How to find the molecular formula for a hydrocarbon compound from known combustion products? What is the molecular formula of a compound containing only carbon and hydrogen if combustion of $1.05~\mathrm{g}$ of the compound produces $3.30~\mathrm{g}~\ce{CO2}$ and $1.35~\mathrm{g}~\ce{H2O}$ and its molar mass is about $70~\mathrm{g}$?


Here is my work:

$$
\begin{array}{cccccc}
                  &amp; \ce{C_{a}H_{b}} &amp; \ce{-&gt;} &amp; \ce{CO2} &amp; + &amp; \ce{H2O} \\
\text{masses (g)} &amp; 1.05            &amp;         &amp; 3.30     &amp;   &amp; 1.35
\end{array}
$$

\begin{align*}
\ce{CO2} &amp;\rightarrow \ce{C} \\
44~\mathrm{g} &amp;\rightarrow 12~\mathrm{g} \\
3.30~\mathrm{g} &amp;\rightarrow x
\end{align*}

$$
x = 0.9~\mathrm{g},~\text{moles of C} = \frac{0.9}{12} = 0.075
$$

\begin{align*}
\ce{H2O} &amp;\rightarrow \ce{2H} \\
18~\mathrm{g} &amp;\rightarrow 2~\mathrm{g} \\
1.35~\mathrm{g} &amp;\rightarrow y
\end{align*}

$$
y = 0.15~\mathrm{g},~\text{moles of H} = \frac{0.15}{1} = 0.15
$$

$$
\text{empirical formula}~\ce{C_{0.075/0.075}H_{0.15/0.075} -&gt; CH2}
$$

$$
\frac{70}{14} = 5
$$

$$
\text{molecular formula is}~\ce{C5H10}
$$
$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -&gt; $a$CO2 + $\frac{b}{2}$ H2O}$$

Suppose you had $n$ moles of hydrocarbon, then we have $a\cdot n$ moles of $\ce{CO2}$ and $\frac{b}{2}\cdot n$ moles of $\ce{H2O}$ dividing their moles we'll get $2\frac{a}{b}$:
$$2\frac ab=\frac{330/44}{135/18}=\frac{7.5}{7.5}=1\implies \frac ab=\frac12$$
So the empirical formula is $\ce{CH2}$
Now for actual formula $\frac{70}{14}=5$
Yes the formula is $\ce{C5H10}$.