# Question #3ef23

##### 1 Answer

#### Answer:

Yes, you need to take the water of hydration into account.

#### Explanation:

As you know, *hydrates* are salts that have **water of hydration** associated with their crystal structure.

This means that you need to take into account the added water molecules when trying to figure out how many moles of

You can think about hydrates as being *impure samples*. In this case, each mole of copper(II) sulfate,

Let's say that your reaction requires the presene of **0.5 moles** of *anhydrous* copper(II) sulfate, this is equivalent to a mass of

#0.5color(red)(cancel(color(black)("moles"))) * "159.6 g"/(1color(red)(cancel(color(black)("mole")))) = "79.8 g"#

The molar mass of the *pentahydrate* is **78.9 g** of pentahydrate, you only get

#78.9color(red)(cancel(color(black)("g"))) * ("1 mole CuSO"""_4 * 5"H"_2"O")/(249.7color(red)(cancel(color(black)("g")))) = "0.320 moles CuSO"""_4 * 5"H"_2"O"#

Since each mole of pentahydrate contains one mole of copper(II) sulfate, adding this much pentahydrate will be equivalent to adding **3.20 moles** of

In this example, to get **79.8 g** of

#(159.6color(red)(cancel(color(black)("g/mole"))))/(249.7color(red)(cancel(color(black)("g/mole")))) * 100 = 63.9%#

This means that you get **63.9 g** of *for every* **100 g** of

#79.8color(red)(cancel(color(black)("g CuSO"""_4))) * ("100 CuSO"""_4 * 5"H"_2"O")/(63.9color(red)(cancel(color(black)("g CuSO"""_4)))) = "124.9 g CuSO"""_4 * 5"H"_2"O"#

So yes, you need to take into account the water of hydration when doing stoichiometric calculations.

If a problem tells you that **100 g** of copper(II) sulfate pentahydrate are dissolved in solution and made to react with a certain mass of sodium carbonate, you need to take into account the water of hydration when you calculate the number of moles of

.