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In this experiment, you used an oxidation-reduction (redox) reaction as a means of analyzing an unknown sample for how much iron(II) the sample contains.
The experiment was performed over two weeks to give you a chance to take your time and get good results. During the first week of the experiment, you were given a solution of potassium permanganate, KMnO4, of an approximate concentration which was to be used as the titrant (the solution in the buret). Potassium permanganate is highly reactive and is not available in a pure form. The solution you were given, therefore, could only be made up to an approximate concentration. Then, in the first week of the experiment, your goal was to determine the exact concentration of the KMnO4 solution by reacting it with a pure, stable iron compound of known composition, ferrous ammonium sulfate (FAS).
Potassium permanganate reacts with iron(II) salts according to the following oxidation-reduction equation
5 X (Fe2+ Fe3+ + e-) oxidation |
MnO4- + 8H+ + 5e- Mn2+ + 4H2O reduction |
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MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O overall process |
By determining the exact mass of the FAS samples taken, and from the volume of KMnO4 solution required to titrate those samples, the exact molarity of the KMnO4 solutions could be calculate.
In the second week of the experiment, you titrated samples of an "unknown" iron(II) salt with the same KMnO4 solution, in order to determine what % by mass of Fe it contained. Knowing the volume of the KMnO4 solution required to reach the endpoint enables you to calculate the number of moles of iron present in each sample and the mass of iron(II) present in each sample. Then, from the mass of iron(II) in a sample and from the mass of the sample itself, you can calculate the % of Fe in the sample.
Potassium permanganate is intensely purple in solutions, whereas the products of the reaction are essentially colorless at the concentrations used, so the "endpoint" of the titration was signaled by the appearance of a pink color in the sample being titrated, as one excess drop of KMnO4 (beyond that needed to react with the Fe2+ present) was added.
Here is some sample data that we will use for the sample calculations:
In the first part of the experiment, you weigh out three samples of known, pure, standard iron(II) compound, ferrous ammonium sulfate hexahydrate. You then titrate the three samples of FAS with your KMnO4 solution.
You weighed out these samples in a sort of unusual manner. The Erlenmeyer flasks in which you perform the titrations weigh too much to be weighed precisely on the electronic balances. Instead you performed a method called "weighing by difference": you placed enough FAS in a small beaker to cover all three of your samples, and then weighed the beaker. Then you transferred a portion of the FAS to your first Erlenmeyer flask, and reweighed the beaker containing the remainder of the FAS. The difference in mass for the beaker represents the mass of the sample transferred to the Erlenmeyer flask. You then repeated the process twice more (weighing the beaker before and after removing a sample) to prepare your additional FAS samples.
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Before Transfer |
After Transfer |
Mass of Sample |
Sample 1 |
15.1249 g |
14.1157 g |
1.0092 g |
Sample 2 |
14.1157 g |
13.0979 g |
1.0178 g |
Sample 3 |
13.0979 g |
12.0052 g |
1.0927 g |
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Initial |
Final |
Vol. of KMnO4 |
Sample 1 |
0.25 mL |
25.36 mL |
25.11 mL |
Sample 2 |
1.24 mL |
26.58 mL |
25.34 mL |
Sample 3 |
1.09 mL |
28.30 mL |
27.21 mL |
From the data in Part I A above, it is possible to calculate the Molarity, M, of the KMnO4 solution (see calculations below) to be used to titrate samples of the unknown iron compound.
The unknown iron compound was also weighed "by difference" directly out of the container (tube or vial) in which it was dispensed.
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Before Transfer |
After Transfer |
Mass of Sample |
Unknown Sample 1 |
35.2453 g |
34.0101 g |
1.2352 g |
Unknown Sample 2 |
34.0101 g |
32.7524 g |
1.2577 g |
Unknown Sample 3 |
32.7524 g |
31.5031 g |
1.2493 g |
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Initial |
Final |
Vol. of KMnO4 |
Sample 1 |
0.29 mL |
26.30 mL |
26.01 mL |
Sample 2 |
1.04 mL |
27.51 mL |
26.47 mL |
Sample 3 |
2.10 mL |
28.40 mL |
26.30 mL |
The number of moles of anything is calculated from the mass of sample and the molar mass. The molar mass of Fe(NH4)2(SO4)2.6H2O is 392.2 g. For sample 1, in which the mas of FAS taken was 1.0092 g, calculate the number of moles of FAS present, then click here to check your answer.
It takes five moles of Fe2+ to react with one mole of KMnO4 according to the balanced chemical equation for the reaction. Each FAS formula unit contains one Fe2+. For Sample 1, calculate the number of moles of KMnO4 required to react with the iron(II) present, then click here to check your answer.
In Question 2, we calculated that it must have required 0.0005146 moles of KMnO4 to react with the iron(II) contained in Sample 1. This number of moles was contained in a volume of 25.11 mL (0.02511 L). Since the Molarity of a solution is defined as the number of moles of soluted contained per Liter of solution, use the information for Sample 1 to calculate its Molarity. Then click here to check your answer.
If you need help on calculating deviations and average deviation, please see Experiment 1 help (or click here).
Using the sample data above, and the same calculation methods used for Sample 1, the individual and average Molarities for the potassium permanganate are:
Results |
Molarity of KMnO4 |
Deviation |
Sample 1 |
0.02049 M |
0.00001 M |
Sample 2 |
0.02048 M |
0.00000 M |
Sample 3 |
0.02047 M |
-0.00001 M |
Average |
0.02048 M |
0.0000067 M » 0.00001 M |
For a solution whose molarity is known, the number of moles contained in a portion is the volume of the portion in liters multiplied by the molarity (moles/L). For Unknown Sample 1, we required 26.01 mL (0.02601 L) of 0.02048 M KMnO4 to reach the endpoint::
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According to the balanced chemical equation for the reaction between iron(II) and permanganate, there must be five times as much iron as permanganate present at the endpoint. If we used 0.0005327 moles KMnO4 to reach the endpoint, the amount of Fe2+ in Sample 1 is given by :
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If we have 0.002634 moles of Fe2+, we just need the atomic molar mass of Fe (55.85 g) to calculate what this iron(II) weighs:
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The percentage of iron in Unknown sample 1 represents what fraction of the total sample is Fe. Unknown Sample 1 had a total mass of 1.2352 g. We have just determined that 0.1488 g of this is Fe. So the %Fe
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moles Fe2+ |
grams Fe |
%Fe |
Deviation |
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Unknown 1 |
0.002634 moles |
0.1488 g Fe |
12.05% Fe |
0.01 |
Unknown 2 |
0.002711 moles |
0.1514 g Fe |
12.04% Fe |
0.00 |
Unknown 3 |
0.002693 moles |
0.1504 g Fe |
12.04% Fe |
0.00 |
Average |
12.04% Fe |
0.003% |
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