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Molar extinction coefficient troublesModerator: BioTeam
7 posts • Page 1 of 1
Molar extinction coefficient troublesHi, i've recieved a piece of work to do with beer's law, i've tried to make some headway with it, but the results i get are always either far too large to be probable or far too small.
The question is stated as follows: You are attempting to determine the amount of haemoglobin present in human blood. To do this, you take 0.005ml of blood and and water to it till the whole volume is 1ml. This causes the cells to lyse releasing theie haemoglobin. You now measure absorbance (540nm) of this solution in a standard 1ml cuvette and obtain a reading of 0.46 The Molar Extinction Coefficient of haemoglobin is 41,700 lotres per mole per cm and the molecular weight of haemoglobin is 64,500. Using your absorbance measurements calculate how many g of haemoglobin are present in a litre of blood. The numbers i have had go from 0.0342 g to over a few thousand kilos, please help :S Thanks in advance
Well i use the formula A = ε l c and re-arrange it to: C = A/ε.l
And so: 0.49/41,700=1.175x10^-5 but the molar co-efficient is for a litre, and we only used 0.005 which goes into a litre 200,000,000 times so i multiply that number to make 1.175x10^-5 x 200,000,000 = 2350.12 But here i get stuck and realise i did something wrong as it couldnt be 2350.12 moles per litre of haemoglobin. :S could you tell me where i am going wrong please?
Re:
This is the basic mistake. You read the units wrong, it is l/mol.cm = l . mol^-1 . cm^-1 (^ means index) = (l . mol^-1) . cm^-1 = (mol . l^-1)^-1 . cm^-1 = M^-1 . cm^-1 (I hope this is not too simple:) so, you can see, it tells you the absorbance per cm of cuvette per molar concentration of solute. So, if you have the 1 ml cuvette, the optical path is probably 1 cm, so the concentration you have calculated should be the actuall concentration. http://www.biolib.cz/en/main/
Cis or trans? That's what matters.
Re: Molar extinction coefficient troubles
7 posts • Page 1 of 1
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