What is the percent by mass of nitrogen in $F e {({N O}_{3})}_{2}$ ?

12

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17

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16

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14

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Solution

The correct option is C $16$ %
The molecular mass of $F e (N O_{3})_{2} = 56 + 2 \times 14 + 2 \times 3 \times 16 = 56 + 28 + 96 = 180 g$
Mass of Nitrogen in $1 m o l e$ of $F e (N O_{3})_{2} = 2 \times 14 = 28 g$
$%$ of Nitrogen in $F e (N O_{3})_{2} = \frac{28}{180} \times 100 \approx 16 %$

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