An ore contains Fe3O4 and no other iron. The iron in a 36.5-gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 26 g. What was the mass of Fe3O4 in the sample of ore?
Answer in units of g
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
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)
figure the percent iron in Fe2o3
percent=2*atommassFe/formulamassFe203
grams in ore: percent*26grams.
okay graham could you possibly put that in a more understandable format because i'm not quite comprehending what you're saying.
Okay I get it now. Thank you.
184g
To find the mass of Fe3O4 in the sample of ore, we need to determine the change in mass due to the conversion of Fe3O4 to Fe2O3.
The conversion between Fe3O4 and Fe2O3 can be represented by the following equation:
3 Fe3O4 → 4 Fe2O3
We are given that the mass of Fe2O3 produced is 26 g. According to the stoichiometry of the equation, for every 4 moles of Fe2O3 produced, we need 3 moles of Fe3O4.
First, let's calculate the molar masses of Fe3O4 and Fe2O3:
Molar mass of Fe3O4 = (3 x atomic mass of Fe) + (4 x atomic mass of O)
= (3 x 55.845 g/mol) + (4 x 16.00 g/mol)
= 231.81 g/mol
Molar mass of Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O)
= (2 x 55.845 g/mol) + (3 x 16.00 g/mol)
= 159.69 g/mol
Next, let's find the number of moles of Fe2O3 produced:
Number of moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
= 26 g / 159.69 g/mol
Now, we can use the stoichiometry of the equation to determine the number of moles of Fe3O4:
Number of moles of Fe3O4 = (3/4) x (moles of Fe2O3)
Finally, we can find the mass of Fe3O4 by multiplying the number of moles of Fe3O4 by its molar mass:
Mass of Fe3O4 = Number of moles of Fe3O4 x molar mass of Fe3O4
= (3/4) x (moles of Fe2O3) x molar mass of Fe3O4
Substituting the known values into the equation:
Mass of Fe3O4 = (3/4) x (26 g / 159.69 g/mol) x 231.81 g/mol
Calculating the result will give us the mass of Fe3O4 in the sample of ore.