Chapter 9 Calculations from Chemical Equations Accurate measurement and calculation of the correct dosage are important in dispensing the correct medicine to patients Introduction to General, Organic, and Biochemistry 10e throughout the John Wiley & Sons, Inc world. Morris Hein, Scott Pattison, and Susan Arena
Chapter Outline 9.1 A Short Review 9.2 Introduction to Stoichiometry 9.3 Mole-Mole Calculations 9.4 Mole-Mass Calculations 9.5 Mass-Mass Calculations 9.6 Limiting Reactant and Yield Calculations Copyright 2012 John Wiley & Sons, Inc 9-2
Molar Mass Molar Mass sum of atomic masses of all atoms in 1 mole of an element or compound ; the units are g/mol. 1 mole = 6.022x10 23 molecules 6.022x10 23 formula units 6.022x10 23 atoms 6.022x10 23 ions Copyright 2012 John Wiley & Sons, Inc 9-3
Molar Mass What is the molar mass of Al(ClO 3 ) 3? Al 1(26.98 g) 3Cl 3(35.45 g) 9O 9(16.00 g) Al(ClO 3 ) 3 277.33 g/mol atomic mass Al 26.98 Cl 35.45 O 16.00 Copyright 2012 John Wiley & Sons, Inc 7-4
Molar Mass Calculate the mass of 2.5 moles of aluminum chlorate. Plan 2.5 mol Al(ClO 3 ) 3 g Al(ClO 3 ) 3 1 mol Al(ClO 3 ) 3 = 277.33 g Al(ClO 3 ) 3 Calculate 277.33 g Al(ClO ) 3 3 2.5 mol Al(ClO 3) 3 = 690 g Al(ClO 3) 3 1 mol Al(ClO 3) 3 Copyright 2012 John Wiley & Sons, Inc 9-5
Molar Mass Calculate the moles of 3.52g of aluminum chlorate. Plan 3.52 g Al(ClO 3 ) 3 mol Al(ClO 3 ) 3 1 mol Al(ClO 3 ) 3 = 277.33 g Al(ClO 3 ) 3 Calculate 3.52 g Al(ClO 3) 3 1 mol Al(ClO ) 3 3 277.33 g Al(ClO ) 3 3 2 = 1.27 10 mol Al(ClO 3) 3 Copyright 2012 John Wiley & Sons, Inc 9-6
Molar Mass Calculate the number of formula units contained in 12.4 g aluminum chlorate. Plan 12.4 g Al(ClO 3 ) 3 formula units Al(ClO 3 ) 3 Calculate 1 mol Al(ClO 3 ) 3 = 277.33 g = 6.022x10 23 formula units 23 6.022 10 formula units 12.4 g Al(ClO 3) 22 3 = 2.69 10 formula units 277.33 g Al(ClO 3) 3 Copyright 2012 John Wiley & Sons, Inc 9-7
Your Turn! What is the mass of 3.61 moles of CaCl 2? a. 3.61 g b. 272 g c. 2.17 10 24 g d. 401 g atomic mass Ca 40.08 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc 9-8
You Turn! How many moles of HCl are contained in 18.2 g HCl? a. 1.00 mol b. 0.500 mol c. 0.250 mol d. 0.125 mol atomic mass H 1.01 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc 9-9
Your Turn! What is the mass of 1.60 10 23 molecules of HCl? a. 9.69 g b. 137 g c. 0.729 g d. 36.5 g atomic mass H 1.01 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc 9-10
Stoichiometry Stoichiometry deals with the quantitative relationships between the reactants and products in a balanced chemical equation. 1N 2(g) + 3I 2(s) 2NI 3(s) 1 mol N 2 + 3 mol I 2 2 mol NI 3 Mole ratios come from the coefficients in the balanced equation: 3 mol I2 3 mol I2 1 mol N2 1 mol N 2 mol NI 2 mol NI 2 The 3 other possibilities are the inverse of these ratios. 3 3 Copyright 2012 John Wiley & Sons, Inc 9-11
