3.3 Empirical formulae

Empirical and molecular formulae

For example, ethane has the molecular formula C₂H₆ but the empirical formula CH₃. Glucose, C₆H₁₂O₆, has the empirical formula CH₂O.

Calculating an empirical formula

1. Write down the mass (or percentage) of each element.
2. Divide by the A_\mathrm{r} of each element to get the number of moles.
3. Divide by the smallest of these numbers.
4. If you don't get whole numbers, multiply to get them (for example, 1 : 1.5 becomes 2 : 3).

Example from masses: 5.6 g of iron reacts with 2.4 g of oxygen.

• Fe: 5.6 ÷ 56 = 0.10 and O: 2.4 ÷ 16 = 0.15
• divide by 0.10 → Fe 1, O 1.5 → multiply by 2 → Fe₂O₃

Example from percentages: a compound is 40.0% C, 6.7% H and 53.3% O.

• C: 40.0 ÷ 12 = 3.33, H: 6.7 ÷ 1 = 6.7, O: 53.3 ÷ 16 = 3.33
• divide by 3.33 → 1 : 2 : 1 → CH₂O

Molecular formula from the empirical formula

Divide the M_\mathrm{r} of the compound by the M_\mathrm{r} of the empirical formula, then multiply the formula by that number.

Example: empirical formula CH₂O (M_\mathrm{r} = 30) and M_\mathrm{r} of compound = 180. 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆.

The magnesium oxide experiment

1. Clean a piece of magnesium ribbon with sandpaper to remove the oxide layer.
2. Weigh a crucible and lid, then add the magnesium and weigh again.
3. Heat strongly, lifting the lid slightly from time to time to let oxygen in, without letting the white magnesium oxide smoke escape.
4. Heat, cool and weigh until the mass is constant, so all the magnesium has reacted.
5. Work out the masses of magnesium and oxygen, then the empirical formula.