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A sample of hydrated copper sulphate is heated to drive off the water of crystallization, cooled and reweighed 0.869 g of CuSO_(4)aH_(2)O gave a residue of 0.556 g. Find the molecular formula of hydrated copper sulphate. |
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Answer» SOLUTION :0.869 g of `CuSO_(4)*aH_(2)O` gave a residue of 0.556 g of Anhydrous `CuSO_(4)` `:` Weight of a `H_(2)O` molecule = 0.869-0.556 = 0.313 g Molecular weight of `H_(2)O = (1xx2)+16=2+16=18` No. of moles of water =`("Mass")/("Molecular mass")` `CuSO_(4)*5H_(2)O` - Molecular mass = 63.5 + 32 + 64 + 90 = 249.5 g 249.5 gof `CuSO_(4) .5H_(2)O` on heating gives 159.5 g of `CuSO_(4)` . 0.869 g of `CuSO_(4)*aH_(2)O` on heating gives `159.5/249.5 xx 0.869` = 0.556 g of anhydrous `CuSO_(4)` `:.` a =5 The molecular formula of hydrated copper SULPHATE = `CuSO_(4)*5H_(2)O.` |
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