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

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$C_{4} H_{9}$

$C H_{3}$

$C H$

$C_{3} H_{8}$

answer is D.

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Detailed Solution

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

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

$4 x / 12 x + y = 12 / 44$

or, $8 x = 3 y$

$x / y = 3 / 8$

Hence, the empirical formula is $C {}_{3}{H_{8}}$ .

Thus, the correct option is D.

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