Question

How do you determine sulphur, Phosphorous are determined quantitiatively in an organic compound ?​

Answer 1

■ Estimation of sulphur :-

A known mass of organic compound is heated with sodium peroxide or fuming nitric acid in a Carius tube. If sulphur is present it is oxidised to sulphuric acid. The acid is precipitated as barium sulphate by adding excess of barium chloride aqueous solution. The precipitate is filtered, washed, dried and weighed.

$\boxed{ \boxed{ \sf{\% \ S = \dfrac{32}{233} \times \dfrac{wt. \: of \: BaSO_4}{wt. \: of \: organic \: compound} \times 100}}}$

■ Estimation of Phosphorus :-

A known mass of organic compound is heated with fumic nitric acid in Carius tube. Phosphorus is oxidised to phosphoric acid. The acid is precipitated as ammonium phosphomolybdate $\sf{(NH_4)_3 PO_4 . 12 MoO_3}$ by adding ammonia and ammonium molybdate solutions.

$\boxed{ \boxed{ \sf{\% P = \dfrac{31 \times wt. \: of \: (NH_4 )_3 PO_4 . 12 MoO_3 \: ppt \times 100}{1887 \times wt \: of organic \: compound}}}}$

Sometimes, the Phosphoric acid is precipitated by adding magnesia mixture. A precipitate of magnesium ammonium phosphate $\sf{(Mg NH_4 PO_4)}$ is formed which on ignition gives magnesium pyrophosphate $\sf{(Mg_2 P_2 O_7)}$.

$\boxed{ \boxed{ \sf{\% P = \dfrac{2 \times 31 \times wt. of \: Mg_2 P_2 O_7}{222 \times wt. of \: organic \: compound} \times 100}}}$

Answer 2

■ Estimation of sulphur :-

A known mass of organic compound is heated with sodium peroxide or fuming nitric acid in a Carius tube. If sulphur is present it is oxidised to sulphuric acid. The acid is precipitated as barium sulphate by adding excess of barium chloride aqueous solution. The precipitate is filtered, washed, dried and weighed.

\begin{gathered} \boxed{ \boxed{ \sf{\% \ S = \dfrac{32}{233} \times \dfrac{wt. \: of \: BaSO_4}{wt. \: of \: organic \: compound} \times 100}}} \\ \end{gathered}

% S=

233

32

×

wt.oforganiccompound

wt.ofBaSO

4

×100

■ Estimation of Phosphorus :-

A known mass of organic compound is heated with fumic nitric acid in Carius tube. Phosphorus is oxidised to phosphoric acid. The acid is precipitated as ammonium phosphomolybdate \sf{(NH_4)_3 PO_4 . 12 MoO_3}[/tex]

by adding ammonia and ammonium molybdate solutions.

\begin{gathered} \boxed{ \boxed{ \sf{\% P = \dfrac{31 \times wt. \: of \: (NH_4 )_3 PO_4 . 12 MoO_3 \: ppt \times 100}{1887 \times wt \: of organic \: compound}}}} \\ \end{gathered} [/tex]

Sometimes, the Phosphoric acid is precipitated by adding magnesia mixture. A precipitate of magnesium ammonium phosphate $\sf{(Mg NH_4 PO_4)}$ is formed which on ignition gives magnesium pyrophosphate \sf{(Mg_2 P_2 O_7)}[/tex]

\begin{gathered}$\boxed{ \boxed{ \sf{\% P = \dfrac{2 \times 31 \times wt. of \: Mg_2 P_2 O_7}{222 \times wt. of \: organic \: compound} \times 100}}} \\ \end{gathered}$