Low level alcohol/spirit analysis GC-FID

[trlizard52]

Hello,

My configuration;
Agilent 6890N GC FID
Column1: CP WAX 57 CB, 50m 0,25mm 0,20um
Column2: DB 624 UI 60m, 0,32mm, 1,80um
Method: Commission Regulation (EC) No 2870/2000

According to REGULATION (EU) 2019/787;

Article 5
Definition of and requirements for ethyl alcohol of agricultural origin

(d) its maximum levels of residues do not exceed the following:

(i) ethyl acetate: 1,3 grams per hectolitre of 100 % vol. alcohol;

(ii) acetaldehyde (sum of ethanal and 1,1-diethoxyethane): 0,5 grams per hectolitre of 100 % vol. alcohol;

(iii) higher alcohols (sum of: propan-1-ol, butan-1-ol, butan-2-ol, 2- methylpropan-1-ol, 2-methylbutan- 1-ol and 3-methylbutan-1-ol): 0,5 grams per hectolitre of 100 % vol. alcohol;

We want higher alcohols to be at or below the LOQ of 0.3 grams per hectolitre of 100 % vol. alcohol - The LOQ of each of the 6 alcohols that make up the higher alcohols is approximately 0.05 grams per hectolitre of 100 % vol. alcohol.

Is it possible to get down to these levels?

[rb6banjo]

0.05 g per hectoliter is 0.5 ppm. Could be tricky with direct injection. That's pretty small. I don't do a lot of direct injection but you might be able to get it all if you use a splitless injection.

Your problem children are those low level materials.

[CharapitsaS]

Dear trlizard52,

the answer to your question can be obtained directly in an experiment on your existing GC Agilent 6890N and with a capillary column, for example, Column2: DB 624 UI 60m, 0,32mm, 1,80um.
You don't have to perform calibration, and for the Agilent 6890N GC FID it is enough to use the tabular values of the relative response factors RRF of FID to the analyzed volatile compound relative to the response of FID to ethanol.

Compound RRFethanol
methanol 1.166
2-butanol 0.631
n-propanol 0.649
2-methylpropan-1-ol 0.548
n-butanol 0.589
2-methylbutan-1-ol 0.540
3-methylbutan-1-ol 0.545

Based on the measured chromatogram data of the prepared standard ethyl alcohol solution with quantitatively added 2-methylpropan-1-ol with a concentration of 0.3 grams per hectolitre of 100 % vol. alcohol, you will receive the value of the experimentally measured concentration of 2-methylpropan-1-ol immediately in the required dimension [grams per hectolitre of 100 % vol] according to the following formula:

Сoncentration (2- methylpropan-1-ol) = Peak Area (2- methylpropan-1-ol)/ Peak Area (ethanol)*RRF (2- methylpropan-1-ol) * density (ethanol),

where density of ethanol is equal to 78927000 grams per hectoliter.

The version of Commission Regulation (EC) No 2870/2000 using the modernized internal standard method is presented here https://elab.inpnet.net/article/355?l=en

I am confident that the Agilent 6890N GC FID equipped with a capillary column DB 624 UI 60m, 0.32mm, 1.80um will provide a confident determination of 2-methylpropan-1-ol with a concentration of 1 mg/L AA.

Best regards,
Siarhei Charapitsa