The percentage composition of a gas is as follow:
Nitrogen $82.35 %$ , Hydrogen $17.64 %$ .
Find the empirical formula for the gas.

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Solution

Total composition = 82.35+17.64=100
Therefore Nitrogen present in 100 g molecule is:
$82.35 g N \times \frac{1 m o l N}{14 g N} = 5.88 m o l N$
and amount of H is: $17.64 g H \times \frac{1 m o l H}{1 g H} = 17.64 m o l H$
Relative ratio of $N$ and $H$ is therefore: $N : H :: 5.88 : 17.64 = 1 : 3 N_{1} H_{3}$
Emperical formula is $N H_{3}$

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The percentage composition of a gas is as follow: Nitrogen 82.35%, Hydrogen 17.64%.Find the empirical formula for the gas.

Total composition = 82.35+17.64=100Therefore Nitrogen present in 100 g molecule is:82.35g N×1 mol N14g N=5.88 mol Nand amount of H is: 17.64 g H×1 mol H1 g H=17.64 mol HRelative ratio of N and H is therefore: N:H::5.88:17.64=1:3N1H3Emperical formula is NH3

The percentage composition of a gas is as follow:
Nitrogen $82.35 %$ , Hydrogen $17.64 %$ .
Find the empirical formula for the gas.