How many grams of $\ce{CuSO4 * 5 H2O}$ are needed to prepare a $20\%$ (w/w%) solution of $\ce{CuSO4}$?

  • $\begingroup$ Welcome to Chemistry.SE. Please note that the homework policy on this site requires you to show some personal effort. $\endgroup$ – Klaus-Dieter Warzecha Apr 10 '15 at 8:54
  • $\begingroup$ here's what i come up with so far .. MA($\ce{CuSO4*5H2O}$)=(65.5+32+64+90)=251.5 g/mol =>20% of ($\ce{CuSO4*5H2O}$)=50g !? i don't know i don't have any clue @KlausWarzecha $\endgroup$ – Moutasem Apr 10 '15 at 9:01
  • $\begingroup$ Yes! Using the molecular weights is the right approach! From the weight percentage in the solution, you can calculate the mass of $\ce{CuSO4}$ in 1 litre. With the molecular weight of $\ce{CuSO4}$, you get the molar concentration. It's the same number of moles you have to add to the water. Now you just have to convert that to a mass of $\ce{CuSO4*5H2O}$ again, using the same relation between moles, mass and molecular weight. Feel free to give the answer to your question yourself, that's perfectly ok! If you do, please describe how you did to. That will help the next with a similar problem :) $\endgroup$ – Klaus-Dieter Warzecha Apr 10 '15 at 9:12

In a $20\%$ solution of $\ce{CuSO4}$ we have $\pu{20 g}$ of $\ce{Cu}$ and $\pu{80 g}$ of $\ce{H2O}$. Then

in every $\pu{259.6 g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{159.6 g}$ $\ce{CuSO4}$,

so for every $x~\pu{g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{20 g}$ $\ce{CuSO4}$

$$x = \pu{31.87 g}$$

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    $\begingroup$ I think there is a very minor mistake in the molar mass of the pentahydrate, which should be 249.6 g/mol. Therefore, your final result is a little bit off. Anyway, the important thing here is to understand the concept. My suggestion is: Always keep the units in your calculations! These are a good measure to see where you have to divide or multiply. $\endgroup$ – Klaus-Dieter Warzecha Apr 10 '15 at 12:27

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