Chemistry Chapter Four:

Making Molecules: Mole-to-Mole Conversions

A balanced chemical equation is a "recipe" for how reactants combine to form products. From our balanced equation for the combustion of octane, for example, we can write the following stoichiometric ratio: 2 mol C8H18(l) : 16 mol CO2(g) We can use this ratio to determine how many moles of CO2 are produced for a given number of moles of C8H18 burned. Suppose that we burn 22.0 moles of C8H18; how many moles of CO2 are produced? We use the ratio from the balanced chemical equation in the same way that we used the ratio from the pizza recipe. This ratio acts as a conversion fac- tor allowing us to convert from the amount in moles of the reactant (C8H18) to the amount in moles of the product (CO2): 22.0 mol C8H18 * 16 mol CO2 2 mol C8H18 = 176 mol CO2 The combustion of 22 moles of C8H18 adds 176 moles of CO2 to the atmosphere.

Making Molecules: Mass-to-Mass Conversion

According to the U.S. Department of Energy, the world burned 3.3 * 1010 barrels of petroleum in 2013, the equivalent of approximately 3.7 * 1015 g of gasoline. We can estimate the mass of CO2 emitted into the atmosphere from burning this much gasoline by using the combustion of 3.7 * 1015 g octane as the representative reaction. This calculation is similar to the one we just did, except that we are given the mass of octane instead of the amount of octane in moles. Consequently, we must first convert the mass (in grams) to the amount (in moles). The general conceptual plan for calculations in which we are given the mass of a reactant or product in a chemical reaction and asked to find the mass of a different reactant or product is Mass A ---> Amount A (in moles) --> Amount B (in moles) ---> Mass B where A and B are two different substances involved in the reaction. We use the molar mass of A to convert from the mass of A to the amount of A (in moles). We use the appropriate ratio from the balanced chemical equation to convert from the amount of A (in moles) to the amount of B (in moles). And finally, we use the molar mass of B to convert from the amount of B (in moles) to the mass of B. To calculate the mass of CO2 emitted upon the combustion of 3.7 * 1015 g of octane, therefore, we use the following conceptual plan: g C8H18 ---> mol C8H18 ---> mol CO2 --> g CO2 We follow the conceptual plan to solve the problem, beginning with g C8H18 and canceling units to arrive at g CO2: 3.7 * 10^15 g C8H18 * 1 mol C8H18/114.22 g C8H18 *16 mol CO2/2 mol C8H18 * 44.01 g CO2/1 mol CO2 = 1.1 * 10^16 g CO2 The world's petroleum combustion produces 1.1 * 1016 g CO2 (1.1 * 1013 kg) per year. In comparison, volcanoes produce about 2.0 * 1011 kg CO2 per year.* In other words, volcanoes emit only 2.0 * 1011 kg 1.1 * 1013 kg * 100% = 1.8% as much CO2 per year as petroleum combustion. The argument that volcanoes emit more carbon dioxide than fossil fuel combustion is blatantly incorrect. Examples 4.1 and 4.2 provide additional practice with stoichiometric calculations.

Gas Evolution Reactions

Aqueous reactions that form a gas when two solutions are mixed are gas-evolution reactions. Some gas-evolution reactions form a gaseous product directly when the cation of one reactant combines with the anion of the other. For example, when sulfuric acid reacts with lithium sulfide, dihydrogen sulfide gas forms: H2SO4(aq) + Li2S(aq) ---> H2S(g) gas + H2SO4(aq) Other gas-evolution reactions form an intermediate product that then decomposes to form a gas. For example, when aqueous hydrochloric acid is mixed with aqueous sodium bicarbonate, the following reaction occurs (FiguRe 4.15▶): HCl(aq) + NaHCO3(aq) ---> H2CO3(aq) + NaCl(aq) ---> H2O(l) + CO2(g) gas + NaCl(aq) The intermediate product, H2CO3, is not stable and decomposes into H2O and gaseous CO2. Other important gas-evolution reactions form H2SO3 or NH4OH as intermediate products: HCl(aq) + NaHSO3(aq) ---> H2SO3(aq) + NaCl(aq) ----> H2O(l) + SO2(g) + NaCl(aq) NH4Cl(aq) + NaOH(aq) ---> NH4OH(aq) + NaCl(aq) ---> H2O(l) + NH3(g) + NaCl(aq)

