Question

Calculate the percentage of phosphorus in the fertilizer superphosphate, Ca(H₂PO₄)₂ (correct to 1 decimal place.) [Ca = 40, H = 1, P = 31, O = 16]

Answer 1

Molecular mass of fertiliser superphosphate, Ca(H₂PO₄)₂ = 40 + 2(2 + 31 + 64) = 234 g
Mass of phosphorus in 234 g of fertiliser = 2(31) g = 62 g

$$\begin{gathered} \mathrm{Percentage}\mathrm{of}\mathrm{phosphorus}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{phosphorus}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{compound}}\times 100 \\ \mathrm{Percentage}\mathrm{of}\mathrm{phosphorus}=\frac{62}{234}\times 100=26.5\% \end{gathered}$$