Chromatography Forum

Got it fixed. It was an easy fix.

https://i.postimg.cc/FsCKy8c2/Methane-b ... alysis.jpg

Here's the 1-page summary. Some example chromatograms. Triplicate analysis of my 25:21 dilution (how I've described above). When I used an old calibration from May, 2024 (from 0 to 100% methane), I say that my samples are 70.1% (v/v) methane. My target was 69.6%. Pretty good agreement (101% recovery).

1.5% RSD on my triplicate measurements of the dilution of the same gas. Pretty good there too.

The water is a waste of water and time. A properly filled empty vial should work just fine.

Thanks for everyone's suggestions. I finally found a leak (it had to be a leak!). The adapter where the column connects to the FID had been bent somehow. I think that I missed it because I had been looking for nitrogen leaks with the FID gasses off. When I finally checked again with the FID gasses on, I picked up hydrogen coming into the oven.

I have a good curve relating propane concentration to FID response.

The method's equation to convert measured concentration (as propane) to organic vapor pressure is:

organic vapor pressure = β * Pbar * Ca

Pbar = Atmospheric pressure at analysis conditions, mm Hg
β = 1.333 × 10−7kPa/[(mm Hg)(ppm)]
Ca = Measured vapor phase organic concentration of sample, ppm as propane.


The remaining questions I have are:

Why does this equation work?

What is β?

Is Pbar the local atmospheric pressure, or that plus the pressure added to the vial by the HS analyzer?

If I put in 1,000,000 ppm into the equation, shouldn't it spit out the vapor pressure for propane? But... at what temperature?

At constant temperature and volume, from the gas law, P1/n1 = P2/n2. What you're shooting for is the partial pressure (P2) of propane.

n2/n1*P1 = P2

What you measure is ppm C3H8. That's µmole C3H8/mole gas. n2/n1 is the concentration of C3H8. There are 7.50062 mmHg/KPa. There are also 1,000,000 µmole C3H8/mole C3H8. If you want P2 in KPa, to convert the ppm C3H8 you measure to KPa you have to multiply the measured ppm C3H8 by 1/7.50062x10^6 and the atmospheric pressure in mmHg. That factor is 1.333 x 10^-7.

KPa = 1.333 x 10^-7 x Atmospheric Pressure (mmHg) x ppm C3H8

Thank you so much for the effort you put into answering my question. This was really helpful.

So to simplify units:

pressure of propane (mmHg) = ppm propane / 1,000,000 * atmospheric pressure (mmHg)

Would you agree that atmospheric pressure should be the pressure in the vial, i.e. ~1 atmosphere plus the pressure the HS adds (in this case, 14 psi)? I'm not sure because it doesn't pressurize it during equilibrium, just at the end for 3 minutes to sample.

... then again the whole point of doing this is to find what the vapor pressure will be of the sample in the real world...

That is correct. I believe your gas is going into the vial with the water at atmospheric pressure so that's it.

You only need the 7.50062 mmHg/KPa if your final result must be in KPa.

You are welcome. It turns out that this might be useful for me too. I have an unknown that I'm working on - might be microbiological in nature and perhaps a methanogenic bacterium - that's causing us some issues. Knowing the methane concentration in a gas sample might lead me to the root cause. We'll see!

Nope. The cause of the pressure increase was not due to methane............... Ugh!