1. Let the hydrocarbon be $C_xH_y$.

2. Use the combustion products to find moles of C and H.

   On complete combustion:

   • All carbon in the hydrocarbon appears in $CO_2$.
   • All hydrogen in the hydrocarbon appears in $H_2O$.

   From $CO_2$

   Given mass of $CO_2 = 3.08\,\text{g}$ $n(CO_2)=\frac{3.08}{44}=0.07\,\text{mol}$ Hence moles of carbon: $n(C)=0.07\,\text{mol}$

   From $H_2O$

   Given mass of $H_2O = 0.72\,\text{g}$ $n(H_2O)=\frac{0.72}{18}=0.04\,\text{mol}$ Each mole of water contains $2$ moles of H atoms, so $n(H)=2\times 0.04=0.08\,\text{mol}$

3. Find the simplest mole ratio of C : H.

   $C:H = 0.07:0.08$

   Multiply by $100$ to simplify: $7:8$

   Therefore, the empirical formula is $C_7H_8$

4. Check options.

   • A: $C_2H_4$ gives ratio $1:2$ ❌
   • B: $C_3H_4$ gives ratio $3:4$ ❌
   • C: $C_6H_5$ gives ratio $6:5$ ❌
   • D: $C_7H_8$ gives ratio $7:8$ ✅

Hence, the correct answer is Option D: $C_7H_8$.