At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% $O_{2}$ by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:

C_{3} H_{8}

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C_{4} H_{8}

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C_{4} H_{10}

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C_{3} H_{6}

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Solution

The correct option is A $C_{3} H_{8}$
Volume of $O_{2} = 375 \times \frac{20}{100} = 75 m L$

Volume of gaseous hydrocarbons = 15 mL

$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O (l)$

$15 (x + \frac{y}{4}) = 75 \dots \dots \dots (1)$

$x + \frac{y}{4} = 5$

Excess air $= 375 - 75 = 300$

300 + volume of $C O_{2} = 330$

Volume of $C O_{2} = 30$

$15 x = 30$

$x = 2$

$2 + \frac{y}{4} = 5$

$\frac{y}{4} = 3$

$y = 12$

$C_{2} H_{12}$ (not possible)

However, considering the equation no. 1 alone, if we put $x$ and $y$ values, 3 and 8 respectively, then equation no. 1 is satisfied and hence the answer will be $C_{3} H_{8}$ .

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