EMPIRICAL AND MOLECULAR
FORMULAS
When chemists perform experiments in the laboratory they are able to
determine the relative masses of different elements in a compound. By using
the masses of the elements in the compound they are then able to determine the
percent composition of the compound and eventually find the empirical
formula (lowest whole-number ratio of elements in a compound) of the
compound. The empirical formula of a compound can be determined by using
the masses of the elements in the compound or the percentages of the elements
in the compound. In either process it is necessary to convert the grams of each
element into moles of each element. Elements combine in mole ratios in
compounds, so by converting the grams of each element into moles we will be
able to determine the ratio that the elements combine in. See the examples
below to see how an empirical formula is determined.
Example 1
A compound in the laboratory is found to contain 2.8 grams of nitrogen and 6.4
grams of oxygen. Determine the compounds empirical formula.
We know that the compound contains nitrogen and oxygen, but we do not
know the ratios that they combine in. The empirical formula will be NxOy but
we still have to determine what numbers go in place of the x and y. The first
step is to convert our grams of nitrogen and oxygen into moles.
1 mole
2.8 grams N x -------------- = .20 moles N
14.0 grams
1 mole
6.4 grmas O x -------------- = .40 moles O
Empirical Formula
16.0 grams
.20
----- = 1 N
.20
NO2
.40
----- = 2 O
.20
By looking at the number of moles of each element we can tell that there
should be twice as many moles (or atoms) of oxygen in the compound than
nitrogen. In order to get the numbers into whole numbers we need to divide
each of the number of moles by the smaller number of moles.
An empirical formula can also be calculated if you know the percent
composition of a compound. This is demonstrated in the following example.
Example 2
What is the empirical formula of a compound that is 80.0 % carbon and 20.0 %
hydrogen?
In this problem we are given percentages instead of grams, but the problem is
solved the same way. If we know the percentages of each element, all we have
to do is solve the problem making the assumption that we have a 100 gram
sample. If we have a 100 gram sample, 80.0 grams of the sample will be
carbon and 20.0 grams of the sample will be hydrogen. Then the problem is
solved using the same method as in Example 1.
1 mole
80.0 grams C x --------------- = 6.67 moles C
6.67
12.0 grams
------- = 1 C
6.67
1 mole
20.0 grams H x --------------- = 20.0 moles H
20.0
Empirical Formula
1.0 grams
------ = 3 H
6.67
CH3
Molecular Formulas
Once the empircal formula for a compound is determined the molecular
formula can also be calculated if the molar mass is known for the molecular
formula. The molecular formula is always some multiple of the empirical
formula. The only way to determine the multiple is to compare the molar mass
of the empirical formula to the molar mass of the molecular formula. The
following example will demonstrate how a molecular formula can be found if
the empirical formula is known.
Example 3
A compound containing hydrogen and oxygen has an empirical formula of
HO. The molar mass of the molecular formula for this compound is known to
be 34.0 grams/mole. What is the molecular formula for this compound?
The first step in solving this problem is to find the molar mass of the empirical
formula from the periodic table.
H = 1.0
O = 16.0
------17.0 grams/mole
Notice that the molar mass of the molecular formula is double the molar mass
of the empirical formula. This tells us that the molecular formula is two times
the empirical formula. The molecular formula is: H2O2