Phosphorus

The Principle: Oxidising Phosphorus to a Precipitate

Phosphorus in an organic compound is not directly measurable by weighing. The method converts it into a solid, weighable compound through a two-step oxidation and precipitation process.

The organic compound is first heated strongly with fuming nitric acid. This harsh oxidising agent breaks down the organic molecule and oxidises any phosphorus present into phosphoric acid ($H_3PO_4$). The phosphoric acid remains dissolved in the reaction mixture.

From this solution, the phosphate ion is then precipitated as a specific, insoluble salt. Two common precipitation routes exist, each leading to a different final weighable form.

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Method 1: Precipitation as Ammonium Phosphomolybdate

Ammonia and ammonium molybdate solution are added to the solution containing phosphoric acid. This produces a bright yellow, crystalline precipitate of ammonium phosphomolybdate, $(NH_4)_3PO_4 \cdot 12MoO_3$.

The precipitate is filtered, washed, dried, and weighed.

Where:

• $m$ = mass of the organic compound taken (in grams)
• $m_1$ = mass of ammonium phosphomolybdate precipitate obtained (in grams)
• $31$ = atomic mass of one phosphorus atom (in g/mol)
• $1877$ = molar mass of $(NH_4)_3PO_4 \cdot 12MoO_3$ (in g/mol)

Derivation of the formula:

One molecule of ammonium phosphomolybdate contains exactly one atom of phosphorus. Therefore, the mass of phosphorus in the precipitate is directly proportional to the mass of the precipitate itself.

Mass of phosphorus in $m_1$ g of precipitate = $\frac{31}{1877} \times m_1$ g

This mass of phosphorus came entirely from the original $m$ g of the organic compound.

Percentage of phosphorus in the compound = $\frac{\text{mass of phosphorus}}{\text{mass of compound}} \times 100$

Substituting the expression for the mass of phosphorus gives the formula above.

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Method 2: Precipitation as Magnesium Pyrophosphate

In this alternative route, the phosphoric acid solution is treated with a "magnesia mixture" (a solution containing magnesium chloride, ammonium chloride, and ammonia). This precipitates white, crystalline magnesium ammonium phosphate, $MgNH_4PO_4$.

The precipitate is filtered, washed, dried, and then strongly ignited (heated to a high temperature). Ignition converts it into magnesium pyrophosphate, $Mg_2P_2O_7$, which is a stable, white powder. This final product is cooled and weighed.

Where:

• $m$ = mass of the organic compound taken (in grams)
• $m_1$ = mass of $Mg_2P_2O_7$ formed (in grams)
• $62$ = mass of two phosphorus atoms (2 × 31 g/mol)
• $222$ = molar mass of $Mg_2P_2O_7$ (in g/mol)

Derivation of the formula:

One molecule of magnesium pyrophosphate, $Mg_2P_2O_7$, contains two atoms of phosphorus. The mass of phosphorus in the precipitate is therefore:

Mass of phosphorus in $m_1$ g of $Mg_2P_2O_7$ = $\frac{62}{222} \times m_1$ g …