A 39 g of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 7.9 grams. What was the percent iron in the sample of ore?

Answer in units of %.
Please help me understand how to solve this problem.

mols Fe2O = grams/molar mass = ?

mols Fe = twice that since there are two mols Fe in each mole of Fe2O3.
Then %Fe = (mass Fe/mass sample)*100 = ?

Thank you!

To solve this problem, we need to determine the percent of iron in the sample of ore.

Step 1: Find the mass of iron in the sample.
- Given: Mass of iron ore = 39 g
- Since the iron is converted to Fe2O3, we need to find the mass of Fe in Fe2O3.
- The molar mass of Fe in Fe2O3 is 55.845 g/mol.
- Using stoichiometry, we find that 1 mole of Fe is equivalent to 2 moles of Fe2O3. So, the mass of Fe in 1 mole of Fe2O3 is (55.845 g/mol) / 2 = 27.9225 g.
- Therefore, the mass of Fe in 7.9 g of Fe2O3 is (27.9225 g/1 mol) * (7.9 g/159.687 g/mol) ≈ 0.1387 g.

Step 2: Calculate the percent of iron in the sample.
- The percent of iron can be calculated using the formula:
Percent Iron = (Mass of Fe / Mass of Iron Ore) * 100
- Substituting the values, we get:
Percent Iron = (0.1387 g / 39 g) * 100 ≈ 0.3559%

Therefore, the percent of iron in the sample of ore is approximately 0.3559%.