Chromatography Theory, Part II: How Separations Occur Illustrated and Quantitated

At the end of the last column (which you can find online at www.cannabissciencetech.com/view/principles-of-chromatography-part-i-theory), we had injected a sample at the head of a chromatography column and were waiting to see what happened next. In this second installment in our discussion on chromatography theory, we discuss via the mechanisms of equilibrium, adsorption, and de-adsorption how molecules move through a chromatography column and how separations occur. A critical variable in determining the retention time of molecules is the fraction of time they spend adsorbed to the stationary phase. Simply put, if you can arrange your chromatography experiment such that the analyte molecules spend different amounts of time adsorbed, a separation is possible.

How a Separation Occurs: Illustrated

For more background on the discussion contained here, I refer you to the references (1,2) and my recent column on chromatographic theory (3). Using a horse race analogy to describe a chromatographic separation may seem silly, but in the end, it works well. At injection time all molecules are “in the starting gate” as illustrated in Figure 1 (note that I left the stationary phase out of the figures below for ease of viewing, but we will assume it is there and doing its job). The purple block at the head of the column in Figure 1 is our sample immediately after injection.

Imagine the sample contains tetrahydrocannabinol (THC) dissolved in ethanol, and that the mobile phase is composed of water and acetonitrile. Thus, this example discusses liquid chromatography (LC), but the principles developed here apply to both gas chromatography (GC) and LC. Both LC and GC are capable of measuring cannabinoid concentrations in samples (4,5). The water–acetonitrile mobile phase mimics the conditions of my favorite LC cannabis potency method (5).

Upon injection some of the THC molecules will adsorb on the stationary phase, indicated in red in Figure 1, and some will stay dissolved in the mobile phase, indicated in blue in Figure 1. The purple rectangle at the head of the column represents what I call a slug of sample, but a scientifically better term is band. The band is purple because the red stationary phase molecules and the blue mobile phase analyte molecules occupy the same length of column, and when you combine red and blue you get purple. Recall that the front edge of the band of sample is known as the solvent front (3).

Above the head of the column in Figure 1 you can see the designation in red of C_s, which denotes the concentration of analyte adsorbed on the stationary phase, and the height of the red rectangle gives the size of C_s. The designation below the column in blue, C_m, is the concentration of analyte in the mobile phase. The height of the blue rectangle below the column gives the magnitude of C_m. The red and blue lines to the right of the red and blue rectangles in Figure 1 show that there is no analyte in the portion of the column just beyond the solvent front.

Now, at this point in the injection we have some THC molecules adsorbed and some in solution, which occupy the same length of column. However, because of the dynamic nature of molecules this situation is not static, and at any given point in time some molecules are in the process of adsorbing on the stationary phase and others are de-adsorbing and entering the mobile phase. The term “equilibrium” at the top of Figure 1 means we are assuming that the rates of adsorption and de-adsorption are the same. We will also assume for the sake of simplicity that THC has an affinity for our stationary phase, and that at equilibrium half the molecules are adsorbed on the stationary phase and half are dissolved in the mobile phase. This is why the red and blue rectangles in Figure 1 are the same size. We will also assume that the ethanol molecules that the THC was dissolved in have no affinity for the stationary phase and spend all their time in the mobile phase. The arrow of time points to the right in Figure 1 denoting the direction of mobile phase flow.

Figure 2 illustrates the situation in our chromatography column, 10 s after injection.

Note that C_s is the same as before because the molecules that were adsorbed are still adsorbed and are indicated by the pink band at the head of the column. The blue rectangle denoting C_m has moved down the column because these molecules are dissolved in the mobile phase. As a result, C_m at the column head is now zero. In Figure 2 we call this situation disequilibrium because the stationary phase and mobile phase analyte molecules are no longer in equilibrium because they are in different sections of the column and no longer in contact with each other.

Figure 3 represents the situation inside our chromatographic column 20 s after injection.

Note that the pink slug of analyte molecules that was at the head of the column in Figure 2 is now gone. These adsorbed molecules had fresh mobile phase above them containing no analyte molecules, and so given the opportunity de-adsorbed and dissolved in the mobile phase. Therefore, C_m has gotten bigger to the left in the column compared to Figure 2.

On the other hand, the molecules that in Figure 2 were in the mobile phase now have a piece of column where no molecules were adsorbed and have taken the opportunity to adsorb. This is why C_s is higher to the right in Figure 3 compared to Figure 2, and the net effect is that the adsorbed molecules have shifted to the right. Now molecules are adsorbing and de-adsorbing so the phrase “adsorption/de-adsorption” appears in Figure 3.

Figure 4 illustrates the situation in our chromatography column 30 s after injection.

C_s is the same as in Figure 3 because the adsorbed molecules have stayed adsorbed, hence the red rectangles in Figures 3 and 4 are in the same place. However, the molecules that were in the mobile phase in Figure 3 have moved to the right due to the flow of the mobile phase and are now above a section of column with adsorbed molecules. Hence the blue rectangle representing C_m has moved to the right in Figure 4 compared to Figure 3. Now that we have mobile phase containing analyte molecules above stationary phase with absorbed analyte molecules, equilibrium is re-established, and we find ourselves in a situation similar to Figure 1, which is why our sample band is purple again. However, compared to Figure 1 the net change is that our band of analyte molecules has moved to the right. Thus, figures 1 through 4 illustrate the mechanism via which an analyte moves down a column.

Figure 5 shows the situation in our chromatography column after 40 s.

