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How will you estimate phosphorous in an organic compound?

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Carius method:


A known mass of organic compound (wg) containing phosphorus is heated with fuming HNO3 in a sealed tube where C is converted into CO2 and H to H2O. Phosphorous present in the organic compound is oxidised to phosphoric acid which is precipitated as ammonium phospho molybdate by heating with conc. HNO3 and by adding ammonium molybdate.

H3PO4 + 12(NH4)2MoO4 + 21 HNO3 \(\overset{\Delta}{\rightarrow}\) (NH4)PO4.12 MoO3 + 21 NH4NO3 + 12 H2O

The precipitate of ammonium phospho molybdate is filtered, washed, dried and weighed.


Mass of organic compound = Wg

Mass of ammonium phospho molybdate = x g

Molar mass of ammonium phospho molybdate 1877 g

1877 g of ammonium phospho molybdate contains 31g of phosphorous

x g of ammonium phospho molybdate contain = \(\frac{31}{1877}\times\) x g of phosphorous

% of phosphorous = \(\big(\frac{31}{1877}\times\frac{x}{w}\times100\big)\%\) of phosphorous

In an alternate method, phosphoric acid is precipitated as magnesium-ammonium phosphate by adding magnesia mixture. The ppt. is washed dried and ignited to get magnesium pyrophosphate which is washed, dried and weighed.

Weight of magnesium pyrophosphate = y g

Molar mass of magnesium pyrophosphate = 222 g

222 g of magnesium pyrophosphate contains 62 g of P

y g of magnesium pyrophosphate contain = \(\frac{62}{222}\times\) y g of P

% of phosphorous = \(\big(\frac{31}{1877}\times\frac{y}{w}\times100\big)\%\)

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