The Solubility of Ionic Compounds

As we have just discussed, when an ionic compound dissolves in water, the resulting solution contains—not the intact ionic compound itself—but its component ions dissolved in water. However, not all ionic compounds dissolve in water. If we add AgCl to water, for example, it remains solid and appears as a white powder at the bottom of the beaker. In general, we say that a compound is soluble if it dissolves in water and insoluble if it does not. However, these classifications are a bit of an oversimplification. In reality, solubility is a continuum. Compounds exhibit a very wide range of solubilities, and even "insoluble" compounds dissolve to some extent, though usually orders of magnitude less than soluble compounds. For example, silver nitrate is soluble. If we mix solid AgNO3 with water, it dissolves and forms a strong electrolyte solution. Silver chloride, on the other hand, is almost completely insoluble. If we mix solid AgCl with water, virtually all of it remains as a solid within the liquid water. Whether a particular compound is soluble or insoluble depends on several factors. In Section 12.3, we will examine more closely the energy associated with solution formation. For now, we can follow a set of empirical rules that has been inferred from observations of many ionic compounds. Table 4.1 summarizes these solubility rules. For example, the solubility rules state that compounds containing the sodium ion are soluble. That means that compounds such as NaBr, NaNO3, Na2SO4, NaOH, and Na2CO3 dissolve in water to form strong electrolyte solutions. Similarly, the solubility rules state that compounds containing the NO3 - ion are soluble. That means that compounds such as AgNO3, Pb(NO3)2, NaNO3, Ca(NO3)2, and Sr(NO3)2 dissolve in water to form strong electrolyte solutions. Notice that when compounds containing polyatomic ions such as NO3 - dissolve, the polyatomic ions remain as intact units when they dissolve. The solubility rules also state that, with some exceptions, compounds containing the CO3 2- ion are insoluble. Therefore, compounds such as CuCO3, CaCO3, SrCO3, and FeCO3 do not dissolve in water. Note that the solubility rules include many exceptions.

Solution Concentration and Solution Stoichiometry

Chemical reactions in which the reactants are dissolved in water are among the most common and important. The reactions that occur in lakes, streams, and oceans, as well as the reactions that occur in every cell of our bodies, take place in water. A homogeneous mixture of two or more substances—such as salt and water—is a solution. The majority component of a solution is the solvent, and the minority component is the solute. A solution in which water acts as the solvent is an aqueous solution. In this section, we first examine how to quantify the concentration of a solution (the amount of solute relative to solvent) and then turn to applying the principles of stoichiometry, which we discussed in the previous section, to reactions occurring in solution.

Representing Aqueous Reactions: Molecular, Ionic and Complete Ionic Equations:

Consider the following equation for a precipitation reaction: Pb(NO3)2(aq) + 2 KCl(aq) ---> PbCl2(s) + 2 KNO3(aq) This equation is a molecular equation, an equation that shows the complete neutral formulas for each compound in the reaction as if they existed as molecules. However, in actual solu- tions of soluble ionic compounds, dissociated substances are present as ions. Equations for reactions occurring in aqueous solution can be written to better show the dissociated nature of dissolved ionic compounds. For example, we can rewrite this molecular equation as: Pb2+(aq) + 2 NO3 (aq) + 2 K+(aq) + 2 Cl-(aq) ---> PbCl2(s) + 2 K+(aq) + 2 NO3 -(aq) Equations such as this, which list individually all of the ions present as either reactants or products in a chemical reaction, are complete ionic equations. Notice that in the complete ionic equation, some of the ions in solution appear unchanged on both sides of the equation. These ions are called spectator ions because they do not participate in the reaction. Pb2+(aq) + 2 NO3 - (aq) + 2 K+(aq) PbCl2(s) + 2 K(aq) + 2 NO3 - (aq) + 2 Cl- (aq) Spectator ions To simplify the equation, and to show more clearly what is happening, we can omit spectator ions. Pb2+(aq) + 2 Cl-(aq) -->PbCl2(s) Equations such as this one, which show only the species that actually change during the reaction, are net ionic equations. As another example, consider the reaction between HCl(aq) and KOH(aq). HCl(aq) + KOH(aq) S H2O(l) + KCl(aq) Since HCl, KOH, and KCl all exist in solution primarily as independent ions, the com- plete ionic equation is: H+(aq) + Cl-(aq) + K+(aq) + OH-(aq) S H2O(l) + K+(aq) + Cl-(aq) To write the net ionic equation, we remove the spectator ions, those that are unchanged on both sides of the equation. H+(aq) + Cl-m(aq) + K+(aq) + OH- (aq) ---> Cl- H2O(l) + K+(aq) The net ionic equation is H+(aq) + OH-(aq) ---> H2O(l). Summarizing: Aqueous Equations ▶ A molecular equation is a chemical equation showing the complete, neutral formulas for every compound in a reaction. ▶ A complete ionic equation is a chemical equation showing all of the species as they are actually present in solution. ▶ A net ionic equation is an equation showing only the species that actually change during the reaction.

