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Chemistry

Calculate the percentage of phosphorus in the fertilizer superphosphate Ca(H2PO4)2. (correct to 1 dp)

[H = 1 ; O = 16 ; P = 31 ; Ca = 40]

Stoichiometry

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Answer

Molecular weight of Ca(H2PO4)2
= 40 + 2[2(1) + 31 + 4(16)]
= 40 + 2[2 + 31 + 64]
= 40 + 2[97]
= 40 + 194
= 234 g

234 g of Ca(H2PO4)2 contains 62 g of P

∴ 100 g of Ca(H2PO4)2 will contain = 62234\dfrac{62}{234} x 100 = 26.49% = 26.5%

Hence, 26.5% phosphorous is present in superphosphate Ca(H2PO4)2

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