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1. Find the molar mass (MM) of the substances.
MM FeSO4 = 151.91g
+MM 7H2O = 126.11g
MM FeSO4 x 7H2O = 278.02g
2. Find the percent water of hydrate.
Divide the mass of water by the mass of the hydrate; multiply result by 100%.
126.11g
278.02g x 100%
Percent water of FeSO4 * 7H2O is 45.36%.
Finding Molar Mass
# atoms Element A x Atomic Mass Element A = Mass A
# atoms Element B x atomic mass Element B = Mass B
... etc.
Add up all the mass values for the substance and you have the molar mass of the substance.
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Ezra
FranTo calculate the percent of water in FeSO4·7H2O, you need to consider the molecular weights of each component. The molecular weight of FeSO4 is 151.91 g/mol, and the molecular weight of 7H2O is 126.14 g/mol. The total molecular weight of FeSO4·7H2O is 277.05 g/mol. The percent of water in FeSO4·7H2O is (126.14/277.05) x 100 = 45.5%.
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