# Calculate the mass of "CuSO"_4 * 5"H"_2"O" needed to prepare "100.00 mL" of a "0.05000-mol/L" solution?

Apr 16, 2018

Here's what I got.

#### Explanation:

For starters, you know that ${\text{0.05000-mol L}}^{- 1}$ solution of copper(II) sulfate contains $0.05000$ moles of copper(II) sulfate, the solute, for every $\text{1 L" = 10^3 quad "mL}$ of the solution.

This means that your sample must contain

100.00 color(red)(cancel(color(black)("mL solution"))) * "0.05000 moles CuSO"_4/(10^3 color(red)(cancel(color(black)("mL solution")))) = "0.005000 moles CuSO"_4

Now, notice that every mole of copper(II) sulfate pentahydrate contains

• $1$ mole of anhydrous copper(II) sulfate, ${\text{CuSO}}_{4}$
• $5$ moles of water of hydration, $\text{H"_2"O}$

This means that in order for you to deliver $0.005000$ moles of anhydrous copper(II) sulfate to the solution, you must deliver $0.005000$ moles of copper(II) sulfate pentahydrate.

In addition to the $0.005000$ moles of anhydrous copper(II) sulfate, this sample will contain

0.005000 color(red)(cancel(color(black)("moles CuSO"_4 * 5"H"_2"O"))) * ("5 moles H"_ 2"O")/(1color(red)(cancel(color(black)("mole CuSO"_4 * 5"H"_2"O")))) = "0.02500 moles H"_2"O"

So the total mass of the copper(II) sulfate pentahydrate will be equal to the mass of $0.005000$ moles of anhydrous copper(II) sulfate and the mass of $0.02500$ moles of water of hydration.

0.005000 color(red)(cancel(color(black)("moles CuSO"_4))) * "159.609 g"/(1color(red)(cancel(color(black)("mole CuSO"_4)))) = "0.798045 g"

0.02500 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.450375 g"

You can thus say that the mass of copper(II) sulfate pentahydrate that will deliver $0.005000$ moles of anhydrous copper(II) sulfate to your solution will be

"0.798045 g + 0.450375 g" = color(darkgreen)(ul(color(black)("1.248 g")))

The answer is rounded to four sig figs, the number of sig figs you have for the molarity of the solution.