How Do You Calculate Molar Absorptivity?

[rwooduk]

Homework Statement

I have calculated the molar absorptivity of phenol, I'm pretty sure it's correct but I can't for the life of me figure out what I've done. Any help would be really appreciated.

Homework Equations

Beers Law:

A = εbc

Where A is absorbance, ε is the molar absorptivity, b is the path length the radiation traverses (1 cm) and c is the concentration of the compound in solution (mol L-1).

The Attempt at a Solution

I have an absorbance of 0.0745. Then to get the number of moles I have done the following:

0.0283 and 10601 are constants as they appear throughout my calculations, but I have absolutely no idea where they have come from. Kicking myself for not keeping this in my lab book... lesson learned!

 

[Ygggdrasil]

There is something wrong with your units. Molar absorptivity is generally reported in units of M^-1 cm^-1 (or L mol^-1 cm^-1). It is possible to convert this figure to an absorption cross section, but this would have units of distance^2.

 

[rwooduk]

There is something wrong with your units. Molar absorptivity is generally reported in units of M^-1 cm^-1 (or L mol^-1 cm^-1). It is possible to convert this figure to an absorption cross section, but this would have units of distance^2.

Sorry maybe I wasn't very clear, the equation has been rearranged for c(=9.7 x 10^-6 moles). It's the constants that are confusing me, would there be any reason to shift the absorbance value by 0.0283? I must have done it for a reason.

 

[Borek]

Blank?

Surprisingly large value and I would expect it to be negative, but that's the only thing I can think of.

 

[HAYAO]

Several points you need to cover:

1) Absorbance of 0.0745 is a little bit lower than the range you should be working in (0.1 to 0.9 is optimal). There must have been some reasons why you had to work in such absorbance range.

2) If you rearranged the equation so that the right side gives you moles of the molecule of whatever solution you used, then you must've multiplied both sides of the original equation by the volume of the solution. Do you remember if you used the entire volume of the original solution or the volume of how much you put in the cuvette? By the way, a typical cuvette can hold around 3 mL.

3) Most of the modern absorption spectrometers come with a software that automatically corrects for blank measurement you did prior to measuring the sample. If not, then I believe Borek is right, although as he said the value is indeed too big and should be negative. Do you remember the manufacturer/model of the spectrometer? A rough description would still help to some degree (does the machine look "modern" or "old"?).

4) Most π-π* transition of low-molecular weight molecules will have a molar absorptivity around minimum of 5000 to 10000 M^-1 cm^-1 at the peak. If it is a symmetry-forbidden transition, then lower (for example benzene). For phenol, I suspect the transition of the longest wavelength would have roughly around 2000 M^-1 cm^-1 due to slight reduction of symmetry compared to benzene. Which peak did you work with?

 

[rwooduk]

Blank?

Surprisingly large value and I would expect it to be negative, but that's the only thing I can think of.

Several points you need to cover:

1) Absorbance of 0.0745 is a little bit lower than the range you should be working in (0.1 to 0.9 is optimal). There must have been some reasons why you had to work in such absorbance range.

2) If you rearranged the equation so that the right side gives you moles of the molecule of whatever solution you used, then you must've multiplied both sides of the original equation by the volume of the solution. Do you remember if you used the entire volume of the original solution or the volume of how much you put in the cuvette? By the way, a typical cuvette can hold around 3 mL.

3) Most of the modern absorption spectrometers come with a software that automatically corrects for blank measurement you did prior to measuring the sample. If not, then I believe Borek is right, although as he said the value is indeed too big and should be negative. Do you remember the manufacturer/model of the spectrometer? A rough description would still help to some degree (does the machine look "modern" or "old"?).

4) Most π-π* transition of low-molecular weight molecules will have a molar absorptivity around minimum of 5000 to 10000 M^-1 cm^-1 at the peak. If it is a symmetry-forbidden transition, then lower (for example benzene). For phenol, I suspect the transition of the longest wavelength would have roughly around 2000 M^-1 cm^-1 due to slight reduction of symmetry compared to benzene. Which peak did you work with?

Thanks for the replies, it definitely wasn't a blank, just very low levels of phenol.

Yes, the low absorbance range was because at higher concentrations the phenol did not show any change during the process. Only used the cuvette volume. The machine is brand new, definitely not an issue.

I'm using the 4-AAP method, basically NH4Cl, NH4OH 30% as buffer solution, 4-AAP, and K3Fe(CN)6. In the presence of phenol it gives a peak at 510 nm.

I think I must have adjusted the absorbance to account for a shift in the peak, i.e. the peak wasn't at 510 nm it was slightly to the left or right.

The only thing I can think of is that a calculative method for phenol was given in a paper somewhere (with the value ε=10601 for phenol) and I used the same thing. I'll take a look through my endnote library today.

