100.5
We can also use scientific notation to express numbers whose magnitudes are less than 1. For example, the quantity 0.006 centimeters can be expressed as follows:
That is, 0.006 centimeters is the same as 6 divided by one thousand, which is the same as 6 divided by 10 x 10 x 10 or 6 times 10–3 (which is read as "ten to the negative third power"). Therefore, 0.006 centimeters can be rewritten as 6 times 10–3, or 6 × 10–3 centimeters.
Recall that:
• 10−1 = 1/10
• 10−2 = 1/100
• 10−3 = 1/1,000
• 10−4 = 1/10,000
• 10−5 = 1/100,000
• and so forth
We use a negative number as the power to indicate the number of places we have to move the decimal point to the right to make it follow the first nonzero digit so that the number is between 1 and 10. This is illustrated as follows:
Note:
In writing scientific notations, the convention is to have only one digit before the decimal point.
• Numbers that are greater than one have a positive power in scientific notation. If the decimal point is moved to the left n places, the power (n) of 10 is positive.
• Numbers that are less than one have a negative power in scientific notation. If the decimal point is moved to the right n places, the power (n) of 10 is negative.
Example $3$
Express each number in scientific notation.
1. 0.000006567
2. −0.0004004
3. 0.000000000000123
Answer a
Move the decimal point six places to the right to get 6.567 × 10−6.
Answer b
Move the decimal point four places to the right to get −4.004 × 10−4. The negative sign on the number itself does not affect how we apply the rules of scientific notation.
Answer c
Move the decimal point 13 places to the right to get 1.23 × 10−13.
Exercise $3$
Express each number in scientific notation.
1. 0.000355
2. 0.314159
3. −0.051204
Answer a
Moving the decimal point four places to the right gives 3.55 × 10−4.
Answer b
Moving the decimal point one place to the right gives 3.14159 × 10−1.
Answer c
Moving the decimal point one place to the right gives −5.1204 × 10−2.
As with numbers with positive powers of 10, when changing from scientific notation to standard or conventional format, we reverse the process.
Note
Changing a number in scientific notation to standard form:
• If the scientific notation has a positive power, the standard number is greater than one. Example: 8 x 104 = 80,000
• If the scientific notation has a negative power, then the standard number is less than one. Example: 8 x 10-2 = 0.08
Example $4$
Change the number in scientific notation to standard form.
1. 6.22 × 10−2
2. 9.9 × 10−9
Answer a
0.0622
Answer b
0.0000000099
Exercise $4$
Change the number in scientific notation to standard form.
1. 9.98 × 10−5
2. 5.109 × 10−8
Answer a
0.0000998
Answer b
0.00000005109
Although calculators can show 8 to 10 digits in their display windows, that is not always enough when working with very large or very small numbers. For this reason, many calculators are designed to handle scientific notation. The method for entering scientific notation differs for each calculator model, so take the time to learn how to do it properly on your calculator, asking your instructor for assistance if necessary. If you do not learn to enter scientific notation into your calculator properly, you will not get the correct final answer when performing a calculation.
Concept Review Exercises
1. Why it is easier to use scientific notation to express very large or very small numbers?
2. What is the relationship between how many places a decimal point moves and the power of 10 used in changing a conventional number into scientific notation?
Answers
1. Scientific notation is more convenient than listing a large number of zeros.
2. The number of places the decimal point moves equals the power of 10. The power of 10 is positive if the decimal point moves to the left and negative if the decimal point moves to the right.
Key Takeaway
• Large or small numbers are expressed in scientific notation, which use powers of 10. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.04%3A_Expressing_Numbers_-_Scientific_Notation.txt |
Learning Objectives
• Understand the importance of significant figures in measured numbers.
• Identify the number of significant figures in a reported value.
• Use significant figures correctly in arithmetical operations.
Scientists have established certain conventions for communicating the degree of precision of a measurement, which is dependent on the measuring device used. Imagine, for example, that you are using a meterstick to measure the width of a table. The centimeters (cm) marked on the meterstick, tell you how many centimeters wide the table is. Many metersticks also have markings for millimeters (mm), so we can measure the table to the nearest millimeter. Most metersticks do not have any smaller (or more precise) markings indicated, so you cannot report the measured width of the table any more precise than to the nearest millimeter. However, you can estimate one past the smallest marking, in this case the millimeter, to the next decimal place in the measurement (Figure \(1\)).
The concept of significant figures takes this limitation into account. The significant figures of a measured quantity are defined as all the digits known with certainty (those indicated by the markings on the measuring device) and the first uncertain, or estimated, digit (one digit past the smallest marking on the measuring device). It makes no sense to report any digits after the first uncertain one, so it is the last digit reported in a measurement. Zeros are used when needed to place the significant figures in their correct positions. Thus, zeros are sometimes counted as significant figures but are sometimes only used as placeholders.
“Sig figs” is a common abbreviation for significant figures.
Consider the earlier example of measuring the width of a table with a meterstick. If the table is measured and reported as being 1,357 mm wide, the number 1,357 has four significant figures. The 1 (thousands place), the 3 (hundreds place), and the 5 (tens place) are certain; the 7 (ones place) is assumed to have been estimated. It would make no sense to report such a measurement as 1,357.0 (five Sig Figs) or 1,357.00 (six Sig Figs) because that would suggest the measuring device was able to determine the width to the nearest tenth or hundredth of a millimeter, when in fact it shows only tens of millimeters and therefore the ones place was estimated.
On the other hand, if a measurement is reported as 150 mm, the 1 (hundreds) and the 5 (tens) are known to be significant, but how do we know whether the zero is or is not significant? The measuring device could have had marks indicating every 100 mm or marks indicating every 10 mm. How can you determine if the zero is significant (the estimated digit), or if the 5 is significant and the zero a value placeholder?
The rules for deciding which digits in a measurement are significant are as follows:
1. All nonzero digits are significant. In 1,357 mm, all the digits are significant.
2. Sandwiched (or embedded) zeros, those between significant digits, are significant. Thus, 405 g has three significant figures.
3. Leading zeros, which are zeros at the beginning of a decimal number less than 1, are not significant. In 0.000458 mL, the first four digits are leading zeros and are not significant. The zeros serve only to put the digits 4, 5, and 8 in the correct decimal positions. This number has three significant figures.
4. Trailing zeros, which are zeros at the end of a number, are significant only if the number has a decimal point. Thus, in 1,500 m, the two trailing zeros are not significant because the number is written without a decimal point; the number has two significant figures. However, in 1,500.00 m, all six digits are significant because the number has a decimal point.
Example \(1\)
How many significant figures does each number have?
1. 6,798,000
2. 6,000,798
3. 6,000,798.00
4. 0.0006798
Answer a
four (by rules 1 and 4)
Answer b
seven (by rules 1 and 2)
Answer c
nine (by rules 1, 2, and 4)
Answer d
four (by rules 1 and 3)
Exercise \(1\)
How many significant figures does each number have?
1. 2.1828
2. 0.005505
3. 55,050
4. 5
5. 500
Answer a
five
Answer b
four
Answer c
four
Answer d
one
Answer e
one
Rounding off numbers
Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1.
Consider the measurement 207.518m. Right now, the measurement contains six significant figures. How would we successively round it to fewer and fewer significant figures? Follow the process as outlined in Table 1.5.1.
Number of Significant Figures Rounded Value Reasoning
Table 1.5.1: Rounding examples
6 207.518 All digits are significant
5 207.52 8 rounds the 1 up to 2
4 207.5 2 is dropped
3 208 5 rounds the 7 up to 8
2 210 8 is replaced by a 0 and rounds the 0 up to 1
1 200 1 is replaced by a 0
Notice that the more rounding that is done, the less reliable the figure is. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail.
It is important to be aware of significant figures when you are mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each? When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer—one rule is for addition and subtraction, and one rule is for multiplication and division.
In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise than the least precise number used to get the answer.
Combining Numbers
For addition or subtraction, the rule is to stack all the numbers with their decimal points aligned and then limit (round to) the answer’s significant figures to the rightmost column for which all the numbers have significant figures. Consider the following:
The arrow points to the rightmost column in which all the numbers have significant figures—in this case, the tenths place. Therefore, we will limit our final answer to the tenths place. Is our final answer therefore 1,459.0? No, because when we drop digits from the end of a number, we also have to round the number. Notice that the first dropped digit, in the hundredths place, is 8. This suggests that the answer is actually closer to 1,459.1 than it is to 1,459.0, so we need to round up to 1,459.1. The standard rules for rounding numbers are simple: If the first dropped digit is 5 or higher, round up. If the first dropped digit is lower than 5, do not round up.
For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows:
The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up.
Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 4,000 has one significant figure and should be written as the number 4 × 104. The number 450 has two significant figures and would be written in scientific notation as 4.5 × 102, whereas 450.0 has four significant figures and would be written as 4.500 × 102. In scientific notation, all reported digits are significant.
Example \(2\)
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
1. 23.096 × 90.300
2. 125 × 9.000
3. 1,027 + 610.0 + 363.06
Answer a
The calculator answer is 2,085.5688, but we need to round it to five significant figures. Because the first digit to be dropped (in the hundredths place) is greater than 5, we round up to 2,085.6, which in scientific notation is 2.0856 × 103.
Answer b
The calculator gives 1,125 as the answer, but we limit it to three significant figures and convert into scientific notation: 1.13 × 103.
Answer c
The calculator gives 2,000.06 as the answer, but because 1,027 has its farthest-right significant figure in the ones column, our answer must be limited to the ones position: 2,000 which in scientific notation is 2.000 × 103.
Exercise \(2\)
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
1. 217 ÷ 903
2. 13.77 + 908.226 + 515
3. 255.0 − 99
4. 0.00666 × 321
Answer a
0.240 = 2.40 x 10-1
Answer b
1437 = 1.437 x 103
Answer c
156 = 1.56 x 102
Answer d
2.14 = 2.14 x 100
Remember that calculators do not understand significant figures. You are the one who must apply the rules of significant figures to a result from your calculator.
Concept Review Exercises
1. Explain why the concept of significant figures is important in scientific measurements.
2. State the rules for determining the significant figures in a measurement.
3. When do you round a number up, and when do you not round a number up?
Answers
1. Significant figures represent all the known digits of a measurement plus the first estimated one. It gives information about how precise the measuring device and measurement is.
2. All nonzero digits are significant; zeros between nonzero digits are significant; zeros at the end of a nondecimal number or the beginning of a decimal number are not significant; zeros at the end of a decimal number are significant.
3. Round up only if the first digit dropped is 5 or higher.
Key Takeaways
• Significant figures properly report the number of measured and estimated digits in a measurement.
• The rule in multiplication and division is that the final answer should have the same number of significant figures as there are in the value with the fewest significant figures.
• The rule in addition and subtraction is that the final answer should have the same number of decimal places as the term with the fewest decimal places. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.05%3A_Expressing_Numbers_-_Significant_Figures.txt |
Learning Objectives
• Recognize the SI base units and explain the system of prefixes used with them.
• Define and calculate density.
People who live in the United States measure weight in pounds, height in feet and inches, and a car’s speed in miles per hour. In contrast, chemistry and other branches of science use the International System of Units (also known as SI after Système Internationale d’Unités), which was established so that scientists around the world could communicate efficiently with each other. Many countries have also adopted SI units for everyday use as well. The United States is one of the few countries that has not.
Base SI Units
Base (or basic) units, are the fundamental units of SI. There are seven base units, which are listed in Table $1$, Chemistry uses five of the base units: the mole for amount, the kilogram for mass, the meter for length, the second for time, and the kelvin for temperature. The degree Celsius (°C) is also commonly used for temperature. The numerical relationship between kelvins and degrees Celsius is as follows:
$K = °C + 273 \label{Eq1}$
Table $1$: The Seven Base SI Units
Property Unit Abbreviation
length meter m
mass kilogram kg
time second s
amount mole mol
temperature kelvin K
electrical current ampere amp
luminous intensity candela cd
The United States uses the English (sometimes called Imperial) system of units for many quantities. Inches, feet, miles, gallons, pounds, and so forth, are all units connected with the English system of units. There have been many mistakes due to the improper conversion of units between the SI and English systems.
The size of each base unit is defined by international convention. For example, the kilogram is defined as the quantity of mass of a special metal cylinder kept in a vault in France (Figure $1$). The other base units have similar definitions and standards. The sizes of the base units are not always convenient for all measurements. For example, a meter is a rather large unit for describing the width of something as narrow as human hair. Instead of reporting the diameter of hair as 0.00012 m or as 1.2 × 10−4 m using scientific notation as discussed in section 1.4, SI also provides a series of prefixes that can be attached to the units, creating units that are larger or smaller by powers of 10.
Common prefixes and their multiplicative factors are listed in Table $2$. (Perhaps you have already noticed that the base unit kilogram is a combination of a prefix, kilo- meaning 1,000 ×, and a unit of mass, the gram.) Some prefixes create a multiple of the original unit: 1 kilogram equals 1,000 grams, and 1 megameter equals 1,000,000 meters. Other prefixes create a fraction of the original unit. Thus, 1 centimeter equals 1/100 of a meter, 1 millimeter equals 1/1,000 of a meter, 1 microgram equals 1/1,000,000 of a gram, and so forth.
Table $2$: Prefixes Used with SI Units
Prefix Abbreviation Multiplicative Factor Multiplicative Factor in Scientific Notation
giga- G 1,000,000,000 × 109 ×
mega- M 1,000,000 × 106 ×
kilo- k 1,000 × 103 ×
deca- D 10 × 101 ×
deci- d 1/10 × 10−1 ×
centi- c 1/100 × 10−2 ×
milli- m 1/1,000 × 10−3 ×
micro- µ* 1/1,000,000 × 10−6 ×
nano- n 1/1,000,000,000 × 10−9 ×
*The letter µ is the Greek lowercase letter for m and is called “mu,” which is pronounced “myoo.”
Both SI units and prefixes have abbreviations, and the combination of a prefix abbreviation with a base unit abbreviation gives the abbreviation for the modified unit. For example, kg is the abbreviation for kilogram. We will be using these abbreviations throughout this book.
The Difference Between Mass and Weight
The mass of a body is a measure of its inertial property or how much matter it contains. The weight of a body is a measure of the force exerted on it by gravity or the force needed to support it. Gravity on earth gives a body a downward acceleration of about 9.8 m/s2. In common parlance, weight is often used as a synonym for mass in weights and measures. For instance, the verb “to weigh” means “to determine the mass of” or “to have a mass of.” The incorrect use of weight in place of mass should be phased out, and the term mass used when mass is meant. The SI unit of mass is the kilogram (kg). In science and technology, the weight of a body in a particular reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame. Thus, the SI unit of the quantity weight defined in this way (force) is the newton (N).
Derived SI Units
Derived units are combinations of SI base units. Units can be multiplied and divided, just as numbers can be multiplied and divided. For example, the area of a square having a side of 2 cm is 2 cm × 2 cm, or 4 cm2 (read as “four centimeters squared” or “four square centimeters”). Notice that we have squared a length unit, the centimeter, to get a derived unit for area, the square centimeter.
Volume is an important quantity that uses a derived unit. Volume is the amount of space that a given substance occupies and is defined geometrically as length × width × height. Each distance can be expressed using the meter unit, so volume has the derived unit m × m × m, or m3 (read as “meters cubed” or “cubic meters”). A cubic meter is a rather large volume, so scientists typically express volumes in terms of 1/1,000 of a cubic meter. This unit has its own name—the liter (L). A liter is a little larger than 1 US quart in volume. Below are approximate equivalents for some of the units used in chemistry.
Approximate Equivalents to Some SI Units
• 1 m ≈ 39.36 in. ≈ 3.28 ft ≈ 1.09 yd
• 1 in. ≈ 2.54 cm
• 1 km ≈ 0.62 mi
• 1 kg ≈ 2.20 lb
• 1 lb ≈ 454 g
• 1 L ≈ 1.06 qt
• 1 qt ≈ 0.946 L
As shown in Figure $3$, a liter is also 1,000 cm3. By definition, there are 1,000 mL in 1 L, so 1 milliliter and 1 cubic centimeter represent the same volume.
$1\; mL = 1\; cm^3 \label{Eq2}$
Example $1$
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kiloliter
2. microsecond
3. decimeter
4. nanogram
Answer a
The abbreviation for a kiloliter is kL. Because kilo means “1,000 ×,” 1 kL equals 1,000 L.
Answer b
The abbreviation for microsecond is µs. Micro implies 1/1,000,000th of a unit, so 1 µs equals 0.000001 s.
Answer c
The abbreviation for decimeter is dm. Deci means 1/10th, so 1 dm equals 0.1 m.
Answer d
The abbreviation for nanogram is ng and equals 0.000000001 g.
Exercise $1$
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kilometer
2. milligram
3. nanosecond
4. centiliter
Answer a
km (1,000 m)
Answer b
mg (0.001 g)
Answer c
ns (0.000000001 s)
Answer d
cL (0.01L)
Energy, another important quantity in chemistry, is the ability to perform work, such as moving a box of books from one side of a room to the other side. It has a derived unit of kg•m2/s2. (The dot between the kg and m units implies the units are multiplied together.) Because this combination is cumbersome, this collection of units is redefined as a joule (J). An older unit of energy, but likely more familiar to you, the calorie (cal), is also widely used. There are 4.184 J in 1 cal. Energy changes occur during all chemical processes and will be discussed in a later chapter.
To Your Health: Energy and Food
The food in our diet provides the energy our bodies need to function properly. The energy contained in food could be expressed in joules or calories, which are the conventional units for energy, but the food industry prefers to use the kilocalorie and refers to it as the Calorie (with a capital C). The average daily energy requirement of an adult is about 2,000–2,500 Calories, which is 2,000,000–2,500,000 calories (with a lowercase c).
If we expend the same amount of energy that our food provides, our body weight remains stable. If we ingest more Calories from food than we expend, however, our bodies store the extra energy in high-energy-density compounds, such as fat, and we gain weight. On the other hand, if we expend more energy than we ingest, we lose weight. Other factors affect our weight as well—genetic, metabolic, behavioral, environmental, cultural factors—but dietary habits are among the most important.
In 2008 the US Centers for Disease Control and Prevention issued a report stating that 73% of Americans were either overweight or obese. More alarmingly, the report also noted that 19% of children aged 6–11 and 18% of adolescents aged 12–19 were overweight—numbers that had tripled over the preceding two decades. Two major reasons for this increase are excessive calorie consumption (especially in the form of high-fat foods) and reduced physical activity. Partly because of that report, many restaurants and food companies are working to reduce the amounts of fat in foods and provide consumers with more healthy food options.
Density is defined as the mass of an object divided by its volume; it describes the amount of matter contained in a given amount of space.
$\mathrm{density=\dfrac{mass}{volume}}\label{Eq3}$
Thus, the units of density are the units of mass divided by the units of volume: g/cm3 or g/mL (for solids and liquids), g/L (for gases), kg/m3, and so forth. For example, the density of water is about 1.00 g/cm3, while the density of mercury is 13.6 g/mL. (Remember that 1 mL equals 1 cm3.) Mercury is over 13 times as dense as water, meaning that it contains over 13 times the amount of matter in the same amount of space. The density of air at room temperature is about 1.3 g/L. Table 1.6.3 shows the densities of some common substances.
Table $2$: Densities of Common Substances
Solids Liquids Gases (at 25 °C and 1 atm)
ice (at 0 °C) 0.92 g/cm3 water 1.0 g/cm3 dry air 1.20 g/L
oak (wood) 0.60–0.90 g/cm3 ethanol 0.79 g/cm3 oxygen 1.31 g/L
iron 7.9 g/cm3 acetone 0.79 g/cm3 nitrogen 1.14 g/L
copper 9.0 g/cm3 glycerin 1.26 g/cm3 carbon dioxide 1.80 g/L
lead 11.3 g/cm3 olive oil 0.92 g/cm3 helium 0.16 g/L
silver 10.5 g/cm3 gasoline 0.70–0.77 g/cm3 neon 0.83 g/L
gold 19.3 g/cm3 mercury 13.6 g/cm3 radon 9.1 g/L
Example $2$: Density of Bone
What is the density of a section of bone if a 25.3 cm3 sample has a mass of 27.8 g?
Solution
Because density is defined as the mass of an object divided by its volume, we can set up the following relationship:
\begin{align*} \mathrm{density} &=\dfrac{mass}{volume} \[4pt] &= \dfrac{27.8\:g}{25.3\:cm^3} \[4pt] &=1.10\:g/cm^3 \end{align*} \nonumber
Note that we have limited our final answer to three significant figures.
Exercise $2$: Density of Oxygen
What is the density of oxygen gas if a 15.0 L sample has a mass of 21.7 g?
Answer
1.45 g/L
Density can be used to convert between the mass and the volume of a substance. This will be discussed in the next section.
Concept Review Exercises
1. What is the difference between a base unit and a derived unit? Give two examples of each type of unit.
2. Do units follow the same mathematical rules as numbers do? Give an example to support your answer.
3. What is density?
Answers
1. Base units are the seven fundamental units of SI; derived units are constructed by making combinations of the base units; Two examples of base units: kilograms and meters (answers will vary); Two examples of derived units: grams per milliliter and joules (answers will vary).
2. yes; $\mathrm{mL\times\dfrac{g}{mL}=g}$ (answers will vary)
3. Density is defined as the mass of an object divided by its volume
Key Takeaways
• Recognize the SI base units and derived units.
• Combining prefixes with base units creates new units of larger or smaller sizes. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.06%3A_The_International_System_of_Units.txt |
Learning Objectives
• Convert a value reported in one unit to a corresponding value in a different unit.
The ability to convert from one unit to another is an important skill. For example, a nurse with 50 mg aspirin tablets who must administer 0.2 g of aspirin to a patient needs to know that 0.2 g equals 200 mg, so 4 tablets are needed. Fortunately, there is a simple way to convert from one unit to another.
Conversion Factors
If you learned the SI units and prefixes described, then you know that 1 cm is 1/100th of a meter.
$1\; \rm{cm} = \dfrac{1}{100} \; \rm{m} \nonumber$
or
$100\; \rm{cm} = 1\; \rm{m} \nonumber$
Suppose we divide both sides of the equation by 1 m (both the number and the unit):
$\mathrm{\dfrac{100\:cm}{1\:m}=\dfrac{1\:m}{1\:m}} \nonumber$
As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the right side of the equation; it now has the same quantity in the numerator (the top) as it has in the denominator (the bottom). Any fraction that has the same quantity in the numerator and the denominator has a value of 1:
We know that 100 cm is 1 m, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units. A fraction that has equivalent quantities in the numerator and the denominator but expressed in different units is called a conversion factor.
Here is a simple example. How many centimeters are there in 3.55 m? Perhaps you can determine the answer in your head. If there are 100 cm in every meter, then 3.55 m equals 355 cm. To solve the problem more formally with a conversion factor, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as $\mathrm{\frac{100\:cm}{1\:m}}$ and multiply:
$3.55 \; \rm{m} \times \dfrac{100 \; \rm{cm}}{1\; \rm{m}} \nonumber$
The 3.55 m can be thought of as a fraction with a 1 in the denominator. Because m, the abbreviation for meters, occurs in both the numerator and the denominator of our expression, they cancel out:
$\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}} \nonumber$
The final step is to perform the calculation that remains once the units have been canceled:
$\dfrac{3.55}{1} \times \dfrac{100 \; \rm{cm}}{1} = 355 \; \rm{cm} \label{Ex1}$
In the final answer, we omit the 1 in the denominator. Thus, by a more formal procedure, we find that 3.55 m equals 355 cm. A generalized description of this process is as follows:
$\text{quantity (in old units)} \times \text{conversion factor} = \text{quantity (in new units)} \nonumber$
You may be wondering why we use a seemingly complicated procedure for a straightforward conversion. In later studies, the conversion problems you will encounter will not always be so simple. If you can master the technique of applying conversion factors, you will be able to solve a large variety of problems.
In the previous example (Equation \ref{Ex1}), we used the fraction $\frac{100 \; \rm{cm}}{1 \; \rm{m}}$ as a conversion factor. Does the conversion factor $\frac{1 \; \rm m}{100 \; \rm{cm}}$ also equal 1? Yes, it does; it has the same quantity in the numerator as in the denominator (except that they are expressed in different units). Why did we not use that conversion factor? If we had used the second conversion factor, the original unit would not have canceled, and the result would have been meaningless. Here is what we would have gotten:
$3.55 \; \rm{m} \times \dfrac{1\; \rm{m}}{100 \; \rm{cm}} = 0.0355 \dfrac{\rm{m}^2}{\rm{cm}} \nonumber$
For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out. Figure $1$ shows a concept map for constructing a proper conversion.
Significant Figures in Conversions
How do conversion factors affect the determination of significant figures? Numbers in conversion factors based on prefix changes, such as kilograms to grams, are not considered in the determination of significant figures in a calculation because the numbers in such conversion factors are exact. Exact numbers are defined or counted numbers, not measured numbers, and can be considered as having an infinite number of significant figures. (In other words, 1 kg is exactly 1,000 g, by the definition of kilo-.) Counted numbers are also exact. If there are 16 students in a classroom, the number 16 is exact. In contrast, conversion factors that come from measurements (such as density, as we will see shortly) or are approximations have a limited number of significant figures and should be considered in determining the significant figures of the final answer.
Example $1$
1. The average volume of blood in an adult male is 4.7 L. What is this volume in milliliters?
2. A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms?
Solution
1. We start with what we are given, 4.7 L. We want to change the unit from liters to milliliters. There are 1,000 mL in 1 L. From this relationship, we can construct two conversion factors:
$\dfrac{1\; \rm{L}}{1,000\; \rm{mL}} \; \text{ or } \; \dfrac{1,000 \; \rm{mL}}{1\; \rm{L}} \nonumber$
We use the conversion factor that will cancel out the original unit, liters, and introduce the unit we are converting to, which is milliliters. The conversion factor that does this is the one on the right.
$4.7 \cancel{\rm{L}} \times \dfrac{1,000 \; \rm{mL}}{1\; \cancel{\rm{L}}} = 4,700\; \rm{mL} \nonumber$
Because the numbers in the conversion factor are exact, we do not consider them when determining the number of significant figures in the final answer. Thus, we report two significant figures in the final answer.
1. We can construct two conversion factors from the relationships between milliseconds and seconds:
$\dfrac{1,000 \; \rm{ms}}{1\; \rm{s}} \; \text{ or } \; \dfrac{1\; \rm{s}}{1,000 \; \rm{ms}} \nonumber$
To convert 18 ms to seconds, we choose the conversion factor that will cancel out milliseconds and introduce seconds. The conversion factor on the right is the appropriate one. We set up the conversion as follows:
$18 \; \cancel{\rm{ms}} \times \dfrac{1\; \rm{s}}{1,000 \; \cancel{\rm{ms}}} = 0.018\; \rm{s} \nonumber$
The conversion factor’s numerical values do not affect our determination of the number of significant figures in the final answer.
Exercise $1$
Perform each conversion.
1. 101,000 ns to seconds
2. 32.08 kg to grams
Answer a
$101,000 \cancel{\rm{ns}} \times \dfrac{1\; \rm{s}}{1,000,000,000\; \cancel{\rm{ns}}} = 0.000101\; \rm{s} \nonumber$
Answer b
$32.08 \cancel{\rm{kg}} \times \dfrac{1,000 \; \rm{g}}{1\; \cancel{\rm{kg}}} = 32,080\; \rm{g} \nonumber$
Conversion Factors From Different Units
Conversion factors can also be constructed for converting between different kinds of units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows:
13.6 g mercury = 1 mL mercury
This relationship can be used to construct two conversion factors:
$\mathrm{\dfrac{13.6\:g}{1\:mL}\:and\:\dfrac{1\:mL}{13.6\:g}} \nonumber$
Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 16 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top so that our final answer has a unit of mass:
\begin{align*} \mathrm{16\:\cancel{mL}\times\dfrac{13.6\:g}{1\:\cancel{mL}}} &= \mathrm{217.6\:g} \[4pt] &\approx \mathrm{220\:g} \end{align*} \nonumber
In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates.
Density can be used as a conversion factor between mass and volume.
Example $2$: Mercury Thermometer
A mercury thermometer for measuring a patient’s temperature contains 0.750 g of mercury. What is the volume of this mass of mercury?
Solution
Because we are starting with grams, we want to use the conversion factor that has grams in the denominator. The gram unit will cancel algebraically, and milliliters will be introduced in the numerator.
\begin{align*} 0.750 \; \cancel{\rm{g}} \times \dfrac{1\; \rm{mL}}{13.6 \; \cancel{\rm{g}}} &= 0.055147 \ldots \; \rm{mL} \[4pt] &\approx 0.0551\; \rm{mL} \end{align*} \nonumber
We have limited the final answer to three significant figures.
Exercise $2$
What is the volume of 100.0 g of air if its density is 1.3 g/L?
Answer
$100.0 \cancel{\rm{g}} \times \dfrac{1\; \rm{L}}{1.3\; \cancel{\rm{g}}} = 76.92307692\; \rm{L} ≈ 77 L \nonumber$
Because the density (1.3 g/L) has only 2 significant figures, we are rounding off the final answer to 2 significant figures.
Looking Closer: Density and the Body
The densities of many components and products of the body have a bearing on our health.
Bones. Bone density is important because bone tissue of lower-than-normal density is mechanically weaker and susceptible to breaking. The density of bone is, in part, related to the amount of calcium in one’s diet; people who have a diet deficient in calcium, which is an important component of bones, tend to have weaker bones. Dietary supplements or adding dairy products to the diet seems to help strengthen bones. As a group, women experience a decrease in bone density as they age. It has been estimated that fully half of women over age 50 suffer from excessive bone loss, a condition known as osteoporosis. Exact bone densities vary within the body, but for a healthy 30-year-old female, it is about 0.95–1.05 g/cm3. Osteoporosis is diagnosed if the bone density is below 0.6–0.7 g/cm3.
Urine. The density of urine can be affected by a variety of medical conditions. Sufferers of diabetes produce an abnormally large volume of urine with a relatively low density. In another form of diabetes, called diabetes mellitus, there is excess glucose dissolved in the urine, so that the density of urine is abnormally high. The density of urine may also be abnormally high because of excess protein in the urine, which can be caused by congestive heart failure or certain renal (kidney) problems. Thus, a urine density test can provide clues to various kinds of health problems. The density of urine is commonly expressed as a specific gravity, which is a unitless quantity defined as
$\dfrac{\text{density of some material}}{\text{density of water}} \nonumber$
Normal values for the specific gravity of urine range from 1.002 to 1.028.
Body Fat. The overall density of the body is one indicator of a person’s total body fat. Fat is less dense than muscle and other tissues, so as it accumulates, the overall density of the body decreases. Measurements of a person’s weight and volume provide the overall body density, which can then be correlated to the percentage of body fat. (The body’s volume can be measured by immersion in a large tank of water. The amount of water displaced is equal to the volume of the body.)
Problem Solving With Multiple Conversions
Sometimes you will have to perform more than one conversion to obtain the desired unit. For example, suppose you want to convert 54.7 km into millimeters. You can either memorize the relationship between kilometers and millimeters, or you can do the conversion in two steps. Most people prefer to convert in steps.
To do a stepwise conversion, we first convert the given amount to the base unit. In this example, the base unit is meters. We know that there are 1,000 m in 1 km:
$54.7\; \cancel{\rm{km}} \times \dfrac{1,000 \; \rm{m}}{1\; \cancel{\rm{km}}} = 54,700\; \rm{m} \nonumber$
Then we take the result (54,700 m) and convert it to millimeters, remembering that there are $1,000\; \rm{mm}$ for every $1\; \rm{m}$:
\begin{align*} 54,700 \; \cancel{\rm{m}} \times \dfrac{1,000 \; \rm{mm}}{1\; \cancel{\rm{m}}} &= 54,700,000 \; \rm{mm} \[4pt] &= 5.47 \times 10^7\; \rm{mm} \end{align*} \nonumber
We have expressed the final answer in scientific notation.
As a shortcut, both steps in the conversion can be combined into a single, multistep expression:
Concept Map
Calculation
\begin{align*} 54.7\; \cancel{\rm{km}} \times \dfrac{1,000 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1,000 \; \rm{mm}}{1\; \cancel{\rm{m}}} &= 54,700,000 \; \rm{mm} \[4pt] &= 5.47 \times 10^7\; \rm{mm} \end{align*} \nonumber
In each step, the previous unit is canceled and the next unit in the sequence is produced, each successive unit canceling out until only the unit needed in the answer is left.
Either method—one step at a time or all the steps together—is acceptable. If you do all the steps together, the restriction for the proper number of significant figures should be done after the last step. As long as the math is performed correctly, you should get the same answer no matter which method you use.
Example $3$
Convert 58.2 ms to megaseconds in one multistep calculation.
Solution
First, convert the given unit (ms) to the base unit—in this case, seconds—and then convert seconds to the final unit, megaseconds:
Calculation
\begin{align*} 58.2 \; \cancel{\rm{ms}} \times \dfrac{\cancel{1 \rm{s}}}{1,000\; \cancel{\rm{ms}}} \times \dfrac{1\; \rm{Ms}}{1,000,000\; \cancel{ \rm{s}}} &=0.0000000582\; \rm{Ms} \[4pt] &= 5.82 \times 10^{-8}\; \rm{Ms} \end{align*} \nonumber
Neither conversion factor affects the number of significant figures in the final answer.
Exercise $3$
Convert 43.007 mg to kilograms in one multistep calculation.
Answer
\begin{align*} 43.007 \; \cancel{\rm{mg}} \times \dfrac{\cancel{1 \rm{g}}}{1,000\; \cancel{\rm{mg}}} \times \dfrac{1\; \rm{kg}}{1,000\; \cancel{ \rm{g}}} &=0.000043007\; \rm{kg} \[4pt] &= 4.3007 \times 10^{-5}\; \rm{kg} \end{align*} \nonumber.
Neither conversion factor affects the number of significant figures in the final answer.
Career Focus: Pharmacist
A pharmacist dispenses drugs that have been prescribed by a doctor. Although that may sound straightforward, pharmacists in the United States must hold a doctorate in pharmacy and be licensed by the state in which they work. Most pharmacy programs require four years of education in a specialty pharmacy school.
Pharmacists must know a lot of chemistry and biology so they can understand the effects that drugs (which are chemicals, after all) have on the body. Pharmacists can advise physicians on the selection, dosage, interactions, and side effects of drugs. They can also advise patients on the proper use of their medications, including when and how to take specific drugs properly. Pharmacists can be found in drugstores, hospitals, and other medical facilities.
Curiously, an outdated name for pharmacist is chemist, which was used when pharmacists formerly did a lot of drug preparation, or compounding. In modern times, pharmacists rarely compound their own drugs, but their knowledge of the sciences, including chemistry, helps them provide valuable services in support of everyone’s health.
Key Takeaway
• A unit can be converted to another unit of the same type with a conversion factor.
Concept Review Exercises
1. How do you determine which quantity in a conversion factor goes in the denominator of the fraction?
2. State the guidelines for determining significant figures when using a conversion factor.
3. Write a concept map (a plan) for how you would convert $1.0 \times 10^{12}$ nanoliters (nL) to kiloliters (kL).
Answers
1. The unit you want to cancel from the numerator goes in the denominator of the conversion factor.
2. Exact numbers that appear in many conversion factors do not affect the number of significant figures; otherwise, the normal rules of multiplication and division for significant figures apply.
3. Concept Map: Convert the given (nanoliters, nL) to liters; then convert liters to kiloliters. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.07%3A_Converting_Units.txt |
Learning Objectives
• Calculate drug dosages using conversion factors.
To Your Health: Dosages
A medicine can be more harmful than helpful if it is not taken in the proper dosage. A dosage (or dose) is the specific amount of a medicine that is known to be therapeutic for an ailment in a patient of a certain size. Dosages of the active ingredient in medications are usually described by units of mass, typically grams or milligrams, and generally are equated with a number of capsules or teaspoonfuls to be swallowed or injected. The amount of the active ingredient in a medicine is carefully controlled so that the proper number of pills or spoonfuls contains the proper dose.
Most drugs must be taken in just the right amount. If too little is taken, the desired effects will not occur (or will not occur fast enough for comfort); if too much is taken, there may be potential side effects that are worse than the original ailment. Some drugs are available in multiple dosages. For example, tablets of the medication levothyroxine sodium, a synthetic thyroid hormone for those suffering from decreased thyroid gland function, are available in 11 different doses, ranging from 25 micrograms (µg) to 300 µg. It is a doctor’s responsibility to prescribe the correct dosage for a patient, and it is a pharmacist’s responsibility to provide the patient with the correct medicine at the dosage prescribed. Thus, proper quantities—which are expressed using numbers and their associated units—are crucial for keeping us healthy.
Effects are dose-dependent
Chemicals are the most common things for which doses are measured, but there are others, such as radiation exposure. For humans, most doses of micronutrients and medications are measured in milligrams (mg), but some are measured in micrograms because of their potency. Nonmedicinal poisons span the measurement scale; some poisons are so dangerous that a single microgram of it could be deadly, whereas other substances take much more. For example, even water is toxic when consumed in large enough quantities.
Dosage (the size of each dose) determines the strength and duration of the health benefits of nutrients, and also of the therapeutic effects of medical treatments. Dosage also determines the severity of adverse effects of treatments and toxins.
Duration of exposure, that is, the period of time over which the dose was received (all at once or gradually) also determines its effects (the body may build tolerance to gradual exposure to a drug, while a large immediate dose could be deadly).
The route by which a dose is exposed to, may affect the outcome, because some medications have different effects depending on whether they are inhaled, ingested, taken transdermally, injected, or inserted.
The dosage, route, concentration, and division over time may all be critical considerations in the administering of drugs, or in responding to exposure to a toxin. In nutrition, the route is usually a given, as nutrients are generally eaten; while dosage and the frequency of ingestion of nutrients are very important variables in preventing disease and promoting overall health.
Calculation of dose
Calculating drug dosages for humans based on the doses used in animal studies can be based on weight (e.g., mg/kg) or surface area (e.g., mg/m2) based on weight2/3.[1]
Drug dosage calculations
Drug dosage calculation is required if the physician’s order is different from what is available.
Example $1$
1. A physician ordered 100 mg of Demerol. Demerol is available as 50 mg per tablet. How many tablets should the nurse administer?
2. The doctor’s order is 1.2 g of Folic Acid. Folic Acid is available as 800 mg per tablet. How many tablets should be taken?
Solution
a. We start with the given, 100 mg. We want to change the unit from mg to tablets. There are 50 mg in 1 tablet (Remember that per tablet means one tablet. From this relationship, we can construct the conversion factor. We use the conversion factor that will cancel out the original unit, mg, and introduce the unit we are converting to, which is tablet.
$100 \cancel{\rm{mg}} \times \dfrac{1\; \rm{tablet}}{50\; \cancel{\rm{mg}}} = 2\; \rm{tablets} \nonumber$
Hence, the nurse should administer 2 tablets.
b. We start with the given, 1.2 g and we want to change grams to number of tablets. First, we convert 1.2 g to mg and then convert mg to tablets. We need a conversion factor for each step.
\begin{align*} 1.2 \; \cancel{\rm{g}} \times \dfrac{1,000 \cancel{\rm{mg}}}{1\; \cancel{\rm{g}}} \times \dfrac{1\; \rm{tablet}}{800\; \cancel{ \rm{mg}}} &=1.5 \; \rm{tablets} \end{align*} \nonumber
Hence, 1.5 tablets should be taken.
Exercise $1$
Calculate each of the following.
1. The physician ordered 20 mg of Valium. Valium is available as 10 mg per tablet. How many tablets should the nurse administer?
2. The doctor’s order is 1 g of Calcium. What is on hand is Calcium as 500 mg per tablet. How many tablets should be taken?
Answer a
Start with 20 mg of Valium. For the conversion factor, we know that 10 mg Valium = 1 tablet
$20 \cancel{\rm{mg}} \times \dfrac{1\; \rm{tablet}}{10\; \cancel{\rm{mg}}} = 2\; \rm{tablets} \nonumber$
Answer b
Start with 1 g (1000 mg) of Calcium. For the conversion factor, we know that 500 mg Calcium = 1 tablet
$1000 \cancel{\rm{mg}} \times \dfrac{1\; \rm{tablet}}{500\; \cancel{\rm{mg}}} = 2\; \rm{tablets} \nonumber$
Drug Dosage Calculation based on Body Weight. Many drugs (especially in children) are dosed according to body weight (mg/kg). These calculations are carried in 3-step conversions. The first step is to convert the body weight from pounds (lbs) to kg. The second step is to convert kg to mg (the total mg dose calculated based on body weight). And, finally, the mg dose is converted to the number of tablets. (as in Example 1.8.1).
Example $2$
1. Demerol is ordered 1.5mg/kg for a patient that is 220 lbs. Demerol is available as 50 mg per tablet. How many tablets should the nurse administer?
2. A doctor prescribes amoxicillin 30mg/kg to a child weighing 73.5 lbs. Amoxicillin is available as 500 mg tablets. How many tablets should the nurse administer?
Solution
a. We start with the given, 220 lbs. We want to change the unit from lbs to kg, and then, from kg to total dose (mg) and then the mg dose to tablets. The first conversion factor will cancel out the original unit, lbs, and introduce the unit we are converting to, which is kg. The second conversion factor will cancel out kg, and introduce the unit of the dose (usually mg) and then mg to tablet.
$220 \cancel{\rm{lbs}} \times \dfrac{1\cancel{\rm{kg}}}{2.2\; \cancel{\rm{lbs}}} \times \dfrac{1.5\cancel{\rm{mg}}}{1\; \cancel{ \rm{kg}}} \times \dfrac{1\; \rm{tablet}}{50\; \cancel{\rm{mg}}}= 3\; \rm{tablets} \nonumber$
Hence, the nurse administers 3 tablets.
b. Start with the given, 73.5 lbs. We want to change the unit from lbs to kg, and then, from kg to total dose (mg) and then mg to tablets. The first conversion factor will cancel out the original unit, lbs, and convert to kg. The second conversion factor will cancel out kg, and convert to the total mg dose and the final conversion will cancel mg to introduce the final unit, tablet.
\begin{align*} 73.5 \; \cancel{\rm{lbs}} \times \dfrac{1\cancel{\rm{kg}}}{2.2\; \cancel{\rm{lbs}}} \times \dfrac{30\cancel{\rm{mg}}}{1\; \cancel{ \rm{kg}}} \times \dfrac{1\; \rm{tablet}}{500\; \cancel{\rm{mg}}}&=2 \; \rm{tablets} \end{align*} \nonumber
Hence, the nurse administers 2 tablets.
Exercise $2$
Calculate each of the following.
1. Vancomycin is ordered 15mg/kg for a patient that is 110 lbs. Vancomycin is available as 250 mg per capsule. How many capsules should the nurse administer? (ans. 3 capsules)
2. A doctor prescribes ampicillin 40mg/kg to a patient weighing 55 lbs. Ampicillin is available as 500 mg tablets. How many tablets should the nurse administer? (ans. 2 tablets)
Answer a
$110 \cancel{\rm{lbs}} \times \dfrac{1\cancel{\rm{kg}}}{2.2\; \cancel{\rm{lbs}}} \times \dfrac{15\cancel{\rm{mg}}}{1\; \cancel{ \rm{kg}}} \times \dfrac{1\; \rm{tablet}}{250\; \cancel{\rm{mg}}}= 3\; \rm{tablets} \nonumber$
Answer b
$55 \cancel{\rm{lbs}} \times \dfrac{1\cancel{\rm{kg}}}{2.2\; \cancel{\rm{lbs}}} \times \dfrac{40\cancel{\rm{mg}}}{1\; \cancel{ \rm{kg}}} \times \dfrac{1\; \rm{tablet}}{500\; \cancel{\rm{mg}}}= 2\; \rm{tablets} \nonumber$
Contributors and Attributions
• Wikipedia
• Leticia Colmenares (UH-WCC) | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.08%3A_Dosage_Calculations.txt |
1. Does each statement refer to a chemical property or a physical property?
1. Balsa is a very light wood.
2. If held in a flame, magnesium metal burns in air.
3. Mercury has a density of 13.6 g/mL.
4. Human blood is red.
2. Does each statement refer to a chemical property or a physical property?
1. The elements sodium and chlorine can combine to make table salt.
2. The metal tungsten does not melt until its temperature exceeds 3,000°C.
3. The ingestion of ethyl alcohol can lead to disorientation and confusion.
4. The boiling point of isopropyl alcohol, which is used to sterilize cuts and scrapes, is lower than the boiling point of water.
3. Define element. How does it differ from a compound?
4. Define compound. How does it differ from an element?
5. Give two examples of a heterogeneous mixture.
6. Give two examples of a homogeneous mixture.
7. Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. xenon, a substance that cannot be broken down into chemically simpler components
2. blood, a substance composed of several types of cells suspended in a salty solution called plasma
3. water, a substance composed of hydrogen and oxygen
8. Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. sugar, a substance composed of carbon, hydrogen, and oxygen
2. hydrogen, the simplest chemical substance
3. dirt, a combination of rocks and decaying plant matter
9. Identify each substance as an element, a compound, a heterogeneous mixture, or a solution.
1. air, primarily a mixture of nitrogen and oxygen
2. ringer’s lactate, a standard fluid used in medicine that contains salt, potassium, and lactate compounds all dissolved in sterile water
3. tartaric acid, a substance composed of carbon, hydrogen, and oxygen
10. Identify each material as an element, a compound, a heterogeneous mixture, or a solution.
1. equal portions of salt and sand placed in a beaker and shaken up
2. a combination of beeswax dissolved in liquid hexane
3. hydrogen peroxide, a substance composed of hydrogen and oxygen
11. What word describes each phase change?
1. solid to liquid
2. liquid to gas
3. solid to gas
12. What word describes each phase change?
1. liquid to solid
2. gas to liquid
3. gas to solid
Answers
1.
1. physical property
2. chemical property
3. physical property
4. physical property
2.
1. chemical property
2. physical property
3. chemical property
4. physical property
1. An element is a substance that cannot be broken down into chemically simpler components. Compounds can be broken down into simpler substances.
4. A compound is composed of two or more elements combined in a fixed ratio. An element is the simplest chemical substance.
1. a salt and pepper mix and a bowl of cereal (answers will vary)
6. vinegar and rubbing alcohol (answers will vary)
1.
1. element
2. heterogeneous mixture
3. compound
8.
1. compound
2. element
3. heterogeneous mixture
1.
1. solution
2. solution
3. compound
10.
1. heterogeneous mixture
2. solution
3. compound
11.
• melting or fusion
• boiling or evaporation
• sublimation
12.
1. freezing
2. condensation
3. deposition | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.E%3A_Chemistry_Matter_and_Measurement_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Chemistry is the study of matter, which is anything that has mass and takes up space. Chemistry is one branch of science, which is the study of the natural universe. Like all branches of science, chemistry relies on the scientific method, which is a process of learning about the world around us. In the scientific method, a guess or hypothesis is tested through experiment and measurement.
Matter can be described in a number of ways. Physical properties describe characteristics of a sample that do not change the chemical identity of the material (size, shape, color, and so on), while chemical properties describe how a sample of matter changes its chemical composition. A substance is any material that has the same physical and chemical properties throughout. An element is a substance that cannot be broken down into chemically simpler components. The smallest chemically identifiable piece of an element is an atom. A substance that can be broken down into simpler chemical components is a compound. The smallest chemically identifiable piece of a compound is a molecule. Two or more substances combine physically to make a mixture. If the mixture is composed of discrete regions that maintain their own identity, the mixture is a heterogeneous mixture. If the mixture is so thoroughly mixed that the different components are evenly distributed throughout, it is a homogeneous mixture. Another name for a homogeneous mixture is a solution. Substances can also be described by their phase: solid, liquid, or gas.
Scientists learn about the universe by making measurements of quantities, which consist of numbers (how many) and units (of what). The numerical portion of a quantity can be expressed using scientific notation, which is based on powers, or exponents, of 10. Large numbers have positive powers of 10, while numbers less than 1 have negative powers of 10. The proper reporting of a measurement requires proper use of significant figures, which are all the known digits of a measurement plus the first estimated digit. The number of figures to report in the result of a calculation based on measured quantities depends on the numbers of significant figures in those quantities. For addition and subtraction, the number of significant figures is determined by position; for multiplication and division, it is decided by the number of significant figures in the original measured values. Nonsignificant digits are dropped from a final answer in accordance with the rules of rounding.
Chemistry uses SI, a system of units based on seven basic units. The most important ones for chemistry are the units for length, mass, amount, time, and temperature. Basic units can be combined with numerical prefixes to change the size of the units. They can also be combined with other units to make derived units, which are used to express other quantities such as volume, density, or energy. A formal conversion from one unit to another uses a conversion factor, which is constructed from the relationship between the two units. Numbers in conversion factors may affect the number of significant figures in a calculated quantity, depending on whether the conversion factor is exact. Conversion factors can be applied in separate computations, or several can be used at once in a single, longer computation. Conversion factors are very useful in calculating dosages. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/01%3A_Chemistry_Matter_and_Measurement/1.S%3A_Chemistry_Matter_and_Measurement_%28Summary%29.txt |
Just as a language has an alphabet from which words are built, chemistry has an alphabet from which matter is described. However, the chemical alphabet is larger than the one we use for spelling. You may have already figured out that the chemical alphabet consists of the chemical elements. Their role is central to chemistry, for they combine to form the millions and millions of known compounds.
• 2.0: Prelude to Elements, Atoms, and the Periodic Table
The hardest material in the human body is tooth enamel. It has to be hard so that our teeth can serve us for a lifetime of biting and chewing; however, tough as it is, tooth enamel is susceptible to chemical attack. Acids found in some foods or made by bacteria that feed on food residues on our teeth are capable of dissolving enamel. Unprotected by enamel, a tooth will start to decay, thus developing cavities and other dental problems.
• 2.1: The Elements
All matter is composed of elements. Chemical elements are represented by a one- or two-letter symbol.
• 2.2: Atomic Theory
Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter.
• 2.3: The Structure of Atoms
Atoms are composed of three main subatomic particles: protons, neutrons, and electrons. Protons and neutrons are grouped together in the nucleus of an atom, while electrons orbit about the nucleus.
• 2.4: Nuclei of Atoms
Elements can be identified by their atomic number and mass number. Isotopes are atoms of the same element that have different masses.
• 2.5: Atomic Masses
Atoms have a mass that is based largely on the number of protons and neutrons in their nucleus.
• 2.6: Arrangements of Electrons
Electrons are organized into shells and subshells about the nucleus of an atom.
• 2.7: The Periodic Table
The chemical elements are arranged in a chart called the periodic table. Some characteristics of the elements are related to their position on the periodic table.
• 2.E: Elements, Atoms, and the Periodic Table (Exercises)
These are homework exercises to accompany Chapter 2 of the Ball et al. "The Basics of GOB Chemistry" Textmap.
• 2.S: Elements, Atoms, and the Periodic Table (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms and ask yourself how they relate to the topics in the chapter.
Thumbnail: Ionization energies superimposed on a periodic table. (CC BY-NC-SA; anonymous by request).
02: Elements Atoms and the Periodic Table
The hardest material in the human body is tooth enamel. It has to be hard so that our teeth can serve us for a lifetime of biting and chewing; however, tough as it is, tooth enamel is susceptible to chemical attack. Acids found in some foods or made by bacteria that feed on food residues on our teeth are capable of dissolving enamel. Unprotected by enamel, a tooth will start to decay, thus developing cavities and other dental problems.
In the early 1900s, a dentist in Colorado Springs, Colorado, noted that many people who lived in the area had brown-stained teeth that, while unsightly, were surprisingly resistant to decay. After years of study, excess fluorine compounds in the drinking water were discovered to be the cause of both these effects. Research continued, and in the 1930s, the US Public Health Service found that low levels of fluorine in water would provide the benefit of resisting decay without discoloring teeth.
The protective effects of fluorine have a simple chemical explanation. Tooth enamel consists mostly of a mineral called hydroxyapatite, which is composed of calcium, phosphorus, oxygen, and hydrogen. We know now that fluorine combines with hydroxyapatite to make fluorapatite, which is more resistant to acidic decay than hydroxyapatite is. Currently about 50% of the US population drinks water that has some fluorine added (in the form of sodium fluoride, NaF) to reduce tooth decay. This intentional fluoridation, coupled with the use of fluoride-containing toothpastes and improved oral hygiene, has reduced tooth decay by as much as 60% in children. The nationwide reduction of tooth decay has been cited as an important public health advance in history. (Another important advance was the eradication of polio.) | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.00%3A_Prelude_to_Elements_Atoms_and_the_Periodic_Table.txt |
Learning Objectives
• Define a chemical element and give examples of the abundance of different elements.
• Represent a chemical element with a chemical symbol.
An element is a substance that cannot be broken down into simpler chemical substances. There are about 90 naturally occurring elements known on Earth. Using technology, scientists have been able to create nearly 30 additional elements that do not occur in nature. Today, chemistry recognizes 118 elements—some of which were created an atom at a time. Figure \(1\) shows some of the chemical elements.
Abundance
The elements vary widely in abundance. In the universe as a whole, the most common element is hydrogen (about 90% of atoms), followed by helium (most of the remaining 10%). All other elements are present in relatively minuscule amounts, as far as we can detect.
Table \(1\): Elemental Composition of Earth
Earth’s Crust Earth (overall)
Element Percentage Element Percentage
oxygen 46.1 iron 34.6
silicon 28.2 oxygen 29.5
aluminum 8.23 silicon 15.2
iron 5.53 magnesium 12.7
calcium 4.15 nickel 2.4
sodium 2.36 sulfur 1.9
magnesium 2.33 all others 3.7
potassium 2.09
titanium 0.565
hydrogen 0.14
phosphorus 0.105
all others 0.174
Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 14–17.
On the planet Earth, however, the situation is rather different. Oxygen makes up 46.1% of the mass of Earth’s crust (the relatively thin layer of rock forming Earth’s surface), mostly in combination with other elements, while silicon makes up 28.2%. Hydrogen, the most abundant element in the universe, makes up only 0.14% of Earth’s crust. Table \(1\) lists the relative abundances of elements on Earth as a whole and in Earth’s crust. Table \(2\) lists the relative abundances of elements in the human body. If you compare Table \(1\) and Table \(2\), you will find disparities between the percentage of each element in the human body and on Earth. Oxygen has the highest percentage in both cases, but carbon, the element with the second highest percentage in the body, is relatively rare on Earth and does not even appear as a separate entry in Table \(1\); carbon is part of the 0.174% representing “other” elements. How does the human body concentrate so many apparently rare elements?
Table \(2\): Elemental Composition of a Human Body
Element Percentage by Mass
oxygen 61
carbon 23
hydrogen 10
nitrogen 2.6
calcium 1.4
phosphorus 1.1
sulfur 0.20
potassium 0.20
sodium 0.14
chlorine 0.12
magnesium 0.027
silicon 0.026
iron 0.006
fluorine 0.0037
zinc 0.0033
all others 0.174
Source: D. R. Lide, ed. CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008–9), 7–24.
The relative amounts of elements in the body have less to do with their abundances on Earth than with their availability in a form we can assimilate. We obtain oxygen from the air we breathe and the water we drink. We also obtain hydrogen from water. On the other hand, although carbon is present in the atmosphere as carbon dioxide, and about 80% of the atmosphere is nitrogen, we obtain those two elements from the food we eat, not the air we breathe.
LOOKING CLOSER: PHOSPHOROUS, THE CHEMICAL BOTTLENECK
There is an element that we need more of in our bodies than is proportionately present in Earth’s crust, and this element is not easily accessible. Phosphorus makes up 1.1% of the human body but only 0.105% of Earth’s crust. We need phosphorus for our bones and teeth, and it is a crucial component of all living cells. Unlike carbon, which can be obtained from carbon dioxide, there is no phosphorus compound present in our surroundings that can serve as a convenient source. Phosphorus, then, is nature’s bottleneck. Its availability limits the amount of life our planet can sustain.
Higher forms of life, such as humans, can obtain phosphorus by selecting a proper diet (plenty of protein); but lower forms of life, such as algae, must absorb it from the environment. When phosphorus-containing detergents were introduced in the 1950s, wastewater from normal household activities greatly increased the amount of phosphorus available to algae and other plant life. Lakes receiving this wastewater experienced sudden increases in growth of algae. When the algae died, concentrations of bacteria that ate the dead algae increased. Because of the large bacterial concentrations, the oxygen content of the water dropped, causing fish to die in large numbers. This process, called eutrophication, is considered a negative environmental impact.
Today, many detergents are made without phosphorus so the detrimental effects of eutrophication are minimized. You may even see statements to that effect on detergent boxes. It can be sobering to realize how much impact a single element can have on life—or the ease with which human activity can affect the environment.
Names and Symbols
Each element has a name. Some of these names date from antiquity, while others are quite new. Today, the names for new elements are proposed by their discoverers but must be approved by the International Union of Pure and Applied Chemistry, an international organization that makes recommendations concerning all kinds of chemical terminology.
Today, new elements are usually named after famous scientists.
The names of the elements can be cumbersome to write in full, especially when combined to form the names of compounds. Therefore, each element name is abbreviated as a one- or two-letter chemical symbol. By convention, the first letter of a chemical symbol is a capital letter, while the second letter (if there is one) is a lowercase letter. The first letter of the symbol is usually the first letter of the element’s name, while the second letter is some other letter from the name. Some elements have symbols that derive from earlier, mostly Latin names, so the symbols may not contain any letters from the English name. Table \(3\) lists the names and symbols of some of the most familiar elements.
Table \(3\): Element Names and Symbols
aluminum Al magnesium Mg
argon Ar manganese Mn
arsenic As mercury Hg*
barium Ba neon Ne
bismuth Bi nickel Ni
boron B nitrogen N
bromine Br oxygen O
calcium Ca phosphorus P
carbon C platinum Pt
chlorine Cl potassium K*
chromium Cr silicon Si
copper Cu* silver Ag*
fluorine F sodium Na*
gold Au* strontium Sr
helium He sulfur S
hydrogen H tin Sn*
iron Fe tungsten W
iodine I uranium U
lead Pb* zinc Zn
lithium Li zirconium Zr
*The symbol comes from the Latin name of element.
The symbol for tungsten comes from its German name—wolfram.
Element names in languages other than English are often close to their Latin names. For example, gold is oro in Spanish and or in French (close to the Latin aurum), tin is estaño in Spanish (compare to stannum), lead is plomo in Spanish and plomb in French (compare to plumbum), silver is argent in French (compare to argentum), and iron is fer in French and hierro in Spanish (compare to ferrum). The closeness is even more apparent in pronunciation than in spelling.
Example \(1\)
Write the chemical symbol for each element without consulting Table \(3\) "Element Names and Symbols".
1. bromine
2. boron
3. carbon
4. calcium
5. gold
Answer a
Br
Answer b
B
Answer c
C
Answer d
Ca
Answer e
Au
Exercise \(1\)
Write the chemical symbol for each element without consulting Table \(3\).
