reacting mass

Question

1. some blue vitriol, CuSO4.5H2O, was heated to constant mass at about 120 degree C, and then at a hogher temp., with results as below:
mass of crucible: 10.00g
mass of crucible+blue vitriol: 14.98g
mass of crucible+ residue (120 degree C): 13.54g
mass of crucible+ residue (higher temp): 13.18g
what stage of dehydration is reached at each of the two temp.?

Answer 1

Molar mass of CuSO4 = 63.5 + 32.1 + 16x4 = 159.6 g mol-1
Molar mass of H2O = 1x2 + 16 = 18 g mol-1
Molar mass of CuSO4 = 159.6 + 18x5 = 249.6 g mol-1

Before heating :
Mass of CuSO4•5H2O = 14.98 - 10 = 4.98 g
Mass of CuSO4 = 4.98 x (159.6/249.6) = 3.18 g

After heating to 120oC，it is dehydrated to CuSO4•nH2O
Mass of CuSO4•nH2O = 13.54 - 10 g = 3.54 g
Mass of CuSO4 = 3.18 g
Mass of H2O = 3.54 – 3.18 = 0.36 g
Mole ratio CuSO4 : H2O = (3.18/159.6) : (0.36/18) = 1 : 1
It is dehydrated to CuSO4•H2O.

After heating to a higher temperature，it is dehydrated to CuSO4•mH2O
Mass of CuSO4•nH2O = 13.18 - 10 g = 3.18 g
Mass of CuSO4 = 3.18 g
Mass of H2O = 3.18 – 3.18 = 0 g
It is dehydrated to anhydrous CuSO4.