Empirical and Molecular Formulae

Empirical Formula

Empirical Formula – the simplest whole number ratio of atoms of each element in a compound.

For example, the empirical formula of ethane (C2H6) would be CH3 as this is the simplest whole number ratio of the atoms.

The empirical formula can be found for both ionic and covalent compounds.

Calculating Empirical Formulae

If we know the mass of each element in a compound, we can find the empirical formula by finding the ratio of each atom. This can be done using the steps below:

Step 1 – Write out the elements.

Step 2 – Write out the mass of each element.

Step 3 – Write out the relative atomic mass of each element.

Step 4 – Divide the mass of each element by its relative atomic mass. This step is essentially finding out the moles of each element in the sample.

Step 5 – Divide each value by the smallest number and you are left with the ratio of each element.

Step 6 – Write the empirical formula using the ratios.

Quite often in questions you will have a whole number ratio, which makes things simple. However, if the ratios aren’t whole numbers you should round if the number is very close to a whole number. But if one of the values is close to .5 double the values. For example, if you get 1:1.5 this would become 2:3.

The best way to do these calculations is to put each step under the previous step and write them out as a ratio from step 1. Let’s look at an example to show how this can be done. 

A compound consists of 20.00g of carbon, 3.35g of hydrogen and 26.65g of oxygen. We can calculate the empirical formula as follows:

Calculating the empirical formula of a compound

IMPORTANT NOTE – if you are given then percentage composition by mass of the atoms, rather than the mass, you can still follow the six steps above. The percentages should add up to 100%, therefore, you effectively have a total mass of 100g and can take the percentage composition of each atom as its mass. There is an example of this in the worked examples below.

Molecular Formula

Molecular Formula – the actual number of atoms of each element in a molecule.

The molecular formula only applies to molecules; small groups of atoms held together by covalent bonds. This means that for ionic compounds, such as sodium iodide (NaI) we cannot use the term molecular formula as no covalent bonds are present.

For example, the molecular formula of butene is C4H8 (whereas the empirical formula would be CH2).

Calculating Molecular Formulae

If we have the empirical formula and the relative molecular mass of a compound, we can calculate the molecular formula. These pieces of information are usually found experimentally.

To calculate the molecular formula from this information we only need 3 simple steps:

Step 1 – Find the relative mass of the empirical formula (by adding up the relative atomic masses of each of the atoms).

Step 2 – divide the relative molecular mass of the compound by the relative mass of the empirical formula.

Step 3 – multiply the number of each atom in the empirical formula by this number.

For example, if the relative molecular mass of the compound used in the example above is 180u we can calculate the molecular mass as follows:

Calculating the molecular formula of a compound

This is the molecular formula of glucose.

IMPORTANT NOTE – the molecular formula does not tell us the order in which the atoms are bonded together. It is simply the number of atoms in the molecule.

Worked Examples

Example 1

A compound is found to have the following percentage composition by mass: 70.9% Potassium (K) and 29.1% Sulphur (S) . Use this information to calculate the empirical formula of the compound. The relative atomic masses of both elements are given below:

K = 39.1u
S = 32.1u

Solution to Example 1

Example 2

The empirical formula of a compound with a relative molecular mass of 92.0u is NO2. Calculate the molecular formula. The relative atomic masses of the elements are given below:

N = 14.0u
O = 16.0u

Solution to Example 2

Example 3

A compound is formed when 171.2g of carbon (C) reacts with 28.8g of hydrogen (H). The relative molecular mass of the compound is 70.0u. Find the molecular formula of the compound. The relative atomic masses of the elements are given below:

C = 12.0u
H = 1.0u

Solution to Example 3

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