Types of Aqueous Solutions and Solubility

Consider two familiar aqueous solutions: salt water and sugar water. Salt water is a homogeneous mixture of NaCl and H2O, and sugar water is a homogeneous mixture of C12H22O11 and H2O. You may have made these solutions yourself by adding solid table salt or solid sugar to water. As you stir either of these two substances into the water, it seems to disappear. However, you know that the original substance is still present because the mixture tastes salty or sweet. How do solids such as salt and sugar dissolve in water? When a solid is put into a liquid solvent, the attractive forces that hold the solid together (the solute-solute interactions) compete with the attractive forces between the solvent molecules and the particles that compose the solid (the solvent-solute interactions), as shown in FiguRe 4.6◀. For example, when sodium chloride is put into water, there is a competition between the attraction of Na+ cations and Cl- anions to each other (due to their opposite charges) and the attraction of Na+ and Cl- to water molecules. The attraction of Na+ and Cl- to water is based on the polar nature of the water molecule. For reasons we discuss later in this book (Section 9.6), the oxygen atom in water is electron rich, giving it a partial negative charge (d -), as shown in FiguRe 4.7◀. The hydrogen atoms, in contrast, are electron-poor, giving them a partial positive charge (d +). As a result, the positively charged sodium ions are strongly attracted to the oxygen side of the water molecule (which has a partial negative charge), and the negatively charged chloride ions are attracted to the hydrogen side of the water molecule (which has a partial positive charge), as shown in FiguRe 4.8▼. In the case of NaCl, the attraction between the separated ions and the water molecules overcomes the attraction of sodium and chloride ions to each other, and the sodium chloride dissolves in the water (FiguRe 4.9▶).

Precipitation Reactions

Have you ever taken a bath in hard water? Hard water contains dissolved ions such as Ca2+ and Mg2+ that diminish the effectiveness of soap. These ions react with soap to form a gray curd that may appear as a "bathtub ring" after you drain the tub. Hard water is particularly troublesome when washing clothes. Imagine how a white shirt would look covered with the gray curd from the bathtub. Consequently, most laundry detergents include substances designed to remove Ca2+ and Mg2+ from the laundry mixture. The most common substance used for this purpose is sodium carbonate, which dissolves in water to form sodium cations (Na+) and carbonate (CO3 2-) anions. Na2CO3(aq) S 2 Na+(aq) + CO3 2-(aq) Sodium carbonate is soluble, but calcium carbonate and magnesium carbonate are not (see the solubility rules in Table 4.1). Consequently, the carbonate anions react with dissolved Mg2+ and Ca2+ ions in hard water to form solids that precipitate from (come out of) solution: Mg2+(aq) + CO3 ---> 2-(aq) ---> MgCO3(s) Ca2+(aq) + CO3 (aq) ---?.CaCO3(s) The precipitation of these ions prevents their reaction with the soap, eliminating curd and preventing white shirts from turning gray. The reactions between CO3 2- and Mg2+ and Ca2+ are examples of precipitation reactions, reactions in which a solid or precipitate forms upon mixing two solutions. Precipitation reactions are common in chemistry. As another example, consider potassium iodide and lead(II) nitrate. Each of these compounds forms colorless, strong electrolyte solutions when dissolved in water. When the two solutions are combined, however, a brilliant yellow precipitate forms (FiguRe 4.12▲). We can describe this precipitation reaction with the chemical equation: 2 KI(aq) + Pb(NO3)2(aq) ---> 2 KNO3(aq) + PbI2(s) Precipitation reactions do not always occur when two aqueous solutions are mixed. For example, if we combine solutions of KI(aq) and NaCl(aq), nothing happens. KI(aq) + NaCl(aq) S NO The key to predicting precipitation reactions is to understand that only insoluble compounds form precipitates. In a precipitation reaction, two solutions containing soluble compounds combine and an insoluble compound precipitates. For example, consider the precipitation reaction described previously: Two new compounds—one or both of which might be insoluble—are now possible. The cation from each compound can pair with the anion from the other compound to form possibly insoluble products (we will learn more about why this happens in Chapter 12). If the possible products are both soluble, then no reaction occurs. If one or both of the possible products are insoluble, a precipitation reaction occurs. In this case, KNO3 is soluble, but PbI2 is insoluble. Consequently, PbI2 precipitates as shown here: To predict whether a precipitation reaction will occur when two solutions are mixed and to write an equation for the reaction, use the procedure that follows. The steps are outlined in the left column, and examples of applying the procedure are shown in the center and right columns in Examples 4.10 and 4.11.

Oxidation States

Identifying a reaction between a metal and a nonmetal as a redox reaction is fairly straightforward because the metal becomes a cation and the nonmetal becomes an anion (electron transfer is obvious). However, how do we identify redox reactions that occur between nonmetals? Chemists have devised a scheme to track electrons before and after a chemical reaction. In this scheme—which is like bookkeeping for electrons—all shared electrons are assigned to the atom that attracts the electrons most strongly. Then a number, called the oxidation state or oxidation number, is given to each atom based on the electron assignments. In other words, the oxidation number of an atom in a compound is the "charge" it would have if all shared electrons were assigned to the atom with a greater attraction for those electrons. For example, consider HCl. Since chlorine attracts electrons more strongly than hydrogen, we assign the two shared electrons in the bond to chlorine; then H (which has lost an electron in our assignment) has an oxidation state of +1, and Cl (which has gained one electron in our assignment) has an oxidation state of -1. Notice that, in contrast to ionic charges, which are usually written with the sign of the charge after the magnitude (1+ and 1-, for example), oxidation states are written with the sign of the charge before the magnitude (+1 and -1, for example). Use the following rules to assign oxidation states to atoms in elements and compounds. When assigning oxidation states, keep these points in mind: The oxidation state of any given element generally depends on what other elements are present in the compound. (The exceptions are the group 1A and 2A metals, which are always +1 and +2, respectively.) Rule 3 must always be followed. Therefore, when following the hierarchy shown in rule 5, give priority to the element(s) highest on the list and then assign the oxidation state of the element lowest on the list using rule 3. When assigning oxidation states to elements that are not covered by rules 4 and 5 (such as carbon) use rule 3 to deduce their oxidation state once all other oxidation states have been assigned. In most cases, oxidation states are positive or negative integers; however, on occasion an atom within a compound can have a fractional oxidation state. For example, consider KO2. The oxidation state is assigned as follows: In KO2, oxygen has a -1, 2 oxidation state. Although this seems unusual, it is accepted because oxidation states are merely an imposed electron bookkeeping scheme, not an actual physical quantity.

