Question

- Solving applied mass percent problems Iron(II) sulfate forms several hydrates with the general formula FeSO, XH,O, where x is an integer. If the hydrate is heated, the water can be driven off, leaving pure FeSO, 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?

Answer 1

ANSWER : x = 6

We can solve given problem in following steps.

Step 1 : Calculation of mass of water and ferrous sulfate present in the hydrate.

Assume mass of FeSO₄. x H₂O is 100 g.

After heating mass of hydrate decreases by 42.0 % by mass.

Therefore, mass of water lost = 42.0 g

Mass of FeSO₄ present in the hydrate = 100 g - 42.0 g = 58.0 g

Step 2 : Calculation of moles of water and moles of ferrous sulfate.

Molar mass of H₂O = 2(1.0079 ) +16.00 = 18.02 g /mol

Molar mass of FeSO₄ = 55.85 + 32.06 + ( 4 (16.00)) = 151.91 g / mol

We have, no. of moles = Mass / Molar mass

$\therefore$ No. of mole of H₂O = 42.0 g / 18.02 g/mol = 2.33 mol

No. of moles of FeSO₄ = 58.0 g / ( 151.91 g/mol ) = 0.382 mol

Step 3 : Calculation of mole ratio of FeSO₄ :  H₂O.

2.33 / 0.382 = 6.01 mol H₂O.

0.382 / .382 = 1 mol FeSO₄

Step 4 : Determination of x

In given hydrate , FeSO₄ and H₂O are present in the ratio 1 : 6.

Hence, formula of hydrate is FeSO₄ 6 H₂O.

Therefore x = 6.