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

2. Chapter Outline Copyright 2012 John Wiley & Sons, Inc9-2 9.1 A Short ReviewA Short Review 9.2 Introduction to StoichiometryIntroduction to Stoichiometry 9.3 Mole-Mole CalculationsMole-Mole Calculations 9.4 Mole-Mass CalculationsMole-Mass Calculations 9.5 Mass-Mass CalculationsMass-Mass Calculations 9.6 Limiting Reactant and Yield CalculationsLimiting Reactant and Yield Calculations

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

4. REVIEWING MOLES 9-4

5. Molar Mass – sum of atomic masses of all atoms in 1 mole of an element or compound; the units are g/mol. 6.022x10 23 molecules 6.022x10 23 formula units 6.022x10 23 atoms 6.022x10 23 ions 9-5 1 mole =

6. What is the molar mass of Al(ClO 3 ) 3 ? Al1(26.98 g) 3Cl3(35.45 g) 9O9(16.00 g) Al(ClO 3 ) 3 277.33 g/mol 7-6 atomic mass Al26.98 Cl35.45 O16.00

7. Calculate the mass of 2.5 moles of aluminum chlorate. 9-7 Plan 2.5 mol Al(ClO 3 ) 3  g Al(ClO 3 ) 3 Calculate 1 mol Al(ClO 3 ) 3 = 277.33 g Al(ClO 3 ) 3

8. Calculate the moles of 3.52g of aluminum chlorate. 9-8 Plan 3.52 g Al(ClO 3 ) 3  mol Al(ClO 3 ) 3 Calculate 1 mol Al(ClO 3 ) 3 = 277.33 g Al(ClO 3 ) 3

9. Calculate the number of formula units contained in 12.4 g aluminum chlorate. 9-9 Plan 12.4 g Al(ClO 3 ) 3  formula units Al(ClO 3 ) 3 1 mol Al(ClO 3 ) 3 = 277.33 g = 6.022x10 23 formula units Calculate

10. 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 9-10 atomic mass Ca40.08 Cl35.45

11. Your 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 9-11 atomic mass H1.01 Cl35.45

12. 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 9-12 atomic mass H1.01 Cl35.45

13. USING STOICHIOMETRY 9-13

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

15. 9-15 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: The 3 other possibilities are the inverse of these ratios.

16. 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 9-16

17. 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) 9-17 Plan Calculate 5.50 mol N 2  mol NI 3 Set-Up

18. 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) Copyright 2012 John Wiley & Sons, Inc9-18 Plan Calculate 5.50 mol N 2  mol I 2 Set-Up

19. 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  2HF a.0.710 b.1.42 c.2.00 d.2.84 9-19

20. 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

21. Copyright 2012 John Wiley & Sons, Inc9-21

22. USING STOICHIOMETRY IN MOLE- MOLE CALCULATIONS 9-22

23. How many moles of Al are needed to make 0.0935 mol of H 2 ? 2Al (s) + 6HCl (aq)  2AlCl 3(aq) + 3H 2(g) 9-23 Plan Calculate 0.0935 mol H 2  mol Al Set-Up

24. 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) 9-24 Plan Calculate 0.0935 mol H 2  mol HCl Set-Up

25. Your Turn! How many moles of H 2 are made by the reaction of 1.5 mol HCl with excess aluminum? 2Al (s) + 6HCl (aq)  2AlCl 3(aq) + 3H 2(g) a.0.75 mol b.3.0 mol c.6.0 mol d.4.5 mol Copyright 2012 John Wiley & Sons, Inc9-25

26. 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, Inc9-26

27. 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? C 3 H 8 + 5O 2  3CO 2 + 4H 2 O a.0.100 b.0.897 c.6.03 × 10 22 d.5.43 × 10 23 Copyright 2012 John Wiley & Sons, Inc9-27

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

29. MOLE-MASS CALCULATIONS 9-29

30. 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) 9-30 Plan Calculate 3.0 mol HCl  mol H 2  g H 2

31. 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) 9-31 Plan Calculate 2.00 g Al  mol Al  mol HCl

32. 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) 9-32 Plan Calculate 0.093 mol Na 2 CO 3  mol Al(NO 3 ) 3  g Al(NO 3 ) 3

33. 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) 9-33 Plan Calculate 3.45g Na 2 CO 3  mol Na 2 CO 3  g Al 2 (CO 3 ) 3

34. Your Turn! How many moles of oxygen are consumed when 38.0g of aluminum oxide are produced in the following equation? 4Al (s) + 3O 2(g)  2Al 2 O 3(s) a. 0.248 b.0.559 c.1.50 d.3.00 Copyright 2012 John Wiley & Sons, Inc9-34 atomic mass Al26.98 O16.00

35. Your Turn! What mass of HCl is produced when 1.81x10 24 molecules of H 2 react completely in the following equation? H 2(g) + Cl 2(g)  2HCl (g) a.54.7g b.72.9g c.109g d.219g Copyright 2012 John Wiley & Sons, Inc9-35 atomic mass H1.01 Cl35.45

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

37. MASS-MASS CALCULATIONS 9-37

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

39. 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) 9-39 Plan Calculate 7.00 g Al  mol Al  mol Br 2  g Br 2

40. 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? 2FeCl 3(aq) + 3Na 2 S (aq)  Fe 2 S 3(s) + 6NaCl (aq) 9-40 Plan Calculate 9.34 g FeCl 3  mol FeCl 3  mol Fe 2 S 3  g Fe 2 S 3

41. Your Turn! What mass of oxygen is consumed when 54.0g of water is produced in the following equation? 2H 2 + O 2  2H 2 O a.0.167 g b.0.667 g c.1.50 g d.47.9 g Copyright 2012 John Wiley & Sons, Inc9-41 atomic mass H1.01 O16.00

42. Your Turn! What mass of H 2 O is produced when 12.0g of HCl react completely in the following equation? 6HCl + Fe 2 O 3  2FeCl 3 + 3H 2 O a.2.97 g b.39.4 g c.27.4 g d.110. g Copyright 2012 John Wiley & Sons, Inc9-42 atomic mass H1.01 O16.00 Cl35.45

43. LIMITING REACTANT 9-43

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

45. 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, Inc9-45 + 

46. 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, Inc9-46 + 

47. 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, Inc9-47

48. 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. 9-48 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

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

50. 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

51. 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) Copyright 2012 John Wiley & Sons, Inc9-51 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.

52. 2Al (s) + 3CuSO 4(aq)  Al 2 (SO 4 ) 3(aq) + 3Cu (s) 9-52 1. Assume Al is limiting and CuSO 4 is in excess. 2. Assume CuSO 4 is limiting and Al is in excess. 53.0 g Cu 9.96 g Cu 3. Compare answers. CuSO 4 is the limiting reagent. The theoretical yield of Cu is 9.96 g.

53. 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, Inc9-53

54. 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, Inc9-54

55. 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, Inc9-55

56. PERCENT YIELD 9-56

57. 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. 9-57

58. 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. 22.2g PCl 3 Compute the % Yield. Copyright 2012 John Wiley & Sons, Inc9-58

59. 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, Inc9-59

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