Solution Stoichiometry

In Section 4.2 we discussed how we can use the coefficients in chemical equations as conversion factors between the amounts of reactants (in moles) and the amounts of products (in moles). In reactions involving aqueous reactants and products, it is often convenient to specify quantities in terms of volumes and concentrations. We can then use these quantities to calculate the amounts in moles of reactants or products and use the stoichiometric coefficients to convert these to amounts of other reactants or products. The general conceptual plan for these kinds of calculations is: Volume A ----> Amount A (in moles) ---> Amount B (in moles) ---> (Volume B) We use molarity to convert between solution volumes and amount of solute, and we use the stoichiometric coefficients from the balanced chemical equation to convert between amounts of A and B. Example 4.8 demonstrates solution stoichiometry.

Limiting Reactant, Theoretical Yield, Percent Yield

Let's return to our pizza analogy to understand three more concepts important in reaction stoichiometry: limiting reactant, theoretical yield, and percent yield. Recall the pizza recipe from Section 4.2: 1 crust + 5 ounces tomato sauce + 2 cups cheese S 1 pizza Suppose that we have 4 crusts, 10 cups of cheese, and 15 ounces of tomato sauce. How many pizzas can we make? We have enough crusts to make: 4 crusts * 1 Pizza/1 crust = 4 pizza 10 cups cheese x 1 pizza/ 2 cups cheese = 5 pizzas We have enough crusts for 4 pizzas, enough cheese for 5 pizzas, but enough tomato sauce for only 3 pizzas. Consequently, unless we get more ingredients, we can make only 3 pizzas. The tomato sauce limits how many pizzas we can make. If the pizza recipe were a chemical reaction, the tomato sauce would be the limiting reactant, the reactant that limits the amount of product in a chemical reaction. Notice that the limiting reactant is simply the reactant that makes the least amount of product. Reactants that do not limit the amount of product—such as the crusts and the cheese in this example—are said to be in excess. If this were a chemical reaction, 3 pizzas would be the theoretical yield, the amount of product that can be made in a chemical reaction based on the amount of limiting reactant. Let us carry this analogy one step further. Suppose we go on to cook our pizzas and accidentally burn one of them. Even though we theoretically have enough ingredients for 3 pizzas, we end up with only 2. If this were a chemical reaction, the 2 pizzas would be our actual yield, the amount of product actually produced by a chemical reaction. (The actual yield is always equal to or less than the theoretical yield because at least a small amount of product is usually lost to other reactions or does not form during a reaction.) Finally, we calculate our percent yield, the percentage of the theoretical yield that was actually attained, as follows: Percent Yield = Actual Yield/Theoretical Yield x 100% Summarizing: Limiting Reactant and Yield ▶ The limiting reactant (or limiting reagent) is the reactant that is completely consumed in a chemical reaction and limits the amount of product. ▶ The reactant in excess is any reactant that occurs in a quantity greater than is required to completely react with the limiting reactant. ▶ The theoretical yield is the amount of product that can be made in a chemical reaction based on the amount of limiting reactant. ▶ The actual yield is the amount of product actually produced by a chemical reaction. ▶ The percent yield is calculated as actual yield/theoretical yield * 100%. We can apply these concepts to a chemical reaction. Recall from Section 3.10 our balanced equation for the combustion of methane: If we start out with 5 CH4 molecules and 8 O2 molecules, what is our limiting reactant? What is our theoretical yield of carbon dioxide molecules? We first calculate the number of CO2 molecules that can be made from 5 CH4 molecules:

