Determine the molecular formula of a hydrocarbon given that its combustion produces 89.6 dm3 of CO2 and 54 g of H2O.

Moles of C atoms = Moles of $\text{CO}_2 = \frac{89.6 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 4 \text{ mol}$
Moles of H atoms = $2 \times \text{Moles of } \text{H}_2\text{O} = 2 \times \frac{54 \text{ g}}{18 \text{ g/mol}} = 2 \times 3 \text{ mol} = 6 \text{ mol}$

The ratio of C to H atoms is $4:6$, which simplifies to $2:3$.
Since these are the total moles, the molecular formula is $\text{C}_4\text{H}_6$.

The formula $\text{C}_4\text{H}_6$ fits the general formula $\text{C}_n\text{H}_{2n-2}$ (where $n=4$), which corresponds to an alkyne or a diene.