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October 6, 2015
Posted by **Sven** on Monday, September 3, 2007 at 4:29pm.

I'm supposed to calculate the moles of permanganate, Fe^2+ and the mass of iron from the sample.

The volume of permanganate is 33.5, and the concentration is 1.475 g/L. Do I multiply those to find the moles?

And then for the iron I divide the sample amount by 55.85 g correct?

Then how do I find the mass of the iron and the percentage of iron in the sample?

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**Sven**, Monday, September 3, 2007 at 4:33pmSorry that Mn is 2+ and I forgot to add that its a redox titration. Thanks!

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**DrBob222**, Monday, September 3, 2007 at 4:45pmConvert concn of MnO4^- to mols/L = Molarity.

M = 1.475 g/L x (1 mol/molar mass KMnO4 in grams) = mols KMnO4.

Mols KMnO4 = M KMnO4 x L used in the titration = M KMnO4 x 0.0335 (I'm surprised you have the grams/L to 4 places and the buret is read to only 3. If that is a 33.50 mL, put the zero there.

Use the equation to convert mols MnO4^- used in the titration to mols Fe in the sample.

mols Fe in sample x molar mass Fe = g iron in sample.

%Fe in sample = (mass Fe in sample/mass sample)*100.

Post your work if you get stuck.

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**Anonymous**, Monday, September 3, 2007 at 6:08pmThanks so much, unfortuantely I did get a little stuck.

Here's my work

1.475 G/L X 1 mole/158.04 = .0093 moles of KMnO4

.0093 moles X 0.03350 L = 3.6 X 10^-4 moles

The actual amount of the pill i transfered was only .209 g because most of it was lost in transfer (we had to crush an iron pill that had an original mass of .381 g

and .065 g of iron.

so .209 g X 1 mol Fe/ 55.85 g = .004 moles Fe

then I'm stuck here....

.004 moles fe X 279.25 g (i got this value by multiplying 5 by 55.85 according to the equation) = 1.117 g Fe

Isn't that a bit much? And I'm so sorry but how do I mind the mass of the whole sample? Do I add 279.25 and 158.04 (the molar masses of potassium permanganate and the molar mass of the iron?)

Thanks again, so much!

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**Anonymous**, Monday, September 3, 2007 at 6:09pmsorry that last part about the .065 g

iron in line 3 isn't supposed to be there.

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**DrBob222**, Monday, September 3, 2007 at 6:34pmThanks so much, unfortuantely I did get a little stuck.

Here's my work

1.475 G/L X 1 mole/158.04 = .0093 moles of KMnO4**You are ok here EXCEPT you are dropping too many of the numbers. If you weighed 1.475 g of the KMnO4, then you have 4 significant figures; therefore, don't drop the 3s on the mols KMnO4. I found 0.009333. And since that is in 1 L of solution, then the molarity of the KMnO4 is 0.009333 M. Personally, I don't like to work with a molarity that small BUT I'm not the one doing the titration AND I'm not the one giving the instructions; therefore, I'll get off my soap box.**

.0093 moles X 0.03350 L = 3.6 X 10^-4 moles**First, I don't think you multiplied correctly. However, you will need to correct for the different M, anyway. I have 0.00031265.</b**

>

The actual amount of the pill i transfered was only .209 g because most of it was lost in transfer (we had to crush an iron pill that had an original mass of .381 g

and .065 g of iron.**OK. I understand.**

so .209 g X 1 mol Fe/ 55.85 g = .004 moles Fe**NO, NO. The mass of the sample has nothing to do with determining how much Fe is in the sample. That is why you did a titration. The mols KMnO4 = 0.00031265. Therefore, the mols of Fe, from the titration equation is**

0.00031265 mols MnO4^- x (5 mols Fe/1 mol MnO4^-) = ?? mols Fe. Then, mols Fe x molar mass Fe = mass Fe in the sample. Then you can find the percent Fe in the sample. The mass of the sample is the mass you took for the work (count only what you took, not the entire mass of the pill since you lost some of it. Post back if this isn't clear.

then I'm stuck here....

.004 moles fe X 279.25 g (i got this value by multiplying 5 by 55.85 according to the equation) = 1.117 g Fe

Isn't that a bit much? And I'm so sorry but how do I mind the mass of the whole sample? Do I add 279.25 and 158.04 (the molar masses of potassium permanganate and the molar mass of the iron?)

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**DrBob222**, Monday, September 3, 2007 at 6:35pmSorry I didn't get the bold face turned off but I think you cn follow through.

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**Sven**, Monday, September 3, 2007 at 7:12pmThank you so much. I was wondering why I didn't have a mole ratio.

But is 5 X 55.85 the molar mass of Fe, because if so, i'm getting a huge percentage error, which isn't right because it's supposed to be around 20%.

because the pills are technically 325 mg and 65mg is iron (according to the package), but when i weighed the pill i got 381 mg. my prof said that was ok and i know from simple algebra that the value should be a little lower or higher than 40.2 mg because 40.2/201 = 20%.

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**DrBob222**, Monday, September 3, 2007 at 7:04pmThe 0.209 g (and I suppose that's 0.2090 g) is the mass of the sample. The mass of Fe IN the sample is determined ONLY by the titration. Be sure and post back if this isn't clear.

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**DrBob222**, Monday, September 3, 2007 at 7:42pmNo, the molar mass of Fe is 55.85 or about that.

OK. So you should get 20.00% if the package label is correct because, as you point out, (65 mg/325 mg)*100 = 20.0%. From your numbers.

M KMnO4 = 1.475g x (1 mol KMnO4/158.04 g) = 0.009333 M/L.

mols KMnO4 in titration = 0.009333 x 0.03350 = 0.00031266

mols Fe = 0.00031266 x (5 mols Fe/1 mol MnO4^-) = 0.001563.

g Fe = 0.001563 x 55.85 = 0.08731 g (which is about twice the number it should be for 0.209 g x 0.20 = 0.0418).

% Fe in sample = (0.08731/0.209)*100 =41.78% from your values.

I know that's too much, but I have check the math and I don't believe there is an error. But you check it anyway.