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# Hydrate problem Copper(II) Sulfate pentahydrate is a well known hydrate. Suppose a sample of the hydrate was heated and 7.24g of anhydrous copper (II) sulfate was recovered. What was the mass of the original hydrated sample?

The mass of the hydrate was **11.3 g**.

The same idea you've used to determine the formula of a hydrate applies in this case, only this time you'll have to work backwards from the formula to determine the mass of the hydrate.

So, you know that you're dealing with copper (II) sulfate pentahydrate, ##CuSO_4 * 5H_2O##. As you can see, for every **1 mole** of anhydrous copper (II) sulfate, you get **5 moles** of water.

You can use this to determine the percent by mass the water has in the total mass of the hydrate. To do this, use the molar masses of water and of the pentahydrate

##(5 * 18.015cancel("g/mol"))/(249.685cancel("g/mol")) * 100 = "36.1% water"##

**SIDE NOTE** You multiply the molar mass of water by 5 because you have 5 moles of water per mole of pentahydrate, as mentioned earlier.

This means that you get **36.1 g** of water for every **100 g** of copper (II) sulfate pentahydrate. Likewise, you get **100 - 36.1 = 63.9 g** of anhydrous copper (II) sulfate for every **100 g** of pentahydrate.

Since you know the mass of the anhydrous salt, and what percentage of the total mass of the hydrate it has (**63.9%**), you can determine the mass of the hydrate by

##7.24cancel("g "CuSO_4) * "100 g pentahydrate"/(63.9cancel("g "CuSO_4)) = color(green)("11.3 g")## ##"pentahydrate"##

Here is a video discussing how to find the empirical formula of copper sulfate.