1. 1 Chapter 6B Chemical Quantities

2. 2 CHAPTER OUTLINE  The Mole Concept The Mole Concept  Molar Mass Molar Mass  Calculations Using the Mole Calculations Using the Mole  Stoichiometry & Molar Ratios Stoichiometry & Molar Ratios  Mole-Mole & Mass-Mole Calculations Mole-Mole & Mass-Mole Calculations  Mass-Mass Calculations Mass-Mass Calculations  Limiting Reactant Limiting Reactant  Percent Yield Percent Yield  Energy Changes in Chemical Reactions Energy Changes in Chemical Reactions

3. 3 THE MOLE CONCEPT  Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present.  In order to relate the mass and number of atoms, chemists use the SI unit mole (abbreviated mol).

4. 4 THE MOLE CONCEPT  The number of particles in a mole is called Avogadro’s number and is 6.02x10 23. 1 mole equals to 6.02 x 10 23 Avogadro’s number (N A )

5. 5 THE MOLE CONCEPT A mole is a very large quantity 6.02x10 23 If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number. MOLE

6. 6 THE MOLE CONCEPT 1 mole H 2 molecules= 6.02x10 23 H 2 molecules = 2 x (6.02x10 23 ) H atoms 1 mole Na + ions 1 mole H 2 O molecules= 6.02x10 23 H 2 O molecules = 2 x (6.02x10 23 ) H atoms = 6.02x10 23 O atoms = 6.02x10 23 Na + ions 1 mole H atoms= 6.02x10 23 H atoms

7. 7 THE MOLE CONCEPT  The atomic mass of one atom expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams. Mass of 1 H atom = 1.008 amu Mass of 1 mol H atoms = 1.008 grams Mass of 1 Mg atom = 24.31 amu Mass of 1 mol Mg atoms = 24.31 g Mass of 1 Cl atom = 35.45 amu Mass of 1 mol Cl atoms = 35.45 g

8. 8 MOLAR MASS 1 mol O atom = 2 mol H atoms = Mass of one mole of H 2 O 2 (1.01 g) = 2.02 g 1 (16.00 g) = 16.00 g 18.02 g Molar mass  The mass of one mole of a substance is called molar mass, and is measured in grams.

9. 9 MOLAR MASS 2 mol O atoms = 2 mol H atoms = Mass of one mole of Ca(OH) 2 2 (1.01 g) = 2.02 g 2 (16.00 g) = 32.00 g 74.10 g Molar mass 1 mol Ca atom =1 (40.08 g) = 40.08 g

10. 10 Calculate the molar mass of each compound shown below: Example 1: Lithium carbonateLi 2 CO 3 Li2 x 6.94= 13.88 g C1 x 12.01= 12.01 g O3 x 16.00= 48.00 g Molar mass= 73.89 g/mol

11. 11 Calculate the molar mass of each compound shown below: Example 2: Salicylic acidC 7 H 6 O 3 C7 x 12.01= 84.07 g H6 x 1.01= 6.06 g O3 x 16.00= 48.00 g Molar mass= 138.13 g/mol

12. 12 CALCULATIONS USING THE MOLE  Conversions between mass, mole and particles can be done using molar mass and Avogadro’s number. Mass of a substance Moles of a substance MM Particles of a substance NANA Molar mass Avogadro’s number

13. 13 How many moles of iron are present in 25.0 g of iron? Example 1: 3 significant figures 55.85 1 Molar mass 0.448

14. 14 What is the mass of 5.00 mol of water? Example 2: 3 significant figures 1 18.02 90.1 Molar mass

15. 15 How many Mg atoms are present in 5.00 g of Mg? Example 3: 1 24.30 mass  mol  atoms Avogadro’s number 1 6.02 x 10 23 1.24x10 23 atoms Mg Molar mass 3 significant figures

16. 16 How many molecules of HCl are present in 25.0 g of HCl? Example 4: 1 36.46 mass  mol  molecules 1 6.02 x 10 23 4.13 x 10 23 molecules HCl 3 significant figures

17. 17 MOLES OF ELEMENTS IN A FORMULA  The subscripts in a chemical formula of a compound indicate the number of atoms of each type of element.  For example, 1 molecule of aspirin contains: C9H8O4C9H8O4 9 C atoms 8 H atoms 4 O atoms

18. 18 MOLES OF ELEMENTS IN A FORMULA  The subscripts also indicate the number of moles of each element in one mole of the compound.  For example, 1 mole of aspirin contains: C9H8O4C9H8O4 9 mole C atoms 8 mole H atoms 4 mole O atoms

