Chem 101 Chapter 4- Chemical Equations and Stoichiometry

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Unformatted text preview: The Basic Tools of Chemistry 4— Chemical Equations and Stoichiometry Black Smokers and the Origin of Life “The origin of life appears almost a miracle, so many are the conditions which would have had to be satisfied to get it going.” National Oceanic and Atmospheric Administration/Department of Commerce Francis Crick, quoted by John Horgan, “In the Beginning,” Scientific American, pp. 116–125, February 1991. A “black smoker” in the East Pacific Rise. 140 The statement by Francis Crick on the origin of life does not mean that chemists and biologists have not tried to find the conditions under which life might have begun. Charles Darwin thought life might have begun when simple molecules combined to produce molecules of greater and greater complexity. Darwin‘s idea lives on in experiments such as those done by Stanley Miller in 1953. Attempting to recreate what was thought to be the atmosphere of the primeval earth, Miller filled a flask with the gases methane, ammonia, and hydrogen and added a bit of water. A discharge of electricity acted like lightning in the mixture. The inside of the flask was soon covered with a reddish slime, a mixture found to contain amino acids, the building blocks of proteins. Chemists thought they would soon know in more detail how living organisms began their development—but it was not to be. As Miller said recently, “The problem of the origin of life has turned out to be much more difficult than I, and most other people, envisioned.” Other theories have been advanced to account for the origin of life. The most recent conjecture relates to the discovery of geologically active sites on the ocean floor. Could life have originated in such exotic environments? The evidence is tenuous. As in Miller‘s experiments, this hypothesis relies on the creation of complex carbon-based molecules from simple ones. In 1977 scientists were exploring the junction of two of the tectonic plates that form the floor of the Pacific Ocean. There they Chapter Goals See Chapter Goals Revisited (page 165). Test your knowledge of these goals by taking the exam-prep quiz on the General ChemistryNow CD-ROM or website. • Balance equations for simple chemical reactions. • Perform stoichiometry calculations using balanced chemical equations. • Understand the meaning of a limiting reactant. • Calculate the theoretical and percent yields of a chemical reaction. Chapter Outline 4.1 Chemical Equations 4.2 Balancing Chemical Equations 4.3 Mass Relationships in Chemical Reactions: Stoichiometry 4.4 Reactions in Which One Reactant Is Present in Limited Supply 4.5 Percent Yield 4.6 Chemical Equations and Chemical Analysis • Use stoichiometry to analyze a mixture of compounds or to determine the formula of a compound. National Oceanic and Atmospheric Administration/Department of Commerce the vents have been called “black smokers.” The solid sulfides settle around the edges of the vent on the sea floor, eventually forming a “chimney” of precipitated minerals. Scientists were amazed to find that the black smoker vents were surrounded by primitive animals living in the hot, sulfide-rich environment. Because smokers lie under hundreds of meters of water and sunlight does not penetrate to these depths, the animals have developed a way to live without energy from sunlight. It is currently believed that they derive the energy needed to survive from the reaction of oxygen with hydrogen sulfide, H2S: H2S(aq) ϩ 2 O2(aq) ¡ H2SO4(aq) ϩ energy Black smoker chimney and shrimp on the Mid-Atlantic Ridge. found thermal springs gushing a hot, black soup of minerals. Water seeping into cracks in the thin surface of the earth is superheated to between 300 and 400 °C by the magma of the earth‘s core. This superhot water dissolves minerals in the crust and provides conditions for the conversion of sulfate ions in sea water to hydrogen sulfide, H2S. When this hot water, now laden with dissolved minerals and rich in sulfides, gushes through the surface, it cools. Metal sulfides, such as those of copper, manganese, iron, zinc, and nickel, then precipitate. Many metal sulfides are black, and the plume of material coming from the sea bottom looks like black “smoke”; for this reason, The hypothesis that life might have originated in this inhospitable location developed out of laboratory experiments by a German lawyer and scientist, G. Wächtershäuser and a colleague, Claudia Huber. They found that metal sulfides such as iron sulfide promote reactions that convert simple carbon-containing molecules to more complex molecules. If this transformation could happen in the laboratory, perhaps similar chemistry might also occur in the exotic environment of black smokers. 141 142 Chapter 4 Chemical Equations and Stoichiometry To Review Before You Begin • Review names and formulas of common compounds and ions (Chapter 3) • Know how to convert mass to moles and moles to mass (Chapters 2 and 3) hen you think about chemistry, you probably think of chemical reactions. The image of a medieval chemist mixing chemicals in hopes of turning lead into gold lingers in the imagination. Of course, there is much more to chemistry. Just reading this sentence involves an untold number of chemical reactions in your body. Indeed, every activity of living things depends on carefully regulated chemical reactions. Our objective in this chapter is to introduce the quantitative study of chemical reactions. Quantitative studies are needed to determine, for example, how much oxygen is required for the complete combustion of a given quantity of gasoline and what mass of carbon dioxide and water can be obtained. This part of chemistry is fundamental to much of what chemists, chemical engineers, biochemists, molecular biologists, geochemists, and many others do. W • • • • Throughout the chapter this icon introduces a list of resources on the General ChemistryNow CD-ROM or website (http://now .brookscole.com/kotz6e) that will: help you evaluate your knowledge of the material provide homework problems allow you to take an exam-prep quiz provide a personalized Learning Plan targeting resources that address areas you should study 4.1—Chemical Equations When a stream of chlorine gas, Cl2, is directed onto solid phosphorus, P4, the mixture bursts into flame, and a chemical reaction produces liquid phosphorus trichloride, PCl3 (Figure 4.1). We can depict this reaction using a balanced chemical equation. P4(s) ϩ 6 Cl2(g) ¡ 4 PCl3(ᐉ) Reactants In a balanced equation, the formulas for the reactants (the substances combined in the reaction) are written to the left of the arrow and the formulas for the products (the substances produced) are written to the right of the arrow. The physical states of reactants and products can also be indicated. The symbol (s) indicates a solid, (g) a gas, and (/) a liquid. A substance dissolved in water—that is, an aqueous solution of a substance—is indicated by (aq). The relative amounts of the reactants and products are shown by numbers, the coefficients, before the formulas. Photos: Charles D. Winters ■ Information from Chemical Equations Chemical equations show the compounds involved in the chemical reaction and their physical state. Equations usually do not show the conditions of the experiment or indicate whether any energy (in the form of heat or light) is involved. Products P4(s) ϩ 6 Cl2(g) Reactants ¡ 4 PCl3(ᐉ) Products Figure 4.1 Reaction of solid white phosphorus with chlorine gas. The product is liquid phosphorus trichloride. 4.1 Chemical Equations Historical Perspectives Antoine Laurent Lavoisier (1743–1794) On Monday, August 7, 1774, the Englishman Joseph Priestley (1733–1804) became the first person to isolate oxygen. He heated solid mercury(II) oxide, HgO, causing the oxide to decompose to mercury and oxygen. Image not available due to copyright restrictions 2 HgO(s) ¡ 2 Hg(/) ϩ O2(g) Priestley did not immediately understand the significance of his discovery, but he mentioned it to the French chemist Antoine Lavoisier in October 1774. One of Lavoisier‘s contributions to science was his recognition of the importance of exact scientific measurements and of carefully planned experiments, and he applied these methods to the study of oxygen. From this work he came to believe Priestley‘s gas was present in all acids, so he named it “oxygen,” from the Greek words meaning “to form an acid.” In addition, Lavoisier observed that the heat produced by a guinea pig when exhaling a given amount of carbon dioxide is similar to the quantity of heat produced by burning carbon to give the same amount of carbon dioxide. From this and other experiments he concluded that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Although he did not understand the details of the process, Because Lavoisier was an aristocrat, he came under suspicion during the Reign of Terror of the French Revolution. He was an investor in the Ferme Générale, the infamous tax-collecting organization in 18thcentury France. Tobacco was a monopoly product of the Ferme Générale, and it was a common occurrence to cheat the purchaser by adding water to the tobacco, a practice that Lavoisier opposed. Nonetheless, because of his involvement with the Ferme, his career was cut short by the guillotine on May 8, 1794, on the charge of “adding water to the people‘s tobacco.” Lavoisier and his wife, as painted in 1788 by Jacques-Louis David. Lavoisier was then 45 and his wife, Marie Anne Pierrette Paulze, was 30. Lavoisier‘s recognition marked an important step in the development of biochemistry. Lavoisier was a prodigious scientist and the principles of naming chemical substances that he introduced are still in use today. Further, he wrote a textbook in which he applied for the first time the principles of the conservation of matter to chemistry and used the idea to write early versions of chemical equations. The decomposition of red mercury (II) oxide. The decomposition reaction gives mercury metal and oxygen gas. The mercury is seen as a film on the surface of the test tube. Photos: (Center) The Metropolitan Museum of Art, Purchase, Mr. and Mrs. Charles Wrightsman gift, in honor of Everett Fahy, 1997. Photograph © 1989 The Metropolitan Museum of Art. (Right) Charles D. Winters. In the 18th century, the great French scientist Antoine Lavoisier (1743–1794) introduced the law of conservation of matter, which states that matter can be neither created nor destroyed. This means that if the total mass of reactants is 10 g, and if the reaction completely converts reactants to products, you must end up with 10 g of products. It also means that if 1000 atoms of a particular element are contained in the reactants, then those 1000 atoms must appear in the products in some fashion. When applied to the reaction of phosphorus and chlorine, the law of conservation of matter tells us that 1 molecule of phosphorus (with 4 phosphorus atoms) and 6 diatomic molecules of Cl2 (with 12 atoms of Cl ) are required to produce 4 molecules of PCl3. Because each PCl3 molecule contains 1 P atom and 3 Cl atoms, the 4 PCl3 molecules are needed to account for 4 P atoms and 12 Cl atoms in the product. 