The combustion reaction are as follows.
[tex]C_{x}H_{y} + O_{2} \rightarrow CO_{2} +H_{2}O[/tex]
From the given:
Amount of carbon dioxide = 33.01 g
Amount of water = 9.02 g
Let's calculate the number of moles of each compound in the reaction
[tex]Moles of CO_{2} = \frac{Given mass of CO_{2}}{Molar mass of CO_{2}}=\frac{33.01 g}{44.01 g/mol}= 0.75 mole[/tex]
[tex]Molesof H_{2}O = \frac{Given massof H_{2}O}{ Molar mass of H_{2}O} = 2 \times \frac{9.02 g}{18 g/ mol} = 1 mole[/tex]
Each value is divided by small value
[tex]Carbon= \frac{0.75}{0.75} = 1[/tex]
[tex]Hydrogen= \frac{1.0}{0.75} = 1.3 \approx 2[/tex]
Therefore, Empirical formula of given hydrocarbon is [tex]CH_{2}[/tex]
