CHAPTER 11 Stoichiometry 11.1 Defining Stoichiometry Stoichiometry is the study of quantitative relationships between amounts of reactants used and products formed by a chemical reaction. Stoichiometry is based on the law of conservation of mass. The law of conservation of mass states that matter is neither created nor destroyed. Thus, in a chemical reaction, the mass of the reactants equals the mass of the products. You can use stoichiometry to answer questions about the amounts of reactants used or products formed by reactions. For example, look at the balanced chemical equation for the formation of table salt (NaCl). Na(s) Cl (g) NaCl(s) You could use stoichiometry to answer the following questions about the chemical reaction. How much sodium is needed to produce a certain amount of table salt? How much chlorine is needed to produce a certain amount of table salt? Given a certain amount of sodium or chlorine, how much table salt can be produced? When you look at a balanced equation, there are two ways to interpret the coefficients. The coefficients tell you how many individual particles are interacting in the chemical reaction. For example, from the chemical equation above, you learn that two sodium atoms react with one chlorine molecule to form two formula units of table salt. You also learn that two moles of sodium react with one mole of chlorine to form two moles of salt. What the coefficients do not tell you directly is the masses of the reactants and the products in the chemical reaction. Example Problem 1 Interpreting Chemical Equations Interpret the balanced chemical equation in terms of particles, moles, and mass. Show that the law of conservation of mass is observed. 1
4NH 3 5O 4NO 6H O The coefficients represent both the numbers of particles and the numbers of moles interacting in the chemical reaction. 4 molecules NH 3 5 molecules O 4 molecules NO 6 molecules H O 4 moles NH 3 + 5 moles O 4 moles NO + 6 moles H O You can calculate the mass of each reactant and product by multiplying the number of moles by the conversion factor molar mass. 4 mol NH 3 17.03 g NH 3 1 mol NH 3 68.1 g NH 3 3.00 g O 5 mol O 160.0 g O 1 mol O 30.01 g NO 4 mol NO 10.0 g NO 1 mol NO 6 mol H O 18.0 g H O 1 mol H O 108.1 g H O The law of conservation of mass is observed because the mass of the reactants (68.1 g NH 3 + 160.0 g O = 8.1 g) equals the mass of the products (10.0 g NO + 108.1 g H O 8.1 g). Practice Problems 1. Interpret each balanced equation in terms of particles, moles, and mass. Show that the law of conservation of mass is observed. a. H O (l) O (g) H O(l) b. H CO 3(aq) H O(l) CO (g) c. 4HCl(aq) O (g) H O(l) Cl (g) As you know, the coefficients in a balanced chemical equation indicate the relationships among the moles of reactants and products in the reaction. You can use the coefficients to write mole ratios. A mole ratio is a ratio between the numbers of moles of any two substances in a balanced chemical equation. What mole ratios can be written for the following chemical equation?
H O (l) O (g) H O(l) mol H O 1 mol O and mol H O mol H O 1 mol O mol H O and 1 mol O mol H O mol H O mol H O and mol H O 1 mol O To determine the number of mole ratios that defines a given chemical reaction, multiply the number of species in the equation by the next lower number. Thus, a chemical reaction with three participating species can be defined by six mole ratios (3 6); a chemical reaction with four species can be defined by 1 mole ratios (4 3 1). Why learn to write mole ratios? They are the key to calculations that are based on chemical equations. Using a balanced chemical equation, mole ratios derived from the equation, and a given amount of one of the reactants or products, you can calculate the amount of any other participant in the reaction. Practice Problems. Determine all the mole ratios for the following balanced chemical equations. a. N (g) O (g) NO(g) b. 4NH 3(aq) 5O (g) 4NO(g) 6H O(l) c. 4HCl(aq) O (g) H O(l) Cl (g) 11. Stoichiometric Calculations There are three basic stoichiometric calculations: mole-to-mole conversions, mole-to-mass conversions, and mass-to-mass conversions. All stoichiometric calculations begin with a balanced equation and mole ratios. Stoichiometric mole-to-mole conversion How can you determine the number of moles of table salt (NaCl) produced from 0.0 moles of chlorine (Cl )? 3
