Concentration of stock standard iron solution: 0.05023 mg/ml
Amount of unknown sample used in the solution: 0.128 grams
Absorbance of unknown solution: 0.220
Table 2. Titration of the Na2C2O4 Solution
Table 3. Titration of the Unknown Iron Solution.
Calculations
Part B:
Concentration of iron in each solution: [Concentration of stock standard solution (mg Fe/mL) x volume of each stock standard iron solution] / final volume after dilution (50mL)
Absorbance of the unknown = 198.2(concentration of the unknown) + 0.0154
Mass of Fe in the unknown sample = ((.00281 mg/mL x 50 mL) / 1 mL) x 100 mL = 14.05 mg Fe
Percent Fe in the unknown sample = (0.014 g Fe / 0.128 g Fe) x 100 = 11.0 % Fe
Part C:
Moles MnO4- = Moles C2O42- (2 MnO4- / 5 C2O42-)
Average Molarity = Moles MnO4- / volume of solution used = 0.021 M
Mass of C2O42- Trial 1: 0.13 grams
Mass of C2O42- Trial 2: 0.12 grams
Mass of C2O42- Trial 3: 0.12 grams
Average percent oxalate in the unknown sample= 61%
Part D:
Percent water in the unknown: (Filled vial before oven – Filled vial after oven) / (Filled vial before oven – empty vial)
Average percent water in the unknown: 13% H2O
Part E:
%K+ = 100% - 11% Fe3+ - 61% C2O42- - 13% H2O = 15% K+
Part F:
11 g iron -> .197 moles
61 g oxalate -> 0.693 moles
13 g water -> 0.722 moles
15 g potassium -> 0.384 moles
2 moles Fe : 7 moles oxalate : 7 moles water : 4 moles potassium
Results and Discussion
Percents of compounds in the unknown sample: 11% Fe3+; 61% C2O42-; 13% H2O; 15% K+
There was some difficulty in taking the mass of the unknown sample after being put in the oven when one of the vials tipped over and spilled into the larger beaker holding all of the vials. By using a rubber scraper all of the unknown was replaced into the vial.
Conclusion
The synthesis and analysis of a compound of iron enabled students to learn how to accurately determine the empirical formula of an unknown compound through various tests.
Questions
B1. The calculated %Fe would be too high because the unknown would not be diluted. The iron would be more concentrated and therefore the absorbance would be affected. The solution in the cuvet would have more iron.
B2. Because the trend in the graph indicates that as absorption increases, concentration increases, a lower absorption would indicate a lower concentration. Therefore the percent iron calculated would be less than if it were set to 510 nm.