Calculate the percentage of nitrogen in $0.25 \ g$ of an organic compound if $38 \ cm^3$ of $N_2$ gas is evolved at $300 \ K$ and $96 \ kPa$ pressure. (Given: $1 \ atm = 101.3 \ kPa$)

(A) Step $1$: Convert pressure to $atm$: $P = \frac{96 \ kPa}{101.3 \ kPa/atm} \approx 0.9477 \ atm$.
Step $2$: Use the Ideal Gas Law $PV = nRT$ to find moles of $N_2$: $n = \frac{PV}{RT} = \frac{0.9477 \ atm \times 0.038 \ L}{0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 300 \ K} \approx 0.00146 \ mol$.
Step $3$: Calculate mass of $N_2$: $Mass = 0.00146 \ mol \times 28 \ g/mol \approx 0.04088 \ g$.
Step $4$: Calculate percentage of $N$: $\frac{0.04088 \ g}{0.25 \ g} \times 100 \approx 16.35 \%$.
The calculated value is approximately $16.38 \%$.