6ϫ2ϭ 12 Cl atoms 4ϫ3ϭ 12 Cl atoms P4(s) ϩ 6 Cl2(g) ¡ 4 PCl3(ᐉ) 4 P atoms 143 4 P atoms ■ More Information from Chemical Equations The same number of atoms must exist after a reaction as before it takes place. However, these atoms are arranged differently. In the phosphorus/chlorine reaction, for example, the P atoms were in the form of P4 molecules before reaction, but appear as PCl3 molecules after reaction. Chapter 4 Chemical Equations and Stoichiometry Photos: Charles D. Winters 144 2 Fe(s) ϩ 3 Cl2(g) ¡ Reactants 2 FeCl3(s) Products Active Figure 4.2 The reaction of iron and chlorine. Hot iron gauze is inserted into a flask of chlorine gas. The heat from the reaction causes the iron gauze to glow, and brown iron(III) chloride forms. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. The numbers in front of each formula in a balanced chemical equation are required by the law of conservation of matter. Review the equation for the reaction of phosphorus and chlorine, and then consider the balanced equation for the reaction of iron and chlorine (Figure 4.2). 2 Fe(s) ϩ 3 Cl2(g) ¡ 2 FeCl3(s) stoichiometric coefficients The number in front of each chemical formula can be read as the number of atoms or molecules (2 atoms of Fe and 3 molecules of Cl2 form 2 formula units of FeCl3). It can refer equally well to amounts of reactants and products: 2 moles of solid iron combine with 3 moles of chlorine gas to produce 2 moles of solid FeCl3. The relationship between the quantities of chemical reactants and products is called stoichiometry (pronounced “stoy-key-AHM-uh-tree”), and the coefficients in a balanced equation are the stoichiometric coefficients. Balanced chemical equations are fundamentally important for understanding the quantitative basis of chemistry. You must always begin with a balanced equation before carrying out a stoichiometry calculation. See the General ChemistryNow CD-ROM or website: • Screen 4.3 The Law of Conservation of Mass, for two exercises on the conservation of mass in several reactions 4.2 Balancing Chemical Equations Exercise 4.1—Chemical Reactions The reaction of aluminum with bromine is shown on page 98. The equation for the reaction is 2 Al(s) ϩ 3 Br2(/) ¡ Al2Br6(s) (a) Name the reactants and products in this reaction and give their states. (b) What are the stoichiometric coefficients in this equation? (c) If you were to use 8000 atoms of Al, how many molecules of Br2 are required to consume the Al completely? 4.2—Balancing Chemical Equations Balancing an equation ensures that the same number of atoms of each element appear on both sides of the equation. Many chemical equations can be balanced by trial and error, although some will involve more trial than others. One general class of chemical reactions is the reaction of metals or nonmetals with oxygen to give oxides of the general formula MxOy. For example, iron can react with oxygen to give iron(III) oxide (Figure 4.3a), 4 Fe(s) ϩ 3 O2(g) ¡ 2 Fe2O3(s) magnesium and oxygen react to form magnesium oxide (Figure 4.3b), 2 Mg(s) ϩ O2(g) ¡ 2 MgO(s) and phosphorus, P4, reacts vigorously with oxygen to give tetraphosphorus decaoxide, P4O10 (Figure 4.3c), P4(s) ϩ 5 O2(g) ¡ P4O10(s) Charles D. Winters The equations written above are balanced. The same number of metal or phosphorus atoms and oxygen atoms occurs on each side of these equations. (a) Reaction of iron and oxygen to give iron(III) oxide, Fe2O3. (b) Reaction of magnesium and oxygen to give magnesium oxide, MgO. (c) Reaction of phosphorus and oxygen to give tetraphosphorus decaoxide, P4O10. Figure 4.3 Reactions of metals and a nonmetal with oxygen. (See General ChemistryNow Screen 4.4 Balancing Chemical Equations, for a video of the phosphorus and oxygen reaction.) 145 146 Chapter 4 Chemical Equations and Stoichiometry The combustion, or burning, of a fuel in oxygen is accompanied by the evolution of heat. You are familiar with combustion reactions such as the burning of octane, C8H18, a component of gasoline, in an automobile engine: 2 C8H18(/) ϩ 25 O2(g) ¡ 16 CO2(g) ϩ 18 H2O(g) Charles D. Winters In all combustion reactions, some or all of the elements in the reactants end up as oxides, compounds containing oxygen. When the reactant is a hydrocarbon (a compound containing only C and H), the products of complete combustion are carbon dioxide and water. When balancing chemical equations, there are two important things to remember: A combustion reaction. Propane, C3H8, burns to give CO2 and H2O. These simple oxides are always the products of the complete combustion of a hydrocarbon. (See General ChemistryNow Screen 4.4 Balancing Chemical Equations, for a animation of this reaction.) • Formulas for reactants and products must be correct or the equation is meaningless. • Subscripts in the formulas of reactants and products cannot be changed to balance equations. Changing the subscripts changes the identity of the substance. For example, you cannot change CO2 to CO to balance an equation; carbon monoxide, CO, and carbon dioxide, CO2, are different compounds. As an example of equation balancing, let us write the balanced equation for the complete combustion of propane, C3H8. Step 1. Write correct formulas for the reactants and products. unbalanced equation uuuuuy CO2(g) ϩ H2O(g) C3H8(g) ϩ O2(g) u Here propane and oxygen are the reactants, and carbon dioxide and water are the products. Step 2. Balance the C atoms. In combustion reactions such as this it is usually best to balance the carbon atoms first and leave the oxygen atoms until the end (because the oxygen atoms are often found in more than one product ). In this case three carbon atoms are in the reactants, so three must occur in the products. Three CO2 molecules are therefore required on the right side: unbalanced equation uuuuuy 3 CO2(g) ϩ H2O(g) C3H8(g) ϩ O2(g) u Step 3. Balance the H atoms. Propane, the reactant, contains 8 H atoms. Each molecule of water has two hydrogen atoms, so four molecules of water account for the required eight hydrogen atoms on the right side: unbalanced equation uuuuuy 3 CO2(g) ϩ 4 H2O(g) C3H8(g) ϩ O2(g) u Step 4. Balance the number of O atoms. Ten oxygen atoms are on the right side (3 ϫ 2 ϭ 6 in CO2 plus 4 ϫ 1 ϭ 4 in water). Therefore, five O2 molecules are needed to supply the required ten oxygen atoms: C3H8(g) ϩ 5 O2(g) ¡ 3 CO2(g) ϩ 4 H2O(g) Step 5. Verify that the number of atoms of each element is balanced. The equation shows three carbon atoms, eight hydrogen atoms, and ten oxygen atoms on each side. 4.2 Balancing Chemical Equations See the General ChemistryNow CD-ROM or website: • Screen 4.4 Balancing Chemical Equations, for a tutorial in which you balance a series of combustion reactions. Example 4.1—Balancing an Equation for a Combustion Reaction Problem Write the balanced equation for the combustion of ammonia (NH3 ϩ O2) to give NO and H2O. Strategy First write the unbalanced equation. Next balance the N atoms, then balance the H atoms, and finally balance the O atoms. Solution Step 1. Write correct formulas for reactants and products. The unbalanced equation for the combustion is unbalanced equation uuuuuy NO(g) ϩ H2O(g) NH3(g) ϩ O2(g) u Step 2. Balance the N atoms. There is one N atom on each side of the equation. The N atoms are in balance, at least for the moment. unbalanced equation uuuuuy NO(g) ϩ H2O(g) NH3(g) ϩ O2(g) u Step 3. Balance the H atoms. There are three H atoms on the left and two on the right. To have the same number on each side, let us use two molecules of NH3 on the left and three molecules of H2O on the right (which gives us six H atoms on each side). unbalanced equation uuuuuy NO(g) ϩ 3 H2O(g) 2 NH3(g) ϩ O2(g) u Notice that when we balance the H atoms, the N atoms are no longer balanced. To bring them into balance, let us use two NO molecules on the right. unbalanced equation uuuuuy 2 NO(g) ϩ 3 H2O(g) 2 NH3(g) ϩ O2(g) u Step 4. Balance the O atoms. After Step 3, there are two O atoms on the left side and five on the right. That is, there are an even number of O atoms on the left and an odd number on the right. Because there cannot be an odd number of O atoms on the left (O atoms are paired in O2 molecules), multiply each coefficient on both sides of the equation by 2 so that an even number of oxygen atoms (ten) can now occur on the right side: unbalanced equation uuuuuy 4 N0(g) ϩ 6 H2O(g) 4 NH3(g) ϩ O2(g) u Now the oxygen atoms can be balanced by having five O2 molecules on the left side of the equation: 4 NH3 1g2 ϩ 5 O2 1g2 uuuuuy 4 NO1g2 ϩ 6 H2O1g2 balanced equation Step 5. Verify the result. Four N atoms, 12 H atoms, and 10 O atoms occur on each side of the equation. 147 148 Chapter 4 Chemical Equations and Stoichiometry Comment An alternative way to write this equation is 2 NH3(g) ϩ 5 O2(g) ¡ 2 NO(g) ϩ 3 H2O(g) 2 where a fractional coefficient has been used. This equation is correctly balanced and will be useful under some circumstances. In general, however, we balance equations with wholenumber coefficients. Exercise 4.2—Balancing the Equation for a Combustion Reaction (a) Butane gas, C4H10, can burn completely in air [use O2(g) as the other reactant] to give carbon dioxide gas and water vapor. Write a balanced equation for this combustion reaction. (b) Write a balanced chemical equation for the complete combustion of liquid tetraethyllead, Pb(C2H5)4 (which was used until the 1970s as a gasoline additive). The products of combustion are PbO(s), H2O(g), and CO2(g). 4.3—Mass Relationships in Chemical Reactions: Stoichiometry A balanced chemical equation shows the quantitative relationship between reactants and products in a chemical reaction. Let us apply this concept to the reaction of phosphorus and chlorine (see Figure 4.1). Suppose you use 1.00 mol of phosphorus (P4, 124 g/mol ) in this reaction. The balanced equation shows that 6.00 mol (ϭ 425 g) of Cl2 must be used for complete reaction with 1.00 mol of P4 and that 4.00 mol (ϭ 549 g) of PCl3 can be produced. ■ Amounts Tables Amounts tables not only are useful here but will also be used extensively when you study chemical equilibria in Chapters 16–18. Equation Initial amount (mol) Change in amount upon reaction (mol) Amount after complete reaction (mol) ■ Mass Balance Mass is always conserved in chemical reactions. The total mass before reaction is always the same as that after reaction. This does not mean, however, that the total amount (moles) of reactants is the same as that of the products. Atoms are rearranged into different “units” (molecules) in the course of a reaction. In the P4 ϩ Cl2 reaction, 7 mol of reactants gives 4 mol of product. P4(s) 1.00 mol (124 g) Ϫ 1.00 mol 0 mol (0 g) ؉ 6 Cl2 (g) ¡ 6.00 mol (425 g) Ϫ 6.00 mol 0 mol (0 g) 4 PCl3 (/) 0 mol (0 g) ϩ 4.00 mol 4.00 mol [549 g ϭ 124 g ϩ 425 g] The mole and mass relationships of reactants and products in a reaction are summarized in an amounts table. Such tables identify the amounts of reactants and products and the changes that occur upon reaction. The balanced equation for a reaction tells us the correct mole ratios of reactants and products. Therefore, the equation for the phosphorus and chlorine reaction, for example, applies no matter how much P4 is used. Suppose 0.0100 mol of P4 (1.24 g) is used. Now only 0.0600 mol of Cl2 (4.25 g) is required, and 0.0400 mol of PCl3 (5.49 g) can form. Following this line of reasoning, let us decide (a) what mass of Cl2 is required to react completely with 1.45 g of phosphorus and (b) what mass of PCl3 is produced. Part (a): Mass of Cl2 Required Step 1. Write the balanced equation (using correct formulas for reactants and products). This is always the first step when dealing with chemical reactions. P4(s) ϩ 6 Cl2(g) ¡ 4 PCl3(/) 4.3 149 Mass Relationships in Chemical Reactions: Stoichiometry Problem-Solving Tip 4.1 Stoichiometry Calculations You are asked to determine what mass of product can be formed from a given mass of reactant. Keep in mind that it is not possible to calculate the mass of product in a single step. Instead, you must follow a route such as that illustrated here for the reaction of a reactant A to give the product B according to an equation such as x A S y B. Here the mass of reactant A is converted to moles of A. Then, using the stoichiometric factor, you find moles of B. Finally, the mass of B is obtained by multiplying moles of B by its molar mass. When solving a chemical stoichiometry problem, remember that you will always use a stoichiometric factor at some point. grams reactant A ϫ ° grams product B 1 mol A ¢ gA direct calculation not possible ϫ moles reactant A ° gB ¢ mol B moles product B y mol product B ¢ ϫ° x mol reactant A ϫ stoichiometric factor Step 2. Calculate moles from masses. From the mass of P4, calculate the amount of P4 available. 