Phosphorus is an essential ingredient in plant fertiliser - VCE - SSCE Chemistry - Question 4 - 2011 - Paper 1

Using the stoichiometry of the precipitate, the moles of P2O5:

Since there is 1 mole of P2O5 in 1 mole of MgNH4PO4·6H2O, we have:

$n(P2O5) = n(MgNH4PO4·6H2O) = 0.0169 \, \text{mol}$

Using the molar mass of P2O5:

$m(P2O5) = n(P2O5) \times 142.0 \, \text{g mol}^{-1} = 0.0169 \, \text{mol} \times 142.0 \, \text{g mol}^{-1} = 2.4028 \, \text{g}$

Now, calculate the percentage by mass of P2O5 in the fertiliser:

$\% P2O5 = \frac{m(P2O5)}{3.256 \, \text{g}} \times 100 = \frac{2.4028 \, \text{g}}{3.256 \, \text{g}} \times 100 = 73.8\%$

If the precipitate collected had been heated above 100°C, it would have been converted to MgNH4PO4, thus removing the water. This would lead to a lower mass being weighed, which results in a higher amount of calculated P2O5 percentage since:

$\% P2O5 = \frac{m(P2O5)}{m(MgNH4PO4)} \times 100$ As the mass of the precipitate decreases (due to loss of water), the percentage by mass of P2O5 would be higher.