[cabaal]

A 5.430 g mixture of FeO and Fe₃O₄ is reacted with excess of oxygen to form 5.779 g Fe₂O₃. Find the masses of FeO and Fe₃O₄ present in the mixture.

I found the answer by doing the following:
Fe₂O₃ = 159.69 g/mol
FeO = 71.745 g/mol
Fe₃O₄ = 231.535 g/mol
5.779 g Fe₂O₃ / 159.69 g/mol = 0.03619 mol Fe₂O₃

x + y 0.03619 mol Fe₂O₃, where x = FeO and y = Fe₃O₄.
This implies that y = 0.03619 - x.
5.430 g = x*71.745 g/mol + (0.03619 - x)*231.535g/mol
x = 0.01846 mol * 71.745 g/mol = 1.324 g FeO
y = 0.03619 - 0.01846 = 0.01773 mol * 231.535 g/mol = 4.105 g Fe₃O₄



My question is why doesn't the following work:



4FeO + O₂ 2Fe₂O₃
4Fe₃O₄ + O₂  6Fe₂O₃
_______________________
2FeO + 2Fe₃O₄ + O₂ 4Fe₂O₃

5.779 g Fe₂O₃ / 159.69 g/mol = 0.03619 mol Fe₂O₃
Product:reactants in question are in a 2:1 ratio as given by the stoichiometric coefficients.
0.03619 mol / 2 = 0.01810 mol of FeO and 0.01810 mol Fe₃O₄.
0.01810 mol FeO * 71.745 g/mol = 1.299 g FeO
0.01810 mol Fe₃O₄ * 231.535 g/mol = 4.191 g Fe₃O₄

Obviously this doesn't work because the original mixture is 5.430 g and not 1.299 g + 4.191 g = 5.490 g. Why doesn't this work the way I think it should? Thank you.

[Borek]

4FeO + O₂ 2Fe₂O₃
4Fe₃O₄ + O₂  6Fe₂O₃
_______________________
2FeO + 2Fe₃O₄ + O₂ 4Fe₂O₃

Adding these reaction equations you are ASSUMING ratio of FeO:Fe₃O₄ to be 1:1 (or 4:4, which is the same).

[DrCMS]

Your solutions do not work because they are BOTH wrong.

You start with a mixture of FeO and Fe₃O₄ that weighs 5.430g.
You end up with Fe₂O₃ that weighs 5.779g

In your first "solution" you assume the moles of product equals the moles of starting materials added together, that is not the case.

In your second "solution" as Borek has pointed out you have assumed the moles of the two starting materials are equal which is not the case.

From the data given you can produce at least two equations with only two unknowns.

For example the weights of FeO and Fe₃O₄ = 5.430g
Xg of FeO + Yg of Fe₃O₄ = 5.430g  so X+Y = 5.43

You can calculate how much Fe is in the final 5.779g of Fe₂O₃. 
That is also the same amount of Fe that is in the FeO and Fe₃O₄ starting mixture.  You can now construct another equation that relates the g of Fe in Xg of FeO and Yg of Fe₃O₄ to the amount in 5.779g of Fe₂O₃.  Rearrange the two equations with two unknowns to one equation and one unknown and you've solved it.

[cabaal]

5.779 g Fe₂O₃ * (111.69 g Fe / 159.69 g Fe₂O₃) = 4.042 g Fe
X = 5.430 - Y, where X = FeO and Y = Fe₃O₄
4.042 g Fe = (grams of Fe in FeO) + (grams of Fe in Fe₃O₄)
4.042 g Fe = (5.430 - Y) * (55.845 g Fe / 71.845 g FeO) + Y * (167.535 g Fe / 231.535 g Fe₃O₄)
Y = 3.327 g Fe₃O₄
X = 5.430 g - 3.327 g = 2.103 g FeO

Is this correct?

[DrCMS]

Yes.

Can you see why this is the correct way to solve it?

[cabaal]

Yes, although it wasn't apparent at first.

It's surprising that I'm 10 weeks into a general chemistry II course, yet I still get stumped by these questions.

Thank you for your help, Borek and DrCMS.

[DrCMS]

Glad we could help you.