Aniline $\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right)$ → it contains ($\mathrm{N}$) element

Benzene sulfonic acid $\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{SO}}_3\mathrm{H}\right)$ → it contains $\left(\mathrm{S}\right)$element

Thiourea $\left(\mathrm{SC}\right({\mathrm{NH}}_2\left)2\right)$ → it  contains $(\mathrm{S} \& \mathrm{N})$ element

Test for nitrogen - The carbon and nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide $\left(\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{N}\right)$ soluble in water. This is converted into sodium ferro cyanide by the addition of sufficient quantity of ferrous sulphate. Ferric ions generated during the process react with Ferro cyanide to form Prussian blue precipitate of ferric ferrocyanide.

$\mathrm{N}\mathrm{a}+\mathrm{C}+\mathrm{N}\to \mathrm{N}\mathrm{a}\mathrm{C}\mathrm{N}$

$\begin{array}{ccccccc}6NaCN& +& {\text{FeSO}}_4& \to & {\text{Na}}_4\left[\text{Fe}{\left(\text{CN}\right)}_6\right]& +& {\text{Na}}_2{\text{SO}}_4\\ & & & & \text{Sodium}& & \\ & & & & \text{ferrocyanide}& & \end{array}$

$\begin{array}{ccccccc}3{Na}_4\left[\text{Fe}{\left(\text{CN}\right)}_6\right]& +& 4{Fe}^{3+}& \to & {\text{Fe}}_4{\left[\text{Fe}{\left(\text{CN}\right)}_6\right]}_3& +& 12{Na}^{+}\\ & & & & \text{Ferric}& & \\ & & & & \text{ferrocyanide}& & \end{array}$

Test of sulphur - If sulphur is present in the organic compound, sodium fusion will convert it into sodium sulphide. Sulphide ions are readily identified using sodium nitroprusside.

$2 \mathrm{N}\mathrm{a}+\mathrm{S}\to \mathrm{N}{\mathrm{a}}_2\mathrm{S}$

$\begin{array}{ccccc}{\text{Na}}_2\text{S}& +& {\text{Na}}_2\left[\text{Fe}{\left(\text{CN}\right)}_5\text{NO}\right]& \to & {\text{Na}}_4\left[\text{Fe}{\left(\text{CN}\right)}_5\text{NOS}\right]\\ & & \text{Sodium}& & \text{Violet}\\ & & \text{nitroprusside}& & \text{colour}\end{array}$

Test for both nitrogen and sulphur - If both nitrogen and sulphur are present in an organic compound, sodium fusion will convert it into sodium thiocyanate which then react with $\mathrm{F}{\mathrm{e}}^{3+}$ to form blood colour complex ${\left[\mathrm{F}\mathrm{e}\left(\mathrm{S}\mathrm{C}\mathrm{N}\right)\right]}^{2+}$

$\mathrm{N}\mathrm{a}+\mathrm{C}+\mathrm{N}+\mathrm{S}\to \mathrm{N}\mathrm{a}\mathrm{S}\mathrm{C}\mathrm{N}$

$\mathrm{F}{\mathrm{e}}^{3+}+3\mathrm{S}\mathrm{C}{\mathrm{N}}^{-}\mathrm{ }\to \mathrm{F}\mathrm{e}{\left(\mathrm{S}\mathrm{C}\mathrm{N}\right)}_3\mathrm{ }\left(\mathrm{Blood}\mathrm{ }\mathrm{red}\right)$

Note: If the Lassaigne's extract containing excess of sodium metal, sodium cyanide and sulphides are formed instead of sodium thiocyanate.

$\mathrm{NaSCN}+2\mathrm{Na}\to \mathrm{NaCN}+{\mathrm{Na}}_2\mathrm{S}$

Here in this type of cases, both sulphur and nitrogen are to be identified in separate tests.