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Question

calculate the percentage of iron in potassium ferro cyanide K4 (Fe(CN)6)

[Fe=56, S=32, O=16, H=1, K=39, C=12, N=14]

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Solution

Molar mass of K4[Fe(CN)6] = 4 X (molar mass of potassium) + molar mass of iron + 6 X (molar mass of carbon) + 6 X (molar mass of nitrogen)

= 4 X (39) + (56) + 6 X (12) + 6 X (14)

= 368 g

One molecule of K4[Fe(CN)6] contains one atom of iron. So the percentage composition of Fe is

= (mass of iron in complex / molar mass of complex) X 100

= (56 X 100) / 368

= 15.217 %


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