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is ELSD similiar to FID in terms of quantification
Discussions about HPLC, CE, TLC, SFC, and other "liquid phase" separation techniques.
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can I use % analyte =100*(analyte peak/SUM(all peaks)) for ELSD if I only need semi-quantification result.
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ELSD's are non-linear.
In ELSD each compound will generate a different resopnse factor (i.e. for the same mass on column, you can get very different peak heights, areas).
I would hesitate to do what you are planning to do.
In ELSD each compound will generate a different resopnse factor (i.e. for the same mass on column, you can get very different peak heights, areas).
I would hesitate to do what you are planning to do.
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ELSD is less linear than UV, linear over a smaller range. But molecules that are different can have closer detector responses.
For example, if a sample contains trace naphthalene, the UV peak could be huge, but just a small amount present, due to its molecular absorptivity. So just realize the limitations of each.
For example, if a sample contains trace naphthalene, the UV peak could be huge, but just a small amount present, due to its molecular absorptivity. So just realize the limitations of each.
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- tom jupille
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As AA and CPG have suggested, the answer is "maybe".
An ELSD essentially detects "dust" left behind when the solvent is evaporated from the droplets that come out of the nebulizer. To a first approximation, "dust is dust" and the response to each dust particle does not depend on the chemical structure of the non-volatile material. So, there is much less variation in response factors with ELSD than with UV.
That said, the presence of analyte may affect the number and/or size distribution of dust particles formed (e.g., by changing the surface tension of the column effluent or by affecting the rate of evaporation of solvent), so it's unrealistic to expect response factors to be exactly the same even for an isocratic separation (with a gradient of course, droplet formation may change quite a bit as a function of the mobile phase composition).
The relatively narrow (actually, nonexistent!) linear range is, to my mind, much less of an issue; I have no problems with appropriate mathematical transforms -- we do it all the time with UV detectors, which actually measure transmittance and then compute absorbance from that.
Bottom line: for "semi-quantitative" estimate, ELSD area% is probably a reasonable choice (certainly better than UV), but don't expect too much!
An ELSD essentially detects "dust" left behind when the solvent is evaporated from the droplets that come out of the nebulizer. To a first approximation, "dust is dust" and the response to each dust particle does not depend on the chemical structure of the non-volatile material. So, there is much less variation in response factors with ELSD than with UV.
That said, the presence of analyte may affect the number and/or size distribution of dust particles formed (e.g., by changing the surface tension of the column effluent or by affecting the rate of evaporation of solvent), so it's unrealistic to expect response factors to be exactly the same even for an isocratic separation (with a gradient of course, droplet formation may change quite a bit as a function of the mobile phase composition).
The relatively narrow (actually, nonexistent!) linear range is, to my mind, much less of an issue; I have no problems with appropriate mathematical transforms -- we do it all the time with UV detectors, which actually measure transmittance and then compute absorbance from that.
Bottom line: for "semi-quantitative" estimate, ELSD area% is probably a reasonable choice (certainly better than UV), but don't expect too much!
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
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Wow!
thanks so much. For UV, A =log(I0/I). This is definitely a mathematical transformation.
Do we have some sort of equation similar to lamb-beer law for ELSD?
thanks so much. For UV, A =log(I0/I). This is definitely a mathematical transformation.
Do we have some sort of equation similar to lamb-beer law for ELSD?
As AA and CPG have suggested, the answer is "maybe".
An ELSD essentially detects "dust" left behind when the solvent is evaporated from the droplets that come out of the nebulizer. To a first approximation, "dust is dust" and the response to each dust particle does not depend on the chemical structure of the non-volatile material. So, there is much less variation in response factors with ELSD than with UV.
That said, the presence of analyte may affect the number and/or size distribution of dust particles formed (e.g., by changing the surface tension of the column effluent or by affecting the rate of evaporation of solvent), so it's unrealistic to expect response factors to be exactly the same even for an isocratic separation (with a gradient of course, droplet formation may change quite a bit as a function of the mobile phase composition).
The relatively narrow (actually, nonexistent!) linear range is, to my mind, much less of an issue; I have no problems with appropriate mathematical transforms -- we do it all the time with UV detectors, which actually measure transmittance and then compute absorbance from that.
Bottom line: for "semi-quantitative" estimate, ELSD area% is probably a reasonable choice (certainly better than UV), but don't expect too much!
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- tom jupille
- Site Admin
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Unfortunately, none that I am aware of.Do we have some sort of equation similar to lamb-beer law for ELSD?
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
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Thanks all the same.
Unfortunately, none that I am aware of.Do we have some sort of equation similar to lamb-beer law for ELSD?
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If I understand your question correctly, for ELSD we have the equations
y = Am^b and Log y = b Log m + Log A
Where:
y = Detector response (peak area)
A = Constant that changes with mobile phase and analyzed compound
m = Injected solute mass
b = Constant that changes with mobile phase and analyzed compound
[Generally, 0.67 - 2 depending on scattering mechanism (refraction-refraction to Rayleigh)]
The above equations can also be found in several reputable sources.
y = Am^b and Log y = b Log m + Log A
Where:
y = Detector response (peak area)
A = Constant that changes with mobile phase and analyzed compound
m = Injected solute mass
b = Constant that changes with mobile phase and analyzed compound
[Generally, 0.67 - 2 depending on scattering mechanism (refraction-refraction to Rayleigh)]
The above equations can also be found in several reputable sources.
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- tom jupille
- Site Admin
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Yes, but those equations don't actually tell us very much; they say that the response "sort of" increases with increasing mass:
it's log-log, but the intercept can be pretty much anything and the slope can vary by a factor of 3 (and, in a gradient, both are likely to be different from one point on the gradient to another).
it's log-log, but the intercept can be pretty much anything and the slope can vary by a factor of 3 (and, in a gradient, both are likely to be different from one point on the gradient to another).
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
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