19. 19 MOLES OF ELEMENTS IN A FORMULA  Using the subscripts from the aspirin formula, one can write the following conversion factors for each of the elements in 1 mole of aspirin. C9H8O4C9H8O4 9 moles C 1 mole C 9 H 8 O 4 8 moles H 1 mole C 9 H 8 O 4 4 moles O 1 mole C 9 H 8 O 4

20. 20 Determine the number of moles of C atoms in 1 mole of each of the following substances: Example 1: a) Acetoaminophen used in TylenolC 8 H 9 NO 2 8 moles C 1 mole C 8 H 9 NO 2 b) Zinc dietary supplementZn(C 2 H 3 O 2 ) 2 4 moles C 1 mole Zn(C 2 H 3 O 2 ) 2

21. 21 How many carbon atoms are present in 1.50 moles of aspirin, C 9 H 8 O 4 ? Example 2: 9 moles C 1 mole C 9 H 8 O 4 1.50 mol C 9 H 8 O 4 xx 6.02x10 23 atoms 1 mole C = 8.13x10 24 atoms

22. 22 SUMMARY OF MASS-MOLE CALCULATIONS

23. 23 STOICHIOMETRY  Stoichiometry is the quantitative relationship between the reactants and products in a balanced chemical equation.  A balanced chemical equation provides several important information about the reactants and products in a chemical reaction.

24. 24 MOLAR RATIOS For example: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 1 molecule3 molecules2 molecules 100 molecules300 molecules200 molecules 10 6 molecules3x10 6 molecules2x10 6 molecules 1 mole3 moles2 moles This is the molar ratios between the reactants and products

25. 25 Example 1: Determine each mole ratio below based on the reaction shown: 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O

26. 26 STOICHIOMETRIC CALCULATIONS  Stoichiometric calculations can be classified as one of the following: MOLES of compound A MOLES of compound B molar ratio Mole-mole calculations MASS of compound A MM Mass-mole calculations MASS of compound B MM Mass-mass calculations

27. 27 MOLE-MOLE CALCULATIONS  Relates moles of reactants and products in a balanced chemical equation MOLES of compound A MOLES of compound B molar ratio

28. 28 How many moles of nitrogen will react with 2.4 moles of hydrogen to produce ammonia as shown in the reaction below? Example 1: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 2.4 mol H 2 1 3 = 0.80 mol N 2 Mole ratio

29. 29 How many moles of ammonia can be produced from 32 moles of hydrogen? (Assume excess N 2 present) Example 2: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 32 mol H 2 2 3 = 21 mol NH 3 Mole ratio

30. 30 In one experiment, 6.80 mol of ammonia are prepared. How many moles of hydrogen were used up in this experiment? Example 3: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 6.80 mol NH 3 3 2 = 10.2 mol H 2 Mole ratio

31. 31 MASS-MOLE CALCULATIONS  Relates moles and mass of reactants or products in a balanced chemical equation MOLES of compound A MOLES of compound B molar ratio MASS of compound A MM

32. 32 How many grams of ammonia can be produced from the reaction of 1.8 moles of nitrogen with excess hydrogen as shown below? Example 1: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 1.8 mol N 2 2 1 = 61 g NH 3 Mole ratio 1 17.04 Molar mass

33. 33 How many moles of hydrogen gas are required to produce 75.0 g of ammonia? Example 2: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 75.0 g NH 3 1 17.04 = 6.60 mol H 2 2 3 Molar mass Mole ratio

34. 34 How many moles of ammonia can be produced from the reaction of 125 g of nitrogen? Example 3: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 125 g N 2 1 28.02 = 8.92 mol NH 3 1 2 Molar mass Mole ratio

35. 35 MASS -MASS CALCULATIONS  Relates mass of reactants and products in a balanced chemical equation

36. 36 What mass of oxygen will be required to react completely with 96.1 g of propane, C 3 H 8, according to the equation below? Example 1: 1 C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) Mass of propane Moles of propane Moles of oxygen Mass of oxygen

37. 37 Example 1: 1 C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) 96.1 g C 3 H 8 44.11 1 = 349 g O 2 1 5 Molar mass Mole ratio 32.00 1 Molar mass

38. 38 What mass of carbon dioxide will be produced from the reaction of 175 g of propane, as shown? Example 2: 1 C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) Mass of propane Moles of propane Moles of carbon dioxide Mass of carbon dioxide

39. 39 Example 2: 1 C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) 175 g C 3 H 8 44.11 1 = 524 g CO 2 1 3 Molar mass Mole ratio 44.01 1 Molar mass

40. 40 LIMITING REACTANT  When 2 or more reactants are combined in non- stoichiometric ratios, the amount of product produced is limited by the reactant that is not in excess.  This reactant is referred to as limiting reactant.  When doing stoichiometric problems of this type, the limiting reactant must be determined first before proceeding with the calculations.

