The mass of Fe₂O₃ contains both iron and oxygen. First, we determine the amount of iron present in this compound. Then, because the mass of iron is conserved, the amount of iron present in the original ore will be the same.
Percentage iron in Fe₂O₃ = (2 * 56) * 100 / (2 * 56 + 3 * 16) = 70%
Therefore, out of the 11.5 grams of Fe₂O₃,
11.5 * 0.7 = 8.05 grams is iron
In the original ore, the percentage of iron is given by:
(8.05 * 100) / 43.6
The original ore had 18.5% iron.
