Question b0054

Apr 23, 2015

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, $C u S {O}_{4} \cdot 5 {H}_{2} O$. As you can see, for every 1 mole of anhydrous copper (II) sulfate, you get 5 moles of water.

You can use this mole ratio 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

$7.24 \cancel{\text{g "CuSO_4) * "100 g pentahydrate"/(63.9cancel("g "CuSO_4)) = color(green)("11.3 g}}$ $\text{pentahydrate}$

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