Calculation of concentration in a gas for GC calibration

Ramon_FiBL · Anmeldungsdatum: 31.07.2023 Beiträge: 2

Meine Frage:
Hi guys

Can you please check and confirm following calculation:

Given:

If we have a Gas containing
1.002 mol-% H2
52.068 mol-% N2/O2
8 mol-% CH4
38.93 mol-% CO2

We take a syringe to inject into the GC system.
Injection volume is 250 µL and split is 1:1

I want to know the amount of CO2 and CH4 in mg that are effectively in the system. I would like to use this gas as a standard for calibration.

Meine Ideen:
Calculation:

Mol in the column:
22.41 L contains 1 mol of gas
22.41 L * 1000000 = 22410000 µL

We only have 125 µL injected (250 µL / 2) (split ratio)
1 mol / 22410000 µL * 125 µL = 0.0000056 mol

Now this is at 0°C, the Lab has approx. 20 °C
(1.013*22.41*293.15)/(273.15*1.013) = 24.05? L at 20°C

0.0000056 mol / 22.41 L * 24.05 L = 0.00000599 mol
Which are on the column/detector

Now calculated with the corresponding gases:
0.00000599 mol * 16.04 g/mol = 0.00000768 g = 0.0077 mg of CH4
0.00000599 mol * 44.01 g/mol = 0.0001026 g = 0.1026 mg of CO2

So the absolute injected content of those 2 gases are the values we use for calibration.

It irritates me that though CH4 is much lower in content, the peak area of above injected sample is higher from CH4 than CO2.
Is that due to the difference in response of different molecules on the detector?

Nobby · Moderator Anmeldungsdatum: 20.10.2014 Beiträge: 6440 Wohnort: Berlin

Your calibration mixture is 50 mol % : 50 mol % CH4 and CO2, what means 5,99 μmol each? Where are the percentage in your calculation. Or does your sample later has this mixture?
1.002 mol-% H2
52.068 mol-% N2/O2
8 mol-% CH4
38.93 mol-% CO2

I would do a calibration first with each gas separate as inert gas Nitrogen and then values which is close to the sample, for CO2 you can keep it for CH4 only 10 mol% to get comparable peaks.

Ramon_FiBL · Anmeldungsdatum: 31.07.2023 Beiträge: 2

Hi Nobby

Sorry for the confusion.
The calibration mixture is:
1.002 mol-% H2
52.068 mol-% N2/O2
8 mol-% CH4
38.93 mol-% CO2

I basically just want to know if my calculation is correct.

Kind regards

Nobby · Moderator Anmeldungsdatum: 20.10.2014 Beiträge: 6440 Wohnort: Berlin

The calculation is correct until here