Question

Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Answer 1

Estimation of halogens by carius method:
A known quantity of organic compound is heated with fuming nitric acid in presence of silver nitrate in a hard glass tube in a furnance. Carbon and hydrogen present in the compound are oxidized to carbon dioxide and water respectively. Halogen present in the compound is converted to silver halide which is filtered, washed, dried and weighed.
Let m g be the mass of organic compound and m1 be the mass of AgX formed. 1 mol of AgX contains 1 mol of X.
Hence, mass of halogen in m1 g of AgX is $\frac{\text{Atomic mass of X}\times m1}{\text{Molecular mass of AgX}}$.
The percentage of halogen is $\frac{\text{Atomic mass of X}\times m1}{\text{Molecular mass of AgX}\times m}\times 100$.
Estimation of sulphur:
A known quantity of organic compound is heated with fuming nitric acid or sodium peroxide in hard glass tube (Carius tube). Sulphur is oxidized to sulphuric acid. Excess barium chloride is added which precipitates barium sulphate. The precipitate is filtered, washed, dried and weighed.
Let m g be the mass of organic compound and m1 g be the mass of barium sulphate formed. The molar mass of barium sulphate is 233 g/mol.
Thus, 233 g of barium sulphate contains 32 g of sulphur. Hence, m1 g of barium sulphate contains $\frac{32\times m1}{233}g$ of sulphur.
The percentage of sulphur is $\frac{32\times m1\times 100}{233\times m}$.
Estimation of phosphorus:
In this method, a known quantity of organic compound is heated with fuming nitric acid. Phosphorous is oxidized to phosphoric acid.
Ammonia and ammonium molybdate is added to the solution, which precipitates phosphorous as ammonium phosphomolybdate. Phosphorous can also be precipitated as $MgN{H}_{4}P{O}_4$ by adding magnesia mixture, which on ignition yields $M{g}_2{P}_2{O}_7$. Let m g be the mass of organic compound and m1 be the mass of ammonium phodphomolybdate (molar mass 1877 g/mol).
The percentage of phosphorous is $\frac{31\times m1\times 100}{1877\times m}$ %.
If P is estimated as $M{g}_2{P}_2{O}_7$, then the percentage of phosphorous is $\frac{62\times m1\times 100}{222\times m}$%.