1. manganese
2. magnesium
3. neon
4. nitrogen
5. silver
Answer a
Mn
Answer b
Mg
Answer c
Ne
Answer d
N
Answer e
Ag
Example \(2\)
What element is represented by each chemical symbol?
1. Na
2. Hg
3. P
4. K
5. I
Answer a
sodium
Answer b
mercury
Answer c
phosphorus
Answer d
potassium
Answer e
iodine
Exercise \(2\)
What element is represented by each chemical symbol?
1. Pb
2. Sn
3. U
4. O
5. F
Answer a
lead
Answer b
tin
Answer c
uranium
Answer d
oxygen
Answer e
fluorine
Concept Review Exercises
1. What is an element?
2. Give some examples of how the abundance of elements varies.
3. Why are chemical symbols so useful? What is the source of the letter(s) for a chemical symbol?
Answers
1. An element is the basic chemical building block of matter; it is the simplest chemical substance.
2. Elements vary from being a small percentage to more than 30% of the atoms around us.
3. Chemical symbols are useful to concisely represent the elements present in a substance. The letters usually come from the name of the element.
Key Takeaways
• All matter is composed of elements.
• Chemical elements are represented by a one- or two-letter symbol. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.01%3A_The_Elements.txt |
Learning Objectives
• Explain how all matter is composed of atoms.
• Describe the modern atomic theory.
• Recognize which elements exist as diatomic molecules.
Take some aluminum foil. Cut it in half. Now you have 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 bec
/
/
'./oming smaller and smaller. But how far can you take this exercise, at least in theory? Can you 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.)
The modern atomic theory, proposed about 1803 by the English chemist John Dalton (Figure \(1\)), is a fundamental concept that states that all elements are composed of atoms. Previously, we defined an atom 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 your little finger (about 1 cm).
Most elements in their pure form exist as individual atoms. For example, a macroscopic chunk of iron (Fe) metal is composed, microscopically, of individual atoms. Some elements, however, exist as groups of atoms called molecules. Several important elements exist as two-atom combinations and are called diatomic molecules. In representing a diatomic molecule, we use the symbol of the element and include the subscript "2" to indicate that two atoms of that element are joined together. The elements that exist as diatomic molecules are hydrogen (H2), nitrogen (N2), fluorine (F2), oxygen (O2), iodine (I2), chlorine (Cl2) and bromine (Br2). A few other elements can exist as 3-atom molecules like ozone (O3) and 4-atom molecules like phosphorus (P4). The most common form of the element sulfur is composed of molecules that consist of eight atoms of sulfur; its molecular formula is S8 (Figure \(2\)).
It is important to note that a subscript following a symbol and a number in front of a symbol do not represent the same thing; for example, H2 and 2H represent distinctly different species. H2 is a molecular formula; it represents a diatomic molecule of hydrogen, consisting of two atoms of the element that are chemically bonded together. The expression 2H, on the other hand, indicates two separate hydrogen atoms that are not combined as a unit. The expression 2H2 represents two molecules of diatomic hydrogen (Figure \(3\)).
Example \(1\)
Write the chemical formula of the following elements:
1. oxygen
2. carbon
3. potassium
Solution
Memorizing the diatomic molecules is worthwhile for our future endeavors. A mnemonic device to help in the memorization of the diatomic elements is as follows: Have No Fear Of Ice Cold Beer. (hydrogen, nitrogen, fluorine, oxygen, iodine, chlorine and bromine).
1. Oxygen is one of the seven diatomic molecular elements. Its chemical formula is O2.
2. Carbon is monatomic, hence its formula is C.
3. Potassium is monatomic hence its formula is K.
Exercise \(1\)
Write the chemical formula of the following elements:
1. hydrogen
2. nitrogen
3. neon
Answer a
Hydrogen is one of the seven diatomic molecular elements. Its chemical formula is H2.
Answer b
Nitrogen is one of the seven diatomic molecular elements. Its chemical formula is N2.
Answer c
Neon is a monatomic element, hence its formula is Ne.
Looking Closer: Atomic Theory
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. So 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.
Concept Review Exercises
1. What is the modern atomic theory?
2. What are atoms?
Answers
1. The modern atomic theory states that all matter is composed of atoms.
2. Atoms are the smallest parts of an element that maintain the identity of that element.
Key Takeaways
• Atoms are the ultimate building blocks of all matter.
• The modern atomic theory establishes the concepts of atoms and how they compose matter.
• The diatomic elements are hydrogen, nitrogen, fluorine, oxygen, iodine, chlorine and bromine. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.02%3A_Atomic_Theory.txt |
Learning Objectives
• Describe the three main subatomic particles.
• State how the subatomic particles are arranged in atoms.
There have been several minor but important modifications to Dalton’s atomic theory. For one thing, Dalton considered atoms to be indivisible. We know now that atoms not only can be divided but also are composed of three different kinds of particle, subatomic particles, with their own properties that are different from the chemical properties of atoms.
Subatomic Particles
The first subatomic particle was identified in 1897 and called the electron. It is an extremely tiny particle, with a mass of about 9.109 × 10−31 kg. Experiments with magnetic fields showed that the electron has a negative electrical charge. By 1920, experimental evidence indicated the existence of a second particle. A proton has the same amount of charge as an electron, but its charge is positive, not negative. Another major difference between a proton and an electron is mass. Although still incredibly small, the mass of a proton is 1.673 × 10−27 kg, which is almost 2,000 times greater than the mass of an electron. Because opposite charges attract each other (while like charges repel each other), protons attract electrons (and vice versa).
Finally, additional experiments pointed to the existence of a third particle. Evidence produced in 1932 established the existence of the neutron, a particle with about the same mass as a proton but with no electrical charge, it is neutral. We understand now that all atoms can be broken down into subatomic particles: protons, neutrons, and electrons. Table \(1\) lists some of their important characteristics and the symbols used to represent each particle.
Table \(1\): Properties of the Subatomic Particles
Particle Symbol Mass (kg) Relative Mass (proton = 1) Relative Charge
proton p+ 1.673 × 10−27 1 +1
neutron n0 1.675 × 10−27 1 0
electron e 9.109 × 10−31 0.00055 −1
The Nucleus
How are these subatomic particles arranged? Between 1909 and 1911, Ernest Rutherford, a Cambridge physicist, and his associates Hans Geiger and Ernest Marsden performed experiments that provided strong evidence concerning the internal structure of an atom. They took a very thin metal foil, such as gold or platinum, and aimed a beam of positively charged particles (called alpha particles, which are combinations of two protons and two neutrons) from a radioactive source toward the foil. Surrounding the foil was a detector—either a scintillator (a material that glows when hit by such particles) or some unexposed film (which is exposed where the particles hit it). The detector allowed the scientists to determine the distribution of the alpha particles after they interacted with the foil. Figure \(1\) shows a diagram of the experimental setup.
Rutherford proposed the following model to explain these experimental results. Protons and neutrons are concentrated in a central region he called the nucleus (plural, nuclei) of the atom. Electrons are outside the nucleus and orbit about it because they are attracted to the positive charge in the nucleus. Most of the mass of an atom is in the nucleus, while the orbiting electrons account for an atom’s size. As a result, an atom consists largely of empty space. Rutherford called his description the “planetary model” of the atom. Figure \(2\) shows how this model explains the experimental results.
The planetary model of the atom replaced the plum pudding model, which had electrons floating around aimlessly like plums in a “pudding” of positive charge.
Rutherford’s model is essentially the same model that we use today to describe atoms but with one important modification. The planetary model suggests that electrons occupy certain specific, circular orbits about the nucleus. We know now that this model is overly simplistic. A better description is that electrons form fuzzy clouds around nuclei. Figure \(3\) shows a more modern version of our understanding of atomic structure.
Concept Review Exercises
1. What are the charges and the relative masses of the three subatomic particles?
2. Describe the structure of an atom in terms of its protons, neutrons, and electrons.
Answers
1. proton: +1, large; neutron: 0, large; electron: −1, small
2. Protons and neutrons are located in a central nucleus, while electrons orbit about the nucleus.
Key Takeaways
• Atoms are composed of three main subatomic particles: protons, neutrons, and electrons.
• Protons and neutrons are grouped together in the nucleus of an atom, while electrons orbit about the nucleus. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.03%3A_The_Structure_of_Atoms.txt |
Learning Objectives
• Define and differentiate between the atomic number and the mass number of an element.
• Explain how isotopes differ from one another.
Now that we know how atoms are generally constructed, what do atoms of any particular element look like? How many protons, neutrons, and electrons are in a specific kind of atom? First, if an atom is electrically neutral overall, then the number of protons equals the number of electrons. Because these particles have the same but opposite charges, equal numbers cancel out, producing a neutral atom.
Atomic Number
In the 1910s, experiments with x-rays led to this useful conclusion: the magnitude of the positive charge in the nucleus of every atom of a particular element is the same. In other words, all atoms of the same element have the same number of protons. Furthermore, different elements have a different number of protons in their nuclei, so the number of protons in the nucleus of an atom is characteristic of a particular element. This discovery was so important to our understanding of atoms that the number of protons in the nucleus of an atom is called the atomic number (Z).
For example, hydrogen has the atomic number 1; all hydrogen atoms have 1 proton in their nuclei. Helium has the atomic number 2; all helium atoms have 2 protons in their nuclei. There is no such thing as a hydrogen atom with 2 protons in its nucleus; a nucleus with 2 protons would be a helium atom. The atomic number defines an element. Table $1$ lists some common elements and their atomic numbers. Based on its atomic number, you can determine the number of protons in the nucleus of an atom. The largest atoms have over 100 protons in their nuclei.
Table $1$: Some Common Elements and Their Atomic Numbers
Element Atomic Number Element Atomic Nmbers
aluminum (Al) 13 magnesium (Mg) 12
americium (Am) 95 manganese (Mn) 25
argon (Ar) 18 mercury (Hg) 80
barium (Ba) 56 neon (Ne) 10
beryllium (Be) 4 nickel (Ni) 28
bromine (Br) 35 nitrogen (N) 7
calcium (Ca) 20 oxygen (O) 8
carbon (C) 6 phosphorus (P) 15
chlorine (Cl) 17 platinum (Pt) 78
chromium (Cr) 24 potassium (K) 19
cesium (Cs) 55 radon (Rn) 86
fluorine (F) 9 silver (Ag) 47
gallium (Ga) 31 sodium (Na) 11
gold (Au) 79 strontium (Sr) 38
helium (He) 2 sulfur (S) 16
hydrogen (H) 1 titanium (Ti) 22
iron (Fe) 26 tungsten (W) 74
iodine (I) 53 uranium (U) 92
lead (Pb) 82 zinc (Zn) 30
lithium (Li) 3 zirconium (Zr) 40
Example $1$
What is the number of protons in the nucleus of each element?
1. aluminum
2. iron
3. carbon
Answer a
According to Table 2.4.1, aluminum has an atomic number of 13. Therefore, every aluminum atom has 13 protons in its nucleus.
Answer b
Iron has an atomic number of 26. Therefore, every iron atom has 26 protons in its nucleus.
Answer c
Carbon has an atomic number of 6. Therefore, every carbon atom has 6 protons in its nucleus.
Exercise $1$
What is the number of protons in the nucleus of each element? Use Table 2.4.1.
1. sodium
2. oxygen
3. chlorine
Answer a
Sodium has 11 protons in its nucleus.
Answer b
Oxygen has 8 protons in its nucleus.
Answer c
Chlorine has 17 protons in its nucleus
How many electrons are in an atom? Previously we said that for an electrically neutral atom, the number of electrons equals the number of protons, so the total opposite charges cancel. Thus, the atomic number of an element also gives the number of electrons in an atom of that element. (Later we will find that some elements may gain or lose electrons from their atoms, so those atoms will no longer be electrically neutral. Thus we will need a way to differentiate the number of electrons for those elements.)
Example $2$
How many electrons are present in the atoms of each element?
1. sulfur
2. tungsten
3. argon
Answer a
The atomic number of sulfur is 16. Therefore, in a neutral atom of sulfur, there are 16 electrons.
Answer b
The atomic number of tungsten is 74. Therefore, in a neutral atom of tungsten, there are 74 electrons.
Answer c
The atomic number of argon is 18. Therefore, in a neutral atom of argon, there are 18 electrons.
Exercise $2$
How many electrons are present in the atoms of each element?
1. magnesium
2. potassium
3. iodine
Answer a
Mg has 12 electrons.
Answer b
K has 19 electrons.
Answer c
I has 53 electrons.
Isotopes
How many neutrons are in atoms of a particular element? At first it was thought that the number of neutrons in a nucleus was also characteristic of an element. However, it was found that atoms of the same element can have different numbers of neutrons. Atoms of the same element (i.e., same atomic number, Z) that have different numbers of neutrons are called isotopes. For example, 99% of the carbon atoms on Earth have 6 neutrons and 6 protons in their nuclei; about 1% of the carbon atoms have 7 neutrons in their nuclei. Naturally occurring carbon on Earth, therefore, is actually a mixture of isotopes, albeit a mixture that is 99% carbon with 6 neutrons in each nucleus.
An important series of isotopes is found with hydrogen atoms. Most hydrogen atoms have a nucleus with only a single proton. About 1 in 10,000 hydrogen nuclei, however, also has a neutron; this particular isotope is called deuterium. An extremely rare hydrogen isotope, tritium, has 1 proton and 2 neutrons in its nucleus. Figure $1$ compares the three isotopes of hydrogen.
The discovery of isotopes required a minor change in Dalton’s atomic theory. Dalton thought that all atoms of the same element were exactly the same.
Most elements exist as mixtures of isotopes. In fact, there are currently over 3,500 isotopes known for all the elements. When scientists discuss individual isotopes, they need an efficient way to specify the number of neutrons in any particular nucleus. The mass number (A) of an atom is the sum of the numbers of protons and neutrons in the nucleus. Given the mass number for a nucleus (and knowing the atomic number of that particular atom), you can determine the number of neutrons by subtracting the atomic number from the mass number.
A simple way of indicating the mass number of a particular isotope is to list it as a superscript on the left side of an element’s symbol. Atomic numbers are often listed as a subscript on the left side of an element’s symbol. Thus, we might see
$\mathrm{^{mass\: number\xrightarrow{\hspace{45px}} 56}_{atomic\: number \xrightarrow{\hspace{35px}} 26}Fe} \label{Eq1}$
which indicates a particular isotope of iron. The 26 is the atomic number (which is the same for all iron atoms), while the 56 is the mass number of the isotope. To determine the number of neutrons in this isotope, we subtract 26 from 56: 56 − 26 = 30, so there are 30 neutrons in this atom.
Example $3$
How many protons and neutrons are in each atom?
1. $\mathrm{^{35}_{17}Cl}$
2. $\mathrm{^{127}_{53}I}$
Answer a
In $\mathrm{^{35}_{17}Cl}$ there are 17 protons, and 35 − 17 = 18 neutrons in each nucleus.
Answer b
In $\mathrm{^{127}_{53}I}$ there are 53 protons, and 127 − 53 = 74 neutrons in each nucleus.
Exercise $3$
How many protons and neutrons are in each atom?
1. $\mathrm{^{197}_{79}Au}$
2. $\mathrm{^{23}_{11}Na}$
Answer a
In $\mathrm{^{197}_{79}Au}$ there are 79 protons, and 197 − 79 = 118 neutrons in each nucleus.
Answer b
In $\mathrm{^{23}_{11}Na}$ there are 11 protons, and 23 − 11 = 12 neutrons in each nucleus.
It is not absolutely necessary to indicate the atomic number as a subscript because each element has its own unique atomic number. Many isotopes are indicated with a superscript only, such as 13C or 235U. You may also see isotopes represented in print as, for example, carbon-13 or uranium-235.
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.
Almost all of the mass of an atom is from the total protons and neutrons contained within a tiny (and therefore very dense) nucleus. The majority of the volume of an atom is the surrounding space in which the electrons reside. A representation of a carbon-12 atom is shown below in Figure $2$.
Concept Review Exercises
1. Why is the atomic number so important to the identity of an atom?
2. What is the relationship between the number of protons and the number of electrons in an atom?
3. How do isotopes of an element differ from each other?
4. What is the mass number of an element?
Answers
1. The atomic number defines the identity of an element. It describes the number of protons in the nucleus.
2. In an electrically neutral atom, the number of protons equals the number of electrons.
3. Isotopes of an element have the same number of protons but have different numbers of neutrons in their nuclei.
4. The mass number is the sum of the numbers of protons and neutrons in the nucleus of an atom.
Key Takeaways
• Each element is identified by its atomic number. The atomic number provides the element's location on the periodic table
• The isotopes of an element have different masses and are identified by their mass numbers. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.04%3A_Nuclei_of_Atoms.txt |
Learning Objectives
• Define atomic mass and atomic mass unit.
• Calculate atomic mass.
Even though atoms are very tiny pieces of matter, they have mass. Their masses are so small, however, that chemists often use a unit other than grams to express them—the atomic mass unit.
Atomic Mass Unit
The atomic mass unit (abbreviated u, although amu is also used) is defined as 1/12 of the mass of a 12C atom:
$\mathrm{1\:u=\dfrac{1}{12}\textrm{ the mass of }^{12}C\:atom} \label{Eq1}$
It is equal to 1.661 × 10−24 g.
Masses of other atoms are expressed with respect to the atomic mass unit. For example, the mass of an atom of 1H is 1.008 u, the mass of an atom of 16O is 15.995 u, and the mass of an atom of 32S is 31.97 u. Note, however, that these masses are for particular isotopes of each element. Because most elements exist in nature as a mixture of isotopes, any sample of an element will actually be a mixture of atoms having slightly different masses (because neutrons have a significant effect on an atom’s mass). How, then, do we describe the mass of a given element? By calculating an average of an element’s atomic masses, weighted by the natural abundance of each isotope, we obtain a weighted average mass called the atomic mass (also commonly referred to as the atomic weight) of an element.
Atomic Mass is the Weighted Average Mass of Isotopes
As stated above, most elements occur naturally as a mixture of two or more isotopes. Listed below (Table $1$) are the naturally occurring isotopes of selected elements along with the percent natural abundance of each.
Element Isotope (Symbol) Percent Natural Abundance Atomic Mass $\left( \text{amu} \right)$ Average Atomic Mass $\left( \text{amu} \right)$
Table $1$: Atomic Masses and Percent Abundances of Some Natural Isotopes
Hydrogen $\ce{_1^1H}$ 99.985 1.0078 1.0079
$\ce{_1^2H}$ 0.015 2.0141
$\ce{_1^3H}$ negligible 3.0160
Carbon $\ce{_6^{12}C}$ 98.89 12.000 12.011
$\ce{_6^{13}C}$ 1.11 13.003
$\ce{_6^{14}C}$ trace 14.003
Oxygen $\ce{_8^{16}O}$ 99.759 15.995 15.999
$\ce{_8^{17}O}$ 0.037 16.995
$\ce{_8^{18}O}$ 0.204 17.999
Chlorine $\ce{_{17}^{35}Cl}$ 75.77 34.969 35.453
$\ce{_{17}^{38}Cl}$ 24.23 36.966
Copper $\ce{_{29}^{63}Cu}$ 69.17 62.930 63.546
$\ce{_{29}^{65}Cu}$ 30.83 64.928
For some elements, one particular isotope is much more abundant than any other isotopes. For example, naturally occurring hydrogen is nearly all hydrogen-1, and naturally occurring oxygen is nearly all oxygen-16. For many other elements, however, more than one isotope may exist in substantial quantities. Chlorine (atomic number 17) is yellowish-green toxic gas. About three quarters of all chlorine atoms have 18 neutrons, giving those atoms a mass number of 35. About one quarter of all chlorine atoms have 20 neutrons, giving those atoms a mass number of 37. Were you to simply calculate the arithmetic average of the precise atomic masses, you would get approximately 36.
$\frac{34.969 \,u + 36.966 \,u}{2} = 35.968 \,u \nonumber$
As you can see, the average atomic mass given in the last column of the table above (35.453) is significantly lower. Why? The reason is that we need to take into account the natural abundance percentages of each isotope in order to calculate what is called the weighted average. The atomic mass of an element is the weighted average of the atomic masses of the naturally occurring isotopes of that element.
$0.7577 \left( 34.969 \,u \right) + 0.2423 \left( 36.966 \,u \right) = 35.453 \,u \nonumber$
The weighted average is determined by multiplying the percent of natural abundance by the actual mass of the isotope. This is repeated until there is a term for each isotope. For chlorine, there are only two naturally occurring isotopes so there are only two terms.
Atomic mass = (%1)(mass 1) + (%2)(mass 2) + ⋯
Another example: oxygen exists as a mixture that is 99.759% 16O, 0.037% 17O and 0.204% 18O. The atomic mass of oxygen (use percent natural abundance data from Table 2.5.1) would be calculated as follows:
Atomic mass = (%1)(mass 1) + (%2)(mass 2) + (%3)(mass 3)
$0.99759 \left( 15.995 u \right) + 0.00037 \left( 16.995 u \right) +0.00204 \left( 17.999 u \right)= 15.999 u \nonumber$
To confirm your answer, compare the calculated value to the weighted mass displayed on the periodic table.
Example $1$
Calculate the atomic mass of oxygen. Oxygen exists as a mixture of 3 isotopes. Their respective masses and natural abundance are shown below.
• 16O: 15.995 u (99.759%)
• 17O: 16.995 u (0.037%)
• 18O: 17.999 u (0.204%)
Solution
Multiply the isotope abundance by the actual mass of the isotope, and then sum up the products.
$0.99759 \left( 15.995\, u \right) + 0.00037 \left( 16.995 \,u \right) +0.00204 \left( 17.999\, u \right)= 15.999\, u \nonumber$
Exercise $1$
Calculate the atomic mass of copper. Copper exists as a mixture of 2 isotopes. Their respective masses and natural abundance are shown below.
• 63Cu: 62.930 u (69.17%)
• 65Cu: 64.928 u (30.83%)
Answer
63.546 u
The atomic mass of each element is found under the element symbol in the periodic table. Examples are shown below. The atomic mass of tin (Sn) is 118.71 u while the atomic mass of carbon (C) is 12.011 u. On the other hand, the atomic number (Z) of each element is found above the atomic symbol.
Atomic mass indicated on entries of the Periodic Table. (public Domain; Pubchem)
The periodic table is found in this link:
https://pubchem.ncbi.nlm.nih.gov/periodic-table/png/Periodic_Table_of_Elements_w_Atomic_Mass_PubChem.png
Example $2$: Mass of Carbon
What is the average mass of a carbon atom in grams? The atomic mass is found in the Periodic Table. Please use two decimal places.
Solution
This is a simple one-step conversion, similar to conversions we did in Chapter 1. We use the fact that 1 u = 1.661 × 10−24 g:
$\mathrm{12.01\:\cancel{u}\times\dfrac{1.661 \times 10^{-24}\:g}{1\:\cancel{u}}=1.995\times 10^{-23}\:g}$
This is an extremely small mass, which illustrates just how small individual atoms are.
Exercise $2$: Mass of Tin
What is the average mass of a tin atom in grams? The average atomic mass is found in the Periodic Table. Please use two decimal places.
Answer
$\mathrm{118.71\:\cancel{u}\times\dfrac{1.661\times10^{-24}\:g}{1\:\cancel{u}}=1.972 \times 10^{-22}\:g}$
Concept Review Exercises
1. Define atomic mass. Why is it considered a weighted average?
2. What is an atomic mass unit?
Answers
1. The atomic mass is an average of an element’s atomic masses, weighted by the natural abundance of each isotope of that element. It is a weighted average because different isotopes have different masses.
2. An atomic mass unit is 1/12th of the mass of a 12C atom.
Key Takeaway
• Atoms have a mass that is based largely on the number of protons and neutrons in their nucleus.
• The atomic mass of each element in the Periodic Table is the weighted average of the mass of all its isotopes. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.05%3A_Atomic_Masses.txt |
Learning Objectives
• Describe how electrons are grouped within atoms.
Although we have discussed the general arrangement of subatomic particles in atoms, we have said little about how electrons occupy the space about the nucleus. Do they move around the nucleus at random, or do they exist in some ordered arrangement?
The modern theory of electron behavior is called quantum mechanics. It makes the following statements about electrons in atoms:
• Electrons in atoms can have only certain specific energies. We say that the energies of the electrons are quantized.
• Electrons are organized according to their energies into sets called shells (labeled by the principle quantum number, n). Generally the higher the energy of a shell, the farther it is (on average) from the nucleus. Shells do not have specific, fixed distances from the nucleus, but an electron in a higher-energy shell will spend more time farther from the nucleus than does an electron in a lower-energy shell.
• Shells are further divided into subsets of electrons called subshells. The first shell has only one subshell, the second shell has two subshells, the third shell has three subshells, and so on. The subshells of each shell are labeled, in order, with the letters s, p, d, and f. Thus, the first shell has only a single s subshell (called 1s), the second shell has 2s and 2p subshells, the third shell has 3s, 3p, and 3d and so forth.
Table $1$: Shells and Subshells
Shell Number of Subshells Names of Subshells
1 1 1s
2 2 2s and 2p
3 3 3s, 3p and 3d
4 4 4s, 4p, 4d and 4f
• Different subshells hold a different maximum number of electrons. Any s subshell can hold up to 2 electrons; p, 6; d, 10; and f, 14.
Table $2$: Number of Electrons
Subshell Maximum Number of Electrons
s 2
p 6
d 10
f 14
It is the arrangement of electrons into shells and subshells that most concerns us here, so we will focus on that.
We use numbers to indicate which shell an electron is in. As shown in Table $1$, the first shell, closest to the nucleus and with the lowest-energy electrons, is shell 1. This first shell has only one subshell, which is labeled 1s and can hold a maximum of 2 electrons. We combine the shell and subshell labels when referring to the organization of electrons about a nucleus and use a superscript to indicate how many electrons are in a subshell. Thus, because a hydrogen atom has its single electron in the s subshell of the first shell, we use 1s1 to describe the electronic structure of hydrogen. This structure is called an electron configuration. Electron configurations are shorthand descriptions of the arrangements of electrons in atoms. The electron configuration of a hydrogen atom is spoken out loud as “one-ess-one.”
Helium atoms have 2 electrons. Both electrons fit into the 1s subshell because s subshells can hold up to 2 electrons; therefore, the electron configuration for helium atoms is 1s2 (spoken as “one-ess-two”).
The 1s subshell cannot hold 3 electrons (because an s subshell can hold a maximum of 2 electrons), so the electron configuration for a lithium atom cannot be 1s3. Two of the lithium electrons can fit into the 1s subshell, but the third electron must go into the second shell. The second shell has two subshells, s and p, which fill with electrons in that order. The 2s subshell holds a maximum of 2 electrons, and the 2p subshell holds a maximum of 6 electrons. Because lithium’s final electron goes into the 2s subshell, we write the electron configuration of a lithium atom as 1s22s1. The shell diagram for a lithium atom is shown below. The shell closest to the nucleus (first shell) has 2 dots representing the 2 electrons in 1s, while the outermost shell (2s) has 1 electron.
The next largest atom, beryllium, has 4 electrons, so its electron configuration is 1s22s2. Now that the 2s subshell is filled, electrons in larger atoms start filling the 2p subshell. Thus, the electron configurations for the next six atoms are as follows:
• B: 1s22s22p1
• C: 1s22s22p2
• N: 1s22s22p3
• O: 1s22s22p4
• F: 1s22s22p5
• Ne: 1s22s22p6
With neon, the 2p subshell is completely filled. Because the second shell has only two subshells, atoms with more electrons now must begin the third shell. The third shell has three subshells, labeled s, p, and d. The d subshell can hold a maximum of 10 electrons. The first two subshells of the third shell are filled in order—for example, the electron configuration of aluminum, with 13 electrons, is 1s22s22p63s23p1. However, a curious thing happens after the 3p subshell is filled: the 4s subshell begins to fill before the 3d subshell does. In fact, the exact ordering of subshells becomes more complicated at this point (after argon, with its 18 electrons), so we will not consider the electron configurations of larger atoms. A fourth subshell, the f subshell, is needed to complete the electron configurations for all elements. An f subshell can hold up to 14 electrons.
Electron filling always starts with 1s, the subshell closest to the nucleus. Next is 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, etc., shown in the electron shell filling order diagram in Figure $2$. Follow each arrow in order from top to bottom. The subshells you reach along each arrow give the ordering of filling of subshells in larger atoms.
Example $1$: Electronic Configuration of Phosphorus Atoms
Using Figure $2$ as your guide, write the electron configuration of a neutral phosphorus atom. The atomic number of P is 15.
Solution
A neutral phosphorus atom has 15 electrons. Two electrons can go into the 1s subshell, 2 can go into the 2s subshell, and 6 can go into the 2p subshell. That leaves 5 electrons. Of those 5 electrons, 2 can go into the 3s subshell, and the remaining 3 electrons can go into the 3p subshell. Thus, the electron configuration of neutral phosphorus atoms is 1s22s22p63s23p3.
Exercise $1$: Electronic Configuration of Chlorine Atoms
Using Figure $2$ as your guide, write the electron configuration of a neutral chlorine atom. The atomic number of Cl is 17.
Answer
A neutral chlorine atom has 17 electrons. Two electrons can go into the 1s subshell, 2 can go into the 2s subshell, and 6 can go into the 2p subshell. That leaves 7 electrons. Of those 7 electrons, 2 can go into the 3s subshell, and the remaining 5 electrons can go into the 3p subshell. Thus, the electron configuration of neutral chlorine atoms is 1s22s22p63s23p5.
Since the arrangement of the periodic table is based on the electron configurations, Figure $3$ provides an alternative method for determining the electron configuration. The filling order simply begins at the top left, with hydrogen (Z=1) and includes each subshell as you proceed in increasing atomic number (Z) order.
For example, the first row (Period 1) contains H and He only, because only two electrons are required to fill the 1s subshell. The second row s-block, contains only two elements, Li and Be, to fill the 2s subshell. This is followed by the second row p-block, containing 6 elements (B through Ne) since six electrons are required to fill the 2p subshell. The third row is similar to the second row elements. Two electrons are needed (Na and Mg) to fill the 3s subshell and six electrons are required (Al through Ar) to complete the 3p subshell. After filling the 3p block up to Ar, we see the next subshell will be 4s (K, Ca), followed by the 3d subshell, which are filled by ten electrons (Sc through Zn). The 4p subshell is filled next by six electrons (Ga through Kr). As you can see, the periodic table shown in Figure $3$ provides a simple way to remember the order of filling the subshells in determining the electron configuration. The order of filling subshells is the same: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, etc.
Example $2$: Aluminum
Using Figure $3$ as your guide, write the electron configuration of neutral aluminum atom. The atomic number of Al is 13.
Solution
Aluminum has 13 electrons.
Start at Period 1 of the periodic table, Figure $3$. Place two electrons in the 1s subshell (1s2).
Proceed to Period 2 (left to right direction). Place the next two electrons in the 2s subshell (2s2) and the next six electrons in the 2p subshell (2p6).
Proceed to Period 3 (left to right direction). Place the next two electrons in the 3s subshell (3s2) and the last one electron in the 3p subshell (3p1).
The electron configuration of Aluminum is 1s22s22p63s23p1
Exercise $2$
Using Figure $3$ as your guide, write the electron configuration of the atom that has 20 electrons
Answer
Start at Period 1 of Figure $3$. Place two electrons in the 1s subshell (1s2).
Proceed to Period 2 (left to right direction). Place the next two electrons in the 2s subshell (2s2) and the next six electrons in the 2p subshell (2p6).
Proceed to Period 3 (left to right direction). Place the next two electrons in the 3s subshell (3s2) and the next six electron in the 3p subshell (3p6).
Proceed to Period 4. Place the remaining two electrons in the 4s subshell (4s2).
The electron configuration is 1s22s22p63s23p64s2
Valence Electrons
In the study of chemical reactivity, we will find that the electrons in the outermost principal energy level are very important and so they are given a special name. Valence electrons are the electrons in the highest occupied principal energy level of an atom.
In the second period elements, the two electrons in the $1s$ sublevel are called inner-shell electrons and are not involved directly in the element's reactivity or in the formation of compounds. Lithium has a single electron in the second principal energy level and so we say that lithium has one valence electron. Beryllium has two valence electrons. How many valence electrons does boron have? You must recognize that the second principal energy level consists of both the $2s$ and the $2p$ sublevels and so the answer is three. In fact, the number of valence electrons goes up by one for each step across a period until the last element is reached. Neon, with its configuration ending in $2s^2 2p^6$, has eight valence electrons.
The alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the 3s orbital, giving a 1s22s22p63s1 configuration. The electrons occupying the outermost shell orbital(s) (highest value of n) are called valence electrons, and those occupying the inner shell orbitals are called core electrons ( Figure \PageIndex4). Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [Ne] represents core electrons, (1s22s22p6) and our abbreviated or condensed configuration is [Ne]3s1.
Similarly, the abbreviated configuration of lithium can be represented as [He]2s1, where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells.
$\ce{Li:[He]}\,2s^1\ \ce{Na:[Ne]}\,3s^1 \nonumber$
A chemical reaction results from electron removal, electron addition, or electron sharing of the valence electrons of the different atoms. The path a specific element will take depends on where the electrons are in the atom and how many there are. Thus, it is convenient to separate electrons into two groups. Valence shell electrons (or, more simply, the valence electrons) are the electrons in the highest-numbered shell, or valence shell, while core electrons are the electrons in lower-numbered shells. We can see from the electron configuration of a carbon atom—1s22s22p2—that it has 4 valence electrons (2s22p2) and 2 core electrons (1s2). You will see in the next chapters that the chemical properties of elements are determined by the number of valence electrons.
Example $3$
Examine the electron configuration of neutral phosphorus atoms in Example $1$, 1s22s22p63s23p3 and write the abbreviated notation.
Solution
Phosphorus has electron configuration, 1s22s22p63s23p3.
The highest-numbered shell is the third shell (3s23p3): 2 electrons in the 3s subshell and 3 electrons in the 3p subshell. That gives a total of 5 valence electrons.
The 10 inner shell (core) electrons, 1s22s22p6 can be replaced by [Ne] (see Figure $3$). Abbreviated notation is : [Ne]3s23p3
Exercise $3$
Examine the electron configuration of neutral calcium atom (Exercise $2$), 1s22s22p63s23p64s2, and write the abbreviated notation.
Answer
The highest-numbered shell is the fourth shell 4s2, which has 2 electrons in the 4s subshell. Hence, Calcium has 2 valence electrons.
The 18 inner-shell (core) electrons, 1s22s22p63s23p6, can be replaced by [Ar], see Figure $3$. The abbreviated notation is: [Ar]4s2
Example $4$
Based on their respective locations in the periodic table (use Figure $3$), determine the number of valence electrons and the valence shell configuration of elements A, B and C.
Solution
Element A is located in Period 2, the 5th position in 2p-block. Before the electrons are placed in 2p subshell, the 2s subshell must be filled first. This means that A has two valence electrons in 2s (2s2) and five valence electrons in 2p (2p5). Answer: 2s22p5. It has 2 + 5 = 7 valence electrons.
Element B is located in Period 3, the 2nd position in 3s-block. This means that B has two valence electrons in 3s (3s2). Answer: 3s2.
Element C is located in Period 5, the 1st position in 5s-block). This means that there is only one valence electron in 5s (5s1). Answer: 5s1.
Exercise $4$
Using the location of Na is the periodic table (Figure $3$), draw the shell diagram of sodium atom.
Answer
Sodium (Na) is the first element in the 3rd row (Period 3) in the periodic table. This means that the first shell and second shells of Na atom are filled to the maximum number of electrons.
The first shell (1s) is filled with 2 electrons. The second shell (2s and 2p) has a total of 8 electrons. And, the third (last) shell has 1 electron.
The shell diagram of the Na atom is shown below. The shell nearest the nucleus (first shell) has 2 electrons (2 dots), the second shell has 8 electrons and the last (outermost) shell has 1 electron. (2.8.1)
Concept Review Exercises
1. How are electrons organized in atoms?
2. What information does an electron configuration convey?
3. What is the difference between core electrons and valence electrons?
Answers
1. Electrons are organized into shells and subshells around nuclei.
2. The electron configuration states the arrangement of electrons in shells and subshells.
3. Valence electrons are in the highest-numbered shell; all other electrons are core electrons.
Key Takeaway
• Electrons are organized into shells and subshells about the nucleus of an atom.
• The valence electrons determine the reactivity of an atom. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.06%3A_Arrangements_of_Electrons.txt |
Learning Objectives
• Explain how elements are organized into the periodic table.
• Describe how some characteristics of elements relate to their positions on the periodic table.
In the 19th century, many previously unknown elements were discovered, and scientists noted that certain sets of elements had similar chemical properties. For example, chlorine, bromine, and iodine react with other elements (such as sodium) to make similar compounds. Likewise, lithium, sodium, and potassium react with other elements (such as oxygen) to make similar compounds. Why is this so?
In 1864, Julius Lothar Meyer, a German chemist, organized the elements by atomic mass and grouped them according to their chemical properties. Later that decade, Dmitri Mendeleev, a Russian chemist, organized all the known elements according to similar properties. He left gaps in his table for what he thought were undiscovered elements, and he made some bold predictions regarding the properties of those undiscovered elements. When elements were later discovered whose properties closely matched Mendeleev’s predictions, his version of the table gained favor in the scientific community. Because certain properties of the elements repeat on a regular basis throughout the table (that is, they are periodic), it became known as the periodic table.
Mendeleev had to list some elements out of the order of their atomic masses to group them with other elements that had similar properties.
The periodic table is one of the cornerstones of chemistry because it organizes all the known elements on the basis of their chemical properties. A modern version is shown in Figure $1$. Most periodic tables provide additional data (such as atomic mass) in a box that contains each element’s symbol. The elements are listed in order of atomic number.
Features of the Periodic Table
Elements that have similar chemical properties are grouped in columns called groups (or families). As well as being numbered, some of these groups have names—for example, alkali metals (the first column of elements), alkaline earth metals (the second column of elements), halogens (the next-to-last column of elements), and noble gases (the last column of elements).
Each row of elements on the periodic table is called a period. Periods have different lengths; the first period has only 2 elements (hydrogen and helium), while the second and third periods have 8 elements each. The fourth and fifth periods have 18 elements each, and later periods are so long that a segment from each is removed and placed beneath the main body of the table.
Metals, Nonmetals and Metalloids
Certain elemental properties become apparent in a survey of the periodic table as a whole. Every element can be classified as either a metal, a nonmetal, or a semimetal, as shown in Figure $2$. A metal is a substance that is shiny, typically (but not always) silvery in color, and an excellent conductor of electricity and heat. Metals are also malleable (they can be beaten into thin sheets) and ductile (they can be drawn into thin wires). A nonmetal is typically dull and a poor conductor of electricity and heat. Solid nonmetals are also very brittle. As shown in Figure $2$, metals occupy the left three-fourths of the periodic table, while nonmetals (except for hydrogen) are clustered in the upper right-hand corner of the periodic table. The elements with properties intermediate between those of metals and nonmetals are called semimetals (or metalloids). Elements adjacent to the bold zigzag line in the right-hand portion of the periodic table have semimetal properties.
Example $1$
Based on its position in the periodic table, do you expect selenium (Se) to be a metal, a nonmetal, or a semimetal?
Solution
The atomic number of selenium is 34, which places it in period 4 and group 16. In Figure 2.7.1, 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 (In) to be a nonmetal, a metal, or a semimetal?
Answer
metal
Representative, Transition and Inner-transition
Another way to categorize the elements of the periodic table is shown in Figure $3$. The first two columns on the left and the last six columns on the right are called the main group or representative elements. The ten-column block between these columns contains the transition metals. The two rows beneath the main body of the periodic table contain the inner transition metals. The elements in these two rows are also referred to as, respectively, the lanthanide metals and the actinide metals.
To Your Health: Transition Metals in the Body
Most of the elemental composition of the human body consists of main group elements. The first element appearing on the list that is not a main group element is iron, at 0.006 percentage by mass. Because iron has relatively massive atoms, it would appear even lower on a list organized in terms of percent by atoms rather than percent by mass.
Iron is a transition metal. Transition metals have interesting chemical properties, partially because some of their electrons are in d subshells. The chemistry of iron makes it a key component in the proper functioning of red blood cells.
Red blood cells are cells that transport oxygen from the lungs to cells of the body and then transport carbon dioxide from the cells to the lungs. Without red blood cells, animal respiration as we know it would not exist. The critical part of the red blood cell is a protein called hemoglobin. Hemoglobin combines with oxygen and carbon dioxide, transporting these gases from one location to another in the body. Hemoglobin is a relatively large molecule, with a mass of about 65,000 u.
The crucial atom in the hemoglobin protein is iron. Each hemoglobin molecule has four iron atoms, which act as binding sites for oxygen. It is the presence of this particular transition metal in your red blood cells that allows you to use the oxygen you inhale.
Other transition metals have important functions in the body, despite being present in low amounts. Zinc is needed for the body’s immune system to function properly, as well as for protein synthesis and tissue and cell growth. Copper is also needed for several proteins to function properly in the body. Manganese is needed for the body to metabolize oxygen properly. Cobalt is a necessary component of vitamin B-12, a vital nutrient. These last three metals are not listed explicitly in Table 2.1.2, so they are present in the body in very small quantities. However, even these small quantities are required for the body to function properly.
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.
To Your Health: Radon
Radon is an invisible, odorless noble gas that is slowly released from the ground, particularly from rocks and soils whose uranium content is high. Because it is a noble gas, radon is not chemically reactive. Unfortunately, it is radioactive, and increased exposure to it has been correlated with an increased lung cancer risk.
Because radon comes from the ground, we cannot avoid it entirely. Moreover, because it is denser than air, radon tends to accumulate in basements, which if improperly ventilated can be hazardous to a building’s inhabitants. Fortunately, specialized ventilation minimizes the amount of radon that might collect. Special fan-and-vent systems are available that draw air from below the basement floor, before it can enter the living space, and vent it above the roof of a house.
After smoking, radon is thought to be the second-biggest preventable cause of lung cancer in the United States. The American Cancer Society estimates that 10% of all lung cancers are related to radon exposure. There is uncertainty regarding what levels of exposure cause cancer, as well as what the exact causal agent might be (either radon or one of its breakdown products, many of which are also radioactive and, unlike radon, not gases). The US Environmental Protection Agency recommends testing every floor below the third floor for radon levels to guard against long-term health effects.
Why do elements in a given group have similar properties?
The periodic table is organized on the basis of similarities in elemental properties, but what explains these similarities? It turns out that the shape of the periodic table reflects the filling of subshells with electrons, as shown in Figure $4$. Starting with the first period and going from left to right, the table reproduces the order of filling of the electron subshells in atoms. Furthermore, elements in the same group share the same valence shell electron configuration. For example, all elements in the first column have a single s electron in their valence shells, so their electron configurations can be described as ns1 (where n represents the shell number). This last observation is crucial. Chemistry is largely the result of interactions between the valence electrons of different atoms. Thus, atoms that have the same valence shell electron configuration will have similar chemistry.
Example $1$
Using the variable n to represent the number of the valence electron shell, write the valence shell electron configuration for each group.
1. the alkaline earth metals
2. the column of elements headed by carbon
Answer a
The alkaline earth metals are in the second column of the periodic table. This column corresponds to the s subshell being filled with 2 electrons. Therefore, the valence shell electron configuration is ns2.
Answer b
The electron configuration of carbon is 1s22s22p2. Its valence shell electron configuration is 2s22p2. Every element in the same column should have a similar valence shell electron configuration, which we can represent as ns2np2.
Exercise $1$
Using the variable n to represent the number of the valence electron shell, write the valence shell electron configuration for each group.
1. the halogens
2. the column of elements headed by oxygen
Answer a
The halogens are in the 17th column (or Group 7A) of the periodic table. This column corresponds to the p subshell being filled with 5 electrons. Therefore, the valence shell electron configuration is ns2np5.
Answer b
The column headed by O is the 16th column (or Group 6A). This column corresponds to the p subshell being filled with 4 electrons. Therefore, the valence shell electron configuration is ns2np4.
Valence Electrons and Group Number
The number of valence electrons of an element can be determined by the periodic table group (vertical column) in which the element is categorized. With the exception of groups 3–12 (the transition metals), the units digit of the group number identifies how many valence electrons are associated with a neutral atom of an element listed under that particular column.
Table $1$. The Group number and the number of valence electrons.
Periodic table group Valence electrons
Group 1 (I) (alkali metals) 1
Group 2 (II) (alkaline earth metals) 2
Groups 3-12 (transition metals) 2*
Group 13 (III) (boron group) 3
Group 14 (IV) (carbon group) 4
Group 15 (V) (pnictogens) 5
Group 16 (VI) (chalcogens) 6
Group 17 (VII) (halogens) 7
Group 18 (VIII or 0) (noble gases) 8**
* The general method for counting valence electrons is generally not useful for transition metals.
** Except for helium, which has only two valence electrons.
Atomic Radius
The periodic table is useful for understanding atomic properties that show periodic trends. One such property is the atomic radius (Figure $5$). The atomic radius is defined as one-half the distance between the nuclei of identical atoms that are bonded together. The units for atomic radii are picometers, equal to $10^{-12}$ meters. As an example, the internuclear distance between the two hydrogen atoms in an $\ce{H_2}$ molecule is measured to be $74 \: \text{pm}$. Therefore, the atomic radius of a hydrogen atom is $\frac{74}{2} = 37 \: \text{pm}$.
As mentioned earlier, the higher the shell number, the farther from the nucleus the electrons in that shell are likely to be. In other words, the size of an atom is generally determined by the number of the valence electron shell. Therefore, as we go down a column on the periodic table, the atomic radius increases. As we go across a period on the periodic table, however, electrons are being added to the same valence shell; meanwhile, more protons are being added to the nucleus, so the positive charge of the nucleus is increasing. The increasing positive charge attracts the electrons more strongly, pulling them closer to the nucleus. Consequently, as we go across a period, from left to right, the atomic radius decreases. These trends are seen clearly in Figure $5$
Example $2$
Using the periodic table (rather than Figure $5$), which atom is larger?
1. N or Bi
2. Mg or Cl
Answer a
Bi is below N in Group 5A in the periodic table and has electrons in higher-numbered shells, hence we expect that Bi atoms are larger than N atoms.
Answer b
Both Mg and Cl are in period 3 of the periodic table, but Cl lies farther to the right. Therefore we expect Mg atoms to be larger than Cl atoms.
Exercise $2$
Using the periodic table (rather than Figure $5$), which atom is larger?
1. Li or F
2. Na or K
Answer a
Li and F are on the same period, but F lies farther to the right. Therefore, we expect Li to be larger than F atoms.
Answer b
K lies below Na in Group 1A, hence has more electron shells, making it larger than Na.
Career Focus: Clinical Chemist
Clinical chemistry is the area of chemistry concerned with the analysis of body fluids to determine the health status of the human body. Clinical chemists measure a variety of substances, ranging from simple elements such as sodium and potassium to complex molecules such as proteins and enzymes, in blood, urine, and other body fluids. The absence or presence, or abnormally low or high amounts, of a substance can be a sign of some disease or an indication of health. Many clinical chemists use sophisticated equipment and complex chemical reactions in their work, so they not only need to understand basic chemistry, but also be familiar with special instrumentation and how to interpret test results.
Concept Review Exercises
1. How are the elements organized into the periodic table?
2. Looking at the periodic table, where do the following elements appear?
1. the metals
2. the nonmetals
3. the halogens
4. the transition metals
5. the noble gases
3. Describe the trends in atomic radii as related to an element’s position on the periodic table.
Answers
1. Elements are organized in order of increasing atomic number.
1. the left three-quarters of the periodic table (to the left of the zigzag band)
2. the right quarter of the periodic table (to the right of the zigzag band)
3. the next-to-last column of the periodic table
4. the middle section of the periodic table
5. the last column of the periodic table
2. As you go across the periodic table, atomic radii decrease; as you go down the periodic table, atomic radii increase.
Key Takeaways
• The chemical elements are arranged in a chart called the periodic table.
• Some characteristics of the elements are related to their position on the periodic table.
• The number of valence electrons of an element can be determined by the group (vertical column) number in the Periodic Table. Elements with the same number of valence electrons have similar chemical properties. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.07%3A_The_Periodic_Table.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms and ask yourself how they relate to the topics in the chapter.
An element is a substance that cannot be broken down into simpler chemical substances. Only about 90 naturally occurring elements are known. They have varying abundances on Earth and in the body. Each element has a one- or two-letter chemical symbol.
The modern atomic theory states that the smallest piece of an element is an atom. Individual atoms are extremely small, on the order of 10−10 m across. Most elements exist in pure form as individual atoms, but some exist as diatomic molecules. Atoms themselves are composed of subatomic particles. The electron is a tiny subatomic particle with a negative charge. The proton has a positive charge and, while small, is much larger than the electron. The neutron is also much larger than an electron but has no electrical charge.
Protons, neutrons, and electrons have a specific arrangement in an atom. The protons and neutrons are found in the center of the atom, grouped together into a nucleus. The electrons are found in fuzzy clouds around the nucleus.
Each element has a characteristic number of protons in its nucleus. This number of protons is the atomic number of the element. An element may have different numbers of neutrons in the nuclei of its atoms; such atoms are referred to as isotopes. Two isotopes of hydrogen are deuterium, with a proton and a neutron in its nucleus, and tritium, with a proton and two neutrons in its nucleus. The sum of the numbers of protons and neutrons in a nucleus is called the mass number and is used to distinguish isotopes from each other.
Masses of individual atoms are measured in atomic mass units. An atomic mass unit is equal to 1/12th of the mass of a single carbon-12 atom. Because different isotopes of an element have different masses, the atomic mass of an element is a weighted average of the masses of all the element’s naturally occurring isotopes.
The modern theory of electron behavior is called quantum mechanics. According to this theory, electrons in atoms can only have specific, or quantized, energies. Electrons are grouped into general regions called shells, and within these into more specific regions called subshells. There are four types of subshells, and each type can hold up to a maximum number of electrons. The distribution of electrons into shells and subshells is the electron configuration of an atom. Chemistry typically occurs because of interactions between the electrons of the outermost shell of different atoms, called the valence shell electrons. Electrons in inner shells are called core electrons.
Elements are grouped together by similar chemical properties into a chart called the periodic table. Vertical columns of elements are called groups or families. Some of the groups of elements have names, like the alkali metals, the alkaline earth metals, the halogens, and the noble gases. A horizontal row of elements is called a period. Periods and groups have differing numbers of elements in them. The periodic table separates elements into metals, nonmetals, and semimetals. The periodic table is also separated into main group elements, transition metals, lanthanide elements, and actinide elements. The lanthanide and actinide elements are also referred to as inner transition metal elements. The shape of the periodic table reflects the sequential filling of shells and subshells in atoms.
The periodic table helps us understand trends in some of the properties of atoms. One such property is the atomic radius of atoms. From top to bottom of the periodic table, atoms get bigger because electrons are occupying larger and bigger shells. From left to right across the periodic table, electrons are filling the same shell but are being attracted by an increasing positive charge from the nucleus, and thus the atoms get smaller. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.S%3A_Elements_Atoms_and_the_Periodic_Table_%28Summary%29.txt |
There are only 118 known chemical elements but tens of millions of known chemical compounds. Compounds can be very complex combinations of atoms, but many important compounds are fairly simple. Table salt, as we have seen, consists of only two elements: sodium and chlorine. Nevertheless, the compound has properties completely different from either elemental sodium (a chemically reactive metal) or elemental chlorine (a poisonous, green gas). We will see additional examples of such differences in this chapter as we consider how atoms combine to form compounds.
• 3.0: Prelude to Ionic Bonding and Simple Ionic Compounds
We will see that the word salt has a specific meaning in chemistry, but to most people, this word refers to table salt. This kind of salt is used as a condiment throughout the world, but it was not always so abundant. Two thousand years ago, Roman soldiers received part of their pay as salt, which explains why the words salt and salary come from the same Latin root (salarium). Today, table salt is either mined or obtained from the evaporation of saltwater.
• 3.1: Two Types of Bonding
Atoms have a tendency to have eight electrons in their valence shell. The attraction of oppositely charged ions is what makes ionic bonds.
• 3.2: Ions
Ions can be positively charged or negatively charged. A Lewis diagram is used to show how electrons are transferred to make ions and ionic compounds.
• 3.3: Formulas for Ionic Compounds
Proper chemical formulas for ionic compounds balance the total positive charge with the total negative charge. Groups of atoms with an overall charge, called polyatomic ions, also exist.
• 3.4: Ionic Nomenclature
Each ionic compound has its own unique name that comes from the names of the ions. After learning a few more details about the names of individual ions, you will be a step away from knowing how to name ionic compounds. This section begins the formal study of nomenclature, the systematic naming of chemical compounds.
• 3.5: Formula Mass
Formula masses of ionic compounds can be determined from the masses of the atoms in their formulas.
• 3.6: Characteristics of Ionic Compounds
Ionic compounds are composed of cations and anions that are strongly attracted to each other. Hence, ionic solids have very high melting points and are extremely hard. When dissolved in water, the ions separate from each other, allowing them to form electrolyte solutions.
• 3.E: Ionic Bonding and Simple Ionic Compounds (Exercises)
These are homework exercises to accompany Chapter 3 of the Ball et al. "The Basics of GOB Chemistry" Textmap.
• 3.S: Ionic Bonding and Simple Ionic Compounds (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms and ask yourself how they relate to the topics in the chapter.
03: Ionic Bonding and Simple Ionic Compounds
We will see that the word salt has a specific meaning in chemistry, but to most people, this word refers to table salt. This kind of salt is used as a condiment throughout the world, but it was not always so abundant. Two thousand years ago, Roman soldiers received part of their pay as salt, which explains why the words salt and salary come from the same Latin root (salarium). Today, table salt is either mined or obtained from the evaporation of saltwater.
Table salt is sodium chloride (NaCl), which is a simple compound of two elements that are necessary for the human body to function properly. Sodium, for example, is important for nerve conduction and fluid balance. In fact, human blood is about a 0.9% sodium chloride solution, and a solution called normal saline is commonly administered intravenously in hospitals.
Although some salt in our diets is necessary to replenish the sodium and chloride ions that we excrete in urine and sweat, too much is unhealthy, and many people may be ingesting more salt than their bodies need. The RDI of sodium is 2,400 mg—the amount in about 1 teaspoon of salt—but the average intake of sodium in the United States is between 4,000 mg and 5,000 mg, partly because salt is a common additive in many prepared foods. Previously, the high ingestion of salt was thought to be associated with high blood pressure, but current research does not support this link. Even so, some doctors still recommend a low-salt diet (never a “no-salt” diet) for patients with high blood pressure, which may include using a salt substitute. Most salt substitutes use potassium instead of sodium, but some people complain that the potassium imparts a slightly bitter taste. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.00%3A_Prelude_to_Ionic_Bonding_and_Simple_Ionic_Compounds.txt |
Learning Objectives
• Define the octet rule.
• Describe how ionic bonds are formed.
Atoms can join together by forming a chemical bond, which is a very strong attraction between two atoms. Chemical bonds are formed when electrons in different atoms interact with each other to make an arrangement that is more stable than when the atoms are apart.
What causes atoms to make a chemical bond with other atoms, rather than remaining as individual atoms? A clue comes by considering the noble gas elements, the rightmost column of the periodic table. These elements—helium, neon, argon, krypton, xenon, and radon—do not form compounds very easily, which suggests that they are especially stable as lone atoms. What else do the noble gas elements have in common? Except for helium, they all have eight valence electrons. Chemists have concluded that atoms are especially stable if they have eight electrons in their outermost shell. This useful rule of thumb is called the octet rule, and it is a key to understanding why compounds form.
Of the noble gases, only krypton, xenon, and radon have been found to make compounds.
There are two ways for an atom that does not have an octet of valence electrons to obtain an octet in its outer shell. One way is the transfer of electrons between two atoms until all atoms have octets. Because some atoms will lose electrons and some atoms will gain electrons, there is no overall change in the number of electrons, but individual atoms acquire a nonzero electric charge. Those that lose electrons become positively charged, and those that gain electrons become negatively charged. Charged atoms are called ions. Because opposite charges attract (while like charges repel), these oppositely charged ions attract each other, forming ionic bonds. The resulting compounds are called ionic compounds and are the primary subject of this chapter.
The second way for an atom to obtain an octet of electrons is by sharing electrons with another atom. These shared electrons simultaneously occupy the outermost shell of more than one atom. The bond made by electron sharing is called a covalent bond.
Despite our focus on the octet rule, we must remember that for small atoms, such as hydrogen, helium, and lithium, the first shell is, or becomes, the outermost shell and hold only two electrons. Therefore, these atoms satisfy a “duet rule” rather than the octet rule.
Example \(1\)
A sodium atom has one valence electron. Do you think it is more likely for a sodium atom to lose one electron or gain seven electrons to obtain an octet?
Solution
Although either event is possible, a sodium atom is more likely to lose its single valence electron. When that happens, it becomes an ion with a net positive charge. This can be illustrated as follows:
Table explaining the solution to the example.
Sodium atom Sodium ion
11 protons 11+ 11 protons 11+
11 electrons 11− 10 electrons 10−
0 overall charge +1 overall charge
Exercise \(1\)
A fluorine atom has seven valence electrons. Do you think it is more likely for a fluorine atom to lose seven electrons or gain one electron to obtain an octet? Write the formula of the resulting ion.
Answer
The process that involves less number of electrons is more favorable. Fluorine would gain one electron. The formula of the resulting ion is F.
Key Takeaways
• Atoms have a tendency to have eight electrons in their valence shell.
• The attraction of oppositely charged ions is what makes ionic bonds.
Exercises
1. What is the octet rule?
2. How are ionic bonds formed?
3. Why is an ionic compound unlikely to consist of two positively charged ions?
4. Why is an ionic compound unlikely to consist of two negatively charged ions?
5. A calcium atom has two valence electrons. Do you think it will lose two electrons or gain six electrons to obtain an octet in its outermost electron shell? Write the formula of the resulting ion.
6. An aluminum atom has three valence electrons. Do you think it will lose three electrons or gain five electrons to obtain an octet in its outermost electron shell? Write the formula of the resulting ion.
7. A selenium atom has six valence electrons. Do you think it will lose six electrons or gain two electrons to obtain an octet in its outermost electron shell? Write the formula of the resulting ion.
8. An iodine atom has seven valence electrons. Do you think it will lose seven electrons or gain one electron to obtain an octet in its outermost electron shell? Write the formula of the resulting ion.
Answers
1. The octet rule is the concept that atoms tend to have eight electrons in their valence electron shell.
2. Ionic bonds are formed by the attraction between oppositely charged ions.
3. Positive charges repel each other, so an ionic compound is not likely between two positively charged ions.
4. Negative charges repel each other also. 5. Ca atom is more likely to lose two electrons. It will become Ca2+ ion. 6. An Al atom is more likely to lose three electrons. It will become Al3+ ion.
