How many moles of MnO4- were added?

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SUMMARY

The discussion centers on calculating the moles of MnO4- ions added during a titration involving an unknown sample containing iron (II). The calculation reveals that 0.0002 moles of MnO4- were added, derived from the formula 0.01522 moles/L multiplied by 0.0126 L. Subsequent calculations determine that 0.001 moles of iron (II) are present, equating to 0.056 grams of iron, which constitutes 10.9% of the sample's mass. The importance of precision in calculations is highlighted, particularly regarding the factor of 5 used in determining the moles of iron (II).

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MrPoison
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Hello,

I have no idea how to solve this...

The initial mass of an unknown sample containing iron (II) is 0.512 g.
12.6 mL of 0.01522 M KMnO4 is required to titrate the unknown to the endpoint.

a) How many moles of MnO4- were added?
b) How many moles of iron (II) must be present in the sample?
c) How many grams of iron (II) must be present in the sample?
d) What is the percent of iron present in the sample?


I would know b, c, d, if I would know how to solve a. b, c, d, questions are simple, but I totally don't understand how to solve a. I wrote the rest so people who don't know similar exercise could get something more out of that topic.
I know I shouldn't ask such a basics here, but I spent like an hour on YouTube and internet with no success. My professor is very unclear on every topic.

Thank you so much for helping.
 
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Hi MrPoison, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif


Would it be any easier for you if the question was:

a) How many moles of KMnO4[/color] were added?[/size]

because I think that's essentially what it's asking. :smile:
 
Last edited by a moderator:
I have no idea, because that's what the question from the professor is. I literally rewrote it from the lab assignment.Is that right?

a) How many moles of MnO4- were added?
0.01522 moles/1 L x 0.0126L ≈ 0.0002 = 2x10-4 moles

b) How many moles of iron (II) must be present in the sample?
2x10-4 x 5 = 0.001

c) How many grams of iron (II) must be present in the sample?
0.001 moles of Fe2+ x 56g Fe/1 mole Fe = 0.056g Fe

d) What is the percent of iron present in the sample?
0.056g Fe/0.512 g unknown sample x 100 = 10.9%
 
Last edited:
MrPoison said:
I have no idea, because that's what the question from the professor is. I literally rewrote it from the lab assignment.


Is that right?

a) How many moles of MnO4- were added?
0.01522 moles/1 L x 0.0126L ≈ 0.0002 = 2x10-4 moles
That's right. I wondered whether you were being tripped up by it asking for the Mn04- ions rather than KMnO4.
b) How many moles of iron (II) must be present in the sample?
2x10-4 x 5 = 0.001
That 5 seems to have sprung from nowhere, without explanation. You'd need to account for it if you want marks for this part.

There are plenty of examples of oxidizing Fe(II) on the web for you to follow, e.g.,
http://faculty.uml.edu/james_hall/84124/16.htm[/color]

Good luck with your studies!
 
NascentOxygen said:
That's right.

It would be perfect, if not for the fact OP rounded it down, losing accuracy of the final result.
 

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