Quantitative Analysis - Phosphorus

Description:

A known mass of organic compound is heated with fuming nitric acid such that phosphorus in the compound is oxidized to phosphoric acid.

By adding ammonia an dammonium molybdate, it is precipitated as ammonium phosphomolybdate (NH₄)₃PO₄.12MoO₃

Alternatively, phosphoric acid can also be precipitated as MgNH₄PO₄ by adding magnesia mixture which on igniting gives Mg₂P₂O₇.

• Let the mass of the organic compound = m g

• Mass of ammonium phosphomolybdate formed = m₁ g

• Molar mass of (NH₄)₃PO₄.12MoO₃ = 1877 g

Percentage of phosphorus =

31 × m₁ × 100

/

1877 × m

%

• If phosphorus is estimated as Mg₂P₂O₇,

• Where molar mass of Mg₂P₂O₇ is 222 𝑢,

• m is the mass of the organic compound,

• m₁ is the mass of Mg₂P₂O₇

• Mass of two phosphorus atoms present in the compound Mg₂P₂O₇.

Percentage of phosphorous =

62 × m₁ × 100

/

222 × m

%