1. APPLICATIONS OF THE MOLE

2. Molar Mass of a Compound
The molar mass of a compound is the mass of a mole of the representative particles of the compound.
Because each representative particle is composed of two or more atoms, the molar mass of the compound is found by adding the molar masses of all of the atoms in the representative particle.

3. Molar Mass of an Element
To determine the molar mass of an element, find the element’s symbol on the periodic table and round the mass so there is one digit beyond the decimal.
The molar mass of carbon (C) is
12.0 g/mol, of chlorine (Cl) is 35.5 g/mol and of iron (Fe) is 55.8 g/mol.

4. Molar Mass of a Compound
In the case of NH3, the molar mass equals the mass of one mole of nitrogen atoms plus the mass of three moles of hydrogen atoms.

5. Molar Mass of a Compound
Molar mass of NH3 = molar mass of N +
3(molar mass of H)
Molar mass of NH3 = 14.0 g +
3(1.0 g)
= 17.0 g/mol

6. Molar Mass of a Compound
You can use the molar mass of a compound to convert between mass and moles, just as you used the molar mass of elements to make these conversions.

7. Molar Mass Conversion
How many moles of magnesium in 56.3 g of Mg?
56.3 g Mg __ 1
mole Mg __
24.3
g Mg
(2.32)

8. Molar Mass Conversion
How many moles are in 146 grams of NH3?
(Answer: 8.59 mol)

9. Molar Mass Conversion
How many moles are in 295 grams of Cr(OH)3?
(Answer: 2.86 mol)

10. Molar Mass Conversion
How many moles are in 22.5 grams of HCl?
(Answer: mol)

11. Molar Mass Conversion
How many grams of sodium chloride in 3.45 moles of NaCl?
3.45 moles NaCl
58.5 __
g NaCl __ 1
mole NaCl
(202)

12. Molar Mass Conversion
How many grams are in moles of AlF3?
(Answer: 10.1 g)

13. Molar Mass Conversion
How many grams are in 13.0 moles of H2SO4?
(Answer: 1280 g)

14. Molar Mass Conversion
How many grams are in 1.6 moles of K2CrO4?
(Answer: 310 g)

15. Percent Composition
Recall that every chemical compound has a definite composition - a composition that is always the same wherever that compound is found.
The composition of a compound is usually stated as the percent by mass of each element in the compound.

16. (molar mass of X) (# X ‘s)
Percent Composition
The percent of an element (X) in a compound can be found in the following way.
% X =
(molar mass of X) (# X ‘s)
molar mass of compound
x 100

17. Percent Composition

18. Example
Determine the percent composition of chlorine in calcium chloride (CaCl2).
First, analyze the information available from the formula.
A mole of calcium chloride consists of one mole of calcium ions and two moles of chloride ions.

19. Calculating Percent Composition
Next, gather molar mass information from the atomic masses on the periodic table.

20. Calculating Percent Composition
To the mass of one mole of CaCl2, a mole of calcium ions contributes 40.1 g, and two moles of chloride ions contribute 2 x 35.5 g = 71.0 g for a total molar mass of g/mol for CaCl2.

21. Calculating Percent Composition
Finally, use the data to set up a calculation to determine the percent by mass of an element in the compound.
The percent by mass of chlorine in CaCl2 can be calculated as follows.

22. Calculating Percent Composition
CaCl2
% X =
(molar mass of X) (# X ions)
molar mass of compound
x 100
( g) ( Cl ions)
35.5 2
% Cl =
X 100
111.1 g
% Cl in CaCl2 = 63.9%

23. (molar mass of X) (# X ions)
Example
Determine the percent composition of carbon in sodium acetate (NaC2H3O2).
% X =
(molar mass of X) (# X ions)
molar mass of compound
x 100

24. Calculating Percent Composition
NaC2H3O2
( g) ( C’s)
12.0 2
% C =
X 100
82.0 g
% C = 29.3%

25. Problem
Calculate the percent composition aluminum of aluminum oxide (Al2O3).
52.9% Al

26. Problem
Determine the percent composition of oxygen in magnesium nitrate, which has the formula Mg(NO3)2.
64.7% O

27. Problem
Determine the percent composition of sulfur in aluminum sulfate, which has the formula Al2(SO4)3.
28.1% S

28. Problem
Determine the percent composition of oxygen in zinc nitrite, which has the formula Zn(NO2)2.
40.7% O

29. Percent Water in a Hydrate
Hydrates are compounds that incorporate water molecules into their fundamental solid structure. In a hydrate (which usually has a specific crystalline form), a defined number of water molecules are associated with each formula unit of the primary material.

30. Percent Water in a Hydrate
Gypsum is a hydrate with two water molecules present for every formula unit of CaSO4. The chemical formula for gypsum is CaSO4 • 2 H2O and the chemical name is calcium sulfate dihydrate. Note that the dot in the formula (or multiplication sign) indicates that the waters are there.

