2. Moles and Compounds You need a calculator for use during these notes!

3. Roses and eggs are conveniently packaged as a dozen. Sheets of paper are packaged as a ream. Small item are packaged in large amounts to make life easier. The same is true for atoms. However, because atoms are really, really, really small, the amounts of them that are packaged together are really, really, really big

4. A package of atoms or compounds that chemists use is called the Mole.

6. A package of atoms or compounds that chemists use is called the Mole. A mole (abbreviated “mol”) contains 602,200,000,000,000,000,000,000 of anything. This really large number is often called Avogadro’s Number and is abbreviated 6.022 × 10 23. Typical representative particles (things) measured in moles: atoms molecules compounds formula units

8. If you know the number of moles, then the amount of atoms can be calculated. How many atoms are in 2.00 moles of Al?

9. If you know the number of moles, then the amount of atoms can be calculated. How many atoms are in 2.00 moles of Al? 2.00 mols Al =

10. If you know the number of moles, then the amount of atoms can be calculated. How many atoms are in 2.00 moles of Al? 2.00 mols Al = 1 mol

11. If you know the number of moles, then the amount of atoms can be calculated. How many atoms are in 2.00 moles of Al? 2.00 mols Al6.022 × 10 23 atoms = 1 mol

12. If you know the number of moles, then the amount of atoms can be calculated. How many atoms are in 2.00 moles of Al? In a calculator, you would put in 2 × 6.022 EE 23. Pushing the button that says EE takes the place of the “ × 10 ” and is the way calculators are meant to be used with scientific notation! 2.00 mols Al6.022 × 10 23 atoms = 1 mol

13. If you know the number of moles, then the amount of atoms can be calculated. How many atoms are in 2.00 moles of Al? 2.00 mols Al6.022 × 10 23 atoms = 1.20 × 10 24 atoms Al 1 mol

14. The opposite can also be done. How many moles is 9.03 × 10 23 atoms of K?

15. The opposite can also be done. How many moles is 9.03 × 10 23 atoms of K? 9.03 × 10 23 atoms K =

16. The opposite can also be done. How many moles is 9.03 × 10 23 atoms of K? 9.03 × 10 23 atoms K = 6.022 × 10 23 atoms

17. The opposite can also be done. How many moles is 9.03 × 10 23 atoms of K? 9.03 × 10 23 atoms K 1 mol = 6.022 × 10 23 atoms

18. The opposite can also be done. How many moles is 9.03 × 10 23 atoms of K? 9.03 × 10 23 atoms K 1 mol = 1.50 mols K 6.022 × 10 23 atoms

19. Notice, in both cases you use that fact that 1 mol = 6.022 × 10 23. Whether the 6.022 × 10 23 is on the top or bottom of the railroad tracks depends on what you start the problem with. Also, this can be done with ANY representative particle, not just atoms. The only difference is the label!

20. Masses and the Mole The mole is not only handy to counting a large amount of atoms or molecules, it is even more useful when measuring out amounts of elements or compounds. The masses on the periodic table have been designed to be the amount of grams in a mole of that element, thus it is called the “molar mass”

21. For example, look at sulfur. The information shown for sulfur tells us that 1 mol of sulfur = 32.066 grams of sulfur. Likewise, 2 mols of sulfur = 2 × 32.066 or 64.132 grams of sulfur, etc. This allows for amounts of elements to be easily measured in the laboratory and turned into moles. 16 S 32.066

22. A student measures out 12.0 g of Mg in lab. How many moles does this student have?

23. 12.0 g Mg =

24. A student measures out 12.0 g of Mg in lab. How many moles does this student have? 12.0 g Mg = 24.31 g

25. A student measures out 12.0 g of Mg in lab. How many moles does this student have? 12.0 g Mg1 mol = 24.31 g

26. A student measures out 12 g of Mg in lab. How many moles does this student have? Notice the molar mass is unique for each element and must come from the periodic table. 12.0 g Mg1 mol = 0.494 mols Mg 24.31 g

27. The opposite can also be done. A student needs 2.30 mols of B for an experiment. How many grams should this student measure out?

28. The opposite can also be done. A student needs 2.30 mols of B for an experiment. How many grams should this student measure out? 2.30 mols B =

29. The opposite can also be done. A student needs 2.30 mols of B for an experiment. How many grams should this student measure out? 2.30 mols B = 1 mol

30. The opposite can also be done. A student needs 2.30 mols of B for an experiment. How many grams should this student measure out? 2.30 mols B10.81 g = 1 mol

31. The opposite can also be done. A student needs 2.30 mols of B for an experiment. How many grams should this student measure out? 2.30 mols B10.81 g = 24.9 g B 1 mol

32. Our previous work with representative particles and molar masses can be put together!

33. A student needs 2.60 × 10 23 atoms of fluorine for an experiment. How many grams of fluorine should the student measure out?

