An ore contains Fe3O4 and no other iron. The iron in a 40.93 gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 30.7 grams. What was the percent Fe3O4 in the sample of ore?

Question

Answer in units of %

Answer 1

Given:
m{sample} = 40.93g
m{Fe2O3} = 30.7g
w{Fe2O3} = 159.6887 ± 0.0002 g/mol
w{Fe3O4} = 231.5333 ± 0.0003 g/mol
And the stoichiometry:
3 m{Fe3O4}/w{Fe3O4} = 2 m{Fe2O3}/w{Fe2O3}

The percentage of magnetite in the sample is:
m{Fe3O4} / m{sample}
= (2/3) m{Fe2O3} w{Fe3O4} / (w{Fe2O3} m{sample})
= (2/3)*(30.7)*(231.5333)/(159.6887)/(40.93)
= 72.5%

Answer 2

thank you