Solubility of salts

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 $\pu{74.47 g}$.




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. 


**GIVEN**


(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


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


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


**SOLUTION**


(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 $= 107.2 - 32.5 = 74.7$


 2

**HINT :**


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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