A sample of CuSO4*5H2O was heated to 110 degrees C, where it lost water and gave another hydrate of copper (II) ion that contains 32.50% Cu. A 98.77-mg sample of this new hydrate gave 116.66 mg or barium sulfate precipitate when treated with barium nitrate solution. What is the formula of the new hydrate?

This is as far as I have got in the past few hours:

CuSo4 x 5H2O --> CuSO4 x ? H2O

CuSO4 x ? H2O + Ba (NO2)2 --> BaSO4

116.66 mg BaSO4
98.77 mg CuSO4 x ? H2O

At some point once I can get the % of the CuSO4 x ? H2O I can turn it into an empirical formula problem and solve but I still cannot get there, any help?

(0.3250)(0.09877g) = 0.03210 g Cu - - - -> also moles of CuSO4
0.03210 g Cu / 63.55 g/mol = 5.05x10^-4 moles Cu
- - - - - - - - - - - - - - - - - - - -
1 mole BaSO4 = 233.4 g
0.11666g BaSO4 / 233.4 g/mole = 5.000x10^-4 moles BaSO4
(confirms mol Cu = moles SO4 as expected in CuSO4
- - - - - - - - - - - - - - - - - - - -
(0.0005 moles CuSO4)(159.6g/mol) = 0.0798g CuSO4
0.09877g hydrate - 0.07980 g CuSO4 = 0.01897 g H2O
0.01897 g H2O /18.015 g/mol H2O = 1.053x10^-3 mol H2O
1.053x10^-3 mol H2O / 5.05x10^-4 mol CuSO4 = 2.00 mol H2O / 1 mole CuSO4
What does the above ratio mean?

Notes on previous posting:
1) Correction: 5.05x10^-4 moles Cu - - - -> also moles of CuSO4
2) The data on BaSO4 confirms that the anhydrous salt is and remains CuSO4 since moles Cu = moles SO4
3) The last set of calculations gives the mole ratio of H2O/CuSO4 in the second hydrate. This ratio confirms its assumed formula.