In the Kjeldahl method,$0.35 \ g$ of an organic compound produces ammonia,which is absorbed in $100 \ mL$ of $0.1 \ M \ H_2SO_4$. The excess $H_2SO_4$ requires $154 \ mL$ of $0.1 \ M \ NaOH$ for titration. Calculate the percentage of nitrogen in the compound. (Given atomic masses: $C=12, H=1, O=16, N=14, S=32$)

(A) $1$. Calculate the total millimoles of $H_2SO_4$ taken: $100 \ mL \times 0.1 \ M = 10 \ mmol$.
$2$. Since $H_2SO_4$ is a dibasic acid,the milliequivalents of $H_2SO_4$ are $10 \ mmol \times 2 = 20 \ meq$.
$3$. Calculate the milliequivalents of $NaOH$ used for titration: $154 \ mL \times 0.1 \ M = 15.4 \ meq$.
$4$. The milliequivalents of $H_2SO_4$ consumed by $NH_3$ are $20 \ meq - 15.4 \ meq = 4.6 \ meq$.
$5$. Since $1 \ meq$ of $NH_3$ contains $1 \ mmol$ of $N$,the mass of nitrogen is $4.6 \ mmol \times 14 \ g/mol = 64.4 \ mg = 0.0644 \ g$.
$6$. The percentage of nitrogen is $\frac{0.0644 \ g}{0.35 \ g} \times 100 = 18.4 \%$.