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.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

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