Determine the percentage of iron, in pure ferrous sulphate FeSO4⋅7H2O

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answer is 20.14.

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Detailed Solution

The formula mass of ferrous sulphate = At. Mass of Fe+ At. Mass of S+4× At. Mass of oxygen +7× Mol. Mass of H2O=56.0+32.0+4×16.0+7×18.0=278.0So, % of water of crystallization =126278×100=45.32% of iron =56278×100=20.14% of sulphur =32278×100=11.51% of oxygen =64278×100=23.02(Oxygen present in water molecules is not taken into account).

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