Calculate:
The percentage of phosphorus in the fertilizer super phosphate Ca(H₂PO₄)₂ correct to 1 decimal point. [At. Wt. H=1, P=31, O=16, Ca=40]
Topic: Mole concept, gas volumes and stoichiometry
Answer
Ca(H₂PO₄)₂ [Ca = 40, H = 1, P = 31, O = 16]
Molar mass: Ca = 40 H₂PO₄ × 2 = 2 × [(2 × 1) + 31 + (4 × 16)] = 2 × [2 + 31 + 64] = 2 × 97 = 194 Total = 40 + 194 = 234 g mol⁻¹
Mass of phosphorus in the formula = 2 × 31 = 62
Percentage of phosphorus = (62 ÷ 234) × 100 = 26.49 %
Percentage of phosphorus = 26.5 % (to 1 decimal place)
The bracket with a subscript outside it is what makes this harder than it looks. (H₂PO₄)₂ means EVERYTHING inside the bracket is doubled — two phosphorus atoms, four hydrogens and eight oxygens. Work out the bracket once (97), then double it.
Forgetting to double gives a molar mass of 137 and a phosphorus mass of 31, and the answer comes out at 22.6 % — wrong twice over. The question asks for one decimal place, so write 26.5, not 26 or 26.49.