How many grams of CuSO4*5H2O yield 10 grams of H2O?

If I want there to be 10.00 g of water in the copper (II) sulphate pentahydrate sample that I'm measuring, what mass of the substance will I need in order to have 10.00 g of water? 


I solved this question by...
$$\ce{CuSO4.5H2O -> CuSO4 + 5H2O}$$


Therefore, since we get five waters for each copper (II) sulphate pentahydrate,
we would only need 2 copper sulphate pentahydrates.


Thus, i think I'll need 2 g of the substance but is it correct?




You need to think about moles first and molecular weights second in order to solve your problem. Although the decomposition reaction you wrote is correct, you dont' need it to work this problem.


(1) The molecular weight of $\ce{CuSO4\cdot 5H2O}$ is 250 $\mathrm{g\cdot mol^{-1}}$. 


(2) Each mole of $\ce{CuSO4\cdot 5H2O}$ yields 5 moles of $\ce{H2O}$, and each mole of $\ce{H2O}$ has a molecular weight of 18 $\mathrm{g\cdot mol^{-1}}$.


(3) For every mole of $\ce{CuSO4\cdot 5H2O}$ you have, there are $5\,\mathrm{mol}\times 18\,\mathrm{g\cdot mol^{-1}} = 90\,\mathrm{g}$ of $\ce{H2O}$ present (the $\mathrm{mol}$ terms cancel here).


(4) You want $10\,\mathrm{g}$ of $\ce{H2O}$, or $10/90 = 1/9$ of the amount you get from 1 mole of $\ce{CuSO4\cdot 5H2O}$, which is the same as saying that $1/9$ $\mathrm{mol}$ $\ce{CuSO4\cdot 5H2O}$ gives you $10\,\mathrm{g}\,\ce{H2O}$.


(5) Finally, you need $250/9 \approx 27.8\,\mathrm{g}$ of $\ce{CuSO4\cdot 5H2O}$ to yield $10\,\mathrm{g}$ of $\ce{H2O}$.


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