10.222 g sample of hydrated barium iodide (Bal) is heated to dry off the water of

crystallization. The dry sample has a mass of 9.520 g. What is the formula of the

hydrate

BaI₂.5H₂O

Explanation:

Given dа ta:

Mass of Hydrated BaI₂ = 10.222 g

Mass of dried BaI₂ = 9.520 g

Mass of Water removed = 10.222-9.520 = 0.702 g

M. Mass of BaI₂ = 391.136 g/mol

M. Mass of Water = 18.02 g/mol

Now,

Calculate moles of dried BaI₂ as,

Moles = Mass / M. Mass

Moles = 9.520 g / 391.136 g/mol

Moles = 0.02434 moles

Calculate moles of Water as,

Moles = Mass / M. Mass

Moles = 0.702 g / 18.02 g/mol

Moles = 0.0389 moles

Then,

Calculate Mole ratio of BaI₂ and water as,

= 0.02434 moles BaI₂ / 0.0389 moles Water

= 0.625

Now,

We will convert this mole ratio to a whole number by multiplying it with a nearest integer,

= 0.625 * 8

= 5

Hence, this means for every one mole of BaI there are 5 moles of Water.

Result:

BaI₂.5H₂O