Following Kjeldahl's method,1 g of organic c | vedclass

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(B) The reaction is: $H_2SO_4 + 2NH_3 ightarrow (NH_4)_2SO_4$. Millimoles of $H_2SO_4$ used = $10 \ mL 	imes 2 \ M = 20 \ mmol$. Since $1 \ mol$ of

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$56$

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(B) The reaction is: $H_2SO_4 + 2NH_3 ightarrow (NH_4)_2SO_4$. Millimoles of $H_2SO_4$ used = $10 \ mL 	imes 2 \ M = 20 \ mmol$. Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NH_3$,the millimoles of $NH_3$ produced = $20 	imes 2 = 40 \ mmol$. Since $1 \ mol$ of $NH_3$ contains $1 \ mol$ of $N$,the millimoles of $N$ = $40 \ mmol$. Mass of $N$ = $\frac{40}{1000} 	imes 14 \ g = 0.56 \ g$. Percentage of $N$ = $\frac{	ext{Mass of } N}{	ext{Mass of compound}} 	imes 100 = \frac{0.56}{1} 	imes 100 = 56 \%$.

Following Kjeldahl's method,$1 \ g$ of organic compound released ammonia,which neutralized $10 \ mL$ of $2 \ M \ H_2SO_4$. The percentage of nitrogen in the compound is . . . . . . $\%$.

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