First, write the balanced equation. Na(s) Cl (g) NaCl(s) Then, use the mole ratio to convert the known number of moles of chlorine to the number of moles of table salt. Use the formula below. moles of known moles of unknown moles of known moles of unknown 0.0 mol Cl mol NaCl 1 mol Cl 0.04 mol Cl Example Problem Stoichiometric Mole-to-Mole Conversion A piece of magnesium burns in the presence of oxygen, forming magnesium oxide (MgO). How many moles of oxygen are needed to produce 1 moles of magnesium oxide? Write the balanced equation and the mole ratio that relates mol O to mol MgO. Mg(s) + O (g) MgO(s) 1 mol O mol MgO Multiply the known number of moles of MgO by the mole ratio. 1 mol MgO 1 mol O mol MgO 6 mol O Six moles of oxygen is needed to produce 1 moles of magnesium oxide. Practice Problems 3. The carbon dioxide exhaled by astronauts can be removed from a spacecraft by reacting it with lithium hydroxide (LiOH). The reaction is as follows: CO(g) LiOH(s) LiCO3(s) HO(l). An average person exhales about 0 moles of CO per day. How many moles of LiOH would be required to maintain two astronauts in a spacecraft for three days? 4
4. Balance the following equation and answer the questions below. KClO 3(s) KCl(s) O (g) a. How many moles of O are produced from 10 moles of KClO 3? b. How many moles of KCl are produced using 3 moles of KClO 3? c. How many moles of KClO 3 are needed to produce 50 moles of O? Stoichiometric mole-to-mass conversion A mole-to-mass conversion allows you to calculate the mass of a product or reactant in a chemical reaction given the number of moles of a reactant or product. Example Problem 3 Stoichiometric Mole-to-Mass Conversion The following reaction occurs in plants undergoing photosynthesis. 6CO (g) 6H O(l) C 6H 1O 6(s) 6O (g) How many grams of glucose (C 6H 1O 6) are produced when 4.0 moles of carbon dioxide reacts in excess water? Determine the number of moles of glucose produced by the given amount of carbon dioxide. 4.0 mol CO 1 mol C 6 H 1 O 6 6 mol CO 4.00 mol C 6 H 1 O 6 Multiply by the molar mass. 4.00 mol C 6 H 1 O 6 180.18 g C 6 H 1 O 6 1 mol C 6 H 1 O 6 71 g C 6 H 1 O 6 71 grams of glucose is produced from 4.0 moles of carbon dioxide. Practice Problems 5. Calculate the mass of sodium chloride (NaCl) produced when 5.50 moles of sodium reacts in excess chlorine gas. 5
6. How many grams of chlorine gas must be reacted with excess sodium iodide (NaI) to produce 6.00 moles of sodium chloride? a. Balance the equation: NaI(aq) Cl (g) NaCl(aq) I (s). b. Perform the calculation. 7. Calculate the mass of hydrochloric acid (HCl) needed to react with 5.00 moles of zinc. a. Balance the equation: Zn(s) HCl(aq) ZnCl (aq) H (g). b. Perform the calculation. Stoichiometric mass-to-mass conversion If you were preparing to carry out a chemical reaction in the laboratory, you would need to know how much of each reactant to use in order to produce a certain mass of product. This is one instance when you would use a mass-tomass conversion. In this calculation, you can find the mass of an unknown substance in a chemical equation if you have the balanced chemical equation and know the mass of one substance in the equation. Example Problem 4 Stoichiometric Mass-to-Mass Conversion How many grams of sodium hydroxide (NaOH) are needed to completely react with 50.0 grams of sulfuric acid (H ) to form sodium sulfate (Na ) and water? Write the balanced equation. NaOH(aq) + H (aq) Na + H O(g) Convert grams of sulfuric acid to moles NaOH. 50.0 g H 1 mol H 98.09 g H 0.510 mol H 0.510 mol H Calculate the mass of NaOH needed. mol NaOH 1 mol H 1.0 mol NaOH 1.0 mol NaOH 40.00 g NaOH 1 mol NaOH 40.8 g NaOH 50.0 grams of H reacts completely with 40.8 grams of NaOH. 6
