# How to find the mass of a solute given a mass fraction? [closed]

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

## closed as off-topic by Jan, andselisk♦, Mithoron, Jon Custer, Todd MinehardtJan 2 '18 at 19:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

• Welcome to Chemistry.SE. Please note that the homework policy on this site requires you to show some personal effort. – Klaus-Dieter Warzecha Apr 10 '15 at 8:54
• 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 – Moutasem Apr 10 '15 at 9:01
• 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 :) – 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 $\pu{20 g}$ $\ce{CuSO4}$
$$x = \pu{31.87 g}$$