A chemist burns 0.230 g of a liquid that holds only carbon, hydrogen and oxygen.

Mass percent of an element $= \dfrac{\text{its mass in 1 mol}}{M} \times 100$. Empirical formula: take $100\ \text{g}$, change each mass to moles, divide by the smallest, and multiply to whole numbers. Molecular formula: $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$. In a combustion analysis, all the carbon ends up in $\text{CO}_2$ ($12\ \text{g}$ in every $44\ \text{g}$), and all the hydrogen in $\text{H}_2\text{O}$ ($2\ \text{g}$ in every $18\ \text{g}$). Oxygen is found by difference.