41. 41 LIMITING REACTANT ANALOGY Consider the following recipe for a sundae:

42. 42 LIMITING REACTANT ANALOGY How many sundaes can be prepared from the following ingredients: The number of sundaes possible is limited by the amount of syrup, the limiting reactant. Limiting reactant Excess reactants

43. 43 GUIDE TO LIMITING REACTANT CALC. Assume reactant 1 is limiting Assume reactant 2 is limiting Compare

44. 44 How many moles of H 2 O can be produced by reacting 4.0 mol of hydrogen and 3.0 mol of oxygen gases as shown below: Example 1: 2 H 2 (g) + 1 O 2 (g)  2 H 2 O (g) Limiting reactant Mole-mole calculations

45. 45 Example 1: 4.0 mol H 2 2 2 = 4.0 mol H 2 O 2 H 2 (g) + 1 O 2 (g)  2 H 2 O (g) Assume H 2 is LR Assume O 2 is LR 3.0 mol O 2 2 1 = 6.0 mol H 2 O Correct answer Correct assumption

46. 46 A fuel mixture used in the early days of rocketry was a mixture of N 2 H 4 and N 2 O 4, as shown below. How many grams of N 2 gas is produced when 100. g of N 2 H 4 and 200. g of N 2 O 4 are mixed? Example 2: 2 N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Limiting reactant Mass-mass calculations

47. 47 Example 2: Assumes N 2 H 4 is LR Assumes N 2 O 4 is LR

48. 48 Example 2: 100. g N 2 H 4 32.06 4.68 mol N 2 2 3 2 N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Assume N 2 H 4 is LR 1

49. 49 Example 2: 200. g N 2 O 4 92.02 6.52 mol N 2 1 3 2 N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Assume N 2 O 4 is LR 1

50. 50 Example 2: 6.52 mol N 2 2 N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Assume N 2 O 4 is LR Assume N 2 H 4 is LR 4.68 mol N 2 Correct amount N 2 H 4 is LR

51. 51 Example 2: 4.68 mol N 2 28.02 = 131 g N 2 2 N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Calculate mass of N 2 1

52. 52 How many moles of Fe 3 O 4 can be produced by reacting 16.8 g of Fe with 10.0 g of H 2 O as shown below: Example 3: 3 Fe (s) + 4 H 2 O (l)  Fe 3 O 4 (s) + 4 H 2 (g) Limiting reactant Mass-mol calculations

53. 53 Example 3: 16.8 g Fe 0.100 mol Fe 3 O 4 3 1 Assume Fe is LR 55.85 1 3 Fe (s) + 4 H 2 O (l)  Fe 3 O 4 (s) + 4 H 2 (g)

54. 54 Example 3: 10.0 g H 2 O 0.139 mol Fe 3 O 4 4 1 Assume H 2 O is LR 18.02 1 3 Fe (s) + 4 H 2 O (l)  Fe 3 O 4 (s) + 4 H 2 (g)

55. 55 Example 3: 0.139 mol Fe 3 O 4 Assume H 2 O is LR Assume Fe is LR 0.100 mol Fe 3 O 4 Correct amount Fe is LR 3 Fe (s) + 4 H 2 O (l)  Fe 3 O 4 (s) + 4 H 2 (g)

56. 56 How many grams of AgBr can be produced when 50.0 g of MgBr 2 is mixed with 100.0 g of AgNO 3, as shown below: Example 4: MgBr 2 + 2 AgNO 3  2 AgBr + Mg(NO 3 ) 2 Limiting Reactant

57. 57 Example 4: 50.0 g MgBr 2 184.101 2 Assume MgBr 2 is LR 1 MgBr 2 + 2 AgNO 3  2 AgBr + Mg(NO 3 ) 2 187.77 1 102 g AgBr

58. 58 Example 4: 100.0 g AgNO 3 169.882 2 Assume AgNO 3 is LR 1 MgBr 2 + 2 AgNO 3  2 AgBr + Mg(NO 3 ) 2 187.77 1 111 g AgBr

59. 59 Example 4: 111 g AgBr Assume AgNO 3 is LR Assume MgBr 2 is LR 102 g AgBr Correct amount MgBr 2 is LR MgBr 2 + 2 AgNO 3  2 AgBr + Mg(NO 3 ) 2

60. 60 PERCENT YIELD  The amount of product calculated through stoichiometric ratios are the maximum amount product that can be produced during the reaction, and is thus called theoretical yield.  The actual yield of a product in a chemical reaction is the actual amount obtained from the reaction.

