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percentage purity

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percentage purity

Postby siwel » Mon Apr 07, 2008 10:12 am

The following calculation is from a lab at uni recently. I had to make a 1:500 dilution of chloroform in 2mg/ml bilirubin. This gives a bilirubin concentration of 1.996mg/ml. The molar extinction coefficient of bilirubin is 60700 and has a MW of 584. I tried to calculate percentage purity of bilirubin using the actual and expected absorptions. Actual abs. obtained is 1.102.

I can't get the calculation to work, where have i went wrong?

Use Beer-Lambert Law to calculate the expected absorption:

A = ε ι c
However, concentration must be converted to mol L-1:

Moles = g ÷ MW
Moles = (1.996 x 10-3) ÷ 584
Moles = 3.417 x 10-6 mol ml-1
Moles = 3.417 x 10-9 mol L-1

Now use Beer-Lambert Law:

A = 60700 x 1 x (3.417 x 10-9)
A = 2.074 x 10-4

To correct for dilution multiply by 500:

2.074 x 10-4 x 500
= 0.104

Thanks in advance

P.s. not sure which group this topic should have been posted, so can the mods please move it to the correct place.
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Re: percentage purity

Postby blcr11 » Mon Apr 07, 2008 2:24 pm

Something’s fishy somewhere. I don’t quite get the connection between the stated concentration and the dilution. If you start off with a 2 mg/ml solution of bilirubin and dilute that 1:500, you expect 0.004 mg/ml of bilirubin (not 2 – 0.004 = 1.996 mg/ml). That converts to 6.848E-6 mole/L of bilirubin. The expected absorbance of the solution is then 0.416 assuming the standard 1 cm pathlength. Now you say that the actual absorbance is 1.102?? I’m assuming that was supposed to be the actual absorbance of the 500-fold diluted sample. If so, then you have more bilirubin than you started with. Was the absobance, perhaps, 0.102? Then the purity would be 0.102/0.416 X 100 = 24.5 %. If the 1.102 absorbance is correct then you have 1.102/0.416 X 100 = 265 %, which means either that you’ve made a mistake, or that the initial concentration of bilirubin was greater than the stated 2 mg/ml.
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