Internal Standard

[Chemist2]

Hello everybody,
I'm trying to use a method for the extraction and quantification of a PAH compound from fruit samples. The extraction is carried out by adding 20 ml of the extractant (n-hexane/Ethanol) containing 10ug/ml internal standard to about 10g of chopped fresh fruit sample. Then, the sample is sonicated for 10 mins, after which it is centrifuged for 5 mins to help separating the two layers. An aliquot of the hexane layer is taken and filtered into HPLC vials for quantification. 10ul is injected for quantification using a standard containing 15ug/ml of both the analyte and the internal standard.

The problem is that the calculations were not clear in the method. Could you please help me to find out how to get the concentration of the analyte in such a method?
Also, what is the purpose of the internal standard? Can you recommend any book or website talking about it.

Thank you in advance.

[DR]

Internal standard helps remove variability as a function of extraction effieiency.

To calculate, compute the ratio of analyte/internal standard peak areas, then treat the resulting quotients as you would the areas for external standard calculations (dilution factors * sample ratio * standard conc/avg std ratio...).

[Chemist2]

Thanks Dr.
Can I ask about the criteria that must be followed in choosing an internal standard?

thanks in advance.

[DR]

Choose something that
1) has a decent chromophore (absorbs at your preferred wavelength)
2) is well resolved from any peaks of interest coming from your samples
3) has no chance of being in your sample in any amount except as a result of your having added it
4) is fairly stable & will not react with your compound(s) of interest (ie - don't throw a primary amine in with a reducing sugar matrix)

[tom jupille]

I'll add to that list:
- behaves like your analyte as far as solubility and extraction
- ideally elutes relatively close to your analytes.

[Chemist2]

Thank you Dr and Tom,
But how can I test the similarity in the behavior of the analyte & th IS?

[tom jupille]

Spike a blank matrix and measure the recovery (should be similar for both the analyte and the proposed IS).

[Chemist2]

Thanks a lot

[Chemist2]

Hi
I tried to design an experiment to check the similarity of the IS and the analyte. However, I stopped again in the calculation step because after the extraction I get 2 layers with unknown volume of the extract. The method calculations depend on using the IS.
How can I get the rcovery of the 2 compounds here?

Thanks

Chemist2

[DR]

Hopefully, the IS you chose will be in the same layer as your analyte of interest. You add IS & analyte together, then extract, wash etc. The point of using IS here is to prevent differences in extract volumes from being a problem. If your extraction efficiency varies from sample to sample, it is not a problem because you're extracting IS w/ the same efficiency as your analyte.

[Chemist2]

Thank you DR
I spiked my matrix with the same amount of the analyte and the IS(same volume and conc.) then extracted the 2 compunds. I calculated the percentage of recoveries as foloows:

%Recovery of IS/Recovery of Analyte= [Recovered IS(ug/ml)/Rec Analyte (ug/ml)]*100

I got values of about 90% in 5 replicates.

Do you think this compound is suitable as an IS for this analyte?
How would this %recovery affect the results obtained by using this IS?
Do you think the differences in the recoveries may be due to differences in the distribution of the 2 compounds between the 2 layers? How can I check that?

Sorry for my too many qustions.

Thank you all in advance.

[tom jupille]

The real question is what was the standard deviation in the relative recovery (i.e., 90% +/- what?). If it was much more than 1%, my guess is that you need to find a different internal standard.

[Chemist2]

Thanks Tom but why should we ask about the standard deviation instead of the real value of the %recovery?What can we know from the two?

[tom jupille]

While it would be nice to have identical recoveries (i.e., 100% relative recovery) for the IS and the analyte, so long as the recoveries are the same, then the IS is usable.

The whole idea behind internal standardization is that the use of an IS will cancel out "correlated" errors (i.e., errors which affect the IS and analyte peaks in the same proportion). The use of an IS will exacerbate uncorrelated (independent) errors.

If there is a lot of variation in your relative recovery, then the errors in IS and analyte are poorly correlated, hence the interest in the variation of the relative recovery.

[Chemist2]

Thanks Tom
The SD was less than 1% but I am still worry about the results I get using this IS. I think I get higher values of the analyte by this method due to the %recovery compared to a method depends on using known extract volumes.
Do you agree with me on this point?