# You weigh a sample of "CusO"_4 * x"H"_2"O" and find that it has a mass of "6.45 g". After you heat the sample, you find that the mass of the anhydrous salt is "4.82 g". What is the value of x ?

Nov 6, 2017

$x = 3$

#### Explanation:

The idea here is that the difference between the mass of the hydrate and the mass of the anhydrous salt is equal to the mass of water of hydration present in the sample.

${m}_{\text{water" = "6.45 g " - " 4.82 g}}$

${m}_{\text{water" = "1.63 g}}$

Use the molar mass of water to convert the mass to moles.

1.63 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.09048 moles H"_2"O"

Do the same for anhydrous copper(II) sulfate, ${\text{CuSO}}_{4}$.

4.82 color(red)(cancel(color(black)("g"))) * "1 mole CuSO"_4/(159.609color(red)(cancel(color(black)("g")))) = "0.03020 moles CuSO"_4

Now, your goal here is to find the number of moles of water of hydration present for every mole of copper(II) sulfate hydrate, so divide both values by the smallest one to get the mole ratio that exists between the anhydrous salt and the water of hydration in the hydrate.

"For CuSO"_4:" " (0.03020 color(red)(cancel(color(black)("moles"))))/(0.03020color(red)(cancel(color(black)("moles")))) = 1

"For H"_2"O": " " (0.09048color(red)(cancel(color(black)("moles"))))/(0.03020color(red)(cancel(color(black)("moles")))) = 2.996 ~~ 3

Since $1 : 3$ represents the smallest whole number ratio that can exist here, you can say that for every $1$ mole of anhydrous copper(II) sulfate, the hydrate contains $3$ moles of water of hydration.

This means that you're dealing with copper(II) sulfate trihydrate, $\text{CuSO"_4 * 3"H"_2"O}$.