1.45 g P4 ϫ 1 mol P4 ϭ 0.0117 mol P4 123.9 g P4 Step 3. Use a stoichiometric factor. The amount of P4 available is related to the amount of the other reactant (Cl2) required by the balanced equation. 0.0117 mol P4 ϫ 6 mol Cl2 required ϭ 0.0702 mol Cl2 required 1 mol P4 available A stoichiometric factor (from balanced equation) To perform this calculation the amount of phosphorus available has been multiplied by a stoichiometric factor, a mole ratio based on the coefficients for the two chemicals in the balanced equation. This is the reason you must balance chemical equations before proceeding with calculations. Here the balanced equation specifies that 6 mol of Cl2 is required for each mole of P4, so the stoichiometric factor is (6 mol Cl2/ 1 mol P4). Calculation shows that 0.0702 mol of Cl2 is required to react with all the available phosphorus (1.45 g, 0.0117 mol ). Step 4. Calculate mass from moles. Convert amount (moles) of Cl2 calculated in Step 3 to quantity (mass in grams) of Cl2 required. 0.0702 mol Cl2 ϫ 70.91 g Cl2 ϭ 4.98 g Cl2 1 mol Cl2 Part (b) Mass of PCl3 Produced from P4 and Cl2 What mass of PCl3 can be produced from the reaction of 1.45 g of phosphorus with 4.98 g of Cl2? Because matter is conserved, the answer can be obtained in this case ■ Stoichiometric Factor The stoichiometric factor is a conversion factor (see page 42). Thus, a stoichiometric factor can also relate moles of a reactant to moles of a product, and vice versa. ■ Amount and Quantity When doing stoichiometry problems, recall from Chapter 2 that the terms “amount” and “quantity” are used in a specific sense by chemists. The amount of a substance is the number of moles of that substance. Quantity refers to the mass of the substance. 150 Chapter 4 Chemical Equations and Stoichiometry by adding the masses of P4 and Cl2 used (giving 1.45 g ϩ 4.98 g ϭ 6.43 g of PCl3 produced). Alternatively, Steps 3 and 4 can be repeated, but with the appropriate stoichiometric factor and molar mass. Step 3b. Use a stoichiometric factor. Convert the amount of available P4 to the amount of PCl3 produced. Here the balanced equation specifies that 4 mol PCl3 is produced for each mole of P4 used, so the stoichiometric factor is (4 mol PCl3/1 mol P4). 0.0117 mol P4 ϫ 4 mol PCl3 produced ϭ 0.0468 mol PCl3 produced 1 mol P4 available A stoichiometric factor (from balanced equation) Step 4b. Calculate mass from moles. Convert the amount of PCl3 produced to a mass in grams. 0.0468 mol PCl3 ϫ 137.3 g PCl3 ϭ 6.43 g PCl3 1 mol PCl3 See the General ChemistryNow CD-ROM or website: • Screen 4.5 Weight Relations in Chemical Reactions (a) for a video and animation of the phosphorus and chlorine reaction discussed in this section (b) for an exercise that examines the reaction between chlorine and elemental phosphorus • Screen 4.6 Calculations in Stoichiometry, for a tutorial on yield Example 4.2—Mass Relations in Chemical Reactions Problem Glucose reacts with oxygen to give CO2 and H2O. C6H12O6(s) ϩ 6 O2(g) ¡ 6 CO2(g) ϩ 6 H2O(/) What mass of oxygen (in grams) is required for complete reaction of 25.0 g of glucose? What masses of carbon dioxide and water (in grams) are formed? Strategy After referring to the balanced equation, you can perform the stoichiometric calculations using the scheme in Problem-Solving Tip 4.1. mass of glucose mass O2 required molar mass of glucose mol glucose molar mass of O2 mol O2 required stoichiometric factor 4.3 Mass Relationships in Chemical Reactions: Stoichiometry First find the amount of glucose available, then relate it to the amount of O2 required using the stoichiometric factor based on the coefficients in the balanced equation. Finally, find the mass of O2 required from the amount of O2. Follow the same procedure to find the masses of carbon dioxide and water. Solution Step 1. Write a balanced equation. C6H12O6(s) ϩ 6 O2(g) ¡ 6 CO2(g) ϩ 6 H2O(/) Step 2. Convert the mass of glucose to moles. 25.0 g glucose ϫ 1 mol ϭ 0.139 mol glucose 180.2 g Step 3. Use the stoichiometric factor. Here we calculate the amount of O2 required. 0.139 mol glucose ϫ 6 mol O2 ϭ 0.832 mol O2 1 mol glucose Step 4. Calculate mass from moles. Convert the required amount of O2 to a mass in grams. 0.832 mol O2 ϫ 32.00 g ϭ 26.6 g O2 1 mol O2 Repeat Steps 3 and 4 to find the mass of CO2 produced in the combustion. First, relate the amount (moles) of glucose available to the amount of CO2 produced using a stoichiometric factor. Then convert the amount of CO2 to the mass in grams. 0.139 mol glucose ϫ 44.01 g CO2 6 mol CO2 ϫ ϭ 36.6 g CO2 1 mol glucose 1 mol CO2 Now, how can you find the mass of H2O produced? You could go through Steps 3 and 4 again. However, recognize that the total mass of reactants 25.0 g C6H12O6 ϩ 26.6 g O2 ϭ 51.6 g of reactants must be the same as the total mass of products. The mass of water that can be produced is therefore Total mass of products ϭ 51.6 g ϭ 36.6 g CO2 produced ϩ ? g H2O Mass of H2O produced ϭ 15.0 g The amounts table for this problem is Equation C6H12O6(s) Initial amount (mol) 0.139 mol ؉ 6 O2(g) ¡ 6(0.139 mol) 6 CO2(g) 0 ؉ 6 H2O(/) 0 ϭ 0.832 mol Change (mol) Ϫ 0.139 mol Amount after reaction (mol) 0 Ϫ 0.832 mol 0 ϩ 0.832 mol ϩ 0.832 mol 0.832 mol 0.832 mol 151 152 Chapter 4 Chemical Equations and Stoichiometry Comment When you know the mass of all but one of the chemicals in a reaction, you can find the unknown mass using the principle of mass conservation (the total mass of reactants must equal the total mass of products). Exercise 4.3—Mass Relations in Chemical Reactions What mass of oxygen, O2, is required to completely combust 454 g of propane, C3H8? What masses of CO2 and H2O are produced? C3H8(g) ϩ 5 O2(g) ¡ 3 CO2(g) ϩ 4 H2O(/) 4.4—Reactions in Which One Reactant Is Present in Limited Supply You may have observed in your laboratory experiments that reactions are often carried out with an excess of one reactant over that required by stoichiometry. This is usually done to ensure that one of the reactants in the reaction is consumed completely, even though some of another reactant remains unused. Suppose you burn a toy “sparkler,” a wire coated with magnesium (Figure 4.3b). The magnesium burns in air, consuming oxygen and producing magnesium oxide, MgO. Mg(s) ϩ O2(g) ¡ 2 MgO(s) The sparkler burns until the magnesium is consumed completely. What about the oxygen? Two moles of magnesium require one mole of oxygen, but there is much, much more O2 available in the air than is needed to consume the magnesium in a sparkler. How much MgO is produced? That depends on the quantity of magnesium in the sparkler, not on the quantity of O2 in the atmosphere. A substance such as the magnesium in this example is called the limiting reactant because its amount determines, or limits, the amount of product formed. Let us look at an example of a limiting reactant situation using the reaction of oxygen and carbon monoxide to give carbon dioxide. The balanced equation for the reaction is 2 CO(g) ϩ O2(g) ¡ 2 CO2(g) Suppose you have a mixture of four CO molecules and three O2 molecules. ■ Comparing Reactant Ratios For the CO/O2 reaction, the stoichiometric ratio of reactants should be (2 mol CO/ 1 mol O2). However, the ratio of amounts of reactants available is (4 mol CO/3 mol O2) or (1.33 mol CO/1 mol O2). Clearly, there is not sufficient CO to react with all of the available O2. Carbon monoxide is the limiting reactant, and some O2 will be left over when all of the CO is consumed. Reactants: 4 CO and 3 O2 ϩ Products: 4 CO2 and 1 O2 ϩ The four CO molecules require only two O2 molecules (and produce four CO2 molecules). This means that one O2 molecule remains after reaction is complete. Because more O2 molecules are available than are required, the number of CO2 molecules produced is determined by the number of CO molecules available. Carbon monoxide, CO, is therefore the limiting reactant in this case. Reactions in Which One Reactant Is Present in Limited Supply a, Charles D. Winters; b, Johnson Matthey 4.4 (b) (a) Active Figure 4.4 Oxidation of ammonia. (a) Burning ammonia on the surface of a platinum wire produces so much heat that the wire glows bright red. (b) Billions of kilograms of HNO3 are made annually starting with the oxidation of ammonia over a wire gauze containing platinum. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. A Stoichiometry Calculation with a Limiting Reactant The first step in the manufacture of nitric acid is the oxidation of ammonia to NO over a platinum-wire gauze (Figure 4.4). 4 NH3(g) ϩ 5 O2(g) ¡ 4 NO(g) ϩ 6 H2O(/) Suppose that equal masses of NH3 and of O2 are mixed (750. g of each). Are these reactants mixed in the correct stoichiometric ratio or is one of them in short supply? That is, will one of them limit the quantity of NO that can be produced? How much NO can be formed if the reaction using this reactant mixture goes to completion? And how much of the excess reactant is left over when the maximum amount of NO has been formed? Step 1. Find the amount of each reactant. 1 mol NH3 ϭ 44.0 mol NH3 available 17.03 g NH3 1 mol O2 750. g O2 ϫ ϭ 23.4 mol O2 available 32.00 g O2 750. g NH3 ϫ Step 2. What is the limiting reactant? Examine the ratio of amounts of reactants. Are the reactants present in the correct stoichiometric ratio as given by the balanced equation? Stoichiometric ratio of reactants required by balanced equation 5 mol O2 1.25 mol O2 ϭ ϭ 4 mol NH3 1 mol NH3 Ratio of reactants actually available ϭ 0.532 mol O2 23.4 mol O2 ϭ 44.0 mol NH3 1 mol NH3 153 154 Chapter 4 Chemical Equations and Stoichiometry Problem-Solving Tip 4.2 2. Quantity of NO produced from 23.4 mol O2 and unlimited NH3 More on Reactions with a Limiting Reactant 23.4 mol O2 ϫ There is another method of solving limiting reactant problems: Calculate the mass of product expected based on each reactant. The limiting reactant is the reactant that gives the smallest quantity of product. For example, refer to the NH3 ϩ O2 reaction on page 153. To confirm that O2 is the limiting reactant, calculate the quantity of NO that can be formed starting with (a) 44.1 mol of NH3 and unlimited O2 and (b) with 23.4 mol of O2 and unlimited NH3. 