Percent Composition
Empirical Formulas and Molecular
Formulas
Quantification in Chemistry
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Homework: Chapter 11
pg 333 problems 46-50
pg 335 problems 51-53
pg 337 problems 54-57
Warm Up-10/17/13
1. Determine the percent composition of
phosphoric acid H3PO4
2. A compound contains 36.84% nitrogen and
63.16% oxygen. What is the empirical formula
for this compound?
3. A liquid composed of 46.68% nitrogen and
53.32% oxygen has a molar mass of 60.01
g/mol. What is the molecular formula?
Percent Composition
Analytical chemist work to identify elements in
compounds and to determine their percent by
mass.
To determine the percentage of an element in a
compound you do the following:
mass of element
X 100
mass of compound
For Example:
A 100-gram sample of a compound contains 55
grams of element X and 45 grams of element
Y. What are the percentages of mass for each
element?
X:
55g element X X 100 = .55 X 100 = 55%
100 g of compound
and,
Y:
45g element Y
X 100 = .45 X 100 = 45%
100 g of compound
Percent Composition: Example 2
Determine the Percent Composition in the Compound
Sodium Hydrogen Carbonate (NaHCO3).
1 mol Na = 22.99 g Na
1 mol H = 1.008 g H
1 mol C = 12.01 g C
1 mol O = 15.99 x 3 g O (48g)
84.01 g/mol NaHCO3
Percent Composition: Example 2 (continued)
Percent Na = 22.99 g Na = .2737 X 100= 27.37% Na
84.01 g/mol NaHCO3
Percent H = 1.008 g H =
.012 X 100 = 1.2% H
84.01 g/mol NaHCO3
Percent C = 12.01 g C =
.143 X 100 = 14.3% C
84.01 g/mol NaHCO3
Percent O = 15.99 x 3 g O =
.5714 X 100 = 57.1% O
84.01 g/mol NaHCO3
Empirical Formulas
An empirical formula is the formula for a
compound with the smallest whole number
ratio of its elements.
• Example 1: The percent composition for sulfur
oxide is 40.05 % sulfur and 59.95 % oxygen.
What is the empirical formula for sulfur oxide?
40.05 g S X 1 mol =
1.29 mol S and
32.07g/mol S
59.95 g O X 1 mol =
3.747 mol O and
15.99g/mol O
Empirical Formulas
Example 1 (continued)
1.29 mol S / 1.29 = 1 mol S
3.747 mol O / 1.29 = 3 mol O
The empirical formula for sulfur oxide is:
SO3
Empirical Formulas
Example 2-Determine the empirical formula for
methyl acetate (CHO) if the compound has the
following percent composition: 48.64% C, 8.16%
H and 43.20% O.
48.64g C x 1 mol C =
4.050 mol C
12.01g/mol C
8.16g H x 1 mol H =
8.10 mol H
1.008g/mol H
43.20g O x 1 mol O =
2.70 mol O
15.99g/mol O
Empirical Formulas Example 2 (continued)
4.050 mol C / 2.70 = 1.5 mol C
8.10 mol H / 2.70 = 3 mol H and
2.70 mol O / 2.70 = 1 mol O
Multiply the numbers of moles to turn them into
whole number ratios
4.050 mol C / 2.70 = 1.5 mol C x 2 = 3 mol
8.10 mol H / 2.70 = 3 mol H x 2 = 6 mol
2.70 mol O / 2.70 = 1 mol O x 2 = 2 mol
• The formula for methyl acetate is: C3H6O2
Warm Up-10/15/13
1. Rubbing alcohol was found to contain 60.0 %
carbon, 13.4 % hydrogen, and the remaining
mass was due to oxygen. What is the
empirical formula of rubbing alcohol?
2. A sample of indium chloride weighing 0.5000
g is found to contain 0.2404 g of chlorine.
What is the empirical formula of the indium
compound?
Molecular Formulas
• Molecular Formulas specify the actual number of
atoms that are in a given compound.
• Example 1: What is the molecular formula for
methyl acetate given that its experimental mass
was determined to be 144 g/mol?
mass of empirical formula : C3H6O2
144 g/mol)/(72 g/mol) = 2 (n)
C = 12.01 x 3 = 36.03
2 x C3H6O2 = C6H12O4
H = 1.008 x 6 = 6.048
O = 15.99 x 2 = 31.98
72 g/mol