Chemical Quantities
Percent Composition, Empirical Formulas, Molecular Formulas
Percent Composition – the percentage by mass of each element in a compound
Percent composition =
mass of the element
Mass of the compound
X 100
Example: What is the percent composition of Potassium Permanganate (KMnO4)?
K = 1(39.10) = 39.1
Mn = 1(54.94) = 54.9
O = 4(16.00) = 64.0
MM = 158.0 g/mol
%K
%Mn
%O
(39.1/158.0)(100) = 24.7%
(54.9/158.0)(100) = 34.8%
(64.0/158.0)(100) = 40.5%
Empirical Formulas
An empirical formula is a “lowest common denominator” molecular formula for covalent molecules. It
represents the ratio in which atoms (or MOLES of atoms) combine to form compounds, but not the
actual numbers of atoms in the compound. Multiple compounds can have the same empirical formula.
Glucose, C6H12O6, contains carbon, hydrogen, and oxygen. The ratio of carbon to hydrogen to oxygen is
1:2:1. CH2O is the empirical formula for glucose.
Since the percent composition of each element in an empirical formula represents the ratio in which
they occur, the percent composition can be used, in conjunction with the molar masses of the elements,
to determine the empirical formula of a compound. It is generally prudent to assume a 100.0 gram
sample of the compound for the purpose of simplifying the calculations.
1. Caffeine has the following percent composition: carbon, 49.48%; hydrogen 5.19%; oxygen, 16.48%;
and nitrogen, 28.85%. What is its empirical formula?
Molecular Formulas
A molecular formula is a formula that represents the actual number of each atom in a compound.
Whereas CH2O is the empirical formula for glucose, C6H12O6 is the molecular formula. An actual
molecule of glucose contains six carbon atoms, twelve hydrogen atoms, and six oxygen atoms. (Or six
moles of carbon atoms, 12 moles of hydrogen atoms, and six moles of oxygen atoms).
Since an empirical formula indicates the ratios of the elements in the compound, it can be used, along
with the molar mass of the compound, to determine the molecular formula. As with empirical formulas,
assume a 100.0 gram sample to simplify the calculations.
Key Question
2. The molar mass of caffeine is 194.19 g/mol. What is its molecular formula?
Hint for determining EF (Empirical Formula):
% to mass, mass to mole, divide by small, multiply till whole
Hint for determining MF (Molecular Formula):
MF=(EF)n
n=MF molar mass (given in problem)
EF molar mass