A 5.012-g sample of an iron chloride hydrate was dried in an oven. The mass of the anhydrous compound was 3.195 g. The compound was then dissolved in water and reacted with an excess of AgNO3. The AgCl precipitate formed weighed 7.225 g. What is the formula of the original compound?

To answer this question we need to calculate the moles (and masses) of each part of the original compound: Moles of water, moles of Fe, and moles of Cl.

1. The mass of water can be calculated using the difference in weight before and after the compound was dried:

5.012 - 3.195 = 1.817 g water

1.817 g water ÷ 18 g/mol = 0.10094 mol H₂O

2. All of the Cl moles in the original compound were transformed into AgCl, so with that mass we can calculate the moles of Cl:

7.225 g AgCl * [tex]\frac{1molAgCl}{143.32g} * \frac{1molCl}{1molAgCl}[/tex] = 0.05041 mol Cl

0.05041 mol Cl * 35.45 g/mol = 1.787 g Cl

3. The original compound is composed of H₂O, Cl and Fe, so the total sum of their masses has to be 5.012 g:

MassH₂O + MassCl + MassFe = 5.012 g
1.817 g + 1.787 g + MassFe = 5.012 g
MassFe = 1.408 g
1.408 g Fe ÷ 55.84 g/mol = 0.02521 mol Fe

To summarize, we have:

0.10094 mol H₂O
0.05041 mol Cl
0.02521 mol Fe

To find the formula, we divide the moles by the lowest value among them (0.02521)

H₂O → 0.10094 / 0.02521 = 4.00

Cl → 0.05041 / 0.02521 = 2.00

Fe → 0.02521 / 0.02521 = 1

So the formula is FeCl₂·4H₂O