Let the gaseous hydrocarbon be $\mathrm{C_xH_y}$.

On complete combustion, $\mathrm{C_xH_y + \left(x + \frac{y}{4}\right)O_2 \to xCO_2 + \frac{y}{2}H_2O}$

For gases at the same temperature and pressure, volume is proportional to number of moles.

So, from the data:

• $10\ \text{mL}$ hydrocarbon gives $40\ \text{mL}$ $\mathrm{CO_2}$
• Therefore, $x = \frac{40}{10} = 4$

Also, $10\ \text{mL}$ hydrocarbon gives $50\ \text{mL}$ water vapour. Since $\frac{y}{2}$ volumes of water vapour are formed per 1 volume of hydrocarbon, $\frac{y}{2} = \frac{50}{10} = 5$ Hence, $y = 10$

Therefore, the hydrocarbon is $\mathrm{C_4H_{10}}$

Total number of carbon and hydrogen atoms in one molecule: $4 + 10 = 14$

Comparison with stored answer:

• Derived answer = $14$
• Stored correct answer = $14$
• They agree.