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Finding amount of iron from absorbance

  • Thread starter Puchinita5
  • Start date
  • #1
183
0
So I had a lab where we are supposed to determine the amount of iron present in a vitamin.

I have a calibration curve already constructed, and it is telling me that the concentration of the iron is about 3.20 ppm.

The vitamin was boiled with acid, filtered, and diluted to 100mL.
Then, 5mL of this solution was diluted to 100mL.
Then, 10mL of this already diluted solution was diluted to 100mL (after being mixed with hyroquinone and o-phenanthroline to create a red complex)

So my question is, how to figure out how much iron was present with all these dilutions.

My guess is to say that

3.20 ppm *(1g/10^6 micro g) * (1000 mL/ L) *(1 mol Fe / 55.8 g Fe) = 5.7 x 10^-5 M (molarity)

and then

5.7 x 10^-5 M * (100 mL / 10 mL) * (100 mL / 5mL) * 100mL = 1.15 mol Fe = 64 g of Fe.

But to me this seems like a lot of iron to be present in one tiny tablet.

What am I doing wrong? Or is this right?
 

Answers and Replies

  • #2
74
2
3.20 ppm of Fe3+ = 320mg/L Fe3+ = 0.320g/L Fe3+ (divide by molar mass)

(0.320g/L)/(55.84g/mol) = 0.0057 mol/L -------> 5.7 * 10^-3 Molar Fe3+ ...(are you doing your math correctly?)
 
  • #3
183
0
I think 3.20 ppm is = to 3.20 mg/L not 320 mg/L .

I did notice that in my second step, I was multiplying by 100 mL instead of .1 L so the units were off. So I think it should be

5.7 x 10^-5 M * (100 mL / 10 mL) * (100 mL / 5mL) * (.100 L) = .001146 mol Fe = 64 mg of Fe.

I think this sounds more reasonable? And I just tried googling it and I found a vitamin on the market that has 65 mg of iron in it (and looks like the pill I used in the experiment). So I'm thinking that maybe this is correct.
 
  • #4
Borek
Mentor
28,360
2,752
I got 64 mg as well, but as I didn't bothered to use paper/calculator I could make some mistake. But it is unlikely that we both made the same mistake - so 64 mg looks quite probable.
 

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