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

Answers (1)

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 = $\frac{62\times m_1\times 100}{222\times m}$%

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