AP Chemistry

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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 %

  • AP Chemistry -

    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%

  • AP Chemistry -

    thank you

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