Oxidation-reduction reactions or redox reactions

Oxidation-reduction reactions or redox reactions are reactions in which electrons are transferred from one reactant to the other. The rusting of iron, the bleaching of hair, and the production of electricity in batteries involve redox reactions. Many redox reactions involve the reaction of a substance with oxygen (FiguRe 4.16▼): 4 Fe(s) + 3 O2(g) ---> 2 Fe2O3(s) (rusting of iron) 2 C8H18(l) + 25 O2(g) ---> 16 CO2(g) + 18 H2O(g)(combustion of octane) 2 H2(g) + O2(g) ---> 2 H2O(g) (combustion of hydrogen) However, redox reactions need not involve oxygen. Consider, for example, the reac- tion between sodium and chlorine to form sodium chloride (NaCl), depicted in FiguRe 4.17▲. 2 Na(s) + Cl2(g) S 2 NaCl(s) This reaction is similar to the reaction between sodium and oxygen to form sodium oxide. 4 Na(s) + O2(g) S 2 Na2O(s) In both cases, a metal (which has a tendency to lose electrons) reacts with a nonmetal (which has a tendency to gain electrons). In both cases, metal atoms lose electrons to nonmetal atoms. A fundamental definition of oxidation is the loss of electrons, and a fundamental definition of reduction is the gain of electrons. The transfer of electrons, however, need not be a complete transfer (as occurs in the formation of an ionic compound) for the reaction to qualify as oxidation-reduction. For example, consider the reaction between hydrogen gas and chlorine gas: H2(g) + Cl2(g) S 2 HCl(g) Even though hydrogen monochloride is a molecular compound with a covalent bond, and even though the hydrogen has not completely transferred its electron to chlorine during the reaction, you can see from the electron density diagrams (FiguRe 4.18◀) that hydrogen has lost some of its electron density—it has partially transferred its electron to chlorine. Therefore, in this reaction, hydrogen is oxidized and chlorine is reduced and the reaction is a redox reaction.

Reaction Stoichiometry: How much Carbon Dioxide

The amount of carbon dioxide emitted by fossil fuel combustion is related to the amount of fossil fuel that is burned—the balanced chemical equations for the combustion reactions give the exact relationships between these amounts. In this discussion, we use octane (a component of gasoline) as a representative fossil fuel. The balanced equation for the combustion of octane is: 2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) The balanced chemical equation shows that 16 CO2 molecules are produced for every 2 molecules of octane burned. We can extend this numerical relationship between molecules to the amounts in moles as follows: The coefficients in a chemical reaction specify the relative amounts in moles of each of the substances involved in the reaction. In other words, from the equation we know that 16 moles of CO2 are produced for every 2 moles of octane burned. The numerical relationship between chemical amounts in a balanced chemical equation is reaction stoichiometry. Stoichiometry allows us to predict the amounts of products that will form in a chemical reaction based on the amounts of reactants that undergo the reaction. Stoichiometry also allows us to determine the amount of reactants necessary to form a given amount of product. These calculations are central to chemistry, allowing chemists to plan and carry out chemical reactions to obtain prod- ucts in the desired quantities.

Solution Concentration

The amount of solute in a solution is variable. For example, we can add just a little salt to water to make a dilute solution, one that contains a small amount of solute relative to the solvent, or we can add a lot of salt to water to make a concentrated solution, one that contains a large amount of solute relative to the solvent. A common way to express solu- tion concentration is molarity (M), the amount of solute (in moles) divided by the vol- ume of solution (in liters). Molarity (M) = amount of solute (in mol)/volume of solution (in L) Note that molarity is a ratio of the amount of solute per liter of solution, not per liter of solvent. To make an aqueous solution of a specified molarity, we usually put the solute into a flask and then add water until we have the desired volume of solution. For example, to make 1 L of a 1 M NaCl solution, we add 1 mol of NaCl to a flask and then add water to make 1 L of solution (FiguRe 4.4▼). We do not combine 1 mol of NaCl with 1 L of water because the resulting solution would have a total volume exceeding 1 L and therefore a molarity of less than 1 M. To calculate molarity, we divide the amount of the solute in moles by the volume of the solution (solute and solvent) in liters, as shown in Example 4.5. The molarity of a solution can be used as a conversion factor between moles of the solute and liters of the solution. For example, a 0.500 M NaCl solution contains 0.500 mol NaCl for every liter of solution. L solution mol NaCl 0.500 mol NaCl converts L solution This conversion factor converts from L solution to mol NaCl. If you want to convert in the other direction, invert the conversion factor. mol NaCl L solution L solution converts 0.500 mol NaCl Example 4.6 illustrates how to use molarity in this way. To save space in laboratory storerooms, solutions are often stored in concentrated forms called stock solutions. For example, hydrochloric acid is often stored as a 12 M stock solution. However, many lab procedures call for hydrochloric acid solutions that are much less concentrated, so we must dilute the stock solution to the required concentration. How do we know how much of the stock solution to use? The easiest way to solve dilution problems is to use the following dilution equation: M₁V₁ = M₂V₂ where M1 and V1 are the molarity and volume of the initial concentrated solution, and M2 and V2 are the molarity and volume of the final diluted solution. This equation works because the molarity multiplied by the volume gives the number of moles of solute, which is the same in both solutions. M1V1 = M2V2 mol1 = mol2 In other words, the number of moles of solute does not change when we dilute a solution. For example, suppose a laboratory procedure calls for 3.00 L of a 0.500 M CaCl2 solution. How should we prepare this solution from a 10.0 M stock solution? We can solve Equation 4.1 for V1, the volume of the stock solution required for the dilution, and substitute in the correct values to calculate it. M1V1 = M2V2 V1 = M2V2 M1 = 0.500 mol/L * 3.00 L 10.0 mol/L = 0.150 L We make the solution by adding enough water to 0.150 L of the stock solution to create a total volume of 3.00 L (V2). The resulting solution will be 0.500 M in CaCl2