Practice Problems 8. Balance each equation and solve the problem. a. If 40.0 g of magnesium reacts with excess hydrochloric acid (HCl), how many grams of magnesium chloride (MgCl ) are produced? Mg(s) HCl(aq) MgCl (aq) H (g) b. Determine the mass of copper needed to react completely with a solution containing 1.0 g of silver nitrate (AgNO 3). Cu(s) AgNO 3(aq) Cu(NO 3) (aq) Ag(s) c. How many grams of hydrogen chloride (HCl) are produced when 15.0 g of sodium chloride (NaCl) reacts with excess sulfuric acid (H )? NaCl(aq) H (aq) Na HCl(g) d. Calculate the mass of silver phosphate (Ag 3PO 4) produced if 30.0 g of silver acetate (AgCH 3COO) reacts with excess sodium phosphate (Na 3PO 4). AgCH 3COO(aq) Na 3PO 4(aq) Ag 3PO 4(s) NaCH 3COO(aq) Steps in stoichiometric calculations Follow these basic steps when performing stoichiometric calculations. 1. Write the balanced equation.. Determine the moles of the given substance using a mass-to-mole conversion. Use the inverse of the molar mass as the conversion factor. 3. Determine the moles of the unknown substance from the moles of the given substance. Use the mole ratio from the balanced equation as the conversion factor. 4. From the moles of the unknown substance, determine the mass of the unknown substance using a mole-tomass conversion. Use the molar mass as the conversion factor. 7
11.3 Limiting Reactants Rarely are the reactants in a chemical reaction present in the exact mole ratios specified in the balanced equation. Usually, one or more of the reactants are present in excess, and the reaction proceeds until all of one reactant is used up. The reactant that is used up is called the limiting reactant. The limiting reactant limits the reaction and, thus, determines how much of the product forms. The left-over reactants are called excess reactants. How can you determine which reactant in a chemical reaction is limited? First, find the number of moles of each reactant by multiplying the given mass of each reactant by the inverse of the molar mass. Next, determine whether the reactants are available in the mole ratio specified in the balanced equation. A reactant that is available in an amount smaller than that required by the mole ratio is a limiting reactant. After the limiting reactant has been determined, calculate the amount of product that can ideally form from the given amount of the limiting reactant. To do this, multiply the given number of moles of the limiting reactant by the mole ratio that relates the limiting reactant to the product. Then, convert moles of product to mass using the molar mass of the product as the conversion factor. Example Problem 5 Determining the Limiting Reactant In the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H ). NaOH(aq) H (aq) Na H O(g) a. Which reactant is the limiting reactant? b. What mass of Na can be produced using the given quantities of the reactants? c. To determine the limiting reactant, calculate the actual ratio of available moles of reactants. 40.0 g NaOH 1 mol NaOH 40.0 g NaOH 1.00 mol NaOH 8
So, 60.0 g H 1 mol H 98.09 g H 0.61 mol H 1.00 mol NaOH 0.61 mol H is available. Compare this ratio with the mole ratio from the balanced equation: mol NaOH 1 mol H, or 1 mol NaOH 0.5 mol H. You can see that when 0.5 mol H has reacted, all of the 1.00 mol of NaOH would be used up. Some H would remain unreacted. Thus, NaOH is the limiting reactant. d. To calculate the mass of Na that can form from the given reactants, multiply the number of moles of the limiting reactant (NaOH) by the mole ratio of the product to the limiting reactant and then multiply by the molar mass of the product. 1.00 mol NaOH 1 mol Na mol NaOH 14.04 g Na 1 mol Na 71.0 g Na 71.0 g of Na can form from the given amounts of the reactants. Practice Problems 9. Ammonia (NH3) is one of the most common chemicals produced in the United States. It is used to make fertilizer and other products. Ammonia is produced by the following chemical reaction. N (g) 3H (g) NH 3(g) a. If you have 1.00 10 3 g of N and.50 10 3 g of H, which is the limiting reactant in the reaction? b. How many grams of ammonia can be produced from the amount of limiting reactant available? c. Calculate the mass of excess reactant that remains after the reaction is complete. 9