34. 2.60 × 10 23 atoms F =

35. A student needs 2.60 × 10 23 atoms of fluorine for an experiment. How many grams of fluorine should the student measure out? 2.60 × 10 23 atoms F = 6.022 × 10 23 atoms

36. A student needs 2.60 × 10 23 atoms of fluorine for an experiment. How many grams of fluorine should the student measure out? 2.60 × 10 23 atoms F 1 mol = 6.022 × 10 23 atoms

37. A student needs 2.60 × 10 23 atoms of fluorine for an experiment. How many grams of fluorine should the student measure out? 2.60 × 10 23 atoms F 1 mol = 6.022 × 10 23 atoms 1 mol

38. A student needs 2.60 × 10 23 atoms of fluorine for an experiment. How many grams of fluorine should the student measure out? 2.60 × 10 23 atoms F 1 mol18.99 g = 6.022 × 10 23 atoms 1 mol

39. A student needs 2.60 × 10 23 atoms of fluorine for an experiment. How many grams of fluorine should the student measure out? 2.60 × 10 23 atoms F 1 mol18.99 g = 8.20 g F 6.022 × 10 23 atoms 1 mol

40. The opposite can also be done. A student has 56.7 grams of aluminum. How many atoms of aluminum does the student have?

41. The opposite can also be done. A student has 56.7 grams of aluminum. How many atoms of aluminum does the student have? 56.7 g Al

42. The opposite can also be done. A student has 56.7 grams of aluminum. How many atoms of aluminum does the student have? 56.7 g Al 26.98 g

43. The opposite can also be done. A student has 56.7 grams of aluminum. How many atoms of aluminum does the student have? 56.7 g Al1 mol 26.98 g

44. The opposite can also be done. A student has 56.7 grams of aluminum. How many atoms of aluminum does the student have? 56.7 g Al1 mol 26.98 g1 mol

45. The opposite can also be done. A student has 56.7 grams of aluminum. How many atoms of aluminum does the student have? 56.7 g Al1 mol6.022 × 10 23 atoms 26.98 g1 mol

46. The opposite can also be done. A student has 56.7 grams of aluminum. How many atoms of aluminum does the student have? 56.7 g Al1 mol6.022 × 10 23 atoms = 1.27 × 10 24 atoms Al 26.98 g1 mol

47. Moles of Compounds

48. A properly written compound shows the ratio of atoms in the compound. For example, sodium carbonate (Na 2 CO 3 ) shows that for every 1 compound there are 2 atoms of Na, 1 atom of C, and 3 atoms of O.

49. This also works with the mole. For 1 mole of the compound, there are 2 moles of Na, 1 mole of C, and 3 moles of O. This information allows us to calculate the molar mass of the entire compound, often called the formula mass.

50. 2 mols Na22.9897 g = 45.9794 g Na 1 mol 1 mol C12.0107 g = 12.0107 g C 1 mol 3 mols O15.9994 g = 47.9982 g O 1 mol The total molar mass of Na 2 CO 3 is 105.9883 g.

51. Once it is known that 1 mol of Na 2 CO 3 = 105.9883 g, conversion factors can be made of this information: These conversion factors can then be used in other calculations. 105.9883 g Na 2 CO 3 or 1 mol 105.9883 g Na 2 CO 3

52. What is the molar mass for potassium oxide (K 2 O)?

53. K 2 O = 2 x K + 1 x O

54. What is the molar mass for potassium oxide (K 2 O)? K 2 O = 2 x K + 1 x O K 2 O = (2 x 39.0983 g) + (15.9994 g)

55. What is the molar mass for potassium oxide (K 2 O)? K 2 O = 2 x K + 1 x O K 2 O = (2 x 39.0983 g) + (15.9994 g) K 2 O = 78.1966 g + 15.9994 g = 94.1960 g