30.01 g NO 4 mol NO ϫ ϭ 562 g NO 5 mol O2 1 mol NO 3. Compare the quantities of NO produced. The available O2 is capable of producing less NO (562 g) than the available NH3 (1320 g), which confirms that O2 is the limiting reactant. As a final note, you may find this approach easier to use when there are more than two reactants, each present initially in some designated quantity. 1. Quantity of NO produced from 44.1 mol of NH3 and unlimited O2 44.0 mol NH3 ϫ 30.01 g NO 4 mol NO ϫ ϭ 1320 g NO 4 mol NH3 1 mol NO Dividing moles of O2 available by moles of NH3 available shows that the ratio of available reactants is much smaller than the 5 mol O2/4 mol NH3 ratio required by the balanced equation. Thus there is not sufficient O2 available to react with all of the NH3. In this case, oxygen, O2, is the limiting reactant. That is, 1 mol of NH3 requires 1.25 mol of O2, but we have only 0.532 mol of O2 available for each mole of NH3. Step 3. Calculate the mass of product. We can now calculate the mass of product, NO, expected based on the amount of the limiting reactant, O2. 23.4 mol O2 ϫ 30.01 g NO 4 mol NO ϫ ϭ 562 g NO 5 mol O2 1 mol NO Step 4. Calculate the mass of excess reactant. Ammonia is the “excess reactant” in this NH3/O2 reaction because more than enough NH3 is available to react with 23.4 mol of O2. Let us calculate the quantity of NH3 remaining after all the O2 has been used. To do so, we first need to know the amount of NH3 required to consume all the limiting reactant, O2. 23.4 mol O2 available ϫ 4 mol NH3 required ϭ 18.8 mol NH3 required 5 mol O2 Because 44.0 mol of NH3 is available, the amount of excess NH3 can be calculated, Excess NH3 ϭ 44.0 mol NH3 available Ϫ 18.8 mol NH3 required ϭ 25.2 mol NH3 remaining and then converted to a mass, 25.2 mol NH3 ϫ 17.03 g NH3 ϭ 429 g NH3 in excess of that required 1 mol NH3 4.4 155 Reactions in Which One Reactant Is Present in Limited Supply Finally, because 429 g of NH3 is left over, this means that 321 g of NH3 has been consumed (ϭ 750. g Ϫ 429 g). It is helpful in limiting reactant problems to summarize your results in an amounts table. Equation 4 NH3(g) Initial amount (mol) 44.0 Ϫ (4/5)(23.4) Change in amount (mol) ϭ Ϫ 18.8 After complete reaction (mol) 25.2 ؉ 5 O2(g) S 4 NO(g) 23.4 0 Ϫ 23.4 Ϫ 23.4 ϩ (4/5)(23.4) ؉ ■ Conservation of Mass Mass is conserved in the NH3 ϩ O2 reaction. The total mass present before reaction (1500. g) is the same as the total mass produced in the reaction plus the mass of NH3 remaining. That is, 562 g of NO (18.8 mol) and 506 g of H2O (28.1 mol) are produced. Because 429 g of NH3 (25.2 mol) remains, the total mass after reaction (562 g ϩ 506 g ϩ 429 g) is the same as the total mass before reaction. 6 H2O(g) 0 ϩ (6/5)(23.4) ϭ ϩ 18.8 ϭ ϩ 28.1 18.8 28.1 0 All of the limiting reactant, O2, has been consumed. Of the original 44.0 mol of NH3, 18.8 mol has been consumed and 25.2 mol remains. The balanced equation indicates that the amount of NO produced is equal to the amount of NH3 consumed, so 18.8 mol of NO is produced from 18.8 mol of NH3. In addition, 28.1 mol of H2O has been produced. See the General ChemistryNow CD-ROM or website: • Screen 4.7 Reactions Controlled by the Supply of One Reactant for a video and animation of the limiting reactant in the methanol and oxygen reaction (a) for an exercise on zinc and hydrochloric acid in aqueous solution (b) for a simulation using limiting reactants Example 4.3—A Reaction with a Limiting Reactant Problem Methanol, CH3OH, which is used as a fuel, can be made by the reaction of carbon monoxide and hydrogen. CO(g) ϩ 2 H2(g) ¡ CH3OH(/) methanol Suppose 356 g of CO and 65.0 g of H2 are mixed and allowed to react. (a) Which is the limiting reactant? (b) What mass of methanol can be produced? (c) What mass of the excess reactant remains after the limiting reactant has been consumed? Strategy There are usually two steps to a limiting reactant problem: (a) After calculating the amount of each reactant, compare the ratio of reactant amounts to the required stoichiometric ratio, 2 mol H2/1 mol CO. • If [mol H2 available/mol CO available] 7 2/1, then CO is the limiting reactant. • If [mol H2 available/mol CO available] 6 2/1, then H2 is the limiting reactant. (b) Use the amount of limiting reactant to find the amount of product. Reuters/Corbis • Screen 4.8 Limiting Reactants A car that uses methanol as a fuel. In this car methanol is converted to hydrogen, which is then combined with oxygen in a fuel cell. The fuel cell generates electric energy to run the car (see Chapter 20). See Example 4.3. 156 Chapter 4 Chemical Equations and Stoichiometry Solution (a) What is the limiting reactant? The amount of each reactant is 1 mol CO ϭ 12.7 mol CO 28.01 g CO 1 mol H2 Amount of H2 ϭ 65.0 g H2 ϫ ϭ 32.2 mol H2 2.016 g H2 Amount of CO ϭ 356 g CO ϫ Are these reactants present in a perfect stoichiometric ratio? Mol H2 available 32.2 mol H2 2.54 mol H2 ϭ ϭ Mol CO available 12.7 mol CO 1.00 mol CO The required mole ratio is 2 mol of H2 to 1 mol of CO. Here we see that more hydrogen is available than is required to consume all the CO. It follows that not enough CO is present to use up all of the hydrogen. CO is the limiting reactant. (b) What is the maximum mass of CH3OH that can be formed? This calculation must be based on the amount of limiting reactant. 12.7 mol CO ϫ 32.04 g CH3OH 1 mol CH3OH formed ϫ ϭ 407 g CH3OH 1 mol CO available 1 mol CH3OH (c) What amount of H2 remains when all the CO has been converted to product? First, we must find the amount of H2 required to react with all the CO. 12.7 mol CO ϫ 2 mol H2 ϭ 25.4 mol H2 required 1 mol CO Because 32.2 mol of H2 is available, but only 25.4 mol is required by the limiting reactant, 32.2 mol Ϫ 25.4 mol ϭ 6.8 mol of H2 is in excess. This is equivalent to 14 g of H2. 6.8 mol H2 ϫ 2.02 g H2 ϭ 14 g H2 remaining 1 mol H2 Comment The amounts table for this reaction is Equation Initial amount (mol) CO(g) ؉ 2 H2(g) ¡ CH3OH(ᐍ) 12.7 32.2 0 Ϫ 12.7 Change (mol) After complete reaction (mol) 0 Ϫ2(12.7) 6.8 ϩ12.7 12.7 The mass of product formed plus the mass of H2 remaining after reaction (407 g CH3OH produced ϩ 14 g H2 remaining ϭ 421 g) is equal to the mass of reactants present before reaction (356 g CO ϩ 65.0 g H2 ϭ 421 g). Exercise 4.4—A Reaction With a Limiting Reactant Titanium is an important structural metal, and a compound of titanium, TiO2, is the white pigment in paint. In the refining process, titanium ore (impure TiO2) is first converted to liquid TiCl4 by the following reaction. TiO2(s) ϩ 2 Cl2(g) ϩ C(s) ¡ TiCl4(/) ϩ CO2(g) 4.5 157 Percent Yield Using 125 g each of Cl2 and C, but plenty of TiO2-containing ore, which is the limiting reactant in this reaction? What mass of TiCl4, in grams, can be produced? Exercise 4.5—A Reaction with a Limiting Reactant The thermite reaction produces iron metal and aluminum oxide from a mixture of powdered aluminum metal and iron(III) oxide. Charles D. Winters Fe2O3(s) ϩ 2 Al(s) ¡ 2 Fe(s) ϩ Al2O3(s) A mixture of 50.0 g each of Fe2O3 and Al is used. (a) Which is the limiting reactant? (b) What mass of iron metal can be produced? Thermite reaction Iron(III) oxide reacts with aluminum metal to produce aluminum oxide and iron metal. The reaction produces so much heat that the iron melts and spews out of the reaction vessel. See Exercise 4.5. 4.5—Percent Yield The maximum quantity of product we calculate can be obtained from a chemical reaction is the theoretical yield. Frequently, however, the actual yield of a compound—the quantity of material that is actually obtained in the laboratory or a chemical plant—is less than the theoretical yield. Some loss of product often occurs during the isolation and purification steps. In addition, some reactions do not go completely to products, and reactions are sometimes complicated by giving more than one set of products. For all these reasons, the actual yield is likely to be less than the theoretical yield (Figure 4.5). To provide information to other chemists who might want to carry out a reaction, it is customary to report a percent yield. Percent yield, which specifies how much of the theoretical yield was obtained, is defined as Percent yield ϭ actual yield ϫ 100% theoretical yield (4.1) (a) Suppose you made aspirin in the laboratory by the following reaction: Charles D. Winters C6H4(OH)CO2H(s) ϩ (CH3CO)2O 1/2 ¡ C6H4(OCOCH3)CO2H(s) ϩ CH3CO2H(/) salicylic acid acetic anhydride aspirin acetic acid and that you began with 14.4 g of salicylic acid and an excess of acetic anhydride. That is, salicylic acid is the limiting reactant. If you obtain 6.26 g of aspirin, what is the percent yield of this product? The first step is to find the amount of the limiting reactant, salicylic acid (C6H4(OH)CO2H). 14.4 g C6H4 1OH2CO2H ϫ 1 mol C6H4 1OH2CO2H ϭ 0.104 mol C6H4 1OH2CO2H 138.1 g C6H4 1OH2CO2H (b) Figure 4.5 Percent yield. Although not a chemical reaction, popping corn is a good analogy to the difference between a theoretical yield and an actual yield. Here we began with 20 popcorn kernels and found that only 16 of them popped. The percent yield from our “reaction” was (16/20) ϫ 100%, or 80%. 158 Chapter 4 Chemical Equations and Stoichiometry Next, use the stoichiometric factor from the balanced equation to find the amount of aspirin expected based on the limiting reactant, C6H4(OH)CO2H. 0.104 mol C6H4 1OH2CO2H ϫ 1 mol aspirin ϭ 0.104 mol aspirin 1 mol C6H4 1OH2CO2H The maximum amount of aspirin that can be produced—the theoretical yield—is 0.104 mol. Because the quantity you measure in the laboratory is the mass of the product, it is customary to express the theoretical yield as a mass in grams. 0.104 mol aspirin ϫ 180.2 g aspirin ϭ 18.7 g aspirin 1 mol aspirin Finally, with the actual yield known to be only 6.26 g, the percent yield of aspirin can be calculated. Percent yield ϭ 6.26 g aspirin obtained 1actual yield2 ϫ 100% ϭ 33.5% yield 18.7 g aspirin expected 1theoretical yield2 See the General ChemistryNow CD-ROM or website: • Screen 4.9 Percent Yield (a) for a tutorial on determining the theoretical yield of a reaction (b) for a tutorial on determining the percent yield of a reaction Exercise 4.6—Percent Yield Methanol, CH3OH, can be burned in oxygen to provide energy, or it can be decomposed to form hydrogen gas, which can then be used as a fuel (see Example 4.3). CH3OH(/) ¡ 2 H2(g) ϩ CO(g) If 125 g of methanol is decomposed, what is the theoretical yield of hydrogen? If only 13.6 g of hydrogen is obtained, what is the percent yield of this gas? Charles D. Winters 4.6—Chemical Equations and Chemical Analysis Figure 4.6 A modern analytical instrument. This nuclear magnetic resonance (NMR) spectrometer is closely related to a magnetic resonance imaging (MRI) instrument found in a hospital. The NMR is used to analyze compounds and to decipher their structure. Analytical chemists use a variety of approaches to identify substances as well as to measure the quantities of components of mixtures. Analytical chemistry is often done now using instrumental methods (Figure 4.6), but classical chemical reactions and stoichiometry play a central role. Quantitative Analysis of a Mixture Quantitative chemical analyses generally depend on one or the other of two basic ideas: • A substance, present in unknown amount, can be allowed to react with a known quantity of another substance. If the stoichiometric ratio for their reaction is known, the unknown amount can be determined. 