[ranma]

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?

[Yggdrasil]

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).

[ranma]

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

[Yggdrasil]

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.

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.

[ranma]

Thanks!!! I was confused about the Dalton measurements at first, but I get it now.