The $%$ of nitrogen is (as nearest integer) :
[Molar mass of ${(N H_{4})}_{2} C_{2} O_{4} = 124.10$ g/mol and atomic mass of $N = 14.0078$ ]

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Solution

Meq. of ${(N H_{4})}_{2} C_{2} O_{4}$ $=$ Meq. of $N H_{3}$ $=$ Meq. of $H_{2} S O_{4}$ used
Meq. of $N H_{3}$ $= \frac{w}{\frac{124.10}{2}} \times 1000$
$= (0.05305 \times 2 \times 50 - 11.3 \times 0.1214)$
$= 3.633 g$
$w_{{(N H_{4})}_{2} C_{2} O_{4}} = 0.2273 g$
Percentage of ${(N H_{4})}_{2} C_{2} O_{4} = \frac{0.2273}{0.4775} \times 100$ $= 47.6 %$
Also, meq. of $N =$ meq. of $N H_{3}$
$\frac{w}{14} \times 1000 = 3.663$ or $w_{N_{2}} = 0.0513 g$
Percentage of $N = \frac{0.0513}{4.775} \times 100 = 10.75 %$
So, the nearest integer is $11$ .

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