4.6 159 Chemical Equations and Chemical Analysis An example of the first type of analysis is the analysis of a sample of vinegar containing an unknown amount of acetic acid, the ingredient that makes vinegar acidic. The acid reacts readily and completely with sodium hydroxide. CH3CO2H(aq) ϩ NaOH(aq) ¡ CH3CO2Na(aq) ϩ H2O(/) Charles D. Winters • A material of unknown composition can be converted to one or more substances of known composition. Those substances can be identified, their amounts determined, and these amounts related to the amount of the original, unknown substance. acetic acid If the exact amount of sodium hydroxide used in the reaction can be measured, the amount of acetic acid present is also known. This type of analysis is the subject of a major portion of Chapter 5 [᭤ Section 5.10]. The second type of analysis is exemplified by the analysis of a sample of a mineral, thenardite, which is largely sodium sulfate, Na2SO4, (Figure 4.7). Sodium sulfate is soluble in water. Therefore, to find the quantity of Na2SO4 in an impure mineral sample, we would crush the rock and then wash it thoroughly with water to dissolve the sodium sulfate. Next, we would treat this solution with barium chloride to form the water-insoluble compound barium sulfate. The barium sulfate is collected on a filter and weighed (Figure 4.8). ■ Analysis and 100% Yield Quantitative analysis requires reactions in which the yield is 100%. Charles D. Winters Na2SO4(aq) ϩ BaCl2(aq) ¡ BaSO4(s) ϩ 2 NaCl(aq) Figure 4.7 Thenardite. The mineral thenardite is sodium sulfate, Na2SO4. It is named after the French chemist Louis Thenard (1777–1857), a co-discoverer (with Gay-Lussac and Davy) of boron. Sodium sulfate is used in making detergents, glass, and paper. (a) (b) Na2SO4(aq), clear solution BaCl2(aq), clear solution (c) BaSO4, white solid NaCl(aq), clear solution (d) BaSO4, NaCl(aq), clear solution white solid caught in filter Active Figure 4.8 Analysis for the sulfate content of a sample. The sulfate ion in a solution of Na2SO4 reacts with barium ion (Ba2ϩ) to form BaSO4. The solid precipitate, barium sulfate (BaSO4), is collected on a filter and weighed. The amount of BaSO4 obtained can be related to the amount of Na2SO4 in the sample. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. Filter paper weighed 160 Chapter 4 Chemical Equations and Stoichiometry We can then find the amount of sulfate in the mineral sample because it is directly related to the amount of BaSO4. 1 mol Na2SO4 ¡ 1 mol BaSO4 This approach to the analysis of a mineral is one of many examples of the use of stoichiometry in chemical analysis. Examples 4.4 and 4.5 further illustrate this method. Example 4.4—Analysis of a Lead-Containing Mineral Problem The mineral cerussite is mostly lead carbonate, PbCO3, but other substances are present. To analyze for the PbCO3 content, a sample of the mineral is first treated with nitric acid to dissolve the lead carbonate. PbCO3(s) ϩ 2 HNO3(aq) ¡ Pb(NO3)2(aq) ϩ H2O(/) ϩ CO2(g) On adding sulfuric acid to the resulting solution, lead sulfate precipitates. Pb(NO3)2(aq) ϩ H2SO4(aq) ¡ PbSO4(s) ϩ 2 HNO3(aq) Solid lead sulfate is isolated and weighed (as in Figure 4.8). Suppose a 0.583-g sample of mineral produced 0.628 g of PbSO4. What is the mass percent of PbCO3 in the mineral sample? Strategy The key is to recognize that 1 mol of PbCO3 will ultimately yield 1 mol of PbSO4. Based on the amount of PbSO4 isolated, we can calculate the amount of PbCO3 (in moles), and its mass, in the original sample. When the mass of PbCO3 is known, this is compared with the mass of the mineral sample to give the percent composition. Solution Let us first calculate the amount of PbSO4. 0.628 g PbSO4 ϫ 1 mol PbSO4 ϭ 0.00207 mol PbSO4 303.3 g PbSO4 From stoichiometry, we can relate the amount of PbSO4 to the amount of PbCO3. (Here the two stoichiometric factors are based on the two balanced equations describing the chemical reactions.) 0.00207 mol PbSO4 ϫ 1 mol Pb1NO3 2 2 1 mol PbSO4 ϫ 1 mol PbCO3 ϭ 0.00207 mol PbCO3 1 mol Pb1NO3 2 2 The mass of PbCO3 is 0.00207 mol PbCO3 ϫ 267.2 g PbCO3 ϭ 0.553 g PbCO3 1 mol PbCO3 Finally, the mass percent of PbCO3 in the mineral sample is Mass percent of PbCO3 ϭ 0.553 g PbCO3 ϫ 100% ϭ 94.9% 0.583 g sample 4.6 161 Chemical Equations and Chemical Analysis Example 4.5—Mineral Analysis Problem Nickel(II) sulfide, NiS, occurs naturally as the relatively rare mineral millerite. One of its occurrences is in meteorites. To analyze a mineral sample for the quantity of NiS, the sample is digested in nitric acid to form a solution of Ni(NO3)2. NiS(s) ϩ 4 HNO3(aq) ¡ Ni(NO3)2(aq) ϩ S(s) ϩ 2 NO2(g) ϩ 2 H2O(/) The aqueous solution of Ni(NO3)2 is then treated with the organic compound dimethylglyoxime (C4H8N2O2, DMG) to give the red solid Ni(C4H7N2O2)2. Ni(NO3)2(aq) ϩ 2 C4H8N2O2(aq) ¡ Ni(C4H7N2O2)2(s) ϩ 2 HNO3(aq) Suppose a 0.468-g sample containing millerite produces 0.206 g of red, solid Ni(C4H7N2O2)2. What is the mass percent of NiS in the sample? Strategy The balanced equations show the following “road map”: Thus, if we know the mass of Ni(C4H7N2O2)2, we can calculate its amount and thus the amount of NiS. The amount of NiS allows us to calculate the mass and mass percent of NiS. Solution The molar mass of Ni(C4H7N2O2)2 is 288.9 g/mol. Thus, the amount of the red solid is 0.206 g Ni1C4H7N2O2 2 2 ϫ 1 mol Ni1C4H7N2O2 2 2 288.9 g Ni1C4H7N2O2 2 2 ϭ 7.13 ϫ 10Ϫ4 mol Ni1C4H7N2O2 2 2 Because 1 mol of Ni(C4H7N2O2)2 is ultimately produced from 1 mol of NiS, the amount of NiS in the sample must have been 7.13 ϫ 10Ϫ4 mol. With the amount of NiS known, we calculate the mass of NiS. 7.13 ϫ 10Ϫ4 mol NiS ϫ 90.76 g NiS ϭ 0.0647 g NiS 1 mol NiS Finally, the mass percent of NiS in the 0.468-g sample is Mass percent NiS ϭ 0.0647 g NiS ϫ 100% ϭ 13.8% NiS 0.468 g sample Exercise 4.7—Analysis of a Mixture One method for determining the purity of a sample of titanium(IV) oxide, TiO2, an important industrial chemical, is to combine the sample with bromine trifluoride. 3 TiO2(s) ϩ 4 BrF3(/) ¡ 3 TiF4(s) ϩ 2 Br2(/) ϩ 3 O2(g) This reaction is known to occur completely and quantitatively. That is, all of the oxygen in TiO2 is evolved as O2. Suppose 2.367 g of a TiO2-containing sample evolves 0.143 g of O2. What is the mass percent of TiO2 in the sample? Photo: Charles D. Winters 1 mol NiS ¡ 1 mol Ni(NO3)2 ¡ 1 mol Ni(C4H7N2O2)2 A precipitate of nickel. Red, insoluble Ni(C4H7N2O2)2 precipitates when dimethylglyoxime (C4H8N2O2) is added to an aqueous solution of nickel(II) ions. (See Example 4.5.) 162 Chapter 4 Chemical Equations and Stoichiometry Determining the Formula of a Compound by Combustion ■ Finding an Empirical Formula by Chemical Analysis Finding the empirical formula of a compound by chemical analysis always uses the following procedure: 1. The unknown but pure compound is decomposed into known products. 2. The reaction products are isolated in pure form and the amount of each is determined. 3. The amount of each product is related to the amount of each element in the original compound to give the empirical formula. The empirical formula of a compound can be determined if the percent composition of the compound is known [᭣ Section 3.6]. But where do the percent composition data come from? One chemical method that works well for compounds that burn in oxygen is analysis by combustion. In this technique, each element in the compound combines with oxygen to produce the appropriate oxide. Consider an analysis of the hydrocarbon methane, CH4, as an example of combustion analysis. A balanced equation for the combustion of methane shows that every mole of carbon in the original compound is converted to a mole of CO2. Every mole of hydrogen in the original compound gives half a mole of H2O. (Here the four moles of H atoms in one mole of CH4 give two moles of H2O.) CH4(g) ϩ 2 O2(g) CO2(g) ϩ 2 H2O(ᐉ) The gaseous carbon dioxide and water are separated (as illustrated in Figure 4.9) and their masses determined. From these masses it is possible to calculate the amounts of C and H in CO2 and H2O, respectively. The ratio of amounts of C and H in a sample of the original compound can then be found. This ratio gives the empirical formula: burn in O2 ϫ 1 mol CO2 44.01 g ϫ g CO2 mol CO2 g H2O 1 mol C 1 mol CO2 mol H2O mol C mol H CxHy ϫ 1 mol H2O 18.02 g ϫ empirical formula 2 mol H 1 mol H2O When using this procedure, a key observation is that every atom of C in the original compound appears as CO2 and every atom of H appears in the form of water. In other words, for every mole of CO2 observed, there must have been one mole of carbon in the unknown compound. Similarly, for every mole of H2O observed from combustion, there must have been two moles of H atoms in the unknown carbonhydrogen compound. Active Figure 4.9 Example 4.6—Using Combustion Analysis to Determine the Formula of a Hydrocarbon Problem When 1.125 g of a liquid hydrocarbon, CxHy, was burned in an apparatus like that shown in Figure 4.9, 3.447 g of CO2 and 1.647 g of H2O were produced. The molar mass of the compound was found to be 86.2 g/mol in a separate experiment. Determine the empirical and molecular formulas for the unknown hydrocarbon, CxHy. Strategy As outlined in the preceding diagram, we first calculate the amounts of CO2 and H2O. These are then converted to amounts of C and H. The ratio (mol H/mol C) gives the empirical formula of the compound. 4.6 Chemical Equations and Chemical Analysis Furnace H2O absorber O2 C x Hy CO2 absorber H2O H2O is absorbed by magnesium perchlorate, CO2 passes through Sample containing hydrogen and carbon CO2 CO2 is absorbed by finely divided NaOH supported on asbestos Active Figure 4.9 Combustion analysis of a hydrocarbon. If a compound containing C and H is burned in oxygen, CO2 and H2O are formed, and the mass of each can be determined. The H2O is absorbed by magnesium perchlorate, and the CO2 is absorbed by finely divided NaOH supported on asbestos. The mass of each absorbent before and after combustion gives the masses of CO2 and H2O. Only a few milligrams of a combustible compound are needed for analysis. See the General ChemistryNow CD-ROM or website to explore an interactive version of this figure accompanied by an exercise. Solution The amounts of CO2 and H2O isolated from the combustion are 3.447 g CO2 ϫ 1 mol CO2 ϭ 0.07832 mol CO2 44.010 g CO2 1.647 g H2O ϫ 1 mol H2O ϭ 0.09142 mol H2O 18.015 g H2O For every mole of CO2 isolated, 1 mol of C must have been present in the compound CxHy. 0.07832 mol CO2 ϫ 1 mol C in CxHy 1 mol CO2 ϭ 0.07832 mol C For every mole of H2O isolated, 2 mol of H must have been present in CxHy. 0.09142 mol H2O ϫ 2 mol H in CxHy 1 mol H2O ϭ 0.1828 mol H in CxHy The original 1.125-g sample of compound therefore contained 0.07832 mol of C and 0.1828 mol of H. To determine the empirical formula of CxHy, we find the ratio of moles of H to moles of C [᭣ Section 3.6]. 