Absorption with UV/VIS spectrometer


by w.shockley
Tags: absorption, spectrometer, uv or vis
w.shockley
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#1
Mar25-13, 06:34 AM
P: 19
I need to characterize some colloids (extract the concentration of material inside).
Why should I dilute the sample before measuring it with spectrometer?
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chemisttree
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#2
Mar25-13, 05:53 PM
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The only time you would need to dilute is when it is too concentrated. Are you asking us why is it too concentrated?
Ben-P
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#3
Mar26-13, 12:27 AM
P: 3
Only at low concentrations is the relation between concentration and absorption linear. Thus, in order to calculate with any precision the concentration (using the Beer-Lambert law), a dilute solution must be used.

w.shockley
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#4
Mar26-13, 05:44 PM
P: 19

Absorption with UV/VIS spectrometer


thank you.
when I multiply the spectrum of the diluted solution, the absorbance is higher than the maximum (1).
how is it possible?
ldc3
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#5
Mar28-13, 08:10 PM
P: 13
Quote Quote by w.shockley View Post
when I multiply the spectrum of the diluted solution, the absorbance is higher than the maximum (1).
how is it possible?
Although the absorbance is the amount of light absorbed by the solution, we can only measure the amount of light that passes through the solution. The scale of the absorbance is set so that 1% transmittance is 1. Values greater than 1 are possible, but rather difficult to accurately determine.
w.shockley
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#6
Mar30-13, 06:03 AM
P: 19
on the machine I read a value for A, from 0 to 10.
What does it mean? does 10 correspond to 100%?
Is it correct to do this?:
A from machine = 1,
that means 10 % for me.
Then A=alpha*length*C
10=alpha*lenght*C -> C=...

I assume that alpha it was calculated assuming A ranging from 0 to 100.

thank you
ldc3
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#7
Mar30-13, 09:13 AM
P: 13
Quote Quote by w.shockley View Post
on the machine I read a value for A, from 0 to 10.
What does it mean? does 10 correspond to 100%?
Is it correct to do this?:
A from machine = 1,
that means 10 % for me.
Then A=alpha*length*C
10=alpha*lenght*C -> C=...

I assume that alpha it was calculated assuming A ranging from 0 to 100.
The absorption is non-linear.
Absorption = -log (Transmittance% / 100)

Here is a table of equal values http://www.fruitsmart.com/PDFs/Techn...sorbTxConv.pdf
ldc3
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#8
Mar30-13, 09:16 AM
P: 13
Quote Quote by ldc3 View Post
The absorption is non-linear.
Absorption = -log (Transmittance% / 100)
Or is it that the transmittance is non-linear?
Obviously, one of them is.
Darwin123
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#9
Mar30-13, 09:53 AM
P: 741
Quote Quote by w.shockley View Post
thank you.
when I multiply the spectrum of the diluted solution, the absorbance is higher than the maximum (1).
how is it possible?
It sounds like you either didn't put a reference in your spectrometer or you used the wrong reference.


Most absorption spectrometers work by using a "reference sample". One generally places two cuvettes in the spectrometer. One cuvette contains the unknown sample and one cuvette contains a reference material.

What one is generally looking for is not the absorbance of the entire unknown cuvette, but an absorbance of one particular component of the unknown sample. For instance, you may not be interested in the component of absorbance of the glass that the cuvette is made of. Therefore, you want to subtract the absorbance of the glass from the absorbance of the unknown cuvette. However, the reflectivity of the glass in the cuvette can also cause a false "gain" in the absorbance of your unknown sample. Therefore, one would want to fill the cuvette with a liquid of the same index of refraction as the index of refraction in your unknown sample.

This logic can continue. If want to know the absorbance of a solute molecule in water, you may not be interested in the absorbance of the water molecules. Therefore, you should fill your reference cuvette with water.

Most spectrometers automatically subtract the absorbance of the reference from the absorbance of the unknown. If the reference isn't chose appropriately, the uninteresting components of the reference will absorb more light than the uninteresting components of the unknown. That will read as a negative absorbance.

You may not need to take all he data again if that is your problem. You can measure the absorbance of an appropriate reference and subtract it from the measured absorbance of your unknown. That would eliminate the absorbance artifact in your absorbance data. You would be doing the arithmetic that the spectrometer does automatically.

One source of "gain artifact" would be if the width of the reference cuvette was narrower than the width of the unknown cuvette. Then, more light can pass by the reference cuvette than passes through the sample cuvette.

All cuvettes are not equal. Another cause of absorbance artifact is photoluminescence is a physical flaw in the cuvettes. Or maybe one of the cuvettes is dirty and the other is clean. Maybe you are using a glass cuvetter on the reference side and a silica cuvette on the sample side. Or maybe one cuvetter has glass one millimeter thick and the other cuvette has glass that is two millimeters thick.

You should chose cuvettes that are as similar as possible with the exception of the analyte that you are studying. It is okay to use a virtual reference if you have a very good idea of what the background in your unknown is going to be. By virtual reference, I mean you use theory to determine part of the background that you can't reproduce otherwise. However, it is usually better to find a physical reference that comes as close as possible to the background absorbance.
ldc3
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#10
Mar30-13, 10:45 AM
P: 13
Quote Quote by Darwin123 View Post
Most absorption spectrometers work by using a "reference sample". One generally places two cuvettes in the spectrometer. One cuvette contains the unknown sample and one cuvette contains a reference material.

...

Most spectrometers automatically subtract the absorbance of the reference from the absorbance of the unknown. If the reference isn't chose appropriately, the uninteresting components of the reference will absorb more light than the uninteresting components of the unknown. That will read as a negative absorbance.

...

All cuvettes are not equal. Another cause of absorbance artifact is photoluminescence is a physical flaw in the cuvettes. Or maybe one of the cuvettes is dirty and the other is clean. Maybe you are using a glass cuvetter on the reference side and a silica cuvette on the sample side. Or maybe one cuvetter has glass one millimeter thick and the other cuvette has glass that is two millimeters thick.

You should chose cuvettes that are as similar as possible with the exception of the analyte that you are studying. It is okay to use a virtual reference if you have a very good idea of what the background in your unknown is going to be. By virtual reference, I mean you use theory to determine part of the background that you can't reproduce otherwise. However, it is usually better to find a physical reference that comes as close as possible to the background absorbance.
A split beam would automatically subtract the absorbance of the reference from the absorbance of the unknown and matched cuvettes would be necessary.

Most spectrophotometer made now are single beam instruments with processor controls which will subtract the blank spectrum from the sample spectrum to display the difference, which we usually call the sample spectrum (since the blank doesn't have any absorbance in the range of interest). Since the same cuvette can be used, one does not need matched cuvettes anymore.
w.shockley
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#11
Mar30-13, 11:28 AM
P: 19
maybe the problem is a confusion between Absorption and Absorbance.
If I got it, Absorption is 1-T and Absorbance is log(1/T) in base 10.
Probably the machine give Absorbance, that goes from 0 to infinity, and it display it only from 0 to 10.
Is it usual?


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