Making Pizza: The Relationships Among Ingredients

The concepts of stoichiometry are similar to those in a cooking recipe. Calculating the amount of carbon dioxide produced by the combustion of a given amount of a fossil fuel is analogous to calculating the number of pizzas that can be made with a given amount of cheese. For example, suppose we use the following pizza recipe: 1 crust + 5 ounces tomato sauce + 2 cups cheese --> 1 pizza The recipe shows the numerical relationships between the pizza ingredients. It says that if we have 2 cups of cheese—and enough of everything else—we can make 1 pizza. We can write this relationship as a ratio between the cheese and the pizza. 2 cups cheese : 1 pizza What if we have 6 cups of cheese? Assuming that we have enough of everything else, we can use the above ratio as a conversion factor to calculate the number of pizzas. 6 cups cheese * 1 pizza 2 cups cheese = 3 pizzas Six cups of cheese are sufficient to make 3 pizzas. The pizza recipe contains numerical ratios between other ingredients as well, including the following: 1 crust : 1 pizza 5 ounces tomato sauce : 1 pizza

Electrolyte and Nonelectrolyte Solutions

The difference in the way that salt (an ionic compound) and sugar (a molecular com- pound) dissolve in water illustrates a fundamental difference between types of solutions. As FiguRe 4.10▼ shows, a salt solution conducts electricity, while a sugar solution does not. As we just saw with sodium chloride, ionic compounds dissociate into their compo- nent ions when they dissolve in water. An NaCl solution, represented as NaCl(aq), does not contain NaCl units, only Na+ ions and Cl- ions. The dissolved ions act as charge carriers, allowing the solution to conduct electricity. Substances that dissolve in water to form solutions that conduct electricity are electrolytes. Substances such as sodium chloride that completely dissociate into ions when they dissolve in water are strong electrolytes, and the resulting solutions are strong electrolyte solutions. In contrast to sodium chloride, sugar is a molecular compound. Most molecular compounds—with the important exception of acids, which we discuss in the next paragraph—dissolve in water as intact molecules. Sugar dissolves because the attraction between sugar molecules and water molecules—both of which contain a distribution of electrons that results in partial positive and partial negative charges—overcomes the attraction of sugar molecules to each other (FiguRe 4.11▲). However, in contrast to a sodium chloride solution (which is composed of dissociated ions), a sugar solution is composed of intact C12H22O11 molecules homogeneously mixed with the water molecules. Compounds such as sugar that do not dissociate into ions when dissolved in water are nonelectrolytes, and the resulting solutions nonelectrolyte solutions—do not conduct electricity. Acids, which we first discussed in Section 3.6, are molecular compounds that ionize—form ions—when they dissolve in water. For example, HCl is a molecular compound that ionizes into H+ and Cl- when it dissolves in water. Hydrochloric acid is an example of a strong acid, one that completely ionizes in solution. Since they completely ionize in solution, strong acids are also strong electrolytes. We represent the complete ionization of a strong acid with a single reaction arrow between the acid and its ionized form: HCl(aq) ---> H+(aq) + Cl-(aq) Many acids are weak acids; they do not completely ionize in water. For example, acetic acid (HC2H3O2), the acid in vinegar, is a weak acid. A solution of a weak acid is composed mostly of the nonionized acid—only a small percentage of the acid molecules ionize. We represent the partial ionization of a weak acid with opposing half arrows between the reactants and products: HC₂H₃O₂(aq) ↔ H⁺ (aq) + C₂H₃O₂ -(aq) Weak acids are weak electrolytes, and the resulting solutions—weak electrolyte solutions—conduct electricity only weakly.

Climate Change and the Combustion of Fossil Fuels

The temperature outside my office today is a cool 48 °F, lower than normal for this time of year on the West Coast. However, today's "chill" pales in comparison with how cold it would be without the presence of greenhouse gases in the atmosphere. These gases act like the glass of a greenhouse, allowing sunlight to enter the atmosphere and warm Earth's surface, but preventing some of the heat generated by the sunlight from escaping, as shown in 4.1▲. The balance between incoming and outgoing energy from the sun determines Earth's average temperature. Without greenhouse gases in the atmosphere, more heat energy would escape, and Earth's average temperature would be about 60 °F colder than it is now. The temperature outside of my office today would be below 0 °F, and even the sunniest U.S. cities would most likely be covered with snow. However, if the concentration of greenhouse gases in the atmosphere were to increase, Earth's average temperature would rise. In recent years scientists have become increasingly concerned because the amount of atmospheric carbon dioxide (CO2)—Earth's most significant greenhouse gas in terms of climate change—is rising. More CO2 enhances the atmosphere's ability to hold heat and may therefore lead to global warming, an increase in Earth's average temperature. Since 1860, atmospheric CO2 levels have risen by 38% (FiguRe 4.2▼), and Earth's average temperature has risen by 0.8 °C (about 1.4 °F), as shown in FiguRe 4.3▶. Most scientists now believe that the primary cause of rising atmospheric CO2 concentration is the burning of fossil fuels (natural gas, petroleum, and coal), which provides 90% of our society's energy. Some people, however, have suggested that fossil fuel combustion does not significantly contribute to global warming and climate change. They argue, for example, that the amount of carbon dioxide emitted into the atmosphere by volcanic eruptions far exceeds that from fossil fuel combustion. Which group is right? To judge the validity of the naysayers' argument, we need to calculate the amount of carbon dioxide emitted by fossil fuel combustion and compare it to that released by volcanic eruptions. As you will see in the next section of the chapter, at this point in your study of chemistry, you have enough knowledge to do just that.

