1. Dr. S. M. Condren Chapter 3 Calculations with Chemical Formulas and Equations

2. Dr. S. M. Condren Molar Mass Sum atomic masses represented by formula atomic masses => gaw molar mass => MM

3. Dr. S. M. Condren 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.

4. Dr. S. M. Condren 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 = 2(12.011) C + 6(1.00794) H + 1(15.9994) O = 24.022 + 6.04764 + 15.9994 = 46.069 g/mol Significant figures rule for multiplication Significant figures rule for addition Sequence – multiplication then addition, apply significant figure rules in proper sequence

5. Dr. S. M. Condren The Mole a unit of measurement, quantity of matter present Avogadro’s Number 6.022 x 10 23 particles Latin for “pile”

6. Dr. S. M. Condren Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? CO 2 MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2 =(10.00g) (1 mol/44.01g)

7. Dr. S. M. Condren Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? CO 2 MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2 =(10.00g) (1 mol/44.01g)

8. Dr. S. M. Condren Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? CO 2 MM = 1(gaw) C + 2(gaw) O = 44.01 g/mol #mol CO 2 =(10.00) (1 mol/44.01) = 0.2272 mol

9. Dr. S. M. Condren Combustion Analysis

10. Dr. S. M. Condren Percentage Composition description of a compound based on the relative amounts of each element in the compound

11. Dr. S. M. Condren 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

12. Dr. S. M. Condren 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

13. Dr. S. M. Condren 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

14. Dr. S. M. Condren 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

15. Dr. S. M. Condren 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

16. Dr. S. M. Condren Simplest (Empirical) Formula formula describing a substance based on the smallest set of subscripts

17. Dr. S. M. Condren 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

18. Dr. S. M. Condren 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? Element P O % 43.7 56.3 Relative Number of Atoms (%/gaw) 43.7/30.97 56.3/15.9994 = 1.41 = 3.52 Divide by Smaller 1.41/1.41 = 1.00 3.52/1.41 = 2.50

19. Dr. S. M. Condren 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

20. Dr. S. M. Condren 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

21. Dr. S. M. Condren 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 Element N O % 30.5 69.5 Relative # Atoms (%/gaw) 30.5/14.0067 = 2.18 69.5/15.9994 = 4.34 Divide by Smaller 2.18/2.18 = 1.00 4.34/2.18 = 1.99

22. Dr. S. M. Condren 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 Element N O % 30.5 69.5 Relative # Atoms (%/gaw) 30.5/14.0067 = 2.18 69.5/15.9994 = 4.34 Divide by Smaller 2.18/2.18 = 1.00 4.34/2.18 = 1.99 Multiply by Integer 1*1.00=>1 1*1.99=>2 Empirical Formula => NO 2

23. Dr. S. M. Condren Molecular Formula the exact proportions of the elements that are formed in a molecule

24. Dr. S. M. Condren Molecular Formula from Simplest Formula empirical formula => EF molecular formula => MF MF = X * EF

25. Dr. S. M. Condren Molecular Formula from Simplest Formula formula mass => FM sum of the atomic weights represented by the formula molar mass = MM = X * FM

26. Dr. S. M. Condren Molecular Formula from Simplest Formula first, knowing MM and FM X = MM/FM then MF = X * EF

27. Dr. S. M. Condren 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

28. Dr. S. M. Condren What is the correct molecular formula for this colorless liquid rocket fuel? 2NO 2 NO 2 N 2 O 4

29. Dr. S. M. Condren 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.

30. Dr. S. M. Condren 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

31. Dr. S. M. Condren 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

32. Dr. S. M. Condren 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 )

33. Dr. S. M. Condren 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 )

34. Dr. S. M. Condren 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)

35. Dr. S. M. Condren Combination Reaction PbNO 3(aq) + K 2 CrO 4(aq)  PbCrO 4(s) + 2 KNO 3(aq) Colorless yellow yellow colorless

36. Dr. S. M. Condren Stoichiometric Roadmap

37. Dr. S. M. Condren 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

38. Dr. S. M. Condren 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

39. Dr. S. M. Condren 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

40. Dr. S. M. Condren 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

41. Dr. S. M. Condren 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?

42. Dr. S. M. Condren 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? #g Fe 2 O 3 = (167 g Fe) (1 mol Fe) * -------------- (55.85 g Fe) (1 mol Fe 2 O 3 ) ----------------- (2 mol Fe) (159.7 g Fe 2 O 3 ) ------------------- (1 mol Fe 2 O 3 ) = 238 g Fe 2 O 3

43. Dr. S. M. Condren 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? #g Al = (167 g Fe) (1 mol Fe) * ---------------- (55.85 g Fe) (2 mol Al) ------------- (2 mol Fe) (26.9815 g Al) ------------------- (1 mol Al) = 80.6 g Al

44. Dr. S. M. Condren 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

45. Dr. S. M. Condren Limiting Reactant reactant that limits the amount of product that can be produced

46. Dr. S. M. Condren 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)

47. Dr. S. M. Condren 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)

48. Dr. S. M. Condren 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)

49. Dr. S. M. Condren 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

50. Dr. S. M. Condren 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

51. Dr. S. M. Condren 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

52. Dr. S. M. Condren 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

53. Dr. S. M. Condren 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

54. Dr. S. M. Condren 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.

55. Dr. S. M. Condren 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.

56. Dr. S. M. Condren 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.

57. Dr. S. M. Condren Theoretical Yield the amount of product produced by a reaction based on the amount of the limiting reactant

58. Dr. S. M. Condren Actual Yield amount of product actually produced in a reaction

59. Dr. S. M. Condren Percent Yield actual yield % yield = --------------------- * 100 theoretical yield

60. Dr. S. M. Condren 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

61. Dr. S. M. Condren 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 = ---------------------

62. Dr. S. M. Condren 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

63. Dr. S. M. Condren 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

64. Dr. S. M. Condren 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)

65. Dr. S. M. Condren 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 )

66. Dr. S. M. Condren 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)

67. Dr. S. M. Condren 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

68. Dr. S. M. Condren 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

69. Dr. S. M. Condren 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)

70. Dr. S. M. Condren 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

71. Dr. S. M. Condren 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 = --------------------------

72. Dr. S. M. Condren 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

73. Dr. S. M. Condren 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

74. Dr. S. M. Condren 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

75. Dr. S. M. Condren 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