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How many grams of water are required to make a $4.00\%$ solution with $15.0~\mathrm{g}$ of copper(II) sulfate pentahydrate?

I calculated the percent composition of water in a molecule of $\ce{CuSO4.5H2O}$ and it came out to be $36\%$. From that I calculated that there was a total of $5.4~\mathrm{g}$ of water in the sample. I'm guessing that we have to calculate how much water is required in order to make it a $4.00\%$ solution, but I don't understand why it wants it in grams. When $$\text{Percent by volume} = \frac{\text{grams of solute}}{\text{mL of water}}?$$

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  • $\begingroup$ For a solute a "4.00% solution" is definitely wt %. For mixing two liquids (e.g. ethanol and water) then it could either be by volume, or by weight. // Be very careful when throwing around 1 ml = 1 gram. For all practical purposes that is ok for pure water, but it doesn't work for solutions. The 4% solution of copper sulfate will be a bit more dense than 1.00 gram per ml. $\endgroup$ – MaxW Sep 20 '16 at 7:56
  • $\begingroup$ It seems like this was asked before and the question deleted. $\endgroup$ – MaxW Sep 20 '16 at 8:01
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A $4~\%$ solution is one that contains $4~\mathrm{g}$ of the solute in $96~\mathrm{g}$ of solvent to give $100~\mathrm{g}$ of solution. Just to get that out of the way.

To calculate the mass of the solute ($m(\ce{CuSO4})$ — not to be confused with $m(\ce{CuSO4 . 5 H2O})$) we can use at least two ways which are mathematically equivalent:

  • Calculate the amount of $\ce{CuSO4 . 5 H2O}$ — this is equivalent to the amount of $\ce{CuSO4}$. Then calculate the latter’s mass.

    $$M(\ce{CuSO4 . 5 H2O}) = 249.72~\mathrm{g\ mol^{-1}}\\[1.2em] n(\ce{CuSO4 . 5 H2O}) = \frac{m}{m} = \frac{15.0~\mathrm{g}}{249.72~\mathrm{g\ mol^{-1}}} = 6.01 \times 10^{-2}~\mathrm{mol}\\[1.5em] M(\ce{CuSO4}) = 159.62~\mathrm{g\ mol^{-1}}\\[1.2em] m(\ce{CuSO4}) = n \times M = 6.01 \times 10^{-2}~\mathrm{mol} \times 159.62~\mathrm{g\ mol^{-1}} = 9.59~\mathrm{g}$$

  • Use the molar weights of copper(II) sulphate and water to determine the mass percentage of copper(II) sulphate in the salt.

    $$\frac{M(\ce{CuSO4})}{M (\ce{CuSO4 . 5 H2O})} = \frac{159.62~\mathrm{g\ mol^{-1}}}{249.72~\mathrm{g\ mol^{-1}}} = 63.9~\%\\[1.5em] \Longrightarrow m(\ce{CuSO4}) = 15.0~\mathrm{g} \times 63.9~\% = 9.59~\mathrm{g}$$

In any case, we will arrive at a mass of the solute of $9.59~\mathrm{g}$. This is equivalent to the $4~\%$ mentioned at the very top. Therefore:

$$m_\mathrm{tot} = \frac{m_\mathrm{solute}}{0.04} = 240~\mathrm{g}$$ $$m_\mathrm{solvent} = 0.96 \times 240~\mathrm{g} = 230~\mathrm{g}$$

You noted that you already have $5.4~\mathrm{g}$ of water in the copper(II) sulphate hydrate present. Thus, you need to add the difference:

$$m_\mathrm{water} = 230~\mathrm{g} - 5.4~\mathrm{g} = 225~\mathrm{g}$$

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