An ore contains Fe3O4 and no other iron.The iron in a 39.5-gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 26.7 g. What was the mass of Fe3O4 in the sample of ore?
Answer in units of g.
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)
To find the mass of Fe3O4 in the sample of ore, we need to use the given information about the mass of Fe2O3 and the relationship between the two compounds.
First, let's write down the balanced chemical equation for the conversion of Fe3O4 to Fe2O3:
Fe3O4 -> Fe2O3
From the equation, we can see that every one mole of Fe3O4 is converted to one mole of Fe2O3.
Now, we need to calculate the number of moles of Fe2O3 in the sample. We can do this by using the molar mass of Fe2O3 (which is 159.69 g/mol):
Number of moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
= 26.7 g / 159.69 g/mol
≈ 0.1672 mol
Since the stoichiometry of the reaction tells us that every one mole of Fe3O4 is converted to one mole of Fe2O3, we can conclude that the number of moles of Fe3O4 in the sample is also 0.1672 mol.
Now we can find the molar mass of Fe3O4 (which is 231.53 g/mol) and calculate the mass of Fe3O4 in the sample:
Mass of Fe3O4 = Number of moles of Fe3O4 x Molar mass of Fe3O4
= 0.1672 mol x 231.53 g/mol
≈ 38.5 g
Therefore, the mass of Fe3O4 in the sample of ore is approximately 38.5 grams.