Discuss the principle of estimation of halogens,
Question:

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

 

Solution:

Estimation of halogens

Halogens are estimated by the Carius method. In this method, a known quantity of organic compound is heated with fuming nitric acid in the presence of silver nitrate, contained in a hard glass tube called the Carius tube, taken in a furnace. Carbon and hydrogen that are present in the compound are oxidized to form COand H2O respectively and the halogen present in the compound is converted to the form of AgX. This AgX is then filtered, washed, dried, and weighed.

Let the mass of organic compound be m g.

Mass of AgX formed = m1 g

1 mol of Agx contains 1 mol of X.

 

Therefore,

Mass of halogen in $m_{1} g$ of $A g X=\frac{\text { Atomic mass of } X \times m_{1} g}{\text { Molecular mass of } A g X}$

Thus, $\%$ of halogen will be $=\frac{\text { Atomic mass of } \mathrm{X} \times m_{1} \times 100}{\text { Molecular mass of } \mathrm{AgX} \times m}$

Estimation of Sulphur

In this method, a known quantity of organic compound is heated with either fuming nitric acid or sodium peroxide in a hard glass tube called the Carius tube. Sulphur, present in the compound, is oxidized to form sulphuric acid. On addition of excess of barium chloride to it, the precipitation of barium sulphate takes place. This precipitate is then filtered, washed, dried, and weighed.

Let the mass of organic compound be m g.

Mass of BaSO4 formed = m1 g

$1 \mathrm{~mol}$ of $\mathrm{BaSO}_{4}=233 \mathrm{~g} \mathrm{BaSO}_{4}=32 \mathrm{~g}$ of Sulphur

Therefore, $m_{1} \mathrm{~g}$ of $\mathrm{BaSO}_{4}$ contains $\frac{32 \times m_{1}}{233} \mathrm{~g}$ of sulphur.

Thus, percentage of sulphur $=\frac{32 \times m_{1} \times 100}{233 \times m}$

Estimation of phosphorus

In this method, a known quantity of organic compound is heated with fuming nitric acid. Phosphorus, present in the compound, is oxidized to form phosphoric acid. By adding ammonia and ammonium molybdate to the solution, phosphorus can be precipitated as ammonium phosphomolybdate.

Phosphorus can also be estimated by precipitating it as $\mathrm{MgNH}_{4} \mathrm{PO}_{4}$ by adding magnesia mixture, which on ignition yields $\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}$.

Let the mass of organic compound be m g.

Mass of ammonium phosphomolybdate formed $=m_{1} \mathrm{~g}$

Molar mass of ammonium phosphomolybdate $=1877 \mathrm{~g}$

Thus, percentage of phosphorus $=\frac{31 \times m_{1} \times 100}{1877 \times m} \%$

If $\mathrm{P}$ is estimated as $\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}$,

Then, percentage of phosphorus $=\frac{62 \times m_{1} \times 100}{222 \times m} \%$

 

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