A bag contains $0.32$ g of oxygen. The same volume of an unknown gas $A$ under similar conditions of temperature and pressure weighs $0.26$ g. The gas $A$ is known to contain only $C$ and $H$ in $1 : 1$ ratio. The molecular formula of the compound is :

C_{2} H_{2}

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C_{4} H_{4}

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C_{3} H_{4}

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C H

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Solution

The correct option is A $C_{2} H_{2}$
Moles of $O_{2}$ in bag $A = \frac{0.32}{32} = 0.01$ mol
Same volume of unknown gas $A$ will also have $0.01$ mol.
$0.01$ mol of known gas $A$ weighs $0.26$ g.
$∴ 1$ mol of unknown gas $A = \frac{0.26}{0.01} = 26$ g
So, the molecular weight of known gas $A = 26$ g.
Given, empirical formula $= C H$ .
Therefore, the empirical formula weight $= 12 + 1 = 13$ g.
$n = \frac{Molecular formula weight}{Empirical formula weight} = \frac{26}{13} = 2$
Molecular formula $= (C H)_{2} = C_{2} H_{2}$ .

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A bag contains 0.32 g of oxygen. The same volume of an unknown gas A under similar conditions of temperature and pressure weighs 0.26 g. The gas A is known to contain only C and H in 1:1 ratio. The molecular formula of the compound is :

A bag contains $0.32$ g of oxygen. The same volume of an unknown gas $A$ under similar conditions of temperature and pressure weighs $0.26$ g. The gas $A$ is known to contain only $C$ and $H$ in $1 : 1$ ratio. The molecular formula of the compound is :