chem question

[Dr.Lecter]

Can't quite get my head around this question and was wondering if anyone could help.

0.6238g of copper (II) sulphate crystals with formula CuSO4.xH2O was dissolved in water, and the black copper (II) oxide was precipitated by treatment with boiling NaOH solution. The precipitate was collected by filtration, washed, dried and weighted.

If the precipitate weighs 0.1988g, calculate the value of x in the formula CuSO4.xH2O

[fady_22]

The wording is terrible, but you gave to realise that the precipitate is CuO.

From there, find the mole of CuO:
n(CuO)=0.0025 mol=n(Cu)=n(CuSO4)
So, m(CuSO4)=0.40g and m(H20)=0.6238-0.4=0.225 g

Now that you have the mass of H20, you can find the mole of H20 and hence determine the ratio of CuSO4:H20.
n(H20)=0.0125 mol

n(CuSO4):n(H20)=1:5, so x=5.

[Dr.Lecter]

thanks for that, yeah tsfx question lol