Question

The action of concentrated sulphuric acid on urea $\left(N{H}_{2}CON{H}_2\right)$ results in the production of a white crystalline solid $X$ of formula $H_{3}N{O}_{3}S$. This is a monobasic acid. On treatment with sodium nitrate and dilute hydrochloride acid at $0^{o}C$, it liberates 1 mole of $N_2$ per mol of $X$ and on addition of aqueous barium chloride, the resulting solution yields 1 mole of $BaS{O}_4$ per mole of $X$ taken.
Deduce the structure of $X$.

Answer 1

$N{H}_{2}CON{H}_2$ (Urea) reacts with conc. $H_{2}S{O}_4$ forming $X\left(H_{3}N{O}_{3}S\right)$.

$X$ contains one ionisation H (since X is a monobasic acid).

Thus, $X$ contains free $-N{H}_2$ as well as free $-S{O}_{3}H$. ($NaN{O}_2+HCl$) converts free $-N{H}_2$ into free $-OH$ group liberating $N_2$.
$NaN{O}_2+HCl⟶HN{O}_2+NaCl$
$\underset{1molX}{N{H}_2-S{O}_{3}H}+HN{O}_2⟶\underset{1molX}{N_2\uparrow }+H_{2}O+\underset{1mol}{H_{2}S{O}_4}$
$\underset{1molX}{H_{2}S{O}_4}+BaC{l}_2⟶\underset{1mol}{BaS{O}_4}+2HCl$

Hence, the formula of $X$ is $H_{2}N-S{O}_{3}H$.