31. Percent Water in a Hydrate
Other examples of hydrates are:
lithium perchlorate trihydrate - LiClO4 • 3 H2O;
magnesium carbonate pentahydrate - MgCO3 • 5 H2O;
and copper (II) sulfate pentahydrate - CuSO4 • 5 H2O.

32. Percent Water in a Hydrate
The water in the hydrate (referred to as "water of hydration") can be removed by heating the hydrate. When all hydrating water is removed, the material is said to be anhydrous and is referred to as an anhydrate.

33. Percent Water in a Hydrate
Experimentally measuring the percent water in a hydrate involves first heating a known mass of the hydrate to remove the waters of hydration and then measuring the mass of the anhydrate remaining. The difference between the two masses is the mass of water lost.

34. Percent Water in a Hydrate
Dividing the mass of the water lost by the original mass of hydrate used is equal to the fraction of water in the compound. Multiplying this fraction by 100 gives the percent water.

35. Example
Determine the percent water in CuSO4 • 5 H2O (s).

36. Calculating Percent Composition
Mass of CuSO4 =
(4) =
Mass of 5 H2O =
1.0 (2) = x 5 = 90.0
Mass of CuSO4 . 5 H2O = =

37. Calculating Percent Composition
( g) ( H2O’s)
18.0 5
% H2O =
X 100
249.7 g
% H2O = 36.0%

38. Problem
Determine the percent water in MgCO3 • 5 H2O (s).
51.6% H2O

39. Problem
Determine the percent water in LiClO4 • 3 H2O (s).
33.7% H2O

40. Empirical Formula
You can use percent composition data to help identify an unknown compound by determining its empirical formula.
The empirical formula is the simplest whole-number ratio of atoms of elements in the compound. In many cases, the empirical formula is the actual formula for the compound.

41. Empirical Formula
For example, the simplest ratio of atoms of sodium to atoms of chlorine in sodium chloride is 1 atom Na : 1 atom Cl.
So, the empirical formula of sodium chloride is Na1Cl1, or NaCl, which is the true formula for the compound.

42. Empirical Formula
The formula for glucose is C6H12O6.
The coefficients in glucose are all divisible by 6.
The empirical formula of glucose is CH2O.

43. Problem
Determine the empirical formula for Tl2C4H4O6.
TlC2H2O3

44. Problem
Determine the empirical formula for N2O4.
NO2

45. Empirical Formula from Percent Composition
The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.
Because percent means “parts per hundred parts,” assume that you have 100 g of the compound.

46. Empirical Formula from Percent Composition
Then calculate the number of moles of each element in the 100 g of compound.

47. Empirical Formula from Percent Composition
The number of moles of manganese may be calculated as follows.
Given: 38.43% Mn
38.43 g Mn
1 mol Mn
= mol Mn
54.9 g Mn

48. Empirical Formula from Percent Composition
The number of moles of carbon may be calculated as follows.
Given: 16.80% C
16.80 g C
1 mol C
= mol C
12.0 g C

49. Empirical Formula from Percent Composition
The number of moles of oxygen may be calculated as follows.
Given: 44.77% O
44.77 g O
1 mol O
= mol O
16.0 g O

50. Empirical Formula from Percent Composition
The results show the following relationship.
mol Mn : mol C : mol O
: : 2.798

51. Empirical Formula from Percent Composition
To obtain the simplest whole-number ratio of moles, divide each number of moles by the smallest number of moles.
mol Mn
0.7000
= 1 mol Mn

52. Empirical Formula from Percent Composition
1.400 mol C
0.7000
= 2 mol C
2.798 mol O
0.7000
= 4 mol O

53. Empirical Formula from Percent Composition
The mole ratio for the compound is:
1 Mn : 2 C : 4 O
The empirical formula for the compound is MnC2O4.

54. Example
Determine the empirical formula of the following compound.
31.9 g Mg, 27.1 g P

55. Example
The number of moles of magnesium may be calculated as follows.
Given: 31.9 g Mg
31.9 g Mg
1 mol Mg
= 1.31 mol Mg
24.3 g Mg

56. Example
The number of moles phosphorous may be calculated as follows.
Given: 27.1 g P
27.1 g P
1 mol P
= mol P
31.0 g P

57. Empirical Formula from Percent Composition
The results show the following relationship.
mol Mg : mol P
1.31 : 0.874

58. Empirical Formula from Percent Composition
To obtain the simplest whole-number ratio of moles, divide each number of moles by the smallest number of moles.
0.874 mol P
0.874
= 1 mol P

59. Empirical Formula from Percent Composition
1.31 mol Mg
0.874
= 1.5 mol Mg

60. Empirical Formula from Percent Composition
The mole ratio for the compound is:
1.5 Mg : 1 P
You cannot have a fraction, so multiply both numbers by 2.
3 Mg : 2 P

61. Empirical Formula from Percent Composition
The empirical formula for the compound is Mg3P2.

62. Problem
The composition of an unknown acid is 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen. Calculate the empirical formula for the acid.
CH20

63. Problem
The composition of an unknown ionic compound is 60.7% nickel and 39.3% fluorine. Calculate the empirical formula for the ionic compound.
NiF2

64. Problem
The composition of a compound is 6.27 g calcium and 1.46 g nitrogen. Calculate the empirical formula for the compound.
Ca3N2

65. Problem
Find the empirical formula for a compound consisting of 63.0% Mn and 37.0% O.
MnO2

66. Molecular Formulas
For many compounds, the empirical formula is not the true formula.
A molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound.

