7(iii)
Calculate:
(a) 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]
(b) Write the empirical formula of C8H18
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(a) 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 = $\frac{62}{234}$ x 100 = 26.49% = 26.5%
Hence, 26.5% phosphorous is present in superphosphate Ca(H2PO4)2
(b) Given,
Molecular Formula = C8H18 = (C4H9)2
Molecular formula = n[Empirical formula]
∴ n = 2
Hence, Empirical formula = C4H9