molar extinction coefficient

[2468]

Hey....

Can someone explain this to me please!

ques: 'from the graph (calibration curve) determine the molar extinction coefficient of the acid form this is the slope of the graph. concentration must be M.'

range of concentrations were used ranging from 0 to 12.5um

How do I do this? i knw its the formula E = A/CL but how do i use the graph for it? Also, i use the formula Y=mx+c? but how exactly will this give me the molar extinction coefficient ..? n fr 'c' concentration...wht do i use? :S

Any help would be greatly appreciated!

Thanks

[jonmoulton]

As you decrease the concentration the absorbance decreases and they should both become zero together so the intercept should be zero; that means from Y=mx+c for c=0 you only need Y=mx. For A=ecl, l=1cm and so except for its contribution to the units you can ignore it. That leaves you with A=ec, which is in the form Y=mx. As you vary the concentration the absorbance changes. The slope of their linear relationship (in the Beer's law range, A<~2) is the extinction coefficient. To lay it out explicitly,

A=ec
Y=mx

Plot A on the Y axis, c on the X axis. The slope e of the plot is also e, the extinction coefficient.

This made my head hurt when I first encountered it. But, it works and once you see it you will see similar tricks you can play with other linear relationships.

[2468]

As you decrease the concentration the absorbance decreases and they should both become zero together so the intercept should be zero; that means from Y=mx+c for c=0 you only need Y=mx. For A=ecl, l=1cm and so except for its contribution to the units you can ignore it. That leaves you with A=ec, which is in the form Y=mx. As you vary the concentration the absorbance changes. The slope of their linear relationship (in the Beer's law range, A<~2) is the extinction coefficient. To lay it out explicitly,

A=ec
Y=mx

Plot A on the Y axis, c on the X axis. The slope e of the plot is also e, the extinction coefficient.

This made my head hurt when I first encountered it. But, it works and once you see it you will see similar tricks you can play with other linear relationships.

Thank you! so if i do y=mx calculation and a right angled triangle....that would be the molar extinction coefficient?

[jonmoulton]

If you plot molar concentration on the X axis and absorbance on the Y axis, the slop is the molar extinction coefficient (assuming a 1cm path spectrometer cell and units of M^-1 cm^-1 for the extinction coefficient).

[2468]

If you plot molar concentration on the X axis and absorbance on the Y axis, the slop is the molar extinction coefficient (assuming a 1cm path spectrometer cell and units of M^-1 cm^-1 for the extinction coefficient).

okay...yes i've already done that...but hw wud i knw what the exact value fr the molar extin coeff/ slope is? as i need a value fr my ques :S

[jonmoulton]

Once you have plotted the data and drawn the line, the slope is the rise/run, the dy/dx. You can count squares on the graph paper.

On the other hand, you can put the values of x and y into a spreadsheet and run a linear regression. If you do that, be sure to go back to the raw data on a graph and make sure the slope makes sense given the graph you see.

[2468]

Thank you so much!

[2468]

Once you have plotted the data and drawn the line, the slope is the rise/run, the dy/dx. You can count squares on the graph paper.

On the other hand, you can put the values of x and y into a spreadsheet and run a linear regression. If you do that, be sure to go back to the raw data on a graph and make sure the slope makes sense given the graph you see.

By the way, using the above i got 0.01871m-1 cm-1 for methyl red....is this value not too small for the molar extinction coefficient though?

[jonmoulton]

Yes, that sounds too low, at least for those wavelengths where methyl red absorbs strongly. Did you correct your concentrations for the dilution you made as you mixed your stock solution into liquid prior to making the spectrometer measurement?

[2468]

Its okay now! got it...had to convert concentrations from uM to M. thanks anyways

[jonmoulton]

Well done -- I promise that the next time you face this sort of graphical calculation, it will make more sense. And as you've found, it's all about the units.