56. If you have 560.0 grams of K 2 O, how many moles do you have?

57. 560.0 g K 2 O =

58. If you have 560.0 grams of K 2 O, how many moles do you have? 560.0 g K 2 O1 mol = 94.196 g

59. If you have 560.0 grams of K 2 O, how many moles do you have? 560.0 g K 2 O1 mol = 5.945 mol K 2 O 94.196 g

60. If you have 3.4 moles of K 2 O, how many grams do you have?

61. 3.4 mols K 2 O =

62. If you have 3.4 moles of K 2 O, how many grams do you have? 3.4 mols K 2 O94.2 g = 1 mol

63. If you have 3.4 moles of K 2 O, how many grams do you have? 3.4 mols K 2 O94.2 g = 320 g K 2 O 1 mol

64. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 54.3 g CoCl 2 =

65. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 54.3 g CoCl 2 = 129.8 g 58.93g + (2 x 35.45 g)

66. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 54.3 g CoCl 2 1 mol CoCl 2 = 129.8 g 58.93g + (2 x 35.45 g)

67. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 54.3 g CoCl 2 1 mol CoCl 2 = 129.8 g1 mol CoCl 2

68. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 54.3 g CoCl 2 1 mol CoCl 2 6.022 x 10 23 formula units = 129.8 g1 mol CoCl 2

69. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 54.3 g CoCl 2 1 mol CoCl 2 6.022 x 10 23 formula units = 2.52 x 10 23 formula units 129.8 g1 mol CoCl 2

70. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 2.52 x 10 23 formula units CoCl 2

71. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 2.52 x 10 23 formula units CoCl 2 1 formula unit CoCl 2

72. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 2.52 x 10 23 formula units CoCl 2 3 atoms 1 formula unit CoCl 2

73. If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this? 2.52 x 10 23 formula units CoCl 2 3 atoms = 7.56 x 10 23 atoms 1 formula unit CoCl 2

74. Percent Composition and Empirical Formulas

75. Percent Composition

76. The percent composition shows the relative percent (by mass) of each element in a compound.

77. Percent Composition The percent composition shows the relative percent (by mass) of each element in a compound. The percent composition is determined by dividing the mass of the individual elements in a compound by the entire formula mass of the compound.

78. Percent Composition = Mass of individual element (g) × 100 %= % of element Formula Mass of Compound (g)

79. For example, when the correct percent composition for HF is determined, the process is as follows:

80. Find the total formula molar mass: 1 mol H = 1.0079 g/mol 1 mol F = + 18.9884 g/mol total = 19.9963 g/mol

81. Take the individual molar mass of each element and divide by the total formula mass, and turn it into a percent:

82. for H 1.0079 g/mol H × 100 %= 5.04% H 19.9963 g/mol HF for F 18.9884 g/mol F × 100 %= 94.96% F 19.9963 g/mol HF

83. A good way to quickly check the answers is to sum the percentages, which should equal 100% (or 1). There will be cases where the percentages might not equal exactly 100% because of rounding, but the total should always be VERY close to 100%.

84. As another example, consider sulfuric acid: H 2 SO 4 :

85. As another example, consider sulfuric acid: H 2 SO 4 : 2 mol H × 1.008 g/mol= 2.016 g/mol 1 mol S × 32.066 g/mol= 32.066 g/mol 4 mol O × 15.999 g/mol= 63.996 g/mol total = 98.078 g/mol

86. As another example, consider sulfuric acid: H 2 SO 4 : for H 2.016 g/mol H × 100%= 2.06% H 98.078 g/mol H 2 SO 4 for S32.066 g/mol S × 100%= 32.69% S 98.078 g/mol H 2 SO 4 for O 63.996 g/mol O × 100%= 65.25% O 98.078 g/mol H 2 SO 4

87. Empirical Formula

88. Once the percent composition of a compound is known, the empirical formula of the compound can be determined. An empirical formula shows the lowest whole-number ratio of the elements in a compound

89. 1.Turn the percent composition information into mass. This is made simple by assuming a theoretical amount of 100 grams. Thus 50% composition is turned into 50 grams, and 36.8% composition is turned into 36.8 grams, etc.

90. 2.Calculate the number of moles for each element that would contain the amount of mass from step 1. This involves dividing the mass from step 1 by the molar mass shown for the element on the periodic table.

91. 3.The simplest whole-number ratio of each element needs to be found. One of the ways to get a good start on this is to divide each number of moles from step 2 by the smallest amount of moles. This will guarantee at least one whole number to start with (a “ 1 ” amount).

92. 3.a. If the other molar amounts are within 0.15 of a whole number, it is usually safe to round up or down to that whole number.

93. 3.b. If the other molar amounts cannot be rounded, it will be necessary to multiply ALL the molar amounts by a whole number to obtain a whole number (or a number close to a whole number.) Thus, if a molar amount had the decimal value of 0.20, it would be necessary to multiply by 5. If the decimal value is 0.25,it would be necessary to multiply by 4, and it would if the decimal value is 0.33, it would be necessary to multiply by 3, etc.