2.335 mol H 0.1828 mol H ϭ 0.07832 mol C 1.000 mol C Atoms combine to form molecules in whole-number ratios. The translation of this ratio (2.335/1) to a whole-number ratio can usually be done quickly by trial and error. Multiplying the numerator and denominator by 3 gives 7/3. So, we know the ratio is 7 mol H to 3 mol C, which means the empirical formula of the hydrocarbon is C3H7. 163 164 Chapter 4 Chemical Equations and Stoichiometry Comparing the experimental molar mass with the molar mass calculated for the empirical formula, Experimental molar mass 86.2 g/mol 2 ϭ ϭ Molar mass of C3H7 43.1 g/mol 1 we find that the molecular formula is twice the empirical formula. That is, the molecular formula is 1C3H7 2 2, or C6H14. Comment As noted in Problem-Solving Tip 3.3 (page 124), for problems of this type be sure to use data with enough significant figures to give accurate atom ratios. Finally, note that the determination of the molecular formula does not end the problem for a chemist. In this case, the formula C6H14 is appropriate for several distinctly different compounds. Two of the five compounds having this formula are shown here: H H3C CH3 C C CH3 CH3 H H H H C C C C H H3C H H H H CH3 To decide finally the identity of the unknown compound, more laboratory experiments will have to be done. Exercise 4.8—Determining the Empirical and Molecular Formulas for a Hydrocarbon A 0.523-g sample of the unknown compound CxHy was burned in air to give 1.612 g of CO2 and 0.7425 g of H2O. A separate experiment gave a molar mass for CxHy of 114 g/mol. Determine the empirical and molecular formulas for the hydrocarbon. Exercise 4.9—Determining the Empirical and Molecular Formulas for a Compound Containing C, H, and O A 0.1342-g sample of a compound with C, H, and O (CxHyOz) was burned in oxygen, and 0.240 g of CO2 and 0.0982 g of H2O were isolated. What is the empirical formula of the compound? If the experimentally determined molar mass was 74.1 g/mol, what is the molecular formula of the compound? (Hint: The carbon atoms in the compound are converted to CO2 and the hydrogen atoms are converted to H2O. The O atoms are found in both CO2 and H2O. To find the mass of O in the original sample, use the masses of CO2 and H2O to find the masses of C and H in the 0.1342 g-sample. Whatever of the 0.1342-g sample is not C and H is the mass of O.) 165 Key Equation Chapter Goals Revisited When you have finished studying this chapter, you should ask whether you have met the chapter goals. In particular, you should be able to Balance equations for simple chemical reactions a. Understand the information conveyed by a balanced chemical equation (Section 4.1). b. Balance simple chemical equations (Section 4.2). General ChemistryNow homework: Study Question(s) 2, 12b Perform stoichiometry calculations using balanced chemical equations a. Understand the principle of the conservation of matter, which forms the basis of chemical stoichiometry (Section 4.3). b. Calculate the mass of one reactant or product from the mass of another reactant or product by using the balanced chemical equation (Section 4.3). General ChemistryNow homework: SQ(s) 8, 16, 47, 53, 70, 72 c. Use amounts tables to organize stoichiometric information. General ChemistryNow homework: SQ(s) 16 Understand the impact of a limiting reactant on a chemical reaction a. Determine which of two reactants is the limiting reactant (Section 4.4). General ChemistryNow homework: SQ(s) 22 b. Determine the yield of a product based on the limiting reactant. General ChemistryNow homework: SQ(s) 20, 24, 26 Calculate the theoretical and percent yields of a chemical reaction a. Explain the differences among actual yield, theoretical yield, and percent yield, and calculate percent yield (Section 4.5). General ChemistryNow homework: SQ(s) 27 Use stoichiometry to analyze a mixture of compounds or to determine the formula of a compound a. Use stoichiometry principles to analyze a mixture (Section 4.6). General ChemistryNow homework: SQ(s) 31, 69, 77 b. Find the empirical formula of an unknown compound using chemical stoichiometry (Section 4.6). General ChemistryNow homework: SQ(s) 37, 42, 66 Key Equation Equation 4.1 (page 157) Calculating percent yield. Percent yield ϭ actual yield 1g2 ϫ 100% theoretical yield 1g2 • • See the General ChemistryNow CD-ROM or website to: Assess your understanding with homework questions keyed to each goal Check your readiness for an exam by taking the exam-prep quiz and exploring the resources in the personalized Learning Plan it provides 166 Chapter 4 Chemical Equations and Stoichiometry Study Questions Mass Relationships in Chemical Reactions: Basic Stoichiometry (See Example 4.2 and General ChemistryNow Screens 4.5 and 4.6.) ▲ denotes more challenging questions. ■ denotes questions available in the Homework and Goals section of the General ChemistryNow CD-ROM or website. Blue numbered questions have answers in Appendix O and fully worked solutions in the Student Solutions Manual. Structures of many of the compounds used in these questions are found on the General ChemistryNow CD-ROM or website in the Models folder. Assess your understanding of this chapter’s topics with additional quizzing and conceptual questions at http://now.brookscole.com/kotz6e 7. Aluminum reacts with oxygen to give aluminum oxide. 4 Al(s) ϩ 3 O2(g) ¡ 2 Al2O3(s) What amount of O2, in moles, is needed for complete reaction with 6.0 mol of Al? What mass of Al2O3, in grams, can be produced? 8. ■ What mass of HCl, in grams, is required to react with 0.750 g of Al(OH)3? What mass of water, in grams, is produced? Al(OH)3(s) ϩ 3 HCl(aq) ¡ AlCl3(aq) ϩ 3 H2O(/) 9. Like many metals, aluminum reacts with a halogen to give a metal halide (see Figure 3.1). 2 Al(s) ϩ 3 Br2(/) ¡ Al2Br6(s) Practicing Skills Balancing Equations (See Example 4.1 and General ChemistryNow Screen 4.4.) 1. Write a balanced chemical equation for the combustion of liquid pentane. 2. ■ Write a balanced chemical equation for the production of ammonia, NH3(g), from N2(g) and H2(g). 3. Balance the following equations: (a) Cr(s) ϩ O2(g) ¡ Cr2O3(s) (b) Cu2S(s) ϩ O2(g) ¡ Cu(s) ϩ SO2(g) (c) C6H5CH3(/) ϩ O2(g) ¡ H2O(/) ϩ CO2(g) 4. Balance the following equations: (a) Cr(s) ϩ Cl2(g) ¡ CrCl3(s) (b) SiO2(s) ϩ C(s) ¡ Si(s) ϩ CO(g) (c) Fe(s) ϩ H2O(g) ¡ Fe3O4(s) ϩ H2(g) 5. Balance the following equations and name each reactant and product: (a) Fe2O3(s) ϩ Mg(s) ¡ MgO(s) ϩ Fe(s) (b) AlCl3(s) ϩ NaOH(aq) ¡ Al(OH)3(s) ϩ NaCl(aq) (c) NaNO3(s) ϩ H2SO4(/) ¡ Na2SO4(s) ϩ HNO3(/) (d) NiCO3(s) ϩ HNO3(aq) ¡ Ni(NO3)2(aq) ϩ CO2(g) ϩ H2O(/) 6. Balance the following equations and name each reactant and product: (a) SF4(g) ϩ H2O(/) ¡ SO2(g) ϩ HF(/) (b) NH3(aq) ϩ O2(aq) ¡ NO(g) ϩ H2O(/) (c) BF3(g) ϩ H2O(/) ¡ HF(aq) ϩ H3BO3(aq) What mass of Br2, in grams, is required for complete reaction with 2.56 g of Al? What mass of white, solid Al2Br6 is expected? 10. The balanced equation for a reaction in the process of reducing iron ore to the metal is Fe2O3(s) ϩ 3 CO(g) ¡ 2 Fe(s) ϩ 3 CO2(g) (a) What is the maximum mass of iron, in grams, that can be obtained from 454 g (1.00 lb) of iron(III) oxide? (b) What mass of CO is required to react with 454 g of Fe2O3? 11. Iron metal reacts with oxygen to give iron(III) oxide, Fe2O3. (a) Write a balanced equation for the reaction. (b) If an ordinary iron nail (assumed to be pure iron) has a mass of 2.68 g, what mass of Fe2O3, in grams, is produced if the nail is converted completely to the oxide? (c) What mass of O2, in grams, is required for the reaction? 12. Methane, CH4, burns in oxygen. (a) What are the products of the reaction? (b) ■ Write the balanced equation for the reaction. (c) What mass of O2, in grams, is required for complete combustion of 25.5 g of methane? (d) What is the total mass of products expected from the combustion of 25.5 g of methane? 13. Sulfur dioxide, a pollutant produced by burning coal and oil in power plants, can be removed by reaction with calcium carbonate. 2 SO2(g) ϩ 2 CaCO3(s) ϩ O2(g) ¡ 2 CaSO4(s) ϩ 2 CO2(g) (a) What mass of CaCO3 is required to remove 155 g of SO2? (b) What mass of CaSO4 is formed when 155 g of SO2 is consumed completely? ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O 167 Study Questions 14. The formation of water-insoluble silver chloride is useful in the analysis of chloride-containing substances. Consider the following unbalanced equation: BaCl2(aq) ϩ AgNO3(aq) ¡ AgCl(s) ϩ Ba(NO3)2(aq) (a) Write the balanced equation. (b) What mass AgNO3, in grams, is required for complete reaction with 0.156 g of BaCl2? What mass of AgCl is produced? Amounts Tables and Chemical Stoichiometry For each question below set up an amounts table that lists the initial amount or amounts of reactants, the changes in amounts of reactants and products, and the amounts of reactants and products after reaction. See page 148 and Example 4.2. 15. A major source of air pollution years ago was the metals industry. One common process involved “roasting” metal sulfides in the air: 2 PbS(s) ϩ 3 O2(g) ¡ 2 PbO(s) ϩ 2 SO2(g) If you heat 2.5 mol of PbS in the air, what amount of O2 is required for complete reaction? What amounts of PbO and SO2 are expected? 16. ■ Iron ore is converted to iron metal in a reaction with carbon. 2 Fe2O3(s) ϩ 3 C(s) ¡ 4 Fe(s) ϩ 3 CO2(g) If 6.2 mol of Fe2O3(s) is used, what amount of C(s) is needed and what amounts of Fe and CO2 are produced? 17. Chromium metal reacts with oxygen to give chromium(III) oxide, Cr2O3. (a) Write a balanced equation for the reaction. (b) If a piece of chromium has a mass of 0.175 g, what mass (in grams) of Cr2O3 is produced if the metal is converted completely to the oxide? (c) What mass of O2 (in grams) is required for the reaction? 18. Ethane, C2H6, burns in oxygen. (a) What are the products of the reaction? (b) Write the balanced equation for the reaction. (c) What mass of O2, in grams, is required for complete combustion of 13.6 of ethane? (d) What is the total mass of products expected from the combustion of 13.6 g of ethane? Limiting Reactants and Amounts Tables (See Example 4.3 and Exercises 4.4 and 4.5. See also the General ChemistryNow Screens 4.7 and 4.8. In each case set up an amounts table.) 19. Sodium sulfide, Na2S, is used in the leather industry to remove hair from hides. (This is the reason these kinds of plants stink!) The Na2S is made by the reaction Na2SO4(s) ϩ 4 C(s) ¡ Na2S(s) ϩ 4 CO(g) Suppose you mix 15 g of Na2SO4 and 7.5 g of C. Which is the limiting reactant? What mass of Na2S is produced? ▲ More challenging 20. ■ Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride. CaO(s) ϩ 2 NH4Cl(s) ¡ 2 NH3(g) ϩ H2O(g) ϩ CaCl2(s) If 112 g of CaO and 224 g of NH4Cl are mixed, what mass of NH3 can be produced? 21. The compound SF6 is made by burning sulfur in an atmosphere of fluorine. The balanced equation is S8(s) ϩ 24 F2(g) ¡ 8 SF6(g) If you begin with 1.6 moles of sulfur, S8, and 35 moles of F2, which is the limiting reagent? 