# What is the hydrate formula for something that is .0243 mol BaI_2 and .098 mol H_2O?

Apr 21, 2016

$\text{BaI"_2 * 4"H"_2"O}$

#### Explanation:

The idea here is that you need to use the number of moles of anhydrous barium iodide, ${\text{BaI}}_{2}$, and of water, $\text{H"_2"O}$, to find the mole ratio that exists between the anhydrous salt and its water of hydration in the hydrate.

So, divide both values by the smallest one to find

"For BaI"_2: color(white)(a)(0.0243 color(red)(cancel(color(black)("moles"))))/(0.0243color(red)(cancel(color(black)("moles")))) = 1

"For H"_2"O: " (0.098 color(red)(cancel(color(black)("moles"))))/(0.0243color(red)(cancel(color(black)("moles")))) = 4.033 ~~ 4

This means that one formula unit of this hydrate will contain $1$ mole of barium iodide and $4$ moles of water. Its chemical formula will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{BaI"_2 * 4"H"_2"O}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ barium iodide tetrahydrate