1. Stoichiometry: Calculations with Chemical Formulas and Equations

2. Formula and Molecular Weights Formula weights (FW): sum of AW for atoms in formula. FW (H 2 SO 4 ) = 2AW(H) + AW(S) + 4AW(O) = 2(1.0 amu) + (32.0 amu) + 4(16.0) = 98.0 amu Molecular weight (MW) is the weight of the molecular formula. MW(C 6 H 12 O 6 ) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) Atomic and Molecular Weights

3. What is the molecular weight of the following molecules? CH 4 H 2 SO 4 Ca(NO 3 ) 2 CuSO 45H 2 O

4. Atomic and Molecular Weights What is the molecular weight of the following molecules? CH 4 12.011 + 4(1.008) = 16.043 amu H 2 SO 4 Ca(NO 3 ) 2 CuSO 45H 2 O

5. Atomic and Molecular Weights What is the molecular weight of the following molecules? CH 4 12.011 + 4(1.008) = 16.043 amu H 2 SO 4 2(1.008) + 32.06 + 4(16.00) = 96.08 amu Ca(NO 3 ) 2 CuSO 45H 2 O

6. Atomic and Molecular Weights What is the molecular weight of the following molecules? CH 4 12.011 + 4(1.008) = 16.043 amu H 2 SO 4 2(1.008) + 32.06 + 4(16.00) = 96.08 amu Ca(NO 3 ) 2 40.08 + 2(14.007) + 6(16.00) = 164.09 amu CuSO 45H 2 O

7. Atomic and Molecular Weights What is the molecular weight of the following molecules? CH 4 12.011 + 4(1.008) = 16.043 amu H 2 SO 4 2(1.008) + 32.06 + 4(16.00) = 96.08 amu Ca(NO 3 ) 2 40.08 + 2(14.007) + 6(16.00) = 164.09 amu CuSO 45H 2 O 63.546+32.06+9(16.00)+10(1.008) = 249.68 amu

8. Percentage Composition from Formulas Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100: Atomic and Molecular Weights

9. Determine the Mass Percent Composition of nitrogen in NH 4 NO 3 ?

10. Atomic and Molecular Weights Determine the Mass Percent Composition of nitrogen in NH 4 NO 3 ? Formula weight 80.043 amu

11. Atomic and Molecular Weights Determine the Mass Percent Composition of nitrogen in NH 4 NO 3 ? Formula weight 80.043 amu

12. The Mole strictly a convenience unit amount that is large enough to see and handle in lab mole = number of things dozen = 12 things mole = 6.022 x 10 23 things Avogadro’s number = 6.022 x 10 23 = N A

13. The Mole how do you know when you have a mole? count it out weigh it out molar mass - mass in grams equal to the atomic weight H is 1.00794 g of H atoms = 6.022 x 10 23 atoms Mg is 24.3050 g of Mg atoms = 6.022 x 10 23 atoms

14. Mole: convenient measure chemical quantities. 1 mole of something = 6.0221367 x 10 23 of that thing. Experimentally, 1 mole of 12 C has a mass of 12 g. Molar Mass Molar mass: mass in grams of 1 mole of substance (units g/mol, g.mol -1 ). Mass of 1 mole of 12 C = 12 g. The Mole

15. Molar Mass Molar mass: sum of the molar masses of the atoms: molar mass of N 2 = 2 x (molar mass of N). Molar masses for elements are found on the periodic table. Formula weights are numerically equal to the molar mass. The Mole

16. Molar Masses of Compounds add atomic weights of each atom The molar mass of propane, C 3 H 8, is: One mole of propane is 44.11 g of propane.

17. Interconverting Masses, Moles, and Numbers of Particles The Mole

18. Calculate the mass of a single Mg atom, in grams, to 3 significant figures.

19. The Mole Calculate the mass of a single Mg atom, in grams, to 3 significant figures.

20. The Mole How many atoms are contained in 1.67 moles of Mg?

21. The Mole How many atoms are contained in 1.67 moles of Mg?

22. The Mole How many moles of Mg atoms are present in 73.4 g of Mg?

23. The Mole How many moles of Mg atoms are present in 73.4 g of Mg?

24. Mole Problems with Compounds Calculate the number of C 3 H 8 molecules in 74.6 g of propane.

25. Mole Problems with Compounds Calculate the number of C 3 H 8 molecules in 74.6 g of propane.

26. Mole Problems with Compounds Calculate the number of O atoms in 26.5 g of Li 2 CO 3.

27. Mole Problems with Compounds Calculate the number of O atoms in 26.5 g of Li 2 CO 3.

28. 1 Mole of Some Common Molecular Substances 1 Mole Br 2 159.81g C 4 H 10 Contains 6.022 x 10 23 Br 2 molecules 2(6.022 x 10 23 ) Br atoms