Many thanks for the suggestions and advice! Just had a little panic yesterday!

 

[HAYAO]

Thanks for the replies, it definitely wasn't a blank, just very low levels of phenol.

Yes, the low absorbance range was because at higher concentrations the phenol did not show any change during the process. Only used the cuvette volume. The machine is brand new, definitely not an issue.

I'm using the 4-AAP method, basically NH4Cl, NH4OH 30% as buffer solution, 4-AAP, and K3Fe(CN)6. In the presence of phenol it gives a peak at 510 nm.

I think I must have adjusted the absorbance to account for a shift in the peak, i.e. the peak wasn't at 510 nm it was slightly to the left or right.

The only thing I can think of is that a calculative method for phenol was given in a paper somewhere (with the value ε=10601 for phenol) and I used the same thing. I'll take a look through my endnote library today.

Many thanks for the suggestions and advice! Just had a little panic yesterday!

Those are some crucial experimental conditions that you should've provided in your OP...why did you decide to leave that out?

The peak around 510 nm is not due to phenol itself. It is due to the product of the reaction between antipyrine with phenol. So whatever value you used for molar absorptivity of phenol is irrelevant here. (You should know that the absorption of phenol in the longest wavelength is around 280 nm.)EDIT: there are some other experimental errors mentioned in your post, but I am going to leave that out for now.

 

[rwooduk]

Those are some crucial experimental conditions that you should've provided in your OP...why did you decide to leave that out?

The peak around 510 nm is not due to phenol itself. It is due to the product of the reaction between antipyrine with phenol. So whatever value you used for molar absorptivity of phenol is irrelevant here. (You should know that the absorption of phenol in the longest wavelength is around 280 nm.)

Fair point. I just thought someone may be able to explain the why I inserted a plus sign into the equation i.e. they had done something similar. But from your reply, it is clear that this is not a standard calculation and I am now 100 % certain it was given in a paper somewhere. You are right I'm not sure why I was looking for the absorptivity of phenol! I'll take this as a slap across the wrist and a lesson learned... always keep a complete lab book! Thanks again.

 

[HAYAO]

Fair point. I just thought someone may be able to explain the why I inserted a plus sign into the equation i.e. they had done something similar. But from your reply, it is clear that this is not a standard calculation and I am now 100 % certain it was given in a paper somewhere. You are right I'm not sure why I was looking for the absorptivity of phenol! I'll take this as a slap across the wrist and a lesson learned... always keep a complete lab book! Thanks again.

Cool.

Then I would speculate that the extra 0.0283 might be some correction factor, e.g. the absorptivity not ideally linear with the concentration because of absorption wavelength shift, interference from other substance that lowers the absorptivity than the idealistic one, or some other issue. These kind of considerations may start you off (to understand what you should be looking for).

 

[rwooduk]

Cool.

Then I would speculate that the extra 0.0283 might be some correction factor, e.g. the absorptivity not ideally linear with the concentration because of absorption wavelength shift, interference from other substance that lowers the absorptivity than the idealistic one, or some other issue. These kind of considerations may start you off (to understand what you should be looking for).

That's it ! I remember now, I plotted the absorption vs. concentration (at the very start of my experiments) and used the gradient as some sort of correction. I just need to find the graph ! Awesome, thanks again !

 

[rwooduk]

EDIT: there are some other experimental errors mentioned in your post, but I am going to leave that out for now.

Hmm, missed this, please could you elaborate a little? I am not a chemist so it is definitely the weaker area of my PhD, unfortunately my examiner will likely be a chemist, so any mistakes I have made in this area could be a problem further down the line. Below is the phenol methodology FYI.

The basis for phenol concentration analysis was via the 4-aminoantipyrine (4-AAP) method in which phenol couples with 4-AAP in an alkaline solution (potassium hexacyanoferrate (III) at PH 7.9 to form a coloured antipyrine complex (AMPH) [19, 50]. This was detectable by UV spectrometry at 510 nm and gave a measurement for the amount of phenol in solution. A blank for UV analysis was made using 2 ml buffer solution containing NH4Cl and NH4OH in 100 ml distilled water, 2 ml 4-AAP reactant in 100 ml distilled water and 2 ml K3Fe(CN)6 catalyst in 100 ml distilled water. For gassed solutions the ‘before’ solution was the phenol solution (0.04 mM) prior to sonication, the ‘after’ solution was the phenol solution after sonication. The following were added to 100 ml of sonicated phenol solution; 2 ml of buffer solution (NH4Cl, NH4OH 30%), 2 ml 4-AAP and 2 ml K3Fe(CN)6 the mixture was then left for 15 minutes to ensure the reaction is complete, followed by spectral analysis.

 

[rwooduk]

Final update. I didn't even use Beers Law, I plotted absorbance vs. concentration. I now have the graph which I made. Thanks again for the help.