How to calculate percentage of hydrate due to CuSO4 and percentage that was due to water?

The mass of the copper II sulfate hydrate was 1.55g and the mass of the anhydrous crystals was .98g. How do I find the percentage composition of the copper II sulfate and then H20? thanks



The empirical formula is CuS04 5H20How to calculate percentage of hydrate due to CuSO4 and percentage that was due to water?
The statement of the problem has at least one problem. The term empirical formula is defined as "The chemical formula of a substance written using the smallest possible integral subscripts that reflect the elemental composition". The formula you have given is called a "molecular formula" The molecular formula shows the actual numbers of atoms in the compound and their arrangements. The copper(II) sulfate pentahydrate molecule is what you have shown.



In order to calculate the % water in the hydrate you would need to divide the mass of the anhydrous crystals, composed of CuSO4, by the given mass of the compound CuSO4.5H2O and multiply by 100. This would give you the % CuSO4. Then you would need to subtract this percentage from 100 to obtain the percentage of water in the hydrate.



An alternative method is to subtract the mass of the anhydrous crystals from the total mass of the copper(II) sulfate pentahydrate. This would give you the mass of water. Next, divide the mass of water by the mass of the CuSO4.5H2O and multiply the quotient by 100 to obtain the percent water in the hydrate.