Iron(II) sulfate forms several hydrates with the general formula FeSO4 ·x H2O, where x is an integer. If the hydrate is heated, the water can be driven off, leaving pure FeSO4 behind.

Question

Iron(II) sulfate forms several hydrates with the general formula FeSO4 ·x H2O, where x is an integer. If the hydrate is heated, the water can be driven off, leaving pure FeSO4 behind.

Suppose a sample of a certain hydrate is heated until all the water is removed, and it's found that the mass of the sample decreases by 42.%.

Which hydrate is it? That is, what is x?

Answer 1

Hi, Yasmine,

We aren't told what amount (mass) of FeSO4 ·x H2O we started with, so we can assume a number that is convenient for us. I vote we start with 100g of FeSO4 ·x H2O. 42% of that is lost as water, so 0.42

*100 g = 42 grams of water. That leaves 58 grams of FeSO4 The molar mass of FeSO4 is 151.95 grams/mole. Calculate the number of moles of FeSO4:

58 grams FeSO4/(151.95 g/mole FeSO4) = 0.382 moles FeSO4.

The molar mass of H2O is 18 g/mole H2O. That means the 40 grams of water represents:

40 g H2O/18 g/mole H2O = 2.222 moles of H2O.

The ratio of the two is 2.222 moles H2O/0.382 moles FeSO4, or = 5.82.

That would mean there are an average 5.82 moles of H2O per mole of FeSO4. We need to round up or down to make this a whole number, so we'll round up to 6,

The formula is FeSO4 ·6 H2O. In reality, the 5.82 number probably represents an average. Hydrated salts can lose or gain water fairly easily, and there may have been a few partially dried molecules in this experiment.

I hope this helps,

Bob

Answer 2

Since we don't have the actual mass of the compound, but do have the % of water, we can simply "assume a 100 g sample" & let the % values equal grams. Hence, of a hypothetical 100 g sample, 42 g were "lost" when it was heated. That would be the mass of the H₂O in the hydrate leaving 58 g of FeSO₄. Now, we will calculate the moles of FeSO₄ & moles of H₂O, find the molar ration, & we will have the formula of the hydrate.

(1 mol FeSO₄/152 g) (58 g FeSO₄) = 0.38 mol FeSO₄

(1 mol H₂O/18 g) (42 g H₂O) = 2.33 mol H₂O

Dividing h each by the smaller number, 0.38, we get a 1:6 ratio.Hence, the formula is FeSO₄ • 6 H₂O