a hydrate of cuso4 has a mass of 12.98 g before heating. after heating, the mass of the anhydrous compound is found to be 9.70 g. explain how you would determine the formula of the hydrate and then write out the full name of the hydrate.

After the water is boiled off, the mass is 9.7 grams, so find the weight of the water.

12.98 - 9.7 = 3.28g H2O

Find how many moles of CuSO4 and H2O you have.

If there are about 18g per mole of H2O

3.28/18 = .1822 moles of H2O

If there are about 160g per mole of CuSO4

9.7/160 = .0606 moles of CuSO4

.1822/.0606 = 3

Ratio of H2O to CuSO4 = 1:3

Formula
CuSO4.3H2O

Full Name
Copper (II) Sulphate Trihydrate

It shows that the experiment is designed properly.