Posted by Gabriella on Friday, August 30, 2013 at 7:31pm.
figure the percent iron in Fe2o3
percent=2*atommassFe/formulamassFe203
grams in ore: percent*26grams.
Amount of Iron in sample:
n[Fe] = 3 m[Fe3O4]/w[Fe3O4]
Amount of Iron in assay:
n[Fe] = 2 m[Fe2O3]/w[Fe2O3]
The amount of iron is conserved, so:
m[Fe3O4] = (2/3) m[Fe2O3] w[Fe3O4] / w[Fe2O3]
Substitute:
m[Fe2O3] = 26g
w[Fe2O3] = 159.6887 ± 0.0002 g/mol
w[Fe3O4] = 231.5333 ± 0.0003 g/mol
okay graham could you possibly put that in a more understandable format because i'm not quite comprehending what you're saying.
Not sure how to be clearer.
From the formula, the mole amount of iron in the magnetite sample is three times the mole amount of iron(II,III) oxide. This amount is the mass of magnetite divided by its molar weight.
n[Fe] = 3 m[Fe3O4]/w[Fe3O4]
The amount of iron in the converted assay is twice the amount of iron(III) oxide. That amount is again the mass of assay divided by its molecular weight.
n[Fe] = 2 m[Fe2O3]/w[Fe2O3]
All of the magnetite was converted, so the amount of iron must remain the same. It is conserved. So we can equate them.
3 m[Fe3O4]/w[Fe3O4] = 2 m[Fe2O3]/w[Fe2O3]
We are given three of these terms and wish to find the fourth. Rearranging gives the mass of magnetite in the sample.
m[Fe3O4] = (2/3) m[Fe2O3] w[Fe3O4] / w[Fe2O3]
Evaluating:
m[Fe3O4] = (2/3)*(26g)*(231.5333)/(159.6887)
Okay I get it now. Thank you.
184g