7. Selenium is more likely to gain two electrons. It will become Se2 ion.
8. Iodine is more likely to gain one electron. It will become I ion. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.01%3A_Two_Types_of_Bonding.txt |
Learning Objectives
• Define the two types of ions.
• Use Lewis diagrams to illustrate ion formation.
Most atoms do not have eight electrons in their valence electron shell. Some atoms have only a few electrons in their outer shell, while some atoms lack only one or two electrons to have an octet. In cases where an atom has three or fewer valence electrons, the atom may lose those valence electrons quite easily until what remains is a lower shell that contains an octet. Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations. Most metals become cations when they make ionic compounds.
Some atoms have nearly eight electrons in their valence shell and can gain additional valence electrons until they have an octet. When these atoms gain electrons, they acquire a negative charge because they now possess more electrons than protons. Negatively charged ions are called anions. Most nonmetals become anions when they make ionic compounds.
The names for positive and negative ions are pronounced CAT-eye-ons (cations) and ANN-eye-ons (anions), respectively.
Electron Transfer
We can use electron configurations to illustrate the electron transfer process between sodium atoms and chlorine atoms.
Na: 1s22s22p63s1
As demonstrated here, a sodium atom (Na) has one valence electron in the third principal energy level. It is likely to achieve an octet in its outermost shell by losing its one valence electron. The cation produced in this way, Na+, is called the sodium ion to distinguish it from the element. The sodium ion, Na+, has the electron configuration with an octet of electrons from the second principal energy level. It is now the same as that of the noble gas neon. The term isoelectronic refers to an atom and an ion of a different atom (or two different ions) that have the same electron configuration. The sodium ion is isoelectronic with the neon atom. The equation below illustrates this process.
$\begin{array}{lcl} \ce{Na} & \rightarrow & \ce{Na^+} + \ce{e^-} \ 1s^2 \: 2s^2 \: 2p^6 \: 3s^1 & & 1s^2 \: 2s^2 \: 2p^6 \: \text{(octet)} \end{array} \nonumber$
Figure $1$ is a graphical depiction of this process.
Now, let's consider chlorine atom, Cl: 1s22s22p63s23p5
Only one more electron is needed to achieve an octet in chlorine’s valence shell. When a chlorine atom gains an electron, its outermost principal energy level achieves an octet. In this case, the ion has the same outermost shell as the original atom, but now that shell has eight electrons in it. Once again, the octet rule has been satisfied. The resulting anion, Cl, is called the chloride ion; note the slight change in the suffix (-ide instead of -ine) to create the name of this anion. This process is illustrated below. (In table salt, this electron comes from the sodium atom.)
$\begin{array}{lcl} \ce{Cl} + \ce{e^-} & \rightarrow & \ce{Cl^-} \ 1s^2 \: 2s^2 \:2p^6 \: 3s^2 \: 3p^5 & & 1s^2 \: 2s^2 \: 2p^6 \: 3s^2 \: 3p^6 \text{(octet)} \end{array} \nonumber$
Figure $2$ is a graphical depiction of this process.
With two oppositely charged ions, there is an electrostatic attraction between them because opposite charges attract. The resulting combination is the compound sodium chloride. Notice that there are no leftover electrons. The number of electrons lost by the sodium atom (one) equals the number of electrons gained by the chlorine atom (one), so the compound is electrically neutral. In macroscopic samples of sodium chloride, there are billions and billions of sodium and chloride ions, although there is always the same number of cations and anions.
Example $1$
Write the electron configuration of aluminum atom (Z=13). How many electrons must Al lose/gain to achieve octet? Write the formula of the resulting ion and its electron configuration.
Solution
The electron configuration of Al atom is 1s22s22p63s23p1. The second shell has octet (2s22p6) while the valence shell has 3 electrons (3s23p1). Mg can achieve octet by losing the 3 valence electrons. The resulting cation is Al3+ with electron configuration, 1s22s22p6.
Exercise $1$
Write the electron configuration of oxygen atom (Z=8). How many electrons must O lose/gain to achieve octet? Write the formula of the resulting ion and its electron configuration.
Answer
The electron configuration of O atom is 1s22s22p4. The second shell has six electrons (2s22p4) and needs two electrons to achieve octet. Oxygen will gain 2 electrons. The resulting anion is O2 with electron configuration, 1s22s22p6.
In many cases, elements that belong to the same group (vertical column) on the periodic table form ions with the same charge because they have the same number of valence electrons. Thus, the periodic table becomes a tool for remembering the charges on many ions. For example, all ions made from alkali metals, the first column on the periodic table, have a 1+ charge. Ions made from alkaline earth metals, the second group on the periodic table, have a 2+ charge. On the other side of the periodic table, the next-to-last column, the halogens, form ions having a 1− charge. Figure $3$ shows how the charge on many ions can be predicted by the location of an element on the periodic table. Note the convention of first writing the number and then the sign on a multiply charged ion. The barium cation is written Ba2+, not Ba+2.
Example $2$
Which of these ions is not likely to form?
1. Mg+
2. K+
Solution
(a) Mg is in Group 2A and has two valence electrons. It achieves octet by losing two electrons to form Mg2+ cation. Losing only one electron to form Mg+ does not make an octet, hence, Mg+ is not likely to form.
Exercise $2$
Which of these ions is not likely to form?
1. S3
2. N3
Answer
(a) S is in Group 6A and has six valence electrons. It achieves octet by gaining two electrons to form S2 anion. Gaining three electrons to form S3does not make it octet, hence, S3 is not likely to form.
Lewis Diagrams
Chemists use simple diagrams to show an atom’s valence electrons and how they transfer. These diagrams have two advantages over the electron shell diagrams. First, they show only valence electrons. Second, instead of having a circle around the chemical symbol to represent the electron shell, they have up to eight dots around the symbol; each dot represents a valence electron. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. For example, the representation for sodium is as follows:
and the representation for chlorine is as follows:
For the above diagrams, it does not matter what sides the dots are placed on in Lewis diagrams as long as each side has a maximum of two dots.
These diagrams are called Lewis electron dot diagrams, or simply Lewis diagrams, after Gilbert N. Lewis, the American chemist who introduced them. To write an element’s Lewis dot symbol, place the dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. In other words, place the dots singly on each side before pairing them. The Lewis electron dot diagram of fluorine, for example, with seven valence electrons, is constructed as follows:
Figure $4$ shows the electron configurations and Lewis diagrams of the elements lithium through neon, which is the entire second period of the periodic table. For the main group elements, the number of valence electrons is the same as the group number listed at the top of the periodic table.
The transfer of electrons can be illustrated easily with Lewis diagrams:
In representing the final formula, the dots are omitted.
Example $3$
Starting with lithium and bromine atoms, use Lewis diagrams to show the formation of the ionic compound LiBr.
Solution
From the periodic table, we see that lithium is in the same column as sodium, so it will have the same valence shell electron configuration. That means that the neutral lithium atom will have the same Lewis diagram that the sodium atom has. Similarly, bromine is in the same column as chlorine, so it will have the same Lewis diagram that chlorine has. Therefore,
Exercise $3$
Starting with magnesium and oxygen atoms, use Lewis diagrams to show the formation of the ionic compound MgO.
Answer
Some ionic compounds have different numbers of cations and anions. In those cases, electron transfer occurs between more than one atom. For example, here is the formation of MgBr2:
Notice that in this example there are two bromide ions (1– charge) needed for every one magnesium ion (2+ charge) in order for the overall charge of the compound to equal zero. This is called charge balance. The number of each type of ion is indicated in the formula by the subscript.
Most of the elements that make ionic compounds form an ion that has a characteristic charge. For example, sodium makes ionic compounds in which the sodium ion always has a 1+ charge. Chlorine makes ionic compounds in which the chloride ion always has a 1− charge. Some elements, especially transition metals, can form ions of multiple charges. Figure $5$ shows the characteristic charges for some of these ions. As we saw in Figure $1$, there is a pattern to the charges on many of the main group ions, but there is no simple pattern for transition metal ions (or for the larger main group elements).
Key Takeaways
• Ions can be positively charged or negatively charged.
• A Lewis diagram is used to show how electrons are transferred to make ions and ionic compounds.
Exercises
1. What are the two types of ions?
2. Use Lewis diagrams to illustrate the formation of an ionic compound from a potassium atom and an iodine atom.
3. When the following atoms become ions, what charges do they acquire?
1. Li
2. S
3. Ca
4. F
4. Identify each as a cation, an anion, or neither.
1. H+
2. Cl
3. O2
4. Ba2+
5. CH4
6. CS2
5. Identify each as a cation, an anion, or neither.
1. NH3
2. Br
3. H
4. Hg2+
5. CCl4
6. SO3
6. Write the electron configuration for each ion.
1. Li+
2. Mg2+
3. F
4. S2−
7. Write the electron configuration for each ion.
1. Na+
2. Be2+
3. Cl
4. O2−
8. Draw Lewis diagrams for the ions listed in Exercise 6. Also include Lewis diagrams for the respective neutral atoms as a comparison.
9. Draw Lewis diagrams for the ions listed in Exercise 7. Also include Lewis diagrams for the respective neutral atoms as a comparison.
10. Using Lewis diagrams, show the electron transfer for the formation of LiF.
11. Using Lewis diagrams, show the electron transfer for the formation of MgO.
12. Using Lewis diagrams, show the electron transfer for the formation of Li2O.
13. Using Lewis diagrams, show the electron transfer for the formation of CaF2.
14. What characteristic charge do atoms in the first column of the periodic table have when they become ions?
15. What characteristic charge do atoms in the second column of the periodic table have when they become ions?
16. What characteristic charge do atoms in the third-to-last column of the periodic table have when they become ions?
17. What characteristic charge do atoms in the next-to-last column of the periodic table have when they become ions?
Answers
1. Cations (positive charged) and anions (negative charged).
1. 1+
2. 2−
3. 2+
4. 1−
4.
1. cation
2. anion
3. neither
4. cation
5. neither
6. neither
5.
1. neither
2. anion
3. anion
4. cation
5. neither
6. neither
6.
1. 1s2
2. 1s22s22p6
3. 1s22s22p6
4. 1s22s22p63s23p6
7.
1. 1s22s22p6
2. 1s2
3. 1s22s22p63s23p6
4. 1s22s22p6
8.
b.
c.
d.
9. 10. 11. 12. 13.
14. 1+
15. 2+
16. 2−
17. 1− | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.02%3A_Ions.txt |
Learning Objectives
• Write the chemical formula for a simple ionic compound.
• Recognize polyatomic ions in chemical formulas.
We have already encountered some chemical formulas for simple ionic compounds. A chemical formula is a concise list of the elements in a compound and the ratios of these elements. To better understand what a chemical formula means, we must consider how an ionic compound is constructed from its ions.
Ionic compounds exist as alternating positive and negative ions in regular, three-dimensional arrays called crystals (Figure $1$). As you can see, there are no individual $\ce{NaCl}$ “particles” in the array; instead, there is a continuous lattice of alternating sodium and chloride ions. However, we can use the ratio of sodium ions to chloride ions, expressed in the lowest possible whole numbers, as a way of describing the compound. In the case of sodium chloride, the ratio of sodium ions to chloride ions, expressed in lowest whole numbers, is 1:1, so we use $\ce{NaCl}$ (one $\ce{Na}$ symbol and one $\ce{Cl}$ symbol) to represent the compound. Thus, $\ce{NaCl}$ is the chemical formula for sodium chloride, which is a concise way of describing the relative number of different ions in the compound. A macroscopic sample is composed of myriads of NaCl pairs; each individual pair called a formula unit. Although it is convenient to think that $\ce{NaCl}$ crystals are composed of individual $\ce{NaCl}$ units, Figure $1$ shows that no single ion is exclusively associated with any other single ion. Each ion is surrounded by ions of opposite charge.
The formula for an ionic compound follows several conventions. First, the cation is written before the anion. Because most metals form cations and most nonmetals form anions, formulas typically list the metal first and then the nonmetal. Second, charges are not written in a formula. Remember that in an ionic compound, the component species are ions, not neutral atoms, even though the formula does not contain charges. Finally, the proper formula for an ionic compound always has a net zero charge, meaning the total positive charge must equal the total negative charge. To determine the proper formula of any combination of ions, determine how many of each ion is needed to balance the total positive and negative charges in the compound.
This rule is ultimately based on the fact that matter is, overall, electrically neutral.
By convention, assume that there is only one atom if a subscript is not present. We do not use 1 as a subscript.
If we look at the ionic compound consisting of lithium ions and bromide ions, we see that the lithium ion has a 1+ charge and the bromide ion has a 1− charge. Only one ion of each is needed to balance these charges. The formula for lithium bromide is $\ce{LiBr}$.
When an ionic compound is formed from magnesium and oxygen, the magnesium ion has a 2+ charge, and the oxygen atom has a 2− charge. Although both of these ions have higher charges than the ions in lithium bromide, they still balance each other in a one-to-one ratio. Therefore, the proper formula for this ionic compound is $\ce{MgO}$.
Now consider the ionic compound formed by magnesium and chlorine. A magnesium ion has a 2+ charge, while a chlorine ion has a 1− charge:
$\ce{Mg^{2+}Cl^{−}} \nonumber$
Combining one ion of each does not completely balance the positive and negative charges. The easiest way to balance these charges is to assume the presence of two chloride ions for each magnesium ion:
$\ce{Mg^{2+} Cl^{−} Cl^{−}} \nonumber$
Now the positive and negative charges are balanced. We could write the chemical formula for this ionic compound as $\ce{MgClCl}$, but the convention is to use a numerical subscript when there is more than one ion of a given type—$\ce{MgCl2}$. This chemical formula says that there are one magnesium ion and two chloride ions in this formula. (Do not read the “Cl2” part of the formula as a molecule of the diatomic elemental chlorine. Chlorine does not exist as a diatomic element in this compound. Rather, it exists as two individual chloride ions.) By convention, the lowest whole number ratio is used in the formulas of ionic compounds. The formula $\ce{Mg2Cl4}$ has balanced charges with the ions in a 1:2 ratio, but it is not the lowest whole number ratio.
By convention, the lowest whole-number ratio of the ions is used in ionic formulas. There are exceptions for certain ions, such as $\ce{Hg2^{2+}}$.
For compounds in which the ratio of ions is not as obvious, the subscripts in the formula can 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 in Figure 3.3.2.
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 Pb4+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Pb2O4. This simplifies to its correct empirical formula PbO2. The empirical formula has one Pb4+ ion and two O2− ions.
Example $1$
Write the chemical formula for an ionic compound composed of each pair of ions.
1. the sodium ion and the sulfur ion
2. the aluminum ion and the fluoride ion
3. the 3+ iron ion and the oxygen ion
Solution
1. To obtain a valence shell octet, sodium forms an ion with a 1+ charge, while the sulfur ion has a 2− charge. Two sodium 1+ ions are needed to balance the 2− charge on the sulfur ion. Rather than writing the formula as $\ce{NaNaS}$, we shorten it by convention to $\ce{Na2S}$.
2. The aluminum ion has a 3+ charge, while the fluoride ion formed by fluorine has a 1− charge. Three fluorine 1− ions are needed to balance the 3+ charge on the aluminum ion. This combination is written as $\ce{AlF3}$.
3. Iron can form two possible ions, but the ion with a 3+ charge is specified here. The oxygen atom has a 2− charge as an ion. To balance the positive and negative charges, we look to the least common multiple—6: two iron 3+ ions will give 6+, while three 2− oxygen ions will give 6−, thereby balancing the overall positive and negative charges. Thus, the formula for this ionic compound is $\ce{Fe2O3}$. Alternatively, use the crossing charges method shown in Figure 3.3.2.
Exercise $1$
Write the chemical formula for an ionic compound composed of each pair of ions.
1. the calcium ion and the oxygen ion
2. the 2+ copper ion and the sulfur ion
3. the 1+ copper ion and the sulfur ion
Answer a:
CaO
Answer b:
CuS
Answer c:
Cu2S
Polyatomic Ions
Some ions consist of groups of atoms covalently bonded together and have an overall electric charge. Because these ions contain more than one atom, they are called polyatomic ions. The Lewis structures, names and formulas of some polyatomic ions are found in Table 3.3.1.
Table $1$: Some Polyatomic Ions
Polyatomic ions have defined formulas, names, and charges that cannot be modified in any way. Table $2$ lists the ion names and ion formulas of the most common polyatomic ions. For example, $\ce{NO3^{−}}$ is the nitrate ion; it has one nitrogen atom and three oxygen atoms and an overall 1− charge. Figure $2$ lists the most common polyatomic ions.
Table $2$: Ion Names and Ion Formulas of Common Polyatomic Ions
Ion Name Ion Formula
ammonium ion NH4+1
hydroxide ion OH−1
cyanide ion CN−1
carbonate ion CO3−2
bicarbonate or hydrogen carbonate HCO3
acetate ion C2H3O2−1 or CH3CO2−1
nitrate ion NO3−1
nitrite ion NO2−1
sulfate ion SO4−2
sulfite ion SO3−2
phosphate ion PO4−3
phosphite ion PO3−3
Note that only one polyatomic ion in this Table, the ammonium ion (NH4+1), is a cation. This polyatomic ion contains one nitrogen and four hydrogens that collectively bear a +1 charge. The remaining polyatomic ions are all negatively-charged and, therefore, are classified as anions. However, only two of these, the hydroxide ion and the cyanide ion, are named using the "-ide" suffix that is typically indicative of negatively-charged particles. The remaining polyatomic anions, which all contain oxygen, in combination with another non-metal, exist as part of a series in which the number of oxygens within the polyatomic unit can vary. As has been repeatedly emphasized in several sections of this text, no two chemical formulas should share a common chemical name. A single suffix, "-ide," is insufficient for distinguishing the names of the anions in a related polyatomic series. Therefore, "-ate" and "-ite" suffixes are employed, in order to denote that the corresponding polyatomic ions are part of a series. Additionally, these suffixes also indicate the relative number of oxygens that are contained within the polyatomic ions. Note that all of the polyatomic ions whose names end in "-ate" contain one more oxygen than those polyatomic anions whose names end in "-ite." Unfortunately, much like the common system for naming transition metals, these suffixes only indicate the relative number of oxygens that are contained within the polyatomic ions. For example, the nitrate ion, which is symbolized as NO3−1, has one more oxygen than the nitrite ion, which is symbolized as NO2−1. However, the sulfate ion is symbolized as SO4−2. While both the nitrate ion and the sulfate ion share an "-ate" suffix, the former contains three oxygens, but the latter contains four. Additionally, both the nitrate ion and the sulfite ion contain three oxygens, but these polyatomic ions do not share a common suffix. Unfortunately, the relative nature of these suffixes mandates that the ion formula/ion name combinations of the polyatomic ions must simply be memorized.
The rule for constructing formulas for ionic compounds containing polyatomic ions is the same as for formulas containing monatomic (single-atom) ions: the positive and negative charges must balance. If more than one of a particular polyatomic ion is needed to balance the charge, the entire formula for the polyatomic ion must be enclosed in parentheses, and the numerical subscript is placed outside the parentheses. This is to show that the subscript applies to the entire polyatomic ion. Two examples are shown below:
Example $2$
Write the chemical formula for an ionic compound composed of each pair of ions.
1. the potassium ion and the sulfate ion
2. the calcium ion and the nitrate ion
Solution
1. Potassium ions have a charge of 1+, while sulfate ions have a charge of 2−. We will need two potassium ions to balance the charge on the sulfate ion, so the proper chemical formula is $\ce{K_2SO_4}$.
2. Calcium ions have a charge of 2+, while nitrate ions have a charge of 1−. We will need two nitrate ions to balance the charge on each calcium ion. The formula for nitrate must be enclosed in parentheses. Thus, we write $\ce{Ca(NO3)2}$ as the formula for this ionic compound.
Exercise $2$
Write the chemical formula for an ionic compound composed of each pair of ions.
1. the magnesium ion and the carbonate ion
2. the aluminum ion and the acetate ion
Answer a:
Mg2+ and CO32- = MgCO3
Answer b:
Al3+ and C2H3O2- = Al(C2H3O2)3
Recognizing Ionic Compounds
There are two ways to recognize ionic compounds. First, compounds between metal and nonmetal elements are usually ionic. For example, CaBr2 contains a metallic element (calcium, a group 2A metal) and a nonmetallic element (bromine, a group 7A nonmetal). Therefore, it is most likely an ionic compound. (In fact, it is ionic.) In contrast, the compound NO2 contains two elements that are both nonmetals (nitrogen, from group 5A, and oxygen, from group 6A). It is not an ionic compound; it belongs to the category of covalent compounds discuss elsewhere. Also note that this combination of nitrogen and oxygen has no electric charge specified, so it is not the nitrite ion.
Second, if you recognize the formula of a polyatomic ion in a compound, the compound is ionic. For example, if you see the formula $\ce{Ba(NO3)2}$, you may recognize the “NO3” part as the nitrate ion, $\rm{NO_3^−}$. (Remember that the convention for writing formulas for ionic compounds is not to include the ionic charge.) This is a clue that the other part of the formula, $\ce{Ba}$, is actually the $\ce{Ba^{2+}}$ ion, with the 2+ charge balancing the overall 2− charge from the two nitrate ions. Thus, this compound is also ionic.
Example $3$
Identify each compound as ionic or not ionic.
1. $\ce{Na2O}$
2. $\ce{PCl3}$
3. $\ce{NH4Cl}$
4. $\ce{OF2}$
Solution
1. Sodium is a metal, and oxygen is a nonmetal; therefore, $\ce{Na2O}$ is expected to be ionic.
2. Both phosphorus and chlorine are nonmetals. Therefore, $\ce{PCl3}$ is not ionic.
3. The $\ce{NH4}$ in the formula represents the ammonium ion, $\ce{NH4^{+}}$, which indicates that this compound is ionic.
4. Both oxygen and fluorine are nonmetals. Therefore, $\ce{OF2}$ is not ionic.
Exercise $3$
Identify each compound as ionic or not ionic.
1. $\ce{N2O}$
2. $\ce{FeCl3}$
3. $\ce{(NH4)3PO4}$
4. $\ce{SOCl2}$
Answer a:
not ionic
Answer b:
ionic
Answer c:
ionic
Answer d:
not ionic
Looking Closer: Blood and Seawater
Science has long recognized that blood and seawater have similar compositions. After all, both liquids have ionic compounds dissolved in them. The similarity may be more than mere coincidence; many scientists think that the first forms of life on Earth arose in the oceans. A closer look, however, shows that blood and seawater are quite different. A 0.9% solution of sodium chloride approximates the salt concentration found in blood. In contrast, seawater is principally a 3% sodium chloride solution, over three times the concentration in blood. Here is a comparison of the amounts of ions in blood and seawater:
Table showing a comparison of the amounts of ions in blood and seawater.
Ion Percent in Seawater Percent in Blood
Na+ 2.36 0.322
Cl 1.94 0.366
Mg2+ 0.13 0.002
SO42 0.09
K+ 0.04 0.016
Ca2+ 0.04 0.0096
HCO3 0.002 0.165
HPO42, H2PO4 0.01
Most ions are more abundant in seawater than they are in blood, with some important exceptions. There are far more hydrogen carbonate ions ($\ce{HCO3^{−}}$) in blood than in seawater. This difference is significant because the hydrogen carbonate ion and some related ions have a crucial role in controlling the acid-base properties of blood. The amount of hydrogen phosphate ions—$\ce{HPO4^{2−}}$ and $\ce{H2PO4^{−}}$—in seawater is very low, but they are present in higher amounts in blood, where they also affect acid-base properties. Another notable difference is that blood does not have significant amounts of the sulfate ion ($\ce{SO4^{2−}}$), but this ion is present in seawater.
Key Takeaways
• Proper chemical formulas for ionic compounds balance the total positive charge with the total negative charge.
• Groups of atoms with an overall charge, called polyatomic ions, also exist.
EXERCISES
1. What information is contained in the formula of an ionic compound?
• Why do the chemical formulas for some ionic compounds contain subscripts, while others do not?
3. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Mg2+ and I
2. Na+ and O2−
4. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Na+ and Br
2. Mg2+ and Br
3. Mg2+ and S2−
5. Write the chemical formula for the ionic compound formed by each pair of ions.
1. K+ and Cl
2. Mg2+ and Cl
3. Mg2+ and Se2
6. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Na+ and N3−
2. Mg2+ and N3−
3. Al3+ and S2−
7. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Li+ and N3−
2. Mg2+ and P3−
3. Li+ and P3−
8. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Fe3+ and Br
2. Fe2+ and Br
3. Au3+ and S2−
4. Au+ and S2−
9. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Cr3+ and O2−
2. Cr2+ and O2−
3. Pb2+ and Cl
4. Pb4+ and Cl
10. Write the chemical formula for the ionic compound formed by each pair of ions.
1. Cr3+ and NO3
2. Fe2+ and PO43
3. Ca2+ and CrO42
4. Al3+ and OH
11. Write the chemical formula for the ionic compound formed by each pair of ions.
1. NH4+ and NO3
2. H+ and Cr2O72
3. Cu+ and CO32
4. Na+ and HCO3
12. For each pair of elements, determine the charge for their ions and write the proper formula for the resulting ionic compound between them.
1. Ba and S
2. Cs and I
13. For each pair of elements, determine the charge for their ions and write the proper formula for the resulting ionic compound between them.
1. K and S
2. Sc and Br
14. Which compounds would you predict to be ionic?
1. Li2O
2. (NH4)2O
3. CO2
4. FeSO3
5. C6H6
6. C2H6O
15. Which compounds would you predict to be ionic?
1. Ba(OH)2
2. CH2O
3. NH2CONH2
4. (NH4)2CrO4
5. C8H18
6. NH3
Answers
1. the ratio of each kind of ion in the compound
2. Sometimes more than one ion is needed to balance the charge on the other ion in an ionic compound.
3.
1. MgI2
2. Na2O
4.
1. NaBr
2. MgBr2
3. MgS
5.
1. KCl
2. MgCl2
3. MgSe
6.
1. Na3N
2. Mg3N2
3. Al2S3
7.
1. Li3N
2. Mg3P2
3. Li3P
8.
1. FeBr3
2. FeBr2
3. Au2S3
4. Au2S
9.
1. Cr2O3
2. CrO
3. PbCl2
4. PbCl4
10.
1. Cr(NO3)3
2. Fe3(PO4)2
3. CaCrO4
4. Al(OH)3
11.
1. NH4NO3
2. H2Cr2O7
3. Cu2CO3
4. NaHCO3
12.
1. Ba2+, S2−, BaS
2. Cs+, I, CsI
13.
1. K+, S2−, K2S
2. Sc3+, Br, ScBr3
14.
1. ionic
2. ionic
3. not ionic
4. ionic
5. not ionic
6. not ionic
15.
1. ionic
2. not ionic
3. not ionic
4. ionic
5. not ionic
6. not ionic | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.03%3A_Formulas_for_Ionic_Compounds.txt |
Learning Objectives
• To use the rules for naming ionic compounds
After learning a few more details about the names of individual ions, you will be a step away from knowing how to name ionic compounds. This section begins the formal study of nomenclature, the systematic naming of chemical compounds.
Naming Ions
The name of a monatomic cation is simply the name of the element followed by the word ion. Thus, Na+ is the sodium ion, Al3+ is the aluminum ion, Ca2+ is the calcium ion, and so forth.
We have seen that some elements lose different numbers of electrons, producing ions of different charges. Iron, for example, can form two cations, each of which, when combined with the same anion, makes a different compound with unique physical and chemical properties. Thus, we need a different name for each iron ion to distinguish Fe2+ from Fe3+. The same issue arises for other ions with more than one possible charge.
There are two ways to make this distinction. In the simpler, more modern approach, called the Stock system, an ion’s positive charge is indicated by a roman numeral in parentheses after the element name, followed by the word ion. Thus, Fe2+ is called the iron(II) ion, while Fe3+ is called the iron(III) ion. This system is used only for elements that form more than one common positive ion. We do not call the Na+ ion the sodium(I) ion because (I) is unnecessary. Sodium forms only a 1+ ion, so there is no ambiguity about the name sodium ion.
Table \(1\): The Common System of Cation Names
Element Stem Charge Name
iron ferr- 2+ ferrous ion
3+ ferric ion
copper cupr- 1+ cuprous ion
2+ cupric ion
tin stann- 2+ stannous ion
4+ stannic ion
lead plumb- 2+ plumbous ion
4+ plumbic ion
chromium chrom- 2+ chromous ion
3+ chromic ion
gold aur- 1+ aurous ion
3+ auric ion
The second system, called the common system, is not conventional but is still prevalent and used in the health sciences. This system recognizes that many metals have two common cations. The common system uses two suffixes (-ic and -ous) that are appended to the stem of the element name. The -ic suffix represents the greater of the two cation charges, and the -ous suffix represents the lower one. In many cases, the stem of the element name comes from the Latin name of the element. Table \(1\) lists the elements that use the common system, along with their respective cation names.
The name of a monatomic anion consists of the stem of the element name, the suffix -ide, and then the word ion. Thus, as we have already seen, Cl is “chlor-” + “-ide ion,” or the chloride ion. Similarly, O2− is the oxide ion, Se2 is the selenide ion, and so forth. Table \(2\) lists the names of some common monatomic ions.
Table \(2\): Some Monatomic Anions
Ion Name
F fluoride ion
Cl chloride ion
Br bromide ion
I iodide ion
O2− oxide ion
S2− sulfide ion
P3− phosphide ion
N3− nitride ion
The polyatomic ions have their own characteristic names, as discussed earlier.
Example \(1\)
Name each ion.
1. Ca2+
2. S2−
3. SO32
4. NH4+
5. Cu+
Answer a
the calcium ion
Answer b
the sulfide ion (from Table \(2\) )
Answer c
the sulfite ion
Answer d
the ammonium ion
Answer e
the copper(I) ion or the cuprous ion (copper can form cations with either a 1+ or 2+ charge, so we have to specify which charge this ion has
Exercise \(1\)
Name each ion.
1. Fe2+
2. Fe3+
3. SO42
4. Ba2+
5. HCO3
Answer a
the iron (II) or ferrous ion
Answer b
the iron (III) or ferric ion
Answer c
the sulfate ion
Answer d
the barium ion
Answer e
the bicarbonate ion or hydrogen carbonate ion
Example \(2\)
Write the formula for each ion.
1. the bromide ion
2. the phosphate ion
3. the cupric ion
4. the magnesium ion
Answer a
Br
Answer b
PO43
Answer c
Cu2+
Answer d
Mg2+
Exercise \(2\)
Write the formula for each ion.
1. the fluoride ion
2. the carbonate ion
3. the ferrous ion
4. the potassium ion
Answer a
F
Answer b
CO32-
Answer c
Fe2+
Answer d
K+
Naming Compounds
Now that we know how to name ions, we are ready to name ionic compounds. We do so by placing the name of the cation first, followed by the name of the anion, and dropping the word ion from both parts. For example, what is the name of the compound whose formula is \(\ce{Ba(NO3)2}\)?
The compound’s name does not indicate that there are two nitrate ions for every barium ion. You must determine the relative numbers of ions by balancing the positive and negative charges.
If you are given a formula for an ionic compound whose cation can have more than one possible charge, you must first determine the charge on the cation before identifying its correct name. For example, consider \(\ce{FeCl2}\) and \(\ce{FeCl3}\). In the first compound, the iron ion has a 2+ charge because there are two \(\ce{Cl^{−}}\) ions in the formula (1− charge on each chloride ion). In the second compound, the iron ion has a 3+ charge, as indicated by the three \(\ce{Cl^{−}}\) ions in the formula. These are two different compounds that need two different names. By the Stock system, the names are iron(II) chloride and iron(III) chloride. If we were to use the stems and suffixes of the common system, the names would be ferrous chloride and ferric chloride, respectively.
Example \(3\)
Name each ionic compound, using both Stock and common systems if necessary.
1. Ca3(PO4)2
2. (NH4)2Cr2O7
3. KCl
4. CuCl
5. SnF2
Answer a
calcium phosphate
Answer b
ammonium dichromate (the prefix di- is part of the name of the anion)
Answer c
potassium chloride
Answer d
copper(I) chloride or cuprous chloride
Answer e
tin(II) fluoride or stannous fluoride
Exercise \(3\)
Name each ionic compound, using both Stock and common systems if necessary.
1. ZnBr2
2. Fe(NO3)3
3. Al2O3
4. CuF2
5. AgF
Answer a
zinc bromide
Answer b
iron (III) nitrate or ferric nitrate
Answer c
aluminum oxide
Answer d
copper (II) fluoride or cupric fluoride
Answer e
silver fluoride
Figure \(1\) is a synopsis of how to name simple ionic compounds.
KEYTAKEAWAY
• Each ionic compound has its own unique name that comes from the names of the ions.
EXERCISES
1. Briefly describe the process for naming an ionic compound.
• In what order do the names of ions appear in the names of ionic compounds?
3. Which ionic compounds can be named using two different systems? Give an example.
4. Name each ion.
1. Ra2+
2. P3−
3. H2PO4
4. Sn4+
5. Name each ion.
1. Cs+
2. As3
3. HSO4
4. Sn2+
6. Name the ionic compound formed by each pair of ions.
1. Na+ and Br
2. Mg2+ and Br
3. Mg2+ and S2−
7. Name the ionic compound formed by each pair of ions.
1. K+ and Cl
2. Mg2+ and Cl
3. Mg2+ and Se2
8. Name the ionic compound formed by each pair of ions.
1. Na+ and N3−
2. Mg2+ and N3−
3. Al3+ and S2−
9. Name the ionic compound formed by each pair of ions.
1. Li+ and N3−
2. Mg2+ and P3−
3. Li+ and P3−
10. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
1. Fe3+ and Br
2. Fe2+ and Br
3. Au3+ and S2−
4. Au+ and S2−
11. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
1. Cr3+ and O2−
2. Cr2+ and O2−
3. Pb2+ and Cl
4. Pb4+ and Cl
12. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
1. Cr3+ and NO3
2. Fe2+ and PO43
3. Ca2+ and CrO42
4. Al3+ and OH
13. Name the ionic compound formed by each pair of ions. Use both the Stock and common systems, where appropriate.
1. NH4+ and NO3
2. K+ and Cr2O72
3. Cu+ and CO32
4. Na+ and HCO3
14. Give two names for each compound.
1. Al(HSO4)3
2. Mg(HSO4)2
15. Give two names for each compound.
1. Co(HCO3)2
2. LiHCO3
AnswerS
1. Name the cation and then the anion but don’t use numerical prefixes.
2. the cation name followed by the anion name
3. Ionic compounds in which the cation can have more than one possible charge have two naming systems. FeCl3 is either iron(III) chloride or ferric chloride (answers will vary).
4.
1. the radium ion
2. the phosphide ion
3. the dihydrogen phosphate ion
4. the tin(IV) ion or the stannic ion
5.
1. the cesium ion
2. the arsenide ion
3. the hydrogen sulfate ion
4. the tin(II) ion or the stannous ion
6.
1. sodium bromide
2. magnesium bromide
3. magnesium sulfide
7.
1. potassium chloride
2. magnesium chloride
3. magnesium selenide
8.
1. sodium nitride
2. magnesium nitride
3. aluminum sulfide
9.
1. lithium nitride
2. magnesium phosphide
3. lithium phosphide
10.
1. iron(III) bromide or ferric bromide
2. iron(II) bromide or ferrous bromide
3. gold(III) sulfide or auric sulfide
4. gold(I) sulfide or aurous sulfide
11.
1. chromium(III) oxide or chromic oxide
2. chromium(II) oxide or chromous oxide
3. lead(II) chloride or plumbous chloride
4. lead(IV) chloride or plumbic chloride
12.
1. chromium(III) nitrate or chromic nitrate
2. iron(II) phosphate or ferrous phosphate
3. calcium chromate
4. aluminum hydroxide
13.
1. ammonium nitrate
2. potassium dichromate
3. copper(I) carbonate or cuprous carbonate
4. sodium hydrogen carbonate or sodium bicarbonate
14.
1. aluminum hydrogen sulfate or aluminum bisulfate
2. magnesium hydrogen sulfate or magnesium bisulfate
15.
1. cobalt hydrogen carbonate or cobalt bicarbonate
2. lithium hydrogen carbonate or lithium bicarbonate | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.04%3A_Ionic_Nomenclature.txt |
Learning Objectives
• To determine the formula mass of an ionic compound.
One skill needed in future chapters is the ability to determine the mass of the formula of an ionic compound. This quantity is called the formula mass. The formula mass is obtained by adding the masses of each individual atom in the formula of the compound. Because a proper formula is electrically neutral (with no net electrons gained or lost), the ions can be considered atoms for the purpose of calculating the formula mass.
Let us start by calculating the formula mass of sodium chloride (NaCl). This formula mass is the sum of the atomic masses of one sodium atom and one chlorine atom, which we find from the periodic table; here, we use the masses to two decimal places:
Table shows how to calculate the formula mass of sodium chloride by using their atomic masses.
Na: 22.99 amu
Cl: + 35.45 amu
Total: 58.44 amu
To two decimal places, the formula mass of NaCl is 58.44 amu.
When an ionic compound has more than one anion or cation, you must remember to use the proper multiple of the atomic mass for the element in question. For the formula mass of calcium fluoride (CaF2), we must multiply the mass of the fluorine atom by 2 to account for the two fluorine atoms in the chemical formula:
Table shows how to find the formula mass of calcium fluoride by using their atomic masses.
Ca: 1 × 40.08 40.08 amu
F: 2 × 19.00 = + 38.00 amu
Total: 78.08 amu
The formula mass of CaF2 is 78.08 amu.
For ionic compounds with polyatomic ions, the sum must include the number and mass of each atom in the formula for the polyatomic ion. For example, potassium nitrate (KNO3) has one potassium atom, one nitrogen atom, and three oxygen atoms:
Table shows how to find the formula mass of potassium nitrate by using their atomic masses.
K: 1 × 39.10 39.10 amu
N: 1 × 14.01 + 14.01 amu
O: 3 × 16.00 = + 48.00 amu
Total: 101.11 amu
The formula mass of KNO3 is 101.11 amu.
Potassium nitrate is a key ingredient in gunpowder and has been used clinically as a diuretic.
When a formula contains more than one polyatomic unit in the chemical formula, as in Ca(NO3)2, do not forget to multiply the atomic mass of every atom inside the parentheses by the subscript outside the parentheses. This is necessary because the subscript refers to the entire polyatomic ion. Thus, for Ca(NO3)2, the subscript 2 implies two complete nitrate ions, so we must sum the masses of two (1 × 2) nitrogen atoms and six (3 × 2) oxygen atoms, along with the mass of a single calcium atom:
Table shows how to find the formula mass of Ca(NO3)2 by using their atomic masses.
Ca: 1 × 40.08 40.08 amu
N: 2 × 14.01 = + 28.02 amu
O: 6 × 16.00 = + 96.00 amu
Total: 164.10 amu
The key to calculating the formula mass of an ionic compound is to correctly count each atom in the formula and multiply the atomic masses of its atoms accordingly.
The Periodic Table is found in this link: Periodic Table of Elements with Atomic Mass [pubchem.ncbi.nlm.nih.gov]
Example \(1\)
Use the atomic masses (rounded to two decimal places) from the Periodic Table to determine the formula mass for each ionic compound.
1. FeCl3
2. (NH4)3PO4
Solution
1.
Table shows how to find the formula mass of FeCl3 by using their atomic masses.
Fe: 55.84 amu
Cl: 3 × 35.45 = + 106.35 amu
Total: 162.19 amu
The formula mass of FeCl3 is 162.19 amu.
1. When we distribute the subscript 3 through the parentheses containing the formula for the ammonium ion, we see that we have 3 nitrogen atoms and 12 hydrogen atoms. Thus, we set up the sum as follows:
Table shows how to find the formula mass of (NH4)3PO4 by using their atomic masses.
N: 3 × 14.01 = 42.03 amu
H: 12 × 1.01 = + 12.12 amu
P: + 30.97 amu
O: 4 × 16.00 = + 64.00 amu
Total: 149.12 amu
The formula mass for (NH4)3PO4 is 149.12 amu.
Exercise \(1\)
Use the atomic masses (rounded to two decimal places) from the Periodic Table to determine the formula mass for each ionic compound.
1. TiO2
2. AgBr
3. Au(NO3)3
4. Fe3(PO4)2
Answer a:
47.87 + 2 (16.00) = 79.87 amu
Answer b:
107.87 + 79.90 = 187.77 amu
Answer c:
196.97 + 3(14.01) + 9(16.00) = 383.00 amu
Answer d:
3(55.84) + 2(30.97) + 8(16.00) = 357.46 amu
To Your Health: Hydrates
Some ionic compounds have water (\(\ce{H2O}\)) incorporated within their formula unit. These compounds, called hydrates, have a characteristic number of water units associated with each formula unit of the compound. Hydrates are solids, not liquids or solutions, despite the water they contain.
To write the chemical formula of a hydrate, write the number of water units per formula unit of compound after its chemical formula. The two chemical formulas are separated by a vertically centered dot. The hydrate of copper(II) sulfate has five water units associated with each formula unit, so it is written as \(\ce{CuSO4 • 5H2O}\). The name of this compound is copper(II) sulfate pentahydrate, with the penta- prefix indicating the presence of five water units per formula unit of copper(II) sulfate.
Magnesium sulfate heptahydrate. (Public Domain; Chemicalinterest).
Hydrates have various uses in the health industry. Calcium sulfate hemihydrate (\(\ce{CaSO4•½H2O}\)), known as plaster of Paris, is used to make casts for broken bones. Epsom salt (\(\ce{MgSO4•7H2O}\)) is used as a bathing salt and a laxative. Aluminum chloride hexahydrate is an active ingredient in antiperspirants. The accompanying table lists some useful hydrates.
Table \(1\): Names and Formulas of Some Widely Used Hydrates
Formula Name Uses
AlCl3•6H2O aluminum chloride hexahydrate antiperspirant
CaSO4•½H2O calcium sulfate hemihydrate (plaster of Paris) casts (for broken bones and castings)
CaSO4•2H2O calcium sulfate dihydrate (gypsum) drywall component
CoCl2•6H2O cobalt(II) chloride hexahydrate drying agent, humidity indicator
CuSO4•5H2O copper(II) sulfate pentahydrate fungicide, algicide, herbicide
MgSO4•7H2O magnesium sulfate heptahydrate (Epsom salts) laxative, bathing salt
Na2CO3•10H2O sodium carbonate decahydrate (washing soda) laundry additive/cleaneKEY TAKEAWAY
KEY TAKEAWAY
• Formula masses of ionic compounds can be determined from the masses of the atoms in their formulas.
EXERCISES
1. What is the relationship between atomic mass and formula mass?
2. How are subscripts used to determine a formula mass when more than one polyatomic ion is present in a chemical formula?
3. What is the formula mass for the ionic compound formed by each pair of ions?
1. Na+ and Br
2. Mg2+ and Br
3. Mg2+ and S2−
4. What is the formula mass for the ionic compound formed by each pair of ions?
1. K+ and Cl
2. Mg2+ and Cl
3. Mg2+ and Se2
5. What is the formula mass for the ionic compound formed by each pair of ions?
1. Na+ and N3−
2. Mg2+ and N3−
3. Al3+ and S2−
6. What is the formula mass for the ionic compound formed by each pair of ions?
1. Li+ and N3−
2. Mg2+ and P3−
3. Li+ and P3−
7. What is the formula mass for each compound?
1. FeBr3
2. FeBr2
3. Au2S3
4. Au2S
8. What is the formula mass for each compound?
1. Cr2O3
2. CrO
3. PbCl2
4. PbCl4
9. What is the formula mass for each compound?
1. Cr(NO3)3
2. Fe3(PO4)2
3. CaCrO4
4. Al(OH)3
10. What is the formula mass for each compound?
1. NH4NO3
2. K2Cr2O7
3. Cu2CO3
4. NaHCO3
11. What is the formula mass for each compound?
1. Al(HSO4)3
2. Mg(HSO4)2
12. What is the formula mass for each compound?
1. Co(HCO3)2
2. LiHCO3
Answers
1. The formula mass is the sum of the atomic masses of the atoms in the formula.
2. The subscript is distributed throughout the parentheses to determine the total number of atoms in the formula.
3.
1. 102.89 amu
2. 184.11 amu
3. 56.38 amu
4.
1. 74.55 amu
2. 95.21 amu
3. 103.28 amu
5.
1. 82.98 amu
2. 100.95 amu
3. 150.17 amu
6.
1. 35.01 amu
2. 134.87 amu
3. 51.79 amu
7.
1. 295.55 amu
2. 215.65 amu
3. 490.15 amu
4. 426.01 amu
8.
1. 152.00 amu
2. 68.00 amu
3. 278.10 amu
4. 349.00 amu
9.
1. 238.03 amu
2. 357.49 amu
3. 156.08 amu
4. 78.01 amu
10.
1. 80.06 amu
2. 294.20 amu
3. 187.11 amu
4. 84.01 amu
11.
1. 318.22 amu
2. 218.47 amu
12.
1. 180.97 amu
2. 67.96 amu | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.05%3A_Formula_Mass.txt |
Learning Objectives
• To describe the physical properties of ionic compounds.
The figure below shows just a few examples of the color and brilliance of naturally occurring ionic crystals. The regular and orderly arrangement of ions in the crystal lattice is responsible for the various shapes of these crystals, while transition metal ions give rise to the colors.
Physical Properties of Ionic Compounds
Melting Points
Because of the many simultaneous attractions between cations and anions that occur, ionic crystal lattices are very strong. The process of melting an ionic compound requires the addition of large amounts of energy in order to break all of the ionic bonds in the crystal. For example, sodium chloride has a melting temperature of about 800oC. As a comparison, the molecular compound water melts at 0 °C.
Shattering
Ionic compounds are generally hard, but brittle. Why? It takes a large amount of mechanical force, such as striking a crystal with a hammer, to force one layer of ions to shift relative to its neighbor. However, when that happens, it brings ions of the same charge next to each other (see below). The repulsive forces between like-charged ions cause the crystal to shatter. When an ionic crystal breaks, it tends to do so along smooth planes because of the regular arrangement of the ions.
Conductivity
Another characteristic property of ionic compounds is their electrical conductivity. The figure below shows three experiments in which two electrodes that are connected to a light bulb are placed in beakers containing three different substances.
In the first beaker, distilled water does not conduct a current because water is a molecular compound. In the second beaker, solid sodium chloride also does not conduct a current. Despite being ionic and thus composed of charged particles, the solid crystal lattice does not allow the ions to move between the electrodes. Mobile charged particles are required for the circuit to be complete and the light bulb to light up. In the third beaker, the NaCl has been dissolved into the distilled water. Now the crystal lattice has been broken apart and the individual positive and negative ions can move. Cations move to one electrode, while anions move to the other, allowing electricity to flow (see figure below). Melting an ionic compound also frees the ions to conduct a current. Ionic compounds conduct an electric current when melted or dissolved in water. The dissolution of ionic compounds in water will be discussed in Section 9.3.
Example \(1\)
Write the dissociation equation of solid NaCl in water.
Solution
NaCl(s) → Na+(aq) + Cl(aq)
Exercise \(1\)
Write the dissociation equation of solid NH4NO3 in water.
Answer
NH4NO3(s) → NH4+(aq) + NO3(aq)
Key Takeaways
• Ionic compounds have high melting points.
• Ionic compounds are hard and brittle.
• Ionic compounds dissociate into ions when dissolved in water.
• Solutions of ionic compounds and melted ionic compounds conduct electricity, but solid materials do not.
• An ionic compound can be identified by its chemical formula: metal + nonmetal or polyatomic ions.
Contributors and Attributions
• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.
Exercises
1. Explain how the bonding in an ionic solid explains some of the properties of these solids.
2. Which type(s) of solid conduct(s) electricity in their liquid state but not in their solid state?
3. Based on chemical formula, identify which of the following is an ionic solid?
a. Hg
b. PH3
c. Ba(NO3)2 | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.06%3A__Characteristics_of_Ionic_Compounds.txt |
Additional Exercises
1. What number shell is the valence electron shell of a sodium atom? What number shell is the valence shell of a sodium ion? Explain the difference.
2. What number shell is the valence electron shell of a bromine atom? What number shell is the valence shell of a bromide ion? Explain the difference between these answers and the answers to Exercise 1.
3. What is the electron configuration of each ion?
1. K+
2. Mg2+
3. F
4. S2−
4. What is the electron configuration of each ion?
1. Li+
2. Ca2+
3. Cl
4. O2−
1. If a sodium atom were to lose two electrons, what would be the electron configuration of the resulting cation?
2. Considering that electron shells are typically separated by large amounts of energy, use your answer to Exercise 5a to suggest why sodium atoms do not form a 2+ cation.
1. If a chlorine atom were to gain two electrons, what would be the electron configuration of the resulting anion?
2. Considering that electron shells are typically separated by large amounts of energy, use your answer to Exercise 6a to suggest why chlorine atoms do not form a 2− anion.
5. Use Lewis diagrams and arrows to show the electron transfer that occurs during the formation of an ionic compound among Mg atoms and F atoms. (Hint: how many atoms of each will you need?)
6. Use Lewis diagrams and arrows to show the electron transfer that occurs during the formation of an ionic compound among K atoms and O atoms. (Hint: how many atoms of each will you need?)
7. Mercury forms two possible cations—Hg2+ and Hg22+, the second of which is actually a two-atom cation with a 2+ charge.
1. Using common names, give the probable names of these ions.
2. What are the chemical formulas of the ionic compounds these ions make with the oxide ion, O2−?
8. The uranyl ion (UO22+) is a common water-soluble form of uranium. What is the chemical formula of the ionic compound uranyl nitrate? What is the chemical formula of the ionic compound uranyl phosphate?
9. The formal chemical name of the mineral strengite is iron(III) phosphate dihydrate. What is the chemical formula of strengite? What is the formula mass of strengite?
10. What is the formula mass of MgSO4·7H2O?
11. What is the formula mass of CaSO4·½H2O?
12. What mass does 20 formula units of NaCl have?
13. What mass does 75 formula units of K2SO4 have?
14. If an atomic mass unit equals 1.66 × 10−24 g, what is the mass in grams of one formula unit of NaCl?
15. If an atomic mass unit equals 1.66 × 10−24 g, what is the mass in grams of 5.00 × 1022 formula units of NaOH?
16. If an atomic mass unit equals 1.66 × 10−24 g, what is the mass in grams of 3.96 × 1023 formula units of (NH4)2SO4?
17. Both tin and lead acquire 2+ or 4+ charges when they become ions. Use the periodic table to explain why this should not surprise you.
18. Which ion would you expect to be larger in size—In3+ or Tl3+? Explain.
19. Which ion would you expect to be smaller in size—I or Br? Explain.
20. Which ion with a 2+ charge has the following electron configuration? 1s22s22p6
21. Which ion with a 3− charge has the following electron configuration? 1s22s22p6
Answers
1. For sodium, the valence shell is the third shell; for the sodium ion, the valence shell is the second shell because it has lost all its third shell electrons.
2. The valence shell for both bromine atom and bromide ion is n=4. This is because the valence shell of bromine atom can accommodate one more electron.
1. 1s22s22p63s23p6
2. 1s22s22p6
3. 1s22s22p6
4. 1s22s22p63s23p6
1. 1s2
2. 1s22s22p63s23p6
3. 1s22s22p63s23p6
4. 1s22s22p6
1. 1s22s22p5
2. It probably requires too much energy to form.
3. 1. 1s22s22p63s23p64s1
2. Gaining the second electron would probably require too much energy.
7.
8.
9. 1. mercuric and mercurous, respectively
2. HgO and Hg2O, respectively
10. uranyl nitrate UO2(NO3)2 and uranyl phosphate (UO2)3(PO4)2
11. FePO4·2H2O; 186.86 u
12. 246.51 u
13. 145.16 u
14. 1,169 u
15. 13,070 u
16. 9.701 x 10-23 g
17. 3.32 g
18. 86.9 g
19. Both tin and lead have two p electrons and two s electrons in their valence shells.
20. Tl3+ is larger because it is found lower on the periodic table
21. Br because it is higher up on the periodic table
22. Mg2+
23. N3− | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.E%3A_Ionic_Bonding_and_Simple_Ionic_Compounds_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms and ask yourself how they relate to the topics in the chapter.
Atoms combine into compounds by forming chemical bonds. A survey of stable atoms and molecules leads to the octet rule, which says that stable atoms tend to have eight electrons in their outermost, or valence, shell. One way atoms obtain eight electrons in the valence shell is for some atoms to lose electrons while other atoms gain them. When this happens, the atoms take on an electrical charge. Charged atoms are called ions. Ions having opposite charges attract each other. This attraction is called ionic bonding, and the compounds formed are called ionic compounds.
Positively charged ions are called cations, while negatively charged ions are called anions. The formation of both cations and anions can be illustrated using electron configurations. Because elements in a column of the periodic table have the same valence shell electron configuration, atoms in the same column of the periodic table tend to form ions having the same charge. Electron dot diagrams, or Lewis diagrams, can also be used to illustrate the formation of cations and anions.
Ionic compounds are represented in writing by a chemical formula, which gives the lowest ratio of cations and anions present in the compound. In a formula, the symbol of the cation is written first, followed by the symbol of the anion. Formula unit is considered the basic unit of an ionic compound because ionic compounds do not exist as discrete units. Instead, they exist as crystals, three-dimensional arrays of ions, with cations surrounded by anions and anions surrounded by cations. Chemical formulas for ionic compounds are determined by balancing the positive charge from the cation(s) with the negative charge from the anion(s). A subscript to the right of the ion indicates that more than one of that ion is present in the chemical formula.
Some ions are groups of atoms bonded together and having an overall electrical charge. These are called polyatomic ions. Writing formulas with polyatomic ions follows the same rules as with monatomic ions, except that when more than one polyatomic ion is present in a chemical formula, the polyatomic ion is enclosed in parentheses and the subscript is outside the right parenthesis. Ionic compounds typically form between metals and nonmetals or between polyatomic ions.
Names of ionic compounds are derived from the names of the ions, with the name of the cation coming first, followed by the name of the anion. If an element can form cations of different charges, there are two alternate systems for indicating the compound’s name. In the Stock system, a roman numeral in parentheses indicates the charge on the cation. An example is the name for FeCl2, which is iron(II) chloride. In the common system, the suffixes -ous and -ic are used to stand for the lower and higher possible charge of the cation, respectively. These suffixes are attached to a stem representing the element (which frequently comes from the Latin form of the element name). An example is the common name for FeCl2, which is ferrous chloride.
The formula mass of an ionic compound is the sum of the masses of each individual atom in the formula. Care must be taken when calculating formula masses for formulas containing multiple polyatomic ions because the subscript outside the parentheses refers to all the atoms in the polyatomic ion.
Ionic compounds are hard, brittle and have very high melting points. When in solution, the ions separate and form electrolyte solutions. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds/3.S%3A_Ionic_Bonding_and_Simple_Ionic_Compounds_%28Summary%29.txt |
Ionic bonding results from the transfer of electrons among atoms or groups of atoms. In this chapter, we will consider another type of bonding—covalent bonding. We will examine how atoms share electrons to form these bonds, and we will begin to explore how the resulting compounds, such as cholesterol, are different from ionic compounds.
• 4.0: Prelude to Covalent Bonding and Simple Molecular Compounds
Cholesterol, a compound that is sometimes in the news, is a white, waxy solid produced in the liver of every animal, including humans. It is important for building cell membranes and in producing certain hormones (chemicals that regulate cellular activity in the body). As such, it is necessary for life, but why is cholesterol the object of attention? Most medical professionals recommend diets that minimize the amount of ingested cholesterol as a way of preventing heart attacks and strokes.
• 4.1: Covalent Bonds
You have already seen examples of substances that contain covalent bonds. One substance mentioned previously was water (H₂O). You can tell from its formula that it is not an ionic compound; it is not composed of a metal and a nonmetal. Consequently, its properties are different from those of ionic compounds. A covalent bond is formed between two atoms by sharing electrons.
• 4.2: Covalent Compounds - Formulas and Names
The chemical formula of a simple covalent compound can be determined from its name. The name of a simple covalent compound can be determined from its chemical formula.
• 4.3: Drawing Lewis Structures
Some molecules must have multiple covalent bonds between atoms to satisfy the octet rule.
• 4.4: Characteristics of Covalent Bonds
Covalent bonds between different atoms have different bond lengths. Covalent bonds can be polar or nonpolar, depending on the electronegativity difference between the atoms involved.
• 4.5: Characteristics of Molecules
A molecule has a certain mass, called the molecular mass. Simple molecules have geometries that can be determined from VSEPR theory.
• 4.6: Organic Chemistry
Organic chemistry is the study of the chemistry of carbon compounds. Organic molecules can be classified according to the types of elements and bonds in the molecules.
• 4.E: Covalent Bonding and Simple Molecular Compounds (Exercises)
These are homework exercises to accompany Chapter 4 of the Ball et al. "The Basics of GOB Chemistry" Textmap.
• 4.S: Covalent Bonding and Simple Molecular Compounds (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
04: Covalent Bonding and Simple Molecular Compounds
Cholesterol (\(\ce{C27H46O}\)), a compound that is sometimes in the news, is a white, waxy solid produced in the liver of every animal, including humans. It is important for building cell membranes and in producing certain hormones (chemicals that regulate cellular activity in the body). As such, it is necessary for life, but why is cholesterol the object of attention?
Besides producing cholesterol, we also ingest some whenever we eat meat or other animal-based food products. People who eat such products in large quantities, or whose metabolisms are unable to handle excess amounts, may experience an unhealthy buildup of cholesterol in their blood. Deposits of cholesterol, called plaque, may form on blood vessel walls, eventually blocking the arteries and preventing the delivery of oxygen to body tissues. Heart attacks, strokes, and other circulatory problems can result.
Most medical professionals recommend diets that minimize the amount of ingested cholesterol as a way of preventing heart attacks and strokes. Tests are available to measure cholesterol in the blood, and there are several drugs capable of lowering cholesterol levels. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.00%3A_Prelude_to_Covalent_Bonding_and_Simple_Molecular_Compounds.txt |
Learning Objectives
• To describe how a covalent bond forms.
• To apply the octet rule to covalent compounds
You have already seen examples of substances that contain covalent bonds. One substance mentioned previously was water (\(\ce{H2O}\)). You can tell from its formula that it is not an ionic compound; it is not composed of a metal and a nonmetal. Consequently, its properties are different from those of ionic compounds.
Electron Sharing
Previously, we discussed ionic bonding where electrons can be transferred from one atom to another so that both atoms have an energy-stable outer electron shell. Because most filled electron shells have eight electrons in them, chemists called this tendency the octet rule. However, there is another way an atom can achieve a full valence shell: atoms can share electrons.
