A gaseous hydrocarbon yields 0.72 text g of water (H_ 2 O) and 3.08 text g of carbon dioxide (CO_ 2 ) after complete combustion. What is the empirical formula of this hydrocarbon?

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EEJ MAIN Chemistry QUESTION #7182

Question 1

A gaseous hydrocarbon yields $0.72 \text{ g}$ of water ($H_{2}O$) and $3.08 \text{ g}$ of carbon dioxide ($CO_{2}$) after complete combustion. What is the empirical formula of this hydrocarbon?

$C_{2}H_{4}$
$C_{3}H_{4}$
$C_{6}H_{5}$
$C_{7}H_{8}$✔️

Correct Answer Explanation

To find the empirical formula, we calculate the moles of $C$ and $H$:

Moles of $C$ from $CO_{2} = 3.08 / 44 = 0.07 \text{ mol}$.
Moles of $H$ from $H_{2}O = 2 \times (0.72 / 18) = 0.08 \text{ mol}$.
Molar ratio $C:H = 0.07 : 0.08 = 7 : 8$.
Empirical Formula = $C_{7}H_{8}$.

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