Question

A compound is 36.1% water. After drying off the water, the rest is 40.1% oxygen, 39.8% copper, and 20.1% sulfur.

Answer 1

My crystal ball is hazy today. It isn't telling me what question you are asking.

Answer 2

what is the empirical formula

Answer 3

You didn't ask a question. I suppose you want to solve for the empirical formula. Take a 100 g sample which will give you
40.1 g oxygen, 39.8 g copper, and 20.1 g sulfur.
mols O atoms = 40.1/16 = 2.51
mols S atoms = 20.1/32.1 = 0.626
mols Cu atoms = 39.8/63.54 = 0.626
Now you want to find the ratio of the Cu to S to O atoms with the smallest whole number being 1.00. The easy way to do this is to divide the three numbers by the smallest value; i.e.,
mols O = 2.51/0.656 = 4.00
mols S = 0.626/0.626 = 1.00
mols Cu = 0.626/0.626 = 1.00
so the empirical formula, without the water, is CuSO4. Now, how much water is in the formula. CuSO4.xH2O
100% - 36.1% = 63.9% water. Again, taking 100 g sample we have
63.9 g CuSO4 and 36.1 g H2O.
mols CuSO4 = 63.9/159.6 = 0.400
mols H2O = 36.1/18.02 = 2
Again, dividing by the smallest number we get
mols CuSO4 = 0.400/0.400 = 1.00
mols H2O = 2/0.400 = 5.00 so x in the formula above is 5.00
and the empirical formula is CuSO4.5H2O