Chapter 9 Calculations from Chemical Equations Accurate measurement

Chapter 9 Calculations from Chemical Equations Accurate measurement and calculation of the correct dosage are important in dispensing the correct medicine to patients throughout the Introduction to General, Organic, and Biochemistry 10 e world. John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena

Chapter Outline 9. 1 A Short Review 9. 5 Mass-Mass Calculations 9. 2 Introduction to Stoichiometry 9. 6 Limiting Reactant and Yield Calculations 9. 3 Mole-Mole Calculations 9. 4 Mole-Mass Calculations Copyright 2012 John Wiley & Sons, Inc 9 -2

Objectives for Today q Review the “mole” concept q Determine relationships between moles using stoichiometry q Develop mole and mass relationships 9 -3

REVIEWING MOLES 9 -4

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. 022 x 1023 molecules 6. 022 x 1023 formula units 6. 022 x 1023 atoms 6. 022 x 1023 ions 9 -5

What is the molar mass of Al(Cl. O 3)3? Al 1(26. 98 g) 3 Cl 3(35. 45 g) 9 O 9(16. 00 g) Al(Cl. O 3)3 277. 33 g/mol atomic mass Al 26. 98 Cl 35. 45 O 16. 00 7 -6

Calculate the mass of 2. 5 moles of aluminum chlorate. Plan 2. 5 mol Al(Cl. O 3)3 g Al(Cl. O 3)3 1 mol Al(Cl. O 3)3 = 277. 33 g Al(Cl. O 3)3 Calculate 9 -7

Calculate the moles of 3. 52 g of aluminum chlorate. Plan 3. 52 g Al(Cl. O 3)3 mol Al(Cl. O 3)3 1 mol Al(Cl. O 3)3 = 277. 33 g Al(Cl. O 3)3 Calculate 9 -8

Calculate the number of formula units contained in 12. 4 g aluminum chlorate. Plan 12. 4 g Al(Cl. O 3)3 formula units Al(Cl. O 3)3 1 mol Al(Cl. O 3)3 = 277. 33 g = 6. 022 x 1023 formula units Calculate 9 -9

Your Turn! What is the mass of 3. 61 moles of Ca. Cl 2? atomic mass a. 3. 61 g Ca 40. 08 b. 272 g Cl 35. 45 c. 2. 17 × 1024 g d. 401 g 9 -10

Your Turn! How many moles of HCl are contained in 18. 2 g atomic mass HCl? H 1. 01 a. 1. 00 mol Cl 35. 45 b. 0. 500 mol c. 0. 250 mol d. 0. 125 mol 9 -11

Your Turn! What is the mass of 1. 60× 1023 molecules of HCl? atomic mass a. 9. 69 g H 1. 01 b. 137 g Cl 35. 45 c. 0. 729 g d. 36. 5 g 9 -12

USING STOICHIOMETRY 9 -13

Stoichiometry deals with the quantitative relationships between the reactants and products in a balanced chemical equation. 9 -14

1 N 2(g) + 3 I 2(s) 2 NI 3(s) 1 mol N 2 + 3 mol I 2 2 mol NI 3 Mole ratios come from the coefficients in the balanced equation: The 3 other possibilities are the inverse of these ratios. 9 -15

Your Turn! Which of these statements is not true about the reaction? 1 N 2(g) + 3 I 2(s) 2 NI 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 9 -16

Calculate the number of moles of NI 3 that can be made from 5. 50 mol N 2 in the reaction: 1 N 2(g) + 3 I 2(s) 2 NI 3(s) Plan 5. 50 mol N 2 mol NI 3 Set-Up Calculate 9 -17

Calculate the number of moles of I 2 needed to react with 5. 50 mol N 2 in the reaction: 1 N 2(g) + 3 I 2(s) 2 NI 3(s) Plan 5. 50 mol N 2 mol I 2 Set-Up Calculate Copyright 2012 John Wiley & Sons, Inc 9 -18

How many moles of HF will be produced by the complete reaction of 1. 42 moles of H 2 in the following equation? H 2 + F 2 2 HF a. b. c. d. 0. 710 1. 42 2. 00 2. 84 9 -19

