The nitrogen in 0.800 g of an organic compound is converted quantitatively to ammonia by the Kjeldahl method. The ammonia is absorbed in 50.0 mL of…

In the Kjeldahl method nitrogen is digested to $\mathrm{(NH_4)_2SO_4}$, then boiled with alkali to liberate $\mathrm{NH_3}$, which is trapped in a known excess of standard sulphuric acid; the leftover acid is back-titrated. Because $\mathrm{H_2SO_4}$ is dibasic, $50.0\ \mathrm{mL}\times0.500\ \mathrm{M}$ gives $25.0$ mmol acid, i.e. $50.0$ mmol of replaceable $\mathrm{H^+}$. The back titration used $28.0\times1.00=28.0$ mmol $\mathrm{NaOH}$, so excess $\mathrm{H^+}=28.0$ mmol. The protons consumed by ammonia are $50.0-28.0=22.0$ mmol, and since $\mathrm{NH_3+H^+\to NH_4^+}$ this equals the moles of $\mathrm{NH_3}$ and hence of N atoms. Mass of N $=22.0\times10^{-3}\times14=0.308$ g. Percentage $=0.308/0.800\times100=38.50\%$. Treating the acid as monobasic would give a wrong, much smaller value.