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Question: The substance was heated so strongly that some of the anhydrous salt was vaporized. Is the influence on % water for a hydrate higher, lower or not affected by the error.

Question:  The substance was heated so strongly that some of the anhydrous salt was vaporized.  

Is the influence on % water for a hydrate higher, lower or not affected by the error.

Grade:12th pass

2 Answers

DINESHKUMAR G
48 Points
6 years ago

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 CuSO43H2O, or CuSO44H2O, or CuSO42H2O, depending on how much water did not evaporate.

DINESHKUMAR G
48 Points
6 years ago
Anhydrous copper(II) sulfate is the chemical compound with the formula CuSO4. This salt exists also in a hydrous form. This is because when copper sulfate is dissolved in water, it exists as the free ions Cu²⁺ and SO₄²ˉ. In solution, the partial charges of water molecules, called dipoles, are attracted to the charged ions, thus forming ion-dipole bonds. As the water is let to evaporate, some of the water molecules remain bonded to the, leading to crystallized water in the now hydrous solid copper(II) sulfate. The crystallized water can be easily removed by heating the compound:

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