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. 9 -20

Copyright 2012 John Wiley & Sons, Inc 9 -21

USING STOICHIOMETRY IN MOLE CALCULATIONS 9 -22

How many moles of Al are needed to make 0. 0935 mol of H 2? 2 Al(s) + 6 HCl (aq) 2 Al. Cl 3(aq) + 3 H 2(g) Plan 0. 0935 mol H 2 mol Al Set-Up Calculate 9 -23

How many moles of HCl are needed to make 0. 0935 mol of H 2? 2 Al(s) + 6 HCl (aq) 2 Al. Cl 3(aq) + 3 H 2(g) Plan 0. 0935 mol H 2 mol HCl Set-Up Calculate 9 -24

Your Turn! How many moles of H 2 are made by the reaction of 1. 5 mol HCl with excess aluminum? 2 Al(s) + 6 HCl (aq) 2 Al. Cl 3(aq) + 3 H 2(g) a. 0. 75 mol b. 3. 0 mol c. 6. 0 mol d. 4. 5 mol Copyright 2012 John Wiley & Sons, Inc 9 -25

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 + 5 O 2 3 CO 2 + 9 H 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 -26

Your Turn! How many moles of C 3 H 8 are consumed when 1. 81 x 1023 molecules of CO 2 are produced in the following equation? C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O a. 0. 100 b. 0. 897 c. 6. 03 × 1022 d. 5. 43 × 1023 Copyright 2012 John Wiley & Sons, Inc 9 -27

Objectives for Today ü Review the “mole” concept ü Determine relationships between moles using stoichiometry ü Develop mole and mass relationships 9 -28

MOLE-MASS CALCULATIONS 9 -29

What mass of H 2 (2. 02 g/mol) is made by the reaction of 3. 0 mol HCl with excess aluminum? 2 Al(s) + 6 HCl (aq) 2 Al. Cl 3(aq) + 3 H 2(g) Plan 3. 0 mol HCl mol H 2 g H 2 Calculate 9 -30

How many moles of HCl are needed to completely consume 2. 00 g Al (26. 98 g/mol)? 2 Al(s) + 6 HCl (aq) 2 Al. Cl 3(aq) + 3 H 2(g) Plan 2. 00 g Al mol HCl Calculate 9 -31

What mass of Al(NO 3)3 (213 g/mol) is needed to react with. 093 mol Na 2 CO 3? 3 Na 2 CO 3(aq) + 2 Al(NO 3)3(aq) Al 2(CO 3)3(s) + 6 Na. NO 3(aq) Plan 0. 093 mol Na 2 CO 3 mol Al(NO 3)3 g Al(NO 3)3 Calculate 9 -32

How many moles of Al 2(CO 3)3 are made by the reaction of 3. 45 g Na 2 CO 3 (105. 99 g/mol) with excess Al(NO 3)3? 3 Na 2 CO 3(aq) + 2 Al(NO 3)3(aq) Al 2(CO 3)3(s) + 6 Na. NO 3(aq) Plan 3. 45 g Na 2 CO 3 mol Na 2 CO 3 g Al 2(CO 3)3 Calculate 9 -33

Your Turn! How many moles of oxygen are consumed when 38. 0 g of aluminum oxide are produced in the following equation? atomic mass 4 Al(s) + 3 O 2(g) 2 Al 2 O 3(s) Al 26. 98 O 16. 00 a. 0. 248 b. 0. 559 c. 1. 50 d. 3. 00 Copyright 2012 John Wiley & Sons, Inc 9 -34

Your Turn! What mass of HCl is produced when 1. 81 x 1024 molecules of H 2 react completely in the following equation? atomic mass H 2(g) + Cl 2(g) 2 HCl(g) H 1. 01 Cl 35. 45 a. 54. 7 g b. 72. 9 g c. 109 g d. 219 g Copyright 2012 John Wiley & Sons, Inc 9 -35

Objectives for Today q Further analyze moles and mass using stoichiometry q Use stoichiometry to examine limiting reagents and yield 9 -36

