Question

What is the amount (in moles) of copper (II) sulfate pentahydrate that you will need to make 100mL of a 0.500 mol/L solution, with all the calculations?

Answer 1

Dissolve 0.05 moles/12.48 g of the pentahydrate salt in the minimum amount of water, then make the solution up to a total volume of 100 mL.

Explanation:

You need to take the waters of hydration into account but you still need this many moles of `CuSO_4`:

`n=C*V=0.500*0.1=0.05" mol"`

In terms of number of moles of `CuSO_4`, the above expression is equivalent to this:

`n(CuSO_4*5H_2O)=C*V=0.500*0.1=0.05" mol"`

Now, you will need the molar mass of the hydrate salt:

`MM(CuSO_4*5H_2O)=(63.5+32+4*16)+(5*18)=249.5" g/mol"`

And so the mass you need to weigh out is:

`m(CuSO_4*5H_2O)=n*MM=0.05*249.5=12.48" g"`

This gives you the required number of moles of `CuSO_4`, so we're all good, but it also gives you some waters that will need to be accounted for.

So...

You need to dissolve this amount of solid in the minimum amount of water (less than than the final volume) first. Then make the total volume up to 100 mL. The waters of hydration have contributed (4.5 g) to the volume of the final solution, meaning that you didn't have to add the full 100 mL. Approximating the density of water to 1 g/mL, you should only have added `100-4.5=95.5" g"=95" mL"` of additional water to get to the required 100 mL.

To see how I got 4.5 g of water from the hydrate, see below.

If you break the chemical formula down you get one mole of `CuSO_4` for every 5 moles of `H_2O`. So another way of thinking about it is that you have this many moles of each component:

`n(CuSO_4)=0.05" mol"`

`n(H_2O)=5*0.05=0.25" mol"`

And multiplying by the molar masses of each component you get:

`m(CuSO_4)=0.05*159.5=7.98" g"`

`m(H_2O)=0.25*18=4.5" g"`

`"total mass"=7.98+4.5=12.48" g"`

This is the same mass that we weighed out before, as it should be.