67. Molecular Formulas
The molecular formula for a compound is either the same as the empirical formula or a whole-number multiple of the empirical formula.

68. Molecular Formulas
In order to determine the molecular formula for an unknown compound, you must know the molar mass of the compound in addition to its empirical formula.

69. Molecular Formulas
Then you can compare the molar mass of the compound with the molar mass represented by the empirical formula as shown in the following example problem.

70. Example
The molecular mass of benzene is g/mol and its empirical formula is CH. What is the molecular formula for benzene?

71. Determining a Molecular Formula
Calculate the molar mass represented by the formula CH.
Here, the molar mass is the sum of the masses of one mole of each element.
Molar mass CH = 12.0 g g
Molar mass CH = 13.0 g/mol

72. Example
As stated in the problem, the molecular mass of benzene is known to be g/mol.

73. Determining a Molecular Formula
To determine the molecular formula for benzene, calculate the whole number multiple, n, to apply to its empirical formula.
n =
78.0 g/mol
13.0 g/mol
= 6

74. Determining a Molecular Formula
This calculation shows that the molar mass of benzene is six times the molar mass of its empirical formula CH.

75. Determining a Molecular Formula
Therefore, the molecular formula must have six times as many atoms of each element as the empirical formula.
Thus, the molecular formula is (CH)6 = C6H6

76. Example
The simplest formula for butane is C2H5 and its molecular mass is about g/mol. What is the molecular formula of butane?

77. Determining a Molecular Formula
To determine the molecular formula for butane, calculate the whole number multiple, n, to apply to its empirical formula.
n =
60.0 g/mol
29.0 g/mol
= 2

78. Determining a Molecular Formula
Therefore, the molecular formula must have two times as many atoms of each element as the empirical formula.
Thus, the molecular formula is (C2H5)2 = C4H10

79. Problem
What is its molecular formula of cyanuric chloride, if the empirical formula is CClN and the molecular mass is g/mol?
C3Cl3N3

80. Problem
The simplest formula for vitamin C is C3H4O3. Experimental data indicates that the molecular mass of vitamin C is about 180. What is the molecular formula of vitamin C?
C6H8O6

81. Example
Maleic acid is a compound that is widely used in the plastics and textiles industries.
The composition of maleic acid is 41.39% carbon, 3.47% hydrogen, and 55.14% oxygen.
Its molar mass is g/mol. Calculate the molecular formula for maleic acid.

82. Determining a Molecular Formula
Start by determining the empirical formula for the compound.

83. Example
The number of moles of carbon may be calculated as follows.
Given: 41.39% C
41.39 g C
1 mol C
= mol C
12.0 g C

84. Example
The number of moles of hydrogen may be calculated as follows.
Given: 3.47% H
3.47 g H
1 mol H
= 3.47 mol H
1.0 g H

85. Example
The number of moles of oxygen may be calculated as follows.
Given: 55.14% O
55.14 g O
1 mol O
= mol O
16.0 g O

86. Determining a Molecular Formula
The numbers of moles of C, H, and O are nearly equal, so it is not necessary to divide through by the smallest value.
You can see by inspection that the smallest whole-number ratio is C : 1H : 1O, and the empirical formula is CHO.

87. Determining a Molecular Formula
Next, calculate the molar mass represented by the formula CHO.
Here, the molar mass is the sum of the masses of one mole of each element.
Molar mass CHO = 12.0 g g g
Molar mass CHO = 29.0 g/mol

88. Example
As stated in the problem, the molar mass of maleic acid is known to be g/mol.

89. Determining a Molecular Formula
To determine the molecular formula for maleic acid, calculate the whole number multiple, n, to apply to its empirical formula.
n =
116.1 g/mol
29.0 g/mol
= 4

90. Determining a Molecular Formula
This calculation shows that the molar mass of maleic acid is four times the molar mass of its empirical formula CHO.

91. Determining a Molecular Formula
Therefore, the molecular formula must have four times as many atoms of each element as the empirical formula.
Thus, the molecular formula is (CHO)4 = C4H4O4

92. Determining a Molecular Formula

93. Determining a Molecular Formula

94. Problem
The composition of silver oxalate is 71.02% silver, 7.91% carbon, and 21.07% oxygen. If the molar mass of silver oxalate is g/mol, what is its molecular formula?
Ag2C2O4

95. Problem
The composition of a compound is 85.6% carbon and 14.4% hydrogen. If the molar mass of the compound is 42.1 g/mol, what is its molecular formula?
C3H6