MASS-MASS CALCULATIONS 9 -37

Mass-Mass Calculations Now we will put it all together. 9 -38

What mass of Br 2 (159. 80 g/mol) is needed to completely consume 7. 00 g Al (26. 98 g/mol)? 2 Al(s) + 3 Br 2(l) 2 Al. Br 3(s) Plan 7. 00 g Al mol Al mol Br 2 g Br 2 Calculate 9 -39

What mass of Fe 2 S 3 (207. 91 g/mol) can be made from the reaction of 9. 34 g Fe. Cl 3 (162. 20 g/mol) with excess Na 2 S? 2 Fe. Cl 3(aq) + 3 Na 2 S(aq) Fe 2 S 3(s) + 6 Na. Cl(aq) Plan 9. 34 g Fe. Cl 3 mol Fe 2 S 3 g Fe 2 S 3 Calculate 9 -40

Your Turn! What mass of oxygen is consumed when 54. 0 g of water is produced in the following atomic mass equation? H 1. 01 2 H 2 + O 2 2 H 2 O O 16. 00 a. 0. 167 g b. 0. 667 g c. 1. 50 g d. 47. 9 g Copyright 2012 John Wiley & Sons, Inc 9 -41

Your Turn! What mass of H 2 O is produced when 12. 0 g of HCl react completely in the following equation? 6 HCl + Fe 2 O 3 2 Fe. Cl 3 + 3 H 2 Oatomic mass H 1. 01 a. 2. 97 g O 16. 00 b. 39. 4 g Cl 35. 45 c. 27. 4 g d. 110. g Copyright 2012 John Wiley & Sons, Inc 9 -42

LIMITING REACTANT 9 -43

Determine the number of that can be made given these quantities of reactants and the reaction equation: + + 9 -44

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 -45

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 -46

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 -47

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 2 H 2(g) + O 2(g) 2 H 2 O(g) 1. Calculate theoretical yield of H 2 O assuming H 2 is the limiting reactant and that O 2 is the excess reactant. 2. Calculate theoretical yield of H 2 O assuming that O 2 is the limiting reactant and that H 2 is the excess reactant. 9 -48

Assuming that H 2 is limiting and O 2 is excess: Assuming that O 2 is limiting and H 2 is excess: So what is the maximum yield of H 2 O? 9 -49

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. 9 -50

Calculate the mass of copper that can be made from the combination of 15. 0 g aluminum with 25. 0 g copper(II) sulfate. 2 Al(s) + 3 Cu. SO 4(aq) Al 2(SO 4)3(aq) + 3 Cu(s) Plan 15 g Al mol Cu g Cu 25 g Cu. SO 4 mol Cu g Cu Compare answers. The smaller number is the right answer. Copyright 2012 John Wiley & Sons, Inc 9 -51

2 Al(s) + 3 Cu. SO 4(aq) Al 2(SO 4)3(aq) + 3 Cu(s) 1. Assume Al is limiting and Cu. SO 4 is in excess. 53. 0 g Cu 2. Assume Cu. SO 4 is limiting and Al is in excess. 9. 96 g Cu 3. Compare answers. Cu. SO 4 is the limiting reagent. The theoretical yield of Cu is 9. 96 g. 9 -52

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 -53

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 + 2 Ag. NO 3 Cu(NO 3)2 + 2 Ag a. Cu b. Ag. NO 3 c. Cu(NO 3)2 d. Ag Copyright 2012 John Wiley & Sons, Inc 9 -54

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 + 2 Ag. NO 3 Cu(NO 3)2 + 2 Ag a. 162 g b. 216 g c. 324 g d. 647 g Copyright 2012 John Wiley & Sons, Inc 9 -55

PERCENT YIELD 9 -56

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 theoretical yield due to experimental losses and errors along the way. 9 -57

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. 2 P + 3 Cl 2 2 PCl 3 Compute the expected yield of PCl 3 from 5. 00 g P with excess Cl 2. 2 g PCl 3 Compute the % Yield. Copyright 2012 John Wiley & Sons, Inc 9 -58

Your Turn! In a reaction to produce ammonia, 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 -59

Objectives for Today ü Further analyze moles and mass using stoichiometry ü Use stoichiometry to examine limiting reagents and yield 9 -60