Empirical Formulas
Salt formulas show the simplest whole number ratio of ions in the formulas. For example, CaCl 2
means Ca2+ ions and Cl ions in the ratio of 1:2. The formulas of covalent compounds are different. A
covalent formula counts the number of each atom in the compound. For example, CH4 literally
means 4 hydrogen atoms joined to 1 carbon atom and moving around as a single molecule.
Let us consider ethane:
The formula of ethane is C2H6 – two carbon atoms joined to each other with 6 hydrogen atoms
attached. If had some ethane but didn’t know which compound it was and separated it into
individual atoms and determined the ratio of those atoms, we could say that the ratio of carbon to
hydrogen was 1:3. The empirical formula for a compound with a carbon to hydrogen ratio of 1:3
would be CH3.
Determining the ratio of carbon to hydrogen atoms doesn’t tell us how many of each atom are
joined together to make the compound. A compound with a ratio of carbon to hydrogen of 1:3 could
have a chemical formula of CH3, C2H6, C3H9, C4H12 and so on. However, the empirical formula does
give us information as to what it could be and could not be. Every hydrocarbon with a carbon to
hydrogen ratio of 1:3 is a possibility while every hydrocarbon with a carbon to hydrogen ratio other
than 1:3 is not a possibility.
Ethane is a possibility but benzene (C6H6) is not a possibility as the carbon to hydrogen ratio of
benzene is 1:1.
When determining the empirical formula of a substance, we are determining the simplest whole
number ratio of atoms or ions in the substance. For a salt, the chemical formula will be the empirical
formula but for covalent compounds it may be but generally will not be.
The method for determining the empirical formula does not generally involve separating the
substance into separate atoms. We generally carry out a chemical reaction to produce different
compounds that contain the atoms in the compound and separate and weigh these substances.
For example, we could combust ethane to produce carbon dioxide and water then separate the
carbon dioxide and water and weigh each substance. All of the carbon in the ethane went into the
carbon dioxide and all of the hydrogen in the ethane went into the water so by weighing the carbon
dioxide and water, we can calculate how much carbon and hydrogen were in the fuel that we
combusted and use this to find the empirical ratio of the fuel.
A simple empirical calculation question could give you the percentage mass of each element in the
compound. For example: a hydrocarbon contains 79.89% carbon and 20.11% hydrogen by mass, find
the empirical formula of the hydrocarbon. To answer this question, we need to convert this to a
mass ratio then a mole ratio.
𝑚𝐶 : 𝑚𝐻 = 79.89: 20.11
𝑛 𝐶 : 𝑛𝐻 =
𝑚𝐶 𝑚𝐻 79.89 20.11
:
=
:
= 6.652: 19.95
𝑀𝐶 𝑀𝐻 12.01 1.008
Next, we need to convert this decimal ratio to a whole number ratio. Dividing the smallest number in
the ratio by itself will turn it into 1. To keep the ratio valid, we must also divide the other number in
the ratio by the smallest number in the ratio.
𝑛𝐶 : 𝑛𝐻 = 6.652: 19.95 =
6.652 19.95
:
= 1: 2.999
6.652 6.652
The ratio 1:2.999 is approximately equal to 1:3. Therefore, the empirical formula is CH3.
There will always be measurement and rounding errors, so don’t expect the empirical formula
calculation to come out to an exact whole number ratio. Be prepared to round it up.
e.g. 2) A compound contains 40.04% calcium, 12.00% carbon and 47.96% oxygen by mass, find the
empirical formula of the compound.
𝑚𝐶𝑎 : 𝑚𝐶 : 𝑚𝑂 = 40.04: 12.00: 47.96
𝑛𝐶𝑎 : 𝑛𝐶 : 𝑛𝑂 =
𝑚𝐶𝑎 𝑚𝐶 𝑚𝑂 40.04 12.00 47.96
:
:
=
:
:
= 0.9990: 0.9992: 2.998
𝑀𝐶𝑎 𝑀𝐶 𝑀𝑂 40.08 12.01 16.00
Next, we need to convert this decimal ratio to a whole number ratio. Dividing the smallest number in
the ratio by itself will turn it into 1. To keep the ratio valid, we must also divide the other number in
the ratio by the smallest number in the ratio.
𝑛𝐶𝑎 : 𝑛𝐶 : 𝑛𝑂 = 0.9990: 0.9992: 2.998 =
0.9990 0.9992 2.998
:
:
= 1: 1.000: 3.001
0.9990 0.9990 0.9990
The ratio a:1.000:3.001 is approximately 1:1:3. Therefore, the empirical formula is CaCO3.