During the drying study, a student mistakenly determined bis-ethylenediamine to have three waters of hydration. How will this mistake affect the percent yield of the reaction?

CuSO4⋅3H2O

Explanation:

So you start with a certain mass of the hydrate. When you heat it, the water of crystalization, i.e. the water that's a part of the compound, evaporates and leaves you with just the anhydrous form - in your case,

CuSO4. If you don't heat the hydrate enough, you won't get all the water to evaporate, which means the final product will still contain some water. The mass of the evaporated water will be smaller than it should be, since not all the water was driven off thorugh heating.

When this happens, you'll get a smaller percentage of water for the hydrate, since the ratio between the remaining mass and the evaporated water will be bigger than it should be.

In this case, you won't get the correct number of moles of water for the chemical formula - you'll get fewer than the actual 5 moles of water per 1 mole of anhydrous

CuSO4. Your formula will turn out to be CuSO4⋅3H2O, or CuSO4⋅4H2O, or CuSO4⋅2H2O, depending on how much water did not evaporate.

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