Hello,
I extracted pigments from a 5g spinach material. To do so, I diluted the leaft content with
100 mL acetone 90% (adjusted with pure acetone to obtain 100 ml).
I measured the absorbance of 10 ml from the extract (diluted 5 times with acetone 90%) :
0.678 for 663 nm
0.338 for 645 nm
In order to find the concentrations in chlorophyl a/b, I used the Arnon equations :
Chl a = 12.7 (0.678) - 2.69 (0.338)
= 7.7 mg/L
Chl b = 22.9 (0.338) - 4.68 (0.678)
= 4.567 mg/L
Are my concentrations correct ?
Finally I struggle to find the dry mass of pigments from the spinach leaf (mg/g).
How can I do this ?
Thank you for your help !
Dear colleague, the equation is correct but you have a small mistake, you did a 5 times dilution that you didn't took in consideration, you have to multiply the equation on the dilution factor (5x), the absorbance values represents the diluted samples not the original ones.
Best regards.
Thank you very much ! I suspected a mistake in the concentrations, but I was confused by the first and second dilution.
Do you think that I should take the 10 ml volume in order to find the dry mass of pigments ?
My pleasure, do you mean the non diluted 10ml? If yes no its not a good idea you ll have very high absirbance values and your calculations will be wrong its better to work with those calculations, and if u want to determinate the concentration in dray weight take in concideration the both dilutions the first and the second one. Use the CiVi=CfVf to calculate both dilutions factors then multiply them together to have the total dilution factor (Fd=Fd1×Fd2), and finally multiply it on the equation.
Your help is very precious, I've got all the results !
I will bother you one last time : How can I consider my spectrophotography of chloro a/b ''pure'' as compared to the reference spectro value ?
My chloro ''a'' aborbs max at : 432/664 nm compared to the reference values : 432/663 nm.
My chloro ''b'' absorbs max at :458/648 nm compared to the reference values : 458/645 nm.
As you can see, they differ slightly along the 640-670 spectrum. Is it enough to consider them contaminated ? I'm not sure what is a pure spectrophotography.
My pleasure dear colleague and no bother at all, well its impossible to consider your extracts as pure chlorophyll because acetone extracts other secondary metabolites, the only way is to purify your extract with column chromatography before you do the spectrophotometric evaluation, else you ll consider your extracts as chlorophyll equivalent if you want to calculate the chlorophyll concentration, another thing its better to do for each sample 3 repetitions and calculate for each repetition the concentration in order to determinate the mean value and the standard deviation. good luck with your experiment and best regards
Your replies are completes, it's such a relief I'm geologist of profession and I'm not at ease with organic chemistry, thank you Doc ! I hope that your answers will help a lot of students/professionals.
I joined a chromatography (to download), and I struggled with the analysis of differents pigments from 3 deposits : spirula (cyanobacteria) /spinach (angiosperm) /chlorella (alguae).
Do you think all 3 pigments on the deposit line are same pigment ? They seem different, but I can't identify them, exept pheothytin for the grey pigment.
It seems cyanobacteria (spirula) lacks chlorophyll b, but I'm no expert.
Again, Thank you for all you have done !
Dear colleague, its hard to say that its the same compound, its better to dilute your extract more and redo two Thin layer chromatography like this one use the solvent systems (dichloromethane-MeOH (80:20) and (90:10)), then you can compare the Rf, here its very difficult to compare them because the spots are too big. If you have an UV lamp (366nm) visualise your plate under it, because chlorophylls give red fluoresent color. Best regards.
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