Consider a 5.430 g mixture of FeO and Fe3O4. You react this mixture with excess of oxygen to form 5.779 g Fe2O3. Calculate the percent mass of FeO in the original mixture.

Explanation:

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 :rarrow: 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 :rarrow: 2Fe2O3

4Fe3O4 + O2  :rarrow: 6Fe2O3

2FeO + 2Fe3O4 + O2 :rarrow: 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.