Your Turn! Which of these statements is not true about the reaction? 1N 2(g) + 3I 2(s) 2NI 3(s) a. 1 mole of nitrogen is needed for every 3 moles of iodine b. 1 gram of nitrogen is needed for every 3 grams of iodine c. Both statements are true Copyright 2012 John Wiley & Sons, Inc 9-12
Using the mole ratio Calculate the number of moles of NI 3 that can be made from 5.50 mol N 2 in the reaction: 1N 2(g) + 3I 2(s) 2NI 3(s) Plan Set-Up 5.50 mol N 2 mol NI 3 moles of desired substance in equation mole ratio = moles of starting substance in equation 2 mol NI mol ratio = 1 mol N 3 2 Calculate 2 mol NI 5.50 mol N 2 3 1 mol N2 = 11.0 mol NI 3 Copyright 2012 John Wiley & Sons, Inc 9-13
Using the mole ratio Calculate the number of moles of I 2 needed to react with 5.50 mol N 2 in the reaction: 1N 2(g) + 3I 2(s) 2NI 3(s) Plan Set-Up 5.50 mol N 2 mol I 2 3 mol I mole ratio = 1 mol N 2 2 Calculate 5.50 mol N 2 3 mol I2 1 mol N 2 = 16.5 mol I 2 Copyright 2012 John Wiley & Sons, Inc 9-14
Your Turn! How many moles of HF will be produced by the complete reaction of 1.42 moles of H 2 in the following equation? a. 0.710 b. 1.42 c. 2.00 d. 2.84 H 2 + F 2 2HF Copyright 2012 John Wiley & Sons, Inc 9-15
Stoichiometry Problem Solving Strategy for stoichiometry problems: 1. Convert starting substance to moles. 2. Convert the moles of starting substance to moles of desired substance. 3. Convert the moles of desired substance to the units specified in the problem. Copyright 2012 John Wiley & Sons, Inc 9-16
Stoichiometry Copyright 2012 John Wiley & Sons, Inc 9-17
Mole-Mole Calculations How many moles of Al are needed to make 0.0935 mol of H 2? Plan Set-Up Calculate 2Al (s) + 6HCl (aq) 2AlCl 3(aq) + 3H 2(g) 0.0935 mol H 2 mol Al mole ratio = 2 mol Al 3 mol H2 2 mol Al 2 3 mol H2 0.0935 mol H =.0623 mol Al Copyright 2012 John Wiley & Sons, Inc 9-18
Mole-Mole Calculations How many moles of HCl are needed to make 0.0935 mol of H 2? 2Al (s) + 6HCl (aq) 2AlCl 3(aq) + 3H 2(g) Plan Set-Up Calculate 0.0935 mol H 2 mol HCl 6 mol HCl mole ratio = 3 mol H2 6 mol HCl 3 mol H2 0.0935 mol H 2 = 0.187 mol HCl Copyright 2012 John Wiley & Sons, Inc 9-19
Your Turn! How many moles of H 2 are made by the reaction of 1.5 mol HCl with excess aluminum? a. 0.75 mol b. 3.0 mol c. 6.0 mol d. 4.5 mol 2Al (s) + 6HCl (aq) 2AlCl 3(aq) + 3H 2(g) Copyright 2012 John Wiley & Sons, Inc 9-20
Your Turn! How many moles of carbon dioxide are produced when 3.00 moles of oxygen react completely in the following equation? C 3 H 8 + 5O 2 3CO 2 + 9H 2 O a. 5.00 mol b. 3.00 mol c. 1.80 mol d. 1.50 mol Copyright 2012 John Wiley & Sons, Inc 9-21
Your Turn! How many moles of C 3 H 8 are consumed when 1.81x10 23 molecules of CO 2 are produced in the following equation? a. 0.100 b. 0.897 c. 6.03 10 22 d. 5.43 10 23 C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Copyright 2012 John Wiley & Sons, Inc 9-22
