[RobertT]

Hey guys, i've been searching the forums and i've seen this question maybe once or twice but i've failed to comprehend the explanation given.

An impure sample of Iron(II) sulfate, weighing 1.545g, was treated to produce a precipitate of Fe203. If the mass of the dried precipitate was 0.315g, calculate the percentage of iron in the sample.

I was wondering if anyone could give me an easier explanation =/ as i always end up with 19.3% rather then the true answer, 14.3%

Appreciate any help that i can get
Cheers, Rob.

[Borek]

Show your work.

[RobertT]

ahh ok,
Firstly i did m/M to find the mol of Fe2O3 = 0.315/159.8 = 0.001971 mol
After that i did a mole ratio with FeSO4 which i think is 1/1
Meaning the mol of FeSO4 would be 0.001971mol (In my calculations)
Then i did m(FeSO4) = n(FeSO4) x M(FeSO4) = 0.001971g x 152g mol^-1
m(FeSO4) = 0.299592g
Then i divided that number with the initial mass and multiplied it with 100

(0.299592g/1.545g) x 100 = 19.39%

[Borek]

After that i did a mole ratio with FeSO4 which i think is 1/1

How many atoms of Fe in Fe₂O₃? In FeSO₄? How many molecules of FeSO₄ needed to produce one molecule of Fe₂O₃?

m(FeSO4) = 0.299592g
Then i divided that number with the initial mass and multiplied it with 100

Mass of FeSO₄ is irrelevant - you are asked about percentage of IRON, not its salt.

[RobertT]

After that i did a mole ratio with FeSO4 which i think is 1/1

How many atoms of Fe in Fe₂O₃? In FeSO₄? How many molecules of FeSO₄ needed to produce one molecule of Fe₂O₃?

m(FeSO4) = 0.299592g
Then i divided that number with the initial mass and multiplied it with 100

Mass of FeSO₄ is irrelevant - you are asked about percentage of IRON, not its salt.

Ahh, just a check is the ratio 2/3? Fe₂O₃ is an Iron (III) compound and FeSO₄ is an Iron(II) compound. I then proceed to multiplying that with 0.001971 mol. I'm not sure of the steps after this however, i tried doing m(Fe) = n(Fe) x M(Fe) but i end up with a totally different answer. Heh I'm pretty slow at this =/

[RobertT]

Ohh, i get it now.

n(Fe₂O₃) = 0.315/(2 x 55.9+ 3 x 16.0) = 1.97 x 10^-3
n(Fe) = 2 x (1.97 x 10^-3) (Because there are 2 Fe atoms in the molecule i'm presuming?)
n(Fe) = 3.94 x 10^-3
m(Fe) = (3.94 x 10^-3) x 55.9
m(Fe) = 0.22
0.22 / 1.545 = 14.3%

Thanks for all the help Borek.

[Borek]

Ahh, just a check is the ratio 2/3?

No, but you have already done it right.