- #1
Puchinita5
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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?
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?