Empirical and molecular formulaeAQA A-Level Chemistry: Revision notes

The empirical formula is the simplest whole number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms of each element in one molecule.

The molecular formula is always a whole number multiple of the empirical formula. For glucose the molecular formula is C₆H₁₂O₆ and the empirical formula is CH₂O, so the multiple is 6. Ethene (C₂H₄) and propene (C₃H₆) share the empirical formula CH₂. For some compounds the two are the same, e.g. water, H₂O.

The method is the same whether you are given masses or percentages by mass (treat percentages as the mass in 100 g):

1. Write the mass of each element
2. Divide each mass by its Ar to get the amount in mol
3. Divide every amount by the smallest one
4. If the numbers are not whole, multiply all by the same small integer
5. Write the formula using the whole number ratio

Compound X is 85.7% C and 14.3% H by mass.

The empirical formula is CH₂.

When a ratio is not whole after dividing by the smallest, multiply to clear the fraction:

• 1 : 1.5 → multiply by 2 (e.g. Fe : O = 2 : 3, Fe₂O₃)
• 1 : 1.33 → multiply by 3
• 1 : 1.25 → multiply by 4
• 1 : 1.67 → multiply by 3

Worked example: 2.00 g of an iron oxide gives 1.40 g of iron, so 0.60 g is oxygen. n(Fe) = 1.40 / 55.8 = 0.0251 mol and n(O) = 0.60 / 16.0 = 0.0375 mol, a ratio of 1 : 1.5, so the empirical formula is Fe₂O₃.

Use the relative molecular mass, Mr:

1. Calculate the empirical formula mass
2. Divide Mr by the empirical formula mass to get a whole number, $n$
3. Multiply every subscript in the empirical formula by $n$

Worked example: compound X has the empirical formula CH₂ (mass 14.0) and Mr = 56.0. $n = \frac{56.0}{14.0} = 4$, so the molecular formula is C₄H₈.