22. ■ Disulfur dichloride, S2Cl2, is used to vulcanize rubber. It can be made by treating molten sulfur with gaseous chlorine: S8(/) ϩ 4 Cl2(g) ¡ 4 S2Cl2(/) Starting with a mixture of 32.0 g of sulfur and 71.0 g of Cl2, which is the limiting reactant? 23. The reaction of methane and water is one way to prepare hydrogen for use as a fuel: CH4(g) ϩ H2O(g) ¡ CO(g) ϩ 3 H2(g) If you begin with 995 g of CH4 and 2510 g of water, (a) Which reactant is the limiting reactant? (b) What is the maximum mass of H2 that can be prepared? (c) What mass of the excess reactant remains when the reaction is completed? 24. ■ Aluminum chloride, AlCl3, is made by treating scrap aluminum with chlorine. 2 Al(s) ϩ 3 Cl2(g) ¡ 2 AlCl3(s) If you begin with 2.70 g of Al and 4.05 g of Cl2, (a) Which reactant is limiting? (b) What mass of AlCl3 can be produced? (c) What mass of the excess reactant remains when the reaction is completed? 25. Hexane (C6H14) burns in air (O2) to give CO2 and H2O. (a) Write a balanced equation for the reaction. (b) If 215 g of C6H14 is mixed with 215 g of O2, what masses of CO2 and H2O are produced in the reaction? (c) What mass of the excess reactant remains after the hexane has been burned? 26. ■ Aspirin, C6H4(OCOCH3)CO2H, is produced by the reaction of salicylic acid, C6H4(OH)CO2H, and acetic anhydride, (CH3CO)2O (page 157). C6H4(OH)CO2H(s) ϩ (CH3CO)2O(/) ¡ C6H4(OCOCH3)CO2H(s) ϩ CH3CO2H(/) If you mix 100. g of each of the reactants, what is the maximum mass of aspirin that can be obtained? ■ In General ChemistryNow Blue-numbered questions answered in Appendix O 168 Chapter 4 Chemical Equations and Stoichiometry Percent Yield (See Exercise 4.6 and General ChemistryNow Screen 4.9) 27. ■ In Example 4.3 you found that a mixture of CO and H2 produced 407 g CH3OH. CO(g) ϩ 2 H2(g) ¡ CH3OH(/) If only 332 g of CH3OH is actually produced, what is the percent yield of the compound? 28. Ammonia gas can be prepared by the following reaction: CaO(s) ϩ 2 NH4Cl(s) ¡ 2 NH3(g) ϩ H2O(g) ϩ CaCl2(s) If 112 g of CaO and 224 g of NH4Cl are mixed, the theoretical yield of NH3 is 68.0 g (Study Question 20). If only 16.3 g of NH3 is actually obtained, what is its percent yield? 29. The deep blue compound Cu(NH3)4SO4 is made by the reaction of copper(II) sulfate and ammonia. CuSO4(aq) ϩ 4 NH3(aq) ¡ Cu(NH3)4SO4(aq) (a) If you use 10.0 g of CuSO4 and excess NH3, what is the theoretical yield of Cu(NH3)4SO4? (b) If you isolate 12.6 g of Cu(NH3)4SO4, what is the percent yield of Cu(NH3)4SO4? 30. A reaction studied by Wächtershäuser and Huber (see “Black Smokers and the Origins of Life”) is 2 CH3SH ϩ CO ¡ CH3COSCH3 ϩ H2S If you begin with 10.0 g of CH3SH, and excess CO, (a) What is the theoretical yield of CH3COSCH3? (b) If 8.65 g of CH3COSCH3 is isolated, what is its percent yield? Analysis of Mixtures (See Examples 4.4 and 4.5 and General ChemistryNow Screen 4.10.) 31. ■ A mixture of CuSO4 and CuSO4 ؒ 5 H2O has a mass of 1.245 g. After heating to drive off all the water, the mass is only 0.832 g. What is the mass percent of CuSO4 ؒ 5 H2O in the mixture? (See page 129.) 32. A 2.634-g sample containing CuCl2 ؒ 2H2O and other materials was heated. The sample mass after heating to drive off the water was 2.125 g. What was the mass percent of CuCl2 ؒ 2H2O in the original sample? 33. A sample of limestone and other soil materials is heated, and the limestone decomposes to give calcium oxide and carbon dioxide. CaCO3(s) ¡ CaO(s) ϩ CO2(g) A 1.506-g sample of limestone-containing material gives 0.558 g of CO2, in addition to CaO, after being heated at a high temperature. What is the mass percent of CaCO3 in the original sample? 34. At higher temperatures NaHCO3 is converted quantitatively to Na2CO3. ▲ More challenging ■ In General ChemistryNow 2 NaHCO3(s) ¡ Na2CO3(s) ϩ CO2(g) ϩ H2O(g) Heating a 1.7184-g sample of impure NaHCO3 gives 0.196 g of CO2. What was the mass percent of NaHCO3 in the original 1.7184-g sample? 35. A pesticide contains thallium(I) sulfate, Tl2SO4. Dissolving a 10.20-g sample of impure pesticide in water and adding sodium iodide precipitates 0.1964 g of thallium(I) iodide, TlI. Tl2SO4(aq) ϩ 2 NaI(aq) ¡ 2 TlI(s) ϩ Na2SO4(aq) What is the mass percent of Tl2SO4 in the original 10.20-g sample? 36. ▲ The aluminum in a 0.764-g sample of an unknown material was precipitated as aluminum hydroxide, Al(OH)3, which was then converted to Al2O3 by heating strongly. If 0.127 g of Al2O3 is obtained from the 0.764-g sample, what is the mass percent of aluminum in the sample? Using Stoichiometry to Determine Empirical and Molecular Formulas (See Example 4.6, Exercise 4.9, and General ChemistryNow Screen 4.11.) 37. ■ Styrene, the building block of polystyrene, consists of only C and H. If 0.438 g of styrene is burned in oxygen and produces 1.481 g of CO2 and 0.303 g of H2O, what is the empirical formula of styrene? 38. Mesitylene is a liquid hydrocarbon. Burning 0.115 g of the compound in oxygen gives 0.379 g of CO2 and 0.1035 g of H2O. What is the empirical formula of mesitylene? 39. Cyclopentane is a simple hydrocarbon. If 0.0956 g of the compound is burned in oxygen, 0.300 g of CO2 and 0.123 g of H2O are isolated. (a) What is the empirical formula of cyclopentane? (b) If a separate experiment gave 70.1 g/mol as the molar mass of the compound, what is its molecular formula? 40. Azulene is a beautiful blue hydrocarbon. If 0.106 g of the compound is burned in oxygen, 0.364 g of CO2 and 0.0596 g of H2O are isolated. (a) What is the empirical formula of azulene? (b) If a separate experiment gave 128.2 g/mol as the molar mass of the compound, what is its molecular formula? 41. An unknown compound has the formula Cx HyOz. You burn 0.0956 g of the compound and isolate 0.1356 g of CO2 and 0.0833 g of H2O. What is the empirical formula of the compound? If the molar mass is 62.1 g/mol, what is the molecular formula? (See Exercise 4.9.) 42. ■ An unknown compound has the formula Cx HyOz. You burn 0.1523 g of the compound and isolate 0.3718 g of CO2 and 0.1522 g of H2O. What is the empirical formula of the compound? If the molar mass is 72.1 g/mol, what is the molecular formula? (See Exercise 4.9.) Blue-numbered questions answered in Appendix O 169 Study Questions 43. Nickel forms a compound with carbon monoxide, Nix(CO)y. To determine its formula, you carefully heat a 0.0973-g sample in air to convert the nickel to 0.0426 g of NiO and the CO to 0.100 g of CO2. What is the empirical formula of Nix(CO)y? pound, is exhaled, giving the breath of untreated diabetics a distinctive odor. The acetone is produced by a breakdown of fats in a series of reactions. The equation for the last step is CH3COCH2CO2H ¡ CH3COCH3 ϩ CO2 44. To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen to give Fe2O3 and CO2. If you burn 1.959 g of Fex(CO)y and obtain 0.799 g of Fe2O3 and 2.200 g of CO2, what is the empirical formula of Fex(CO)y? General Questions on Stoichiometry These questions are not designated as to type or location in the chapter. They may combine several chapters. 45. Balance the following equations: (a) The synthesis of urea, a common fertilizer acetone, CH3COCH3 What mass of acetone can be produced from 125 mg of acetoacetic acid (CH3COCH2CO2H)? CO2(g) ϩ NH3(g) ¡ NH2CONH2(s) ϩ H2O(/) (b) Reactions used to make uranium(VI) fluoride for the enrichment of natural uranium UO2(s) ϩ HF(aq) ¡ UF4(s) ϩ H2O(/) UF4(s) ϩ F2(g) ¡ UF6(s) 50. Your body deals with excess nitrogen by excreting it in the form of urea, NH2CONH2. The reaction producing it is the combination of arginine (C6H14N4O2) with water to give urea and ornithine (C5H12N2O2). C6H14N4O2 ϩ H2O ¡ NH2CONH2 ϩ C5H12N2O2 (c) The reaction to make titanium(IV) chloride, which is then converted to titanium metal Arginine TiO2(s) ϩ Cl2(g) ϩ C(s) ¡ TiCl4(/) ϩ CO(g) TiCl4(/) ϩ Mg(s) ¡ Ti(s) ϩ MgCl2(s) NaBH4(s) ϩ H2SO4(aq) ¡ B2H6(g) ϩ H2(g) ϩ Na2SO4(aq) (c) Reaction to produce tungsten metal from tungsten(VI) oxide WO3(s) ϩ H2(g) ¡ W(s) ϩ H2O(/) (d) Decomposition of ammonium dichromate (NH4)2Cr2O7(s) ¡ N2(g) ϩ H2O(/) ϩ Cr2O3(s) 47. ■ Suppose 16.04 g of benzene, C6H6, is burned in oxygen. (a) What are the products of the reaction? (b) What is the balanced equation for the reaction? (c) What mass of O2, in grams, is required for complete combustion of benzene? (d) What is the total mass of products expected from 16.04 g of benzene? 48. If 10.0 g of carbon is combined with an exact, stoichiometric amount of oxygen (26.6 g) to produce carbon dioxide, what is the theoretical yield of CO2, in grams? Ornithine If you excrete 95 mg of urea, what mass of arginine must have been used? What mass of ornithine must have been produced? 51. In the Figure 4.2, you see the reaction of iron metal and chlorine gas to give iron(III) chloride. (a) Write the balanced chemical equation for the reaction. (b) Beginning with 10.0 g of iron, what mass of Cl2, in grams, is required for complete reaction? What mass of FeCl3 can be produced? (c) If only 18.5 g of FeCl3 is obtained from 10.0 g of iron and excess Cl2, what is the percent yield? (d) If equal masses of iron and chlorine are combined (10.0 g of each), what is the theoretical yield of iron(III) chloride? 52. Two beakers sit on a balance; the total mass is 161.170 g. Charles D. Winters 46. Balance the following equations: (a) Reaction to produce “superphosphate” fertilizer Ca3(PO4)2(s) ϩ H2SO4(aq) ¡ Ca(H2PO4)2(aq) ϩ CaSO4(s) (b) Reaction to produce diborane, B2H6 Urea Solutions of KI and Pb(NO3)2 before reaction. 49. The metabolic disorder diabetes causes a buildup of acetone, CH3COCH3, in the blood. Acetone, a volatile com▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O 170 Chapter 4 Chemical Equations and Stoichiometry One beaker contains a solution of KI; the other contains a solution of Pb(NO3)2. When the solution in one beaker is poured completely into the other, the following reaction occurs: 2 KI(aq) ϩ Pb(NO3)2(aq) ¡ 2 KNO3(aq) ϩ PbI2(s) 55. Sodium azide, the explosive chemical used in automobile airbags, is made by the following reaction: NaNO3 ϩ 3 NaNH2 ¡ NaN3 ϩ 3 NaOH ϩ NH3 If you combine 15.0 g of NaNO3 (85.0 g/mol ) with 15.0 g of NaNH2, what mass of NaN3 is produced? 56. Iodine is made by the reaction Charles D. Winters 2 NaIO3(aq) ϩ 5 NaHSO3(aq) ¡ 3 NaHSO4(aq)+ 2 Na2SO4(aq) ϩ H2O(/) ϩ I2(aq) (a) Name the two reactants. (b) If you wish to prepare 1.00 kg of I2, what mass of NaIO3 is required? What mass of NaHSO3? Solutions after reaction. What is the total mass of the beakers and solutions after reaction? Explain completely. (See the General ChemistryNow Screen 4.3, Exercise 1.) 53. ■ Some metal halides react with water to produce the metal oxide and the appropriate hydrogen halide (see photo). For example, TiCl4(/) ϩ 2 H2O(/) ¡ TiO2(s) ϩ 4 HCl(g) 57. Copper(I) sulfide reacts with O2 upon heating to give copper metal and sulfur dioxide. (a) Write a balanced equation for the reaction. (b) What mass of copper metal can be obtained from 500. g of copper(I) sulfide? 58. Saccharin, an artificial sweetener, has the formula C7H5NO3S. Suppose you have a sample of a saccharincontaining sweetener with a mass of 0.2140 g. After decomposition to free the sulfur and convert it to the SO42Ϫ ion, the sulfate ion is trapped as water-insoluble BaSO4 (see Figure 4.8). The quantity of BaSO4 obtained is 0.2070 g. What is the mass percent of saccharin in the sample of sweetener? 