I've been stumped by a question in my textbook concerning combustion analysis.
A $\pu{0.5438g}$ sample of a liquid consisting of only $\ce{C}$, $\ce{H}$, and $\ce{O}$ was burned in pure oxygen, and $\pu{1.039g}$ of $\ce{CO2}$ and $\ce{0.6369g}$ of $\ce{H2O}$ were obtained. What is the empirical formula of the compound?
Firstly, I calculated the amount of substance of carbon and hydrogen by dividing the masses of each product by their respective molar masses. \begin{align} (1.0390/44.01) = n(\ce{C}) &= 0.02360\\ 2\cdot(0.6369/18.02) = n(\ce{H}) &= 0.0706\\ \end{align}
My textbook then states that I must find the sum of the masses of carbon and hydrogen, and then deduct this sum from the mass of the liquid compound to get the mass of oxygen: \begin{align} 0.02360 \cdot 12.01 &= 0.283535\\ 0.0706 \cdot 1.01 &= 0.28637\\ 0.5438 - (0.283535 + 0.28637) &= m(\ce{O}) \\ m(\ce{O}) &= \pu{0.1888g}\\ \end{align}
I have no qualms with this approach, but why can't I find the amount of oxygen directly from the products? \begin{align} \text{I.e.}&& 2\cdot(1.0390/44.01) &= n(\ce{O}) &\text{in }\ce{CO2}\\ \text{and} && (0.6369/18.02) &= n(\ce{O}) &\text{in }\ce{H_2O}.\\ \end{align}
$$\text{Amount of oxygen in liquid compound} = n(\ce{O})\text{ in }\ce{CO2} + n(\ce{O})\text{ in }\ce{H2O}.$$
I know this doesn't work, but my question is why? It seems perfectly intuitive to me, and yet it doesn't work.