2. Chapter 3 Calculations with Chemical Formulas and Equations HSTMr. Watson

3. HST Molar Mass Sum atomic masses represented by formula atomic masses => gaw molar mass => MM

4. Mr. WatsonHST One Mole of each Substance Clockwise from top left: 1-Octanol, C 8 H 17 OH; Mercury(II) iodide, HgI 2 ; Methanol, CH 3 OH; and Sulfur, S 8.

5. Mr. WatsonHST Example What is the molar mass of ethanol, C 2 H 5 OH?

6. Mr. WatsonHST Example What is the molar mass of ethanol, C 2 H 5 O 1 H 1 ? MM = 2(gaw) C + (5 + 1)(gaw) H + 1(gaw) O

7. Mr. WatsonHST Example What is the molar mass of ethanol, C 2 H 5 OH? MM = 2(gaw) C + (5 + 1)(gaw) H + 1(gaw) O = 2(12.011) C + 6(1.00794) H + 1(15.9994) O

8. Mr. WatsonHST Example What is the molar mass of ethanol, C 2 H 5 OH? MM = 2(gaw) C + (5 + 1)(gaw) H + 1(gaw) O = 2(12.011) C + 6(1.00794) H + 1(15.9994) O = 46.069 g/mol

9. Mr. WatsonHST The Mole a unit of measurement, quantity of matter present Avogadro’s Number 6.022 x 10 23 particles Latin for “pile”

10. Mr. WatsonHST Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide?

11. Mr. WatsonHST Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol

12. Mr. WatsonHST Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2

13. Mr. WatsonHST Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2 = (10.00g)

14. Mr. WatsonHST Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2 = (10.00g)(1 mol/44.01g)

15. Mr. WatsonHST Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2 = (10.00g)(1 mol/44.01g)

16. Mr. WatsonHST Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2 = (10.00)(1 mol/44.01)

17. Mr. WatsonHST Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2 = (10.00)(1 mol/44.01) = 0.2272 mol

18. Mr. WatsonHST Combustion Analysis

19. Mr. WatsonHST Percentage Composition description of a compound based on the relative amounts of each element in the compound

20. Mr. WatsonHST EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu

21. Mr. WatsonHST EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(gaw) %C = ------------ X 100 MM

22. Mr. WatsonHST EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(gaw) %C = ------------ X 100 MM 1(12.011) %C = -------------- X 100 = 10.061% C 119.377

23. Mr. WatsonHST EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(1.00797) %H = ---------------- X 100 = 0.844359% H 119.377

24. Mr. WatsonHST EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 3(35.453) %Cl = -------------- X 100 = 89.095% Cl 119.377

25. Mr. WatsonHST EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = 119.377amu %C = 10.061% C %H = 0.844359% H %Cl = 89.095% Cl

26. Mr. WatsonHST Simplest (Empirical) Formula formula describing a substance based on the smallest set of subscripts

27. Mr. WatsonHST Acetylene, C 2 H 2, and benzene, C 6 H 6, have the same empirical formula. Is the correct empirical formula: C 2 H 2 CH C 6 H 6

28. Mr. WatsonHST EXAMPLE: A white compound is formed when phosphorus burns in air. Analysis shows that the compound is composed of 43.7% P and 56.3% O by mass. What is the empirical formula of the compound? Relative Number of Atoms Element % (%/gaw) P 43.7 43.7/30.97 = 1.41 O 56.3 56.3/15.9994 = 3.52

29. Mr. WatsonHST EXAMPLE: A white compound is formed when phosphorus burns in air. Analysis shows that the compound is composed of 43.7% P and 56.3% O by mass. What is the empirical formula of the compound? Relative Number of Atoms Element % (%/gaw) Divide by Smaller P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00 O 56.3 56.3/15.9994 = 3.52 3.52/1.41 = 2.50

30. Mr. WatsonHST EXAMPLE: A white compound is formed when phosphorus burns in air. Analysis shows that the compound is composed of 43.7% P and 56.3% O by mass. What is the empirical formula of the compound? Relative Number of Atoms Multiply % (%/gaw) Divide by Smaller by Integer P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00 2*1.00 => 2 O 56.3 56.3/15.9994 = 3.52 3.52/1.41 = 2.50 2*2.50 => 5

31. Mr. WatsonHST EXAMPLE: A white compound is formed when phosphorus burns in air. Analysis shows that the compound is composed of 43.7% P and 56.3% O by mass. What is the empirical formula of the compound? Relative Number of Atoms Multiply % (%/gaw) Divide by Smaller by Integer P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00 2*1.00 => 2 O 56.3 56.3/15.9994 = 3.52 3.52/1.41 = 2.50 2*2.50 => 5 Empirical Formula => P 2 O 5

