# Molar Extinction Coefficient

• Author
Posts
• #6865
Miss_Me
Participant

Hi.

Please can someone help me here, as I am confused and I don’t want to end up more confused! Thanks in advance!

The extinction coefficient is present in the beer-lambert law (A=ECL). However, how would I calculate the extinction coefficient from a graph where concentration is on the x-axis and optical density on the y-axis? Would I have to work out a gradient? And so, where do I go next to work out the extinction coefficient??

I’ve tried reading up on the extiction coefficient but I don’t seem to find anything which would help me work the extiction coefficient from a graph.

Thankss!!
ðŸ™‚

• #68383
cbourne
Participant

Hi there-

If you solve Beer’s law for E you get = A/cl
In your graph you have c on the x-axis, and A is on the y-axis
So the slope of your line (rise over run) is A / c
You’ll have to compensate if you’ve used a path length besides 1cm, otherwise you can ignore l.
If your c on the x-axis is in molar, then your units for E are M-1
You can then take any given A reading, divide by E, and get M

You can also calculate the extinction coefficient- if you can find it check out Protein Science (1995) 4:2411-2423; Pace et al "How to measure and predict the molar adsorption coefficient of a protein".

Good luck!

• #68405
sdekivit
Participant

omg:

A = ecl with e = molar extinction coefficient, c = concentration and l = length of the light through the sample

to calculate e you chose on the x-axis the concentration and on the y-axis the absorbance/extinction or whatever you call it.

–> than the slope of that grapgh will be equal to delta y/delta x = A/c = e*l

–> thus your slope in the curve is equal to 2 constants: the molar extinction coefficient and the path of the light through the sample –> e * l

• #68406
sdekivit
Participant
quote cbourne:

Hi there-

If your c on the x-axis is in molar, then your units for E are M-1
You can then take any given A reading, divide by E, and get M

not true: absorbance/extinction has no unit (because it is also equal to I/I(0) –> intensity transmitted/blanc transmittance), thus when c in molarity and l in cm, than the molar extinction coefficient will have the unist M^-1 * cm^-1

• #89769
r6barfly
Participant

So if I have this data:

Absorbance @ 405nm:
.875

Concentration (M) of Cr(NO3)3 (aq):
.25M

Since you’re saying that if l = 1cm (standard curvette), then E = A/C
so then:
.875/.25 = E
E = 3.5?