(N/A) Procedure: $A$ known mass of an organic compound is heated in a Carius tube with fuming nitric acid.
Principle: Sulphur is oxidized to sulphuric acid $(H_{2}SO_{4})$. It is then precipitated as barium sulphate $(BaSO_{4})$ by adding an excess of barium chloride $(BaCl_{2})$ solution. The precipitate is filtered,washed,dried,and weighed.
Chemical reaction: $S$ $\xrightarrow[\Delta]{HNO_{3}} H_{2}SO_{4}$ $\xrightarrow{BaCl_{2}} BaSO_{4(s)}$
Calculation:
Let mass of organic compound = $m \ g$
Let mass of $BaSO_{4}$ formed = $m_{1} \ g$
Since $1 \ mol$ of $BaSO_{4}$ $(233 \ g)$ contains $32 \ g$ of sulphur,
Mass of sulphur in $m_{1} \ g$ of $BaSO_{4} = \frac{32 \times m_{1}}{233} \ g$
Percentage of sulphur = $\frac{32}{233} \times \frac{m_{1}}{m} \times 100$