32. Mr. WatsonHST EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? 2.34 %N = ----------------- X 100 = 30.5% N 2.34 + 5.34 5.34 %O = ----------------- X 100 = 69.5% O 2.34 + 5.34

33. Mr. WatsonHST EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? %N = 30.5% N%O = 69.5% O Relative # Atoms Element %(%/gaw) N 30.5 30.5/14.0067 = 2.18 O 69.5 69.5/15.9994 = 4.34

34. Mr. WatsonHST EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? %N = 30.5% N%O = 69.5% O Relative # Atoms %(%/gaw) Divide by Smaller N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 O 69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99

35. Mr. WatsonHST EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? %N = 30.5% N%O = 69.5% O Relative # Atoms Multiply %(%/gaw) Divide by Smallerby Integer N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 1*1.00=>1 O 69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99 1*1.99=>2

36. Mr. WatsonHST EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? %N = 30.5% N%O = 69.5% O Relative # Atoms Multiply %(%/gaw) Divide by Smallerby Integer N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 1*1.00=>1 O 69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99 1*1.99=>2 Empirical Formula => NO 2

37. Mr. WatsonHST Molecular Formula the exact proportions of the elements that are formed in a molecule

38. Mr. WatsonHST Molecular Formula from Simplest Formula empirical formula => EF molecular formula => MF MF = X * EF

39. Mr. WatsonHST Molecular Formula from Simplest Formula formula mass => FM sum of the atomic weights represented by the formula molar mass = MM = X * FM

40. Mr. WatsonHST Molecular Formula from Simplest Formula first, knowing MM and FM X = MM/FM then MF = X * EF

41. Mr. WatsonHST EXAMPLE: A colorless liquid used in rocket engines, whose empirical formula is NO 2, has a molar mass of 92.0. What is the molecular formula? FM = 1(gaw) N + 2(gaw) O = 46.0 MM 92.0 X = ------- = -------- = 2 FM 46.0

42. Mr. WatsonHST EXAMPLE: A colorless liquid used in rocket engines, whose empirical formula is NO 2, has a molar mass of 92.0. What is the molecular formula? FM = 1(gaw) N + 2(gaw) O = 46.0 MM 92.0 X = ------- = -------- = 2 FM 46.0 thus MF = 2 * EF

43. Mr. WatsonHST What is the correct molecular formula for this colorless liquid rocket fuel? 2NO NO N 2 O 4

44. Mr. WatsonHST Stoichiometry stoi·chi·om·e·try noun 1. Calculation of the quantities of reactants and products in a chemical reaction. 2. The quantitative relationship between reactants and products in a chemical reaction.

45. Mr. WatsonHST The Mole and Chemical Reactions: The Macro-Nano Connection 2 H 2 + O 2 -----> 2 H 2 O 2 H 2 molecules 1 O 2 molecule 2 H 2 O molecules 2 H 2 moles molecules 1 O 2 mole molecules 2 H 2 O moles molecules 4 g H 2 32 g O 2 36 g H 2 O

46. Mr. WatsonHST EXAMPLE How much H 2 O, in moles results from burning an excess of H 2 in 3.3 moles of O 2 ? H 2 + O 2 -----> H 2 O 2 H 2 + O 2 -----> 2 H 2 O

47. Mr. WatsonHST EXAMPLE How much H 2 O, in moles results from burning an excess of H 2 in 3.3 moles of O 2 ? H 2 + O 2 -----> H 2 O 2 H 2 + O 2 -----> 2 H 2 O (3.3 mol O 2 ) (2 mol H 2 O) #mol H 2 O = ------------------------------------ (1 mol O 2 )

48. Mr. WatsonHST EXAMPLE How much H 2 O, in moles results from burning an excess of H 2 in 3.3 moles of O 2 ? H 2 + O 2 -----> H 2 O 2 H 2 + O 2 -----> 2 H 2 O (3.3 mol O 2 ) (2 mol H 2 O) #mol H 2 O = ------------------------------------ (1 mol O 2 )

49. Mr. WatsonHST EXAMPLE How much H 2 O, in moles results from burning an excess of H 2 in 3.3 moles of O 2 ? H 2 + O 2 -----> H 2 O 2 H 2 + O 2 -----> 2 H 2 O (3.3) (2 mol H 2 O) #mol H 2 O = ------------------------ = 6.6 mol H 2 O (1)

50. Mr. WatsonHST Combination Reaction PbNO 3(aq) + K 2 CrO 4(aq)  PbCrO 4(s) + 2 KNO 3(aq) Colorless yellow yellow colorless