This concept can be illustrated by using two hydrogen atoms, each of which has a single electron in its valence shell. (For small atoms such as hydrogen atoms, the valence shell will be the first shell, which holds only two electrons.) We can represent the two individual hydrogen atoms as follows:
In contrast, when two hydrogen atoms get close enough together to share their electrons, they can be represented as follows:
By sharing their valence electrons, both hydrogen atoms now have two electrons in their respective valence shells. Because each valence shell is now filled, this arrangement is more stable than when the two atoms are separate. The sharing of electrons between atoms is called a covalent bond, and the two electrons that join atoms in a covalent bond are called a bonding pair of electrons. A discrete group of atoms connected by covalent bonds is called a molecule—the smallest part of a compound that retains the chemical identity of that compound.
Chemists frequently use Lewis diagrams to represent covalent bonding in molecular substances. For example, the Lewis diagrams of two separate hydrogen atoms are as follows:
The Lewis diagram of two hydrogen atoms sharing electrons looks like this:
This depiction of molecules is simplified further by using a dash to represent a covalent bond. The hydrogen molecule is then represented as follows:
Remember that the dash, also referred to as a single bond, represents a pair of electrons.
The bond in a hydrogen molecule, measured as the distance between the two nuclei, is about 7.4 × 10−11 m, or 74 picometers (pm; 1 pm = 1 × 10−12 m). This particular bond length represents a balance between several forces: the attractions between oppositely charged electrons and nuclei, the repulsion between two negatively charged electrons, and the repulsion between two positively charged nuclei. If the nuclei were closer together, they would repel each other more strongly; if the nuclei were farther apart, there would be less attraction between the positive and negative particles.
Fluorine is another element whose atoms bond together in pairs to form diatomic (two-atom) molecules. Two separate fluorine atoms have the following electron dot diagrams:
Each fluorine atom contributes one valence electron, making a single bond and giving each atom a complete valence shell, which fulfills the octet rule:
The circles show that each fluorine atom has eight electrons around it. As with hydrogen, we can represent the fluorine molecule with a dash in place of the bonding electrons:
Each fluorine atom has six electrons, or three pairs of electrons, that are not participating in the covalent bond. Rather than being shared, they are considered to belong to a single atom. These are called nonbonding pairs (or lone pairs) of electrons.
Covalent Bonds between Different Atoms
Now that we have looked at electron sharing between atoms of the same element, let us look at covalent bond formation between atoms of different elements. Consider a molecule composed of one hydrogen atom and one fluorine atom:
Each atom needs one additional electron to complete its valence shell. By each contributing one electron, they make the following molecule:
In this molecule, the hydrogen atom does not have nonbonding electrons, while the fluorine atom has six nonbonding electrons (three lone electron pairs). The circles show how the valence electron shells are filled for both atoms.
Example \(1\)
Use Lewis diagrams to indicate the formation of the following:
1. Cl2
2. HBr
Solution
a. When two chlorine atoms form a chlorine molecule, they share one pair of electrons. In Cl2 molecule, each chlorine atom is surrounded by an octet number of electrons.
The Lewis diagram for a Cl2 molecule is similar to the one for F2 (shown above).
b. When a hydrogen atom and a bromine atom form HBr, they share one pair of electrons. In the HBr molecule, H achieves a full valence of two electrons (duet) while Br achieves an octet. The Lewis diagram for HBr is similar to that for HF shown above.
Exercise \(1\)
Draw the Lewis diagram for each compound.
1. a molecule composed of one chlorine atom and one fluorine atom
2. a molecule composed of one hydrogen atom and one iodine atom
Answer a:
Answer b:
Covalent Bonds in Larger Molecules
The formation of a water molecule from two hydrogen atoms and an oxygen atom can be illustrated using Lewis dot symbols (shown below).
The structure on the right is the Lewis electron structure, or Lewis structure, for \(\ce{H2O}\). With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown (below).
Other large molecules are constructed in a similar fashion, with some atoms participating in more than one covalent bond. For example, methane (\(\ce{CH4}\)), the central carbon atom bonded to four hydrogen atoms, can be represented using either of the Lewis structures below. Again, sharing electrons between C and H atoms results in C achieving and octet while H achieving a duet number of electrons.
How Many Covalent Bonds Are Formed?
The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons). In the Lewis structure, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For example, each atom of a group 4A (14) element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CH4 (methane). Group 5A (15) elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH3 (ammonia). Oxygen and other atoms in group 6A (16) obtain an octet by forming two covalent bonds. Fluorine and the other halogens in group 7A (17) have seven valence electrons and can obtain an octet by forming one covalent bond.
Typically, the atoms of group 4A form 4 covalent bonds; group 5A form 3 bonds; group 6A form 2 bonds; and group 7A form one bond. The number of electrons required to obtain an octet determines the number of covalent bonds an atom can form. This is summarized in the table below. In each case, the sum of the number of bonds and the number of lone pairs is 4, which is equivalent to eight (octet) electrons.
This table shows atoms and their group numbers, and how many bonds and lone pairs each has.
Atom (Group number) Number of Bonds Number of Lone Pairs
Carbon (Group 14 or 4A) 4 0
Nitrogen (Group 15 or 5A) 3 1
Oxygen (Group 16 or 6A) 2 2
Fluorine (Group 17 or 7A) 1 3
Because hydrogen only needs two electrons to fill its valence shell, it follows the duet rule. It is an exception to the octet rule. Hydrogen only needs to form one bond. This is the reason why H is always a terminal atom and never a central atom. Figure \(1\) shows the number of covalent bonds various atoms typically form.
The transition elements and inner transition elements also do not follow the octet rule since they have d and f electrons involved in their valence shells.
Example \(2\)
Examine the Lewis structure of OF2 below. Count the number of bonds formed by each element. Based on the element's location in the periodic table, does it correspond to the expected number of bonds shown in Table 4.1? Does the Lewis structure below follow the octet rule?
Solution
Yes. F (group 7A) forms one bond and O (group 6A) forms 2 bonds. Each atom is surrounded by 8 electrons. This structure satisfies the octet rule.
Exercise \(2\)
Examine the Lewis structure of NCl3 below. Count the number of bonds formed by each element. Based on the element's location in the periodic table, does it correspond to the expected number of bonds shown in Table 4.1? Does the Lewis structure below follow the octet rule?
Answer
Both Cl and N form the expected number of bonds. Cl (group 7A) has one bond and 3 lone pairs. The central atom N (group 5A) has 3 bonds and one lone pair. Yes, the Lewis structure of NCl3 follows the octet rule.
Key Takeaways
• A covalent bond is formed between two atoms by sharing electrons.
• The number of bonds an element forms in a covalent compound is determined by the number of electrons it needs to reach octet.
• Hydrogen is an exception to the octet rule. H forms only one bond because it needs only two electrons. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.01%3A_Covalent_Bonds.txt |
Learning Objectives
• Identify covalent and ionic compounds.
• Determine the chemical formula of a simple covalent compound from its name.
• Determine the name of a simple covalent compound from its chemical formula.
COVALENTANDIONICCOMPOUNDS
What elements make covalent bonds? Covalent bonds form when two or more nonmetals combine. For example, both hydrogen and oxygen are nonmetals, and when they combine to make water, they do so by forming covalent bonds. Compounds that are composed of only non-metals or semi-metals with non-metals will display covalent bonding and will be classified as molecular compounds.
As a general rule of thumb, compounds that involve a metal binding with either a non-metal or a semi-metal will display ionic bonding. Thus, the compound formed from sodium and chlorine will be ionic (a metal and a non-metal). Nitrogen monoxide (NO) will be a covalently bound molecule (two non-metals), silicon dioxide (SiO2) will be a covalently bound molecule (a semi-metal and a non-metal) and MgCl2 will be ionic (a metal and a non-metal).
A polyatomic ion is an ion composed of two or more atoms that have a charge as a group (poly = many). The ammonium ion (see figure below) consists of one nitrogen atom and four hydrogen atoms. Together, they comprise a single ion with a 1+ charge and a formula of NH4+. The carbonate ion (see figure below) consists of one carbon atom and three oxygen atoms and carries an overall charge of 2−. The formula of the carbonate ion is CO32.
The atoms of a polyatomic ion are tightly bonded together and so the entire ion behaves as a single unit. Several examples are found in Table 3.3.1. Nonmetal atoms in polyatomic ions are joined by covalent bonds, but the ion as a whole participates in ionic bonding. For example, ammonium chloride (NH4Cl) has ionic bonding between a polyatomic ion, \(\ce{NH_4^{+}}\), and \(\ce{Cl^{−}}\) ions, but within the ammonium ion (NH4+), the nitrogen and hydrogen atoms are connected by covalent bonds (shown above).
Both ionic and covalent bonding are also found in calcium carbonate. Calcium carbonate (CaCO3) has ionic bonding between calcium ion \(\ce{Ca^{2+}}\) and a polyatomic ion, \(\ce{CO_3^{2-}}\), but within the carbonate ion (CO32-), the carbon and oxygen atoms are connected by covalent bonds (shown above).
Characteristics of Covalent (Molecular) Compounds
Compounds that contain covalent bonds (also called molecular compounds) exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds (discussed in Section 3.6). For example, water (molecular compound) boils at 100 °C while sodium chloride (ionic compound) boils at 1413 °C. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds, being electrically neutral, are poor conductors of electricity in any state. The attraction between molecules (called intermolecular forces) will be discussed in more detail in Section 8.1
Example \(1\)
Is each compound formed from ionic bonds, covalent bonds, or both?
1. \(\ce{Na_2O}\)
2. \(\ce{Na_3PO_4}\)
3. \(\ce{N_2O_4}\)
Answer a
The elements in \(\ce{Na_2O}\) are a metal and a nonmetal, which form ionic bonds.
Answer b
Because sodium is a metal and we recognize the formula for the phosphate ion, we know that this compound is ionic. However, within the polyatomic phosphate ion, the atoms are held together by covalent bonds, so this compound contains both ionic and covalent bonds.
Answer c
The elements in \(\ce{N_2O_4}|\) are both nonmetals, rather than a metal and a nonmetal. Therefore, the atoms form covalent bonds.
Exercise \(1\)
Is each compound are formed from ionic bonds, covalent bonds, or both?
1. \(\ce{Ba(OH)_2}\)
2. \(\ce{F_2}\)
3. \(\ce{PCl_3}\)
Answer a:
both
Answer b:
covalent
Answer c:
covalent
MOLECULARFORMULAS
The chemical formulas for covalent compounds are referred to as molecular formulas because these compounds exist as separate, discrete molecules. Typically, a molecular formula begins with the nonmetal that is closest to the lower left corner of the periodic table, except that hydrogen is almost never written first (H2O is the prominent exception). Then the other nonmetal symbols are listed. Numerical subscripts are used if there is more than one of a particular atom. For example, we have already seen CH4, the molecular formula for methane. Below is the molecular formula of ammonia, NH3.
NAMING COVALENT COMPOUNDS
Naming binary (two-element) covalent compounds is similar to naming simple ionic compounds. The first element in the formula is simply listed using the name of the element. The second element is named by taking the stem of the element name and adding the suffix -ide. A system of numerical prefixes is used to specify the number of atoms in a molecule. Table \(1\) lists these numerical prefixes. Normally, no prefix is added to the first element’s name if there is only one atom of the first element in a molecule. If the second element is oxygen, the trailing vowel is usually omitted from the end of a polysyllabic prefix but not a monosyllabic one (that is, we would say “monoxide” rather than “monooxide” and “trioxide” rather than “troxide”).
Table \(1\): Numerical Prefixes for Naming Binary Covalent Compounds
Number of Atoms in Compound Prefix on the Name of the Element
1 mono-*
2 di-
3 tri-
4 tetra-
5 penta-
6 hexa-
7 hepta-
8 octa-
9 nona-
10 deca-
*This prefix is not used for the first element’s name.
Let us practice by naming the compound whose molecular formula is CCl4. The name begins with the name of the first element—carbon. The second element, chlorine, becomes chloride, and we attach the correct numerical prefix (“tetra-”) to indicate that the molecule contains four chlorine atoms. Putting these pieces together gives the name carbon tetrachloride for this compound.
Example \(2\)
Write the molecular formula for each compound.
1. chlorine trifluoride
2. phosphorus pentachloride
3. sulfur dioxide
4. dinitrogen pentoxide
Solution
If there is no numerical prefix on the first element’s name, we can assume that there is only one atom of that element in a molecule.
1. ClF3
2. PCl5
3. SO2
4. N2O5 (The di- prefix on nitrogen indicates that two nitrogen atoms are present.)
Exercise \(2\)
Write the molecular formula for each compound.
1. nitrogen dioxide
2. dioxygen difluoride
3. sulfur hexafluoride
4. selenium monoxide
Answer a:
a. NO2
Answer b:
O2F2
Answer c:
SF6
Answer d:
SeO
Because it is so unreactive, sulfur hexafluoride is used as a spark suppressant in electrical devices such as transformers.
Example \(3\)
Write the name for each compound.
1. BrF5
2. S2F2
3. CO
Solution
1. bromine pentafluoride
2. disulfur difluoride
3. carbon monoxide
Exercise \(3\)
Write the name for each compound.
1. CF4
2. SeCl2
3. SO3
Answer a:
carbon tetrafluoride
Answer b:
selenium dichloride
Answer c:
sulfur trioxide
For some simple covalent compounds, we use common names rather than systematic names. We have already encountered these compounds, but we list them here explicitly:
• H2O: water
• NH3: ammonia
• CH4: methane
Methane is the simplest organic compound. Organic compounds are compounds with carbon atoms and are named by a separate nomenclature system that we will introduce in in a separate section.
Key Takeaways
• The chemical formula of a simple covalent compound can be determined from its name.
• The name of a simple covalent compound can be determined from its chemical formula. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.02%3A_Covalent_Compounds_-_Formulas_and_Names.txt |
Learning Objectives
• To draw Lewis structures.
• To recognize molecules that are likely to have multiple covalent bonds.
DRAWINGLEWISSTRUCTURES
For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:
For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:
1. Determine the total number of valence (outer shell) electrons among all the atoms. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
4. Place all remaining electrons on the central atom.
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
Let us determine the Lewis structures of OF2 and HCN as examples in following this procedure:
1. Determine the total number of valence (outer shell) electrons in the molecule or ion. For a molecule, we add the number of valence electrons (use the main group number) on each atom in the molecule. This is the total number of electrons that must be used in the Lewis structure.
O + 2 (F) = OF2
6e- + (2 x 7e-) = 20e-
H + C + N = HCN
1e-+ 4e-+ 5e-= 10e-
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. Note that H and F can only form one bond, and are always on the periphery rather than the central atom.
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.
• In OF2, six electrons are placed on each F.
• In HCN, six electrons placed on N
4. Place all remaining electrons on the central atom.
• In OF2, 4 electrons are placed on O.
• In HCN: no electrons remain (the total valence of 10e-is reached) so nothing changes.
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
• In OF2, each atom has an octet as drawn, so nothing changes.
• In HCN, form two more C–N bonds
Finally, check to see if the total number of valence electrons are present in the Lewis structure. And then, inspect if the H atom has 2 electrons surrounding it and if each of the main group atoms is surrounded by 8 electrons.
MULTIPLEBONDS
In many molecules, the octet rule would not be satisfied if each pair of bonded atoms shares only two electrons. Review HCN in Step 5 above. Another example is carbon dioxide (CO2). CO2 has a total valence of 4e- + (2 x 6e-) = 16e-. Following steps 1 to 4, we draw the following:
This does not give the carbon atom a complete octet; only four electrons are in its valence shell. This arrangement of shared electrons is far from satisfactory.
In this case, more than one pair of electrons must be shared between two atoms for both atoms to have an octet. A second electron pair from each oxygen atom must be shared with the central carbon atom shown by the arrows above. A lone pair from each O must be converted into a bonding pair of electrons.
In this arrangement, the carbon atom shares four electrons (two pairs) with the oxygen atom on the left and four electrons with the oxygen atom on the right. There are now eight electrons around each atom. Two pairs of electrons shared between two atoms make a double bond between the atoms, which is represented by a double dash:
Some molecules contain triple bonds (like HCN, shown above). Triple bonds are covalent bonds in which three pairs of electrons are shared by two atoms. Another compound that has a triple bond is acetylene (C2H2), whose Lewis diagram is as follows:
Example \(1\)
Draw the Lewis diagram for each molecule.
1. \(\ce{N2}\)
2. \(\ce{CH2O}\) (The carbon atom is the central atom.) One application of CH2O, also called formaldehyde, is the preservation of biological specimens. Aqueous solutions of CH2O are called formalin and have a sharp, characteristic (pungent) odor.
Solution
1. The total number of electrons is 2 x 5 = 10 electrons. The bond between the two nitrogen atoms is a triple bond. The Lewis diagram for N2is as follows:
1. The total number of electrons is 4 x 2(1) + 6 = 12 electrons. In CH2O, the central atom is surrounded by two different types of atoms. The Lewis diagram that fills each atom’s valence electron shell is as follows:
Exercise \(1\)
Draw the Lewis diagram for each molecule.
1. \(\ce{O2}\)
2. \(\ce{C2H4}\)
Answer a:
or
Answer b:
or or
Example \(2\)
Which is the correct Lewis structure for N2H2?
A.
B.
C.
Solution
Lewis structure A is the correct answer. It has a total of (2 x 5e-) + (2 x 1e-) = 12e-. Each of the N atoms satisfy the octet requirement and the H atoms follow the duet rule.
Structure B is electron deficient. It has only 10e- instead of 12.
Structure C has 14 (2 extra) electrons. The N atoms do not satisfy the octet.
Exercise \(2\)
Which is the correct Lewis structure for NOCl?
A.
B.
C.
Answer
Structure A violates the octet rule; N is surrounded by only 6e-.
Structure B violates the octet rule; Cl has 10e- around it. Furthermore, there are a total of 20e- instead of 18e-.
Structure C is the correct structure. It has a total of 6e- + 5e- + 7e- = 18e-. Each atom is surrounded by 8 electrons (octet rule).
Key Takeaways
• A Lewis structure shows the bonding and nonbonding electrons around individual atoms in a molecule.
• Some molecules must have multiple covalent bonds between atoms to satisfy the octet rule.
• A double bond contains four electrons and a triple bond contains six electrons. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.03%3A_Drawing_Lewis_Structures.txt |
Learning Objectives
• Recognize bond characteristics of covalent compounds: bond length and bond polarity.
• Use electronegativity values to predict bond polarity.
Bond Length
We previously stated that the covalent bond in the hydrogen molecule (H2) has a certain length (about 7.4 × 10−11 m). Other covalent bonds also have known bond lengths, which are dependent on both the identities of the atoms in the bond and whether the bonds are single, double, or triple bonds. Table \(1\) lists the approximate bond lengths for some single covalent bonds. The exact bond length may vary depending on the identity of the molecule but will be close to the value given in the table.
Table \(1\): Approximate Bond Lengths of Some Single Bonds
Bond Length (× \(10^{−12}\, m\))
H–H 74
H–C 110
H–N 100
H–O 97
H–I 161
C–C 154
C–N 147
C–O 143
N–N 145
O–O 145
Table \(2\) compares the lengths of single covalent bonds with those of double and triple bonds between the same atoms. Without exception, as the number of covalent bonds between two atoms increases, the bond length decreases. With more electrons between the two nuclei, the nuclei can get closer together before the internuclear repulsion is strong enough to balance the attraction.
Table \(2\): Comparison of Bond Lengths for Single and Multiple Bonds
Bond Length (× \(10^{−12}\, m\))
C–C 154
C=C 134
C≡C 120
C–N 147
C=N 128
C≡N 116
C–O 143
C=O 120
C≡O 113
N–N 145
N=N 123
N≡N 110
O–O 145
O=O 121
Electronegativity and Bond Polarity
Although we defined covalent bonding as electron sharing, the electrons in a covalent bond are not always shared equally by the two bonded atoms. Unless the bond connects two atoms of the same element, as in H2, there will always be one atom that attracts the electrons in the bond more strongly than the other atom does, as in HCl, shown in Figure \(1\). A covalent bond that has an equal sharing of electrons (Figure \(\PageIndex{1a}\)) is called a nonpolar covalent bond. A covalent bond that has an unequal sharing of electrons, as in Figure \(\PageIndex{1b}\), is called a polar covalent bond.
The distribution of electron density in a polar bond is uneven. It is greater around the atom that attracts the electrons more than the other. For example, the electrons in the H–Cl bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Note that the shaded area around Cl in Figure \(\PageIndex{1b}\) is much larger than it is around H.
This imbalance in electron density results in a buildup of partial negative charge (designated as δ−) on one side of the bond (Cl) and a partial positive charge (designated δ+) on the other side of the bond (H). This is seen in Figure \(\PageIndex{2a}\). The separation of charge in a polar covalent bond results in an electric dipole (two poles), represented by the arrow in Figure \(\PageIndex{2b}\). The direction of the arrow is pointed toward the δ− end while the + tail of the arrow indicates the δ+ end of the bond.
Any covalent bond between atoms of different elements is a polar bond, but the degree of polarity varies widely. Some bonds between different elements are only minimally polar, while others are strongly polar. Ionic bonds can be considered the ultimate in polarity, with electrons being transferred rather than shared. To judge the relative polarity of a covalent bond, chemists use electronegativity, which is a relative measure of how strongly an atom attracts electrons when it forms a covalent bond. There are various numerical scales for rating electronegativity. Figure \(3\) shows one of the most popular—the Pauling scale.
Looking Closer: Linus Pauling
Arguably the most influential chemist of the 20th century, Linus Pauling (1901–94) is the only person to have won two individual (that is, unshared) Nobel Prizes. In the 1930s, Pauling used new mathematical theories to enunciate some fundamental principles of the chemical bond. His 1939 book The Nature of the Chemical Bond is one of the most significant books ever published in chemistry.
By 1935, Pauling’s interest turned to biological molecules, and he was awarded the 1954 Nobel Prize in Chemistry for his work on protein structure. (He was very close to discovering the double helix structure of DNA when James Watson and James Crick announced their own discovery of its structure in 1953.) He was later awarded the 1962 Nobel Peace Prize for his efforts to ban the testing of nuclear weapons.
In his later years, Pauling became convinced that large doses of vitamin C would prevent disease, including the common cold. Most clinical research failed to show a connection, but Pauling continued to take large doses daily. He died in 1994, having spent a lifetime establishing a scientific legacy that few will ever equal.
The polarity of a covalent bond can be judged by determining the difference in the electronegativities of the two atoms making the bond. The greater the difference in electronegativities, the greater the imbalance of electron sharing in the bond. Although there are no hard and fast rules, the general rule is if the difference in electronegativities is less than about 0.4, the bond is considered nonpolar; if the difference is greater than 0.4, the bond is considered polar. If the difference in electronegativities is large enough (generally greater than about 1.8), the resulting compound is considered ionic rather than covalent. An electronegativity difference of zero, of course, indicates a nonpolar covalent bond.
Example \(1\)
Describe the electronegativity difference between each pair of atoms and the resulting polarity (or bond type).
1. C and H
2. H and H
3. Na and Cl
4. O and H
Solution
1. Carbon has an electronegativity of 2.5, while the value for hydrogen is 2.1. The difference is 0.4, which is rather small. The C–H bond is therefore considered nonpolar.
2. Both hydrogen atoms have the same electronegativity value—2.1. The difference is zero, so the bond is nonpolar.
3. Sodium’s electronegativity is 0.9, while chlorine’s is 3.0. The difference is 2.1, which is rather high, and so sodium and chlorine form an ionic compound.
4. With 2.1 for hydrogen and 3.5 for oxygen, the electronegativity difference is 1.4. We would expect a very polar bond. The sharing of electrons between O and H is unequal with the electrons more strongly drawn towards O.
Exercise \(1\)
Describe the electronegativity (EN) difference between each pair of atoms and the resulting polarity (or bond type).
1. C and O
2. K and Br
3. N and N
4. Cs and F
Answer a:
The EN difference is 1.0 , hence polar. The sharing of electrons between C and O is unequal with the electrons more strongly drawn towards O.
Answer b:
The EN difference is greater than 1.8, hence ionic.
Answer c:
Identical atoms have zero EN difference, hence nonpolar.
Answer d:
The EN difference is greater than 1.8, hence ionic.
Bond Polarity and Molecular Polarity
If there is only one bond in the molecule, the bond polarity determines the molecular polarity. Any diatomic molecule in which the two atoms are the same element must be a nonpolar molecule. A diatomic molecule that consists of a polar covalent bond, such as HF, is a polar molecule. A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. The two electrically charged regions on either end of the molecule are called poles, similar to a magnet having a north and a south pole. Hence, a molecule with two poles is called a dipole. A simplified way to depict polar molecules like HF is pictured below (see figure below).
When placed between oppositely charged plates, polar molecules orient themselves so that their positive ends are closer to the negative plate and their negative ends are closer to the positive plate (see Figure 4.4.6 below).
Experimental techniques involving electric fields can be used to determine if a certain substance is composed of polar molecules and to measure the degree of polarity.
For molecules with more than two atoms, the molecular geometry must also be taken into account when determining if the molecule is polar or nonpolar. (Figure \(7\)) is a comparison between carbon dioxide and water. Carbon dioxide (CO2) is a linear molecule. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the C atom to each O atom. However, since the dipoles are of equal strength and are oriented directly opposite each other, they cancel each other out, and the overall molecular polarity of CO2 is zero. CO2 is a nonpolar molecule. In H2O, the orientation of the two O–H bonds is bent: one end of the molecule has a partial positive charge, and the other end has a partial negative charge. In short, the molecule itself is polar. The polarity of water has an enormous impact on its physical and chemical properties. For example, the boiling point of water (100°C) is high for such a small molecule due to the fact that polar molecules attract each other strongly. On the other hand, the nonpolar carbon dioxide becomes a gas at −77°C, almost 200° lower than the temperature at which water boils. More details will be presented in the next section.
Key Takeaways
• Covalent bonds between different atoms have different bond lengths.
• Covalent bonds can be polar or nonpolar, depending on the electronegativity difference between the atoms involved. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.04%3A_Characteristics_of_Covalent_Bonds.txt |
Learning Objectives
• Determine the molecular mass of a molecule.
• Predict the general shape of a simple covalent molecule.
• Predict the polarity of the molecule.
• Compare the properties of ionic and molecular compounds.
Unlike the ions in ionic compounds, which are arranged in lattices called crystals, molecules of covalent compounds exist as discrete units with a characteristic mass and a certain three-dimensional shape.
Molecular Mass
The mass of a molecule—the molecular mass (sometimes called the molecular weight)—is simply the sum of the masses of its atoms. As with formula masses, it is important that you keep track of the number of atoms of each element in the molecular formula to obtain the correct molecular mass.
Example \(1\)
What is the molecular mass of each covalent compound?
1. H2O
2. C6H6
3. NO2
4. N2O5
Solution
Use the atomic masses from the Periodic Table in Section 2.7.
The molecular formula H2O indicates that there are two hydrogen atoms and one oxygen atom in each molecule. Summing the masses of these atoms,
This table finds the mass of atoms.
2 H: 2 × 1.01 = 2.02 amu
1 O: + 16.00 amu
Total: 18.02 amu
The molecular mass of H2O is 18.02 amu.
This table finds the mass of atoms.
6 C: 6 × 12.01 = 72.06 amu
6 H: 6 × 1.01 = + 6.06 amu
Total: 78.12 amu
The molecular mass of C6H6 is 78.12 amu.
This table finds the mass of atoms.
1 N: 14.01 amu
2 O: 2 × 16.00 = + 32.00 amu
Total: 46.01 amu
The molecular mass of NO2 is 46.01 amu.
This table finds the mass of atoms.
2 N: 2 × 14.01 = 28.02 amu
5 O: 5 × 16.00 = + 80.00 amu
Total: 108.02 amu
The molecular mass of N2O5 is 108.02 amu.
Note that the two different nitrogen and oxygen compounds in these examples have different molecular masses.
Exercise \(1\)
What is the molecular mass of each covalent compound?
1. C2H2
2. CO
3. CO2
4. BF3
Answer a:
26.04 amu
Answer b:
28.01 amu
Answer c:
44.01 amu
Answer d:
67.81 amu
Molecular Shape: VSEPR Theory
Unlike ionic compounds, with their extended crystal lattices, covalent molecules are discrete units with specific three-dimensional shapes. The shape of a molecule is determined by the fact that covalent bonds, which are composed of negatively charged electrons, tend to repel one another. This concept is called the valence shell electron pair repulsion (VSEPR) theory. For example, the two covalent bonds in \(\ce{BeCl2}\) stay as far from each other as possible, ending up 180° apart from each other. The result is a linear molecule:
The three covalent bonds in BF3 repel each other to form 120° angles in a plane, in a shape called trigonal planar:
The molecules \(\ce{BeCl2}\) and \(\ce{BF3}\) actually violate the octet rule; however, such exceptions are rare and will not be discussed in this text.
Try sticking three toothpicks into a marshmallow or a gumdrop and see if you can find different positions where your “bonds” are farther apart than the planar 120° orientation.
The four covalent bonds in CCl4 arrange themselves three dimensionally, pointing toward the corner of a tetrahedron and making bond angles of 109.5°. CCl4 is said to have a tetrahedral shape:
Table describes how to use the number of atoms around the central atom to determine the geometry of a molecule, as well as examples.
Atoms Around Central Atom Geometry Example
2 \(\ce{AB_2}\) Linear \(\ce{BeCl_2}\)
3 \(\ce{AB_3}\) Trigonal Planar \(\ce{BF_3}\)
4 \(\ce{AB_4}\) Tetrahedral \(\ce{CCl_4}\)
In determining the shapes of molecules, it is useful to first determine the Lewis diagram for a molecule. The shapes of molecules with multiple bonds are determined by treating the multiple bonds as one bond. Thus, formaldehyde (CH2O) has a shape similar to that of BF3. It is trigonal planar.
Molecules With Lone Pairs Around Central Atom
Molecules with lone electron pairs around the central atom have a shape based on the position of the atoms, not the electron pairs. For example, NH3 has one lone electron pair and three bonded electron pairs. These four electron pairs repel each other and adopt a tetrahedral arrangement. However, the shape of the molecule is described in terms of the positions of the atoms, not the lone electron pairs. Thus, NH3 is said to have a trigonal pyramidal shape, not a tetrahedral one.
Similarly, H2O has two lone pairs of electrons around the central oxygen atom and two bonded electron pairs. Although the four electron pairs adopt a tetrahedral arrangement, the shape of the molecule is described by the positions of the atoms only. The shape of H2O is bent with an approximate 109.5° angle.
In summary, to determine the molecular geometry:
Step 1: Draw the Lewis structure.
Step 2: Count the number of bonds (a double/triple bond counts as one) and lone pairs around the central atom.
Step 3: Use Table 4.5.1 to determine the molecular geometry.
Table 4.5.1: The molecular geometry depends on the number of bonds and lone pairs around the central atom, A.
Example \(1\)
What is the geometry of the ammonium ion, NH4+? Its Lewis structure is shown below. How is this different from ammonia, NH3?
Solution
In ammonium ion, the central atom N has 4 bonds and no lone pair. It is equivalent to the below in Table 4.5.1. Hence, this is tetrahedral.
In ammonia (NH3), shown below, N has 3 bonds and one lone pair.
It is equivalent to the below in Table 4.5.1. Hence, the shape of this molecule is trigonal pyramid.
Exercise \(1\)
What is the molecular shape of nitrosyl chloride, a highly corrosive, reddish-orange gas? Its Lewis structure is shown below.
Answer
Focus on the central atom, N. It has a double bond to O, count this as one bond. It also has a single bond to Cl. Thus, N has 2 bonds and one lone pair. These 3 electron pairs will spread out 120 degrees from each other. But, since the shape is defined by the arrangement of the atoms only, the shape is bent or angular. If you consult Table 4.5.1, this molecule is equivalent to the below. Hence, two bonds and one lone pair has a bent or angular shape.
Molecular Polarity
In general, a molecule is nonpolar if all its bonds are nonpolar. Examples are I2, O2, H2, CH4, C2H6 and C3H8.
In general, a molecule is polar if it contains polar bonds EXCEPT when the bond polarities cancel each other. As mentioned in Section 4.4, the shape of the CO2 molecule (linear) orients the two C=O polar bonds directly opposite each other, thus cancelling each other's effect. Carbon dioxide (CO2) is a nonpolar molecule.
On the other hand, water (as discussed above) is a bent molecule because of the two lone pairs on the central oxygen atom. Because of the bent shape, the dipoles do not cancel each other out and the water molecule is polar. In the figure below, the individual H-O polar bonds represented by the two red arrows are not directly opposite each other. These two dipoles don't cancel each other out. In fact, the net dipole (blue arrow) points upward. There is a resultant partial positive charge at one end (between the two H atoms) and a partial negative charge on the other end (where O is located). The uneven distribution of charge or the overall dipole is shown by the blue arrow below (Figure 4.5.1). Hence, water is polar (has + and - poles) while carbon dioxide is nonpolar.
Similarly, in BF3 (trigonal planar), the effect of a B-F bond is cancelled by the sum of the other two B-F bonds (see video). Hence, a trigonal planar molecule (BF3) is nonpolar because the bond polarities cancel each other, but a trigonal pyramidal molecule (NH3) is polar.
Some other molecules are shown in the figure below. Notice that a tetrahedral molecule such as CCl4 is nonpolar. However, if the peripheral atoms are not of the same electronegativity, the bond polarities don't cancel and the molecule becomes polar, as in CH3Cl.
Physical Properties of Molecular Compounds
The physical state and properties of a particular compound depend in large part on the type of chemical bonding it displays. Molecular compounds, sometimes called covalent compounds, display a wide range of physical properties due to the different types of intermolecular attractions such as different kinds of polar interactions. The melting and boiling points of molecular compounds are generally quite low compared to those of ionic compounds. This is because the energy required to disrupt the intermolecular forces (discussed in Section 8.1) between molecules is far less than the energy required to break the ionic bonds in a crystalline ionic compound. Since molecular compounds are composed of neutral molecules, their electrical conductivity is generally quite poor, whether in the solid or liquid state. Ionic compounds do not conduct electricity in the solid state because of their rigid structure, but conduct well when either molten or dissolved into a solution. The water solubility of molecular compounds is variable and depends primarily on the type of intermolecular forces involved. Substances that exhibit hydrogen bonding or dipole-dipole forces are generally water soluble, whereas those that exhibit only London dispersion forces are generally insoluble. Most, but not all, ionic compounds are quite soluble in water. The table below summarizes some of the differences between ionic and molecular compounds.
Table 4.5.2: Comparison of Ionic and Molecular Compounds
Property Ionic Compounds Molecular Compounds
Type of elements Metal and nonmetal Nonmetals only
Bonding Ionic - transfer of electron(s) between atoms Covalent - sharing of pair(s) of electrons between atoms
Representative unit Formula unit Molecule
Physical state at room temperature Solid Gas, liquid, or solid
Water solubility Usually high Variable
Melting and boiling temperatures Generally high Generally low
Electrical conductivity Good when molten or in solution Poor
In summary, covalent compounds are softer, have lower boiling and melting points, are more flammable, are less soluble in water and do not conduct electricity compared to ionic compounds. The individual melting and boiling points, solubility and other physical properties of molecular compounds depend on molecular polarity.
Example \(2\)
Describe the shape of each molecule. Is it polar or nonpolar?
1. PCl3
2. CO2
Solution
1. The Lewis diagram for PCl3 is as follows:
Focus on the central atom, P that has 3 bonds and one lone pair. The four electron pairs arrange themselves tetrahedrally, but the lone electron pair is not considered in describing the molecular shape. Like NH3, this molecule is pyramidal. The 3 P-Cl bonds don't cancel each other. This is polar.
• The Lewis diagram for CO2 is as follows:
Focus on the central atom, C. The multiple bonds are treated as one group, hence C has 2 bonds and zero lone pair. CO2 has only two groups of electrons that repel each other. They will direct themselves 180° apart from each other, so CO2 molecules are linear. This is highly symmetrical, with the two opposite dipoles cancelling each other. The CO2 molecule is nonpolar.
Exercise \(2\)
Describe the shape of each molecule. Is it polar or nonpolar?
1. CBr4
2. BCl3
Answer a:
or
The Lewis structure shows 4 groups attached to the central atom, hence tetrahedral. All the 4 groups are identical and the shape is symmetrical. Hence, it is nonpolar.
Answer b:
The Lewis diagram shows 3 groups attached to the central atom, hence trigonal planar. All the 3 groups are identical and shape is symmetrical, hence, it is nonpolar.
Key Takeaways
• A molecule has a certain mass, called the molecular mass.
• Simple molecules have geometries that can be determined from VSEPR theory.
• Polar molecules result from differences in electronegativity of the atoms in the molecule.
• Dipoles that are directly opposite one another cancel each other out. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.05%3A_Characteristics_of_Molecules.txt |
Learning Objectives
• Define organic chemistry.
• Identify organic molecules as alkanes, alkenes, alkynes, alcohols, or carboxylic acids.
When methane was mentioned previously, we described it as the simplest organic compound. In this section, we introduce organic chemistry more formally. Organic chemistry is the study of the chemistry of carbon compounds. Carbon is singled out because it has a chemical diversity unrivaled by any other chemical element. Its diversity is based on the following:
• Carbon atoms bond reasonably strongly with other carbon atoms.
• Carbon atoms bond reasonably strongly with atoms of other elements.
• Carbon atoms make a large number of covalent bonds (four).
Curiously, elemental carbon is not particularly abundant. It does not even appear in the list of the most common elements in Earth’s crust. Nevertheless, all living things consist of organic compounds. Most organic chemicals are covalent compounds, which is why we introduce organic chemistry here. By convention, compounds containing carbonate ions and bicarbonate ions, as well as carbon dioxide and carbon monoxide, are not considered part of organic chemistry, even though they contain carbon.
Alkanes
The simplest organic compounds are the hydrocarbons, compounds composed of carbon and hydrogen atoms only. Some hydrocarbons have only single bonds and appear as a chain (which can be a straight chain or can have branches) of carbon atoms also bonded to hydrogen atoms. These hydrocarbons are called alkanes (saturated hydrocarbons). Each alkane has a characteristic, systematic name depending on the number of carbon atoms in the molecule. These names consist of a stem that indicates the number of carbon atoms in the chain plus the ending -ane. The stem meth- means one carbon atom, so methane is an alkane with one carbon atom. Similarly, the stem eth- means two carbon atoms; ethane is an alkane with two carbon atoms. Continuing, the stem prop- means three carbon atoms, so propane is an alkane with three carbon atoms. The stem but- means four carbon atoms; butane is an alkane with four carbon atoms. Figure \(1\) gives the Lewis structures, condensed structural formulas and molecular formulas of the four simplest alkanes. In the condensed structural formula, the covalent bonds are understood to exist between each carbon and the hydrogens associated with it, as well as between carbon atoms.
Alkenes
Some hydrocarbons have one or more carbon–carbon double bonds (denoted C=C). These hydrocarbons are called alkenes. Figure \(2\) shows the formulas and the molecular models of the two simplest alkenes. Note that the names of alkenes have the same stem as the alkane with the same number of carbon atoms in its chain but have the ending -ene. Thus, ethene is an alkene with two carbon atoms per molecule, and propene is a compound with three carbon atoms and one double bond.
Alkynes
Alkynes are hydrocarbons with a carbon–carbon triple bond (denoted C≡C) as part of their carbon skeleton. Below is the formula and the molecular model of the simplest alkyne and its systematic name. Its common name is acetylene. Its chemical formula is C2H2.
Ethyne
The names for alkynes have the same stems as for alkanes but with the ending -yne.
To Your Health: Saturated and Unsaturated Fats
Hydrocarbons are not the only compounds that can have carbon–carbon double bonds. A group of compounds called fats can have them as well, and their presence or absence in the human diet is becoming increasingly correlated with health issues.
Fats are combinations of long-chain organic compounds (fatty acids) and glycerol (C3H8O3). (For more information on fats, see Chapter 17) The long carbon chains can have either all single bonds, in which case the fat is classified as saturated, or one or more double bonds, in which case it is a monounsaturated or a polyunsaturated fat, respectively. Saturated fats are typically solids at room temperature; beef fat (tallow) is one example. Mono- or polyunsaturated fats are likely to be liquids at room temperature and are often called oils. Olive oil, flaxseed oil, and many fish oils are mono- or polyunsaturated fats.
Studies have linked higher amounts of saturated fats in people’s diets with a greater likelihood of developing heart disease, high cholesterol, and other diet-related diseases. In contrast, increases in unsaturated fats (either mono- or polyunsaturated) have been linked to a lower incidence of certain diseases. Thus, there have been an increasing number of recommendations by government bodies and health associations to decrease the proportion of saturated fat and increase the proportion of unsaturated fat in the diet. Most of these organizations also recommend decreasing the total amount of fat in the diet.
Recently, certain fats called trans fats have been implicated in the presence of heart disease. These are fats from animal sources and are also produced when liquid oils are exposed to partial hydrogenation, an industrial process that increases their saturation. Trans fats are used in many prepared and fried foods. Because they bring with them the health risks that naturally occurring saturated fats do, there has been some effort to better quantify the presence of trans fats in food products. US law now requires that food labels list the amount of trans fat in each serving.
Since the early 1900's, the US Department of Agriculture has been providing science-based dietary guidelines for the public. The most current version is the MyPlate illustration that gives a simple, visual picture of how much of what kind of foods make up a good, balanced diet. It recommends minimizing daily intake of sugars, the "bad fats", trans and saturated fat, and sodium. "Good fats", unsaturated fats or oils, are not considered a food group but do contain essential nutrients and therefore are included as part of a healthy eating pattern. The difference as simple as the difference between a single and double carbon–carbon bond, good and bad fats, can have a significant impact on health.
Functional Groups
The carbon–carbon double and triple bonds are examples of functional groups in organic chemistry. A functional group is a specific structural arrangement of atoms or bonds that imparts a characteristic chemical reactivity to a molecule. Alkanes have no functional group. A carbon–carbon double bond is considered a functional group because carbon–carbon double bonds chemically react in specific ways that differ from reactions of alkanes (for example, under certain circumstances, alkenes react with water); a carbon–carbon triple bond also undergoes certain specific chemical reactions. In the remainder of this section, we introduce two other common functional groups.
If an OH group (also called a hydroxyl group) is substituted for a hydrogen atom in a hydrocarbon molecule, the compound is an alcohol. Alcohols are named using the parent hydrocarbon name but with the final -e dropped and the suffix -ol attached. The two simplest alcohols are methanol and ethanol. Figure \(4\) shows their formulas along with a molecular model of each.
Cholesterol, described in the chapter-opening essay, has an alcohol functional group, as its name implies.
Alcohol
Ethanol (also called ethyl alcohol) is the alcohol in alcoholic beverages. Other alcohols include methanol (or methyl alcohol), which is used as a solvent and a cleaner, and isopropyl alcohol (or rubbing alcohol), which is used as a medicinal disinfectant. Neither methanol nor isopropyl alcohol should be ingested, as they are toxic even in small quantities.
Another important family of organic compounds has a carboxyl group, in which a carbon atom is double-bonded to an oxygen atom and to an OH group. Compounds with a carboxyl functional group are called carboxylic acids, and their names end in -oic acid. Figure \(5\) shows the formulas and the molecular models of the two simplest carboxylic acids, perhaps best known by the common names formic acid and acetic acid. The carboxyl group is sometimes written in molecules as COOH.
The condensed structures of methanoic acid and ethanoic acid are HCOOH and CH3COOH, respectively.
Many organic compounds are considerably more complex than the examples described here. Many compounds, such as cholesterol discussed in the chapter-opening essay, contain more than one functional group. The formal names can also be quite complex.
Example \(1\)
Identify the functional group(s) in each molecule as a double bond, a triple bond, an alcohol, or a carboxyl.
b.
c.
d. \(\ce{CH3CH2CH2CH2OH} \)
Answer a
This molecule has a double bond and a carboxyl functional group.
Answer b
This molecule has an alcohol functional group.
Answer c
This molecule has a carbon-carbon double bond and a carboxyl functional group.
Answer d
This molecule has an alcohol functional group.
Exercise \(1\)
Identify the functional group(s) in each molecule as a double bond, a triple bond, an alcohol, or a carboxyl.
d.
Answer a:
triple bond (alkyne)
Answer b:
carboxyl group
Answer c:
alcohol group
Answer d:
double bond (alkene) and carboxyl group
Career Focus: Forensic Chemist
The main job of a forensic chemist is to identify unknown materials and their origins. Although forensic chemists are most closely associated in the public mind with crime labs, they are employed in pursuits as diverse as tracing evolutionary patterns in living organisms, identifying environmental contaminants, and determining the origin of manufactured chemicals.
In a crime lab, the forensic chemist has the job of identifying the evidence so that a crime can be solved. The unknown samples may consist of almost anything—for example, paint chips, blood, glass, cloth fibers, drugs, or human remains. The forensic chemist subjects them to a variety of chemical and instrumental tests to discover what the samples are. Sometimes these samples are extremely small, but sophisticated forensic labs have state-of-the-art equipment capable of identifying the smallest amount of unknown sample.
Another aspect of a forensic chemist’s job is testifying in court. Judges and juries need to be informed about the results of forensic analyses, and it is the forensic chemist’s job to explain those results. Good public-speaking skills, along with a broad background in chemistry, are necessary to be a successful forensic chemist.
Key Takeaways
• Organic chemistry is the study of the chemistry of carbon compounds.
• Organic molecules can be classified according to the types of elements and bonds in the molecules. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.06%3A_Organic_Chemistry.txt |
Concept Review Exercises
1. How is a covalent bond formed between two atoms?
2. How does covalent bonding allow atoms in group 6A to satisfy the octet rule?
Answers
1. Covalent bonds are formed by two atoms sharing electrons.
2. The atoms in group 6A make two covalent bonds.
Exercises
1. Define covalent bond.
2. What is electron sharing?
3. Draw the Lewis diagram for the covalent bond in the H2 molecule.
4. Draw the Lewis diagram for the covalent bond in the Br2 molecule.
5. Draw the Lewis diagram for the covalent bond in the HCl molecule.
6. What is the difference between a molecule and a formula unit?
7. Why do hydrogen atoms not follow the octet rule when they form covalent bonds?
8. Draw the Lewis diagram for the covalent bonding in H2S. How many bonding electrons and nonbonding electrons are in the molecule?
9. Draw the Lewis diagram for the covalent bonding in CF4. How many bonding electrons and nonbonding electrons are in the molecule?
10. Draw the Lewis diagram for the covalent bonding in PCl3. How many bonding electrons and nonbonding electrons are in the molecule?
11. How many covalent bonds does a hydrogen atom typically form? Why?
12. How many covalent bonds does an oxygen atom typically form? Why?
13. Tellurium atoms make covalent bonds. How many covalent bonds would a tellurium atom make? Predict the formula of a compound between tellurium and hydrogen.
14. Tin atoms make covalent bonds. How many covalent bonds would a tin atom make? Predict the formula of a compound between tin and hydrogen.
15. Astatine is a synthetic element, made one atom at a time in huge “atom-smasher” machines. It is in the halogen group on the periodic table. How many covalent bonds would an atom of this element form?
16. There have been reports that atoms of element 116 (Lv) were made by smashing smaller atoms together. Using the periodic table, determine what column element 116 would be in and suggest how many covalent bonds an atom of this element would form.
Answers
1. A covalent bond is formed when two atoms share electrons.
2. Electron sharing joins two atoms in a covalent bond. This is a more stable arrangement than 2 individual atoms.
3.
4.
5.
6. A molecule is a discrete combination of atoms; a formula unit is the lowest ratio of ions in a crystal. 7. Hydrogen atoms follow the duet rule (not the octet rule). This is because it has only one shell and this shell can only hold 2 electrons.
8.
bonding electrons: 4; nonbonding electrons: 4
9.
bonding electrons: 8; nonbonding electrons: 24
10.
bonding electrons: 6; nonbonding electrons: 20
11. Hydrogen atoms form only one covalent bond because they have only one valence electron to pair.
12. Oxygen atoms form 2 covalent bonds because oxygen atoms have 6 valence electrons (2 lone pairs plus 2 unpaired electrons that are shared to achieve octet).
13. two; H2Te
14. four: SnH4
15. one
16. two
Concept Review Exercises
1. How do you recognize a covalent compound?
2. What are the rules for writing the molecular formula of a simple covalent compound?
3. What are the rules for naming a simple covalent compound?
Answers
1. A covalent compound is usually composed of two or more nonmetal elements.
2. It is just like an ionic compound except that the element further down and to the left on the periodic table is listed first and is named with the element name.
3. Name the first element first and then the second element by using the stem of the element name plus the suffix -ide. Use numerical prefixes if there is more than one atom of the first element; always use numerical prefixes for the number of atoms of the second element.
Exercises
1. Identify whether each compound has covalent bonds.
1. NaI
2. Na2CO3
3. N2O
4. SiO2
2. Identify whether each compound has covalent bonds.
1. C2H6
2. C6H5Cl
3. KC2H3O2
4. Ca(OH)2
3. Identify whether each compound has ionic bonds, covalent bonds, or both.
1. Na3PO4
2. K2O
3. COCl2
4. CoCl2
4. Identify whether each compound has ionic bonds, covalent bonds, or both.
1. FeCl3
2. Fe(NO3)3
3. (NH2)2CO
4. SO3
5. Which is the correct molecular formula—H4Si or SiH4? Explain.
6. Which is the correct molecular formula—SF6 or F6S? Explain.
7. Write the name for each covalent compound.
1. SiF4
2. NO2
3. CS2
4. P2O5
8. Write the name for each covalent compound.
1. CO
2. S2O3
3. BF3
4. GeS2
9. Write the formula for each covalent compound.
1. iodine trichloride
2. disulfur dibromide
3. arsenic trioxide
4. xenon hexafluoride
10. Write the formula for each covalent compound.
1. boron trichloride
2. carbon dioxide
3. tetraphosphorus decoxide
4. germanium dichloride
11. Write two covalent compounds that have common rather than systematic names.
12. What is the name of the simplest organic compound? What would its name be if it followed the nomenclature for binary covalent compounds?
Answers
1. no
2. yes
3. yes
4. yes
2.
1. yes
2. yes
3. yes
4. yes
1. both
2. ionic
3. covalent
4. ionic
4.
1. ionic
2. both
3. covalent
4. covalent
1. SiH4; except for water, hydrogen is almost never listed first in a covalent compound.
6. SF6; the less electronegative atom (S) is written first
1. silicon tetrafluoride
2. nitrogen dioxide
3. carbon disulfide
4. diphosphorus pentoxide
8.
1. carbon monoxide
2. disulfur trioxide
3. boron trifluoride
4. germanium disulfide
1. ICl3
2. S2Br2
3. AsO3
4. XeF6
10.
1. BCl3
2. CO2
3. P4O10
4. GeCl2
1. H2O and NH3 (water and ammonia) (answers will vary)
2. CH4; carbon tetrahydride
Exercises
1. What is one clue that a molecule has a multiple bond?
2. Draw the Lewis diagram for each of the following.
a. H2O
b. NH3
c. C2H6
d. CCl4
3. Each molecule contains double bonds. Draw the Lewis diagram for each. The first element is the central atom.
1. CS2
2. C2F4
3. COCl2
4. Each molecule contains multiple bonds. Draw the Lewis diagram for each. Assume that the first element is the central atom, unless otherwise noted.
1. N2
2. CO
3. HCN (The carbon atom is the central atom.)
4. POCl (The phosphorus atom is the central atom.)
5. Explain why hydrogen atoms do not form double bonds.
6. Why is it incorrect to draw a double bond in the Lewis diagram for MgO?
Answers
1. If single bonds between all atoms do not give all atoms (except hydrogen) an octet, multiple covalent bonds may be present.
2. a.
b.
c.
d.
3. a.
b.
c.
4. a.
b.
c.
d.
5. Hydrogen can accept only one more electron; multiple bonds require more than one electron pair to be shared.
6. MgO is an ionic compound (Mg transfers two electrons to O). The electrons are not shared hence it's incorrect to draw a double bond.
This is the Lewis dot structure of MgO.
Concept Review Exercises
1. What is the name for the distance between two atoms in a covalent bond?
2. What does the electronegativity of an atom indicate?
3. What type of bond is formed between two atoms if the difference in electronegativities is small? Medium? Large?
Answers
1. bond length
2. Electronegativity is a qualitative measure of how much an atom attracts electrons in a covalent bond.
3. nonpolar; polar; ionic
Exercises
1. Which is longer—a C–H bond or a C–O bond? (Refer to Table \(1\).)
2. Which is shorter—an N–H bond or a C–H bond? (Refer to Table \(1\).)
3. A nanometer is 10−9 m. Using the data in Table \(1\) and Table \(2\), determine the length of each bond in nanometers.
1. a C–O bond
2. a C=O bond
3. an H–N bond
4. a C≡N bond
4. An angstrom (Å) is defined as 10−10 m. Using Table \(1\) and Table \(2\), determine the length of each bond in angstroms.
1. a C–C bond
2. a C=C bond
3. an N≡N bond
4. an H–O bond
5. Refer to Exercise 3. Why is the nanometer unit useful as a unit for expressing bond lengths?
6. Refer to Exercise 4. Why is the angstrom unit useful as a unit for expressing bond lengths?
7. Using Figure \(3\), determine which atom in each pair has the higher electronegativity.
1. H or C
2. O or Br
3. Na or Rb
4. I or Cl
8. Using Figure \(3\), determine which atom in each pair has the lower electronegativity.
1. Mg or O
2. S or F
3. Al or Ga
4. O or I
9. Will the electrons be shared equally or unequally across each covalent bond? If unequally, to which atom are the electrons more strongly drawn?
1. a C–O bond
2. an F–F bond
3. an S–N bond
4. an I–Cl bond
10. Will the electrons be shared equally or unequally across each covalent bond? If unequally, to which atom are the electrons more strongly drawn?
• a C–C bond
• a S–Cl bond
• an O–H bond
• an H–H bond
11. Arrange the following bonds from least polar to most polar: H-F, H-N, H-O, H-C
12. Arrange the following bonds from least polar to most polar: C-F, C-N, C-O, C-C
Answers
1. A C–O bond is longer.
2. An H-N bond is shorter than an H-C bond.
1. 0.143 nm
2. 0.120 nm
3. 0.100 nm
4. 0.116 nm
4.
1. 1.54 Å
2. 1.34 Å
3. 1.10 Å
4. 0.97 Å
1. Actual bond lengths are very small, so the nanometer unit makes the expression of length easier to understand.
6. Actual bond lengths are very small, so the angstrom unit makes the expression of length easier to understand.
1. C
2. O
3. Na
4. Cl
8.
1. Mg
2. S
3. Al
4. I
9.
1. unequally toward the O
2. equally
3. unequally toward the N
4. unequally toward the Cl
10.
• equally
• unequally toward the Cl
• unequally toward the O
• equally
11. The electronegativity difference increases from 0.4; 0.9; 1.4; 1.9. Hence, the least to most polar: H-C, H-N, H-O, H-F
12. The electronegativity difference increases from 0; 0.5; 1.0; 1.5. Hence, the least to most polar: C-C, C-N, C-O, C-F
Concept Review Exercises
1. How do you determine the molecular mass of a covalent compound?
2. How do you determine the shape of a molecule?
3. How do you determine whether a molecule is polar or nonpolar?
Answers
1. The molecular mass is the sum of the masses of the atoms in the formula.
2. The shape of a molecule is determined by the position of the atoms, which in turn is determined by the repulsion of the bonded and lone electron pairs around the central atom.
3. If all the bonds in a molecule are nonpolar, the molecule is nonpolar. If it contains identical polar bonds that are oriented symmetrically opposite each other (linear, trigonal planar or tetrahedral) then the molecule is nonpolar. If it contains polar bonds that don't cancel each other's effects, the molecule is polar.
Exercises
1. What is the molecular mass of each compound?
1. H2S
2. N2O4
3. ICl3
4. HCl
2. What is the molecular mass of each compound?
1. O2F2
2. CCl4
3. C6H6
4. SO3
3. Aspirin (C9H8O4) is a covalent compound. What is its molecular mass?
4. Cholesterol (C27H46O) is a biologically important compound. What is its molecular mass?
5. What is the shape of each molecule? State whether it is polar or nonpolar.
1. H2S
2. COCl2
3. SO2
6. What is the shape of each molecule? State whether it is polar or nonpolar.
1. NBr3
2. SF2
3. SiH4
7. Predict the shape of nitrous oxide (N2O), which is used as an anesthetic. A nitrogen atom is in the center of this three-atom molecule. Is this polar?
8. Predict the shape of acetylene (C2H2), which has the two carbon atoms in the middle of the molecule with a triple bond. What generalization can you make about the shapes of molecules that have more than one central atom?
Answers
1. 34.08 amu
2. 92.02 amu
3. 233.25 amu
4. 36.46 amu
2. What is the molecular mass of each compound?
1. 70.00 amu
2. 153.81 amu
3. 78.12 amu
4. 80.06 amu
1. 180.17 amu
4. 386.73 amu
1. bent; polar
2. trigonal planar; nonpolar
3. bent; polar
6.
1. pyramidal; polar
2. bent; polar
3. tetrahedral; nonpolar
7. linear; polar
8. linear; in a molecule with more than one central atom, the geometry around each central atom needs to be examined.
Concept Review Exercises
1. What is organic chemistry?
2. What is a functional group? Give at least two examples of functional groups.
Answers
1. Organic chemistry is the study of the chemistry of carbon compounds.
2. A functional group is a specific structural arrangement of atoms or bonds that imparts a characteristic chemical reactivity to the molecule; alcohol group and carboxylic group (answers will vary).
Exercises
1. Give three reasons why carbon is the central element in organic chemistry.
2. Are organic compounds based more on ionic bonding or covalent bonding? Explain.
3. Identify the type of hydrocarbon in each structure.
4. Identify the type of hydrocarbon in each structure.
5. Identify the functional group(s) in each molecule.
6. Identify the functional group(s) in each molecule.
7. How many functional groups described in this section contain carbon and hydrogen atoms only? Name them.
8. What is the difference in the ways the two oxygen atoms in the carboxyl group are bonded to the carbon atom?
Answers
1. Carbon atoms bond reasonably strongly with other carbon atoms. Carbon atoms bond reasonably strongly with atoms of other elements. Carbon atoms make a large number of covalent bonds (four).
2. Organic compounds are based on covalent bonding or electron sharing. The atoms C, H, O, N that make up organic compounds are all nonmetals.
3.
a. alkane
b. alkene
c. alkene
d. alkyne
4.
a. alkene
b. alkane
c. alkyne
d. alkene
5.
a. alcohol
b. carboxyl
c. alcohol
d. alkene and alkyne
6.
1. a carbon-carbon double bond and alcohol
2. carboxyl group
3. carbon-carbon double bond and alcohol
4. carbon-carbon double bond; alcohol and carboxyl group
7. two; carbon-carbon double bonds and carbon-carbon triple bonds
8. There are two oxygen atoms in a carboxyl group: one is double-bonded while the other is OH, single bonded to the same carbon atom.
Additional Exercises
Use the atomic masses found in Figure 2.7.1
1. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of each molecule of (a) H2S (b) N2O4 (c) ICl3 (d) NCl3?
2. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of (a) O2F2 (b) CCl4 (c) C6H6 (d) SO3?
3. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of 5.00 × 1022 molecules of C9H8O4?
4. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of 1.885 × 1020 molecules of C27H46O?
5. Acetic acid has the following structure:
This molecule can lose a hydrogen ion (H+) and the resulting anion can combine with other cations, such as Na+:
Name this ionic compound.
6. Formic acid (HCOOH) loses a hydrogen ion to make the formate ion (HCOO). Write the formula for each ionic compound: potassium formate, calcium formate, and ferric formate.
7. Cyanogen has the formula C2N2. Propose a bonding scheme that gives each atom the correct number of covalent bonds. (Hint: the two carbon atoms are in the center of a linear molecule.)
8. How many carbon–carbon single bonds, linked together, are needed to make a carbon chain that is 1.000 cm long?
9. How many carbon–carbon double bonds, linked together, are needed to make a carbon chain that is 1.000 cm long?
10. In addition to themselves, what other atoms can carbon atoms bond with and make covalent bonds that are nonpolar (or as nonpolar as possible)?
11. What is the greatest possible electronegativity difference between any two atoms? Use Figure 4.4 to find the answer.
12. Acetaminophen, a popular painkiller, has the following structure:
Name the recognizable functional groups in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups?
13. Glutamic acid is the parent compound of monosodium glutamate (known as MSG), which is used as a flavor enhancer. Glutamic acid has the following structure:
Name the functional groups you recognize in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups?
Answers
1.
a: 5.661 × 10−23 g
b: 1.528 × 10−22 g
c: 3.874 × 10−22 g
d: 1.999 × 10−22 g
2.
a: 1.163 × 10−22 g
b: 2.555 × 10−22 g
c: 1.298 × 10−22 g
d: 1.330 × 10−22 g
3. 14.96 g
4. 0.1211 g
5. sodium acetate
6.
a. KHCOO
b. Ca(HCOO)2
c. Fe(HCOO)3
1. :N≡C–C≡N:
2. 6.49 × 107 C-C bonds
3. 7.46 × 107 C=C bonds
4. Hydrogen atoms make relatively nonpolar bonds with carbon atoms.
5. The greatest electronegativity difference is 3.2, between F and Rb.
6. alcohol; the ring with double bonds, and the O=C-NH are also likely functional groups.
7. carboxyl and -NH2 functional groups
Additional Questions
1. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of each molecule of (a) H2S (b) N2O4 (c) ICl3 (d) NCl3?
2. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of (a) O2F2 (b) CCl4 (c) C6H6 (d) SO3?
3. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of 5.00 × 1022 molecules of C9H8O4?
4. An atomic mass unit equals 1.661 × 10−24 g. What is the mass in grams of 1.885 × 1020 molecules of C27H46O?
5. Acetic acid has the following structure:
This molecule can lose a hydrogen ion (H+) and the resulting anion can combine with other cations, such as Na+:
Name this ionic compound.
6. Formic acid (HCOOH) loses a hydrogen ion to make the formate ion (HCOO). Write the formula for each ionic compound: potassium formate, calcium formate, and ferric formate.
7. Cyanogen has the formula C2N2. Propose a bonding scheme that gives each atom the correct number of covalent bonds. (Hint: the two carbon atoms are in the center of a linear molecule.)
8. The molecular formula C3H6 represents not only propene, a compound with a carbon–carbon double bond, but also a molecule that has all single bonds. Draw the molecule with formula C3H6 that has all single bonds.
9. How many carbon–carbon single bonds, linked together, are needed to make a carbon chain that is 1.000 cm long?
10. How many carbon–carbon double bonds, linked together, are needed to make a carbon chain that is 1.000 cm long?
11. In addition to themselves, what other atoms can carbon atoms bond with and make covalent bonds that are nonpolar (or as nonpolar as possible)?
12. What is the greatest possible electronegativity difference between any two atoms? Use Figure 4.4 to find the answer.
13. Acetaminophen, a popular painkiller, has the following structure:
Name the recognizable functional groups in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups?
14. Glutamic acid is the parent compound of monosodium glutamate (known as MSG), which is used as a flavor enhancer. Glutamic acid has the following structure:
Name the functional groups you recognize in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups?
Answers
1. 1a: 5.75 × 10−23 g; 1b: 1.53 × 10−22 g; 1c: 3.88 × 10−22 g; 1d: 6.06 × 10−23 g
1. 14.96 g
1. sodium acetate
1. N≡C–C≡N
1. 6.49 × 107 bonds
1. Hydrogen atoms make relatively nonpolar bonds with carbon atoms.
1. alcohol; the N–H group, the ring with double bonds, and the C=O are also likely functional groups. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.E%3A_Covalent_Bonding_and_Simple_Molecular_Compounds_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Atoms can share pairs of valence electrons to obtain a valence shell octet. This sharing of electrons is a covalent bond. A species formed from covalently bonded atoms is a molecule and is represented by a molecular formula, which gives the number of atoms of each type in the molecule. The two electrons shared in a covalent bond are called a bonding pair of electrons. The electrons that do not participate in covalent bonds are called nonbonding pairs (or lone pairs) of electrons. A covalent bond consisting of one pair of shared electrons is called a single bond.
Covalent bonds occur between nonmetal atoms. Naming simple covalent compounds follows simple rules similar to those for ionic compounds. However, for covalent compounds, numerical prefixes are used as necessary to specify the number of atoms of each element in the compound.
In some cases, more than one pair of electrons is shared to satisfy the octet rule. Two pairs of electrons are shared by two atoms to make a double bond. Three pairs of atoms are shared to make a triple bond. Single, double, and triple covalent bonds may be represented by one, two, or three dashes, respectively, between the symbols of the atoms.
The distance between two covalently bonded atoms is the bond length. Bond lengths depend on the types of atoms participating in the bond as well as the number of electron pairs being shared. A covalent bond can be a polar covalent bond if the electron sharing between the two atoms is unequal. If the sharing is equal, the bond is a nonpolar covalent bond. Because the strength of an atom’s attraction for electrons in a bond is rated by the atom’s electronegativity, the difference in the two atoms’ electronegativities indicates how polar a covalent bond between those atoms will be.
The mass of a molecule is called its molecular mass and is the sum of the masses of the atoms in the molecule. The shape of a molecule can be predicted using valence shell electron pair repulsion (VSEPR), which uses the fact that the negative electrons in covalent bonds repel each other as much as possible. Molecules with polar bonds are polar except when the bond polarities cancel due to symmetry.
Organic chemistry is the chemistry of carbon compounds. Carbon forms covalent bonds with other carbon atoms and with the atoms of many other elements. The simplest organic compounds are hydrocarbons, which consist solely of carbon and hydrogen. Hydrocarbons containing only single bonds are called alkanes (saturated hydrocarbons). Hydrocarbons containing carbon–carbon double bonds are alkenes, while hydrocarbons with carbon–carbon triple bonds are alkynes. Carbon-carbon double and triple bonds are examples of functional groups, atoms or bonds that impart a characteristic chemical function to the molecule. Other functional groups include the alcohol functional group (OH) and the carboxyl functional group (COOH). They are the characteristic functional group in organic compounds called alcohols and carboxylic acids. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/04%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/4.S%3A_Covalent_Bonding_and_Simple_Molecular_Compounds_%28Summary%29.txt |
Chemical change is a central concept in chemistry. The goal of chemists is to know how and why a substance changes in the presence of another substance or even by itself. Because there are tens of millions of known substances, there are a huge number of possible chemical reactions. In this chapter, we will find that many of these reactions can be classified into a small number of categories according to certain shared characteristics.
• 5.0: Prelude to Introduction to Chemical Reactions
Although yeast has been used for thousands of years, its true nature has been known only for the last two centuries. Yeasts are single-celled fungi. About 1,000 species are recognized, but the most common species is Saccharomyces cerevisiae, which is used in bread making. Other species are used for the fermentation of alcoholic beverages. Some species can cause infections in humans.
• 5.1: The Law of Conservation of Matter
One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved. The amount of matter in a closed system is conserved.
• 5.2: Chemical Equations
Chemical reactions are represented by chemical equations that list reactants and products. Proper chemical equations are balanced; the same number of each element’s atoms appears on each side of the equation.
• 5.3: Quantitative Relationships Based on Chemical Equations
A balanced chemical equation not only describes some of the chemical properties of substances—by showing us what substances react with what other substances to make what products—but also shows numerical relationships between the reactants and the products. The study of these numerical relationships is called stoichiometry. A balanced chemical equation gives the ratios in which molecules of substances react and are produced in a chemical reaction.
• 5.4: Some Types of Chemical Reactions
Although there are untold millions of possible chemical reactions, most can be classified into a small number of general reaction types. Classifying reactions has two purposes: it helps us to recognize similarities among them, and it enables us to predict the products of certain reactions. A particular reaction may fall into more than one of the categories that we will define in this book.
• 5.5: Oxidation-Reduction (Redox) Reactions
Chemical reactions in which electrons are transferred are called oxidation-reduction, or redox, reactions. Oxidation is the loss of electrons. Reduction is the gain of electrons. Oxidation and reduction always occur together, even though they can be written as separate chemical equations.
• 5.6: Redox Reactions in Organic Chemistry and Biochemistry
Redox reactions are common in organic and biological chemistry, including the combustion of organic chemicals, respiration, and photosynthesis.
• 5.E: Introduction to Chemical Reactions (Exercises)
• 5.S: Introduction to Chemical Reactions (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
http://en.Wikipedia.org/wiki/Exother...teReaction.jpg
05: Introduction to Chemical Reactions
Although yeast has been used for thousands of years, its true nature has been known only for the last two centuries. Yeasts are single-celled fungi. About 1,000 species are recognized, but the most common species is Saccharomyces cerevisiae, which is used in bread making. Other species are used for the fermentation of alcoholic beverages. Some species can cause infections in humans.
Yeasts live primarily on sugars, such as glucose ($\ce{C6H12O6}$). They convert glucose into carbon dioxide ($\ce{CO2}$) and ethanol ($\ce{C2H5OH}$) in a chemical transformation that is represented as follows:
$\ce{C6H12O6(s) → 2CO2(g) + 2C2H5OH(ℓ)} \nonumber$
Bread making depends on the production of carbon dioxide. The gas, which is produced in tiny pockets in bread dough, acts as a leavening agent: it expands during baking and makes the bread rise. Leavened bread is softer, lighter, and easier to eat and chew than unleavened bread. The other major use of yeast, fermentation, depends on the production of ethanol, which results from the same chemical transformation. Some alcoholic beverages, such as champagne, can also be carbonated using the carbon dioxide produced by the yeast.
Yeast is among the simplest life forms on Earth, yet it is absolutely necessary for at least two major food industries. Without yeast to turn dough into bread and juice into wine, these foods and food industries would not exist today. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.00%3A_Prelude_to__Introduction_to_Chemical_Reactions.txt |
Learning Objectives
• Correctly define a law as it pertains to science.
• State the law of conservation of matter.
In science, a law is a general statement that explains a large number of observations. Before being accepted, a law must be verified many times under many conditions. Laws are therefore considered the highest form of scientific knowledge and are generally thought to be inviolable. Scientific laws form the core of scientific knowledge. One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved.
With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These eventually formed the basis of Dalton's Atomic Theory of Matter.
Law of Conservation of Mass
According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.
$\overbrace{\underbrace{\ce{HgO (s)}}_{\text{100 g}}}^{\text{Mercuric oxide}} \rightarrow \underbrace{\overbrace{\ce{Hg (l) }}^{\text{Mercury}}}_{\text{92.6 g}} + \underbrace{\overbrace{\ce{O2 (g)}}^{\text{Oxygen}}}_{\text{7.4 g}} \nonumber$
Another way of stating this is, "In a chemical reaction, matter is neither created nor destroyed." The law of conservation of mass is also known as the "law of indestructibility of matter."
Example $1$
If heating 10 grams of $\ce{CaCO3}$ produces 4.4 g of $\ce{CO2}$ and 5.6 g of $\ce{CaO}$, show that these observations are in agreement with the law of conservation of mass.
Solution
• Mass of the reactants, $\ce{CaCO3}$ : $10 \,g$
• Mass of the products, $\ce{CO2}$ and $\ce{CaO}$: $4.4 \,g+ 5.6\, g = 10\, g$.
Because the mass of the reactants = the mass of the products, the observations are in agreement with the law of conservation of mass.
What does this mean for chemistry? In any chemical change, one or more initial substances change into a different substance or substances. Both the initial and final substances are composed of atoms because all matter is composed of atoms. According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and kind of atoms after the chemical change as were present before the chemical change.
It may seem as though burning destroys matter, but the same amount, or mass, of matter still exists after a campfire as before. Figure 5.1.1 shows that when wood burns, it combines with oxygen and changes not only to ashes, but also to carbon dioxide and water vapor. The gases float off into the air, leaving behind just the ashes. Suppose we had measured the mass of the wood before it burned and the mass of the ashes after it burned. Also suppose we had been able to measure the oxygen used by the fire and the gases produced by the fire. What would we find? The total mass of matter after the fire would be the same as the total mass of matter before the fire.
Exercise $1$
1. What is the law of conservation of matter?
2. How does the law of conservation of matter apply to chemistry?
Answer a:
The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant
Answer b:
The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.
Key Takeaway
The amount of matter in a closed system is conserved. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.01%3A_The_Law_of_Conservation_of_Matter.txt |
Learning Objectives
• Define chemical reaction.
• Use a balanced chemical equation to represent a chemical reaction.
Water (H2O) is composed of hydrogen and oxygen. Suppose we imagine a process in which we take some elemental hydrogen (H2) and elemental oxygen (O2) and let them react to make water. The statement
"hydrogen and oxygen react to make water"
is one way to represent that process, which is called a chemical reaction. Figure $1$ shows a rather dramatic example of this very reaction.
To simplify the writing of reactions, we use formulas instead of names when we describe a reaction. We can also use symbols to represent other words in the reaction. A plus sign connects the initial substances (and final substances, if there is more than one), and an arrow (→) represents the chemical change:
$\ce{H_2 + O_2 \rightarrow H_2O} \label{Eq1}$
This statement is one example of a chemical equation, an abbreviated way of using symbols to represent a chemical change. The substances on the left side of the arrow are called reactants, and the substances on the right side of the arrow are called products. It is not uncommon to include a phase label with each formula—(s) for solid, (ℓ) for liquid, (g) for gas, and (aq) for a substance dissolved in water, also known as an aqueous solution. If we included phase labels for the reactants and products, under normal environmental conditions, the reaction would be as follows:
$\ce{H2(g) + O2(g) \rightarrow H2O (ℓ)} \label{Eq2}$
Chemical equations can also be used to describe physical changes. We will see examples of this soon.
This equation is still not complete because it does not satisfy the law of conservation of matter. Count the number of atoms of each element on each side of the arrow. On the reactant side, there are two H atoms and two O atoms; on the product side, there are two H atoms and only one oxygen atom. The equation is not balanced because the number of oxygen atoms on each side is not the same (Figure $2$).
To make this chemical equation conform to the law of conservation of matter, we must revise the amounts of the reactants and the products as necessary to get the same number of atoms of a given element on each side. Because every substance has a characteristic chemical formula, we cannot change the chemical formulas of the individual substances. For example, we cannot change the formula for elemental oxygen to O. However, we can assume that different numbers of reactant molecules or product molecules may be involved. For instance, perhaps two water molecules are produced, not just one:
$\ce{H2(g) + O2 (g) \rightarrow 2H2O (ℓ)} \label{Eq3}$
The 2 preceding the formula for water is called a coefficient. It implies that two water molecules are formed. There are now two oxygen atoms on each side of the equation.
This point is so important that we should repeat it. You cannot change the formula of a chemical substance to balance a chemical reaction! You must use the proper chemical formula of the substance.
Unfortunately, by inserting the coefficient 2 in front of the formula for water, we have also changed the number of hydrogen atoms on the product side as well. As a result, we no longer have the same number of hydrogen atoms on each side. This can be easily fixed, however, by putting a coefficient of 2 in front of the diatomic hydrogen reactant:
$\ce{2H2(g) + O2(g) \rightarrow 2H2O (ℓ)} \label{Eq4}$
Now we have four hydrogen atoms and two oxygen atoms on each side of the equation. The law of conservation of matter is satisfied because we now have the same number of atoms of each element in the reactants and in the products. We say that the reaction is balanced (Figure $3$). The diatomic oxygen has a coefficient of 1, which typically is not written but assumed in balanced chemical equations.
Proper chemical equations should be balanced. Writing balanced reactions is a chemist’s way of acknowledging the law of conservation of matter.
Example $1$
Is each chemical equation balanced?
1. 2Na(s) + O2(g) → 2Na2O(s)
2. CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)
3. AgNO3(aq) + 2KCl(aq) → AgCl(s) + KNO3(aq)
Solution
1. By counting, we find two sodium atoms and two oxygen atoms in the reactants and four sodium atoms and two oxygen atoms in the products. This equation is not balanced.
2. The reactants have one carbon atom, four hydrogen atoms, and four oxygen atoms. The products have one carbon atom, four hydrogen atoms, and four oxygen atoms. This equation is balanced.
3. The reactants have one silver atom, one nitrogen atom, three oxygen atoms, two potassium atoms, and two chlorine atoms. The products have one silver atom, one chlorine atom, one potassium atom, one nitrogen atom, and three oxygen atoms. Because there are different numbers of chlorine and potassium atoms, this equation is not balanced.
Exercise $1$
Is each chemical equation balanced?
1. $2Hg_{(ℓ)} + O_{2(g)} \rightarrow Hg_2O_{2(s)}$
2. $C_2H_{4(g)} + 2O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(ℓ)}$
3. $Mg(NO_3)_{2(s)} + 2Li_{(s)} \rightarrow Mg_{(s)} + 2LiNO_{3(s)}$.
Answer a:
balanced
Answer b:
O is not balanced; the 4 atoms of oxygen on the left does not balance with the 6 oxygen atoms on the right
Answer c:
balanced
How does one balance a chemical equation, starting with the correct formulas of the reactants and products? Basically, a back-and-forth approach is adopted, counting the number of atoms of one element on one side, checking the number of atoms of that element on the other side, and changing a coefficient if necessary. Then check another element, going back and forth from one side of the equation to another, until each element has the same number of atoms on both sides of the arrow. In many cases, it does not matter which element is balanced first and which is balanced last, as long as all elements have the same number of atoms on each side of the equation.
Below are guidelines for writing and balancing chemical equations.
1. Determine the correct chemical formulas for each reactant and product. Write the skeleton equation.
2. Count the number of atoms of each element that appears as a reactant and as a product. If a polyatomic ion is unchanged on both sides of the equation, count it as a unit.
3. Balance each element one at a time by placing coefficients in front of the formulas. No coefficient is written for a 1. It is best to begin by balancing elements that only appear in one chemical formula on each side of the equation. NEVER change the subscripts in a chemical formula - you can only balance equations by using coefficients.
4. Check each atom or polyatomic ion to be sure that they are equal on both sides of the equation.
5. Make sure that all coefficients are in the lowest possible ratio. If necessary, reduce to the lowest ratio.
For example, to balance the equation
Step 1: Write the skeleton equation with the correct formulas.
$\ce{CH4 + Cl2 \rightarrow CCl4 + HCl} \label{Eq5}$
Step 2: Count the number of each atom or polyatomic ion on both sides of the equation.
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \ 4 \: \ce{H} \: \text{ions} & 1 \: \ce{H} \: \text{ions} \ 2 \: \ce{Cl} \: \text{atom} & 5 \: \ce{Cl} \: \text{atoms} \end{array} \nonumber$
Step 3: We find that both sides are already balanced with one carbon atom. So we proceed to balance the hydrogen atoms. We find that the reactant side has four hydrogen atoms, so the product side must also have four hydrogen atoms. This is balanced by putting a 4 in front of the HCl:
$\ce{CH4 + Cl2 \rightarrow CCl4 + 4HCl } \label{Eq6}$
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \ 4 \: \ce{H} \: \text{ions} & 4 \: \ce{H} \: \text{ions} \ 2 \: \ce{Cl} \: \text{atom} & 8 \: \ce{Cl} \: \text{atoms} \end{array} \nonumber$
Now each side has four hydrogen atoms. The product side has a total of eight chlorine atoms (four from the $CCl_4$ and four from the four molecules of HCl), so we need eight chlorine atoms as reactants. Because elemental chlorine is a diatomic molecule, we need four chlorine molecules to get a total of eight chlorine atoms. We add another 4 in front of the Cl2 reactant:
$\ce{CH4 + 4Cl2 \rightarrow CCl4 + 4HCl } \label{Eq7}$
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \ 4 \: \ce{H} \: \text{ions} & 4 \: \ce{H} \: \text{ions} \ 8 \: \ce{Cl} \: \text{atom} & 8 \: \ce{Cl} \: \text{atoms} \end{array} \nonumber$
Step 3: Now we check: each side has one carbon atom, four hydrogen atoms, and eight chlorine atoms. The chemical equation is balanced. And, the coefficients are in the lowest possible ratio.
Example $2$
Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the balanced chemical equation for this reaction.
Solution
Step 1: Write the skeleton equation with the correct formulas.
$\ce{Pb(NO_3)_2} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right) \nonumber$
Step 2: Count the number of each atom or polyatomic ion on both sides of the equation.
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \ 2 \: \ce{NO_3^-} \: \text{ions} & 1 \: \ce{NO_3^-} \: \text{ions} \ 1 \: \ce{Na} \: \text{atom} & 1 \: \ce{Na} \: \text{atom} \ 1 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array} \nonumber$
Step 3: Solve. The nitrate ions and the chlorine atoms are unbalanced. Start by placing a 2 in front of the $\ce{NaCl}$. This increases the reactant counts to 2 $\ce{Na}$ atoms and 2 $\ce{Cl}$ atoms. Then place a 2 in front of the $\ce{NaNO_3}$. The result is:
$\ce{Pb(NO_3)_2} \left( aq \right) + 2 \ce{NaCl} \left( aq \right) \rightarrow 2 \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right) \nonumber$
Step 4: The new count for each atom and polyatomic ion becomes:
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \ 2 \: \ce{NO_3^-} \: \text{ions} & 2 \: \ce{NO_3^-} \: \text{ions} \ 2 \: \ce{Na} \: \text{atom} & 2 \: \ce{Na} \: \text{atom} \ 2 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array} \nonumber$
Step 5: Think about the result.
The equation is now balanced since there are equal numbers of atoms of each element on both sides of the equation. And, the coefficients are in the lowest possible ratio.
Exercise $2$
Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Hum ans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose (C6H12O6) are converted to ethanol (C2H5OH) and carbon dioxide CO2. Write a balanced chemical reaction for the fermentation of glucose.
Answer
$\ce{C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)} \nonumber$
Key Takeaway
Chemical reactions are represented by chemical equations that list reactants and products. Proper chemical equations are balanced; the same number of each element’s atoms appears on each side of the equation. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.02%3A_Chemical_Equations.txt |
Learning Objectives
• To calculate the amount of one substance that will react with or be produced from a given amount of another substance.
A balanced chemical equation not only describes some of the chemical properties of substances—by showing us what substances react with what other substances to make what products—but also shows numerical relationships between the reactants and the products. The study of these numerical relationships is called stoichiometry. The stoichiometry of chemical equations revolves around the coefficients in the balanced chemical equation because these coefficients determine the molecular ratio in which reactants react and products are made.
The word stoichiometry is pronounced “stow-eh-key-OM-et-tree.” It is of mixed Greek and English origins, meaning roughly “measure of an element.”
Looking Closer: Stoichiometry in Cooking
Let us consider a stoichiometry analogy from the kitchen. A recipe that makes 1 dozen biscuits needs 2 cups of flour, 1 egg, 4 tablespoons of shortening, 1 teaspoon of salt, 1 teaspoon of baking soda, and 1 cup of milk. If we were to write this as a chemical equation, we would write
2 c flour + 1 egg + 4 tbsp shortening + 1 tsp salt + 1 tsp baking soda + 1 c milk → 12 biscuits
(Unlike true chemical reactions, this one has all 1 coefficients written explicitly—partly because of the many different units here.) This equation gives us ratios of how much of what reactants are needed to make how much of what product. Two cups of flour, when combined with the proper amounts of the other ingredients, will yield 12 biscuits. One teaspoon of baking soda (when also combined with the right amounts of the other ingredients) will make 12 biscuits. One egg must be combined with 1 cup of milk to yield the product food. Other relationships can also be expressed.
We can use the ratios we derive from the equation for predictive purposes. For instance, if we have 4 cups of flour, how many biscuits can we make if we have enough of the other ingredients? It should be apparent that we can make a double recipe of 24 biscuits.
But how would we find this answer formally, that is, mathematically? We would set up a conversion factor, much like we did in Chapter 1. Because 2 cups of flour make 12 biscuits, we can set up an equivalency ratio:
$\mathrm{\dfrac{12\: biscuits}{2\: c\: flour}} \nonumber$
We then can use this ratio in a formal conversion of flour to biscuits:
$\mathrm{4\: c\: flour\times\dfrac{12\: biscuits}{2\: c\: flour}=24\: biscuits} \nonumber$
Similarly, by constructing similar ratios, we can determine how many biscuits we can make from any amount of ingredient. When you are doubling or halving a recipe, you are doing a type of stoichiometry. Applying these ideas to chemical reactions should not be difficult if you use recipes when you cook.
Consider the following balanced chemical equation:
$\ce{2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O} \label{Eq1}$
The coefficients on the chemical formulas give the ratios in which the reactants combine and the products form. Thus, we can make the following statements and construct the following ratios:
Table uses a statement from the balanced chemical reaction to show ratios and inverse ratios.
Statement from the Balanced Chemical Reaction Ratio Inverse Ratio
two C2H2 molecules react with five O2 molecules $\mathrm{\dfrac{2C_2H_2}{5O_2}}$ $\mathrm{\dfrac{5O_2}{2C_2H_2}}$
two C2H2 molecules react to make four CO2 molecules $\mathrm{\dfrac{2C_2H_2}{4CO_2}}$ $\mathrm{\dfrac{4CO_2}{2C_2H_2}}$
five O2 molecules react to make two H2O molecules $\mathrm{\dfrac{5O_2}{2H_2O}}$ $\mathrm{\dfrac{2H_2O}{5O_2}}$
four CO2 molecules are made at the same time as two H2O molecules $\mathrm{\dfrac{2H_2O}{4CO_2}}$ $\mathrm{\dfrac{4CO_2}{2H_2O}}$
Other relationships are possible; in fact, 12 different conversion factors can be constructed from this balanced chemical equation. In each ratio, the unit is assumed to be molecules because that is how we are interpreting the chemical equation.
Any of these fractions can be used as a conversion factor to relate an amount of one substance to an amount of another substance. For example, suppose we want to know how many CO2 molecules are formed when 26 molecules of C2H2 are reacted. As usual with a conversion problem, we start with the amount we are given—26C2H2—and multiply it by a conversion factor that cancels out our original unit and introduces the unit we are converting to—in this case, CO2. That conversion factor is $\mathrm{\dfrac{4CO_2}{2C_2H_2}}$, which is composed of terms that come directly from the balanced chemical equation. Thus, we have
$\mathrm{26C_2H_2\times\dfrac{4CO_2}{2C_2H_2}} \nonumber$
The molecules of C2H2 cancel, and we are left with molecules of CO2. Multiplying through, we get
$\mathrm{26C_2H_2\times\dfrac{4CO_2}{2C_2H_2}= 52CO_2} \nonumber$
Thus, 52 molecules of CO2 are formed.
This application of stoichiometry is extremely powerful in its predictive ability, as long as we begin with a balanced chemical equation. Without a balanced chemical equation, the predictions made by simple stoichiometric calculations will be incorrect.
Example $1$
Start with this balanced chemical equation.
$\ce{KMnO4 + 8HCl + 5FeCl2 → 5 FeCl3 + MnCl2 + 4H2O + KCl} \nonumber$
1. Verify that the equation is indeed balanced.
2. Give 2 ratios that give the relationship between HCl and FeCl3.
Solution
1. Each side has 1 K atom and 1 Mn atom. The 8 molecules of HCl yield 8 H atoms, and the 4 molecules of H2O also yield 8 H atoms, so the H atoms are balanced. The Fe atoms are balanced, as we count 5 Fe atoms from 5 FeCl2 reactants and 5 FeCl3 products. As for Cl, on the reactant side, there are 8 Cl atoms from HCl and 10 Cl atoms from the 5 FeCl2 formula units, for a total of 18 Cl atoms. On the product side, there are 15 Cl atoms from the 5 FeCl3 formula units, 2 from the MnCl2 formula unit, and 1 from the KCl formula unit. This is a total of 18 Cl atoms in the products, so the Cl atoms are balanced. All the elements are balanced, so the entire chemical equation is balanced.
2. Because the balanced chemical equation tells us that 8 HCl molecules react to make 5 FeCl3 formula units, we have the following 2 ratios: $\mathrm{\dfrac{8HCl}{5FeCl_3}\:and\: \dfrac{5FeCl_3}{8HCl}}$. There are a total of 42 possible ratios. Can you find the other 40 relationships?
Exercise $1$
Start with this balanced chemical equation.
$\mathrm{2KMnO_4+3CH_2\textrm{=C}H_2+4H_2O\rightarrow 2MnO_2+3HOCH_2CH_2OH+2KOH} \nonumber$
1. Verify that the equation is balanced.
2. Give 2 ratios that give the relationship between KMnO4 and CH2=CH2. (A total of 30 relationships can be constructed from this chemical equation. Can you find the other 28?)
Answer a:
Each side has 2 K atoms and 2 Mn atoms. On the reactant side, 3 CH2=CH2 yield 6 C atoms and on the product side, 3 HOCH2CH2OH also yield 6 C atoms, so the C atoms are balanced. There are 20 H atoms on the reactants side: 12 H atoms from 3 CH2=CH2 and 8 H atoms from 4 H2O. On the product side, there are also 20 H atoms: 18 H atoms from 3 HOCH2CH2OH and 2 H atoms from 2 KOH. So, the H atoms are balanced. As for O, on the reactant side, there are 8 O atoms from 2 KMnO4 and 4 O atoms from 4 H2O, for a total of 12 O atoms. On the product side, there are 4 O atoms from the 2 MnO2 formula units, 6 O atoms from 3 HOCH2CH2OH, and 2 O atoms from the 2 KOH formula units. This is a total of 12 O atoms in the products, so the O atoms are balanced. All the elements are balanced, so the entire chemical equation is balanced.
Answer b:
Because the balanced chemical equation tells us that 2 KMnO4 formula units react with 3 CH2=CH2 molecules, we have the following 2 ratios: $\mathrm{\dfrac{2KMnO_4}{3CH_2\textrm{=C}H_2}\:and\: \dfrac{3CH_2\textrm{=C}H_2}{2KMnO_4}}$. There are a total of 30 possible ratios. Can you find the other 28 relationships?
Key Takeaway
The coefficients in a balanced chemical equation give the ratios in which molecules of substances react and are produced in a chemical reaction. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.03%3A_Quantitative_Relationships_Based_on_Chemical_Equations.txt |
Learning Objectives
• To classify a given chemical reaction into a variety of types.
Although there are untold millions of possible chemical reactions, most can be classified into a small number of general reaction types. Classifying reactions has two purposes: it helps us to recognize similarities among them, and it enables us to predict the products of certain reactions. A particular reaction may fall into more than one of the categories that we will define in this book.
Combination (composition) Reactions
A combination (composition) reaction is a chemical reaction that makes a single substance from two or more reactants. There may be more than one molecule of product in the balanced chemical equation, but there is only one substance produced. For example, the equation
$\ce{4Fe + 3O_2 \rightarrow 2Fe_2O_3} \label{Eq1}$
is a combination reaction that produces $\ce{Fe2O3}$ from its constituent elements—Fe and O2. Combination reactions do not have to combine elements, however. The chemical equation
$\ce{Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3} \label{Eq2}$
shows a combination reaction in which $\ce{Fe2O3}$ combines with three molecules of $\ce{SO3}$ to make $\ce{Fe2(SO4)3}$.
Example $1$
Which equations are combination reactions?
1. Co(s) + Cl2(g) → CoCl2(s)
2. CO(g) + Cl2(g) → COCl2(g)
3. N2H4(ℓ) + O2(g) → N2(g) + 2H2O(ℓ)
Solution
1. This is a combination reaction.
2. This is a combination reaction. (The compound $\ce{COCl2}$ is called phosgene and, in the past, was used as a gassing agent in chemical warfare.)
3. This is not a combination reaction.
Exercise $1$
Which equations are combination reactions?
1. P4(s) + 6Cl2(g) → 4PCl3(g)
2. SO3(ℓ) + H2O(ℓ) → H2SO4(ℓ)
3. NaOH(s) + HCl(g) → NaCl(s) + H2O(ℓ)
Answer
a and b are combination reactions
Decomposition Reactions
A decomposition reaction is the reverse of a combination reaction. In a decomposition reaction, a single substance is converted into two or more products. There may be more than one molecule of the reactant, but there is only one substance initially. For example, the equation
$\ce{2NaHCO3(s) \rightarrow Na_2CO3(s) + CO2(g) + H_2O (ℓ) } \label{Eq3}$
is a decomposition reaction that occurs when $\ce{NaHCO3}$ is exposed to heat. Another example is the decomposition of $\ce{KClO3}$:
$\ce{2KClO3 (s) \rightarrow 2KCl (s) + 3O2 (g)} \label{Eq4}$
This reaction was once commonly used to generate small amounts of oxygen in the chemistry lab.
The decomposition reaction of $\ce{NaHCO3}$ is the reaction that occurs when baking soda is poured on a small kitchen fire. The intent is that the $\ce{H2O}$ and $\ce{CO2}$ produced by the decomposition will smother the flames.
Combustion Reactions
A combustion reaction occurs when a substance combines with molecular oxygen to make oxygen-containing compounds of other elements in the reaction. Many combustion reactions occur with a hydrocarbon, a compound made up solely of carbon and hydrogen. The products of the combustion of hydrocarbons are carbon dioxide and water. Many hydrocarbons are used as fuel because their combustion releases very large amounts of heat energy. An example is propane (C3H8), a gaseous hydrocarbon that is commonly used as the fuel source in gas grills.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Another example is the burning of acetylene ($\ce{C2H2}$) in welding torches:
$\ce{2C2H2 + 5O2 \rightarrow 4CO2 + 2H2O } \label{Eq5}$
Oxygen (in its elemental form) is a crucial reactant in combustion reactions.
Energy in the form of heat is usually given off as a product in a combustion reaction as well.
Example $2$
Carbon dioxide (CO2) is an important heat-trapping (greenhouse) gas, which is released through human activities such as deforestation and burning fossil fuels, as well as natural processes such as respiration and volcanic eruptions. The graph above shows CO2 levels during the last three glacial cycles, as reconstructed from ice cores.
Carbon dioxide is the primary greenhouse gas emitted through human activities. In 2015, CO2 accounted for about 82.2% of all U.S. greenhouse gas emissions from human activities. Carbon dioxide is naturally present in the atmosphere as part of the Earth's carbon cycle (the natural circulation of carbon among the atmosphere, oceans, soil, plants, and animals). Human activities are altering the carbon cycle–both by adding more CO2 to the atmosphere and by influencing the ability of natural sinks, like forests, to remove CO2 from the atmosphere. While CO2 emissions come from a variety of natural sources, human-related emissions are responsible for the increase that has occurred in the atmosphere since the industrial revolution.
The main human activity that emits CO2 is the combustion of fossil fuels (coal, natural gas, and oil) for energy and transportation, although certain industrial processes and land-use changes also emit CO2. An example of how CO2 can be generated is the combustion of octane (C8H10), a component of gasoline.
Write the balanced equation to represent the combustion of octane.
Solution
2C8H10(l) + 25O2(g) → 16CO2(g) + 18H2O(g)
The balanced reaction argues that for every two molecules of octane that are burned, 16 molecules of CO2 are generated.
Example $2$
Identify each type of reaction.
1. $\ce{2K(s) + S(s) + 2O2(g) → K2SO4(s)}$
2. $\ce{(NH4)2Cr2O7(s) → N2(g) + Cr2O3(s) + 4H2O(ℓ)}$
3. $\ce{CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)}$
Solution
1. Multiple reactants are combining to make a single product, so this reaction is a combination reaction.
2. A single substance reacts to make several products, so we have a decomposition reaction.
3. Oxygen reacts with a compound to make carbon dioxide (an oxide of carbon) and water (an oxide of hydrogen). This is a combustion reaction.
Exercise $2$
Identify each type of reaction.
1. $\ce{C2H5OH + 3O2 → 2CO2 + 3H2O}$
2. $\ce{2Mg(s) + O2(g) → 2MgO(s)}$
3. $\ce{CaCO3(s) → CaO(s) + CO2(g)}$
Answer a
combustion
Answer b
combination (also combustion)
Answer c
decomposition
Key Takeaway
There are several recognizable types of chemical reactions: combination, decomposition, and combustion reactions are examples. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.04%3A_Some_Types_of_Chemical_Reactions.txt |
Learning Objectives
• To identify a chemical reaction as an oxidation-reduction reaction.
When zinc metal is submerged into a quantity of aqueous $\ce{HCl}$, the following reaction occurs (Figure $1$):
$\ce{Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2(aq)} \label{eq:1}$
This is one example of what is sometimes called a single replacement reaction because $\ce{Zn}$ replaces $\ce{H}$ in combination with $\ce{Cl}$.
Because some of the substances in this reaction are aqueous, we can separate them into ions:
$\ce{Zn (s) + 2H^{+} (aq) + 2Cl^{−} (aq) → H2(g) + Zn^{2+} (aq) + 2Cl^{−} (aq)} \nonumber$
Viewed this way, the net reaction seems to be a charge transfer between zinc and hydrogen atoms. (There is no net change experienced by the chloride ion.) In fact, electrons are being transferred from the zinc atoms to the hydrogen atoms (which ultimately make a molecule of diatomic hydrogen), changing the charges on both elements.
To understand electron-transfer reactions like the one between zinc metal and hydrogen ions, chemists separate them into two parts: one part focuses on the loss of electrons, and one part focuses on the gain of electrons. The loss of electrons is called oxidation. The gain of electrons is called reduction. Because any loss of electrons by one substance must be accompanied by a gain in electrons by something else, oxidation and reduction always occur together. As such, electron-transfer reactions are also called oxidation-reduction reactions, or simply redox reactions. The atom that loses electrons is oxidized, and the atom that gains electrons is reduced. Also, because we can think of the species being oxidized as causing the reduction, the species being oxidized is called the reducing agent, and the species being reduced is called the oxidizing agent.
Because batteries are used as sources of electricity (that is, of electrons), all batteries are based on redox reactions.
Although the two reactions occur together, it can be helpful to write the oxidation and reduction reactions separately as half reactions. In half reactions, we include only the reactant being oxidized or reduced, the corresponding product species, any other species needed to balance the half reaction, and the electrons being transferred. Electrons that are lost are written as products; electrons that are gained are written as reactants. For example, in our earlier equation, now written without the chloride ions,
$\ce{Zn (s) + 2H^{+} (aq) → Zn^{2+}(aq) + H2(g)} \nonumber$
zinc atoms are oxidized to Zn2+. The half reaction for the oxidation reaction, omitting phase labels, is as follows:
$\ce{Zn → Zn^{2+} + 2e^{−}} \nonumber$
This half reaction is balanced in terms of the number of zinc atoms, and it also shows the two electrons that are needed as products to account for the zinc atom losing two negative charges to become a 2+ ion. With half reactions, there is one more item to balance: the overall charge on each side of the reaction. If you check each side of this reaction, you will note that both sides have a zero net charge.
Hydrogen is reduced in the reaction. The balanced reduction half reaction is as follows:
$\ce{2H^{+} + 2e^{−} → H2} \nonumber$
There are two hydrogen atoms on each side, and the two electrons written as reactants serve to neutralize the 2+ charge on the reactant hydrogen ions. Again, the overall charge on both sides is zero.
The overall reaction is simply the combination of the two half reactions and is shown by adding them together.
Because we have two electrons on each side of the equation, they can be canceled. This is the key criterion for a balanced redox reaction: the electrons have to cancel exactly. If we check the charge on both sides of the equation, we see they are the same—2+. (In reality, this positive charge is balanced by the negative charges of the chloride ions, which are not included in this reaction because chlorine does not participate in the charge transfer.)
Redox reactions are often balanced by balancing each individual half reaction and then combining the two balanced half reactions. Sometimes a half reaction must have all of its coefficients multiplied by some integer for all the electrons to cancel. The following example demonstrates this process.
Example $1$: Reducing Silver Ions
Write and balance the redox reaction that has silver ions and aluminum metal as reactants and silver metal and aluminum ions as products. Identify the substance oxidized, substance reduced, reducing agent and reducing agent.
Solution
We start by using symbols of the elements and ions to represent the reaction:
$\ce{Ag^{+} + Al → Ag + Al^{3+}} \nonumber$
The equation looks balanced as it is written. However, when we compare the overall charges on each side of the equation, we find a charge of +1 on the left but a charge of +3 on the right. This equation is not properly balanced. To balance it, let us write the two half reactions. Silver ions are reduced, and it takes one electron to change Ag+ to Ag:
Reduction half-reaction: $\ce{Ag^{+} + e^{−} → Ag} \nonumber$
Aluminum is oxidized, losing three electrons to change from Al to Al3+:
Oxidation half-reaction: $\ce{Al → Al^{3+} + 3e^{−}} \nonumber$
To combine these two half reactions and cancel out all the electrons, we need to multiply the silver reduction reaction by 3:
Now the equation is balanced, not only in terms of elements but also in terms of charge.
• The substance oxidized is the reactant that had undergone oxidation: Al
• The substance reduced is the reactant that had undergone reduction: Ag+
• The reducing agent is the same as the substance oxidized: Al
• The oxidizing agent is the same as the substance reduced: Ag+
Exercise $1$
Write and balance the redox reaction that has calcium ions and potassium metal as reactants and calcium metal and potassium ions as products. Identify the substance oxidized, substance reduced, reducing agent and reducing agent.
Answer
Reduction: Ca2+ + 2e→ Ca
Oxidation: 2 (K → K+ + e)
Combined: Ca2+ + 2K → Ca + 2K+
• The substance oxidized is the reactant that had undergone oxidation: K
• The substance reduced is the reactant that had undergone reduction: Ca2+
• The reducing agent is the same as the substance oxidized: K
• The oxidizing agent is the same as the substance reduced: Ca2+
Potassium has been used as a reducing agent to obtain various metals in their elemental form.
To Your Health: Redox Reactions and Pacemaker Batteries
All batteries use redox reactions to supply electricity because electricity is basically a stream of electrons being transferred from one substance to another. Pacemakers—surgically implanted devices for regulating a person’s heartbeat—are powered by tiny batteries, so the proper operation of a pacemaker depends on a redox reaction.
Pacemakers used to be powered by NiCad batteries, in which nickel and cadmium (hence the name of the battery) react with water according to this redox reaction:
$\ce{Cd(s) + 2NiOOH(s) + 2H2O(ℓ) → Cd(OH)2(s) + 2Ni(OH) 2(s)} \nonumber$
The cadmium is oxidized, while the nickel atoms in NiOOH are reduced. Except for the water, all the substances in this reaction are solids, allowing NiCad batteries to be recharged hundreds of times before they stop operating. Unfortunately, NiCad batteries are fairly heavy batteries to be carrying around in a pacemaker. Today, the lighter lithium/iodine battery is used instead. The iodine is dissolved in a solid polymer support, and the overall redox reaction is as follows:
$\ce{2Li(s) + I2(s) → 2LiI (s)} \nonumber$
Lithium is oxidized, and iodine is reduced. Although the lithium/iodine battery cannot be recharged, one of its advantages is that it lasts up to 10 years. Thus, a person with a pacemaker does not have to worry about periodic recharging; about once per decade a person requires minor surgery to replace the pacemaker/battery unit. Lithium/iodine batteries are also used to power calculators and watches.
Oxidation and reduction can also be defined in terms of changes in composition. The original meaning of oxidation was “adding oxygen,” so when oxygen is added to a molecule, the molecule is being oxidized. The reverse is true for reduction: if a molecule loses oxygen atoms, the molecule is being reduced. For example, the acetaldehyde ($\ce{CH3CHO}$) molecule takes on an oxygen atom to become acetic acid ($\ce{CH3COOH}$).
$\ce{2CH3CHO + O2 → 2CH_3COOH} \nonumber$
Thus, acetaldehyde is being oxidized.
Similarly, reduction and oxidation can be defined in terms of the gain or loss of hydrogen atoms. If a molecule adds hydrogen atoms, it is being reduced. If a molecule loses hydrogen atoms, the molecule is being oxidized. For example, in the conversion of acetaldehyde into ethanol ($\ce{CH3CH2OH}$), hydrogen atoms are added to acetaldehyde, so the acetaldehyde is being reduced:
$\ce{CH3CHO + H2 → CH3CH2OH} \nonumber$
Table shows how each process affects the change in oxygen and change in hydrogen.
Process Change in oxygen (some reactions) Change in hydrogen (some reactions)
Oxidation gain lose
Reduction lose gain
Example $2$
In each conversion, indicate whether oxidation or reduction is occurring.
1. N2 → NH3
2. CH3CH2OHCH3 → CH3COCH3
3. HCHOHCOOH
Solution
1. Hydrogen is being added to the original reactant molecule, so reduction is occurring.
2. Hydrogen is being removed from the original reactant molecule, so oxidation is occurring.
3. Oxygen is being added to the original reactant molecule, so oxidation is occurring.
Exercise $2$
In each conversion, indicate whether oxidation or reduction is occurring.
1. CH4 → CO2 + H2O
2. NO2 → N2
3. CH2=CH2 → CH3CH3
Answer a:
Oxygen is being added. Oxidation is occurring.
Answer b:
Oxygen is being removed. Reduction is occurring.
Answer a:
Hydrogen is being added. Reduction is occurring.
Key Takeaway
Chemical reactions in which electrons are transferred are called oxidation-reduction, or redox, reactions. Oxidation is the loss of electrons. Reduction is the gain of electrons. Oxidation and reduction always occur together, even though they can be written as separate chemical equations. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.05%3A_Oxidation-Reduction_%28Redox%29_Reactions.txt |
Learning Objectives
• To identify oxidation-reduction reactions with organic compounds.
Oxidation-reduction reactions are of central importance in organic chemistry and biochemistry. The burning of fuels that provides the energy to maintain our civilization and the metabolism of foods that furnish the energy that keeps us alive both involve redox reactions.
All combustion reactions are also redox reactions. A typical combustion reaction is the burning of methane, the principal component of natural gas (Figure $1$).
$\ce{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O }\label{Eq1}$
In respiration, the biochemical process by which the oxygen we inhale in air oxidizes foodstuffs to carbon dioxide and water, redox reactions provide energy to living cells. A typical respiratory reaction is the oxidation of glucose ($\ce{C6H12O6}$), the simple sugar we encountered in the chapter-opening essay that makes up the diet of yeast:
$\ce{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O} \label{Eq2}$
Organic chemists use a variety of redox reactions. For example, potassium dichromate ($\ce{K2Cr2O7}$) is a common oxidizing agent that can be used to oxidize alcohols (symbolized by the general formula ROH). The product of the reaction depends on the location of the OH functional group in the alcohol molecule, the relative proportions of alcohol and the dichromate ion, and reaction conditions such as temperature. If the OH group is attached to a terminal carbon atom and the product is distilled off as it forms, the product is an aldehyde, which has a terminal carbonyl group (C=O) and is often written as RCHO. One example is the reaction used by the Breathalyzer to detect ethyl alcohol ($\ce{CH3CH2OH}$) in a person’s breath:
$\ce{3CH_3CH_2OH + Cr_2O_7^{2−} + 8H^+ \rightarrow 3CH_3CHO + 2Cr^{3+} + 7H_2O } \label{Eq3}$
If the product acetaldehyde ($\ce{CH3CHO}$) is not removed as it forms, it is further oxidized to acetic acid ($\ce{CH3COOH}$). In this case, the overall reaction is as follows:
$\ce{3CH_3CH_2OH + 2Cr_2O_7^{2−} + 16H^+ \rightarrow 3CH_3COOH + 4Cr^{3+} + 11H_2O } \label{Eq4}$
In this reaction, the chromium atom is reduced because it lost oxygen atoms from $Cr_2O_7^{2−}$ to $Cr^{3+}$. On the other hand, the carbon atom in ethanol is oxidized. In the oxidation of ethyl alcohol (CH3CH2OH, a.k.a. ethanol) to form acetaldehayde (CH3CHO, a.k.a. ethanal), the number of bonds to oxygen has increased and the number of hydrogen atoms has decreased from six to four. Either or both of these indicate that an oxidation has occurred.
In the oxidation of acetaldehyde to acetic acid (a.k.a. ethanoic acid), the carbon atom that gained an additional oxygen is the element oxidized.
When the OH group of the alcohol is bonded to an interior carbon atom, the oxidation of an alcohol will produce a ketone (the formulas of ketones are often written as RCOR, and the carbon–oxygen bond is a double bond). The simplest ketone is derived from 2-propanol ($\ce{CH3CHOHCH3}$). It is the common solvent acetone [$\ce{(CH3)2CO}$], which is used in varnishes, lacquers, rubber cement, and nail polish remover. Acetone can be formed by the following redox reaction:
$\ce{3CH_3CHOHCH_3 + Cr_2O_7^{2−} + 8H^+ \rightarrow 3(CH_3)_2CO + 2Cr^{3+} + 7H_2O} \label{Eq5}$
As we have just seen, aldehydes and ketones can be formed by the oxidation of alcohols. Conversely, aldehydes and ketones can be reduced to alcohols. Reduction of the carbonyl group is important in living organisms. For example, in anaerobic metabolism, in which biochemical processes take place in the absence of oxygen, pyruvic acid ($\ce{CH3COCOOH}$) is reduced to lactic acid ($\ce{CH3CHOHCOOH}$) in the muscles.
$\ce{CH_3COCOOH \rightarrow CH_3CHOHCOOH } \label{Eq6}$
(Pyruvic acid is both a carboxylic acid and a ketone; only the ketone group is reduced.) The buildup of lactic acid during vigorous exercise is responsible in large part for the fatigue that we experience.
In food chemistry, the substances known as antioxidants are reducing agents. Ascorbic acid (vitamin C; $\ce{C6H8O6}$) is thought to retard potentially damaging oxidation of living cells. In the process, it is oxidized to dehydroascorbic acid ($\ce{C6H6O6}$). In the stomach, ascorbic acid reduces the nitrite ion ($\ce{NO_2^{−}}$) to nitric oxide ($\ce{NO}$):
$\ce{C_6H_8O_6 + 2H^{+} + 2NO_2^{−} \rightarrow C_6H_6O_6 + 2H_2O + 2NO} \label{Eq7}$
If reaction in Equation $\ref{Eq7}$ did not occur, nitrite ions from foods would oxidize the iron in hemoglobin, destroying its ability to carry oxygen.
Tocopherol (vitamin E) is also an antioxidant. In the body, vitamin E is thought to act by scavenging harmful by-products of metabolism, such as the highly reactive molecular fragments called free radicals. In foods, vitamin E acts to prevent fats from being oxidized and thus becoming rancid. Vitamin C is also a good antioxidant (Figure $2$).
Finally, and of greatest importance, green plants carry out the redox reaction that makes possible almost all life on Earth. They do this through a process called photosynthesis, in which carbon dioxide and water are converted to glucose ($\ce{C6H12O6}$). The synthesis of glucose requires a variety of proteins called enzymes and a green pigment called chlorophyll that converts sunlight into chemical energy (Figure $3$). The overall change that occurs is as follows:
$\ce{6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2} \label{Eq8}$
In this reaction, carbon dioxide is reduced to glucose, and water is oxidized to oxygen gas. Other reactions convert the glucose to more complex carbohydrates, plant proteins, and oils.
Summary
Redox reactions are common in organic and biological chemistry, including the combustion of organic chemicals, respiration, and photosynthesis. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.06%3A_Redox_Reactions_in_Organic_Chemistry_and_Biochemistry.txt |
5.1: The Law of Conservation of Matter
1. What is the law of conservation of matter?
2. How does the law of conservation of matter apply to chemistry?
ANSWERS
1. The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant
2. The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.
Exercises
1. Express the law of conservation of matter in your own words.
2. Explain why the concept of conservation of matter is considered a scientific law.
3. Potassium hydroxide ($\ce{KOH}$) readily reacts with carbon dioxide ($\ce{CO2}$) to produce potassium carbonate ($\ce{K2CO3}$) and water ($\ce{H2O}$). How many grams of potassium carbonate is produced if 224.4 g of $\ce{KOH}$ reacted with 88.0 g of $\ce{CO2}$. The reaction also produced 36.0 g of water.
Answers
1. Matter may not be created or destroyed.
2. The concept is a scientific law because it is based on experimentation.
3. 276.4 g
Concept Review Exercises
1. What are the parts of a chemical equation?
2. Explain why chemical equations need to be balanced.
Answers
a. reactants and products
b. Chemical equations need to be balanced to satisfy the law of conservation of matter.
Exercises
1. Write a chemical equation to express the fact that hydrogen gas and solid iodine react to make gaseous hydrogen iodide. Make sure the equation satisfies the law of conservation of matter.
2. Write a chemical equation to express the fact that sodium metal and chlorine gas react to make solid sodium chloride. Make sure the equation satisfies the law of conservation of matter.
3. Write an equation expressing the fact that hydrogen gas and fluorine gas react to make gaseous hydrogen fluoride. Make sure the equation satisfies the law of conservation of matter.
4. Write an equation expressing the fact that solid potassium and fluorine gas react to make solid potassium fluoride. Make sure the equation satisfies the law of conservation of matter.
5. Mercury reacts with oxygen to make mercury(II) oxide. Write a balanced chemical equation that summarizes this reaction.
6. Octane (C8H18) reacts with oxygen to make carbon dioxide and water. Write a balanced chemical equation that summarizes this reaction.
7. Propyl alcohol (C3H7OH) reacts with oxygen to make carbon dioxide and water. Write a balanced chemical equation that summarizes this reaction.
8. Sulfuric acid (H2SO4) reacts with iron metal to make iron(III) sulfate and hydrogen gas. Write a balanced chemical equation that summarizes this reaction.
9. Balance each equation.
1. MgCl2 + K → KCl + Mg
2. C6H12O6 + O2 → CO2 + H2O
3. NaN3 → Na + N2 (This is the reaction used to inflate airbags in cars.)
10. Balance each equation.
1. NH4NO3 → N2O + H2O
2. TiBr4 + H2O → TiO2 + HBr
3. C3H5N3O9 → CO2 + N2 + O2 + H2O (This reaction represents the decomposition of nitroglycerine.)
11. Balance each equation.
1. NH3 + O2 → NO + H2O
2. Li + N2 → Li3N
3. AuCl → Au + AuCl3
12. Balance each equation.
1. NaOH + H3PO4 → Na3PO4 + H2O
2. N2H4 + Cl2 → N2 + HCl
3. Na2S + H2S → NaSH
13. Chromium(III) oxide reacts with carbon tetrachloride to make chromium(III) chloride and phosgene (COCl2). Write the balanced chemical equation for this reaction.
14. The reaction that occurs when an Alka-Seltzer tablet is dropped into a glass of water has sodium bicarbonate reacting with citric acid (H3C6H5O7) to make carbon dioxide, water, and sodium citrate (Na3C6H5O7). Write the balanced chemical equation for this reaction.
15. When sodium hydrogen carbonate is used to extinguish a kitchen fire, it decomposes into sodium carbonate, carbon dioxide and water. Write a balanced chemical equation for this reaction.
16. Elemental bromine gas can be generated by reacting sodium bromide with elemental chlorine. The other product is sodium chloride. Write a balanced chemical equation for this reaction.
Answers
1. H2(g) + I2(s) → 2HI(g)
2. 2Na(s) + Cl2(g) → 2NaCl(s)
1. H2(g) + F2(g) → 2HF(g)
4. 2K(s) + F2(g) → 2KF(s)
1. 2Hg + O2 → 2HgO
6. 2C8H18 + 25O2 → 16CO2 + 18H2O
1. 2C3H7OH + 9O2 → 6CO2 + 8H2O
8. 3H2SO4 + 2Fe → Fe2(SO4)3 + 3H2
1. MgCl2 + 2K → 2KCl + Mg
2. C6H12O6 + 6O2 → 6CO2 + 6H2O
3. 2NaN3 → 2Na + 3N2
10.
1. NH4NO3 → N2O + 2H2O
2. TiBr4 + 2H2O → TiO2 + 4HBr
3. 4C3H5N3O9 → 12CO2 + 6N2 + O2 + 10H2O
11.
1. 4NH3 + 5O2 → 4NO + 6H2O
2. 6Li + N2 → 2Li3N
3. 3AuCl → 2Au + AuCl3
12.
1. 3NaOH + H3PO4 → Na3PO4 + 3H2O
2. N2H4 + 2Cl2 → N2 + 4HCl
3. Na2S + H2S → 2NaSH
13. Cr2O3 + 3CCl4 → 2CrCl3 + 3COCl2
14. 3NaHCO3 + H3C6H5O7 → 3CO2 + 3H2O + Na3C6H5O7
15. 2NaHCO3 → Na2CO3 + CO2 + H2O
16. 2NaBr + Cl2 → Br2 + 2NaCl
Concept Review Exercises
1. Explain how stoichiometric ratios are constructed from a chemical equation.
2. Why is it necessary for a chemical equation to be balanced before it can be used to construct conversion factors?
Answers
1. Stoichiometric ratios are made using the coefficients of the substances in the balanced chemical equation.
2. A balanced chemical equation is necessary so one can construct the proper stoichiometric ratios.
Exercises
1. Balance this equation and write every stoichiometric ratio you can from it.
NH4NO3 → N2O + H2O
2. Balance this equation.
N2 + H2 → NH3
3. Balance this equation and write every stoichiometric ratio you can from it.
Fe2O3 + C → Fe + CO2
4. Balance this equation.
Fe2O3 + CO → Fe + CO2
5. Balance this equation and determine how many molecules of CO2 are formed if 15 molecules of C6H6 are reacted.
C6H6 + O2 → CO2 + H2O
6. Balance this equation and determine how many formula units of Ag2CO3(s) are produced if 20 formula units of Na2CO3 are reacted.
