Question

Find the percentage of phosphorus in the fertilizer superphosphate Ca(H₂PO₄)₂.

Answer 1

The molecular mass of Ca(H₂PO₄)₂ = 40 + 2(1 × 2 + 31 + 16 × 64)

= 234 g

Mass of phosphorus in 234 g superphosphate = 62 g

∴ Percentage of phosphorus in superphosphate = `62/234 xx 100`

= 26.5%