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]

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Solution

## 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 $\mathrm{Percentage}\mathrm{of}\mathrm{phosphorus}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{phosphorus}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{compound}}×100\phantom{\rule{0ex}{0ex}}\mathrm{Percentage}\mathrm{of}\mathrm{phosphorus}=\frac{62}{234}×100=26.5%$

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