Mole-Mass Calculations What mass of H 2 (2.02 g/mol) is made by the reaction of 3.0 mol HCl with excess aluminum? 2Al (s) + 6HCl (aq) 2AlCl 3(aq) + 3H 2(g) Plan 3.0 mol HCl mol H 2 g H 2 Calculate 3.0 mol HCl 3 mol H2 =1.5 mol H 2 6 mol HCl 2.02 g H 1 mol H2 1.5 mol H 2 2 2 = 3.0 g H Copyright 2012 John Wiley & Sons, Inc 9-23
Mole-Mass Calculations How many moles of HCl are needed to completely consume 2.00 g Al (26.98g/mol)? 2Al (s) + 6HCl (aq) 2AlCl 3(aq) + 3H 2(g) Plan 2.00 g Al mol Al mol HCl Calculate 2.00 g Al 1 mol Al = 0.0741 mol Al 26.98 g Al 6 mol HCl 2 mol Al 0.0741 mol Al = 0.0222 mol HCl Copyright 2012 John Wiley & Sons, Inc 9-24
Mole-Mass Calculations What mass of Al(NO 3 ) 3 (213g/mol) is needed to react with.093 mol Na 2 CO 3? 3Na 2 CO 3(aq) + 2Al(NO 3 ) 3(aq) Al 2 (CO 3 ) 3(s) + 6NaNO 3(aq) Plan 0.093 mol Na 2 CO 3 mol Al(NO 3 ) 3 g Al(NO 3 ) 3 Calculate.093 mol Na CO 3 3 2 3 2 mol Al(NO ) 3 mol Na 2CO3 3 3 0.062 mol Al(NO 3) 3 1 mol Al(NO 3) 3 3 3 213.00g Al(NO ) =.062 mol Al(NO 3) 3 =13 g Al(NO ) Copyright 2012 John Wiley & Sons, Inc 9-25
Mole-Mass Calculations How many moles of Al 2 (CO 3 ) 3 are made by the reaction of 3.45g Na 2 CO 3 (105.99 g/mol) with excess Al(NO 3 ) 3? 3Na 2 CO 3(aq) + 2Al(NO 3 ) 3(aq) Al 2 (CO 3 ) 3(s) + 6NaNO 3(aq) Plan 3.45g Na 2 CO 3 mol Na 2 CO 3 g Al 2 (CO 3 ) 3 Calculate 3.45g Na CO 2 3 0.0326 mol Na 2CO3 2 3 1 mol Na CO 105.99g Na CO 2 3 1 mol Al (CO ) 3 mol Na 2CO3 2 3 3 = 0.0326 mol Na 2CO3 = 0.0109 mol Na 2CO3 Copyright 2012 John Wiley & Sons, Inc 9-26
Your Turn! How many moles of oxygen are consumed when 38.0g of aluminum oxide are produced in the following equation? atomic mass 4Al (s) + 3O 2(g) 2Al 2 O 3(s) Al 26.98 a. 0.248 O 16.00 b. 0.559 c. 1.50 d. 3.00 Copyright 2012 John Wiley & Sons, Inc 9-27
Your Turn! What mass of HCl is produced when 1.81x10 24 molecules of H 2 react completely in the following equation? a. 54.7g b. 72.9g c. 109g d. 219g H 2(g) + Cl 2(g) 2HCl (g) H 1.01 atomic mass Cl 35.45 Copyright 2012 John Wiley & Sons, Inc 9-28
Mass-Mass Calculations Now we will put it all together. What mass of Br 2 (159.80 g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? 2Al (s) + 3Br 2(l) 2AlBr 3(s) 7.00 g Al mol Al mol Br 2 g Br 2 Copyright 2012 John Wiley & Sons, Inc 9-29
Mass-Mass Calculations What mass of Br 2 (159.80 g/mol) is needed to completely consume 7.00 g Al (26.98 g/mol)? Plan Calculate 2Al (s) + 3Br 2(l) 2AlBr 3(s) 7.00 g Al mol Al mol Br 2 g Br 2 7.00 g Al 1 mol Al 26.98g Al 3 mol Br 2 0.259 mol Al = 0.389 mol Br2 2 mol Al 159.80g Br 1 mol Br2 = 0.259 mol Al = 62.2 g Br 0.389 mol Br 2 2 2 Copyright 2012 John Wiley & Sons, Inc 9-30
Mass-Mass Calculations What mass of Fe 2 S 3 (207.91g/mol) can be made from the reaction of 9.34 g FeCl 3 (162.20 g/mol) with excess Na 2 S? Plan Calculate 2FeCl 3(aq) + 3Na 2 S (aq) Fe 2 S 3(s) + 6NaCl (aq) 9.34 g FeCl 3 mol FeCl 3 mol Fe 2 S 3 g Fe 2 S 3 1 mol FeCl 162.20g FeCl 9.34 g FeCl 3 3 1 mol Fe S 0.0576 mol FeCl 2 3 3 2 3 2 mol FeCl 3 3 207.91g Fe S = 0.0576 mol FeCl 3 = 0.0288 mol Fe S 0.0288 mol Fe 2 3 2S3 2 3 1 mol Fe S 2 3 = 5.99 g Fe S Copyright 2012 John Wiley & Sons, Inc 9-31
