What is the percentage composition of the elements in ammonia?
$(N = 14, H = 1)$

N = 82.35 %; H = 17.65 %

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N = 83.35 %; H = 16.65 %

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N = 81.35 %; H = 18.65 %

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N = 80.35 %; H = 19.65 %

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Solution

The correct option is A $N = 82.35 %; H = 17.65 %$
The atomic masses of $N$ and $H$ are $14 g/mol$ and $1 g/mol$ respectively.

The molecular weight of $N H_{3}$ is $14 + 3 (1) = 17 g/mol$ .

Thus $17 g$ of $N H_{3}$ will contain $14 g$ of $N$ and $3 g$ of $H$

$Percentage of N in N H_{3} = \frac{14}{17} \times 100 = 82.35 %$

$Percentage of H in N H_{3} = 100 - 82.35 = 17.65 %$

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