51. Mr. WatsonHST Stoichiometric Roadmap

52. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. http://www.cbu.edu/~mcondren/Demos/Thermite-Welding.ppt

53. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67 X 10 3 g/in

54. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132) (1/36 in) (454 g) = 1.67 X 10 3 g/in The mass of iron in a weld adding 10% mass: #g = (1.67 X 10 3 g) (0.10) = 167 g

55. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67 X 10 3 g/in The mass of iron in a weld adding 10% mass: #g = (1.67 X 10 3 g) (0.10) = 167 g Balanced chemical equation: Fe 2 O 3 + 2 Al ---> 2 Fe + Al 2 O 3

56. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67 X 10 3 g/in The mass of iron in a weld adding 10% mass: #g = (1.67 X 10 3 g) (0.10) = 167 g Balanced chemical equation: Fe 2 O 3 + 2 Al ---> 2 Fe + Al 2 O 3 What mass of Fe 2 O 3 is required for the thermite process?

57. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g = (1.67 X 10 3 g) (0.10) = 167 g Balanced chemical equation: Fe 2 O 3 + 2 Al ---> 2 Fe + Al 2 O 3 What mass of Fe 2 O 3 is required for the thermite process? (1 mol Fe) (1 mol Fe 2 O 3 ) (159.7 g Fe 2 O 3 ) #g Fe 2 O 3 = (167 g Fe) * ------------------------------------------------------- (55.85 g Fe) (2 mol Fe) (1 mol Fe 2 O 3 ) = 238 g Fe 2 O 3

58. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe Balanced chemical equation: Fe 2 O 3 + 2 Al ---> 2 Fe + Al 2 O 3 What mass of Fe 2 O 3 is required for the thermite process? #g Fe 2 O 3 = 238 g Fe 2 O 3 What mass of Al is required for the thermite process?

59. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe Balanced chemical equation: Fe 2 O 3 + 2 Al ---> 2 Fe + Al 2 O 3 What mass of Fe 2 O 3 is required for the thermite process? #g Fe 2 O 3 = 238 g Fe 2 O 3 What mass of Al is required for the thermite process? (1 mol Fe) (2 mol Al) (26.9815 g Al) #g Al = (167 g Fe) * ----------------------------------------------- (55.85 g Fe) (2 mol Fe) (1 mol Al) = 80.6 g Al

60. Mr. WatsonHST EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe #g Fe 2 O 3 = 238 g Fe 2 O 3 #g Al = 80.6 g Al

61. Mr. WatsonHST Limiting Reactant reactant that limits the amount of product that can be produced

62. Mr. WatsonHST EXAMPLE What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S)

63. Mr. WatsonHST EXAMPLE What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) balanced equation relates: 2Fe 2 S 3(S) 6H 2 O (l) 3O 2(g)

64. Mr. WatsonHSTEXAMPLE What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) balanced equation relates: 2Fe 2 S 3(S) 6H 2 O (l) 3O 2(g) have only: 1Fe 2 S 3(S) 2H 2 O (l) 3O 2(g)

65. Mr. WatsonHST EXAMPLE What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) balanced equation relates: 2Fe 2 S 3(S) 6H 2 O (l) 3O 2(g) have only: 1Fe 2 S 3(S) 2H 2 O (l) 3O 2(g) not enough H 2 O to use all Fe 2 S 3 plenty of O 2

66. Mr. WatsonHST EXAMPLE What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) if use all Fe 2 S 3 : (1.0 mol Fe 2 S 3 ) (4 mol Fe(OH) 3 ) #mol Fe(OH) 3 = ------------------------------------------ (2 mol Fe 2 S 3 ) = 2.0 mol Fe(OH) 3

67. Mr. WatsonHST EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) if use all H 2 O: (2.0 mol H 2 O) (4 mol Fe(OH) 3 ) #mol Fe(OH) 3 = ----------------------------------------- (6 mol H 2 O) = 1.3 mol Fe(OH) 3

68. Mr. WatsonHST EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) if use all O 2 (3.0 mol O 2 ) (4 mol Fe(OH) 3 ) #mol Fe(OH) 3 = --------------------------------------- (3 mol O 2 ) = 4.0 mol Fe(OH) 3

69. Mr. WatsonHST EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) 1.0 mol Fe 2 S 3 => 2.0 mol Fe(OH) 3 2.0 mol H 2 O => 1.3 mol Fe(OH) 3 3.0 mol O 2 => 4.0 mol Fe(OH) 3

70. Mr. WatsonHST EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) 1.0 mol Fe 2 S 3 => 2.0 mol Fe(OH) 3 2.0 mol H 2 O => 1.3 mol Fe(OH) 3 3.0 mol O 2 => 4.0 mol Fe(OH) 3 Since 2.0 mol H 2 O will produce only 1.3 mol Fe(OH) 3, then H 2 O is the limiting reactant. Thus the correct number of moles of Fe(OH) 3 is 1.33 moles.

