Question

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?
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