59. ▲ Boron forms an extensive series of compounds with hydrogen, all with the general formula Bx Hy. Bx Hy(s) ϩ excess O2(g) ¡ x 2 y B2O3(s) ϩ 2 H2O(g) If 0.148 g of Bx Hy gives 0.422 g of B2O3 when burned in excess O2, what is the empirical formula of Bx Hy? Charles D. Winters 60. ▲ Silicon and hydrogen form a series of compounds with the general formula Six Hy. To find the formula of one of them, a 6.22-g sample of the compound is burned in oxygen. All of the Si is converted to 11.64 g of SiO2, and all of the H is converted to 6.980 g of H2O. What is the empirical formula of the silicon compound? (a) Name the four compounds involved in this reaction. (b) If you begin with 14.0 mL of TiCl4 (d ϭ 1.73 g/mL), what mass of water, in grams, is required for complete reaction? (c) What mass of each product is expected? 54. The reaction of 750. g each of NH3 and O2 was found to produce 562 g of NO (see pages 153–155). 4 NH3(g) ϩ 5 O2(g) ¡ 4 NO(g) ϩ 6 H2O(g) (a) What mass of water is produced by this reaction? (b) What quantity of O2 is required to consume 750. g of NH3? ▲ More challenging ■ In General ChemistryNow 61. ▲ Menthol, from oil of mint, has a characteristic odor. The compound contains only C, H, and O. If 95.6 mg of menthol burns completely in O2, and gives 269 mg of CO2 and 110 mg of H2O, what is the empirical formula of menthol? 62. ▲ Quinone, a chemical used in the dye industry and in photography, is an organic compound containing only C, H, and O. What is the empirical formula of the compound if 0.105 g of the compound gives 0.257 g of CO2 and 0.0350 g of H2O when burned completely in oxygen? 63. ▲ In the Simulation portion of Screen 4.8 of the General ChemistryNow CD-ROM or website, choose the reaction of FeCl2 and Na2S. (a) Write the balanced equation for the reaction. (b) Choose 40 g of Na2S as one reactant and add 40 g of FeCl2.What is the limiting reactant? Blue-numbered questions answered in Appendix O 171 Study Questions (c) What mass of FeS is produced? (d) What mass of Na2S or FeCl2 remains after the reaction? (e) What mass of FeCl2 is required to react completely with 40 g of Na2S? 64. Sulfuric acid can be prepared starting with the sulfide ore, cuprite (Cu2S). If each S atom in Cu2S leads to one molecule of H2SO4, what mass of H2SO4 can be produced from 3.00 kg of Cu2S? 65. ▲ In an experiment 1.056 g of a metal carbonate, containing an unknown metal M, is heated to give the metal oxide and 0.376 g CO2. MCO3(s) ϩ heat ¡ MO(s) ϩ CO2(g) What is the identity of the metal M? (a) M ϭ Ni (c) M ϭ Zn (b) M ϭ Cu (d) M ϭ Ba 66. ■ ▲ An unknown metal reacts with oxygen to give the metal oxide, MO2. Identify the metal based on the following information: Mass of metal ϭ 0.356 g Mass of sample after converting metal completely to oxide ϭ 0.452 g 67. ▲ Titanium(IV) oxide, TiO2, is heated in hydrogen gas to give water and a new titanium oxide, TixOy . If 1.598 g of TiO2 produces 1.438 g of TixOy, what is the formula of the new oxide? 68. ▲ Thioridazine, C21H26N2S2, is a pharmaceutical used to regulate dopamine. (Dopamine, a neurotransmitter, affects brain processes that control movement, emotional response, and ability to experience pleasure and pain.) A chemist can analyze a sample of the pharmaceutical for the thioridazine content by decomposing it to convert the sulfur in the compound to sulfate ion. This is then “trapped” as water-insoluble barium sulfate (see Figure 4.8). SO42Ϫ(aq, from thioridazine) ϩ BaCl2(aq) ¡ BaSO4(s) ϩ 2 ClϪ(aq) Suppose a 12-tablet sample of the drug yielded 0.301 g of BaSO4. What is the thioridazine content, in milligrams, of each tablet? 69. ■ ▲ A herbicide contains 2,4-D (2,4-dichlorophenoxyacetic acid), C8H6Cl2O3. A 1.236-g sample of the herbicide was decomposed to liberate the chlorine as ClϪ ion. This was precipitated as AgCl, with a mass of 0.1840 g. What is the mass percent of 2,4-D in the sample? C C Cl C C H What mass of Cl2(g) is required to produce 234 kg of KClO4? 71. ▲ Commercial sodium “hydrosulfite” is 90.1% pure Na2S2O4. The sequence of reactions used to prepare the compound is Zn(s) ϩ 2 SO2(g) ¡ ZnS2O4(s) ZnS2O4(s) ϩ Na2CO3(aq) ¡ ZnCO3(s) ϩ Na2S2O4(aq) (a) What mass of pure Na2S2O4 can be prepared from 125 kg of Zn, 500 g of SO2, and an excess of Na2CO3? (b) What mass of the commercial product would contain the Na2S2O4 produced using the amounts of reactants in part (a)? 72. ■ What mass of lime, CaO, can be obtained by heating 125 kg of limestone that is 95.0% by mass CaCO3? CaCO3(s) ¡ CaO(s) ϩ CO2(g) 73. Sulfuric acid can be produced from a sulfide ore such as iron pyrite by the following sequence of reactions: 4 FeS2(s) ϩ 11 O2(g) ¡ 2 Fe2O3(s) ϩ 8 SO2(g) 2 SO2(g) ϩ O2(g) ¡ 2 SO3(g) SO3(g) ϩ H2O(/) ¡ H2SO4(/) Starting with 525 kg of FeS2 (and an excess of other reactants), what mass of pure H2SO4 can be prepared? 74. ▲ The elements silver, molybdenum, and sulfur combine to form Ag2MoS4. What is the maximum mass of Ag2MoS4 that can be obtained if 8.63 g of silver, 3.36 g of molybdenum, and 4.81 g of sulfur are combined? 75. ▲ A mixture of butene, C4H8, and butane, C4H10, is burned in air to give CO2 and water. Suppose you burn 2.86 g of the mixture and obtain 8.80 g of CO2 and 4.14 g of H2O. What is the weight percents of butene and butane in the mixture? 76. ▲ Cloth can be waterproofed by coating it with a silicone layer. This is done by exposing the cloth to (CH3)2SiCl2 vapor. The silicon compound reacts with OH groups on the cloth to form a waterproofing film (density ϭ 1.0 g/cm3)of [(CH3)2SiO]n, where n is a large integer number. The coating is added layer by layer, each layer of [(CH3)2SiO]n being 0.60 nm thick. Suppose you want to waterproof a piece of cloth that is 3.00 m square, and you want 250 layers of waterproofing compound on the cloth. What mass of (CH3)2SiCl2 do you need? C C Cl2(g) ϩ 2 KOH(aq) ¡ KCl(aq) ϩ KClO(aq) ϩ H2O(/) 3 KClO(aq) ¡ 2 KCl(aq) ϩ KClO3(aq) 4 KClO3(aq) ¡ 3 KClO4(aq) ϩ KCl(aq) n(CH3)2SiCl2 ϩ 2n OHϪ ¡ 2n ClϪ ϩ n H2O ϩ [(CH3)2SiO]n OCH2CO2H H 70. ■ ▲ Potassium perchlorate is prepared by the following sequence of reactions: H Cl ▲ More challenging ■ In General ChemistryNow Blue-numbered questions answered in Appendix O 172 Chapter 4 Chemical Equations and Stoichiometry 77. ▲ Sodium hydrogen carbonate, NaHCO3, can be decomposed quantitatively by heating. (iii) When 2.50 g of Fe is added to the Br2, both reactants are used up completely. (iv) When 2.00 g of Fe is added to the Br2, 10.0 g of product is formed. The percent yield must therefore be 20.0%. 2 NaHCO3(s) ¡ Na2CO3(s) ϩ CO2(g) ϩ H2O(g) A 0.682-g sample of impure NaHCO3 yielded a solid residue (consisting of Na2CO3 and other solids) with a mass of 0.467 g. What was the mass percent of NaHCO3 in the sample? 80. Chlorine and iodine react according to the balanced equation I2(g) ϩ 3 Cl2(g) ¡ 2 ICl3(g) 78. ▲ Copper metal can be prepared by roasting copper ore, which can contain cuprite (Cu2S) and copper(II) sulfide. Suppose that you mix I2 and Cl2 in a flask and that the mixture is represented by the diagram below. Cu2S(s) ϩ O2(g) ¡ 2 Cu(s) ϩ SO2(g) CuS(s) ϩ O2(g) ¡ Cu(s) ϩ SO2(g) ¡ I2 Suppose an ore sample contains 11.0% impurity in addition to a mixture of CuS and Cu2S. Heating 100.0 g of the mixture produces 75.4 g of copper metal with a purity of 89.5%. What is the weight percent of CuS in the ore? The weight percent of Cu2S? ¡ Cl2 Summary and Conceptual Questions When the reaction between the Cl2 and I2 (according to the balanced equation above) is complete, which panel below represents the outcome? Which compound is the limiting reactant? The following questions use concepts from the preceding chapters. 79. ▲ A weighed sample of iron (Fe) is added to liquid bromine (Br2) and allowed to react completely. The reaction produces a single product, which can be isolated and weighed. The experiment was repeated a number of times with different masses of iron but with the same mass of bromine. (See the graph below.) Mass of product (g) 12 10 (a) 8 (b) (d) (c) (e) 6 4 2 0 0 1 2 3 4 Mass of Fe (g) (a) What mass of Br2 is used when the reaction consumes 2.0 g of Fe? (b) What is the mole ratio of Br2 to Fe in the reaction? (c) What is the empirical formula of the product? (d) Write the balanced chemical equation for the reaction of iron and bromine. (e) What is the name of the reaction product? (f ) Which statement or statements best describe the experiments summarized by the graph? (i) When 1.00 g of Fe is added to the Br2, Fe is the limiting reagent. (ii) When 3.50 g of Fe is added to the Br2, there is an excess of Br2. ▲ More challenging ■ In General ChemistryNow 81. Cisplatin [Pt(NH3)2Cl2] is a cancer chemotherapy agent. Notice that it contains NH3 groups attached to platinum. (a) What is the weight percent of Pt, N, and Cl in the cisplatin? (b) Cisplatin is made by reacting K2PtCl4 with ammonia. K2PtCl4(aq) ϩ 2 NH3(aq) ¡ Pt(NH3)2Cl2(aq) ϩ 2 KCl(aq) If you begin with 16.0 g of K2PtCl4, what mass of ammonia should be used to completely consume the K2PtCl4? What mass of cisplatin will be produced? Blue-numbered questions answered in Appendix O 173 Study Questions When the reactants are combined, the H2 inflates the balloon attached to the flask. The results are as follows: Flask 1: Balloon inflates completely but some Zn remains when inflation ceases. Flask 2: Balloon inflates completely. No Zn remains. Flask 3: Balloon does not inflate completely. No Zn remains. 82. Iron(III) chloride is produced by the reaction of iron and chlorine (Figure 4.2). (a) If you place 1.54 g of iron gauze in chlorine gas, what mass of chlorine is required for complete reaction? What mass of iron(III) chloride is produced? (b) Iron(III) chloride reacts readily with NaOH to produce iron(III) hydroxide and sodium chloride. If you mix 2.0 g of iron(III) chloride with 4.0 g of NaOH, what mass of iron(III) hydroxide is produced? (See the General ChemistryNow Screen 4.8 Simulation.) 83. Let us explore a reaction with a limiting reactant. (See the General ChemistryNow Screen 4.8.) Here zinc metal is added to a flask containing aqueous HCl, and H2 gas is a product. Zn(s) ϩ 2 HCl(aq) ¡ ZnCl2(aq) ϩ H2(g) Explain these results completely. Perform calculations that support your explanation. 84. The reaction of aluminum and bromine is pictured in Figure 3.1 and below. The white solid on the lip of the beaker at the end of the reaction is Al2Br6. In the reaction pictured below, which was the limiting reactant, Al or Br2? (See General ChemistryNow Screen 4.2.) Charles D. Winters The three flasks each contain 0.100 mol of HCl. Zinc is added to each flask in the following quantities. Flask 1: 7.00 g Zn Flask 2: 3.27 g Zn Flask 3: 1.31 g Zn Charles D. Winters Before reaction Flask 1 Flask 2 Flask 3 ▲ More challenging After reaction Do you need a live tutor for homework problems? Access vMentor at General ChemistryNow at http://now.brookscole.com/kotz6e for one-on-one tutoring from a chemistry expert ■ In General ChemistryNow Blue-numbered questions answered in Appendix O ...
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