Topic 9 – Percent Composition, Empirical and Molecular Formulas
I. Percent composition.
A. Many of the metals we use cannot be mined directly, but must be extracted from minerals
called ores.
B. Ores containing a higher percent by mass of the metal are typically more profitable to process.
C. Percent by mass can also be used to identify compounds with the help of a mass spectrometer.
The mass spectrometer can identify the percent of each element present in a compound.
D. Calculating percent by mass.
1. Determine the molar mass of the compound.
2. Divide the mass of the individual element in the compound by the molar mass.
% by mass = (mass of element / molar mass) * 100
3. Repeat for each element in the compound.
4. Check to see that the percentages add up to 100%.
II. Empirical formulas.
A. An empirical formula is the chemical formula with the subscripts simplified.
Example: CH2 is the empirical formula for C3H6.
B. To determine an empirical formula, either the gram amounts or percent by mass of each
element must be known. Example: 48.6% C, 8.2% H, 43.2% O by mass.
C. If the starting amounts are percents, simply treat the numbers as gram amounts for a 100 gram
sample. Example: 48.6 g C, 8.2 g H, 43.2 g O
D. Convert all of the grams to moles.
Example: 48.6 g C = 4.047 mol C
8.2 g H = 8.119 mol H
43.2 g O = 2.700 mol O
E. Divide each mole amount by the smallest mole value.
Example: 4.047 mol C/2.700 = 1.50
8.119 mol H/2.700 = 3.01
2.700 mol O/2.700 = 1
F. These are your subscripts in the empirical formula. Example: C1.5H3O
G. The subscripts must be whole numbers. If necessary, multiply each number by a common multiple
to make all the subscripts whole numbers. Each subscript gets multiplied by the same number
to keep the overall ratio intact. Example: C3H6O2
1. If all the subscripts are whole numbers, they don’t need to be multiplied by anything.
2. If a subscript has a decimal value of .50, multiply all the subscripts by 2.
3. If a subscript has a decimal value of .33 or .67, multiply all the subscripts by 3.
4. If a subscript has a decimal value of .25, multiply all the subscripts by 4.
III. Molecular formulas from empirical formulas.
A. If both the empirical formula and molar mass are known, the molecular formula can easily be
determined.
B. Divide the molar mass of the compound by the empirical mass (molar mass of the empirical
formula). Example: Suppose a compound has a molar mass of 148 grams and an empirical
formula of C3H6O2.
148 / (3*12 + 6*1 + 2*16) = 2
C. The result of the division from the previous step should be a whole number or something very close
to it. Multiply all subscripts by this number to get the molecular formula.
Example: C3H6O2 x 2 = C6H12O4 (this is the molecular formula).
Assignments:
o read p.341-350
o p.344 54-61
o p.350 62-69,73
o Lab: Determining an empirical formula
Grades:
Lab Report – Determining an Empirical Formula
Lab Quiz – Determining an Empirical Formula
Topic 9 Quiz - % by mass, empirical formulas
p.344 54-61
54. 3.08% H, 31.61% P, 65.31% O
55. H2SO3
56. 36.11% Ca, 63.89% Cl
57. a. Na2SO4
b. ionic
c. 32.37% Na, 22.58% S, 45.05% O
58. N2O3
59. Al2S3
60. C3H8
61. C9H8O4
p.350 62-69, 73
62. C4H10
63. N2O2
64. K2O
65. C6H6O2
66. C17H19O3N
67. No. Molecular mass will always be a whole number multiple of empirical mass.
68. Fe2O3
69. Al2O3
73. Hematite is 69.94% Fe, while magnetite is 72.36% Fe. Magnetite provides more iron per kg.