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

Using the stoichiometry of the formula, we find the moles of P2O5 present:

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

Now, convert moles of P2O5 to grams:

$m(P2O5) = 0.0169 \, \text{mol} \times 142.0 \, \text{g mol}^{-1} = 2.398 \text{ g}$

Finally, calculate the percentage of P2O5 in the original sample:

$\text{Percentage of } P2O5 = \left(\frac{m(P2O5)}{3.256 \, \text{g}}\right) \times 100 = \left(\frac{2.398}{3.256}\right) \times 100 = 73.7\%$

If MgNH4PO4·6H2O is heated above 100°C, it will lose water and convert to MgNH4PO4. Since the water mass is not included in this calculation, the total mass of the precipitate would decrease. Consequently, when calculating the percentage of P2O5, the numerator (mass of P2O5) remains the same but the denominator (total mass after heating) decreases, leading to a higher percentage value for P2O5 in the sample that was heated compared to the sample that retained its water of crystallization.