Na2CO3(aq) + AgNO3(aq) → NaNO3(aq) + Ag2CO3(s)
7. Copper metal reacts with nitric acid according to this equation:
3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(ℓ)
1. Verify that this equation is balanced.
2. How many Cu atoms will react if 488 molecules of aqueous HNO3 are reacted?
8. Gold metal reacts with a combination of nitric acid and hydrochloric acid according to this equation:
Au(s) + 3HNO3(aq) + 4HCl(aq) → HAuCl4(aq) + 3NO2(g) + 3H2O(ℓ)
1. Verify that this equation is balanced.
2. How many Au atoms react with 639 molecules of aqueous HNO3?
9. Sulfur can be formed by reacting sulfur dioxide with hydrogen sulfide at high temperatures according to this equation:
SO2(g) + 2H2S(g) → 3S(g) + 2H2O(g)
1. Verify that this equation is balanced.
2. How many S atoms will be formed from by reacting 1,078 molecules of H2S?
10. Nitric acid is made by reacting nitrogen dioxide with water:
3NO2(g) + H2O(ℓ) → 2HNO3(aq) + NO(g)
1. Verify that this equation is balanced.
2. How many molecules of NO will be formed by reacting 2,268 molecules of NO2?
Answers
1. NH4NO3 → N2O + 2H2O; the stoichiometric ratios are $\mathrm{\dfrac{1NH_4NO_3}{1N_2O}\:,\: \dfrac{1NH_4NO_3}{2H_2O}\:,\: \dfrac{1N_2O}{2H_2O}\:,}$ and their reciprocals.
2. N2 + 3H2 → 2NH3
1. 2Fe2O3 + 3C → 4Fe + 3CO2; the stoichiometric ratios are $\mathrm{\dfrac{2Fe_2O_3}{3C}\:,\: \dfrac{2Fe_2O_3}{4Fe}\:,\: \dfrac{2Fe_2O_3}{3CO_2}\:,\: \dfrac{3C}{4Fe}\:,\: \dfrac{3C}{3CO_2}\:,\: \dfrac{4Fe}{3CO_2}\:,\: }$ and their reciprocals.
4. Fe2O3 + 3CO → 2Fe + 3CO2
1. 2C6H6 + 15O2 → 12CO2 + 6H2O; 90 molecules
6. Na2CO3(aq) + 2AgNO3(aq) → 2NaNO3(aq) + Ag2CO3(s); 20 formula units
7.
1. It is balanced.
2. 183 atoms
8.
1. It is balanced.
2. 213 atoms
9.
1. It is balanced.
2. 1,617 atoms
10.
1. It is balanced.
2. 756 molecules
Concept Review Exercises
1. What is the difference between a combination reaction and a combustion reaction?
2. Give the distinguishing characteristic(s) of a decomposition reaction.
3. How do we recognize a combustion reaction?
Answers
1. A combination reaction produces a certain substance; a combustion reaction is a vigorous reaction, usually a combination with oxygen, that is accompanied by the production of light and/or heat.
2. In a decomposition reaction, a single substance reacts to make multiple substances as products.
3. A combustion reaction is typically a vigorous reaction accompanied by light and/or heat, usually because of reaction with oxygen.
Exercises
1. Identify each type of reaction.
1. C6H5CH3 + 9O2 → 7CO2 + 4H2O
2. 2NaHCO3 → Na2CO3 + H2O + CO2
3. C + 2H2 → CH4
2. Identify each type of reaction.
1. P4O10 + 6H2O → 4H3PO4
2. FeO + SO3 → FeSO4
3. CaCO3(s) → CO2(g) + CaO(s)
3. Identify each type of reaction.
1. 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
2. Hg(ℓ) + ½O2 (g) → HgO(s)
3. CH2CH2(g) + Br2(ℓ) → CH2BrCH2Br
4. Identify each type of reaction.
1. Ti(s) + O2(g) → TiO2(s)
2. H2SO3(aq) → H2O(ℓ) + SO2(g)
3. 3O2(g) → 2O3(g)
Answers
1.
a. combustion
b. decomposition
c. combination
2.
a. combination
b. combination
c. decomposition
3.
a. decomposition
b. combustion (also combination)
c. combination
4.
a. combination
b. decomposition
c. combination
Concept Review Exercises
1. Give two different definitions for oxidation and reduction.
2. Give an example of each definition of oxidation and reduction.
Answers
1. Oxidation is the loss of electrons or the addition of oxygen; reduction is the gain of electrons or the addition of hydrogen.
2. Zn → Zn2+ +2e (oxidation); C2H4 + H2 → C2H6 (reduction) (answers will vary)
Exercises
1. Which reactions are redox reactions? For those that are redox reactions, identify the oxidizing and reducing agents.
1. NaOH + HCl → H2O + NaCl
2. 3Mg + 2AlCl3 → 2Al + 3MgCl2
3. H2O2 + H2 → 2H2O
4. KCl + AgNO3 → AgCl + KNO3
2. Which reactions are redox reactions? For those that are redox reactions, identify the oxidizing and reducing agents.
1. 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O
2. 2C2H6 + 7O2 → 4CO2 + 6H2O
3. 2NaHCO3 → Na2CO3 + CO2 + H2O
4. 2K + 2H2O → 2KOH + H2
3. Balance each redox reaction by writing appropriate half reactions and combining them to cancel the electrons.
1. Ca(s) + H+(aq) → Ca2+(aq) + H2(g)
2. I(aq) + Br2(ℓ) → Br(aq) + I2(s)
4. Balance each redox reaction by writing appropriate half reactions and combining them to cancel the electrons.
1. Fe(s) + Sn4+(aq) → Fe3+(aq) + Sn2+(aq)
2. Pb(s) + Pb4+(aq) → Pb2+(aq) (Hint: both half reactions will yield the same product.)
Answers
1.
a. no
b. yes; oxidizing agent: AlCl3; reducing agent: Mg
c. yes; oxidizing agent: H2O2; reducing agent: H2
d. no
2.
a. yes; oxidizing agent: HNO3; reducing agent: Cu
b. yes; oxidizing agent: O2; reducing agent: C2H6
c. no
d. yes; oxidizing agent: H2O; reducing agent: K
3.
1. Combined: Ca + 2H+ → Ca2+ + H2
2. Combined: Br2 + 2I → 2Br + I2
4.
a. (Fe → Fe3+ + 3e) x 2
(Sn4++ 2e→ Sn2+) x 3
Combined: 2Fe + 3Sn4+ → 2Fe3+ + 3Sn2+
b. Pb → Pb2+ + 2e
Pb4++ 2e→ Pb2+
Combined: Pb + Pb4+ → 2Pb2+
Concept Review Exercise
1. Give some biochemical examples of oxidation and reduction reactions.
Answer
1. photosynthesis and antioxidants in foods (answers will vary)
Exercises
1. A typical respiratory reaction discussed in the text is the oxidation of glucose (C6H12O6):
C6H12O6 + 6O2 → 6CO2 + 6H2O
Is this a redox reaction? If so, what are the oxidizing and reducing agents?
2. The major net reaction in photosynthesis is as follows:
6CO2 + 6H2O → C6H12O6 + 6O2
Is this a redox reaction? If so, what are the oxidizing and reducing agents?
3. What would be the ultimate organic product if CH3CH2CH2OH were to react with a solution of K2Cr2O7?
4. What would be the ultimate organic product if CH3CH2CH2CH2OH were to react with a solution of K2Cr2O7?
5. What would be the final organic product if CH3CH2CHOHCH3 were to react with a solution of K2Cr2O7?
6. What would be the major organic product if CH3CH2CHOHCH2CH3 were to react with a solution of K2Cr2O7?
7. What alcohol is produced in the reduction of acetone [(CH3)2CO]?
8. What alcohol is produced in the reduction of propanal (CH3CH2CHO)?
Answers
1. yes; oxidizing agent: O2; reducing agent: C6H12O6
2. yes; oxidizing agent: CO2; reducing agent: H2O
1. CH3CH2COOH
4. CH3CH2CH2COOH
1. CH3CH2C(O)CH3, where the carbon is double bonded to the oxygen
6. CH3CH2C(O)CH2CH3, carbon #3 is double bonded to the oxygen
7. CH3CHOHCH3, or isopropyl alcohol
8. CH3CH2CH2OH
1.
Additional Exercises
1. Isooctane (C8H18) is used as a standard for comparing gasoline performance. Write a balanced chemical equation for the combustion of isooctane.
2. Heptane (C7H16), like isooctane (see Exercise 1), is also used as a standard for determining gasoline performance. Write a balanced chemical equation for the combustion of heptane.
3. What is the difference between a combination reaction and a redox reaction? Are all combination reactions also redox reactions? Are all redox reactions also combination reactions?
4. Are combustion reactions always redox reactions as well? Explain.
5. A friend argues that the equation
Fe2+ + Na → Fe + Na+
is balanced because each side has one iron atom and one sodium atom. Explain why your friend is incorrect.
6. Some antacids contain aluminum hydroxide [Al(OH)3]. This compound reacts with excess hydrochloric acid (HCl) in the stomach to neutralize it. If the products of this reaction are water and aluminum chloride, what is the balanced chemical equation for this reaction? What is the stoichiometric ratio between the number of HCl molecules made to the number of H2O molecules made?
7. Sulfuric acid is made in a three-step process: (1) the combustion of elemental sulfur to produce sulfur dioxide, (2) the continued reaction of sulfur dioxide with oxygen to produce sulfur trioxide, and (3) the reaction of sulfur trioxide with water to make sulfuric acid (H2SO4). Write balanced chemical equations for all three reactions.
8. If the products of glucose metabolism are carbon dioxide and water, what is the balanced chemical equation for the overall process? What is the stoichiometric ratio between the number of CO2 molecules made to the number of H2O molecules made?
9. Historically, the first true battery was the Leclanché cell, named after its discoverer, Georges Leclanché. It was based on the following reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Identify what is being oxidized, what is being reduced, and the respective reducing and oxidizing agents.
Answers
1. 2C8H18 + 25O2 → 16CO2 + 18H2O
2. C7H16 + 11O2→ 7CO2+ 8H2O
1. A combination reaction makes a new substance from more than one reactant; a redox reaction rearranges electrons. Most (not all) combination reactions are redox reactions. Not all redox reactions are combination reactions.
4. All combustion reactions are redox reactions. In combustion a chemical is combining with oxygen and that chemical is being oxidized. Oxygen, on the other hand, is being reduced.
1. Your friend is incorrect because the number of electrons transferring is not balanced. A balanced equation must not only have the same number of atoms of each element on each side of the equation but must also have the same charge on both sides.
6. Al(OH)3 + 3HCl → 3H2O + AlCl3; 1:1
1. (1) S + O2 → SO2; (2) 2SO2 + O2 → 2SO3; (3) SO3 + H2O → H2SO4
8. C6H12O6 + 6O2→ 6CO2+ 6H2O; 1:1
1. oxidized and reducing agent: Zn; reduced and oxidizing agent: Cu2+
5.S: Introduction to Chemical Reactions (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Scientific laws are general statements that apply to a wide variety of circumstances. One important law in chemistry is the law of conservation of matter, which states that in any closed system, the amount of matter stays constant.
Chemical equations are used to represent chemical reactions. Reactants change chemically into products. The law of conservation of matter requires that a proper chemical equation be balanced. Coefficients are used to show the relative numbers of reactant and product molecules.
In stoichiometry, quantities of reactants and/or products can be related to each other using the balanced chemical equation. The coefficients in a balanced chemical reaction are used to devise the proper ratios that relate the number of molecules of one substance to the number of molecules of another substance.
Chemical reactions can be classified by type. Combination reactions (also called composition reactions) make a substance from other substances. Decomposition reactions break one substance down into multiple substances. Combustion reactions combine molecular oxygen with the atoms of another reactant.
Oxidation reactions are reactions in which an atom loses an electron. Reduction reactions are reactions in which an atom gains an electron. These two processes always occur together, so they are collectively referred to as oxidation-reduction (or redox) reactions. The species being oxidized it called the reducing agent, while the species being reduced is the oxidizing agent. Alternate definitions of oxidation and reduction focus on the gain or loss of oxygen atoms, or the loss or gain of hydrogen atoms. Redox reactions are easily balanced if the overall reaction is first separated into half reactions, which are individually balanced.
Oxidation-reduction reactions are common in organic and biological chemistry. Respiration, the process by which we inhale and metabolize oxygen, is a series of redox reactions. In the absence of oxygen, redox reactions still occur in a process called anaerobic metabolism. Antioxidants such as ascorbic acid also play a part in the human diet, acting as reducing agents in various biochemical reactions. Photosynthesis, the process by which plants convert water and carbon dioxide to glucose, is also based on redox reactions. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.E%3A_Introduction_to_Chemical_Reactions_%28Exercises%29.txt |
So far, we have talked about chemical reactions in terms of individual atoms and molecules. Although this works, most of the reactions occurring around us involve much larger amounts of chemicals. Even a tiny sample of a substance will contain millions, billions, or a hundred billion billions of atoms and molecules. How do we compare amounts of substances to each other in chemical terms when it is so difficult to count to a hundred billion billion? Actually, there are ways to do this, which we will explore in this chapter. In doing so, we will increase our understanding of stoichiometry, which is the study of the numerical relationships between the reactants and the products in a balanced chemical reaction.
• 6.0: Prelude to Quantities in Chemical Reactions
Amounts do matter and in a variety of circumstances. The chapter-opening essay in Chapter 1 tells the story of a nurse who mistakenly read “2–3 mg” as “23 mg” and administered the higher and potentially fatal dose of morphine to a child. Food scientists who work in test kitchens must keep track of specific amounts of ingredients as they develop new products for us to eat. Quality control technicians measure amounts of substances in manufactured products to ensure that the products meet company o
• 6.1: The Mole
Chemists use the term mole to represent a large number of atoms or molecules. Just as a dozen implies 12 things, a mole (abbreviated as mol) represents 6.022 x 10²³ things. The number 6.022 x 10²³, called Avogadro’s number after the 19th-century chemist Amedeo Avogadro, is the number we use in chemistry to represent macroscopic amounts of atoms and molecules.
• 6.2: Atomic and Molar Masses
One mole of a substance has the same mass in grams that one atom or molecule has in atomic mass units. The numbers in the periodic table that we identified as the atomic masses of the atoms not only tell us the mass of one atom in u but also tell us the mass of 1 mol of atoms in grams.
• 6.3: Mole-Mass Conversions
The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversion (or its reverse, a mass-mole conversion). In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).
• 6.4: Mole-Mole Relationships in Chemical Reactions
The balanced chemical reaction can be used to determine molar relationships between substances.
• 6.5: Mole-Mass and Mass-Mass Problems
A balanced chemical equation can be used to relate masses or moles of different substances in a reaction.
• 6.E: Quantities in Chemical Reactions (Exercise)
Select problems and solution.
• 6.S: Quantities in Chemical Reactions (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
06: Quantities in Chemical Reactions
When the disengaged gasses are carefully examined, they are found to weigh 113.7 grs.; these are of two kinds, viz. 144 cubical inches of carbonic acid gas, weighing 100 grs. and 380 cubical inches of a very light gas, weighing only 13.7 grs.…and, when the water which has passed over into the bottle [labeled] H is carefully examined, it is found to have lost 85.7 grs. of its weight. Thus, in this experiment, 85.7 grs. of water, joined to 28 grs. of charcoal, have combined in such a way as to form 100 grs. of carbonic acid, and 13.7 grs. of a particular gas capable of being burnt. (Bold emphasis added.)
In this paragraph from the Elements of Chemistry, Antoine Lavoisier (1743–94) is explaining an experiment in which he was trying to demonstrate that water is not an element but instead is composed of hydrogen (the gas “capable of being burnt”) and oxygen. This is a historical account of a groundbreaking experiment and illustrates the importance of amounts in chemistry. Lavoisier was pointing out that the initial total mass of water and charcoal, 85.7 g plus 28 g, equals the final total mass of carbonic acid and the particular gas, 100 g plus 13.7 g. In this way, he was illustrating the law of conservation of matter. It is another way of saying that amounts matter.
Amounts do matter and in a variety of circumstances. The chapter-opening essay in Chapter 1 tells the story of a nurse who mistakenly read “2–3 mg” as “23 mg” and administered the higher and potentially fatal dose of morphine to a child. Food scientists who work in test kitchens must keep track of specific amounts of ingredients as they develop new products for us to eat. Quality control technicians measure amounts of substances in manufactured products to ensure that the products meet company or government standards. Supermarkets routinely weigh meat and produce and charge consumers by the ounce or the pound. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/06%3A_Quantities_in_Chemical_Reactions/6.00%3A_Prelude_to_Quantities_in_Chemical_Reactions.txt |
Learning Objectives
• To define the mole unit.
Figure $1$ shows that we need 2 hydrogen atoms and 1 oxygen atom to make 1 water molecule. If we want to make 2 water molecules, we will need 4 hydrogen atoms and 2 oxygen atoms. If we want to make 5 molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom.
One problem we have, however, is that it is extremely difficult, if not impossible, to organize atoms one at a time. As stated in the introduction, we deal with billions of atoms at a time. How can we keep track of so many atoms (and molecules) at a time? We do it by using mass rather than by counting individual atoms.
A hydrogen atom has a mass of approximately 1 u. An oxygen atom has a mass of approximately 16 u. The ratio of the mass of an oxygen atom to the mass of a hydrogen atom is therefore approximately 16:1.
If we have 2 atoms of each element, the ratio of their masses is approximately 32:2, which reduces to 16:1—the same ratio. If we have 12 atoms of each element, the ratio of their total masses is approximately (12 × 16):(12 × 1), or 192:12, which also reduces to 16:1. If we have 100 atoms of each element, the ratio of the masses is approximately 1,600:100, which again reduces to 16:1. As long as we have equal numbers of hydrogen and oxygen atoms, the ratio of the masses will always be 16:1.
The same consistency is seen when ratios of the masses of other elements are compared. For example, the ratio of the masses of silicon atoms to equal numbers of hydrogen atoms is always approximately 28:1, while the ratio of the masses of calcium atoms to equal numbers of lithium atoms is approximately 40:7.
So we have established that the masses of atoms are constant with respect to each other, as long as we have the same number of each type of atom. Consider a more macroscopic example. If a sample contains 40 g of Ca, this sample has the same number of atoms as there are in a sample of 7 g of Li. What we need, then, is a number that represents a convenient quantity of atoms so we can relate macroscopic quantities of substances. Clearly even 12 atoms are too few because atoms themselves are so small. We need a number that represents billions and billions of atoms.
Chemists use the term mole to represent a large number of atoms or molecules. Just as a dozen implies 12 things, a mole (abbreviated as mol) represents 6.022 × 1023 things. The number 6.022 × 1023, called Avogadro’s number after the 19th-century chemist Amedeo Avogadro, is the number we use in chemistry to represent macroscopic amounts of atoms and molecules. Thus, if we have 6.022 × 1023 Na atoms, we say we have 1 mol of Na atoms. If we have 2 mol of Na atoms, we have 2 × (6.022 × 1023) Na atoms, or 1.2044 × 1024 Na atoms. Similarly, if we have 0.5 mol of benzene (C6H6) molecules, we have 0.5 × (6.022 × 1023) C6H6 molecules, or 3.011 × 1023 C6H6 molecules.
A mole represents a very large number! If 1 mol of quarters were stacked in a column, it could stretch back and forth between Earth and the sun 6.8 billion times.
Notice that we are applying the mole unit to different types of chemical entities. The word mole represents a number of things—6.022 × 1023 of them—but does not by itself specify what “they” are. The chemical entities can be atoms, molecules, formula units and ions. This specific information needs to be specified accurately. Most students find this confusing hence, we need to review the composition of elements, covalent and ionic compounds.
Most elements are made up of individual atoms, such as helium. However, some elements consist of molecules, such as the diatomic elements, nitrogen, hydrogen, oxygen, etc. One mole of He consists of 6.022 × 1023 He atoms but one mole of nitrogen contains 6.022 × 1023 N2 molecules. The basic units of covalent (molecular) compounds are molecules as well. The molecules of "compounds" consist of different kinds of atoms while the molecules of "elements" consist of only one type of atom. For example, the molecules of ammonia (NH3) consist of nitrogen and hydrogen atoms while N2 molecules have N atoms only. Compounds that are ionic, like NaCl, are represented by ionic formulas. One mole of NaCl, for example, refers to 6.022 × 1023 formula units of NaCl. And, one formula unit of NaCl consists of one sodium ion and one chloride ion. Figure 6.1.2 summarizes the basic units of elements, covalent and ionic compounds
Conversion Between Moles and Atoms, Molecules and Ions
Using our unit conversion techniques learned in Chapter 1, we can use the mole relationship and the chemical formula to convert back and forth between the moles and the number of chemical entities (atoms, molecules or ions).
Because 1 N2 molecule contains 2 N atoms, 1 mol of N2 molecules (6.022 × 1023 molecules) has 2 mol of N atoms. Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of ethanol (C2H6O), we can construct the following relationships (Table $1$):
Table $1$: Molecular Relationships
1 Molecule of $C_2H_6O$ Has 1 Mol of $C_2H_6O$ Has Molecular Relationships
2 C atoms 2 mol of C atoms $\mathrm{\dfrac{2\: mol\: C\: atoms}{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{2\: mol\: C\: atoms}}$
6 H atoms 6 mol of H atoms $\mathrm{\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6\: mol\: H\: atoms}}$
1 O atom 1 mol of O atoms $\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{1\: mol\: O\: atoms}}$
The following example illustrates how we can use these relationships as conversion factors.
Example $1$
If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles of carbon atoms, hydrogen atoms, and oxygen atoms does it have?
Solution
Using the relationships in Table $1$, we apply the appropriate conversion factor for each element:
Note how the unit mol C2H6O molecules cancels algebraically. Similar equations can be constructed for determining the number of H and O atoms:
$\mathrm{2.5\: mol\: C_2H_6O\: molecules\times\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}=15\: mol\: H\: atoms}$
$\mathrm{2.5\: mol\: C_2H_6O\: molecules\times\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}=2.5\: mol\: O\: atoms}$
Exercise $1$
If a sample contains 6.75 mol of Na2SO4, how many moles of sodium atoms, sulfur atoms, and oxygen atoms does it have?
Answer
13.5 mol Na, 6.75 mol S and 27 mol O.
We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. For example, if we are dealing with element X, the mole relationship is expressed as follows:
$\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms} \nonumber$
We can convert this relationship into two possible conversion factors shown below:
$\mathrm{\dfrac{1\: mol\: X\: }{6.022\times 10^{23}\: X\: atoms}}$ or $\mathrm{\dfrac{6.022\times 10^{23}\: X\: atoms}{1\: mol\: X\: }}$
If the number of "atoms of element X" is given, we can convert it into "moles of X" by multiplying the given value with the conversion factor at the left. However, if the number of "mol of X" is given, the appropriate conversion factor to use is the one at the right.
If we are dealing with a molecular compound (such as C4H10), the mole relationship is expressed as follows:
$\text{1 mol C4H10} = 6.022 \times 10^{23} \text{ C4H10 molecules} \nonumber$
If working with ionic compounds (such as NaCl), the mole relationship is expressed as follows:
$\text{1 mol NaCl} = 6.022 \times 10^{23} \text{ NaCl formula units} \nonumber$
Example $2$
How many formula units are present in 2.34 mol of NaCl? How many ions are in 2.34 mol?
Solution
Typically in a problem like this, we start with what we are given and apply the appropriate conversion factor. Here, we are given a quantity of 2.34 mol of NaCl, to which we can apply the definition of a mole as a conversion factor:
$\mathrm{2.34\: mol\: NaCl\times\dfrac{6.022\times10^{23}\: NaCl\: units}{1\: mol\: NaCl}=1.41\times10^{24}\: NaCl\: units}$
Because there are two ions per formula unit, there are
$\mathrm{1.41\times10^{24}\: NaCl\: units\times\dfrac{2\: ions}{NaCl\: units}=2.82\times10^{24}\: ions}$
in the sample.
Exercise $2$
How many molecules are present in 16.02 mol of C4H10? How many atoms are in 16.02 mol?
Answer
9.647 x 1024molecules, 1.351 x 1026 atoms.
Key Takeaway
• A mole is 6.022 × 1023 things. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/06%3A_Quantities_in_Chemical_Reactions/6.01%3A_The_Mole.txt |
Learning Objectives
• To learn how the masses of moles of atoms and molecules are expressed.
Now that we have introduced the mole and practiced using it as a conversion factor, we ask the obvious question: why is the mole that particular number of things? Why is it $6.022 \times 10^{23}$ and not $1 \times 10^{23}$ or even $1 \times 10^{20}$?
The number in a mole, Avogadro’s number, is related to the relative sizes of the atomic mass unit and gram mass units. Whereas one hydrogen atom has a mass of approximately 1 u, 1 mol of H atoms has a mass of approximately 1 gram. And whereas one sodium atom has an approximate mass of 23 u, 1 mol of Na atoms has an approximate mass of 23 grams.
One mole of a substance has the same mass in grams that one atom or molecule has in atomic mass units. The numbers in the periodic table that we identified as the atomic masses of the atoms not only tell us the mass of one atom in u but also tell us the mass of 1 mol of atoms in grams.
One mole of a substance has the same mass in grams that one atom or molecule has in atomic mass units.
Example $1$: Moles to Mass Conversion with Elements
What is the mass of each quantity?
1. 1 mol of Al atoms
2. 2 mol of U atoms
Solution
1. One mole of Al atoms has a mass in grams that is numerically equivalent to the atomic mass of aluminum. The periodic table shows that the atomic mass (rounded to two decimal points) of Al is 26.98, so 1 mol of Al atoms has a mass of 26.98 g.
2. According to the periodic table, 1 mol of U has a mass of 238.0 g, so the mass of 2 mol is twice that, or 476.0 g.
Exercise $1$: Moles to Mass Conversion with Elements
What is the mass of each quantity?
1. 1 mol of Au atoms
2. 5 mol of Br atoms
Answer a:
197.0 g
Answer b:
5 mol Br atoms x 79.90 g/mol = 399.5 g
The mole concept can be extended to masses of formula units and molecules as well. The mass of 1 mol of molecules (or formula units) in grams is numerically equivalent to the mass of one molecule (or formula unit) in atomic mass units. For example, a single molecule of O2 has a mass of 32.00 u, and 1 mol of O2 molecules has a mass of 32.00 g. As with atomic mass unit–based masses, to obtain the mass of 1 mol of a substance, we simply sum the masses of the individual atoms in the formula of that substance. The mass of 1 mol of a substance is referred to as its molar mass, whether the substance is an element, an ionic compound, or a covalent compound.
Example $2$: Moles to Mass Conversion with Compounds
What is the mass of 1 mol of each substance?
1. NaCl
2. bilirubin (C33H36N4O6), the principal pigment present in bile (a liver secretion)
Solution
1. Summing the molar masses of the atoms in the NaCl formula unit gives
Solutions to Example 6.2.2, Summing the molar masses of the atoms in the NaCl formula unit gives
1 Na molar mass: 22.99 g
1 Cl molar mass: 35.45 g
Total: 58.44 g
The mass of 1 mol of NaCl is 58.44 g.
1. Multiplying the molar mass of each atom by the number of atoms of that type in bilirubin’s formula and adding the results, we get
Solutions to Example 6.2.2, Multiplying the molar mass of each atom by the number of atoms of that type in bilirubin’s formula and adding the results, we get
33 C molar mass: 33 × 12.01 g 396.33 g
36 H molar mass: 36 × 1.01 = 36.36 g
4 N molar mass: 4 × 14.01 = 56.04 g
6 O molar mass: 6 × 16.00 = 96.00 g
Total: 584.73 g
The mass of 1 mol of bilirubin is 584.73 g.
Exercise $2$: Moles to Mass Conversion with Compounds
What is the mass of 1 mol of each substance?
1. barium sulfate (BaSO4), used to take X rays of the gastrointestional tract
2. adenosine (C10H13N5O4), a component of cell nuclei crucial for cell division
Answer a:
233.36 g
Answer b:
267.28 g
Be careful when counting atoms. In formulas with polyatomic ions in parentheses, the subscript outside the parentheses is applied to every atom inside the parentheses. For example, the molar mass of Ba(OH)2 requires the sum of 1 mass of Ba, 2 masses of O, and 2 masses of H:
The molar mass of Ba(OH)2 requires the sum of 1 mass of Ba, 2 masses of O, and 2 masses of H:
1 Ba molar mass: 1 × 137.3 g = 137.3 g
2 O molar mass: 2 × 16.00 g = 32.00 g
2 H molar mass: 2 × 1.01 g = 2.02 g
Total: 171.32 g
Because molar mass is defined as the mass for 1 mol of a substance, we can refer to molar mass as grams per mole (g/mol). The division sign (/) implies “per,” and “1” is implied in the denominator. Thus, the molar mass of bilirubin can be expressed as 584.73 g/mol, which is read as “five hundred eighty four point seventy three grams per mole.”
Key Takeaway
• The mass of moles of atoms and molecules is expressed in units of grams. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/06%3A_Quantities_in_Chemical_Reactions/6.02%3A_Atomic_and_Molar_Masses.txt |
Learning Objectives
• To convert quantities between mass units and mole units.
A previous Example stated that the mass of 2 mol of U is twice the molar mass of uranium. Such a straightforward exercise does not require any formal mathematical treatment. Many questions concerning mass are not so straightforward, however, and require some mathematical manipulations.
The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversion (or its reverse, a mass-mole conversion). In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).
We also established that 1 mol of Al has a mass of 26.98 g (Example $1$). Stated mathematically,
1 mol Al = 26.98 g Al
We can divide both sides of this expression by either side to get one of two possible conversion factors:
$\mathrm{\dfrac{1\: mol\: Al}{26.98\: g\: Al}\quad and \quad \dfrac{26.98\: g\: Al}{1\: mol\: Al}} \nonumber$
The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.”
Example $1$
What is the mass of 3.987 mol of Al?
Solution
The first step in a conversion problem is to decide what conversion factor to use. Because we are starting with mole units, we want a conversion factor that will cancel the mole unit and introduce the unit for mass in the numerator. Therefore, we should use the $\mathrm{\dfrac{26.98\: g\: Al}{1\: mol\: Al}}$ conversion factor. We start with the given quantity and multiply by the conversion factor:
$\mathrm{3.987\: mol\: Al\times\dfrac{26.98\: g\: Al}{1\: mol\: Al}}$
Note that the mol units cancel algebraically. (The quantity 3.987 mol is understood to be in the numerator of a fraction that has 1 in the unwritten denominator.) Canceling and solving gives
$\mathrm{3.987\: mol\: Al\times \dfrac{26.98\: g\: Al}{1\: mol\: Al}=107.6\: g\: Al}$
Our final answer is expressed to four significant figures.
Exercise $1$
How many moles are present in 100.0 g of Al? (Hint: you will have to use the other conversion factor we obtained for aluminum.)
Answer
$\mathrm{100.0\: g\: Al\times \dfrac{1\: mol\: Al}{26.98\: g\: Al}=3.706\: mol\: Al}$
Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure $1$ is a chart for determining what conversion factor is needed, and Figure $2$ is a flow diagram for the steps needed to perform a conversion.
Example $2$
A biochemist needs 0.00655 mol of bilirubin (C33H36N4O6) for an experiment. How many grams of bilirubin will that be?
Solution
To convert from moles to mass, we need the molar mass of bilirubin, which we can determine from its chemical formula:
Solutions to Example 6.3.2
33 C molar mass: 33 × 12.01 g = 396.33 g
36 H molar mass: 36 × 1.01 g = 36.36 g
4 N molar mass: 4 × 14.01 g = 56.04 g
6 O molar mass: 6 × 16.00 g = 96.00 g
Total: 584.73 g
The molar mass of bilirubin is 584.73 g. Using the relationship
1 mol bilirubin = 584.73 g bilirubin
we can construct the appropriate conversion factor for determining how many grams there are in 0.00655 mol. Following the steps from Figure $2$:
$\mathrm{0.00655\: mol\: bilirubin \times \dfrac{584.73\: g\: bilirubin}{mol\: bilirubin}=3.83\: g\: bilirubin}$
The mol bilirubin unit cancels. The biochemist needs 3.83 g of bilirubin.
Exercise $2$
A chemist needs 457.8 g of KMnO4 to make a solution. How many moles of KMnO4 is that?
Answer
$\mathrm{457.8\: g\: KMnO_4\times \dfrac{1\: mol\: KMnO_4}{158.04\: g\: KMnO_4}=2.897\: mol\: KMnO_4}$
To Your Health: Minerals
For our bodies to function properly, we need to ingest certain substances from our diets. Among our dietary needs are minerals, the noncarbon elements our body uses for a variety of functions, such developing bone or ensuring proper nerve transmission. The US Department of Agriculture has established some recommendations for the RDIs of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet.
Table $1$: Essential Minerals and their Composition in Humans
Mineral Male (age 19–30 y) Female (age 19–30 y)
Ca 1,000 mg 0.025 mol 1,000 mg 0.025 mol
Cr 35 µg 6.7 × 10−7 mol 25 µg 4.8 × 10−7 mol
Cu 900 µg 1.4 × 10−5 mol 900 µg 1.4 × 10−5 mol
F 4 mg 2.1 × 10−4 mol 3 mg 1.5 × 10−4 mol
I 150 µg 1.2 × 10−6 mol 150 µg 1.2 × 10−6 mol
Fe 8 mg 1.4 × 10−4 mol 18 mg 3.2 × 10−4 mol
K 3,500 mg 9.0 × 10−2 mol 3,500 mg 9.0 × 10−2 mol
Mg 400 mg 1.6 × 10−2 mol 310 mg 1.3 × 10−2 mol
Mn 2.3 mg 4.2 × 10−5 mol 1.8 mg 3.3 × 10−5 mol
Mo 45 mg 4.7 × 10−7 mol 45 mg 4.7 × 10−7 mol
Na 2,400 mg 1.0 × 10−1 mol 2,400 mg 1.0 × 10−1 mol
P 700 mg 2.3 × 10−2 mol 700 mg 2.3 × 10−2 mol
Se 55 µg 7.0 × 10−7 mol 55 µg 7.0 × 10−7 mol
Zn 11 mg 1.7 × 10−4 mol 8 mg 1.2 × 10−4 mol
Table $1$ illustrates several things. First, the needs of men and women for some minerals are different. The extreme case is for iron; women need over twice as much as men do. In all other cases where there is a different RDI, men need more than women.
Second, the amounts of the various minerals needed on a daily basis vary widely—both on a mass scale and a molar scale. The average person needs 0.1 mol of Na a day, which is about 2.5 g. On the other hand, a person needs only about 25–35 µg of Cr per day, which is under one millionth of a mole. As small as this amount is, a deficiency of chromium in the diet can lead to diabetes-like symptoms or neurological problems, especially in the extremities (hands and feet). For some minerals, the body does not require much to keep itself operating properly.
Although a properly balanced diet will provide all the necessary minerals, some people take dietary supplements. However, too much of a good thing, even minerals, is not good. Exposure to too much chromium, for example, causes a skin irritation, and certain forms of chromium are known to cause cancer (as presented in the movie Erin Brockovich).
Key Takeaway
• It is possible to convert between moles of material and mass of material. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/06%3A_Quantities_in_Chemical_Reactions/6.03%3A_Mole-Mass_Conversions.txt |
Learning Objectives
• To use a balanced chemical reaction to determine molar relationships between the substances.
Previously, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). Here, we will extend the meaning of the coefficients in a chemical equation.
Consider the simple chemical equation
$\ce{2H_2 + O_2 → 2H_2O}\nonumber$
The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as
$\ce{4H_2 + 2O_2 → 4H_2O}\nonumber$
The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as
$\ce{22H_2 + 11O_2 → 22H_2O}\nonumber$
because 22:11:22 also reduces to 2:1:2.
Suppose we want to use larger numbers. Consider the following coefficients:
$12.044 \times 10^{23}\; H_2 + 6.022 \times 10^{23}\; O_2 → 12.044 \times 10^{23}\; H_2O\nonumber$
These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 × 1023 is 1 mol, while 12.044 × 1023 is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as
$\ce{2 \;mol\; H_2 + 1\; mol\; O_2 → 2 \;mol\; H_2O}\nonumber$
We can leave out the word mol and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is
$\ce{2H_2 + O_2 → 2H_2O}\nonumber$
Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level but also in terms of molar amounts of reactants and products. Thus, we can read this reaction as “two moles of hydrogen react with one mole of oxygen to produce two moles of water.”
2 molecules H2 1 molecule O2 2 molecules H2O
2 moles H2 1 mole O2 2 moles H2O
2 x 2.02 g=4.04 g H2 32.0 g O2 2 x 18.02 g=36.04 g H2O
Figure $1$: This representation of the production of water from oxygen and hydrogen show several ways to interpret the quantitative information of a chemical reaction.
By the same token, the ratios we constructed to describe molecules reaction can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios:
$\mathrm{\dfrac{2\: mol\: H_2}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2}}\nonumber$
$\mathrm{\dfrac{2\: mol\: H_2O}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2O}}\nonumber$
$\mathrm{\dfrac{2\: mol\: H_2}{2\: mol\: H_2O}\: or\: \dfrac{2\: mol\: H_2O}{2\: mol\: H_2}}\nonumber$
We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry. The ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products is called the stoichiometric factor.
Example $1$
How many moles of oxygen react with hydrogen to produce 27.6 mol of H2O? The balanced equation is as follows:
$\ce{2H2 + O2 -> 2H2O} \nonumber \nonumber$
Solution
Because we are dealing with quantities of H2O and O2, we will use the stoichiometric ratio that relates those two substances. Because we are given an amount of H2O and want to determine an amount of O2, we will use the ratio that has H2O in the denominator (so it cancels) and O2 in the numerator (so it is introduced in the answer). Thus,
$\mathrm{27.6\: mol\: H_2O\times\dfrac{1\: mol\: O_2}{2\: mol\: H_2O}=13.8\: mol\: O_2}$
To produce 27.6 mol of H2O, 13.8 mol of O2 react.
Exercise $1$
Using 2H2 + O2 → 2H2O, how many moles of hydrogen react with 3.07 mol of oxygen to produce H2O?
Answer
$\mathrm{3.07\: mol\: O_2\times\dfrac{2\: mol\: H_2}{1\: mol\: O_2}=6.14\: mol\: H_2}$
Key Takeaway
• The balanced chemical reaction can be used to determine molar relationships between substances. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/06%3A_Quantities_in_Chemical_Reactions/6.04%3A_Mole-Mole_Relationships_in_Chemical_Reactions.txt |
Learning Objectives
• To convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.
We have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:
Collectively, these conversions are called mole-mass calculations.
As an example, consider the balanced chemical equation
$\ce{Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3} \nonumber \nonumber$
If we have 3.59 mol of Fe2O3, how many grams of SO3 can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO3 in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO3 needed. Then using the molar mass of SO3 as a conversion factor, we determine the mass that this number of moles of SO3 has.
The first step resembles the exercises we did in Section 6.4. As usual, we start with the quantity we were given:
$\mathrm{3.59\: mol\: Fe_2O_3\times\dfrac{3\: mol\: SO_3}{1\: mol\: Fe_2O_3}=10.77\: mol\: SO_3} \nonumber \nonumber$
The mol Fe2O3 units cancel, leaving mol SO3 unit. Now, we take this answer and convert it to grams of SO3, using the molar mass of SO3 as the conversion factor:
$\mathrm{10.77\: mol\: SO_3\times\dfrac{80.07\: g\: SO_3}{1\: mol\: SO_3}=862.4\: g\: SO_3} \nonumber \nonumber$
Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO3 will react with 3.59 mol of Fe2O3. Many problems of this type can be answered in this manner.
The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:
We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time.
Example $1$
How many grams of CO2 are produced if 2.09 mol of HCl are reacted according to this balanced chemical equation?
$\ce{CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O} \nonumber$
Solution
Our strategy will be to convert from moles of HCl to moles of CO2 and then from moles of CO2 to grams of CO2. We will need the molar mass of CO2, which is 44.01 g/mol. Performing these two conversions in a single-line gives 46.0 g of CO2:
The molar ratio between CO2 and HCl comes from the balanced chemical equation.
Exercise
How many grams of glucose (C6H12O6) are produced if 17.3 mol of H2O are reacted according to this balanced chemical equation?
$\ce{6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2} \nonumber$
Answer
$\mathrm{17.3\: mol\: H_2O\times\dfrac{1\: mol\: C_6H_{12}O_6}{6\: mol\: H_2O}\times\dfrac{180.18\: g\: C_6H_{12}O_6}{1\: mol\: C_6H_{12}O_6}=520\: g\: C_6H_{12}O_6}$
It is a small step from mole-mass calculations to mass-mass calculations. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:
This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques.
Example $2$: Chlorination of Carbon
Methane can react with elemental chlorine to make carbon tetrachloride ($\ce{CCl4}$). The balanced chemical equation is as follows:
$\ce{CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl} \nonumber$
How many grams of HCl are produced by the reaction of 100.0 g of $\ce{CCl4}$?
Solution
First, let us work the problem in stepwise fashion. We begin by converting the mass of $\ce{CCl4}$ to moles of $\ce{CCl4}$, using the molar mass of $\ce{CCl4}$ (16.05 g/mol) as the conversion factor:
$\mathrm{100.0\: g\: CH_4\times\dfrac{1\: mol\: CH_4}{16.05\: g\: CH_4}=6.231\: mol\: CH_4}$
Note that we inverted the molar mass so that the gram units cancel, giving us an answer in moles. Next, we use the balanced chemical equation to determine the ratio of moles $\ce{CCl4}$ and moles HCl and convert our first result into moles of HCl:
$\mathrm{6.231\: mol\: CH_4\times\dfrac{4\: mol\: HCl}{1\: mol\: CH_4}=24.92\: mol\: HCl}$
Finally, we use the molar mass of HCl (36.46 g/mol) as a conversion factor to calculate the mass of 24.92 mol of HCl:
$\mathrm{24.92\: mol\: HCl\times\dfrac{36.46\: g\: HCl}{1\: mol\: HCl}=908.5\: g\: HCl}$
In each step, we have limited the answer to the proper number of significant figures. If desired, we can do all three conversions on a single line:
$\mathrm{100.0\: g\: CH_4\times\dfrac{1\: mol\: CH_4}{16.05\: g\: CH_4}\times\dfrac{4\: mol\: HCl}{1\: mol\: CH_4}\times\dfrac{36.46\: g\: HCl}{1\: mol\: HCl}=908.7\: g\: HCl}$
This final answer is slightly different from our first answer because only the final answer is restricted to the proper number of significant figures. In the first answer, we limited each intermediate quantity to the proper number of significant figures. As you can see, both answers are essentially the same.
Exercise $2$: Oxidation of Propanal
The oxidation of propanal (CH3CH2CHO) to propionic acid (CH3CH2COOH) has the following chemical equation:
CH3CH2CHO + 2K2Cr2O7 → CH3CH2COOH + other products
How many grams of propionic acid are produced by the reaction of 135.8 g of K2Cr2O7?
Answer
$\mathrm{135.8\: g\: K_2Cr_2O_7\times\dfrac{1\: mol\: K_2Cr_2O_7}{294.20\: g\: K_2Cr_2O_7}\times\dfrac{1\: mol\: CH_3CH_2COOH}{2\: mol\: K_2Cr_2O_7}\times\dfrac{74.09\: g\: CH_3CH_2COOH}{1\: mol\: CH_3CH_2COOH}=17.10\: g\: CH_3CH_2COOH}$
To Your Health: The Synthesis of Taxol
Taxol is a powerful anticancer drug that was originally extracted from the Pacific yew tree (Taxus brevifolia). As you can see from the accompanying figure, taxol is a very complicated molecule, with a molecular formula of $\ce{C47H51NO14}$. Isolating taxol from its natural source presents certain challenges, mainly that the Pacific yew is a slow-growing tree, and the equivalent of six trees must be harvested to provide enough taxol to treat a single patient. Although related species of yew trees also produce taxol in small amounts, there is significant interest in synthesizing this complex molecule in the laboratory.
After a 20-year effort, two research groups announced the complete laboratory synthesis of taxol in 1994. However, each synthesis required over 30 separate chemical reactions, with an overall efficiency of less than 0.05%. To put this in perspective, to obtain a single 300 mg dose of taxol, you would have to begin with 600 g of starting material. To treat the 26,000 women who are diagnosed with ovarian cancer each year with one dose, almost 16,000 kg (over 17 tons) of starting material must be converted to taxol. Taxol is also used to treat breast cancer, with which 200,000 women in the United States are diagnosed every year. This only increases the amount of starting material needed.
Clearly, there is intense interest in increasing the overall efficiency of the taxol synthesis. An improved synthesis not only will be easier but also will produce less waste materials, which will allow more people to take advantage of this potentially life-saving drug.
Key Takeaway
• A balanced chemical equation can be used to relate masses or moles of different substances in a reaction. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/06%3A_Quantities_in_Chemical_Reactions/6.05%3A_Mole-Mass_and_Mass-Mass_Problems.txt |
Concept Review Exercise
1. What is a mole?
Answer
1. A mole is 6.022 × 1023 things.
Exercises
1. How many dozens are in 1 mol? Express your answer in proper scientific notation.
2. A gross is a dozen dozen, or 144 things. How many gross are in 1 mol? Express your answer in proper scientific notation.
3. How many moles of each type of atom are in 1.0 mol of C6H12O6?
4. How many moles of each type of atom are in 1.0 mol of K2Cr2O7?
5. How many moles of each type of atom are in 2.58 mol of Na2SO4?
6. How many moles of each type of atom are in 0.683 mol of C34H32FeN4O4? (This is the formula of heme, a component of hemoglobin.)
7. How many molecules are in 16.8 mol of H2O?
8. How many formula units are in 0.778 mol of iron(III) nitrate?
9. A sample of gold contains 7.02 × 1024 atoms. How many moles of gold is this?
10. A flask of mercury contains 3.77 × 1022 atoms. How many moles of mercury are in the flask?
11. An intravenous solution of normal saline may contain 1.72 mol of sodium chloride (NaCl). How many sodium and chlorine atoms are present in the solution?
12. A lethal dose of arsenic is 1.00 × 1021 atoms. How many moles of arsenic is this?
Answers
1. 5.018 × 1022 dozens
2. 4.18 x 1021 grosses
1. 6.0 mol of C atoms, 12.0 mol of H atoms, and 6.0 mol of O atoms
4. 2.0 mol of K atoms, 2.0 mol of Cr atoms, and 7.0 mol of O atoms
1. 5.16 mol of Na atoms, 2.58 mol of S atoms, and 10.32 mol of O atoms
6. 23.2 mol of C atoms, 21.9 mol of H atoms, 0.683 mol of Fe, 2.73 mol of N and 2.73 mol of O atoms
1. 1.012 × 1025 molecules
8. 4.69 x 1023 formula units
9. 11.7 mol
10. 0.0626 mol
11. 1.04 × 1024 Na atoms and 1.04 × 1024 Cl atoms
12. 0.00166 mol
Concept Review Exercises
1. How are molar masses of the elements determined?
2. How are molar masses of compounds determined?
Answers
1. Molar masses of the elements are the same numeric value as the masses of a single atom in atomic mass units but in units of grams instead.
2. Molar masses of compounds are calculated by adding the molar masses of their atoms.
Exercises
1. What is the molar mass of Si? What is the molar mass of U?
2. What is the molar mass of Mn? What is the molar mass of Mg?
3. What is the molar mass of FeCl2? What is the molar mass of FeCl3?
4. What is the molar mass of C6H6? What is the molar mass of C6H5CH3?
5. What is the molar mass of (NH4)2S? What is the molar mass of Ca(OH)2?
6. What is the molar mass of (NH4)3PO4? What is the molar mass of Sr(HCO3)2?
7. Aspirin (C9H8O4) is an analgesic (painkiller) and antipyretic (fever reducer). What is the molar mass of aspirin?
8. Ibuprofen (C13H18O2) is an analgesic (painkiller). What is the molar mass of ibuprofen?
9. Morphine (C17H19NO3) is a narcotic painkiller. What is the mass of 1 mol of morphine?
10. Heroin (C21H23NO5) is a narcotic drug that is a derivative of morphine. What is the mass of 1 mol of heroin?
Answers
1. 28.09 g/mol; 238.0 g/mol
2. 54.94 g/mol; 24.31 g/mol
3. 126.75 g/mol; 162.20 g/mol
4. 78.12 g/mol; 92.15 g/mol
5. 68.16 g/mol; 74.10 g/mol
6. 149.12 g/mol; 209.64 g/mol
7. 180.17 g/mol
8. 206.31 g/mol
9. 285.37 g
10. 369.45 g
Concept Review Exercises
1. What relationship is needed to perform mole-mass conversions?
2. What information determines which conversion factor is used in a mole-mass conversion?
Answers
1. The atomic or molar mass is needed for a mole-mass conversion.
2. The unit of the initial quantity determines which conversion factor is used.
Exercises
1. What is the mass of 8.603 mol of Fe metal?
2. What is the mass of 0.552 mol of Ag metal?
3. What is the mass of 6.24 × 104 mol of Cl2 gas?
4. What is the mass of 0.661 mol of O2 gas?
5. What is the mass of 20.77 mol of CaCO3?
6. What is the mass of 9.02 × 10−3 mol of the hormone epinephrine (C9H13NO3)?
7. How many moles are present in 977.4 g of NaHCO3?
8. How many moles of erythromycin (C37H67NO13), a widely used antibiotic, are in 1.00 × 103 g of the substance?
9. Cortisone (C21H28O5) is a synthetic steroid that is used as an anti-inflammatory drug. How many moles of cortisone are present in one 10.0 mg tablet?
10. Recent research suggests that the daily ingestion of 85 mg of aspirin (also known as acetylsalicylic acid, C9H8O4) will reduce a person’s risk of heart disease. How many moles of aspirin is that?
Answers
1. 480.5 g
2. 59.6 g
3. 4.42 × 106 g
4. 21.2 g
5. 2,079 g
6. 1.65 g
7. 11.63 mol
8. 1.36 mol
9. 2.77 × 10−5 mol
10. 4.7 x 10−4 mol
Concept Review Exercise
1. How do we relate molar amounts of substances in chemical reactions?
Answer
1. Amounts of substances in chemical reactions are related by their coefficients in the balanced chemical equation.
Exercises
1. List the molar ratios you can derive from this balanced chemical equation:
NH3 + 2O2 → HNO3 + H2O
2. List the molar ratios you can derive from this balanced chemical equation
2C2H2 + 5O2 → 4CO2 + 2H2O
3. Given the following balanced chemical equation,
6NaOH + 3Cl2 → NaClO3 + 5NaCl + 3H2O
how many moles of NaCl can be formed if 3.77 mol of NaOH were to react?
4. Given the following balanced chemical equation,
C5H12 + 8O2 → 5CO2 + 6H2O
how many moles of H2O can be formed if 0.0652 mol of C5H12 were to react?
5. Balance the following unbalanced equation and determine how many moles of H2O are produced when 1.65 mol of NH3 react.
NH3 + O2 → N2 + H2O
6. Trinitrotoluene [C6H2(NO2)3CH3], also known as TNT, is formed by reacting nitric acid (HNO3) with toluene (C6H5CH3):
HNO3 + C6H5CH3 → C6H2(NO2)3CH3 + H2O
Balance the equation and determine how many moles of TNT are produced when 4.903 mol of HNO3 react.
7. Chemical reactions are balanced in terms of molecules and in terms of moles. Are they balanced in terms of dozens? Defend your answer.
8. Explain how a chemical reaction balanced in terms of moles satisfies the law of conservation of matter.
Answers
1. 1 mol NH3:2 mol O2:1 mol HNO3:1 mol H2O
2. 2 mol C2H2:5 mol O2:4 mol CO2:2 mol H2O
1. 3.14 mol
4. 0.3912 mol
1. 4NH3 + 3O2 → 2N2 + 6H2O; 2.48 mol
6. 3HNO3 + C6H5CH3 → C6H2(NO2)3CH3 + 3H2O; 1.634 mol
7. Yes, they are still balanced.
8. A chemical reaction, balanced in terms of moles, contains the same number of atoms of each element, before and after the reaction. This means that all the atoms and its masses are conserved.
Concept Review Exercises
1. What is the general sequence of conversions for a mole-mass calculation?
2. What is the general sequence of conversions for a mass-mass calculation?
Answers
1. mol first substance → mol second substance → mass second substance
2. mass first substance → mol first substance → mol second substance → mass second substance
Exercises
1. Given the following unbalanced chemical equation,
H3PO4 + NaOH → H2O + Na3PO4
what mass of H2O is produced by the reaction of 2.35 mol of H3PO4?
2. Given the following unbalanced chemical equation,
C2H6 + Br2 → C2H4Br2 + HBr
what mass of HBr is produced if 0.884 mol of C2H6 is reacted?
3. Certain fats are used to make soap, the first step being to react the fat with water to make glycerol (also known as glycerin) and compounds called fatty acids. One example is as follows:
$\mathrm{\underset{\large{a\: fat}}{C_3H_5(OOC(CH_2)_{14}CH_3)_3}+3H_2O\rightarrow \underset{\large{glycerol}}{C_3H_5(OH)_3}+\underset{\large{fatty\: acid}}{3CH_3(CH_2)_{14}COOH}}$
How many moles of glycerol can be made from the reaction of 1,000.0 g of C3H5(OOC(CH2)14CH3)3?
4. Photosynthesis in plants leads to the general overall reaction for producing glucose (C6H12O6):
6CO2 + 6H2O → C6H12O6 + 6O2
How many moles of glucose can be made from the reaction of 544 g of CO2?
5. Precipitation reactions, in which a solid (called a precipitate) is a product, are commonly used to remove certain ions from solution. One such reaction is as follows:
Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)
How many grams of Na2SO4 are needed to precipitate all the barium ions produced by 43.9 g of Ba(NO3)2?
6. Nitroglycerin [C3H5(ONO2)3] is made by reacting nitric acid (HNO3) with glycerol [C3H5(OH)3] according to this reaction:
C3H5(OH)3 + 3HNO3 → C3H5(ONO2)3 + 3H2O
If 87.4 g of HNO3 are reacted with excess glycerol, what mass of nitroglycerin can be made?
7. Antacids are bases that neutralize acids in the digestive tract. Magnesium hydroxide [Mg(OH)2] is one such antacid. It reacts with hydrochloric acid in the stomach according to the following reaction:
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
How many grams of HCl can a 200 mg dose of Mg(OH)2 neutralize?
8. Acid rain is caused by the reaction of nonmetal oxides with water in the atmosphere. One such reaction involves nitrogen dioxide (NO2) and produces nitric acid (HNO3):
3NO2 + H2O → 2HNO3 + NO
If 1.82 × 1013 g of NO2 enter the atmosphere every year due to human activities, potentially how many grams of HNO3 can be produced annually?
9. A simplified version of the processing of iron ore into iron metal is as follows:
2Fe2O3 + 3C → 4Fe + 3CO2
How many grams of C are needed to produce 1.00 × 109 g of Fe?
10. The SS Hindenburg contained about 5.33 × 105 g of H2 gas when it burned at Lakehurst, New Jersey, in 1937. The chemical reaction is as follows:
2H2 + O2 → 2H2O
How many grams of H2O were produced?
1. 127 g
2. 143 g
1. 1.238 mol
4. 2.06 mol
5. 23.9 g
6. 105 g
7. 0.250 g
8. 1.66 x 1013 g
9. 1.61 × 108 g
10. 4.75 x 106 g
Additional Exercises
1. If the average male has a body mass of 70 kg, of which 60% is water, how many moles of water are in an average male?
2. If the average female is 60.0 kg and contains 0.00174% iron, how many moles of iron are in an average female?
3. How many moles of each element are present in 2.67 mol of each compound?
1. HCl
2. H2SO4
3. Al(NO3)3
4. Ga2(SO4)3
4. How many moles of each element are present in 0.00445 mol of each compound?
1. HCl
2. H2SO4
3. Al2(CO3)3
4. Ga2(SO4)3
5. What is the mass of one hydrogen atom in grams? What is the mass of one oxygen atom in grams? Do these masses have a 1:16 ratio, as expected?
6. What is the mass of one sodium atom in grams?
7. If 6.63 × 10−6 mol of a compound has a mass of 2.151 mg, what is the molar mass of the compound?
8. Hemoglobin (molar mass is approximately 64,000 g/mol) is the major component of red blood cells that transports oxygen and carbon dioxide in the body. How many moles are in 0.034 g of hemoglobin?
Answers
1. 2,330 mol
2. 0.0187 mol
1. 2.67 mol of H and 2.67 mol of Cl
2. 5.34 mol of H, 2.67 mol of S, and 10.68 mol of O
3. 2.67 mol of Al, 8.01 mol of N, and 24.03 mol of O
4. 5.34 mol of Ga, 8.01 mol of S, and 32.04 mol of O
4.
1. 0.00445 mol of H and 0.00445 mol of Cl
2. 0.00890 mol of H, 0.00445 mol of S, and 0.0178 mol of O
3. 0.00890 mol of Al, 0.0134 mol of C, and 0.0401 mol of O
4. 0.00890 mol of Ga, 0.0134 mol of S, and 0.0534 mol of O
5. H = 1.68 × 10−24 g and O = 2.66 × 10−23 g; yes, they are in a 1:16 ratio.
6. 3.819 x 10-23 g
7. 324 g/mol
8. 5.3 x 10-7 mol
6.S: Quantities in Chemical Reactions (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Chemical reactions relate quantities of reactants and products. Chemists use the mole unit to represent 6.022 × 1023 things, whether the things are atoms of elements or molecules of compounds. This number, called Avogadro’s number, is important because this number of atoms or molecules has the same mass in grams as one atom or molecule has in atomic mass units. Molar masses of substances can be determined by summing the appropriate masses from the periodic table; the final molar mass will have units of grams.
Because one mole of a substance will have a certain mass, we can use that relationship to construct conversion factors that will convert a mole amount into a mass amount, or vice versa. Such mole-mass conversions typically take one algebraic step.
Chemical reactions list reactants and products in molar amounts, not just molecular amounts. We can use the coefficients of a balanced chemical equation to relate moles of one substance in the reaction to moles of other substances (stoichiometry). Chemical reactions obey the Law of Conservation of Mass. To balance a chemical reaction, the coefficients in front of each compound can be adjusted until the total number of atoms of each elements is equal on both sides of the reaction arrow.
Collision Theory can be used to described the energetic aspects of chemical reactions. The reactants and products of a chemical reaction store potential energy in the form of chemical bonds and intermolecular forces. The energy difference in the bond energies of the reactants and products is called "Enthalpy". When the products are lower in potential energy than the reactants, then this excess energy is released as heat and the reaction is described as exothermic. When the products are higher in potential energy than the reactants, then energy must be added to the reaction as heat for it to occur and the reaction is described as endothermic.
The rate of a chemical reaction is influenced by its activation energy. The larger the activation energy, the slower the reaction rate. Catalysts can be added to reactions to lower the activation energy and increase the reaction rate.
Not all reactions go to completion. For some reactions, both the forward and reverse reaction can occur simultaneously. When the forward and reverse reaction rates are equal, the reaction is described as being "at equilibrium". Another way to recognize when a system is at equilibrium is when the concentration of reactants and products remain constant. Le Chatlier's principle states that the direction of an equilibrium reaction will shift to reduce the stress. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/06%3A_Quantities_in_Chemical_Reactions/6.E%3A_Quantities_in_Chemical_Reactions_%28Exercise%29.txt |
Energy is a vital component of the world around us. Nearly every physical and chemical process, including all the chemical reactions discussed in previous chapters, occurs with a simultaneous energy change. In this chapter, we will explore the nature of energy and how energy and chemistry are related.