29. 1 Mole of Some Common Molecular Substances 1 Mole Br 2 159.81g C 4 H 10 58.04 g Contains 6.022 x 10 23 Br 2 molecules 2(6.022 x 10 23 ) Br atoms 6.022 x 10 23 C 4 H 10 molecules 4 (6.022 x 10 23 ) C atoms 10 (6.022 x 10 23 ) H atoms

30. Formulas from Elemental Composition empirical formula - simplest molecular formula, shows ratios of elements but not actual numbers of elements molecular formula - actual numbers of atoms of each element in the compound determine empirical & molecular formulas of a compound from percent composition percent composition is determined experimentally

31. Empirical Formulas Simplest whole number ratio of atoms or ions present in a compound Empirical Formula H 2 O C 6 H 12 O 6 H 2 O 2

32. Empirical Formulas Simplest whole number ratio of atoms or ions present in a compound Empirical Formula H 2 OH 2 O C 6 H 12 O 6 H 2 O 2

33. Empirical Formulas Simplest whole number ratio of atoms or ions present in a compound Empirical Formula H 2 OH 2 O C 6 H 12 O 6 CH 2 O H 2 O 2

34. Empirical Formulas Simplest whole number ratio of atoms or ions present in a compound Empirical Formula H 2 OH 2 O C 6 H 12 O 6 CH 2 O H 2 O 2 HO

35. Once we know the empirical formula, we need the MW to find the molecular formula. Subscripts in the molecular formula are always whole- number multiples of subscripts in the empirical formula. Empirical Formulas and Molecular Formula

36. Combustion Analysis Empirical Formulas from Analyses

37. Percent Composition mass of that element divided by the mass of the compound x 100% percent composition of C in C 3 H 8

38. Percent Composition percent composition of H in C 3 H 8

39. Percent Composition percent composition of H in C 3 H 8

40. Percent Composition Calculate the percent composition of Fe 2 (SO 4 ) 3 to 3 sig. fig.

41. Percent Composition Calculate the percent composition of Fe 2 (SO 4 ) 3 to 3 sig. fig.

42. Start with mass % of elements (i.e. empirical data) and calculate a formula, or Start with the formula and calculate the mass % elements. Empirical Formulas from Analyses

43. Empirical Formulas from Analysis Determine the empirical formula (Cr x O y ) for a compound with the percent composition: 68.42% Cr, 31.58% O. 1.) Assume 100g 2.) Determine mol of each element 3.) Divide through by smallest subscript 4.) Convert all subscripts to integers

44. Empirical Formulas from Analysis Determine the empirical formula (Cr x O y ) for a compound with the percent composition: 68.42% Cr, 31.58% O. 1.) Assume 100g 68.42g Cr and31.58 g O

45. Empirical Formulas from Analysis Determine the empirical formula (Cr x O y ) for a compound with the percent composition: 68.42% Cr, 31.58% O. 1.) Assume 100g 68.42g Cr and31.58 g O 2.) Determine mol of each element

46. Empirical Formulas from Analysis Divide through by smallest whole number 1.316/1.316 = 1 Cr 1.974/1.316 = 1.50 O Convert to whole numbers Cr 1 O 1.5 x 2 = Cr 2 O 3

47. Empirical Formulas from Analysis A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? Make the simplifying assumption that you have 100.0 g of compound. in 100.0 g of compound there is 24.74 g of K 34.76 g of Mn 40.50 g of O

48. Empirical Formulas from Analysis

49. A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? Empirical Formulas from Analysis

50. A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? Empirical Formulas from Analysis

51. A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? Empirical Formulas from Analysis

52. A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? Empirical Formulas from Analysis

53. Chemical Equations Lavoisier: mass is conserved in a chemical reaction. (Law of Conservation of Mass) Chemical equations: symbolic representation of a chemical reaction. Two parts to an equation: reactants and products: 2H 2 + O 2  2H 2 O reactants on left side of reaction products on right side of equation relative amounts of each with stoichiometric coefficients

54. Chemical Equations attempt to show on paper what is happening at the molecular level

55. Chemical Equations look at what an equation tells us reactants yields products 1 form. unit 3 mol. 2 atoms 3 mol. 1 mole 3 moles 2 moles 3 moles 159.7g 84g 111.7g 132g

56. Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products. Chemical Equations

58. Law of Conservation of Matter no detectable change in quantity of matter in an ordinary chemical reaction discovered by Lavoisier balance chemical reactions to obey this law propane,C 3 H 8, burns in oxygen to give carbon dioxide and water C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O note that there are equal numbers of atoms of each element on both sides of equation

59. Law of Conservation of Matter NH 3 burns in oxygen to form NO & water

60. Law of Conservation of Matter C 5 H 12 burns in oxygen to form carbon dioxide & water balancing equations is a skill acquired only with lots of practice work many problems