10. Aluminum reacts with chlorine to produce aluminum chloride. a. Balance the equation: Al(s) Cl (g) AlCl 3(s). b. If you begin with 3. g of aluminum and 5.4 g of chlorine, which is the limiting reactant? c. How many grams of aluminum chloride can be produced from the amount of limiting reactant available? d. Calculate the mass of excess reactant that remains after the reaction is complete. Reactions do not always continue until all of the reactants are used up. Using an excess of the least expensive reactant in a reaction can ensure that all of the more expensive reactant is used up, making the chemical reaction more efficient and cost-effective. Using an excess of one reactant can also speed up some reactions. 11.4 Percent Yield Most chemical reactions do not produce the predicted amount of product. Although your work so far with stoichiometric problems may have led you to believe that chemical reactions proceed according to the balanced equation and always produce the calculated amount of product, it s not true. Many reactions stop before all the reactants are used up, so less product is formed than expected. Also, products other than those expected sometimes form from competing chemical reactions, thereby reducing the amount of the desired product. The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant under ideal circumstances. This is the amount you have been calculating in practice problems so far. Chemical reactions hardly ever produce the theoretical yield. The actual yield is the amount of product that is actually produced when a chemical reaction is carried out in an experiment. It is determined by measuring the mass of the product. Percent yield of product is the ratio of the actual yield to the theoretical yield expressed as a percent. Percent yield actual yield (from an experiment) theoretical yield (from stoichiometric calculations) 100 10
Percent yield tells you how efficient a chemical reaction is in producing the desired product. Example Problem 6 Calculating Percent Yield Aspirin (C 9H 8O 4) can be made from salicylic acid (C 7H 6O 3) and acetic anhydride (C 4H 6O 3). Suppose you mix 13. g of salicylic acid with an excess of acetic anhydride and obtain 5.9 g of aspirin and some water. Calculate the percent yield of aspirin in this reaction. Write the balanced equation. C 7H 6O 3(s) C 4H 6O 3(l) C 9H 8O 4(s) + H O(l) Calculate the theoretical yield. Salicylic acid is the limiting reactant. 13. g C 7 H 6 O 3 1 mol C 7 H 6 O 3 138.1 g C 7 H 6 O 3 0.0956 mol C 7 H 6 O 3 0.0956 mol C 7 H 6 O 3 mol C 9 H 8 O 4 mol C 7 H 6 O 3 0.0956 mol C 9 H 8 O 4 0.0956 mol C 9 H 8 O 4 180. g C 9 H 8 O 4 1 mol C 9 H 8 O 4 17. g C 9 H 8 O 4 Calculate the percent yield. Practice Problems 5.9 g C 9 H 8 O 4 17. g C 9 H 8 O 4 100 34% 11. Calculate the percent yield for each chemical reaction based on the data provided. a. theoretical yield: 5 g; actual yield: 0 g b. theoretical yield: 55 g; actual yield: 4 g c. theoretical yield: 5. g; actual yield: 4.9 g 1. Calculate the actual yield for each chemical reaction based on the data provided. a. theoretical yield: 0 g; percent yield: 95% b. theoretical yield: 75 g; percent yield: 88% c. theoretical yield: 9. g; percent yield: 6% 11
13. In an experiment, 10.0 g of magnesium reacted with excess hydrochloric acid forming magnesium chloride. Mg(s) HCl(aq) MgCl (aq) H (g) At the completion of the reaction, 9.5 g of magnesium chloride was produced. Calculate the theoretical yield and the percent yield. Percent yield is important in the calculation of overall cost effectiveness in industrial processes. Manufacturers must reduce the cost of making products to the lowest level possible. For example, sulfuric acid (H ) is a raw material for many products, including fertilizers, detergents, pigments, and textiles. The cost of sulfuric acid affects the cost of