# Predict the effect of each of the following by mass of water in the hydrate and the chemical of the hydrate? a) not heating the hydrate enough b) overheating the hydrate and producing some copper (ii) oxide

Mar 25, 2015

I assume you're doing a lab in which you have to determine the empirical formula of copper (II) sulfate pentahydrate, $C u S {O}_{4} \cdot 5 {H}_{2} O$.

#### Explanation:

I won't go into too much detail about hydrates and anhydrous compounds.

$\textcolor{g r e e n}{\text{Point a)}}$

So, you start with a certain mass of the hydrate. When you heat it, the water of crystalization, i.e. the water that's a part of the compound, evaporates and leaves you with just the anhydrous form - in your case, $C u S {O}_{4}$.

If you don't heat the hydrate enough, you won't get all the water to evaporate, which means the final product will still contain some water. The mass of the evaporated water will be smaller than it should be, since not all the water was driven off thorugh heating.

When this happens, you'll get a smaller percentage of water for the hydrate, since the ratio between the remaining mass and the evaporated water will be bigger than it should be.

In this case, you won't get the correct number of moles of water for the chemical formula - you'll get fewer than the actual 5 moles of water per 1 mole of anhydrous $C u S {O}_{4}$.

Your formula will turn out to be $C u S {O}_{4} \cdot 3 {H}_{2} O$, or $C u S {O}_{4} \cdot 4 {H}_{2} O$, or $C u S {O}_{4} \cdot 2 {H}_{2} O$, depending on how much water did not evaporate.

$\textcolor{g r e e n}{\text{Point b)}}$

When you overheat the hydrate, you drive off all the water, but lose some of the anhydrous $C u S {O}_{4}$, which breaks down to form copper (II) oxide.

The effect will be opposite of what you saw in point a), meaning that the percentage of water will appear bigger than it really is because you have less anhydrous $C u S {O}_{4}$ than you should.

As a result, the percentage of water in the hydrate will seem bigger than it actually is, and you'll get a formula that could be $C u S {O}_{4} \cdot 7 {H}_{2} O$, $C u S {O}_{4} \cdot 6 {H}_{2} O$, and so on, depending on how much oxide was produced, i.e. how much anhydrous $C u S {O}_{4}$ was lost.

Think of it like this - the ratio between the number of moles of anhydrous $C u S {O}_{4}$ and the number of moles of water will change if you either don't evaporate all the water, or break down some of the $C u S {O}_{4}$.

You don't evaporate all the water, it will seem like you've got more $C u S {O}_{4}$ than you actually have; on the other hand, if you produce copper (II) oxide, it will seem like you've got more water than actually present.

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