How to find the molecular formula for a hydrocarbon compound from known combustion products?
$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -&gt; $a$CO2 + $\frac{b}{2}$ H2O}$$

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 $\ce{C5H10}$.