Balanced chemical equation for the combustion of cyanogen (C_nN₂) in excess oxygen:

                    C_nN₂ + (n+1)O₂ → nCO₂ + $\frac{\mathrm{n}}{2}$N₂

given               500mL              1000mL   500mL

                          1 L                       2 L           1 L

The volume is directly proportional to the number of moles.

Thus, 1 mole of cyanogen on combustion will give 2 moles of CO₂ and 1 mole of N₂.

Thus, the molecular formula of the hydrocarbon is C₂N₂.