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percentage purityModerator: BioTeam
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percentage purityThe 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 BeerLambert Law to calculate the expected absorption: A = ε ι c However, concentration must be converted to mol L1: Moles = g ÷ MW Moles = (1.996 x 103) ÷ 584 Moles = 3.417 x 106 mol ml1 Moles = 3.417 x 109 mol L1 Now use BeerLambert Law: A = 60700 x 1 x (3.417 x 109) A = 2.074 x 104 To correct for dilution multiply by 500: 2.074 x 104 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.
Re: percentage puritySomething’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 6.848E6 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 500fold 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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