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

Answers (1)
M manish

Estimation of halogen is done by Carius method. In this method, a known quantity of organic compound is heated with fuming HNO_{3} (nitric acid) with the presence of silver nitrate, contained in a hard glass tube, known as carius tube. C and H present in the compound are oxidised to carbon dioxide and water. And halogens are into to form AgX and then it is filtered, dried, and weighed.

Let the mass of the organic compound be m gram.
Mass of AgX formed = m_1 gram
1 mol of AgX contains 1 mol of X.
Therefore, Mass of halogen inm1 g of AgX = (Atomic mass of X*m_1) /(Molecular mass of AgX)

Thus, % of halogen will be  =   (Atomic mass of X(m_1\times 100) /mol. wt of AgX*(m)

Estimation of sulphur- In this method, Organic compound is heated either fuming nitric acid or sodium peroxide in a hard glass tube called carius tube. Sulphur present in the compound is oxidised to form sulphuric acid. It is precipitated by as barium sulphate by adding barium chloride solution in water. Then ppt is filtered, washed and weighed.

Let the mass of organic compound taken =  m g
and the mass of barium sulphate formed = m_1 g

1 mol of BaSO_{4} = 233 g BaSO_{4} = 32 g sulphur

 m_1 g  BaSO_{4} contains = \frac{32\times m_1}{233}g sulphur 
Percentage (%)of sulphur =  
                                              \frac{32\times m_1\times 100}{233\times m}

Estimation of phosphorus-  In this process, a known mass of an organic compound is heated with fuming nitric acid and the phosphorus gets oxidised to phosphoric acid. By adding ammonia and ammonium molybdate, phosphorus can be precipitated as ammonium phosphomolybdate, (NH_4)_3PO_4.12MoO_{3}. It can be also estimated as by precipitating it as MgNH_{4}PO_{4} by adding magnesia mixture which on ignition yields Mg_2P_2O_{7}.
 

Let the mass of organic compound taken = m g and mass of ammonium phosphomolybdate = m_1 g

Molar mass of (NH_4)_3PO_4.12MoO_{3}=1877 g 
Percentage(%) of phosphorus = \frac{31\times m_1\times 100}{1877\times m}

If phosphorus is estimated as Mg_2P_2O_{7},
Percentage(%) of phosphorus = \frac{62\times m_1\times 100}{222\times m}%

Preparation Products

Knockout KCET 2021

An exhaustive E-learning program for the complete preparation of KCET exam..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout KCET JEE Main 2021

It is an exhaustive preparation module made exclusively for cracking JEE & KCET.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Exams
Articles
Questions