Your Turn! What mass of oxygen is consumed when 54.0g of water is produced in the following equation? a. 0.167 g b. 0.667 g c. 1.50 g d. 47.9 g 2H 2 + O 2 2H 2 O H 1.01 atomic mass O 16.00 Copyright 2012 John Wiley & Sons, Inc 9-32
Your Turn! What mass of H 2 O is produced when 12.0g of HCl react completely in the following equation? a. 2.97 g b. 39.4 g c. 27.4 g d. 110. g 6HCl + Fe 2 O 3 2FeCl 3 + 3H 2 O H 1.01 atomic mass O 16.00 Cl 35.45 Copyright 2012 John Wiley & Sons, Inc 9-33
Limiting Reactant Determine the number of that can be made given these quantities of reactants and the reaction equation: + + Copyright 2012 John Wiley & Sons, Inc 9-34
Limiting Reactant The limiting reactant is the reactant that limits the amount of product that can be made. The reaction stops when the limiting reactant is used up. What was the limiting reactant in the reaction: + The small blue balls. Copyright 2012 John Wiley & Sons, Inc 9-35
Excess Reactant The excess reactant is the reactant that remains when the reaction stops. There is always left over excess reactant. What was the excess reactant in the reaction: + The excess reactant was the larger blue ball. Copyright 2012 John Wiley & Sons, Inc 9-36
Limiting reactant Figure 9.2 The number of bicycles that can be built from these parts is determined by the limiting reactant (the pedal assemblies). Copyright 2012 John Wiley & Sons, Inc 9-37
Limiting Reactant Calculations Technique for solving limiting reactant problems: 1. Convert reactant 1 to moles or mass of product 2. Convert reactant 2 to moles or mass of product 3. Compare answers. The smaller answer is the maximum theoretical yield. Copyright 2012 John Wiley & Sons, Inc 9-38
Limiting Reactant Calculation Calculate the number of moles of water that can be made by the reaction of 1.51 mol H 2 with 0.932 mol O 2. 2H 2(g) + O 2(g) 2H 2 O (g) 1. Calculate the theoretical yield of H 2 O assuming H 2 is the limiting reactant and that O 2 is the excess reactant. 2. Calculate the theoretical yield of H 2 O assuming that O 2 is the limiting reactant and that H 2 is the excess reactant. Copyright 2012 John Wiley & Sons, Inc 9-39
Limiting Reactant Calculation continued Assuming that H 2 is limiting and O 2 is excess: 2 mol H2O 1.51 mol H 2 =1.51 mol H2O 2 mol H2 Assuming that O 2 is limiting and H 2 is excess: 2 mol H2O 0.932 mol O 2 =1.86 mol H2O 1 mol 02 So what is the maximum yield of H 2 O? Copyright 2012 John Wiley & Sons, Inc 9-40
