# How do you know when copper(II) sulfate hydrated water evaporates?

## When considering $C u S {O}_{4} \cdot x {H}_{2} O$, does the number of waters, or $x$ have to be a whole number? If you were to get rid of the water molecules and just have the copper sulfate, how would I know when it has evaporated?

Nov 11, 2016

The color changes!

#### Explanation:

To answer your first question, yes, $x$ is usually a whole number. For a general form copper(II) sulfate hydrate formula

$\text{CuSO"_4 * x"H"_2"O}$

we use $x$ to represent the number of water molecules present as water of crystallization per formula unit of anhydrous copper(II) sulfate, ${\text{CuSO}}_{4}$.

More often than not, $x$ is a whole number, but keep in mind that there are some exceptions. In the case of copper(II) sulfate hydrates, you only get $x$ as a whole number.

Now, the difference between hydrated copper(II) sulfate and anhydrous copper(II) sulfate, except for the fact that the former contains water of crystallization, is the color.

For example, copper(II) pentahydrate, $\text{CuSO"_4 * 5"H"_2"O}$, is blue. On the other hand, anhydrous copper(II) sulfate, ${\text{CuSO}}_{4}$, is white.

This means that when copper(II) sulfate pentahydrate is heated, its color will change from blue to white as the water of crystallization evaporates.

The reaction is reversible, i.e. if you add water to the white anhydrous copper(II) sulfate you'll get the blue copper(II) sulfate pentahydrate again.

Nov 11, 2016

The solid turns white.

#### Explanation:

The equation for the reaction is

underbrace("CuSO"_4·x"H"_2"O")_color(red)("bright blue") stackrelcolor(blue)("heat"color(white)(m))(→) underbrace("CuSO"_4)_color(red)("white") + x"H"_2"O"

(From www.yooniqimages.com)

In hydrates, $x$ is usually a whole number (there are exceptions).

You probably started with the pentahydrate, $\text{CuSO"_4·"5H"_2"O}$, which is the most common form.

The hydrate is blue because the copper is present as the blue tetraaquocopper(II) ion, $\text{Cu"("H"_2"O")_4^"2+}$.

As you heat the blue hydrate and drive off the water, the ion is converted to the colourless copper(II) ion, $\text{Cu"^"2+}$

The dehydration actually occurs in steps.

underbrace("Cu"^"2+"·"4H"_2"O")_color(red)("bright blue"),"SO"_4^"2-"·"H"_2"O" stackrelcolor(blue)("63 °C"color(white)(m))(→) ("Cu"^"2+"·2"H"_2"O"),"SO"_4^"2-"·"H"_2"O" + 2"H"_2"O"

("Cu"^"2+"·"2H"_2"O"),"SO"_4^"2-"·"H"_2"O" stackrelcolor(blue)("109 °C"color(white)(m))(→) "Cu"^"2+","SO"_4^"2-"·"H"_2"O" + 2"H"_2"O"

$\text{Cu"^"2+","SO"_4^"2-"·"H"_2"O" stackrelcolor(blue)("200 °C"color(white)(m))(→) underbrace("Cu"^"2+" ,"SO"_4^"2-")_color(red)("white") + "H"_2"O}$