# A 5.25 gram sample of a cobalt chloride hydrate is heated until dried. The anhydrous sample has a mass of 3.00 gram. What is the percent water by mass of the original hydrate?

Dec 27, 2015

42.9%

#### Explanation:

So, you're dealing with a $\text{5.25-g}$ sample of a cobalt(II) chloride hydrate, let's say $\text{CoCl"_2 * color(blue)(n)"H"_2"O}$.

You know that the mass of the anhydrous salt, which is what remains after all the water of crystallization was driven off by heating, is equal to $\text{3.00 g}$.

Assuming the all the water of crystallization was indeed driven off, you can say that since

${m}_{\text{hydrate" = m_"anhydrous salt" + m_"water}}$

you will have

${m}_{\text{water" = "5.25 g" - "3.00 g" = "2.25 g}}$

This means that your initial sample of cobalt(II) chloride hydrate contained $\text{3.00 g}$ of anhydrous cobalt(II) chloride and $\text{2.25 g}$ of water of crystallization.

The percent composition of water in the hydrate will thus be

(2.25 color(red)(cancel(color(black)("g"))))/(5.25color(red)(cancel(color(black)("g")))) xx 100 = color(green)("42.9% H"_2"O")

Here is a similar lab with analysis conducted using copper (II) sulfate.

Hope this helps!