Four grams of hydrocarbon $(C_{x} H_{y})$ on complete combustion gave $12 g$ of $C O_{2}$ . What is the empirical formula of the hydrocarbon?

C H_{3}

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C_{4} H_{9}

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C H

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C_{3} H_{8}

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Solution

The correct option is D $C_{3} H_{8}$
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} ⟶ x C O_{2} + \frac{y}{2} H_{2} O$
$1$ mole of $C_{x} H_{y}$ gives $x$ moles of $C O_{2}$
Given $\frac{4}{12 x + y}$ moles of $C_{x} H_{y}$ gives $\frac{12}{44}$ moles of $C O_{2}$
$∴ \frac{4 x}{12 x + y} = \frac{12}{44}$
or, $8 x = 3 y$
$\Rightarrow \frac{x}{y} = \frac{3}{8}$
Therefore, Empirical formula of the hydrocarbon is $C_{3} H_{8}$ .

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