1. The problem statement, all variables and given/known data A 0.6500 g sample of an impure FeSO4 sample is titrated with 33.25 mL of 0.0190 M KMnO4 to an endpoint. What is the % of iron in the sample? 2. Relevant equations KMnO4 + FeSO4 --> KSO4 + FeMnO4 % Fe = (grams of Fe / grams of sample) * 100% 3. The attempt at a solution 0.03325 L sol'n * (0.0190 M KMnO4 / 1 L sol'n) = 6.3175*10-4 M KMnO4 g FeSO4 = 6.3175*10-4 M KMnO4 * (1 mol FeSO4 / 1 mol KMnO4) * (151.903 g FeSO4 / 1 mol FeSO4 = 0.09596 g FeSO4 But now I'm stuck. How do I find g of Fe from 0.09596 g FeSO4? % Fe = (grams Fe alone/ 0.6500g FeSO4)*100% = final answer.