Notes On Organic Compounds : Quantitative Analysis (S,P,O) - CBSE Class 11 Chemistry
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 BaSO4 formed = m1 g      Molecular mass of BaSO4 = 32 g Sulphur m1 g BaSO4 contains (m1 x 32)/233 g Sulphur Percentage of Sulphur = (32 x m1 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 = m1 x 62/222 x 100/m The mass of the organic compound = m g                      The mass of Mg2 P2O7 = m1 g    Molecular mass of Mg2 P2O7 = 222 g Mass of the two phosphorous atoms present in the coumpound Mg2 P2O7 = 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 CO2 formed = m1 g m1 g of CO2 is obtained if (32 x m1)/88 g of O2 is liberated Therefore,            Amount of oxygen liberated from m grams of organic compound = (32 x m1 g)/(88 x m) 100 grams of the coumpound produces = (32 x m1 x 100g)/(88 x m)

#### 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 BaSO4 formed = m1 g      Molecular mass of BaSO4 = 32 g Sulphur m1 g BaSO4 contains (m1 x 32)/233 g Sulphur Percentage of Sulphur = (32 x m1 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 = m1 x 62/222 x 100/m The mass of the organic compound = m g                      The mass of Mg2 P2O7 = m1 g    Molecular mass of Mg2 P2O7 = 222 g Mass of the two phosphorous atoms present in the coumpound Mg2 P2O7 = 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 CO2 formed = m1 g m1 g of CO2 is obtained if (32 x m1)/88 g of O2 is liberated Therefore,            Amount of oxygen liberated from m grams of organic compound = (32 x m1 g)/(88 x m) 100 grams of the coumpound produces = (32 x m1 x 100g)/(88 x m)

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