An ore contains Fe3O4 and no other iron. The iron in a 60.67 gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 36.1 grams. What was the percent Fe3O4 in the sample of ore?
Answer in units of %
Convert 36.1 g Fe2O3 to g Fe3O4.
36.1 x (2*molar mass Fe3O4/3*molar mass Fe2O3) = ? g Fe3O4.
% Fe3O4 = (g Fe3O4/60.67)*100 = ?
To find the percent Fe3O4 in the sample of ore, we need to compare the mass of Fe3O4 to the total mass of the sample.
We are given that the mass of Fe2O3 (product) formed is 36.1 grams. Since the iron in the sample is completely converted to Fe2O3, we can assume that the mass of Fe3O4 in the original ore is equal to the mass of Fe2O3.
Therefore, the mass of Fe3O4 is also 36.1 grams.
The total mass of the sample is given as 60.67 grams.
To calculate the percent Fe3O4, we can use the following formula:
Percent Fe3O4 = (Mass of Fe3O4 / Total mass of the sample) x 100%
Substituting the values into the formula, we get:
Percent Fe3O4 = (36.1 grams / 60.67 grams) x 100%
Calculating this expression gives us:
Percent Fe3O4 = 0.5949 x 100%
Therefore, the percent Fe3O4 in the sample of ore is approximately 59.49%.