Question

the percent of N in 66% pure (NH4)2 SO4 sample is-
1)32
2)28
3)14
4)22

Answer 1

Molar mass of $(NH_{4})_{2} SO_{4}$ = 132 g

Mass % of N in 1 mol $(NH_{4})_{2}SO_{4}$ = $\frac{ 2 * 14g }{132g} *100$ = 21.2 %

Percentage of N in 66 % pure compound = $21.2 % * \frac{66}{100} = 14 %$