In Kjeldahl's method, the ammonia from 0.600 g of an organic compound is absorbed in 40.0 mL of 0.100 M H2SO4. The excess acid needs 18.0 mL of 0.200 M NaOH. Taking N=14, the percentage of nitrogen in the compound is:

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In Kjeldahl's method, the ammonia from $0.600 g$ of an organic compound is absorbed in $40.0 mL$ of $0.100 M$ $H X_{2} SO X_{4}$ . The excess acid needs $18.0 mL$ of $0.200 M$ $NaOH$ . Taking $N = 14$ , the percentage of nitrogen in the compound is:

Acid taken

= 40.0 \times 0.100 \times 2 = 8.0

meq; excess

= 18.0 \times 0.200 = 3.6

meq; acid used by

NH X_{3}

= 4.4

meq

= 4.4

mmol of

NH X_{3}

(one N each). Mass of N

= 4.4 \times 14 = 61.6

mg. Percentage

= 61.6/600 \times 100 = 10.27%