Question

A sample of hydrated copper (II) sulfate (CuSO4•nH2O) is heated to 150°C and produces 103.74 g anhydrous copper (II) sulfate and 58.55 g water. How many moles of water molecules are present in 1.0 mol of hydrated copper (II) sulfate?

Answer 1

Answer:

5

Explanation:

Firstly, we convert what we have to percentage compositions.

There are two parts in the molecule, the sulphate part and the water part.

The percentage compositions is as follows:

Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%

The water part = 100 - 64 = 36%

Now, we divide the percentages by the molar masses.

For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol

For the H2O = 2(1) + 16 = 18g/mol

Now we divide the percentages by these masses

Sulphate = 64/160 = 0.4

Water = 36/18 = 2

The ratio is thus 0.4:2 = 1:5

Hence, there are 5 water molecules.