What is the percent water in the compound barium chloride dihydrate?

Mar 11, 2016

14.751%

Explanation:

As its name suggests, barium chloride dihydrate, $\text{BaCl"_2 * color(red)(2)"H"_2"O}$, is an ionic compound that contains water molecules in its crystal structure, i.e a hydrate.

The water that's part of the solid's crystal structure is called water of crystallization.

In the case of barium chloride dihydrate, you get $\textcolor{red}{2}$ moles of water of crystallization for every $1$ mole of hydrate.

In other words, if you were to drive the water of crystallization off by heating the hydrate, you would be left with $1$ mole of anhydrous barium chloride, ${\text{BaCl}}_{2}$, for every $1$ mole of hydrate.

The find the hydrate's percent composition of water in barium chloride dihydrate, use the mass of one mole of hydrate and of two moles of water.

Barium chloride dihydrate has a molar mass of ${\text{244.26 g mol}}^{- 1}$, which means that one mole of hydrate has a mass of $\text{244.26 g}$.

Since one mole of hydrate contains $\textcolor{red}{2}$ moles of water, and since water has a molar mass of ${\text{18.015 g mol}}^{- 1}$, it follows that you get $\textcolor{red}{2} \times \text{18.015 g}$ of water for every $\text{244.26 g}$ of hydrate.

This means that the percent composition of water is

(color(red)(2) xx 18.015color(red)(cancel(color(black)("g"))))/(244.26color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"14.751%"color(white)(a/a)|)))

This means that you get $\text{14.751 g}$ of water for every $\text{100 g}$ of $\text{BaCl"_2 * 2"H"_2"O}$.

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

Hope this helps!