many consumer items that use sulfuric acid as a raw material. The manufacture of sulfuric acid is sometimes achieved using a two-step process called the contact process. The two steps are as follows. S 8(s) 8O (g) 8SO (g) SO (g) O (g) SO 3(g) The last step results in sulfuric acid as the product. SO 3(g) H O(l) H (aq) The first step produces almost 100 percent yield. The second step will also produce a high yield if a catalyst is used. A catalyst is a substance that speeds up a chemical reaction but is not used up in the chemical reaction and does not appear in the chemical equation. Chapter 11 Review 14. What is stoichiometry? 15. Write two questions that stoichiometry can help you answer about the following chemical equation. 4HCl(aq) O (g) H O(l) Cl (g) 16. Relate the law of conservation of mass to stoichiometry. 17. What is the difference between a limiting reactant and an excess reactant? 1
Answer Key Chapter 11 Practice Problems 1. a. HO(l) O(g) HO(l) molecules H O 1 molecule O molecules H O moles H O 1 mole O moles H O 68.04 g H O 3.00 g O 36.04 g H O, as shown below. 68.04 g reactant 68.04 g products 34.0 g HO mol HO 1 mol H O 1 mol O mol H O 68.04 g HO 3.00 g O 3.00 g O 1 mol O 18.0 g H O 1 mol H O 36.04 g H O b. H CO 3(aq) H O(l) CO (g) 1 formula unit H CO 3 1 molecule H O 1 molecule CO 1 mole H CO 3 1 mole H O 1 mole CO 6.03 g H CO 3 18.0 g H O 44.01 g CO, as shown below. 6.03 g reactant 6.03 g products 6.03 g HCO3 1 mol HCO3 6.03 g HCO3 1 mol H CO 1 mol H O 1 mol CO 18.0 g H O 3 1 mol H O 44.01 g CO 18.0 g H O 44.01 g CO 1 mol CO 13
Answer Key (continued) c. 4HCl(aq) O (g) H O(l) Cl (g) 4 molecules HCl 1 molecule O molecules H O molecules Cl 4 mole HCl 1 mol O mole H O mole Cl 146 g HCl 3.00 g O 36.04 g H O 14 g Cl, as shown below. 177.84 g reactants 177.84 g products 36.46 g HCl 4 mol HCl 145.84 g HCl 1 mol HCl 3.00 g O 1 mol O 3.00 g O 1 mol O 18.0 g HO mol HO 1 mol H O mol Cl 70.90 g Cl 36.04 g H O 141.80 g Cl 1 mol Cl. a. N(g) O(g) NO(g) 1 mol N 1 mol N 1 mol O 1 mol O ; ; ; ; 1 mol O mol NO 1 mol N mol NO mol NO mol NO ; 1 mol N 1 mol O 14
Answer Key (continued) b. 4NH 3(aq) 5O (g) 4NO(g) 6H O(l) 4 mol NH3 4 mol NH3 4 mol NH3 5 mol O,,,, 5 mol O 4 mol NO 6 mol H O 4 mol NH 3 5 mol O 5 mol O 4 mol NO 4 mol NO,,,, 4 mol NO 6 mol H O 4 mol NH 5 mol O 3 3 4 mol NO 6 mol HO 6 mol HO,,, 6 mol H O 4 mol NH 5 mol O 6 mol HO 4 mol NO c. 4HCl(aq) O(g) HO(l) Cl 4 mol HCl 4 mol HCl 4 mol HCl 1 mol O,,,, 1 mol O mol H O mol Cl 4 mol HCl 1 mol O 1 mol O mol HO mol HO,,,, mol Cl mol H O 4 mol HCl 1 mol O mol HO mol Cl mol Cl mol Cl,,, mol Cl 4 mol HCl 1 mol O mol H O 3. 40 mol LiOH is needed. 4. KClO3(s) KCl(s) 3O(g) a. 15 mol O b. 3 mol KCl c. 33 mol KClO 3, or 30 mol KClO 3 using significant figures 5. From the reaction Na(s) Cl(g) NaCl(s), 31 g NaCl is produced. 6. a. NaI(aq) Cl(g) NaCl(aq) I(s) b. 13 g Cl 7. a. Zn(s) HCl(aq) ZnCl(aq) H(g) b. 365 g HCl 8. a. Mg(s) HCl(aq) MgCl(aq) H(g); 157 g MgCl b. Cu(s) AgNO 3(aq) Cu(NO 3) (aq) Ag(s);.4 g Cu c. NaCl(aq) H (aq) Na HCl(g); 9.36 g HCl d. 3AgCH 3COO(aq) Na 3PO 4(aq) Ag 3PO 4(s) 3NaCH 3COO(aq); 5.1 g Ag 3PO 4 9. a. N 15
Answer Key (continued) b. 1. 10 3 g NH 3 c. 34 g H 10. a. Al(s) 3Cl(g) AlCl3(s) b. Cl c. 6.8 g AlCl 3 d. 1.8 g Al 11. a. 80% b. 76% c. 94% 1. a. 19 g b. 66 g c. 5.7 g 13. theoretical yield: 39. g; percent yield: 75.3% Chapter 11 Review 14. Stoichiometry is the study of quantitative relationships between amounts of reactants used and products formed by a chemical reaction. 15. Sample questions: What mass of HCl is needed to react completely with a known mass of O? How much water will be produced if a given mass of HCl is used in the reaction? 16. The law of conservation of mass states that matter is neither created nor destroyed; thus, in a chemical reaction, the mass of the reactants must equal the mass of the products. 17. The limiting reactant limits the amount of product that can form from the reaction. An excess reactant is left over after the reaction is complete and all of the limiting reactant has been used up. 16