• 7.0: Prelude to Energy and Chemical Processes
Metabolism is the collective term for the chemical reactions that occur in cells and provide energy to keep cells alive. Some of the energy from metabolism is in the form of heat, and some animals use this heat to regulate their body temperatures. Such warm-blooded animals are called endotherms. In endotherms, problems with metabolism can lead to fluctuations in body temperature. When humans get sick, for instance, our body temperatures can rise higher than normal; we develop a fever.
• 7.1: Energy and Its Units
Energy is the ability to do work. Heat is the transfer of energy due to temperature differences. Energy and heat are expressed in units of joules.
• 7.2: Heat and Temperature
Heat transfer is related to temperature change. Heat is equal to the product of the mass, the change in temperature, and a proportionality constant called the specific heat.
• 7.3: Phase Changes
There is an energy change associated with any phase change. There is an energy change associated with any phase change.
• 7.4: Bond Energies and Chemical Reactions
Atoms are held together by a certain amount of energy called bond energy. Chemical processes are labeled as exothermic or endothermic based on whether they give off or absorb energy, respectively.
• 7.5: The Energy of Biochemical Reactions
Energy to power the human body comes from chemical reactions.
• 7.E: Energy and Chemical Processes (Exercises)
Problems and Solutions to accompany the chapter.
• 7.S: Energy and Chemical Processes (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
07: Energy and Chemical Processes
Metabolism is the collective term for the chemical reactions that occur in cells and provide energy to keep cells alive. Some of the energy from metabolism is in the form of heat, and some animals use this heat to regulate their body temperatures. Such warm-blooded animals are called endotherms. In endotherms, problems with metabolism can lead to fluctuations in body temperature. When humans get sick, for instance, our body temperatures can rise higher than normal; we develop a fever. When food is scarce (especially in winter), some endotherms go into a state of controlled decreased metabolism called hibernation. During hibernation, the body temperatures of these endotherms actually decrease. In hot weather or when feverish, endotherms will pant or sweat to rid their bodies of excess heat.
Table \(1\): Average Body Temperatures of Selected Endotherms
Endotherm Body Temperature (°F) Body Temperature (°C)
bird up to 110 up to 43.5
cat 101.5 38.6
dog 102 38.9
horse 100.0 37.8
human 98.6 37.0
pig 102.5 39.2
Ectotherms, sometimes called cold-blooded animals, do not use the energy of metabolism to regulate body temperature. Instead, they depend on external energy sources, such as sunlight. Fish, for example, will seek out water of different temperatures to regulate body temperature. The amount of energy available is directly related to the metabolic rate of the animal. When energy is scarce, ectotherms may also hibernate.
The connection between metabolism and body temperature is a reminder that energy and chemical reactions are intimately related. A basic understanding of this relationship is especially important when those chemical reactions occur within our own bodies. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/07%3A_Energy_and_Chemical_Processes/7.00%3A_Prelude_to_Energy_and_Chemical_Processes.txt |
Learning Objectives
• To define energy and heat.
• To relate calories to nutrition and exercise.
Energy is the ability to do work. You can understand what this means by thinking about yourself when you feel “energetic.” You feel ready to go—to jump up and get something done. When you have a lot of energy, you can perform a lot of work. By contrast, if you do not feel energetic, you have very little desire to do much of anything. This description is not only applicable to you but also to all physical and chemical processes. The quantity of work that can be done is related to the quantity of energy available to do it.
Energy can be transferred from one object to another if the objects have different temperatures. The transfer of energy due to temperature differences is called heat. For example, if you hold an ice cube in your hand, the ice cube slowly melts as energy in the form of heat is transferred from your hand to the ice. As your hand loses energy, it starts to feel cold.
Because of their interrelationships, energy, work, and heat have the same units. The SI unit of energy, work, and heat is the joule (J). A joule is a tiny amount of energy. For example, it takes about 4 J to warm 1 mL of H2O by 1°C. Many processes occur with energy changes in thousands of joules, so the kilojoule (kJ) is also common. Another unit of energy, used widely in the health professions and everyday life, is the calorie (cal). The calorie was initially defined as the amount of energy needed to warm 1 g of H2O by 1°C, but in modern times, the calorie is related directly to the joule, as follows:
$1\; cal = 4.184\; J \nonumber$
We can use this relationship to convert quantities of energy, work, or heat from one unit to another.
Although the joule is the proper SI unit for energy, we will use the calorie or the kilocalorie (or Calorie) in this chapter because they are widely used by health professional
Example $1$
The energy content of a single serving of bread is 70.0 Cal. What is the energy content in calories? In joules?
Solution
This is a simple conversion-factor problem. Using the relationship 1 Cal = 1,000 cal, we can answer the first question with a one-step conversion:
$\mathrm{70.0 \: Cal\times\dfrac{1,000\: cal}{1\: Cal}= 70,000\: cal}$
Then we convert calories into joules
$\mathrm{70,000 \: cal \times \dfrac{4.184\: J}{1\: cal}=293,000\: J}$
and then kilojoules
$\mathrm{293,000\: J\times\dfrac{1\: kJ}{1,000\: J}=293\: kJ}$
The energy content of bread comes mostly from carbohydrates.
Exercise $1$
The energy content of one cup of honey is 1,030 Cal. What is its energy content in calories and joules?
Answer
1,030,000 (1.03 x 106) cal; 4,309,520 (4.31 x 106) J
The calorie is used in nutrition to express the energy content of foods. However, because a calorie is a rather small quantity, nutritional energies are usually expressed in kilocalories (kcal), also called Calories (capitalized; Cal). For example, a candy bar may provide 120 Cal (nutritional calories) of energy, which is equal to 120,000 cal. Figure $1$ shows an example.
The caloric content of foods is determined by analyzing the food for protein, carbohydrate, fat, water, and “minerals” and then calculating 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.
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
The compositions and caloric contents of some common foods are given in $2$.
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. 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.
To Your Health: Energy Expenditures
Most health professionals agree that exercise is a valuable component of a healthy lifestyle. Exercise not only strengthens the body and develops muscle tone but also expends energy. After obtaining energy from the foods we eat, we need to expend that energy somehow, or our bodies will store it in unhealthy ways (e.g., fat). Like the energy content in food, the energy expenditures of exercise are also reported in kilocalories, usually kilocalories per hour of exercise. These expenditures vary widely, from about 440 kcal/h for walking at a speed of 4 mph to 1,870 kcal/h for mountain biking at 20 mph. Table $3$ lists the energy expenditure for a variety of exercises.
Table $3$: Energy Expenditure of a 180-Pound Person during Selected Exercises
Exercise Energy Expended (kcal/h)
aerobics, low-level 325
basketball 940
bike riding, 20 mph 830
golfing, with cart 220
golfing, carrying clubs 425
jogging, 7.5 mph 950
racquetball 740
skiing, downhill 520
soccer 680
walking upstairs 1,200
yoga 280
Because some forms of exercise use more energy than others, anyone considering a specific exercise regimen should consult with his or her physician first.
Summary
Energy is the ability to do work. Heat is the transfer of energy due to temperature differences. Energy and heat are expressed in units of joules. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/07%3A_Energy_and_Chemical_Processes/7.01%3A_Energy_and_Its_Units.txt |
Learning Objectives
• To relate heat transfer to temperature change.
• Memorize temperature equations for Celsius, Fahrenheit and Kelvin conversions.
• Understand how body temperature can vary.
The concept of temperature may seem familiar to you, but many people confuse temperature with heat. Temperature is a measure of how hot or cold an object is relative to another object (its thermal energy content), whereas heat is the flow of thermal energy between objects with different temperatures. When we touch a hot object, energy flows from the hot object into our fingers, and we perceive that incoming energy as the object being “hot.” Conversely, when we hold an ice cube in our palms, energy flows from our hand into the ice cube, and we perceive that loss of energy as “cold.” In both cases, the temperature of the object is different from the temperature of our hand, so we can conclude that differences in temperatures are the ultimate cause of heat transfer.
Temperature Scales
Temperature is a measure of the average amount of energy of motion, or kinetic energy, a system contains. Temperatures are expressed using scales that use units called degrees. There are three scales used for reporting temperatures. Figure $1$ compares the three temperature scales: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). Thermometers measure temperature by using materials that expand or contract when heated or cooled. Mercury or alcohol thermometers, for example, have a reservoir of liquid that expands when heated and contracts when cooled, so the liquid column lengthens or shortens as the temperature of the liquid changes.
In the United States, the commonly used temperature scale is the Fahrenheit scale (symbolized by °F and spoken as “degrees Fahrenheit”). On this scale, the freezing point of liquid water (the temperature at which liquid water turns to solid ice) is 32 °F, and the boiling point of water (the temperature at which liquid water turns to steam) is 212 °F.
Science also uses other scales to express temperature. For example, the Celsius scale (symbolized by °C and spoken as “degrees Celsius”) defines 0°C as the freezing point of water and 100 °C as the boiling point of water. This scale is divided into 100 divisions between these two landmarks and extended higher and lower as well. By comparing the Fahrenheit and Celsius scales, a conversion between the two scales can be determined:
$^{\circ}C= \left(^{\circ}F-32\right )\times \dfrac{5}{9} \label{F2C}$
$^{\circ}F= \left(^{\circ}C\times \dfrac{9}{5}\right)+32 \label{C2F}$
Note that science uses the Celsius and Kelvin scales almost exclusively; virtually no practicing chemist expresses laboratory-measured temperatures with the Fahrenheit scale. (In fact, the United States is one of the few countries in the world that still uses the Fahrenheit scale on a daily basis. People driving near the borders of Canada or Mexico may pick up local radio stations on the other side of the border that express the daily weather in degrees Celsius, so do not get confused by their weather reports.)
Example $1$: Conversions
1. What is 98.6°F in degrees Celsius?
2. What is 25.0°C in degrees Fahrenheit?
Solution
1. Using Equation \ref{F2C}, we have
\begin{align*} ^{\circ}C &= (98.6-32)\times \dfrac{5}{9} \[5pt] &=66.6\times \dfrac{5}{9} \[5pt] &= 37.0^{\circ}C \end{align*} \nonumber
1. Using Equation \ref{C2F}, we have
\begin{align*} ^{\circ}F &= (25.0\times \dfrac{9}{5})+32 \[5pt] &= 45.0+32 \[5pt] &=77.0^{\circ}F \end{align*} \nonumber
For more examples of how to perform these types of problems, click on this video to see your professor in action.
Exercise $1$
1. Convert 0°F to degrees Celsius.
2. Convert 212°C to degrees Fahrenheit.
Answer a
−17.8°C
Answer b
414°F
The fundamental unit of temperature in SI is the Kelvin (K). The Kelvin temperature scale (note that the name of the scale capitalizes the word Kelvin, but the unit itself is lowercase) uses degrees that are the same size as the Celsius degree, but the numerical scale is shifted up by 273.15 units. That is, the conversion between the Kelvin and Celsius scales is as follows:
$K=^{\circ}C+273.15 \label{C2K}$
For most purposes, it is acceptable to use 273 instead of 273.15 in Equation \ref{C2K}.
Note that the Kelvin scale does not use the word degrees; a temperature of 295 K is spoken of as “two hundred ninety-five kelvin” and not “two hundred ninety-five degrees Kelvin.”
The reason that the Kelvin scale is defined this way is that there exists a minimum possible temperature called absolute zero (zero Kelvin). The Kelvin temperature scale is set so that 0 K is absolute zero, and the temperature is counted upward from there. Normal room temperature is about 295 K, as seen in the following example.
Example $2$: Room Temperature
If the normal room temperature is 72.0°F, what is room temperature in degrees Celsius and kelvin?
Solution
First, we use Equation \ref{F2C} to determine the temperature in degrees Celsius:
\begin{align*} ^{\circ}C &= (72.0-32)\times \dfrac{5}{9} \[5pt] &= 40.0\times \dfrac{5}{9} \[5pt] &= 22.2^{\circ}C \end{align*} \nonumber
Then we use Equation \ref{C2K} to determine the temperature in the Kelvin scale:
\begin{align*} K &= 22.2^{\circ}C+273.15 \[5pt] &= 295.4K \end{align*} \nonumber
So, room temperature is about 295 K.
Health Application: Body temperature
Normal body temperature is defined as being 98.6°F (+/- 1.0°F). To determine body temperature, thermometers can be placed inside or on the surface of the body. The two best methods of obtaining body temperature are placing the thermometer either under the tongue or inside the rectum. Typically, children are capable of holding a thermometer in their mouths around the age of four (have fun before that age).
Fever is defined as body temperature being above 100°F (adults). High fever status occurs at and above 104°F. For adults, these adults should seek medical attention immediately if the fever exceeds 104°F. For children, these values are much lower.
Hyperthermia (inability of the body to regulate heat) occurs when normal body temperature is exceeded. Conditions that can cause hyperthermia are fever (infection), heat stroke, thyroid disorders, heart attack, or traumatic injury. Medications for cancer, arthritis, and thyroid patients can cause the body temperature to rise. Symptoms of hyperthermia include sweating, confusion, nausea, and dizziness.
Hypothermia (exposure to cold environments) occurs when normal body temperature dips below 95.0°F. When this occurs, the affected person(s) should seek immediate medical attention. During hypothermia, the body has problems producing heat. Medical conditions like diabetes, infection, and thyroid dysfunction can cause hypothermia. Watch this video of American Marines attempting to survive extreme environments. Symptoms of this condition involve shivering, confusion, and sluggish behavior.
Treating hyperthermia could involve hydrating a patient. If a fever is due to infection, analgesics (like Tylenol, Advil, aspirin, or Aleve) can help alleviate fever. Placing someone in a cool bath can also relieve symptoms.
To combat hypothermia, one must remove wet clothing, redress in warm materials, and participate in physical movement.
Heat transfer
Heat is a familiar manifestation of energy. When we touch a hot object, energy flows from the hot object into our fingers, and we perceive that incoming energy as the object being “hot.” Conversely, when we hold an ice cube in our palms, energy flows from our hand into the ice cube, and we perceive that loss of energy as “cold.” In both cases, the temperature of the object is different from the temperature of our hand, so we can conclude that differences in temperatures are the ultimate cause of heat transfer.
Suppose we consider the transfer of heat from the opposite perspective—namely, what happens to a system that gains or loses heat? Generally, the system’s temperature changes. (We will address a few exceptions later.) The greater the original temperature difference, the greater the transfer of heat, and the greater the ultimate temperature change. The relationship between the amount of heat transferred and the temperature change can be written as
$\text{heat} \propto ΔT \label{Eq1}$
where ∝ means “is proportional to” and ΔT is the change in temperature of the system. Any change in a variable is always defined as “the final value minus the initial value” of the variable, so ΔT is TfinalTinitial. In addition, the greater the mass of an object, the more heat is needed to change its temperature. We can include a variable representing mass (m) to the proportionality as follows:
$\text{heat} \propto mΔT \label{Eq2}$
To change this proportionality into an equality, we include a proportionality constant. The proportionality constant is called the specific heat and is commonly symbolized by $c$:
$\text{heat} = mcΔT \label{Eq3}$
Every substance has a characteristic specific heat, which is reported in units of cal/g•°C or cal/g•K, depending on the units used to express ΔT. The specific heat of a substance is the amount of energy that must be transferred to or from 1 g of that substance to change its temperature by 1°. Table $1$ lists the specific heats for various materials.
Table $1$: Specific Heats of Selected Substances
Substance c (cal/g•°C)
aluminum (Al) 0.215
aluminum oxide (Al2O3) 0.305
benzene (C6H6) 0.251
copper (Cu) 0.092
ethanol (C2H6O) 0.578
hexane (C6H14) 0.394
hydrogen (H2) 3.419
ice [H2O(s)] 0.492
iron (Fe) 0.108
iron(III) oxide (Fe2O3) 0.156
mercury (Hg) 0.033
oxygen (O2) 0.219
sodium chloride (NaCl) 0.207
steam [H2O(g)] 0.488
water [H2O(ℓ)] 1.00
The proportionality constant c is sometimes referred to as the specific heat capacity or (incorrectly) the heat capacity.
The direction of heat flow is not shown in heat = mcΔT. If energy goes into an object, the total energy of the object increases, and the values of heat ΔT are positive. If energy is coming out of an object, the total energy of the object decreases, and the values of heat and ΔT are negative.
Example $3$
What quantity of heat is transferred when a 150.0 g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of heat flow?
Solution
We can use Equation \ref{Eq3} to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the iron is 73.3°C and the initial temperature is 25.0°C, ΔT is as follows:
\begin{align*} ΔT = T_{final} − T_{initial} \[4pt] &= 73.3^oC − 25.0^oC \[4pt] &= 48.3^oC \end{align*} \nonumber
The mass is given as 150.0 g, and Table $1$ gives the specific heat of iron as 0.108 cal/g•°C. Substitute the known values into Equation \ref{Eq3} and solve for amount of heat:
$\mathrm{heat=(150.0\: g)\left(0.108\: \dfrac{cal} {g\cdot {^\circ C}}\right)(48.3^\circ C) = 782\: cal} \nonumber$
Note how the gram and °C units cancel algebraically, leaving only the calorie unit, which is a unit of heat. Because the temperature of the iron increases, energy (as heat) must be flowing into the metal.
Exercise $3$
What quantity of heat is transferred when a 295.5 g block of aluminum metal is cooled from 128.0°C to 22.5°C? What is the direction of heat flow?
Answer
6,700 cal
Example $4$
A 10.3 g sample of a reddish-brown metal gave off 71.7 cal of heat as its temperature decreased from 97.5°C to 22.0°C. What is the specific heat of the metal? Can you identify the metal from the data in Table $1$?
Solution
The question gives us the heat, the final and initial temperatures, and the mass of the sample. The value of ΔT is as follows:
ΔT = TfinalTinitial = 22.0°C − 97.5°C = −75.5°C
If the sample gives off 71.7 cal, it loses energy (as heat), so the value of heat is written as a negative number, −71.7 cal. Substitute the known values into heat = mcΔT and solve for c:
−71.7 cal = (10.3 g)(c)(−75.5°C)
$c \,\mathrm{=\dfrac{-71.7\: cal}{(10.3\: g)(-75.5^\circ C)}}$
c = 0.0923 cal/g•°C
This value for specific heat is very close to that given for copper in Table $1$.
Exercise $4$
A 10.7 g crystal of sodium chloride (NaCl) had an initial temperature of 37.0°C. What is the final temperature of the crystal if 147 cal of heat were supplied to it?
Answer
103.4°C
Notice that water has a very high specific heat compared to most other substances. Water is commonly used as a coolant for machinery because it is able to absorb large quantities of heat (see table above). Coastal climates are much more moderate than inland climates because of the presence of the ocean. Water in lakes or oceans absorbs heat from the air on hot days and releases it back into the air on cool days.
Summary
• Heat transfer is related to temperature change.
• The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/07%3A_Energy_and_Chemical_Processes/7.02%3A_Heat_and_Temperature.txt |
Learning Objectives
• Determine the heat associated with a phase change.
Matter can exist in one of several different states, including a gas, liquid, or solid state. The amount of energy in molecules of matter determines the state of matter.
• A gas is a state of matter in which atoms or molecules have enough energy to move freely. The molecules come into contact with one another only when they randomly collide.
• A liquid is a state of matter in which atoms or molecules are constantly in contact but have enough energy to keep changing positions relative to one another.
• A solid is a state of matter in which atoms or molecules do not have enough energy to move. They are constantly in contact and in fixed positions relative to one another.
The following are the changes of state:
Changes of State
Solid → Liquid Melting or fusion
Liquid → Gas Vaporization
Liquid → Solid Freezing
Gas → Liquid Condensation
Solid → Gas Sublimation
• If heat is added to a substance, such as in melting, vaporization, and sublimation, the process is endothermic. In this instance, heat is increasing the speed of the molecules causing them move faster (examples: solid to liquid; liquid to gas; solid to gas).
• If heat is removed from a substance, such as in freezing and condensation, then the process is exothermic. In this instance, heat is decreasing the speed of the molecules causing them move slower (examples: liquid to solid; gas to liquid). These changes release heat to the surroundings.
• The amount of heat needed to change a sample from solid to liquid would be the same to reverse from liquid to solid. The only difference is the direction of heat transfer.
Example $1$
Label each of the following processes as endothermic or exothermic.
1. water boiling
2. ice forming on a pond
Solution
1. endothermic - you must put a pan of water on the stove and give it heat in order to get water to boil. Because you are adding heat/energy, the reaction is endothermic.
2. exothermic - think of ice forming in your freezer instead. You put water into the freezer, which takes heat out of the water, to get it to freeze. Because heat is being pulled out of the water, it is exothermic. Heat is leaving.
Exercise $1$
Label each of the following processes as endothermic or exothermic.
1. water vapor condensing
2. gold melting
Answer
a. exothermic
b. endothermic
A phase change is a physical process in which a substance goes from one phase to another. Usually the change occurs when adding or removing heat at a particular temperature, known as the melting point or the boiling point of the substance. The melting point is the temperature at which the substance goes from a solid to a liquid (or from a liquid to a solid). The boiling point is the temperature at which a substance goes from a liquid to a gas (or from a gas to a liquid). The nature of the phase change depends on the direction of the heat transfer. Heat going into a substance changes it from a solid to a liquid or a liquid to a gas. Removing heat from a substance changes a gas to a liquid or a liquid to a solid.
Two key points are worth emphasizing. First, at a substance’s melting point or boiling point, two phases can exist simultaneously. Take water (H2O) as an example. On the Celsius scale, H2O has a melting point of 0°C and a boiling point of 100°C. At 0°C, both the solid and liquid phases of H2O can coexist. However, if heat is added, some of the solid H2O will melt and turn into liquid H2O. If heat is removed, the opposite happens: some of the liquid H2O turns into solid H2O. A similar process can occur at 100°C: adding heat increases the amount of gaseous H2O, while removing heat increases the amount of liquid H2O (Figure $1$).
Water is a good substance to use as an example because many people are already familiar with it. Other substances have melting points and boiling points as well.
Second, as shown in Figure $1$, the temperature of a substance does not change as the substance goes from one phase to another. In other words, phase changes are isothermal (isothermal means “constant temperature”). Again, consider H2O as an example. Solid water (ice) can exist at 0°C. If heat is added to ice at 0°C, some of the solid changes phase to make liquid, which is also at 0°C. Remember, the solid and liquid phases of H2O can coexist at 0°C. Only after all of the solid has melted into liquid does the addition of heat change the temperature of the substance.
For each phase change of a substance, there is a characteristic quantity of heat needed to perform the phase change per gram (or per mole) of material. The heat of fusion (ΔHfus) is the amount of heat per gram (or per mole) required for a phase change that occurs at the melting point. The heat of vaporization (ΔHvap) is the amount of heat per gram (or per mole) required for a phase change that occurs at the boiling point. If you know the total number of grams or moles of material, you can use the ΔHfus or the ΔHvap to determine the total heat being transferred for melting or solidification using these expressions:
$\text{heat} = n \times ΔH_{fus} \label{Eq1a}$
where $n$ is the number of moles and $ΔH_{fus}$ is expressed in energy/mole or
$\text{heat} = m \times ΔH_{fus} \label{Eq1b}$
where $m$ is the mass in grams and $ΔH_{fus}$ is expressed in energy/gram.
For the boiling or condensation, use these expressions:
$\text{heat} = n \times ΔH_{vap} \label{Eq2a}$
where $n$ is the number of moles) and $ΔH_{vap}$ is expressed in energy/mole or
$\text{heat} = m \times ΔH_{vap} \label{Eq2b}$
where $m$ is the mass in grams and $ΔH_{vap}$ is expressed in energy/gram.
Remember that a phase change depends on the direction of the heat transfer. If heat transfers in, solids become liquids, and liquids become solids at the melting and boiling points, respectively. If heat transfers out, liquids solidify, and gases condense into liquids. At these points, there are no changes in temperature as reflected in the above equations.
Example $2$
How much heat is necessary to melt 55.8 g of ice (solid H2O) at 0°C? The heat of fusion of H2O is 79.9 cal/g.
Solution
We can use the relationship between heat and the heat of fusion (Equation $1$) to determine how many cal of heat are needed to melt this ice:
\begin{align*} \ce{heat} &= \ce{m \times ΔH_{fus}} \[4pt] \mathrm{heat} &= \mathrm{(55.8\: \cancel{g})\left(\dfrac{79.9\: cal}{\cancel{g}}\right)=4,460\: cal} \end{align*} \nonumber
Exercise $2$
How much heat is necessary to vaporize 685 g of H2O at 100°C? The heat of vaporization of H2O is 540 cal/g.
Answer
\begin{align*} \ce{heat} &= \ce{m \times ΔH_{vap}} \[4pt] \mathrm{heat} &= \mathrm{(685\: \cancel{g})\left(\dfrac{540\: cal}{\cancel{g}}\right)=370,000\: cal} \end{align*} \nonumber
Table $1$ lists the heats of fusion and vaporization for some common substances. Note the units on these quantities; when you use these values in problem solving, make sure that the other variables in your calculation are expressed in units consistent with the units in the specific heats or the heats of fusion and vaporization.
Table $1$: Heats of Fusion and Vaporization for Selected Substances
Substance ΔHfus (cal/g) ΔHvap (cal/g)
aluminum (Al) 94.0 2,602
gold (Au) 15.3 409
iron (Fe) 63.2 1,504
water (H2O) 79.9 540
sodium chloride (NaCl) 123.5 691
ethanol (C2H5OH) 45.2 200.3
benzene (C6H6) 30.4 94.1
Sublimation
There is also a phase change where a solid goes directly to a gas:
$\text{solid} \rightarrow \text{gas} \label{Eq3}$
This phase change is called sublimation. Each substance has a characteristic heat of sublimation associated with this process. For example, the heat of sublimation (ΔHsub) of H2O is 620 cal/g.
We encounter sublimation in several ways. You may already be familiar with dry ice, which is simply solid carbon dioxide (CO2). At −78.5°C (−109°F), solid carbon dioxide sublimes, changing directly from the solid phase to the gas phase:
$\mathrm{CO_2(s) \xrightarrow{-78.5^\circ C} CO_2(g)} \label{Eq4}$
Solid carbon dioxide is called dry ice because it does not pass through the liquid phase. Instead, it does directly to the gas phase. (Carbon dioxide can exist as liquid but only under high pressure.) Dry ice has many practical uses, including the long-term preservation of medical samples.
Even at temperatures below 0°C, solid H2O will slowly sublime. For example, a thin layer of snow or frost on the ground may slowly disappear as the solid H2O sublimes, even though the outside temperature may be below the freezing point of water. Similarly, ice cubes in a freezer may get smaller over time. Although frozen, the solid water slowly sublimes, redepositing on the colder cooling elements of the freezer, which necessitates periodic defrosting (frost-free freezers minimize this redeposition). Lowering the temperature in a freezer will reduce the need to defrost as often.
Under similar circumstances, water will also sublime from frozen foods (e.g., meats or vegetables), giving them an unattractive, mottled appearance called freezer burn. It is not really a “burn,” and the food has not necessarily gone bad, although it looks unappetizing. Freezer burn can be minimized by lowering a freezer’s temperature and by wrapping foods tightly so water does not have any space to sublime into.
Key Takeaway
• There is an energy change associated with any phase change. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/07%3A_Energy_and_Chemical_Processes/7.03%3A_Phase_Changes.txt |
Learning Objectives
The Learning Objectives of this Module are to:
• Define bond energy.
• Calculate enthalpy change or heat of reaction.
• Determine if a chemical process is exothermic or endothermic.
What happens when you take a basketball, place it halfway up a playground slide, and then let it go? The basketball rolls down the slide. What happens if you do it again? Does the basketball roll down the slide? It should.
If you were to perform this experiment over and over again, do you think the basketball would ever roll up the slide? Probably not. Why not? Well, for starters, in all of our experience, the basketball has always moved to a lower position when given the opportunity. The gravitational attraction of Earth exerts a force on the basketball, and given the chance, the basketball will move down. We say that the basketball is going to a lower gravitational potential energy. The basketball can move up the slide, but only if someone exerts some effort (that is, work) on the basketball. A general statement, based on countless observations over centuries of study, is that all objects tend to move spontaneously to a position of minimum energy unless acted on by some other force or object.
Bond Energy
A similar statement can be made about atoms in compounds. Atoms bond together to form compounds because in doing so they attain lower energies than they possess as individual atoms. A quantity of energy, equal to the difference between the energies of the bonded atoms and the energies of the separated atoms, is released, usually as heat. That is, the bonded atoms have a lower energy than the individual atoms do. When atoms combine to make a compound, energy is always given off, and the compound has a lower overall energy. In making compounds, atoms act like a basketball on a playground slide; they move in the direction of decreasing energy.
We can reverse the process, just as with the basketball. If we put energy into a molecule, we can cause its bonds to break, separating a molecule into individual atoms. Bonds between certain specific elements usually have a characteristic energy, called the bond energy, that is needed to break the bond. The same amount of energy was liberated when the atoms made the chemical bond in the first place. The term bond energy is usually used to describe the strength of interactions between atoms that make covalent bonds. A C–C bond has an approximate bond energy of 80 kcal/mol, while a C=C has a bond energy of about 145 kcal/mol. The C=C bond is stronger than C-C (as discussed in relation to bond length in Section 4.4). For atoms in ionic compounds attracted by opposite charges, the term lattice energy is used. For now, we will deal with covalent bonds in molecules.
Although each molecule has its own characteristic bond energy, some generalizations are possible. For example, although the exact value of a C–H bond energy depends on the particular molecule, all C–H bonds have a bond energy of roughly the same value because they are all C–H bonds. It takes roughly 100 kcal of energy to break 1 mol of C–H bonds, so we speak of the bond energy of a C–H bond as being about 100 kcal/mol. Table $1$ lists the approximate bond energies of various covalent bonds.
Table $1$: Approximate Bond Energies
Bond Bond Energy (kcal/mol)
C–H 100
C–O 86
C=O 190
C–N 70
C–C 85
C=C 145
C≡C 200
N–H 93
H–H 105
Br-Br 46
Cl–Cl 58
O–H 110
O=O 119
H–Br 87
H–Cl 103
When a chemical reaction occurs, the atoms in the reactants rearrange their chemical bonds to make products. The new arrangement of bonds does not have the same total energy as the bonds in the reactants. Therefore, when chemical reactions occur, there will always be an accompanying energy change. The enthalpy change, for a given reaction can be calculated using the bond energy values from Table $1$.
Enthalpy Change or Heat of Reaction, ΔH
Energy changes in chemical reactions are usually measured as changes in enthalpy. In this process, one adds energy to the reaction to break bonds, and extracts energy for the bonds that are formed.
$\text{enthalpy change} ≈ \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \nonumber$
The ≈ sign is used because we are adding together average bond energies (i.e., over many different molecules). Hence, this approach does not give exact values for the enthalpy change, ΔH.
Let’s consider the reaction of 2 mols of hydrogen gas (H2) with 1 mol of oxygen gas (O2) to give 2 mol water:
$\ce{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \nonumber$
or shown graphically:
with these average bond energies involved:
• $\ce{H–H}$ = 105 kcal/mol
• $\ce{O=O}$ = 119 kcal/mol
• $\ce{O–H}$ = 110 kcal/mol
In this reaction, 2 H–H bonds and 1 O=O bonds are broken, while 4 O–H bonds (two for each H2O) are formed. The energy changes can be tabulated and calculated as follows:
2 H–H bonds and 1 O=O bonds are broken, while 4 O–H bonds (two for each H2O) are formed. The energy changes can be tabulated and calculated as follows:
Bonds Broken (kcal/mol) Bonds Formed (kcal/mol)
2 H–H 2 mol x 105 kcal/mol = 210 kcal 4 O–H 4 mols x 110 kcal/mol = 440 kcal
1 O=O 1 mol x 119 kcal/mol = 119 kcal
Total = 329 kcal Total = 440 kcal
\begin{align*} \Delta H &≈ \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \[4pt] &≈ 329 \: \text{kcal} - 440 \: \text{kcal} \[4pt] &≈ −111 \: \text{kcal} \end{align*} \nonumber
The enthalpy change (ΔH) of the reaction is approximately −111 kcal/mol. This means that bonds in the products (440 kcal) are stronger than the bonds in the reactants (329 kcal) by about 111 kcal/mol. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes. This excess energy is released as heat, so the reaction is exothermic. Hence, we can re-write the reaction with the heat released (111 kcal) on the product side of the equation, as follows:
We can also re-write the reaction equation with the ΔH information (see below). Note that an exothermic reaction has a negative ΔH value.
$\ce{2H_2(g)+O_2(g) \rightarrow 2H_2O(g)} \ \: \: \: \: \: \Delta H = -111 \: \text{kcal} \nonumber$
Example $1$
What is the enthalpy change for this reaction? Is the reaction exothermic or endothermic?
$\ce{H_2(g) + Br_2(g) \rightarrow 2HBr(g)} \nonumber$
Solution
Step 1- First look at the equation and identify which bonds exist on in the reactants (bonds broken).
• one H-H bond and
• one Br-Br bond
Step 2- Do the same for the products (bonds formed)
• two H-Br bonds
Step 3- Identify the bond energies of these bonds from Table $1$:
• H-H bonds: 105 kcal/mol
• Br-Br bonds: 46 kcal/mol
Step 4- Set up the table (see below) and apply the formula for enthalpy change.
Solutions to Example 7.4.1
Bonds Broken (kcal/mol) Bonds Formed (kcal/mol)
1 H–H 1 mol x 105 kcal/mol = 105 kcal 2 H–Br 2 mols x 87 kcal/mol = 174 kcal
1 Br–Br 1 mol x 46 kcal/mol = 46 kcal
Total = 151 kcal Total = 174 kcal
\begin{align*} \Delta H &≈ 151 \: \text{kcal} - 174 \: \text{kcal} \[4pt] &≈ −23 \: \text{kcal} \end{align*} \nonumber
Step 5- Since $ΔH$ is negative (−23 kcal), the reaction is exothermic.
Exercise $1$
Using the bond energies given in the chart above, find the enthalpy change for the thermal decomposition of water:
$\ce{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)} \nonumber$
Is the reaction written above exothermic or endothermic? Explain.
Answer
ΔH = −43 kcal
Since ΔH is negative (−43 kcal), the reaction is exothermic.
Endothermic and Exothermic Reactions
Endothermic and exothermic reactions can be thought of as having energy as either a reactant of the reaction or a product. Endothermic reactions require energy, so energy is a reactant. Heat flows from the surroundings to the system (reaction mixture) and the enthalpy of the system increases (ΔH is positive). As discussed in the previous section, in an exothermic reaction, heat is released (considered a product) and the enthalpy of the system decreases (ΔH is negative).
In the course of an endothermic process, the system gains heat from the surroundings and so the temperature of the surroundings decreases (gets cold). A chemical reaction is exothermic if heat is released by the system into the surroundings. Because the surroundings is gaining heat from the system, the temperature of the surroundings increases. See Figure $1$.
Endothermic Reaction: When $1 \: \text{mol}$ of calcium carbonate decomposes into $1 \: \text{mol}$ of calcium oxide and $1 \: \text{mol}$ of carbon dioxide, $177.8 \: \text{kJ}$ of heat is absorbed. Because the heat is absorbed by the system, the $177.8 \: \text{kJ}$ is written as a reactant. The ΔH is positive for an endothermic reaction.
$\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = +177.8 \: \text{kJ} \nonumber$
Exothermic Reaction: When methane gas is combusted, heat is released, making the reaction exothermic. Specifically, the combustion of $1 \: \text{mol}$ of methane releases 890.4 kilojoules of heat energy. This information can be shown as part of the balanced equation in two ways. First, the amount of heat released can be written in the product side of the reaction. Another way is to write the heat of reactionH) information with a negative sign, $-890.4 \: \text{kJ}$.
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -890.4 \: \text{kJ} \nonumber$
Example $2$
Is each chemical reaction exothermic or endothermic?
1. CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ) + 213 kcal
2. N2(g) + O2(g) + 45 kcal → 2NO(g)
Solution
1. Because energy (213 kcal) is a product, energy is given off by the reaction. Therefore, this reaction is exothermic.
2. Because energy (45 kcal) is a reactant, energy is absorbed by the reaction. Therefore, this reaction is endothermic.
Exercise $2$
Is each chemical reaction exothermic or endothermic?
1. H2(g) + F2(g) → 2HF (g) + 130 kcal
2. 2C(s) + H2(g) + 5.3 kcal → C2H2(g)
Answer
a. The energy (130 kcal) is produced, hence the reaction is exothermic
b. The energy (5.3 kcal) is supplied or absorbed to react, hence, the reaction is endothermic
Energy Diagrams
Endothermic and exothermic reactions can be visually represented by energy-level diagrams like the ones in Figure $2$. In endothermic reactions, the reactants have higher bond energy (stronger bonds) than the products. Strong bonds have lower potential energy than weak bonds. Hence, the energy of the reactants is lower than that of the products. This type of reaction is represented by an "uphill" energy-level diagram shown in Figure $\PageIndex{2A}$. For an endothermic chemical reaction to proceed, the reactants must absorb energy from their environment to be converted to products.
In an exothermic reaction, the bonds in the product have higher bond energy (stronger bonds) than the reactants. In other words, the energy of the products is lower than the energy of the reactants, hence is energetically downhill, shown in Figure $\PageIndex{2B}$. Energy is given off as reactants are converted to products. The energy given off is usually in the form of heat (although a few reactions give off energy as light). In the course of an exothermic reaction, heat flows from the system to its surroundings, and thus, gets warm.
Table $2$: Endothermic and Exothermic Reactions
Endothermic Reactions Exothermic Reactions
Heat is absorbed by reactants to form products. Heat is released.
Heat is absorbed from the surroundings; as a result, the surroundings get cold. Heat is released by the reaction to surroundings; surroundings feel hot.
ΔHrxn is positive ΔHrxn is negative
The bonds broken in the reactants are stronger than the bonds formed in the products The bonds formed in the products are stronger than the bonds broken in the reactants
The reactants are lower in energy than the products The products are lower in energy than the reactants
Represented by an "uphill" energy diagram Represented by an "downhill" energy diagram
Key Takeaways
• Atoms are held together by a certain amount of energy called bond energy.
• Energy is required to break bonds. Energy is released when chemical bonds are formed because atoms become more stable.
• Chemical processes are labeled as exothermic or endothermic based on whether they give off or absorb energy, respectively.
7.05: The Energy of Biochemical Reactions
Learning Objectives
• To relate the concept of energy change to chemical reactions that occur in the body.
The chemistry of the human body, or any living organism, is very complex. Even so, the chemical reactions found in the human body follow the same principles of energy that other chemical reactions follow.
Where does the energy that powers our bodies come from? The details are complex, but we can look at some simple processes at the heart of cellular activity.
An important reaction that provides energy for our bodies is the oxidation of glucose ($\ce{C6H12O6}$):
$\ce{ C6H12O6(s) + 6O2(g) \rightarrow 6CO2(g) + 6H2O(ℓ)} + \text{670 kcal} \label{Eq1}$
Considering that 1 mol of $\ce{C6H12O6(s)}$ has a volume of about 115 mL, we can see that glucose is a compact source of energy.
Glucose and other sugars are examples of carbohydrates, which are one of the three main dietary components of a human diet. All carbohydrates supply approximately 4 kcal/g. You can verify that by taking the heat of reaction for glucose oxidation and dividing it by its molar mass. Proteins, the building blocks of structural tissues like muscle and skin, also supply about 4 kcal/g. Other important energy sources for the body are fats, which are largely hydrocarbon chains. Fats provide even more energy per gram, about 9 kcal/g.
Another important reaction is the conversion of adenosine triphosphate (ATP) to adenosine diphosphate (ADP), which is shown in Figure $1$. Under physiological conditions, the breaking of an O–P bond and the formation of an O–P and two O–H bonds gives off about 7.5 kcal/mol of ATP. This may not seem like much energy, especially compared to the amount of energy given off when glucose reacts. It is enough energy, however, to fuel other biochemically important chemical reactions in our cells.
Even complex biological reactions must obey the basic rules of chemistry.
Career Focus: Dietitian
A dietitian is a nutrition expert who communicates food-related information to the general public. In doing so, dietitians promote the general well-being among the population and help individuals recover from nutritionally related illnesses.
Our diet does not just supply us with energy. We also get vitamins, minerals, and even water from what we eat. Eating too much, too little, or not enough of the right foods can lead to a variety of problems. Dietitians are trained to make specific dietary recommendations to address particular issues relating to health. For example, a dietitian might work with a person to develop an overall diet that would help that person lose weight or control diabetes. Hospitals employ dietitians in planning menus for patients, and many dietitians work with community organizations to improve the eating habits of large groups of people.
Key Takeaway
• Energy to power the human body comes from chemical reactions. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/07%3A_Energy_and_Chemical_Processes/7.04%3A_Bond_Energies_and_Chemical_Reactions.txt |
Concept Review Exercises
1. What is the relationship between energy and heat?
2. What units are used to express energy and heat?
Answers
1. Heat is the exchange of energy from one part of the universe to another. Heat and energy have the same units.
2. Joules and calories are the units of energy and heat.
Exercises
1. Define energy.
2. What is heat?
3. What is the relationship between a calorie and a joule? Which unit is larger?
4. What is the relationship between a calorie and a kilocalorie? Which unit is larger?
5. Express 1,265 cal in kilocalories and in joules.
6. Express 9,043.3 J in calories and in kilocalories.
7. One kilocalorie equals how many kilojoules?
8. One kilojoule equals how many kilocalories?
9. Many nutrition experts say that an average person needs 2,000 Cal per day from his or her diet. How many joules is this?
10. Baby formula typically has 20.0 Cal per ounce. How many ounces of formula should a baby drink per day if the RDI is 850 Cal?
Answers
1. Energy is the ability to do work.
2. Heat is a form of enery (thermal) that can be transferred from one object to another.
1. 1 cal = 4.184 J; the calorie is larger.
4. 1 kilocalorie(1 Cal) = 1000 cal; the kcal is larger.
1. 1.265 kcal; 5,293 J
6. 2161.4 cal; 2.1614 kcal
1. 1 kcal = 4.184 kJ
8. 1 kJ = 0.239 kcal
9. 8.4 × 106 J
10. 42.5 oz
Concept Review Exercise
1. Describe the relationship between heat transfer and the temperature change of an object.
2. Describe what happens when two objects that have different temperatures come into contact with one another.
Answer
1. Heat is equal to the product of the mass, the change in temperature, and a proportionality constant called the specific heat.
2. The temperature of the hot object decreases and the temperature of the cold object increases as heat is transferred from the hot object to the cold object. The change in temperature of each depends on the identity and properties of each substance.
Exercises
1. The melting point of mercury is −38.84oC. Convert this value to degrees Fahrenheit and Kelvin.
2. A pot of water is set on a hot burner of a stove. What is the direction of heat flow?
3. Some uncooked macaroni is added to a pot of boiling water. What is the direction of heat flow?
4. How much energy in calories is required to heat 150 g of H2O from 0°C to 100°C?
5. How much energy in calories is required to heat 125 g of Fe from 25°C to 150°C?
6. If 250 cal of heat were added to 43.8 g of Al at 22.5°C, what is the final temperature of the aluminum?
7. If 195 cal of heat were added to 33.2 g of Hg at 56.2°C, what is the final temperature of the mercury?
8. A sample of copper absorbs 145 cal of energy, and its temperature rises from 37.8°C to 41.7°C. What is the mass of the copper?
9. A large, single crystal of sodium chloride absorbs 98.0 cal of heat. If its temperature rises from 22.0°C to 29.7°C, what is the mass of the NaCl crystal?
10. If 1.00 g of each substance in Table 7.3 were to absorb 100 cal of heat, which substance would experience the largest temperature change?
11. If 1.00 g of each substance in Table 7.3 were to absorb 100 cal of heat, which substance would experience the smallest temperature change?
12. Determine the heat capacity of a substance if 23.6 g of the substance gives off 199 cal of heat when its temperature changes from 37.9°C to 20.9°C.
13. What is the heat capacity of gold if a 250 g sample needs 133 cal of energy to increase its temperature from 23.0°C to 40.1°C?
Answers
1. -37.910F and 234.31 K
2. Heat flows into the pot of water.
3. Heat flows to the macaroni.
4. 15,000 cal
5. 1,690 cal
6. 49.0°C
7. 234°C
8. 404 g
9. 61 g
10. Mercury would experience the largest temperature change.
11. hydrogen (H2)
12. 0.496 cal/g•°C
13. 0.031 cal/g•°C
Concept Review Exercises
1. Explain what happens when heat flows into or out of a substance at its melting point or boiling point.
2. How does the amount of heat required for a phase change relate to the mass of the substance?
3. What is the direction of heat transfer in boiling water?
4. What is the direction of heat transfer in freezing water?
5. What is the direction of heat transfer in sweating?
Answers
1. The energy goes into changing the phase, not the temperature.
2. The amount of heat is a constant per gram of substance.
3. Boiling. Heat is being added to the water to get it from the liquid state to the gas state.
4. Freezing. Heat is exiting the system in order to go from liquid to solid. Another way to look at it is to consider the opposite process of melting. Energy is consumed (endothermic) to melt ice (solid to liquid) so the opposite process (liquid to solid) must be exothermic.
5. Sweating. Heat is consumed to evaporate the moisture on your skin which lowers your temperature.
Exercises
1. How much energy is needed to melt 43.8 g of Au at its melting point of 1,064°C?
2. How much energy is given off when 563.8 g of NaCl solidifies at its freezing point of 801°C?
3. What mass of ice can be melted by 558 cal of energy?
4. How much ethanol (C2H5OH) in grams can freeze at its freezing point if 1,225 cal of heat are removed?
5. What is the heat of vaporization of a substance if 10,776 cal are required to vaporize 5.05 g? Express your final answer in joules per gram.
6. If 1,650 cal of heat are required to vaporize a sample that has a heat of vaporization of 137 cal/g, what is the mass of the sample?
7. What is the heat of fusion of water in calories per mole?
8. What is the heat of vaporization of benzene (C6H6) in calories per mole?
9. What is the heat of vaporization of gold in calories per mole?
10. What is the heat of fusion of iron in calories per mole?
Answers
1. 670 cal
2. 69,630 cal
3. 6.98 g
4. 27.10 g
1. 8,930 J/g
6. 12.0 g
1. 1,440 cal/mol
8. 7,350 cal/mol
9. 80,600 cal/mol
10. 3,530 cal/mol
Concept Review Exercises
1. What is the connection between energy and chemical bonds?
2. Why does energy change during the course of a chemical reaction?
3. Two different reactions are performed in two identical test tubes. In reaction A, the test tube becomes very warm as the reaction occurs. In reaction B, the test tube becomes cold. Which reaction is endothermic and which is exothermic? Explain.
4. Classify "burning paper" as endothermic or exothermic processes.
Answers
1. Chemical bonds have a certain energy that is dependent on the elements in the bond and the number of bonds between the atoms.
2. Energy changes because bonds rearrange to make new bonds with different energies.
3. Reaction A is exothermic because heat is leaving the system making the test tube feel hot. Reaction B is endothermic because heat is being absorbed by the system making the test tube feel cold.
4. "Burning paper" is exothermic because burning (also known as combustion) releases heat
Exercises
1. Using the data in Table 7.4.1, calculate the energy of one C–H bond (as opposed to 1 mol of C–H bonds). Recall that 1 mol = 6.022 x 1023 C–H bonds
2. Using the data in Table 7.4.1, calculate the energy of one C=C bond (as opposed to 1 mol of C=C bonds). Recall that 1 mol = 6.022 x 1023 C=C bonds
3. Is a bond-breaking process exothermic or endothermic?
4. Is a bond-making process exothermic or endothermic?
5. Is each chemical reaction exothermic or endothermic?
1. 2SnCl2(s) + 33 kcal → Sn(s) + SnCl4(s)
2. CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ) + 213 kcal
6. Is each chemical reaction exothermic or endothermic?
1. C2H4(g) + H2(g) → C2H6(g) + 137 kJ
2. C(s, graphite) + 1.9 kJ → C(s, diamond)
Answers
1. 1.661 × 10−19 cal
2. 2.408 x 10-19 cal
3. endothermic
4. exothermic
5.
a. endothermic
b. exothermic
6.
a. exothermic
b. endothermic
Concept Review Exercise
1. What is the energy content per gram of proteins, carbohydrates, and fats?
Answer
1. proteins and carbohydrates: 4 kcal/g; fats: 9 kcal/g
Exercises
1. An 8 oz serving of whole milk has 8.0 g of fat, 8.0 g of protein, and 13 g of carbohydrates. Approximately how many kilocalories does it contain?
2. A serving of potato chips has 160 kcal. If the chips have 15 g of carbohydrates and 2.0 g of protein, about how many grams of fat are in a serving of potato chips?
3. The average body temperature of a person is 37°C, while the average surrounding temperature is 22°C. Is overall human metabolism exothermic or endothermic?
4. Cold-blooded animals absorb heat from the environment for part of the energy they need to survive. Is this an exothermic or an endothermic process?
5. If the reaction ATP → ADP gives off 7.5 kcal/mol, then the reverse process, ADP → ATP requires 7.5 kcal/mol to proceed. How many moles of ADP can be converted to ATP using the energy from 1 serving of potato chips (see Exercise 2)?
6. If the oxidation of glucose yields 670 kcal of energy per mole of glucose oxidized, how many servings of potato chips (see Exercise 2) are needed to provide the same amount of energy?
1. 156 kcal
2. 10.2 g
3. exothermic
4. endothermic
5. 21.3 mol
6. 4.2 servings
Additional Exercises
1. Sulfur dioxide (SO2) is a pollutant gas that is one cause of acid rain. It is oxidized in the atmosphere to sulfur trioxide (SO3), which then combines with water to make sulfuric acid (H2SO4).
1. Write the balanced reaction for the oxidation of SO2 to make SO3. (The other reactant is diatomic oxygen.)
2. When 1 mol of SO2 reacts to make SO3, 23.6 kcal of energy are given off. If 100 lb (1 lb = 454 g) of SO2 were converted to SO3, what would be the total energy change?
2. Ammonia (NH3) is made by the direct combination of H2 and N2 gases according to this reaction:
N2(g) + 3H2(g) → 2NH3(g) + 22.0 kcal
1. Is this reaction endothermic or exothermic?
2. What is the overall energy change if 1,500 g of N2 are reacted to make ammonia?
3. A 5.69 g sample of iron metal was heated in boiling water to 99.8°C. Then it was dropped into a beaker containing 100.0 g of H2O at 22.6°C. Assuming that the water gained all the heat lost by the iron, what is the final temperature of the H2O and Fe?
4. A 5.69 g sample of copper metal was heated in boiling water to 99.8°C. Then it was dropped into a beaker containing 100.0 g of H2O at 22.6°C. Assuming that the water gained all the heat lost by the copper, what is the final temperature of the H2O and Cu?
5. When 1 g of steam condenses, 540 cal of energy is released. How many grams of ice can be melted with 540 cal?
6. When 1 g of water freezes, 79.9 cal of energy is released. How many grams of water can be boiled with 79.9 cal?
7. The change in energy is +65.3 kJ for each mole of calcium hydroxide [Ca(OH)2] according to the following reaction:
Ca(OH)2(s) → CaO(s) + H2O(g)
How many grams of Ca(OH)2 could be reacted if 575 kJ of energy were available?
8. The thermite reaction gives off so much energy that the elemental iron formed as a product is typically produced in the liquid state:
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(ℓ) + 204 kcal
How much heat will be given off if 250 g of Fe are to be produced?
9. A normal adult male requires 2,500 kcal per day to maintain his metabolism.
1. Nutritionists recommend that no more than 30% of the calories in a person’s diet come from fat. At 9 kcal/g, what is the maximum mass of fat an adult male should consume daily?
2. At 4 kcal/g each, how many grams of protein and carbohydrates should an adult male consume daily?
10. A normal adult male requires 2,500 kcal per day to maintain his metabolism.
1. At 9 kcal/g, what mass of fat would provide that many kilocalories if the diet was composed of nothing but fats?
2. At 4 kcal/g each, what mass of protein and/or carbohydrates is needed to provide that many kilocalories?
11. The volume of the world’s oceans is approximately 1.34 × 1024 cm3.
1. How much energy would be needed to increase the temperature of the world’s oceans by 1°C? Assume that the heat capacity of the oceans is the same as pure water.
2. If Earth receives 6.0 × 1022 J of energy per day from the sun, how many days would it take to warm the oceans by 1°C, assuming all the energy went into warming the water?
12. Does a substance that has a small specific heat require a small or large amount of energy to change temperature? Explain.
13. Some biology textbooks represent the conversion of adenosine triphosphate (ATP) to adenosine diphosphate (ADP) and phosphate ions as follows:
ATP → ADP + phosphate + energy
What is wrong with this reaction?
14. Assuming that energy changes are additive, how much energy is required to change 15.0 g of ice at −15°C to 15.0 g of steam at 115°C? (Hint: you will have five processes to consider.)
Answers
1.
1. 2SO2 + O2 → 2SO3
2. 16,700 kcal
2.
1. exothermic
2. 1177 kcal
1. about 23.1°C
4. about 23.0°C
5. 6.76 g
6. 0.148 g
1. 652 g
8. 457 kcal
1.
1. 83.3 g
2. 438 g
10.
a. 278 g
b. 625 g
11.
1. 1.34 × 1024 cal
2. 93 days
12. A substance with smaller specific heat requires less energy per unit of mass to raise its temperature,
13. A reactant is missing: H2O is missing.
14. Total energy = 11,019 cal
7.S: Energy and Chemical Processes (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Energy is the ability to do work. The transfer of energy from one place to another is heat. Heat and energy are measured in units of joules, calories, or kilocalories (equal to 1,000 calories). The amount of heat gained or lost when the temperature of an object changes can be related to its mass and a constant called the specific heat of the substance.
The transfer of energy can also cause a substance to change from one phase to another. During the transition, called a phase change, heat is either added or lost. Despite the fact that heat is going into or coming out of a substance during a phase change, the temperature of the substance does not change until the phase change is complete; that is, phase changes are isothermal. Analogous to specific heat, a constant called the heat of fusion of a substance describes how much heat must be transferred for a substance to melt or solidify (that is, to change between solid and liquid phases), while the heat of vaporization describes the amount of heat transferred in a boiling or condensation process (that is, to change between liquid and gas phases).
Every chemical change is accompanied by an energy change. This is because the interaction between atoms bonding to each other has a certain bond energy, the energy required to break the bond (called lattice energy for ionic compounds), and the bond energies of the reactants will not be the same as the bond energies of the products. Reactions that give off energy are called exothermic, while reactions that absorb energy are called endothermic. Energy-level diagrams can be used to illustrate the energy changes that accompany chemical reactions.
Even complex biochemical reactions have to follow the rules of simple chemistry, including rules involving energy change. Reactions of carbohydrates and proteins provide our bodies with about 4 kcal of energy per gram, while fats provide about 9 kcal per gram. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/07%3A_Energy_and_Chemical_Processes/7.E%3A_Energy_and_Chemical_Processes_%28Exercises%29.txt |
Most of us are familiar with the three phases of matter: solid, liquid, and gas. Indeed, we addressed the energy changes involved in phase changes. The substance we are probably most familiar with has those three phases: water. In everyday life, we commonly come in contact with water as a solid (ice), as a liquid, and as a gas (steam). All we have to do is change the conditions of the substance—typically temperature—and we can change the phase from solid to liquid to gas and back again. Under the proper conditions of temperature and pressure, many substances—not only water—can experience the three different phases. An understanding of the phases of matter is important for our understanding of all matter. In this chapter, we will explore the three phases of matter.
• 8.0: Prelude to Solids, Liquids, and Gases
Solid carbon dioxide is called dry ice because it converts from a solid to a gas directly, without going through the liquid phase, in a process called sublimation. Thus, there is no messy liquid phase to worry about. Although it is a novelty, dry ice has some potential dangers. Because it is so cold, it can freeze living tissues very quickly, so people handling dry ice should wear special protective gloves.
• 8.1: Intermolecular Interactions
A phase is a form of matter that has the same physical properties throughout. Molecules interact with each other through various forces: ionic and covalent bonds, dipole-dipole interactions, hydrogen bonding, and dispersion forces.
• 8.2: Solids and Liquids
Solids and liquids are phases that have their own unique properties.
• 8.3: Gases and Pressure
The gas phase is unique among the three states of matter in that there are some simple models we can use to predict the physical behavior of all gases—independent of their identities. We cannot do this for the solid and liquid states. Initial advances in the understanding of gas behavior were made in the mid 1600s by Robert Boyle, an English scientist who founded the Royal Society (one of the world’s oldest scientific organizations).
• 8.4: Gas Laws
The physical properties of gases are predictable using mathematical formulas known as gas laws.
• 8.E: Solids, Liquids, and Gases (Exercises)
Problems and select solutions to this chapter.
• 8.S: Solids, Liquids, and Gases (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Thumbnail: A water drop. (CC BY 2.0; José Manuel Suárez via Wikipedia).
08: Solids Liquids and Gases
We normally experience carbon dioxide (\(\ce{CO2}\)) as a gas, but if it were cooled down to about −78°C, it would become a solid. The everyday term for solid carbon dioxide is dry ice.
Why “dry” ice? Solid carbon dioxide is called dry ice because it converts from a solid to a gas directly, without going through the liquid phase, in a process called sublimation. Thus, there is no messy liquid phase to worry about. Although it is a novelty, dry ice has some potential dangers. Because it is so cold, it can freeze living tissues very quickly, so people handling dry ice should wear special protective gloves. The cold carbon dioxide gas is also heavier than air (because it is cold and more dense), so people in the presence of dry ice should be in a well-ventilated area.
Dry ice has several common uses. Because it is so cold, it is used as a refrigerant to keep other things cold or frozen (e.g., meats or ice cream). In the medical field, dry ice is used to preserve medical specimens, blood products, and drugs. It also has dermatological applications (e.g., freezing off warts). Organs for transplant are kept cool with dry ice until the recipient of the new organ is ready for surgery. In this respect, carbon dioxide is much like water—more than one phase of the same substance has significant uses in the real world. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.00%3A_Prelude_to_Solids_Liquids_and_Gases.txt |
Learning Objectives
• Define phase.
• Identify the types of interactions between molecules.
A phase is a certain form of matter that includes a specific set of physical properties. That is, the atoms, the molecules, or the ions that make up the phase do so in a consistent manner throughout the phase. Science recognizes three stable phases: the solid phase, in which individual particles can be thought of as in contact and held in place; the liquid phase, in which individual particles are in contact but moving with respect to each other; and the gas phase, in which individual particles are separated from each other by relatively large distances (see Figure 8.1.1). Not all substances will readily exhibit all phases. For example, carbon dioxide does not exhibit a liquid phase unless the pressure is greater than about six times normal atmospheric pressure. Other substances, especially complex organic molecules, may decompose at higher temperatures, rather than becoming a liquid or a gas.
