A saturated solution is prepared at 70 °C containing $\pu{32.0 g}~ \ce{CuSO4.5H2O}$ per 100 g solution. A 335 g sample of this solution is then cool to 0 °C so that $\ce{CuSO4.5H2O}$ crystallises out. If the concentration of solution of a saturated solution at 0 °C is $\pu{12.5 g}~\ce{CuSO4.5H2O}$ per 100 g solution how much $\ce{CuSO4.5H2O}$ is crystallised?

I found that in 335 g saturated solution at 70 °C there is $\pu{107.2 g}~ \ce{CuSO4.5H2O}$ and then found that there is $\pu{41.875 g}~\ce{CuSO4.5H2O}$ in 335 g saturated solution at 0 °C . Then I found that $$\pu{107.2 g} - \pu{41.875 g} = \pu{65.325 g}~ \ce{CuSO4.5H2O}$$ is crystallised, but the correct answer should be $\pu{74.47 g}$.

  • 1
    $\begingroup$ (1) Bless you for writing the whole problem. (2) In the future please add your calculations below the question and format them so that the calcs are easy to follow. $\endgroup$
    – MaxW
    Commented Feb 15, 2017 at 19:11
  • $\begingroup$ Hello and welcome to Chemistry.SE! As noted above, you did some very good things with your question. At this point, I suggest you take the short tour to better familiarize yourself with this site, and in particular take a look at this discussion. Also, it would be much easier to read the question if you would edit it to include the information in your first comment, then just delete the comment. Again, welcome and good luck! $\endgroup$
    – airhuff
    Commented Feb 15, 2017 at 20:19

2 Answers 2


There is a twist in the calculations...

I would encourage you to write out all the details in the steps instead of just equations with a bunch of numbers. It is so easy to get lost in the numbers alone.


(1) For the 70 °C solution.

% $\ce{CuSO4\cdot5H2O}$ = $\dfrac{32.0}{100.0} = 32.0\% $

% $\ce{H2O} = 100 - 32.0 = 68\%$

(2) For the 0 °C solution

% $\ce{CuSO4\cdot5H2O} = 12.5\%$

% $\ce{H2O} = 100 - 12.5 = 87.5\%$


(1) Calculate the g of $\ce{CuSO4\cdot5H2O}$ and water in 335 g the 70 °C solution.

g $\ce{CuSO4\cdot5H2O} = 0.320 \times 335 = 107.2 $

g $\ce{H2O} = 0.680 \times 335 = 227.8$

(2) Calculate g of solution at 0 °C which has 227.8 g water.

g 0 °C solution = $\dfrac{227.8}{87.5} = 260.3$

(3) Calculate g $\ce{CuSO4\cdot5H2O}$ in 0 °C solution

g $\ce{CuSO4\cdot5H2O} = 0.125\times260.3 = 32.5 $

(4) Calculate g $\ce{CuSO4\cdot5H2O}$ crystals

g $\ce{CuSO4\cdot5H2O}$ crystals $= 107.2 - 32.5 = 74.7$



Let me call the mass of the whole solution $\text{S}$. I'll call $\text{L}$ the mass of the liquid and $\text{C}$ the mass of the crystals, so

$$\begin{cases} \text{A = L + C} \\ \text{A}x_\text{A} = \text{L}x_\text{L} +\text{C}x_\text{C} \end{cases}$$

So if we assume only $\ce{CuSO4, 5H2O}$ cristalise then $x_\text{C}=1$ and then

$$\text{C}=\text{A}\left(\frac{x_\text{A}-x_\text{L}}{1 - x_\text{L}}\right)$$

Now all you have to do is to find the values of the $x_i$ (which are the mass fraction of $\ce{CuSO4, 5H2O}$ in the considered phase.)


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