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 = 80.35

H = 19.65

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N = 81.35

H = 18.65

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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.
The percentage of N in $N H_{3}$ $= \frac{14}{17} \times 100 = 82.35 %$
The percentage of H in $N H_{3}$ $= 100 - 82.35 = 17.65 %$

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