An acid found in sour milk contains only carbon, hydrogen and oxygen…

Step 1. Carbon. $\dfrac{0.792\ \text{g}}{44\ \text{g mol}^{-1}} = 0.0180\ \text{mol}$ of $\text{CO}_2$, so $0.0180\ \text{mol}$ of C. Mass: $0.0180\ \text{mol} \times 12\ \text{g mol}^{-1} = 0.216\ \text{g}$.

Step 2. Hydrogen. $\dfrac{0.324\ \text{g}}{18\ \text{g mol}^{-1}} = 0.0180\ \text{mol}$ of water. Each molecule has $2$ H atoms: $0.0360\ \text{mol}$ of H, that is $0.036\ \text{g}$.

Step 3. Oxygen by difference. $0.540\ \text{g} - 0.216\ \text{g} - 0.036\ \text{g} = 0.288\ \text{g}$, that is $53.3\%$ of the sample. In moles: $\dfrac{0.288\ \text{g}}{16\ \text{g mol}^{-1}} = 0.0180\ \text{mol}$.

Step 4. Empirical formula. C : H : O $= 0.0180 : 0.0360 : 0.0180 = 1 : 2 : 1$. The empirical formula is $\text{CH}_2\text{O}$, with mass $30\ \text{g mol}^{-1}$.

Step 5. Molar mass and $n$. Equal volumes hold equal numbers of molecules, so $\dfrac{M}{M_{\text{H}_2}} = 45$. We want $M$, so $M = 45 \times 2\ \text{g mol}^{-1} = 90\ \text{g mol}^{-1}$. Then $n = \dfrac{90}{30} = 3$, and the molecular formula is $\text{C}_3\text{H}_6\text{O}_3$.

Step 6. Oxygen for burning. $\text{C}_3\text{H}_6\text{O}_3 + 3\text{O}_2 \rightarrow 3\text{CO}_2 + 3\text{H}_2\text{O}$. The sample is $\dfrac{0.540\ \text{g}}{90\ \text{g mol}^{-1}} = 0.0060\ \text{mol}$, so it needs $0.0180\ \text{mol}$ of $\text{O}_2$. Volume: $0.0180\ \text{mol} \times 22.7\ \text{L mol}^{-1} = 409\ \text{mL}$.