# How many grams of CuSO_4 * 5H_2O are needed to prepare 100 mL of a 0.10 M solution?

Mar 11, 2016

$\text{2.5 g CuSO"_4 * 5"H"_2"O}$

#### Explanation:

You're dealing with copper(II) sulfate pentahydrate, $\text{CuSO"_4 * 5"H"_2"O}$, an ionic compound that contains water of crystallization in its structure.

More specifically, you have five moles of water of crystallization for every one mole of anhydrous copper(II) sulfate. This means that you're going to have to account for the mass of this water of crystallization in your calculations.

Now, you need your target solution to have a molarity of $\text{0.10 M}$ and a volume of $\text{100. mL}$. Since molarity is defined as moles of solute per liter of solution, you can say that the target solution must contain

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = \frac{n}{V} \implies n = c \cdot V \textcolor{w h i t e}{\frac{a}{a}} |}}}$

${n}_{C u S {O}_{4}} = {\text{0.10 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100. * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.0100 moles CuSO}}_{4}$

It's important to realize that your solute is anhydrous copper(II) sulfate, ${\text{CuSO}}_{4}$. Use the anhydrous salt's molar mass to determine how many grams of copper(II) sulfate would contain that many moles

0.0100 color(red)(cancel(color(black)("moles CuSO"_4))) * "159.61 g"/(1color(red)(cancel(color(black)("mole CuSO"_4)))) = "1.596 g CuSO"_4

Now you have to determine how many grams of copper(II) sulfate pentahydrate would contain this many grams of copper(II) sulfate.

To do that, calculate the hydrate's percent composition by using its molar mass and the molar mass of anhydrous copper(II) sulfate.

Remember that there are $\textcolor{red}{5}$ moles of water for every $1$ mole of hydrate

(159.61 color(red)(cancel(color(black)("g mol"^(-1)))))/( (159.61 + color(red)(5) xx 18.015)color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "63.92% CuSO"_4

This tells you that for every $\text{100 g}$ of copper(II) sulfate pentahydrate, you get $\text{63.92 g}$ of anhydrous copper(II) sulfate. Use this as a conversion factor to get

1.596color(red)(cancel(color(black)("g CuSO"_4))) * ("100 g CuSO"_4 * 5"H"_2"O")/(63.92color(red)(cancel(color(black)("g CuSO"_4)))) = "2.497 g CuSO"_4 * 5"H"_2"O"

Rounded to two sig figs, the answer will be

${m}_{C u S {O}_{4} \cdot 5 {H}_{2} O} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{2.5 g CuSO"_4 * 5"H"_2"O} \textcolor{w h i t e}{\frac{a}{a}} |}}}$