Limiting Reactant Calculation continued How much H 2 and O 2 remain when the reaction stops? H 2 : Limiting Reactant None remains. It was used up in the reaction. O 2 : Excess Reactant Calculate the amount of O 2 used in the reaction with H 2. Then subtract that from the original amount. 1.51 mol H 1 mol O x =0.755 mol O 2 2 2 2 mol H2 0.932 mol O 2 to start - 0.755 mol O 2 = 0.177 mol of excess O2 Copyright 2012 John Wiley & Sons, Inc 9-41
Limiting Reactant Calculation Calculate the mass of copper that can be made from the combination of 15.0 g aluminum with 25.0 g copper(ii) sulfate. 2Al (s) + 3CuSO 4(aq) Al 2 (SO 4 ) 3(aq) + 3Cu (s) Plan 15 g Al mol Al mol Cu g Cu 25 g CuSO 4 mol CuSO 4 mol Cu g Cu Compare answers. The smaller number is the right answer. Copyright 2012 John Wiley & Sons, Inc 9-42
Limiting Reactant Calculation continued 2Al (s) + 3CuSO 4(aq) Al 2 (SO 4 ) 3(aq) + 3Cu (s) 1. Assume Al is limiting and CuSO 4 is in excess. 15.0 g Al x 1 mol Al 26.98 g Al 3 mol Cu 2 mol Al 63.55 g Cu = 1 mol Cu 2. Assume CuSO 4 is limiting and Al is in excess. 1 mol CuSO 3 mol Cu 3 mol CuSO4 4 25.0 g CuSO4 159.58 g CuSO 4 3. Compare answers. 63.55 g Cu = 1 mol Cu 53.0 g Cu CuSO 4 is the limiting reagent. The theoretical yield of Cu is 9.96 g. 9.96 g Cu Copyright 2012 John Wiley & Sons, Inc 9-43
Your Turn! True/False: You can compare the quantities of reactant when you work a limiting reactant problem. The reactant you have the least of is the limiting reactant. a. True b. False Copyright 2012 John Wiley & Sons, Inc 9-44
Your Turn! Which is the limiting reactant when 3.00 moles of copper are reacted with 3.00 moles of silver nitrate in the following equation? Cu + 2AgNO 3 Cu(NO 3 ) 2 + 2Ag a. Cu b. AgNO 3 c. Cu(NO 3 ) 2 d. Ag Copyright 2012 John Wiley & Sons, Inc 9-45
Your Turn! What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate? Cu + 2AgNO 3 Cu(NO 3 ) 2 + 2Ag a. 162g b. 216g c. 324g d. 647g Copyright 2012 John Wiley & Sons, Inc 9-46
Percent Yield To determine the efficiency of a process for making a compound, chemists compute the percent yield of the reaction. Actual Yield % Yield = 100 Theoretical Yield The theoretical yield is the result calculated using stoichiometry. The actual yield of a chemical reaction is the experimental result, which is often less than the theoretical yield due to experimental losses and errors along the way. Copyright 2012 John Wiley & Sons, Inc 9-47
Percent Yield Calculate the % yield of PCl 3 that results from reacting 5.00 g P with excess Cl 2 if only 17.2 g of PCl 3 were recovered. 2P + 3Cl 2 2PCl 3 Compute the expected yield of PCl 3 from 5.00 g P with excess Cl 2. 5.00 g P x 1 mol P 30.97 g P Compute the % Yield. 2 mol PCl3 2 mol P 137.33 g PCl 1 mol PCl3 3 = 22.2g PCl 3 Actual Yield 17.2 g % Yield = 100%= 100%=77.5% Theoretical Yield 22.2 g Copyright 2012 John Wiley & Sons, Inc 9-48
Your Turn! In a reaction to produce ammonia, the theoretical yield is 420. g. What is the percent yield if the actual yield is 350. g? A. 83.3% B. 20.0% C. 16.7% D. 120.% Copyright 2012 John Wiley & Sons, Inc 9-49