Calculate the mass of P2O5 in 1.00 kg of fertiliser.

Question

Phosphorus is an essential ingredient in plant fertiliser. The phosphorus content in fertiliser can be determined as a percentage, by mass, of P2O5. A 3.256 g sample of fertiliser is mixed with 40.0 mL of deionised water and the insoluble residue is then removed using vacuum filtration. 45.0 mL of 10% MgSO4.7H2O solution is added to the filtrate followed by 150.0 mL of 2 M NH3 solution. A white precipitate forms. This precipitate is filtered and washed with three 5 mL portions of deionised water. The final mass of the precipitate, once thoroughly dried, was 4.141 g. The formula of the precipitate is known to be MgNH4PO4.6H2O. Assume that the experiment was conducted at 25 °C and that all the phosphorus had been precipitated as MgNH4PO4.6H2O.

(Molar mass of MgNH4PO4.6H2O = 245.3 g mol–1. Molar mass of P2O5 = 142.0 g mol–1.)

Answer 1

The answer is 368.11 grams. Round it to 368.1 grams (4 significant figures).

The fertilizer is 36.811% P₂O₅ by mass.

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STOICHIOMETRY CALCULATION:

1. There is 1 mol of P in 1 mol of MgNH₄PO₄·6H₂O(s)
2. 1 mole of P₂O₅ contains 2 moles of P

4.141 g MgNH4PO4·6H2O • [ 1 mole MgNH₄PO₄·6H₂O / 245.3 g MgNH₄PO₄·6H₂O ] • [ 1 mole P / 1 mole MgNH₄PO₄·6H₂O ] • [ 1 mole P₂O₅ / 2 moles P ] • [142 g P₂O₅ / 1 mole P₂O₅] =

588.022 / 490.6 = 1.19857725 g P₂O₅

1.199 grams of P₂O₅ (phosphorous) were present in the fertilizer sample.

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% P₂O₅ CALCULATION:

[1.19857725 grams P₂O₅ / 3.256 g fertilizer] x 100% =

36.811%

The fertilizer is 36.81% phosphorus (P₂O₅) by mass.

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AMOUNT IN 1 KILOGRAM OF FERTILIZER:

1. 1 kilogram = 1000 grams
2. 36.811% = 0.36811 in decimal form

Simply multiply the 1000 grams by the [decimal form of] the percent mass by phosphorous:

(1000 grams) • (0.36811) = 368.11

A kilogram of fertilizer contains 368.1 grams of P₂O₅ (phosphorous).