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

Asked by Abhisek | 1 year ago |  61

1 Answer

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 CO2 and 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 = m1g

1 mol of Agx contains 1 mol of X.

Therefore,

Mass of halogen in m1g of AgX

\(\)\( \dfrac{Atomic\; mass\; of\; X \times m_1g}{Molecular\; mass\; of \;AgX}\)

Thus % of halogen will be

\( \dfrac{Atomic\; mass\; of\; X \times m_1\times 100}{Molecular\; mass\; of \;AgX\times m}\)

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 mg.

Mass of BaSO4 formed = m1g

1 mol of BaSO4 

= 233 g BaSO4 

= 32g of Sulphur

Therefore, m1g of BaSO4 contains

\( \dfrac{32\times m_1}{233}g\; of\; sulphur\)

Therefore, percentage of sulphur

\( \dfrac{32\times m_1\times 100}{233\times m}\)

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 phosphor molybdate.

Phosphorus can also be estimated by precipitating it as MgNH4PO4 by adding magnesia mixture, which on ignition yields Mg2P2O7.

Let the mass of organic compound be m g.

Mass of ammonium phosphor molybdate formed = m1g

Molar mass of ammonium phosphor molybdate = 1877g

Thus, percentage of phosphorus

\( \dfrac{31\times m_1\times 100}{1877\times m}\%\)

If P is estimated as Mg2P2O7,

Thus, percentage of phosphorus

\( \dfrac{62\times m_1\times 100}{222\times m}\%\)

Answered by Pragya Singh | 1 year ago

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