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Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: ranma on July 25, 2009, 11:13:37 PM

Title: The absorbance of a solution
Post by: ranma on July 25, 2009, 11:13:37 PM
Hi, here's my question:

The absorbance A of a solution is defined as:

A=log10(I0/I)

In which I0 is the incident light intensity, and I is the transmitted light intensity. The absorbace is related to the molar absorption coefficient(extinction coefficient) ε (in M-1 cm-1), concentration c (in M) and path length l (in cm) by

A= εlc

The absorption coefficient of myoglobin at 580nm is 15,000 M-1 cm-1. What is the absorbance of a 1mg/ml solution across a 1cm light path? What percentage of the incident light is transmitted by this solution?

The first step that comes to mind is to convert mg/ml to moles/ml (since the absorption coefficient is in units of M-1 cm-1), but I'm not sure how to do that! Does anyone have any ideas on how to solve this problem?


Title: Re: The absorbance of a solution
Post by: Yggdrasil on July 26, 2009, 01:59:56 AM
You'd want to convert the concentration to M (moles per liter), but to do that you would need to know the molecular weight of myoglobin (if that was not provided to you, you can find it on wikipedia, http://en.wikipedia.org/wiki/Myoglobin).
Title: Re: The absorbance of a solution
Post by: ranma on July 26, 2009, 04:08:32 PM
Okay, I found the molecular weight of myoglobin to be 16,700D=16,700u.

Since 1Dalton=1.6605*10-24 grams, I did the calculation to get the mass:
(1D/1.6605*10-24 g)=(16,700D/x), so x=2.8*10-20 grams.

From there, I calculated the molar mass:
(2.8*10-20) * (6.02*1023)= 1.78*104 grams/mole.

Then, I converted that to the molar concentration:
1mol/ (1.7*104 grams)*(1 gram/1L)= 5.9*10-5 mol/L,

Putting the numbers together:
ε= 15,000M-1 cm-1
l= 1cm
c= 5.9*10-5

A=εlc= (15,000)(1)(5.9*10-5)= 0.882


Title: Re: The absorbance of a solution
Post by: Yggdrasil on July 26, 2009, 04:28:11 PM
Quote from: ranma on July 26, 2009, 04:08:32 PM
Okay, I found the molecular weight of myoglobin to be 16,700D=16,700u.

Since 1Dalton=1.6605*10-24 grams, I did the calculation to get the mass:
(1D/1.6605*10-24 g)=(16,700D/x), so x=2.8*10-20 grams.

From there, I calculated the molar mass:
(2.8*10-20) * (6.02*1023)= 1.78*104 grams/mole.

A molecular weight of 16,700 Da means that the molar mass is 16,700 g/mol.  These two values are always equivalent.

Quote
Then, I converted that to the molar concentration:
1mol/ (1.7*104 grams)*(1 gram/1L)= 5.9*10-5 mol/L,

Putting the numbers together:
ε= 15,000M-1 cm-1
l= 1cm
c= 5.9*10-5

A=εlc= (15,000)(1)(5.9*10-5)= 0.882

These calculations are basically correct.
Title: Re: The absorbance of a solution
Post by: ranma on July 26, 2009, 05:36:34 PM
Thanks!!! I was confused about the Dalton measurements at first, but I get it now.  :)