A gaseous hydrocarbon on combustion gives 0.72g of water and 3.08 g of \[C{O_2}\]. The empirical formula of the hydrocarbon is:
\[\begin{array}{*{20}{l}}
  {a.{\text{ }}-{C_3}{H_4}} \\
  {b.- {\text{ }}{C_6}{H_5}} \\
  {c.{\text{ }}-{C_7}{H_8}} \\
  {d.- {\text{ }}{C_2}{H_4}}
\end{array}\]

Hint: In combustion, hydrocarbon reacts with oxygen to give energy. Other products of this reaction are carbon dioxide and water. General representation for such reaction is:
\[{C_x}{H_Y}{\text{ }} + {\text{ }}(x + \dfrac{y}{4}){O_2}\; \to {\text{ }}\;xC{O_2}\; + {\text{ }}2y{H_2}O\]

Complete answer:
Step-1:
First of all find the given identities and what to find out.
Given: Amount of water obtained: 0.72 g
Amount of carbon dioxide obtained: 3.08 g
Empirical formula of hydrocarbon: ?
Step-2
Find the number of moles of water and carbon dioxide obtained:
Moles of \[{H_2}O\] obtained=$\dfrac{W}{{MW}}$
= $\dfrac{{0.72}}{{18}}$
= 0.04 mole
Moles of\[\;C{O_2}\]obtained​=$\dfrac{W}{{MW}}$
= $\dfrac{{3.08}}{{44}}$
= 0.07 mole
Step-3
Now apply general representation equation and find x, y
\[{C_x}{H_Y}{\text{ }} + {\text{ }}(x + \dfrac{y}{4}){O_2}\; \to {\text{ }}\;xC{O_2}\; + {\text{ }}2y{H_2}O\]
2y= moles of water obtained
2y=0.04
$\dfrac{y}{2}$ =0.04
 y= 0.08 …..(i)
x= moles of carbon dioxide obtained
x=0.07 …. (ii)
Use the value of x and y from equation (i) and (ii) and try to make formula of hydrocarbon
Here y= 0.08 and x=0.07
\[{C_x}{H_Y} = {C_{0.07}}{H_{0.08}}\]
Or we can write ${C_7}{H_8}$.

Hence, the empirical formula of the hydrocarbon is ${C_7}{H_8}$i.e. option (C).

Note:
Direct prediction of molecular formula is not possible. We have to convert the grams of product into mole first then frame the required hydrocarbon.