61. 61 PERCENT YIELD  The percent yield of a reaction is obtained as follows:

62. 62 In an experiment forming ethanol, the theoretical yield is 50.5 g and the actual yield is 46.8 g. What is the percent yield for this reaction? Example 1: 92.7 %

63. 63 Silicon carbide can be formed from the reaction of sand (SiO 2 ) with carbon as shown below: Example 2: 1 SiO 2 (s) + 3 C (s)  1 SiC (s) + 2 CO (g) When 125 g of sand are processed, 68.4 of SiC is produced. What is the percent yield of SiC in this reaction? Actual yield

64. 64 Example 2: 125 g SiO 2 60.09 83.4 g SiC 1 1 Calculate theoretical yield 1 1 SiO 2 (s) + 3 C (s)  1 SiC (s) + 2 CO (g) 1 40.10

65. 65 Example 2: 82.0 % Calculate percent yield

66. 66 In an experiment to prepare aspirin, the theoretical yield is 153.7 g and the actual yield is 124.0 g. What is the percent yield of this reaction? Example 3: 80.68 %

67. 67 Carbon tetrachloride (CCl 4 ) was prepared by reacting 100.0 g of Cl 2 with excess carbon disulfide (CS 2 ), as shown below. If 65.0 g was prepared in this reaction, calculate the percent yield. Example 4: CS 2 + 3 Cl 2  CCl 4 + S 2 Cl 2

68. 68 Example 4: 100.0 g Cl 2 70.90 3 1 Calculate theoretical yield 1 1 153.81 CS 2 + 3 Cl 2  CCl 4 + S 2 Cl 2 72.31 g

69. 69 Example 4: 89.9 % Calculate percent yield

70. 70 ENERGY CHANGES IN CHEMICAL REACTIONS  Heat is energy change that is lost or gained when a chemical reaction takes place.  The system is the reactants and products that we are observing. The surroundings are all the things that contain and interact with the system, such as the reaction flask, the laboratory room and the air in the room.  The direction of heat flow depends whether the products in a reaction have more or less energy than the reactants.

71. 71 ENERGY CHANGES IN CHEMICAL REACTIONS  For a chemical reaction to occur, the molecules of the reactants must collide with each other with the proper energy and orientation.  The minimum amount of energy required for a chemical reaction to occur is called the activation energy.  The heat of reaction is the amount of heat absorbed or released during a reaction and is designated by the symbol  H.

72. 72 ENERGY CHANGES IN CHEMICAL REACTIONS  When heat is released during a chemical reaction, it is said to be exothermic.  For exothermic reactions,  H is negative and is included on the right side of the equation. 3 H 2 (g) + N 2 (g) 2 NH 3 (g) + 22.0 kcal  H=  22.0 kcal

73. 73 ENERGY CHANGES IN CHEMICAL REACTIONS  When heat is gained during a chemical reaction, it is said to be endothermic.  For endothermic reactions,  H is positive and is included on the left side of the equation. 2 H 2 O (l) + 137 kcal 2 H 2 (g) + O 2 (g)  H= +137 kcal

74. 74 CALCULATING HEAT IN A REACTION  The value of  H in a chemical reaction refers to the heat lost or gained for the number of moles of reactants and products in a balanced chemical equation.  For example, based on the chemical equation shown below: 2 H 2 O (l) 2 H 2 (g) + O 2 (g)  H= +137 kcal 137 kcal 2 mol H 2 or 137 kcal 2 mol H 2 O or 137 kcal 1 mol O 2

75. 75 Given the reaction shown below, how much heat, in kJ, is released when 50.0 g of NH 3 form? Example 1: 3 H 2 (g) + N 2 (g) 2 NH 3 (g)  H=  92.2 kJ 50.0 g NH 3 x 1 mol NH 3 17.04 g NH 3 92.2 kJ 2 mol NH 3 x = 135 kJ

76. 76 How many kJ of heat are needed to react 25.0 g of HgO according to the reaction shown below: Example 2: 2 HgO (s) 2 Hg (l) + O 2 (g)  H= 182 kJ 25.0 g HgO x 1 mol HgO 216.59 g HgO 182 kJ 2 mol HgO x = 10.5 kJ

77. 77 THE END