Acid-Base Reactions

Two other important classes of reactions that occur in aqueous solution are acid-base reactions and gas-evolution reactions. In an acid-base reaction (also called a neutralization reaction), an acid reacts with a base and the two neutralize each other, producing water (or in some cases a weak electrolyte). In a gas-evolution reaction, a gas forms, resulting in bubbling. In both cases, as in precipitation reactions, the reactions occur when the anion from one reactant combines with the cation of the other. In addition, many gas-evolution reactions are also acid-base reactions. Our stomachs contain hydrochloric acid, which acts in the digestion of food. Certain foods or stress, however, can increase the stomach's acidity to uncomfortable levels, causing acid stomach or heartburn. Antacids are over-the-counter medicines that work by reacting with, and neutralizing, stomach acid. Antacids employ different bases—substances that produce hydroxide (OH-) ions in water—as neutralizing agents. Milk of magnesia, for example, contains Mg(OH)2 and Mylanta contains Al(OH)3. All antacids, no matter what base they contain, have the same effect of neutralizing stomach acid through acid-base reactions and relieving heartburn. Recall from Chapter 3 that an acid forms H+ ions in solution, and we just saw that a base is a substance that produces OH- ions in solution: Acid Substance that produces H+ ions in aqueous solution Base Substance that produces OH- ions in aqueous solution These definitions of acids and bases are called the Arrhenius definitions, after Swedish chemist Svante Arrhenius (1859-1927). In Chapter 15, we will learn more general definitions of acid-base behavior, but these are sufficient to describe neutralization reactions. According to the Arrhenius definition, HCl is an acid because it produces H+ ions in solution: HCl(aq) ---> H+(aq) + Cl-(aq) An H+ ion is a bare proton. Protons associate with water molecules in solution to form hydronium ions (FiguRe 4.13◀): H+(aq) + H2O(l) ---> H3O+(aq) Chemists use H+(aq) and H3O+(aq) interchangeably to mean the same thing—an H+ ion dissolved in water. The ionization of HCl and other acids is often written to show the association of the proton with a water molecule to form the hydronium ion. HCl(aq) + H2O(l) ---> H3O+(aq) + Cl-(aq) As we discussed in Section 4.6, some acids are weak acids—they do not completely ionize in solution. We represent the partial ionization of a weak acid with opposing half arrows. HC2H3O2(aq) L H+(aq) + C2H3O2 -(aq) Some acids—called polyprotic acids—contain more than one ionizable proton and release them sequentially. For example, sulfuric acid, H2SO4, is a diprotic acid. It is strong in its first ionizable proton, but weak in its second. H2SO4(aq) ---> H+(aq) + HSO4 -(aq) HSO4 -(aq) L H+(aq) + SO4 2-(aq) According to the Arrhenius definition, NaOH is a base because it produces OH- ions in solution. NaOH(aq) ---> Na+(aq) + OH-(aq) In analogy to diprotic acids, some bases, such as Sr(OH)2, for example, produce two moles of OH- per mole of the base. Sr(OH)2(aq) --> Sr2+(aq) + 2 OH-(aq) When an acid and a base are mixed, the H+(aq) from the acid—whether it is weak or strong—combines with the OH-(aq) from the base to form H2O(l) (FiguRe 4.14▼). For example, consider the reaction between hydrochloric acid and sodium hydroxide: HCl(aq) + NaOH( aq) ----> H2O(l) NaCl(aq) Acid-base reactions generally form water and an ionic compound—called a salt— that usually remains dissolved in the solution. The net ionic equation for acid-base reactions involving a strong acid is H+(aq) + OH-(aq) --->H2O(l) However, if the acid is a weak acid, the net ionic equation is slightly different. For example, consider the acid-base equation between hydrofluoric acid and sodium hydroxide: HF(aq) acid + NaOH(aq) base ---> H2O(l) water + NaF(aq) salt The complete ionic equation and net ionic equation for this reaction are: HF(aq) + Na+(aq) + OH-(aq) ---> H2O(l) + Na+(aq) + F-(aq) (Complete ionic eqution) HF(aq) + OH-(aq) --->H2O(l) + F -(aq) (Net ionic equation) Notice that, since HF is a weak acid, it is not shown as being ionized in the ionic equations. Another example of an acid-base reaction is the reaction between sulfuric acid and potassium hydroxide: H2SO4(aq) acid+ 2 KOH(aq) base ---> 2 H2O(l) water + K2SO4(aq) salt Again, notice the pattern of acid and base reacting to form water and a salt. Acid ∙ Base S Water ∙ Salt (acid-base reactions) When writing equations for acid-base reactions, write the formula of the salt using the procedure for writing formulas of ionic compounds from Section 3.5.