94. Example: White gold is 75.0% gold, 10.0% palladium, 10.0% nickel, and 5.00% zinc. What would be the empirical formula of white gold?

95. 75.0% Au →75.0 g Au1 mole Au = 0.3807 moles Au 197.0 g 10.0% Pd →10.0 g Pd1 mole Pd = 0.09398 moles Pd 106.4 g 10.0% Ni →10.0 g Ni1 mole Ni = 0.1704 moles Ni 58.69 g 5.00% Zn →5.00 g Zn1 mole Zn = 0.07646 moles Zn 65.39 g

96. Dividing by the lowest amount of moles from above (0.07646 mol): 0.3807 moles Au = 4.979 moles Au ≈ 5 moles Au 0.07646 0.09398 moles Pd = 1.229 moles Pd 0.07646 0.1704 moles Ni = 2.229 moles Ni 0.07646 0.07646 moles Zn = 1 moles Zn 0.07646

97. The gold and zinc are already expressed in a whole number, but to express the palladium and nickel as a whole number, it will be necessary to multiply everything by 4. This would make the palladium and nickel 4.916 moles and 8.916 moles (respectively), which are now close enough to round. Do not forget to multiply everything, even the ones that are already whole numbers!

98. The gold and zinc are already expressed in a whole number, but to express the palladium and nickel as a whole number, it will be necessary to multiply everything by 4. This would make the palladium and nickel 4.916 moles and 8.916 moles (respectively), which are now close enough to round. Do not forget to multiply everything, even the ones that are already whole numbers! Thus the final relative amount of moles is 20 Au, 5 Pd, 9 Ni, 4 Zn. The empirical formula is Au 20 Pd 5 Ni 9 Zn 4.

99. Practice Exercise: Find the empirical formula for purple gold, Purple Gold = 80% Au, 20% Al

100. mass1 mol = answer = ×factor=whole # P.T. masslowest # mass1 mol = answer = ×factor=whole # P.T. masslowest #

101. 80 g1 mol = 0.4062 mol = answer × factor = whole # 196.97 glowest # 20 g1 mol = 0.7413 mol = answer × factor = whole # 26.98 glowest #

102. 80 g1 mol = 0.4062 mol =1× factor = whole # 196.97 g0.4062 mol 20 g1 mol = 0.7413 mol =1.825× factor = whole # 26.98 g0.4062 mol

103. 80 g1 mol = 0.4062 mol =1×5=5 196.97 g0.4062 mol 20 g1 mol = 0.7413 mol =1.825×5=9 26.98 g0.4062 mol

104. 80 g1 mol = 0.4062 mol =1×5=5 196.97 g0.4062 mol 20 g1 mol = 0.7413 mol =1.825×5=9 26.98 g0.4062 mol Empirical Formula = Au 5 Al 9

105. Molecular Formulas

106. An empirical formula shows the lowest whole-number ratio of the elements in a compound, but may not be the actual formula for the molecular compound, called the molecular formula. The molecular formula is always some whole number multiple of the empirical formula.

107. For Example CompoundE.F.M.F.Multiple WaterH2OH2OH2OH2O1 Hydrogen PeroxideHOH2O2H2O2 2 GlucoseCH 2 OC 6 H 12 O 6 6

109. To find the molecular formula it is necessary to know the empirical formula and the multiple. Often the empirical formula is calculated from the percent composition, and then the multiple is calculated by knowing the actual molar mass of the molecular compound.

110. For example: A compound has a formula mass of 78.11 g/mol and is 92.24% carbon and 7.76% hydrogen. What is the molecular formula?

111. Step 1: Determine the empirical formula:

112. 92.24 g C = 7.68 mols 12.01 g/mol 7.76 g H = 7.68 mols 1.01 g/mol

113. Step 1: Determine the empirical formula: Dividing both by 7.68 moles gives an empirical formula of CH. 92.24 g C = 7.68 mols 12.01 g/mol 7.76 g H = 7.68 mols 1.01 g/mol

114. Step 2: Determine the molar mass of the empirical formula:

115. CH = 13.02 g/mol

116. Step 3: Determine the factor:

117. molecular formula mass = 78.11 empirical formula mass =13.02 = 5.999 or 6

118. Step 3: Determine the factor: Thus the actual molecular formula is 6 times more than the empirical formula, or C 6 H 6. molecular formula mass = 78.11 empirical formula mass =13.02 = 5.999 or 6