61. Balance the following Equations CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) CaCO 3 (s)  CaO(s) + CO 2 (g) AgNO 3 (aq) + NaI(aq)  NaNO 3 (aq) + AgI(s) H 2 SO 4 (aq) + KOH  K 2 SO 4 (aq) + H 2 O(l)

62. Balance the following Equations CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)

63. Balance the following Equations CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) CaCO 3 (s)  CaO(s) + CO 2 (g)

64. Balance the following Equations CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) CaCO 3 (s)  CaO(s) + CO 2 (g) AgNO 3 (aq) + NaI(aq)  NaNO 3 (aq) + AgI(s)

65. Balance the following Equations CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) CaCO 3 (s)  CaO(s) + CO 2 (g) AgNO 3 (aq) + NaI(aq)  NaNO 3 (aq) + AgI(s) H 2 SO 4 (aq) + 2KOH  K 2 SO 4 (aq) + 2H 2 O(l)

66. Balanced chemical equation gives number of molecules that react to form products. Interpretation: ratio of number of moles of reactant required to give the ratio of number of moles of product. These ratios are called stoichiometric ratios. NB: Stoichiometric ratios are ideal proportions Real ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles). Quantitative Information from Balanced Equations

67. The ratio of grams of reactant cannot be directly related to the grams of product. Quantitative Information from Balanced Equations

68. Calculations Based on Chemical Equations can do this in moles, formula units, etc. most often work in grams (or kg or pounds or tons) How many CO molecules are required to react with 25 formula units of Fe 2 O 3 ?

69. Calculations Based on Chemical Equations How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide?

70. Calculations Based on Chemical Equations How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide?

71. Calculations Based on Chemical Equations What mass of CO is required to react with 146 g of iron (III) oxide?

72. Calculations Based on Chemical Equations What mass of CO is required to react with 146 g of iron (III) oxide?

73. Calculations Based on Chemical Equations What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

74. Calculations Based on Chemical Equations What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

75. Calculations Based on Chemical Equations What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

76. Calculations Based on Chemical Equations What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

77. Calculations Based on Chemical Equations

78. How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide?

79. Calculations Based on Chemical Equations 159.6 g84.0 g 111.6 g132.0 g or 159.6 lb84.0 lb 111.6 lb132.0 lb

80. Other Interpretations of Chemical Formulas What mass of ammonium phosphate, (NH 4 ) 3 PO 4, would contain 15.0 g of N?

81. Other Interpretations of Chemical Formulas What mass of ammonium phosphate, (NH 4 ) 3 PO 4, would contain 15.0 g of N?

82. Other Interpretations of Chemical Formulas What mass of ammonium phosphate, (NH 4 ) 3 PO 4, would contain 15.0 g of N?

83. Using the Periodic Table Properties of compounds vary systematically because of good ordering in the periodic table. 2K(s) + 2H 2 O(l)  2KOH(aq) + H 2 (g) 2M(s) + 2H 2 O(l)  2MOH(aq) + H 2 (g) Combustion in Air Combustion is the burning of a substance in oxygen from air: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l) Patterns of Chemical Reactivity

85. Combination and Decomposition Reactions 2Mg(s) + O 2 (g)  2MgO(s) There are fewer products than reactants; the Mg has combined with O 2 to form MgO. 2NaN 3 (s)  2Na(s) + 3N 2 (g) (the reaction that occurs in an air bag) There are more products than reactants; the sodium azide has decomposed into Na and nitrogen gas. Patterns of Chemical Reactivity

86. Combination and Decomposition Reactions Combination reactions: fewer reactants than products. Decomposition reactions: more products than reactants. Patterns of Chemical Reactivity

87. If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess). Limiting Reactant: one reactant that is consumed. Limiting Reactant Concept

88. What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110 g of oxygen?

89. Limiting Reactant Concept What do we do next?

90. Limiting Reactant Concept What can we conclude from our data now? Which is limiting reactant? What is maximum mass of sulfur dioxide?

91. Limiting Reactant Concept limiting reactant is O 2 maximum amount of SO 2 is 147 g

92. The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield. The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield: Theoretical yield

93. Percent Yields from Reactions theoretical yield is what we have been calculating actual yield is what you have in your flask A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What is the percent yield?

94. Percent Yields from Reactions

96. Synthesis Problem In 1986, Bednorz and Muller succeeded in making the first of a series of chemical compounds that were superconducting at relatively high temperatures. This first compound was La 2 CuO 4 which superconducts at 35K. In their initial experiments, Bednorz and Muller made only a few mg of this material. How many La atoms are present in 3.56 mg of La 2 CuO 4 ?

97. Synthesis Problem

98. Group Activity Within a year after Bednorz and Muller’s initial discovery of high temperature superconductors, Wu and Chu had discovered a new compound, YBa 2 Cu 3 O 7, that began to superconduct at 100 K. If we wished to make 1.00 pound of YBa 2 Cu 3 O 7, how many grams of yttrium must we buy?