For many substances, there are different arrangements the particles can take in the solid phase, depending on temperature and pressure.
Which phase a substance adopts depends on the pressure and the temperature it experiences. Of these two conditions, temperature variations are more obviously related to the phase of a substance. When it is very cold, H2O exists in the solid form as ice. When it is warmer, the liquid phase of H2O is present. At even higher temperatures, H2O boils and becomes steam.
Pressure changes can also affect the presence of a particular phase (as we indicated for carbon dioxide), but its effects are less obvious most of the time. We will mostly focus on the temperature effects on phases, mentioning pressure effects only when they are important. Most chemical substances follow the same pattern of phases when going from a low temperature to a high temperature: the solid phase, then the liquid phase, and then the gas phase. However, the temperatures at which these phases are present differ for all substances and can be rather extreme. Table $1$ shows the temperature ranges for solid, liquid, and gas phases for three substances. As you can see, there is extreme variability in the temperature ranges.
Table $1$: Temperature Ranges for the Three Phases of Various Substances
Substance Solid Phase Below Liquid Phase Above Gas Phase Above
hydrogen (H2) −259°C −259°C −253°C
water (H2O) 0°C 0°C 100°C
sodium chloride (NaCl) 801°C 801°C 1413°C
The melting point of a substance is the temperature that separates a solid and a liquid. The boiling point of a substance is the temperature that separates a liquid and a gas.
What accounts for this variability? Why do some substances become liquids at very low temperatures, while others require very high temperatures before they become liquids? It all depends on the strength of the intermolecular forces (IMF) between the particles of substances and the kinetic energies (KE) of its molecules. (Although ionic compounds are not composed of discrete molecules, we will still use the term intermolecular to include interactions between the ions in such compounds.) Substances that experience strong intermolecular interactions require higher temperatures to become liquids and, finally, gases. Substances that experience weak intermolecular interactions do not need much energy (as measured by temperature) to become liquids and gases and will exhibit these phases at lower temperatures.
Covalent Network Materials
Substances with the highest melting and boiling points have covalent network bonding. This type of intermolecular interaction is actually a covalent bond. In these substances, all the atoms in a sample are covalently bonded to one another; in effect, the entire sample is essentially one giant molecule. Many of these substances are solid over a large temperature range because it takes a lot of energy to disrupt all the covalent bonds at once. One example of a substance that shows covalent network bonding is diamond (Figure $2$). Diamond is composed entirely of carbon atoms, each bonded to four other carbon atoms in a tetrahedral geometry. Melting a covalent network solid is not accomplished by overcoming the relatively weak intermolecular forces. Rather, all of the covalent bonds must be broken, a process that requires extremely high temperatures. Diamond, in fact, does not melt at all. Instead, it vaporizes to a gas at temperatures above 3,500°C. Diamond is extremely hard and is one of the few materials that can cut glass.
Ionic Compounds
The strongest force between any two particles is the ionic bond, in which two ions of opposing charge are attracted to each other. Thus, ionic interactions between particles are another type of intermolecular interaction. Substances that contain ionic interactions are relatively strongly held together, so these substances typically have high melting and boiling points. Sodium chloride (Figure $3$) is an example of a substance whose particles experience ionic interactions (Table $1$). These attractive forces are sometimes referred to as ion-ion interactions.
Covalent Molecular Compounds
There are two different covalent structures: molecular and network. Covalent network compounds like SiO2 (quartz) have structures of atoms in a network like diamond described earlier. In this section, we are dealing with the molecular type that contains individual molecules. The bonding between atoms in the individual molecule is covalent but the attractive forces between the molecules are called intermolecular forces (IMF).
In contrast to intramolecular forces (see Figure 8.1.4), such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.
In this section, we will discuss the three types of IMF in molecular compounds: dipole-dipole, hydrogen bonding and London dispersion forces.
Dipole-dipole Intermolecular Forces
As discussed in Section 4.4, covalent bond that has an unequal sharing of electrons is called a polar covalent bond. (A covalent bond that has an equal sharing of electrons, as in a covalent bond with the same atom on each side, is called a nonpolar covalent bond.) A molecule with a net unequal distribution of electrons in its covalent bonds is a polar molecule. HF is an example of a polar molecule (see Figure 8.1.5).
The charge separation in a polar covalent bond is not as extreme as is found in ionic compounds, but there is a related result: oppositely charged ends of different molecules will attract each other. This type of intermolecular interaction is called a dipole-dipole interaction. Many molecules with polar covalent bonds experience dipole-dipole interactions.
The covalent bonds in some molecules are oriented in space in such a way that the bonds in the molecules cancel each other out. The individual bonds are polar, but due to molecular symmetry, the overall molecule is not polar; rather, the molecule is nonpolar. Such molecules experience little or no dipole-dipole interactions. Carbon dioxide (CO2) and carbon tetrachloride (CCl4) are examples of such molecules (Figure $6$).
Recall from the Sections 4.4 and 4.5, on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule—a separation of charge called a dipole. Consider a polar molecule such as hydrogen chloride, HCl. In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a dipole-dipole attraction—the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure $7$.
The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar F2 molecules. Both HCl and F2 consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K, molecules of both substances would have the same average kinetic energy. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to “stick together” to form a liquid, whereas the relatively weaker dispersion forces between nonpolar F2 molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl (188 K) compared to F2 (85 K) is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar F2 molecules. We will often use values such as boiling or freezing points as indicators of the relative strengths of IMFs of attraction present within different substances.
Example $1$
Predict which will have the higher boiling point: N2 or CO. Explain your reasoning.
Solution
CO and N2 are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N2 is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N2 molecules, so CO is expected to have the higher boiling point.
Exercise $1$
Predict which will have the higher boiling point: $\ce{ICl}$ or $\ce{Br2}$. Explain your reasoning.
Answer
ICl. ICl and Br2 have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br2 is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point.
Hydrogen Bonding Intermolecular Forces
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F tend to exhibit unusually strong intermolecular interactions due to a particularly strong type of dipole-dipole attraction called hydrogen bonding. The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded (4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of F, O, or N atoms, leads to highly concentrated partial charges with these atoms. Because the hydrogen atom does not have any electrons other than the ones in the covalent bond, its positively charged nucleus is almost completely exposed, allowing strong attractions to other nearby lone pairs of electrons.
Examples of hydrogen bonds include HF⋯HF, H2O⋯HOH, and H3N⋯HNH2, in which the hydrogen bonds are denoted by dots. Figure $8$ illustrates hydrogen bonding between water molecules. The physical properties of water, which has two O–H bonds, are strongly affected by the presence of hydrogen bonding between water molecules. Most molecular compounds that have a mass similar to water are gases at room temperature. However, because of the strong hydrogen bonds, water molecules are able to stay condensed in the liquid state.
A hydrogen bond is an intermolecular attractive force in which a hydrogen atom, that is covalently bonded to a small, highly electronegative atom, is attracted to a lone pair of electrons on an atom in a neighboring molecule. Figure $9$ shows how methanol (CH3OH) molecules experience hydrogen bonding. Methanol contains both a hydrogen atom attached to O; methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor (lone pair). The hydrogen-bonded structure of methanol is as follows:
Despite use of the word “bond,” keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, but are generally much stronger than other dipole-dipole attractions and dispersion forces.
Effect of Hydrogen Bonding on Boiling Points
Consider the compounds dimethylether (CH3OCH3), ethanol (CH3CH2OH), and propane (CH3CH2CH3). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning.
Solution
The shapes of CH3OCH3, CH3CH2OH, and CH3CH2CH3 are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since CH3CH2CH3 is nonpolar, it may exhibit only dispersion forces. Because CH3OCH3 is polar, it will also experience dipole-dipole attractions. Finally, CH3CH2OH has an −OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH3CH2CH3 < CH3OCH3 < CH3CH2OH. The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C.
Exercise $2$
Ethane (CH3CH3) has a melting point of −183 °C and a boiling point of −89 °C. Predict the melting and boiling points for methylamine (CH3NH2). Explain your reasoning.
Answer
The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. CH3CH3 and CH3NH2 are similar in size and mass, but methylamine possesses an −NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of −93 °C and a boiling point of −6 °C.
Hydrogen Bonding and DNA
Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism’s characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism’s offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure $10$.
Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and T share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure $11$
The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily “unzip” down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication.
London Dispersion Forces
Finally, there are forces between all molecules that are caused by electrons being in different places in a molecule at any one time, which sets up a temporary separation of charge that disappears almost as soon as it appears. These are very weak intermolecular interactions and are called dispersion forces (or London forces). (An alternate name is London dispersion forces.) Molecules that experience no other type of intermolecular interaction will at least experience dispersion forces. Substances that experience only dispersion forces are typically soft in the solid phase and have relatively low melting points. Examples include waxes, which are long hydrocarbon chains that are solids at room temperature because the molecules have so many electrons. The resulting dispersion forces between these molecules make them assume the solid phase at normal temperatures.
Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. F2 and Cl2 are gases at room temperature (reflecting weaker attractive forces); Br2 is a liquid, and I2 is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table 8.1.2.
Table $2$: Melting and Boiling Points of the Halogens
Halogen Molar Mass Atomic Radius Melting Point Boiling Point
fluorine, F2 38 g/mol 72 pm 53 K 85 K
chlorine, Cl2 71 g/mol 99 pm 172 K 238 K
bromine, Br2 160 g/mol 114 pm 266 K 332 K
iodine, I2 254 g/mol 133 pm 387 K 457 K
astatine, At2 420 g/mol 150 pm 575 K 610 K
The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.
Example $3$
Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH4, SiH4, GeH4, and SnH4. Explain your reasoning.
Solution
Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH4, SiH4, GeH4, and SnH4 are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, CH4 is expected to have the lowest boiling point and SnH4 the highest boiling point. The ordering from lowest to highest boiling point is expected to be
CH4 < SiH4 < GeH4 < SnH4
A graph of the actual boiling points of these compounds versus the period of the group 14 elementsshows this prediction to be correct:
Exercise $3$
Order the following hydrocarbons from lowest to highest boiling point: C2H6, C3H8, and C4H10.
Answer
All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore
C2H6 < C3H8 < C4H10.
Applications: Geckos and Intermolecular Forces
Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way.
Geckos’ toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko, shown in Figure 8.1.12, with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko’s weight.
In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Further investigations may eventually lead to the development of better adhesives and other applications.
Boiling Points and Bonding Types
In order for a substance to enter the gas phase, its particles must completely overcome the intermolecular forces holding them together. Therefore, a comparison of boiling points is essentially equivalent to comparing the strengths of the attractive intermolecular forces exhibited by the individual molecules. For small molecular compounds, London dispersion forces are the weakest intermolecular forces. Dipole-dipole forces are somewhat stronger, and hydrogen bonding is a particularly strong form of dipole-dipole interaction. However, when the mass of a nonpolar molecule is sufficiently large, its dispersion forces can be stronger than the dipole-dipole forces in a lighter polar molecule. Thus, nonpolar Cl2 has a higher boiling point than polar HCl.
Table $3$: Intermolecular Forces and Boiling Points
Substance Strongest Intermolecular Force Boiling Point $\left( ^\text{o} \text{C} \right)$
$\ce{H_2}$ dispersion -253
$\ce{Ne}$ dispersion -246
$\ce{O_2}$ dispersion -183
$\ce{Cl_2}$ dispersion -34
$\ce{HCl}$ dipole-dipole -85
$\ce{HBr}$ dipole-dipole -66
$\ce{H_2S}$ dipole-dipole -61
$\ce{NH_3}$ hydrogen bonding -33
$\ce{HF}$ hydrogen bonding 20
$\ce{H_2O}$ hydrogen bonding 100
Example $4$: Intermolecular Forces
What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature?
1. potassium chloride (KCl)
2. ethanol (C2H5OH)
3. bromine (Br2)
Solution
1. Potassium chloride is composed of ions, so the intermolecular interaction in potassium chloride is ionic forces. Because ionic interactions are strong, it might be expected that potassium chloride is a solid at room temperature.
2. Ethanol has a hydrogen atom attached to an oxygen atom, so it would experience hydrogen bonding. If the hydrogen bonding is strong enough, ethanol might be a solid at room temperature, but it is difficult to know for certain. (Ethanol is actually a liquid at room temperature.)
3. Elemental bromine has two bromine atoms covalently bonded to each other. Because the atoms on either side of the covalent bond are the same, the electrons in the covalent bond are shared equally, and the bond is a nonpolar covalent bond. Thus, diatomic bromine does not have any intermolecular forces other than dispersion forces. It is unlikely to be a solid at room temperature unless the dispersion forces are strong enough. Bromine is a liquid at room temperature.
Exercise $4$
What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature?
1. methylamine (CH3NH2)
2. calcium sulfate (CaSO4)
3. carbon monoxide (CO)
Answer
a. dipole-dipole, hydrogen bonding
b. ionic forces (solid at room temperature)
c. dipole-dipole
Key Takeaways
• A phase is a form of matter that has the same physical properties throughout.
• Molecules interact with each other through various forces: dipole-dipole interactions, hydrogen bonding, and dispersion forces.
• Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one dipolar molecule for the partial positive end of another.
• Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: F, O, or N.
• The temporary dipole that results from the motion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.01%3A_Intermolecular_Interactions.txt |
Learning Objectives
• To describe the solid and liquid phases.
Solids and liquids are collectively called condensed phases because their particles are in virtual contact. The two states share little else, however.
Solids
In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume. Most solids are hard, but some (like waxes) are relatively soft. Many solids composed of ions can also be quite brittle.
Solids usually have their constituent particles arranged in a regular, three-dimensional array of alternating positive and negative ions called a crystal. The effect of this regular arrangement of particles is sometimes visible macroscopically, as shown in Figure \(1\). Some solids, especially those composed of large molecules, cannot easily organize their particles in such regular crystals and exist as amorphous (literally, “without form”) solids. Glass is one example of an amorphous solid.
Liquids
If the particles of a substance have enough energy to partially overcome intermolecular interactions, then the particles can move about each other while remaining in contact. This describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container.
Gases
If the particles of a substance have enough energy to completely overcome intermolecular interactions, then the particles can separate from each other and move about randomly in space. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either.
The change from solid to liquid usually does not significantly change the volume of a substance. However, the change from a liquid to a gas significantly increases the volume of a substance, by a factor of 1,000 or more. Figure \(3\) shows the differences among solids, liquids, and gases at the molecular level, while Table \(1\) lists the different characteristics of these states.
Table \(1\): Characteristics of the Three States of Matter
Characteristic Solid Liquid Gas
shape definite indefinite indefinite
volume definite definite indefinite
relative intermolecular interaction strength strong moderate weak
relative particle positions in contact and fixed in place in contact but not fixed not in contact, random positions
Example \(1\)
What state or states of matter does each statement, describe?
1. This state has a definite volume.
2. This state has no definite shape.
3. This state allows the individual particles to move about while remaining in contact.
Solution
1. This statement describes either the liquid state or the solid state.
2. This statement describes either the liquid state or the gas state.
3. This statement describes the liquid state.
Exercise \(1\)
What state or states of matter does each statement describe?
1. This state has individual particles in a fixed position with regard to each other.
2. This state has individual particles far apart from each other in space.
3. This state has a definite shape.
Answer
a. solid
b. gas
c. solid
Looking Closer: Water, the Most Important Liquid
Earth is the only known body in our solar system that has liquid water existing freely on its surface. That is a good thing because life on Earth would not be possible without the presence of liquid water.
Water has several properties that make it a unique substance among substances. It is an excellent solvent; it dissolves many other substances and allows those substances to react when in solution. In fact, water is sometimes called the universal solvent because of this ability. Water has unusually high melting and boiling points (0°C and 100°C, respectively) for such a small molecule. The boiling points for similar-sized molecules, such as methane (BP = −162°C) and ammonia (BP = −33°C), are more than 100° lower. Though a liquid at normal temperatures, water molecules experience a relatively strong intermolecular interaction that allows them to maintain the liquid phase at higher temperatures than expected.
Unlike most substances, the solid form of water is less dense than its liquid form, which allows ice to float on water. The most energetically favorable configuration of H2O molecules is one in which each molecule is hydrogen-bonded to four neighboring molecules. Owing to the thermal motions, this ideal is never achieved in the liquid, but when water freezes to ice, the molecules settle into exactly this kind of an arrangement in the ice crystal. This arrangement requires that the molecules be somewhat farther apart than would otherwise be the case; as a consequence, ice, in which hydrogen bonding is at its maximum, has a more open structure, and thus a lower density than water.
Here are three-dimensional views of a typical local structure of water (left) and ice (right.) Notice the greater openness of the ice structure which is necessary to ensure the strongest degree of hydrogen bonding in a uniform, extended crystal lattice. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion. Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Water also requires an unusually large amount of energy to change temperature. While 100 J of energy will change the temperature of 1 g of Fe by 230°C, this same amount of energy will change the temperature of 1 g of H2O by only 100°C. Thus, water changes its temperature slowly as heat is added or removed. This has a major impact on weather, as storm systems like hurricanes can be impacted by the amount of heat that ocean water can store.
Water’s influence on the world around us is affected by these properties. Isn’t it fascinating that such a small molecule can have such a big impact?
Key Takeaway
• Solids and liquids are phases that have their own unique properties. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.02%3A_Solids_and_Liquids.txt |
Learning Objectives
• To describe the gas phase.
The gas phase is unique among the three states of matter in that there are some simple models we can use to predict the physical behavior of all gases—independent of their identities. We cannot do this for the solid and liquid states. In fact, the development of this understanding of the behavior of gases represents the historical dividing point between alchemy and modern chemistry. Initial advances in the understanding of gas behavior were made in the mid 1600s by Robert Boyle, an English scientist who founded the Royal Society (one of the world’s oldest scientific organizations).
How is it that we can model all gases independent of their chemical identity? The answer is in a group of statements called the kinetic theory of gases:
• Gases are composed of tiny particles that are separated by large distances.
• Gas particles are constantly moving, experiencing collisions with other gas particles and the walls of their container.
• The velocity of gas particles is related to the temperature of a gas.
• Gas particles do not experience any force of attraction or repulsion with each other.
Did you notice that none of these statements relates to the identity of the gas? This means that all gases should behave similarly. A gas that follows these statements perfectly is called an ideal gas. Most gases show slight deviations from these statements and are called real gases. However, the existence of real gases does not diminish the importance of the kinetic theory of gases.
One of the statements of the kinetic theory mentions collisions. As gas particles are constantly moving, they are also constantly colliding with each other and with the walls of their container. There are forces involved as gas particles bounce off the container walls (Figure $1$). The force generated by gas particles divided by the area of the container walls yields pressure. Pressure is a property we can measure for a gas, but we typically do not consider pressure for solids or liquids.
The basic unit of pressure is the newton per square meter (N/m2). This combined unit is redefined as a pascal (Pa). One pascal is not a very large amount of pressure. A more useful unit of pressure is the bar, which is 100,000 Pa (1 bar = 100,000 Pa). Other common units of pressure are the atmosphere (atm), which was originally defined as the average pressure of Earth’s atmosphere at sea level; and mmHg (millimeters of mercury), which is the pressure generated by a column of mercury 1 mm high. The unit millimeters of mercury is also called a torr, named after the Italian scientist Evangelista Torricelli, who invented the barometer in the mid-1600s. A more precise definition of atmosphere, in terms of torr, is that there are exactly 760 torr in 1 atm. A bar equals 1.01325 atm. Given all the relationships between these pressure units, the ability to convert from one pressure unit to another is a useful skill.
Example $1$: Converting Pressures
Write a conversion factor to determine how many atmospheres are in 1,547 mmHg.
Solution
Because 1 mmHg equals 1 torr, the given pressure is also equal to 1,547 torr. Because there are 760 torr in 1 atm, we can use this conversion factor to do the mathematical conversion:
$\mathrm{1,547\: torr\times \dfrac{1\: atm}{760\: torr}=2.04\: atm}$
Note how the torr units cancel algebraically.
Exercise $1$: Converting Pressures
Write a conversion factor to determine how many millimeters of mercury are in 9.65 atm.
Answer
$\mathrm{9.65\: atm\times \dfrac{760\: mm Hg}{1\: atm}=7,334 \: mm Hg}$.
The kinetic theory also states that there is no interaction between individual gas particles. Although we know that there are, in fact, intermolecular interactions in real gases, the kinetic theory assumes that gas particles are so far apart that the individual particles don’t “feel” each other. Thus, we can treat gas particles as tiny bits of matter whose identity isn’t important to certain physical properties.
Key Takeaway
• The gas phase has certain general properties characteristic of that phase. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.03%3A_Gases_and_Pressure.txt |
Learning Objectives
• To predict the properties of gases using the gas laws.
Experience has shown that several properties of a gas can be related to each other under certain conditions. The properties are pressure (P), volume (V), temperature (T, in kelvins), and amount of material expressed in moles (n). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur.
Boyle’s Law
The first simple relationship, referred to as a gas law, is between the pressure of a gas and its volume. If the amount of gas in a sample and its temperature are kept constant, then as the pressure of a gas is increased, the volume of the gas decreases proportionately. Mathematically, this is written as
$\mathrm{P \propto \dfrac{1}{V}} \nonumber$
where the “∝” symbol means “is proportional to.” This is one form of Boyle’s law, which relates the pressure of a gas to its volume.
A more useful form of Boyle’s law involves a change in conditions of a gas. For a given amount of gas at a constant temperature, if we know the initial pressure and volume of a gas sample and the pressure or volume changes, we can calculate what the new volume or pressure will be. That form of Boyle’s law is written
$P_iV_i = P_fV_f \label{Eq1}$
where the subscript $i$ refers to initial conditions and the subscript $f$ refers to final conditions.
To use $\ref{Eq1}$, you need to know any three of the variables so that you can algebraically calculate the fourth variable. Also, the pressure quantities must have the same units, as must the two volume quantities. If the two similar variables don’t have the same variables, one value must be converted to the other value’s unit.
Example $1$: Increasing Pressure in a Gas
What happens to the volume of a gas if its pressure is increased? Assume all other conditions remain the same.
Solution
If the pressure of a gas is increased, the volume decreases in response.
Exercise $1$: Increasing Volume in a Gas
What happens to the pressure of a gas if its volume is increased? Assume all other conditions remain the same.
Answer
If the volume of a gas is increased, the pressure decreases.
Example $2$: Gas Compression
If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is reduced to 0.987 atm? Assume that the amount and the temperature of the gas remain constant.
Solution
The key in problems like this is to be able to identify which quantities represent which variables from the relevant equation. The way the question is worded, you should be able to tell that 1.56 atm is Pi, 7.02 L is Vi, and 0.987 atm is Pf. What we are looking for is the final volume—Vf. Therefore, substituting these values into PiVi = PfVf:
(1.56 atm)(7.02 L) = (0.987 atm) × Vf
The expression has atmospheres on both sides of the equation, so they cancel algebraically:
(1.56)(7.02 L) = (0.987) × Vf
Now we divide both sides of the expression by 0.987 to isolate Vf, the quantity we are seeking:
$\mathrm{\dfrac{(1.56)(7.02\: L)}{0.987}=V_f}$
Performing the multiplication and division, we get the value of Vf, which is 11.1 L. The volume increases. This should make sense because the pressure decreases, so pressure and volume are inversely related.
Exercise $2$
If a sample of gas has an initial pressure of 3.66 atm and an initial volume of 11.8 L, what is the final pressure if the volume is reduced to 5.09 L? Assume that the amount and the temperature of the gas remain constant.
Answer
8.48 atm
If the units of similar quantities are not the same, one of them must be converted to the other quantity’s units for the calculation to work out properly. It does not matter which quantity is converted to a different unit; the only thing that matters is that the conversion and subsequent algebra are performed properly. The following example illustrates this process.
Example $3$
If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is changed to 1,775 torr? Does the answer make sense? Assume that the amount and the temperature of the gas remain constant.
Solution
This example is similar to Example $2$, except now the final pressure is expressed in torr. For the math to work out properly, one of the pressure values must be converted to the other unit. Let us change the initial pressure to torr:
$\mathrm{1.56\: atm\times\dfrac{760\: torr}{1\: atm}=1,190\: torr}$
Now we can use Boyle’s law:
(1,190 torr)(7.02 L) = (1,775 torr) × Vf
Torr cancels algebraically from both sides of the equation, leaving
(1,190)(7.02 L) = (1,775) × Vf
Now we divide both sides of the equation by 1,775 to isolate Vf on one side. Solving for the final volume,
$\mathrm{V_f=\dfrac{(1,190)(7.02\: L)}{1,775}=4.71\: L}$
Because the pressure increases, it makes sense that the volume decreases.
The answer for the final volume is essentially the same if we converted the 1,775 torr to atmospheres: $\mathrm{1,775\: torr\times\dfrac{1\: atm}{760\: torr}=2.336\: atm}$. Using Boyle’s law: (1.56 atm)(7.02 L) = (2.335 atm) × Vf; $\mathrm{V_f=\dfrac{(1.56\: atm)(7.02\: L)}{2.336\: atm}=4.69\: L}$.
Exercise $3$
If a sample of gas has an initial pressure of 375 torr and an initial volume of 7.02 L, what is the final pressure if the volume is changed to 4,577 mL? Does the answer make sense? Assume that amount and the temperature of the gas remain constant.
Answer
575 torr
To Your Health: Breathing
Breathing certainly is a major contribution to your health! Without breathing, we could not survive. Curiously, the act of breathing itself is little more than an application of Boyle’s law.
The lungs are a series of ever-narrowing tubes that end in a myriad of tiny sacs called alveoli. It is in the alveoli that oxygen from the air transfers to the bloodstream and carbon dioxide from the bloodstream transfers to the lungs for exhalation. For air to move in and out of the lungs, the pressure inside the lungs must change, forcing the lungs to change volume—just as predicted by Boyle’s law.
The pressure change is caused by the diaphragm, a muscle that covers the bottom of the lungs. When the diaphragm moves down, it expands the size of our lungs. When this happens, the air pressure inside our lungs decreases slightly. This causes new air to rush in, and we inhale. The pressure decrease is slight—only 3 torr, or about 0.4% of an atmosphere. We inhale only 0.5–1.0 L of air per normal breath.
Exhaling air requires that we relax the diaphragm, which pushes against the lungs and slightly decreases the volume of the lungs. This slightly increases the pressure of the air in the lungs, and air is forced out; we exhale. Only 1–2 torr of extra pressure is needed to exhale. So with every breath, our own bodies are performing an experimental test of Boyle’s law.
Charles’s Law
Another simple gas law relates the volume of a gas to its temperature. Experiments indicate that as the temperature of a gas sample is increased, its volume increases as long as the pressure and the amount of gas remain constant. The way to write this mathematically is
$\mathrm V \propto T \nonumber$
At this point, the concept of temperature must be clarified. Although the Kelvin scale is the preferred temperature scale, the Celsius scale is also a common temperature scale used in science. The Celsius scale is based on the melting and boiling points of water and is actually the common temperature scale used by most countries around the world (except for the United States, which still uses the Fahrenheit scale). The value of a Celsius temperature is directly related to its Kelvin value by a simple expression:
Kelvin temperature = Celsius temperature + 273
Thus, it is easy to convert from one temperature scale to another.
The Kelvin scale is sometimes referred to as the absolute scale because the zero point on the Kelvin scale is at absolute zero, the coldest possible temperature. On the other temperature scales, absolute zero is −260°C or −459°F.
The expression relating a gas volume to its temperature begs the following question: to which temperature scale is the volume of a gas related? The answer is that gas volumes are directly related to the Kelvin temperature. Therefore, the temperature of a gas sample should always be expressed in (or converted to) a Kelvin temperature.
Example $4$: Increasing Temperature
What happens to the volume of a gas if its temperature is decreased? Assume that all other conditions remain constant.
Solution
If the temperature of a gas sample is decreased, the volume decreases as well.
Exercise $4$
What happens to the temperature of a gas if its volume is increased? Assume that all other conditions remain constant.
Answer
The temperature increases.
As with Boyle’s law, the relationship between volume and temperature can be expressed in terms of initial and final values of volume and temperature, as follows:
$\mathrm{\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}} \nonumber$
where Vi and Ti are the initial volume and temperature, and Vf and Tf are the final volume and temperature. This is Charles’s law. The restriction on its use is that the pressure of the gas and the amount of gas must remain constant. (Charles’s law is sometimes referred to as Gay-Lussac’s law, after the scientist who promoted Charles’s work.)
Example $5$
A gas sample at 20°C has an initial volume of 20.0 L. What is its volume if the temperature is changed to 60°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant.
Solution
Although the temperatures are given in degrees Celsius, we must convert them to the kelvins before we can use Charles’s law. Thus,
20°C + 273 = 293 K = Ti 60°C + 273 = 333 K = Tf
Now we can substitute these values into Charles’s law, along with the initial volume of 20.0 L:
$\mathrm{\dfrac{20.0\: L}{293\: K}=\dfrac{V_f}{333\: K}}$
Multiplying the 333 K to the other side of the equation, we see that our temperature units will cancel:
$\mathrm{\dfrac{(333\: K)(20.0\: L)}{293\: K}=V_f}$
Solving for the final volume, Vf = 22.7 L. So, as the temperature is increased, the volume increases. This makes sense because volume is directly proportional to the absolute temperature (as long as the pressure and the amount of the remain constant).
Exercise $5$
A gas sample at 35°C has an initial volume of 5.06 L. What is its volume if the temperature is changed to −35°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant.
Answer
3.91 L
Combined Gas Law
Other gas laws can be constructed, but we will focus on only two more. The combined gas law brings Boyle’s and Charles’s laws together to relate pressure, volume, and temperature changes of a gas sample:
$\mathrm{\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}} \nonumber$
To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically.
Example $6$
A sample of gas has Pi = 1.50 atm, Vi = 10.5 L, and Ti = 300 K. What is the final volume if Pf = 0.750 atm and Tf = 350 K?
Solution
Using the combined gas law, substitute for five of the quantities:
$\mathrm{\dfrac{(1.50\: atm)(10.5\: L)}{300\: K}=\dfrac{(0.750\: atm)(V_f)}{350\: K}}$
We algebraically rearrange this expression to isolate Vf on one side of the equation:
$\mathrm{V_f=\dfrac{(1.50\: atm)(10.5\: L)(350\: K)}{(300\: K)(0.750\: atm)}=24.5\: L}$
Note how all the units cancel except the unit for volume.
Exercise $6$
A sample of gas has Pi = 0.768 atm, Vi = 10.5 L, and Ti = 300 K. What is the final pressure if Vf = 7.85 L and Tf = 250 K?
Answer
0.856 atm
Example $7$
A balloon containing a sample of gas has a temperature of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The balloon has a volume of 1,070 mL. The balloon is transported by plane to Denver, where the temperature is 11°C and the pressure is 655 torr. What is the new volume of the balloon?
Solution
The first task is to convert all quantities to the proper and consistent units. The temperatures must be expressed in kelvins, and the pressure units are different so one of the quantities must be converted. Let us convert the atmospheres to torr:
22°C + 273 = 295 K = Ti
11°C + 273 = 284 K = Tf
$\mathrm{1.09\: atm\times\dfrac{760\: torr}{1\: atm}=828\: torr = P_i}$
Now we can substitute the quantities into the combined has law:
$\mathrm{\dfrac{(828\: torr)(1,070\: mL)}{295\: K}=\dfrac{(655\: torr)\times V_f}{284\: K}}$
To solve for Vf, we multiply the 284 K in the denominator of the right side into the numerator on the left, and we divide 655 torr in the numerator of the right side into the denominator on the left:
$\mathrm{\dfrac{(828\: torr)(1,070\: mL)(284\: K)}{(295\: K)(655\: torr)}=V_f}$
Notice that torr and kelvins cancel, as they are found in both the numerator and denominator. The only unit that remains is milliliters, which is a unit of volume. So Vf = 1,300 mL. The overall change is that the volume of the balloon has increased by 230 mL.
Exercise $7$
A balloon used to lift weather instruments into the atmosphere contains gas having a volume of 1,150 L on the ground, where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this gas has a pressure of 6.88 torr and a temperature of −15°C. What is the new volume of the gas?
Answer
110,038 L
The Ideal Gas Law
So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law. The formula of this law is as follows:
$\color{red} PV = nRT \nonumber$
In this equation, P is pressure, V is volume, n is amount of moles, and T is temperature. R is called the ideal gas law constant and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts i and f to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under any conditions.
The value of R depends on what units are used to express the other quantities. If volume is expressed in liters and pressure in atmospheres, then the proper value of R is as follows:
$\mathrm{R=0.08205 \: \dfrac{L\cdot atm}{mol\cdot K}} \nonumber$
This may seem like a strange unit, but that is what is required for the units to work out algebraically.
Example $8$
What is the volume in liters of 1.45 mol of N2 gas at 298 K and 3.995 atm?
Solution
Using the ideal gas law where P = 3.995 atm, n = 1.45, and T = 298,
$\mathrm{(3.995\: atm)\times V=(1.45\: mol)\left(0.08205\: \dfrac{L\cdot atm}{mol\cdot K}\right)(298\: K)}$
On the right side, the moles and kelvins cancel. Also, because atmospheres appear in the numerator on both sides of the equation, they also cancel. The only remaining unit is liters, a unit of volume. So
3.995 × V = (1.45)(0.08205)(298) L
Dividing both sides of the equation by 3.995 and evaluating, we get V = 8.87 L. Note that the conditions of the gas are not changing. Rather, the ideal gas law allows us to determine what the fourth property of a gas (here, volume) must be if three other properties (here, amount, pressure, and temperature) are known.
Exercise $8$
What is the pressure of a sample of CO2 gas if 0.557 mol is held in a 20.0 L container at 451 K?
Answer
1.03 atm
For convenience, scientists have selected 273 K (0°C) and 1.00 atm pressure as a set of standard conditions for gases. This combination of conditions is called standard temperature and pressure (STP). Under these conditions, 1 mol of any gas has about the same volume. We can use the ideal gas law to determine the volume of 1 mol of gas at STP:
$\mathrm{(1.00\: atm)\times V=(1.00\: mol)\left(0.08205\: \dfrac{L\cdot atm}{mol\cdot K}\right)(273\: K)} \nonumber$
This volume is 22.4 L. Because this volume is independent of the identity of a gas, the idea that 1 mol of gas has a volume of 22.4 L at STP makes a convenient conversion factor:
1 mol gas = 22.4 L (at STP)
Example $9$
Cyclopropane (C3H6) is a gas that formerly was used as an anesthetic. How many moles of gas are there in a 100.0 L sample if the gas is at STP?
Solution
We can set up a simple, one-step conversion that relates moles and liters:
$\mathrm{100.0\: L\: C_3H_6\times \dfrac{1\: mol}{22.4\: L}=4.46\: mol\: C_3H_6}$
There are almost 4.5 mol of gas in 100.0 L.
Note: Because of its flammability, cyclopropane is no longer used as an anesthetic gas.
Exercise $9$
Freon is a trade name for a series of fluorine- and chlorine-containing gases that formerly were used in refrigeration systems. What volume does 8.75 mol of Freon have at STP?
Note: Many gases known as Freon are no longer used because their presence in the atmosphere destroys the ozone layer, which protects us from ultraviolet light from the sun.
Answer
196 L
Airbags
Airbags (Figure $3$) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN3. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN3 to initiate its decomposition:
$\ce{2NaN3}(s)\rightarrow \ce{3N2}(g)+\ce{2Na}(s) \nonumber$
This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN3 will generate approximately 50 L of N2.
Dalton's Law of Partial Pressures
The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only new concept we need in order to deal with gas mixtures is the partial pressure, a concept invented by the famous English chemist John Dalton (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist mostly of empty space; from this it follows that when two or more different gases occupy the same volume, they behave entirely independently. The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the partial pressure of that gas.
The definition of Dalton's Law of Partial Pressures that address this is:
The total pressure of a gas is the sum of the partial pressures of its components
which is expressed algebraically as
$P_{total}=P_1+P_2+P_3 ... = \sum_i P_i \nonumber$
or, equivalently
$P_{total} = \dfrac{RT}{V} \sum_i n_i \nonumber$
There is also a similar relationship based on volume fractions, known as Amagat's law of partial volumes. It is exactly analogous to Dalton's law, in that it states that the total volume of a mixture is just the sum of the partial volumes of its components. But there are two important differences: Amagat's law holds only for ideal gases which must all be at the same temperature and pressure. Dalton's law has neither of these restrictions. Although Amagat's law seems intuitively obvious, it sometimes proves useful in chemical engineering applications. We will make no use of it in this course.
Example $10$
Three flasks having different volumes and containing different gases at various pressures are connected by stopcocks as shown. When the stopcocks are opened,
1. What will be the pressure in the system?
2. Which gas will be most abundant in the mixture?
Assume that the temperature is uniform and that the volume of the connecting tubes is negligible.
Solution
The trick here is to note that the total number of moles nT and the temperature remain unchanged, so we can make use of Boyle's law PV = constant. We will work out the details for CO2 only, denoted by subscripts a.
For CO2,
$P_aV_a = (2.13\; atm)(1.50\; L) = 3.19\; L \cdot atm \nonumber$
Adding the PV products for each separate container, we obtain
$\sum_i P_iV_i = 6.36\; L \cdot atm = n_T RT \nonumber$
We will call this sum P1V1. After the stopcocks have been opened and the gases mix, the new conditions are denoted by P2V2.
From Boyle's law ($\ref{Eq1}$,
$P_1V_1 = P_2V_2 = 6.36\; L \cdot atm \nonumber$
$V_2 = \sum_i V_i = 4.50\; L \nonumber$
Solving for the final pressure P2 we obtain (6.36 L-atm)/(4.50 L) = 1.41 atm.
For part (b), note that the number of moles of each gas is n = PV/RT. The mole fraction of any one gas is Xi = ni /nT . For CO2, this works out to (3.19/RT) / (6.36/RT) = 0.501. Because this exceeds 0.5, we know that this is the most abundant gas in the final mixture.
Dalton’s law states that in a gas mixture ($P_{total}$) each gas will exert a pressure independent of the other gases ($P_n$) and each gas will behave as if it alone occupies the total volume. By extension, the partial pressure of each gas can be calculated by multiplying the total pressure ($P_{total}$) by the gas percentage (%).
$P_{Total} = P_1 + P_2 + P_3 + P_4 + ... + P_n \nonumber$
or
$P_n = \dfrac{\text{% of individual gas}_n}{P_{Total}} \nonumber$
Table $1$: Partial Pressures for the gases in air on a typical day
Gas Partial Pressure (mm Hg) Percentage (%)
Nitrogen, (N_2\) $P_{N_2}$ = 594 78
Oxygen, $O_2$ $P_{O_2}$= 160 21
Carbon Dioxide, $CO_2$ $P_{CO_2}$ = 0.25 0.033
Water Vapor, $H_2O$ $P_{H_2O}$ = 5.7 0.75
Other trace gases $P_{Other}$ = 0.05 0.22
Total air $P_{Total}$ = 760 1
Application of Dalton's Law: Collecting Gases over Water
A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a pneumatic trough, and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part.
The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H2O. The partial pressure of H2O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton's Law to find the partial pressure of that gas.
Example $11$
Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of $O_2$ produced.
Solution
From Dalton's law,
$P_{O_2} = P_{total} – P_{H_2O} = 754 – 19.8 = 734 \; torr = 0.966\; atm \nonumber$
Now use the Ideal Gas Law to convert to moles
$n =\dfrac{PV}{RT} = \dfrac{(0.966\; atm)(0.155\;L)}{(0.082\; L atm mol^{-1} K^{-1})(295\; K)}= 0.00619 \; mol \nonumber$
Henry’s Law
Henry's law is one of the gas laws formulated by William Henry in 1803. It 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.
To explain this law, Henry derived the equation:
$C =k P_{gas} \nonumber$
where
Henry’s Law tells us that the greater the pressure of gas above the surface of a liquid, the higher the concentration of the gas in the liquid. Also, Henry’s law tells us that gases diffuse from areas of high gas concentration to areas of low gas concentration.
Applicability of Henry's Law
• Henry's law only works if the molecules are at equilibrium.
• Henry's law does not work for gases at high pressures (e.g., $N_{2\;(g)}$ at high pressure becomes very soluble and harmful when in the blood supply).
• Henry's law does not work if there is a chemical reaction between the solute and solvent (e.g., $HCl_{(g)}$ reacts with water by a dissociation reaction to generate $H_3O^+$ and $Cl^-$ ions).
Application of Henry's Law: Scuba diving
Our respiratory systems are designed to maintain the proper oxygen concentration in the blood when the partial pressure of O2 is 0.21 atm, its normal sea-level value. Below the water surface, the pressure increases by 1 atm for each 10.3 m increase in depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on the body. In order to prevent the lungs from collapsing, the air the diver breathes should also be at about the same pressure.
But at a total pressure of 2 atm, the partial pressure of $O_2$ in ordinary air would be 0.42 atm; at a depth of 100 ft (about 30 m), the $O_2$ pressure of 0.8 atm would be far too high for health. For this reason, the air mixture in the pressurized tanks that scuba divers wear must contain a smaller fraction of $O_2$. This can be achieved most simply by raising the nitrogen content, but high partial pressures of N2 can also be dangerous, resulting in a condition known as nitrogen narcosis. The preferred diluting agent for sustained deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures.
Career Focus: Respiratory Therapist
Certain diseases—such as emphysema, lung cancer, and severe asthma—primarily affect the lungs. Respiratory therapists help patients with breathing-related problems. They can evaluate, help diagnose, and treat breathing disorders and even help provide emergency assistance in acute illness where breathing is compromised.
Most respiratory therapists must complete at least two years of college and earn an associate’s degree, although therapists can assume more responsibility if they have a college degree. Therapists must also pass state or national certification exams. Once certified, respiratory therapists can work in hospitals, doctor’s offices, nursing homes, or patient’s homes. Therapists work with equipment such as oxygen tanks and respirators, may sometimes dispense medication to aid in breathing, perform tests, and educate patients in breathing exercises and other therapy.
Because respiratory therapists work directly with patients, the ability to work well with others is a must for this career. It is an important job because it deals with one of the most crucial functions of the body.
Key Takeaway
• The physical properties of gases are predictable using mathematical formulas known as gas laws.
• $C$ is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L)
• $k$ is Henry's law constant (often in units of M/atm)
• $P_{gas}$ is the partial pressure of the gas (often in units of Atm) | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.04%3A_Gas_Laws.txt |
Concept Review Exercise
1. What types of intermolecular interactions can exist in compounds?
2. What is the difference between covalent network and covalent molecular compounds?
Answer
1. ionic bonding, network covalent, dispersion forces, dipole-dipole interactions, and hydrogen bonding.
2. Covalent network compounds contain atoms that are covalently bonded to other individual atoms in a giant 3-dimensional network. Covalent molecular compounds contain individual molecules that are attracted to one another through dispersion, dipole-dipole or hydrogen bonding.
Exercises
1. List the three common phases in the order you are likely to find them—from lowest temperature to highest temperature.
2. List the three common phases in the order they exist from lowest energy to highest energy.
3. List these intermolecular interactions from weakest to strongest: London forces, hydrogen bonding, and ionic interactions.
4. List these intermolecular interactions from weakest to strongest: covalent network bonding, dipole-dipole interactions, and dispersion forces.
5. What type of intermolecular interaction is predominate in each substance?
1. water (H2O)
2. sodium sulfate (Na2SO4)
3. decane (C10H22)
6. What type of intermolecular interaction is predominate in each substance?
1. diamond (C, crystal)
2. helium (He)
3. ammonia (NH3)
7. Explain how a molecule like carbon dioxide (CO2) can have polar covalent bonds but be nonpolar overall.
8. Sulfur dioxide (SO2) has a formula similar to that of carbon dioxide (see Exercise 7) but is a polar molecule overall. What can you conclude about the shape of the SO2 molecule?
9. What are some of the physical properties of substances that experience covalent network bonding?
10. What are some of the physical properties of substances that experience only dispersion forces?
Answers
1. solid, liquid, and gas
2. solid, liquid, and gas
1. London forces, hydrogen bonding, and ionic interactions
4. dispersion, dipole-dipole, network covalent
1. hydrogen bonding
2. ionic interactions
3. dispersion forces
6. a. network covalent
b. dispersion
c. hydrogen bonding
1. The two covalent bonds are oriented in such a way that their dipoles cancel out.
8. SO2 is not a linear molecule. It has a bent or V-shape.
9. very hard, high melting point
10. very soft, very low melting point
Concept Review Exercise
1. How do the strengths of intermolecular interactions in solids and liquids differ?
Answer
1. Solids have stronger intermolecular interactions than liquids do.
Exercises
1. What are the general properties of solids?
2. What are the general properties of liquids
3. What are the general properties of gases?
4. What phase or phases have a definite volume? What phase or phases do not have a definite volume?
5. Name a common substance that forms a crystal in its solid state.
6. Name a common substance that forms an amorphous solid in its solid state.
7. Are substances with strong intermolecular interactions likely to be solids at higher or lower temperatures? Explain.
8. Are substances with weak intermolecular interactions likely to be liquids at higher or lower temperatures? Explain.
9. State two similarities between the solid and liquid states.
10. State two differences between the solid and liquid states.
11. If individual particles are moving around with respect to each other, a substance may be in either the _______ or ________ state but probably not in the _______ state.
12. If individual particles are in contact with each other, a substance may be in either the ______ or _______ state but probably not in the ______ state.
Answers
1. hard, specific volume and shape, high density, cannot be compressed
2. fixed volume, no definite shape, high density, individual molecules touch each other but in a random way
1. variable volume and shape, low density, compressible
4. solid and liquid have definite volume; gas has no definite volume
1. sodium chloride (answers will vary)
6. glass
1. At higher temperatures, their intermolecular interactions are strong enough to hold the particles in place.
8. Substances with weak intermolecular interactions are likely to be liquids at lower temperatures. Their attractive forces are more easily broken hence they melt more readily.
1. high density; definite volume
10. Solids have definite shape while liquids don't. In solids, molecules occupy fixed positions in a pattern, while in liquids, the molecules are moving in a random arrangement.
11. liquid; gas; solid
12. solid; liquid; gas
Concept Review Exercise
1. What is pressure, and what units do we use to express it?
Answer
1. Pressure is the force per unit area; its units can be pascals, torr, millimeters of mercury, or atmospheres.
Exercises
1. What is the kinetic theory of gases?
2. According to the kinetic theory of gases, the individual gas particles are (always, frequently, never) moving.
3. Why does a gas exert pressure?
4. Why does the kinetic theory of gases allow us to presume that all gases will show similar behavior?
5. Arrange the following pressure quantities in order from smallest to largest: 1 mmHg, 1 Pa, and 1 atm.
6. Which unit of pressure is larger—the torr or the atmosphere?
7. How many torr are there in 1.56 atm?
8. Convert 760 torr into pascals.
9. Blood pressures are expressed in millimeters of mercury. What would be the blood pressure in atmospheres if a patient’s systolic blood pressure is 120 mmHg and the diastolic blood pressure is 82 mmHg? (In medicine, such a blood pressure would be reported as “120/82,” spoken as “one hundred twenty over eighty-two.”)
10. In weather forecasting, barometric pressure is expressed in inches of mercury (in. Hg), where there are exactly 25.4 mmHg in every 1 in. Hg. What is the barometric pressure in millimeters of mercury if the barometric pressure is reported as 30.21 in. Hg?
Answers
1. Gases are composed of tiny particles that are separated by large distances. Gas particles are constantly moving, experiencing collisions with other gas particles and the walls of their container. The velocity of gas particles is related to the temperature of a gas. Gas particles do not experience any force of attraction or repulsion with each other.
2. always
1. A gas exerts pressure as its particles rebound off the walls of its container.
4. Because the molecules are far apart and don't have attractive forces between them
1. 1 Pa, 1 mmHg, and 1 atm
6. atm
1. 1,190 torr
8. 98,700 Pa
9. 0.158 atm; 0.108 atm
10. 767.3 mm Hg
Concept Review Exercises
1. What properties do the gas laws help us predict?
2. What makes the ideal gas law different from the other gas laws?
Answers
1. Gas laws relate four properties: pressure, volume, temperature, and number of moles.
2. The ideal gas law does not require that the properties of a gas change.
Exercises
1. What conditions of a gas sample should remain constant for Boyle’s law to be used?
2. What conditions of a gas sample should remain constant for Charles’s law to be used?
3. Does the identity of a gas matter when using Boyle’s law? Why or why not?
4. Does the identity of a gas matter when using Charles’s law? Why or why not?
5. A sample of nitrogen gas is confined to a balloon that has a volume of 1.88 L and a pressure of 1.334 atm. What will be the volume of the balloon if the pressure is changed to 0.662 atm? Assume that the temperature and the amount of the gas remain constant.
6. A sample of helium gas in a piston has a volume of 86.4 mL under a pressure of 447 torr. What will be the volume of the helium if the pressure on the piston is increased to 1,240 torr? Assume that the temperature and the amount of the gas remain constant.
7. If a gas has an initial pressure of 24,650 Pa and an initial volume of 376 mL, what is the final volume if the pressure of the gas is changed to 775 torr? Assume that the amount and the temperature of the gas remain constant.
8. A gas sample has an initial volume of 0.9550 L and an initial pressure of 564.5 torr. What would the final pressure of the gas be if the volume is changed to 587.0 mL? Assume that the amount and the temperature of the gas remain constant.
9. A person draws a normal breath of about 1.00 L. If the initial temperature of the air is 18°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and amount of the gas remain constant.
10. A person draws a normal breath of about 1.00 L. If the initial temperature of the air is −10°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and the amount of the gas remain constant.
11. An air/gas vapor mix in an automobile cylinder has an initial temperature of 450 K and a volume of 12.7 cm3. The gas mix is heated to 565°C. If pressure and amount are held constant, what is the final volume of the gas in cubic centimeters?
12. Given the following conditions for a gas: Vi = 0.665 L, Ti = 23.6°C, Vf = 1.034 L. What is Tf in degrees Celsius and kelvins?
13. Assuming the amount remains the same, what must be the final volume of a gas that has an initial volume of 387 mL, an initial pressure of 456 torr, an initial temperature of 65.0°C, a final pressure of 1.00 atm, and a final temperature of 300 K?
14. When the nozzle of a spray can is depressed, 0.15 mL of gas expands to 0.44 mL, and its pressure drops from 788 torr to 1.00 atm. If the initial temperature of the gas is 22.0°C, what is the final temperature of the gas?
15. Use the ideal gas law to show that 1 mol of a gas at STP has a volume of about 22.4 L.
16. Use a standard conversion factor to determine a value of the ideal gas law constant R that has units of L•torr/mol•K.
17. How many moles of gas are there in a 27.6 L sample at 298 K and a pressure of 1.44 atm?
18. How many moles of gas are there in a 0.066 L sample at 298 K and a pressure of 0.154 atm?
19. A 0.334 mol sample of carbon dioxide gas is confined to a volume of 20.0 L and has a pressure of 0.555 atm. What is the temperature of the carbon dioxide in kelvins and degrees Celsius?
20. What must V be for a gas sample if n = 4.55 mol, P = 7.32 atm, and T = 285 K?
21. What is the pressure of 0.0456 mol of Ne gas contained in a 7.50 L volume at 29°C?
22. What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of −86.0°C?
23. A mixture of the gases $N_2$, $O_2$, and $Ar$ has a total pressure of 760 mm Hg. If the partial pressure of $N_2$ is 220 mm Hg and of $O_2$ is 470 mm Hg, What is the partial pressure of $Ar$?
24. What percent of the gas above is Ar?
25. Apply Henry’s Law to the diagram below to explain:
why oxygen diffuses from the alveoli of the lungs into the blood and from the blood into the tissues of the body. why carbon dioxide diffuses from the tissues into the blood and from the blood into the alveoli and then finally out into the atmosphere.
Answers
1. temperature and amount of the gas
2. pressure and amount of the gas
3. The identity does not matter because the variables of Boyle’s law do not identify the gas.
4. The identity does not matter because the variables of Charles law do not identify the gas.
5. 3.79 L
6. 31.1 mL
7. 92.1 mL
8. 918.4 torr
9. 1.07 L
10. 1.18 L
11. 23.7 cm3
12. 461 K; 1880C
13. 206 mL
14. 835 K; 5620C
15. The ideal gas law confirms that 22.4 L equals 1 mol.
16. $\dfrac{760\: torr}{1\: atm}$
17. 1.63 mol
18. 4.2 x 10-4 mol
19. 405 K; 132°C
20. 14.5 L
21. 0.151 atm
22. 18.2 atm
23. 70 mm Hg
24. 9.2%
25. Gases diffuse from high concentration to low concentration (Henry's Law). The partial pressure of oxygen is high in the alveoli and low in the blood of the pulmonary capillaries. As a result, oxygen diffuses across the respiratory membrane from the alveoli into the blood. It's also higher partial pressure in the blood than in the tissues, hence it transfers to the tissues. On the other hand, carbon dioxide diffuses from the tissues (highest CO2 partial pressure) and across the respiratory membrane from the blood into the alveoli and out to the atmosphere.
Additional Exercises
1. How many grams of oxygen gas are needed to fill a 25.0 L container at 0.966 atm and 22°C?
2. A breath of air is about 1.00 L in volume. If the pressure is 1.00 atm and the temperature is 37°C, what mass of air is contained in each breath? Use an average molar mass of 28.8 g/mol for air.
3. The balanced chemical equation for the combustion of propane is as follows:
$C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(ℓ)}$
1. If 100.0 g of propane are combusted, how many moles of oxygen gas are necessary for the reaction to occur?
2. At STP, how many liters of oxygen gas would that be?
4. The equation for the formation of ammonia gas (NH3) is as follows:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
At 500°C and 1.00 atm, 10.0 L of N2 gas are reacted to make ammonia.
1. If the pressures and temperatures of H2 and NH3 were the same as those of N2, what volume of H2 would be needed to react with N2, and what volume of NH3 gas would be produced?
2. Compare your answers to the balanced chemical equation. Can you devise a “shortcut” method to answer Exercise 4a?
5. At 20°C, 1 g of liquid H2O has a volume of 1.002 mL. What volume will 1 g of water vapor occupy at 20°C if its pressure is 17.54 mmHg? By what factor has the water expanded in going from the liquid phase to the gas phase?
6. At 100°C, 1 g of liquid H2O has a volume of 1.043 mL. What volume will 1 g of steam occupy at 100°C if its pressure is 760.0 mmHg? By what factor has the water expanded in going from the liquid phase to the gas phase?
7. Predict whether NaCl or NaI will have the higher melting point. Explain. (Hint: consider the relative strengths of the intermolecular interactions of the two compounds.)
8. Predict whether CH4 or CH3OH will have the lower boiling point. Explain. (Hint: consider the relative strengths of the intermolecular interactions of the two compounds.)
9. A standard automobile tire has a volume of about 3.2 ft3 (where 1 ft3 equals 28.32 L). Tires are typically inflated to an absolute pressure of 45.0 pounds per square inch (psi), where 1 atm equals 14.7 psi. Using this information with the ideal gas law, determine the number of moles of air needed to fill a tire if the air temperature is 18.0°C.
10. Another gas law, Amontons’s law, relates pressure and temperature under conditions of constant amount and volume:
$\mathrm{\dfrac{P_i}{T_i}=\dfrac{P_f}{T_f}}$
If an automobile tire (see Exercise 9) is inflated to 45.0 psi at 18.0°C, what will be its pressure if the operating temperature (i.e., the temperature the tire reaches when the automobile is on the road) is 45.0°C? Assume that the volume and the amount of the gas remain constant.
Answers
1. 31.9 g
2. 1.13 g
3.
1. 11.4 mol
2. 255 L
4.
a. 30.0 L H2 and 20.0 L NH3
b. the mole ratio in the balanced equation is the same as the volume ratio.
$\mathrm{10.0\: L\: N_2\times\dfrac{3\: L\ H_2}{1\: L\ N_2}=30.0\: L\ H_2}$
$\mathrm{10.0\: L\: N_2\times\dfrac{2\: L\ NH_3}{1\: L\ N_2}=20.0\: L\ NH_3}$
1. 57.75 L; an expansion of 57,600 times
6. 1.698 L; an expansion of 1,628 times
1. NaCl; with smaller anions, NaCl likely experiences stronger ionic bonding.
8. CH4 will have the lower boiling point because its intermolecular force (London dispersion force only) is weaker than those in CH3OH. Aside from London dispersion, CH3OH has dipole-dipole and hydrogen bonding.
9. 11.6 mol
10. 49.2 psi | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.E%3A_Solids_Liquids_and_Gases_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
A phase is a certain form of matter that has the same physical properties throughout. Three phases are common: the solid, the liquid, and the gas phase. What determines the phase of a substance? Generally, the strength of the intermolecular interactions determines whether a substance is a solid, liquid, or gas under any particular conditions. Covalent network bonding is a very strong form of intermolecular interaction. Diamond is one example of a substance that has this intermolecular interaction. Ionic interactions, the forces of attraction due to oppositely charged ions, are also relatively strong. Covalent bonds are another type of interaction within molecules, but if the bonds are polar covalent bonds, then the unequal sharing of electrons can cause charge imbalances within molecules that cause interactions between molecules. These molecules are described as polar, and these interactions are called dipole-dipole interactions. A certain rather strong type of dipole-dipole interaction, involving a hydrogen atom, is called hydrogen bonding. On the other hand, equal sharing of electrons forms nonpolar covalent bonds, and the interactions between different molecules is less because the molecules are nonpolar. All substances have very weak dispersion forces (also called London forces) caused by the movement of electrons within the bonds themselves.
In the solid phase, intermolecular interactions are so strong that they hold the individual atoms or molecules in place. In many solids, the regular three-dimensional arrangement of particles makes a crystal. In other solids, the irregular arrangement of particles makes an amorphous solid. In liquids, the intermolecular interactions are strong enough to keep the particles of substance together but not in place. Thus, the particles are free to move over each other but still remain in contact.
In gases, the intermolecular interactions are weak enough that the individual particles are separated from each other in space. The kinetic theory of gases is a collection of statements that describe the fundamental behavior of all gases. Among other properties, gases exert a pressure on their container. Pressure is measured using units like pascal, bar, atmosphere, or mmHg (also called a torr).
There are several simple relationships between the variables used to describe a quantity of gas. These relationships are called gas laws. Boyle’s law relates the pressure and volume of a gas, while Charles’s law relates the volume and absolute temperature of a gas. The combined gas law relates the volume, pressure, and absolute temperature of a gas sample. All of these gas laws allow us to understand the changing conditions of a gas. The ideal gas law relates the pressure, volume, amount, and absolute temperature of a gas under any conditions. These four variables are related to the ideal gas law constant, which is the proportionality constant used to calculate the conditions of a gas. Because the conditions of a gas can change, a set of benchmark conditions called standard temperature and pressure (STP) is defined. Standard temperature is 0ºC, and standard pressure is 1.00 atm. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.S%3A_Solids_Liquids_and_Gases_%28Summary%29.txt |
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