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 %
thank you
To solve this problem, we need to determine the percent of Fe3O4 in the sample of ore.
First, let's find the mass of Fe3O4 in the sample. Since the sample originally contained only Fe3O4 and no other iron, the mass of Fe3O4 is equal to the total mass of the ore minus the mass of Fe2O3 produced.
Mass of Fe3O4 = Total mass of the sample - Mass of Fe2O3
Mass of Fe3O4 = 40.93 grams - 30.7 grams
Mass of Fe3O4 = 10.23 grams
Next, we calculate the percentage of Fe3O4 in the sample by dividing the mass of Fe3O4 by the total mass of the sample and multiplying by 100.
% Fe3O4 = (Mass of Fe3O4 / Total mass of the sample) * 100
% Fe3O4 = (10.23 grams / 40.93 grams) * 100
% Fe3O4 ≈ 24.98 %
Therefore, the percent Fe3O4 in the sample of ore is approximately 24.98%.
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%