71. Mr. WatsonHST EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) 1.0 mol Fe 2 S 3 => 2.0 mol Fe(OH) 3 2.0 mol H 2 O => 1.3 mol Fe(OH) 3 least amount 3.0 mol O 2 => 4.0 mol Fe(OH) 3 Since 2.0 mol H 2 O will produce only 1.3 mol Fe(OH) 3, then H 2 O is the limiting reactant.

72. Mr. WatsonHST EXAMPLE: What is the number of moles of Fe(OH) 3 (S) that can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O, and 3.0 mol O 2 to react? 2Fe 2 S 3(S) + 6H 2 O (l) + 3O 2(g) -----> 4Fe(OH) 3(S) + 6S (S) 1.0 mol Fe 2 S 3 => 2.0 mol Fe(OH) 3 2.0 mol H 2 O => 1.3 mol Fe(OH) 3 least amount 3.0 mol O 2 => 4.0 mol Fe(OH) 3 Since 2.0 mol H 2 O will produce only 1.3 mol Fe(OH) 3, then H 2 O is the limiting reactant. Thus the maximum number of moles of Fe(OH) 3 that can be produced by this reaction is 1.3 moles.

73. Mr. WatsonHST Theoretical Yield the amount of product produced by a reaction based on the amount of the limiting reactant

74. Mr. WatsonHST Actual Yield amount of product actually produced in a reaction

75. Mr. WatsonHST Percent Yield actual yield % yield = --------------------- * 100 theoretical yield

76. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process

77. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00 kg Cl 2 ) #kg N 2 H 4 = ---------------------

78. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00 kg Cl 2 ) (1000 g Cl 2 ) #kg N 2 H 4 = ----------------------------------- (1 kg Cl 2 ) metric conversion

79. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00) (1000 g Cl 2 ) (1 mol Cl 2 ) #kg N 2 H 4 = ----------------------------------------- (1) (70.9 g Cl 2 ) molar mass

80. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1 mol Cl 2 ) #kg N 2 H 4 = ----------------------------------------- (1)(70.9)

81. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1 mol Cl 2 )(1 mol N 2 H 4 ) #kg N 2 H 4 = ------------------------------------------------- (1) (70.9) (1 mol Cl 2 )

82. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1 mol N 2 H 4 ) #kg N 2 H 4 = ------------------------------------------------- (1) (70.9)(1)

83. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1 mol N 2 H 4 ) (32.0 g N 2 H 4 ) #kg N 2 H 4 = -------------------------------------------------------- (1)(70.9) (1) (1 mol N 2 H 4 ) molar mass

84. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0 g N 2 H 4 )(1 kg N 2 H 4 ) #kg N 2 H 4 = ---------------------------------------------------------- (1)(70.9)(1)(1) (1000 g N 2 H 4 ) metric conversion

85. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0)(1 kg N 2 H 4 ) #kg N 2 H 4 = ---------------------------------------------------------- (1)(70.9)(1)(1)(1000)

86. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00)(1000)(1)(1) (32.0)(1 kg N 2 H 4 ) #kg N 2 H 4 = ---------------------------------------------------------- (1)(70.9)(1)(1)(1000) = 0.451 kg N 2 H 4

87. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = 0.451 kg N 2 H 4 (b) actual yield (0.299 kg product) # kg N 2 H 4 = --------------------------

88. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = 0.451 kg N 2 H 4 (b) actual yield (0.299 kg product) (98.0 kg N 2 H 4 ) # kg N 2 H 4 = -------------------------------------------- (100 kg product) purity factor

89. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = 0.451 kg N 2 H 4 (b) actual yield (0.299 kg product) (98.0 kg N 2 H 4 ) # kg N 2 H 4 = -------------------------------------------- (100 kg product) = 0.293 kg N 2 H 4

90. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = 0.451 kg N 2 H 4 (b) actual yield# kg N 2 H 4 = 0.293 kg N 2 H 4

91. Mr. WatsonHST EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N 2 H 4 for every 1.00 kg of Cl 2 that is reacted with excess NaOH and NH 3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3 -----> N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = 0.451 kg N 2 H 4 (b) actual yield# kg N 2 H 4 = 0.293 kg N 2 H 4 (c) percent yield 0.293 kg % yield = -------------- X 100 = 65.0 % yield 0.451kg