Process Gas Chromatographs. Tony Waters
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to the migration rate of the peak along the column. The word “migration” infers a gradual movement by alternately stopping and going, to remind us what is really going on in there.
It should now be clear why the propane peak in Figure 2.9 is exactly in the middle of the column. We assumed that 50 % of the propane molecules dissolve in the liquid phase and 50 % remain in the gas phase. Remember that those molecules are rapidly moving between the two phases. So the average propane molecule spends 50 % of its time in the gas phase moving at carrier speed and 50 % of its time in the liquid phase going nowhere. Therefore, when the carrier gas that was present during sample injection reaches the end of the column, the propane peak is exactly half‐way.
Figure 3.1 applies this logic to two additional peaks, one more soluble and one less soluble than propane. The 1‐butene peak is more soluble, so we would expect it to spend more time in the liquid phase and elute from the column later than the propane. And so it does. For instance, if the solubility of 1‐butene is 75 %, only 25 % of the molecules move each time the carrier gas moves, and the peak stays in the column twice as long as the propane peak does. The carbon dioxide peak is less soluble: If its solubility is 25 %, then 75 % of its molecules move with the carrier gas.
Drawing peaks above a column is a common way to show the position of component molecules within the column at a certain instant of time. Here, an air peak (which doesn't dissolve in the liquid phase) has moved along with the carrier gas and has just arrived at the column end. At that instant, the three colored peak drawings indicate the position within the column of the other component molecules, predictable from their solubility in the liquid phase.
Figure 3.1 Effect of Component Solubility.
This is the true cause of separation. When in the gas phase, all sample molecules move along the column at the same speed as the carrier gas. But when in the liquid phase, the molecules stop moving and the more soluble ones stop longer than the less soluble ones do.
Figure 3.1 illustrates the net effect of solubility difference. It shows the location of four components peaks at the exact moment the air peak reaches the end of the column. A small white‐and‐blue equilibrium diagram indicates the solubility of each peak.
Figure 3.1 assumes that:
The stationary phase is a liquid, and the air peak is not soluble at all. The air peak therefore moves with the carrier gas, and its retention time is a good indicator of average carrier gas velocity.
The carbon dioxide (CO2) solubility is 25 %, so an average CO2 molecule spends 75 % of the time traveling and only 25 % of the time stopped. Therefore, the CO2 peak moves 75 % of the distance that the carrier gas moves.
The propane solubility is 50 %, so an average propane molecule spends half the time traveling and half the time stopped. Therefore, the propane peak moves 50 % of the distance that the carrier gas moves.
The 1‐butene solubility is 75 %, so an average molecule spends only 25 % of the time traveling and 75 % of the time stopped. It follows that the 1‐butene peak moves only 25 % of the distance that the carrier gas moves.
These convenient values for the component solubilities are just simple examples assumed for discussion purposes, but every substance has a real solubility in a given liquid that depends only on the temperature and pressure.
Figure 3.1 shows the location of each peak when the air peak has reached the end of the column and is about to flow into the detector. At that moment, the four peaks in our example are equally spaced along the column. Similar spacing is a common occurrence in real columns, but a very strange thing then happens to the chromatogram. The exercise that follows challenges you to discover what comes next.
A challenge question
Imagine that the effluent from the column in Figure 3.1 flows into a detector and the detector signal is recorded in the form of a chromatogram. For this exercise, assume Figure 3.1 shows the correct position of the four peaks in the column at three minutes after sample injection.
Now predict what the chromatogram will look like after all the peaks have exited the column.
It's worth your time to stop for a moment and try to do this; you will learn a lot from the exercise. Be careful; it's more difficult than it looks!
You can draw your chromatogram on Figure 3.2. Sketch the chromatogram you would expect to get, showing your chosen time base in minutes and the exact positions of the four peaks.
Based on the information given in Figure 3.1, draw the chromatogram you would expect to see after all four peaks have passed through the detector. Your chromatogram should show the four peaks at their correct retention times. The injection marker is time‐zero on your chromatogram, but you must decide the time scale. In this exercise, don't worry about peak widths.
Figure 3.2 Draw Your Own Chromatogram.
How confident are you of your chromatogram? People rarely get it right at first attempt. The most common mistakes are:
Inserting all four peaks between 0 and 3 mins on the chromatogram
Assuming all four peaks are equally spaced
Assuming three of the peaks are equally spaced
Assuming any peak has a fractional retention time (none do)
Remember to measure retention times from sample injection to the apex of each peak; i.e., the retention of the average molecule.
Want to try again? We'll reveal the correct answer later.
Significance of the air peak
The air peak is a valuable indicator of column performance. Since air doesn't dissolve in column liquids to any significant extent, the air peak remains in the gas phase all the way through the column – traveling at full gas velocity. We call it an unretained peak. Its position on the chromatogram indicates the elapsed time for the carrier gas to travel from one end of the column to the other end.
If using a detector that doesn't respond to air, another unretained peak can act as a surrogate. For instance, methane often serves as an adequate “air peak” on a flame ionization detector.
Since we know the column length (L), the air peak retention time (tM) allows us to calculate the average carrier gas velocity (uM) in m/s:
(3.1)