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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.
To 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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