6.7 THE EMPERICAL FORMULAS
If we go back to John Dalton’s Atomic Theory, we read that atoms combine with others in whole
number ratios when forming compounds.
His theory was correct and there are two ways in which we can write chemical formulas with
whole number ratios:
Table 1: Definitions of Empirical and Molecular Formulae
Empirical Formula
A formula that shows the simplest wholenumber rations of elements in a compound
Molecular Formula
A formula that shows the element symbols and
actual number of each type of atom in a
molecular compound
Table 2: Comparing Empirical, Molecular and Structural Formulae (Alternate Example on Pg. 290)
Compound
Nitrogen
dioxide
Empirical Formula
Molecular Formula
Structural Formula and
Structural Diagram
(Space-Filling Model)
N
NO2
NO2
O
O
O
Dinitrogen
tetroxide
O
NO2
N2O4
N
N
O
O
It is possible for two substances to have the same empirical formula, but different molecular
formulas. Dinitrogen tetroxide, N2O4 is a pungent brown gas with a molar mass of 92 g/mol and
a boiling point of 20.9 C and nitrogen dioxide, NO2 is a reddish-brown gas with a molar mass of
46 g/mol and it is used for the industrial synthesis of nitric acid. It is also one of the pollutants
that lead to acid rain. It is also possible for a substance to have the same empirical and
molecular formula (e.g. NH3)
Molecular Formula = n x Emperical Formula
Example 1: What is the empirical formula of a compound that is 39.10% carbon, 8.77% hydrogen
and 52.13% oxygen?
Step 1 – Understand what percentage means. When you are given percentage composition,
this is always a mass percent. It’s safe to assume that we can deal with a 100 g sample of
the compound.
Table 3 on Page 291
39.10 g C
8.77 g H
52.13 g O
Decimal or
Fraction
Whole
Number
Multiply all
Subscripts by
0.25 =
4(0.25) = 1
4
Moles of C in 100 g sample = 39.10 g
= 3.26 mol
12.01 g/mol
0.33 =
3(0.33) = 1
3
0.50 =
2(0.5) = 1
2
Moles of H in 100 g sample = 8.77 g
= 8.68 mol
1.01 g/mol
0.67 =
3(0.67) = 2
3
0.75 =
4(0.75) = 3
4
Step 2 – Convert grams into moles
Moles of O in 100g sample = 52.13 g = 3.26 mol
16.00 g/mol
Step 3 – Find the simplest mole ratio. Divide the number of moles of each element by the
smallest of the mole ratios calculated.
C: 3.26 mol/3.26 mol = 1
H: 8.68 mol/3.26 mol = 2.66
O: 3.26 mol/3.26 mol = 1
Step 4 – Convert the mole ration to the simplest whole number mole ratio. In this example,
we should multiply each ratio by 3.
C: 1 x 3 = 3
H: 2.66 x 3 = 8
O: 1 x 3 = 3
We can now conclude that the empirical formula is C3H8O3.
Determining the Empirical Formula by Experiments
If you know the mass of one of the reactants and you have measured the mass of the product in
an experiment, it is possible to get the percentage composition. This would have to be a
synthesis reaction. If I know that I have 5.0 g of Cu and by resulting compound has a mass of
6.3 g, how can I find the empirical formula?