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NCERT Class XI Chemistry Organic Chemistry Some basic principles and Techniques Solutions

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Question : 23 of 40
Marks: +1, -0
Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Solution:  
Principles of estimation
(1) Estimation of halogens by Carius method : Upon heating an organic compound with fuming HNO3\mathrm{HNO}_3 in the presence of AgNO3\mathrm{AgNO}_3, the halogen present in the compound forms the corresponding silver halide which is collected filtered; washed, dried and weighed. This is the amount of halogen present in the compound.
Let the mass of organic compound taken = m g
Mass of AgX formed = m1m_1 g
1 mol of AgX contains 1 mol of X
Mass of halogen in m1m_1 g of AgX = atomic mass of X×m1 gmolecular mass of AgX\frac{\text{atomic mass of X} \times m_1\,\text{g}}{\text{molecular mass of AgX}}
Percentage of halogen = atomic mass of X×m1×100molecular mass of AgX×m\frac{\text{atomic mass of X} \times m_1 \times 100}{\text{molecular mass of AgX} \times m}
(2) Estimation of sulphur : An organic compound containing sulphur is heated with fuming nitric acid. This oxidises the sulphur to sulphuric acid which is precipitated as BaSO4 upon reaction with Ba(OH)2. The mass of BaSO4 tells the % of sulphur present in the compound.
Let the mass of organic compound taken = m g
and the mass of barium sulphate formed = m1m_1 g
1 mol of BaSO4\mathrm{BaSO}_4 = 233 g BaSO4\mathrm{BaSO}_4 = 32 g sulphur
m1m_1 g BaSO4\mathrm{BaSO}_4 contains 32×m1233\frac{32 \times m_1}{233} g sulphur , % of sulphur = 32×m1×100233×m\frac{32 \times m_1 \times 100}{233 \times m}
(3) Estimation of phosphorus : A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH4)3PO4â‹…12MoO3\mathrm{(NH_4)_3PO_4 \cdot 12MoO_3}, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as MgNH4PO4\mathrm{MgNH_4PO_4} by adding magnesia mixture which on ignition yields Mg2P2O7\mathrm{Mg_2P_2O_7}.
Let the mass of organic compound taken = m g
and mass of ammonium phosphomolybdate = m1m_1 g
Mg2P2O7222 g\underset{222\,\text{g}}{\mathrm{Mg_2P_2O_7}} = 2P62 g\underset{62\,\text{g}}{2\mathrm{P}}
Percentage of phosphorus = 62×m1×100222×m\frac{62 \times m_1 \times 100}{222 \times m}
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