Identifying Redox Reactions

We can use oxidation states to identify redox reactions, even between nonmetals. For example, is the following reaction between carbon and sulfur a redox reaction? C + 2 S --> CS2 If so, what element is oxidized? What element is reduced? We can use the oxidation state rules to assign oxidation states to all elements on both sides of the equation. C + 2 ---> CS2 Oxidation states: 0 - 0 +4 2 Oxidation Reduction Carbon changed from an oxidation state of 0 to an oxidation state of +4. In terms of our electron bookkeeping scheme (the assigned oxidation state), carbon lost electrons and was oxidized. Sulfur changed from an oxidation state of 0 to an oxidation state of -2. In terms of our electron bookkeeping scheme, sulfur gained electrons and was reduced. In terms of oxidation states, oxidation and reduction are defined as follows: Oxidation: An increase in oxidation state Reduction: A decrease in oxidation state Oxygen, for example, is an excellent oxidizing agent because it causes the oxidation of many other substances. In a redox reaction, the oxidizing agent is always reduced. A substance that causes the reduction of another substance is a reducing agent. Hydrogen, for example, as well as the group 1A and group 2A metals (because of their tendency to lose electrons) are excellent reducing agents. In a redox reaction, the reducing agent is always oxidized. In Section 18.2 you will learn more about redox reactions, including how to balance them. For now, be able to identify redox reactions, as well as oxidizing and reducing agents, according to these guidelines. Redox reactions include:Any reaction in which there is a change in the oxidation states of atoms between the reactants and the products. In a redox reaction: The oxidizing agent oxidizes another substance (and is itself reduced). The reducing agent reduces another substance (and is itself oxidized).

Combustion Reactions

We encountered combustion reactions, a type of redox reactions, in the opening section of this chapter. Combustion reactions are important because most of our society's energy is derived from them (FiguRe 4.19▶). Combustion reactions are characterized by the reaction of a substance with O2 to form one or more oxygen-containing compounds, often including water. Combustion reactions also emit heat. For example, as we saw earlier in this chapter, natural gas (CH4) reacts with oxygen to form carbon dioxide and water. CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(g) Oxidation state: -4 +1 0 +4 -2 +1 -2 In this reaction, carbon is oxidized and oxygen is reduced. Ethanol, the alcohol in alcoholic beverages, also reacts with oxygen in a combustion reaction to form carbon dioxide and water. C2H5OH(l) + 3 O2(g) ---> 2 CO2(g) + 3 H2O(g) Compounds containing carbon and hydrogen—or carbon, hydrogen, and oxygen—always form carbon dioxide and water upon complete combustion. Other combustion reactions include the reaction of carbon with oxygen to form carbon dioxide C(s) + O2(g) S CO2(g) and the reaction of hydrogen with oxygen to form water. 2 H2(g) + O2(g) S 2 H2O(g)

Limiting Reactant, Theoretical Yield, Percent Yield from Initial Reactant Masses

When working in the laboratory, we normally measure the initial quantities of reactants in grams, not in number of molecules. To find the limiting reactant and theoretical yield from initial masses, we must first convert the masses to amounts in moles. Consider, for example, the following reaction: 2 Mg(s) + O2(g) ---> 2 MgO(s) If we have 42.5 g Mg and 33.8 g O2, what is the limiting reactant and theoretical yield? To solve this problem, we must determine which of the reactants makes the least amount of product. Conceptual Plan We find the limiting reactant by calculating how much product can be made from each reactant. Since we are given the initial quantities in grams, and stoichiometric relation- ships are between moles, we must first convert to moles. We then convert from moles of the reactant to moles of product. The reactant that makes the least amount of product is the limiting reactant. The conceptual plan is: In this conceptual plan, we compare the number of moles of MgO made by each reac- tant and convert only the smaller amount to grams. (Alternatively, we can convert both quantities to grams and determine the limiting reactant based on the mass of the product.) Relationships Used molar mass Mg = 24.31 g>mol molar mass O2 = 32.00 g>mol 2 mol Mg : 2 mol MgO 1 mol O2 : 2 mol MgO molar mass MgO = 40.31 g>mol Solution Beginning with the masses of each reactant, we follow the conceptual plan to calculate how much product can be made from each. Since Mg makes the least amount of product, it is the limiting reactant and O2 is in excess. Notice that the limiting reactant is not necessarily the reactant with the least mass. In this case, the mass of O2 is less than the mass of Mg, yet Mg is the limiting reactant because it makes the least amount of MgO. The theoretical yield is therefore 70.5 g of MgO, the mass of product possible based on the limiting reactant. Now suppose that when the synthesis is carried out, the actual yield of MgO is 55.9 g. What is the percent yield? We calculate the percent yield as follows: