Question

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

Answer 1

`"BaI"_2 * 4"H"_2"O"`

Explanation:

The idea here is that you need to use the number of moles of anhydrous barium iodide, `"BaI"_2`, and of water, `"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

`color(green)(|bar(ul(color(white)(a/a)color(black)("BaI"_2 * 4"H"_2"O")color(white)(a/a)|))) ->` barium iodide tetrahydrate