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Organic Compounds : Quantitative Analysis (S,P,O)

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Organic Compounds : Quantitative Analysis (S,P,O) - Lesson Summary

To detect the presence of sulphur, a known mass of an organic compound is taken in a Carius tube and heated with sodium peroxide (or) fuming nitric acid. The sulphur present in the compound is oxidised to sulphuric acid.



The  acid  is cooled and treated with excess of aqueous  barium chloride solution.
It gives a precipitate of barium sulphate.

Let the mass of the organic compound = m g
      The mass of BaSO 4 formed = m 1 g
     Molecular mass of BaSO 4 = 32 g Sulphur
m 1 g BaSO 4 contains (m 1 x 32)/233 g Sulphur

Percentage of Sulphur = (32 x m 1 x 100)/(233 x m)

The quantitative analysis of phosphorus in the given organic compound is done by heating a known mass of organic compound with fuming nitric acid in a carius tube. Thus, phosphorus is oxidised to phosphoric acid.

When this phosphoric acid is treated with ammonia and ammonium molybdate, a precipitate of ammonium phosphomolybdate is formed. Sometimes phosphoric acid is precipitated by adding magnesia mixture.

On adding magnesia mixture to phosphoric acid, a precipitate of magnesium ammonium phosphate is formed.

This precipitate is filtered, washed, dried, and then ignited to produce magnesium pyrophosphate.
Phosphorous is estimated as magnesium pyrophosphate then the percentage of phosphorous

Percentage of phosphorous in the compound = m 1 x 62/222 x 100/m

The mass of the organic compound = m g
                     The mass of Mg 2 P 2O 7 = m 1 g
   Molecular mass of Mg 2 P 2O 7 = 222 g
Mass of the two phosphorous atoms present in the coumpound Mg 2 P 2O 7 = 31 x 2
                                                                                                                     = 62

For the quantitative analysis of oxygen in an organic compound, the compound is heated in a stream of nitrogen gas. Therefore, the organic compound gets decomposed.

The oxygen containing mixture is passed over red-hot coke where all the oxygen is converted to carbon monoxide.


This gaseous mixture is passed through warm iodine pentoxide when carbon monoxide is oxidised to carbon dioxide producing iodine.

The mass of the organic compound = m g
       The mass of CO 2 formed = m 1 g
m 1 g of CO 2 is obtained if (32 x m 1)/88 g of O 2 is liberated
Therefore,
           Amount of oxygen liberated from m grams of organic compound = (32 x m 1 g)/(88 x m)

100 grams of the coumpound produces = (32 x m 1 x 100g)/(88 x m)

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