Now, we begin the whole process over again. Molecules in the mobile phase have moved to the right in the figure hence the blue rectangle representing C_m has moved compared to Figure 4, whereas the adsorbed molecules have stayed put, giving the pink rectangle in the column in the figure and the fact that the red column representing C_s has not moved. Thus, in this fashion adsorbed and mobile phase molecules “leapfrog” each other down the column until they reach the end.

How a Separation Occurs: Quantitated

Our discussion so far explains how a band of analyte molecules moves down a chromatography column, but how does chromatography achieve a separation when there are multiple analytes present?

The total time it takes for any molecule to exit the column, the elution time, is the sum of the time it spends adsorbed on the stationary phase and the time it spends moving with the mobile phase, as given by Equation 1:

where t_e is the elution time, t_m is the time spent in mobile phase, and t_s is the time spent in the stationary phase.

Remember that our sample is a mixture of ethanol and THC, and we assumed ethanol had no affinity for the stationary phase. This means for ethanol t_s is zero, and it will move through the column at the flow rate (3) of the mobile phase. The time it takes a nonadsorbing molecule to exit a column is t₀ or the void time, and is given by the flow rate and void volume (the amount of space in the column not occupied by stationary phase), as seen in Equation 2:

where t₀ is the void time in minutes, V₀ is the void volume in mL, and f is the flow rate in mL/min.

Assuming for our THC–ethanol separation the void volume is 2 mL, and the flow rate is 1 mL/min, the ethanol molecules will elute at (2 mL/1 mL/min) or 2 min, which is t₀ for this separation. Figure 6 shows a chromatogram with a peak with a t₀ of 2 min that we will assume is from unabsorbed ethanol molecules.

Now, t_m for all molecules is the same because when they are in the mobile phase they all flow at the same speed, the flow rate. What needs to vary then for molecules to elute from a column at different times is for them to have different values of time spent in the stationary phase, t_s. To use our horse race analogy, imagine one horse runs a straight, unobstructed path to the finish line while another has fences, trees, and barriers to go over or around. Naturally, the horse with the unobstructed path finishes first, or in our case the molecule that spends the least amount of time adsorbed on the stationary phase elutes first.

What determines t_s? In a nutshell, the affinity of the molecule for the stationary phase. This then is the necessary condition that needs to be met for a chromatographic separation to take place. A large part of the art and science of chromatographic method development is making sure all the analytes in a mixture have different elution times. More on that later.

All this means there hopefully is a mathematical relationship between elution time, the quantity we care about in a separation, and t_s. To derive this equation, we first note that the fraction of time analyte molecules spend in the mobile phase and the fraction of time spent adsorbed on the stationary phase equals 1 as follows:

where F_m is the fraction of time spent in mobile phase and F_s is the fraction of time spent in the stationary phase.

Equation 3 is true because an analyte molecule can partition between two phases, stationary and mobile, so the total time spent in the column is the sum of the time spent in each phase, or 1 in this case.

The fractions of time spent mobile and stationary can be obtained from a chromatogram as follows:

where all these variables are defined above, and realizing that for a given separation t_m is t₀.

If we take the relationship for F_m from Equations 4 and substitute it into Equation 3 we obtain

Rearranging Equation 5 to solve for t_e gives:

where t_e is the retention time, t₀ is the void time in minutes, and F_s is the fraction of time spent in the stationary phase.

Note that the denominator in equation 6 contains the quantity (1-F_s). As the fraction of time an analyte spends stationary F_s goes up, (1-F_s) goes down and hence te goes up. This is the mathematical relationship between F_s and retention time we have been seeking. Given our example for THC, F_s is 0.5, (1 – F_s) = 0.5 and t₀ is 2 min. This means in this case THC will elute at (2 min/0.5) or 4 min, which is the second peak seen in Figure 6.

Equation 6 is in my opinion the most important equation in chromatography because it relates elution time, the x-axis quantity we care about that is derived from a chromatogram, to the fraction of time an analyte spends adsorbed, which is determined in turn by the chemical properties of the analyte, stationary phase, and mobile phase. Simply put, if you can get F_s to be different for each analyte in a mixture, you can in theory achieve a separation.

Drawing Analogies to Spectroscopy

For those of you who are spectroscopically inclined, such as myself, I draw the analogy between Equation 6 and the formula for the peak positions in infrared spectra seen in Equation 7 (6):

where W is the peak position in wavenumber (cm^-1), c is the speed of light, k is the force constant for a chemical bond, and µis the reduced mass for a chemical bond.

The force constant, k, is related to chemical bond strength. The reduced mass, µ, is calculated from the mass of the two atoms in a bond. Note that k and µ are chemical properties that are different for different chemical bonds. For example, the values of k and µ for a C-H bond are different than for a C=O bond, which is why they have different peak positions in infrared spectra (6). Equation 7 is in my opinion the most important equation in infrared spectroscopy because it relates peak position, the x-axis quantity we care about that is derived from a spectrum, to the chemical properties of the analyte. Similarly, equation 6 relates elution time, the x-axis quantity we care about that is derived from a chromatogram, to the chemical properties of the analyte, stationary phase, and mobile phase. Thus, the two most important equations in spectroscopy and chromatography relate peak position on the x-axis to molecular properties. Note that equation 6 depends upon the experimental setup, such as the nature of the mobile phase, stationary phase, and analyte, whereas infrared peak positions depend upon the chemical structure of the analyte only. More on this later.

Conclusions

The passage of analyte molecules through a chromatographic column is analogous to a horse race, with different molecules passing through the column at different speeds. Via words, pictures, and equations we found that a key quantity that determines elution time is the fraction of time an analyte spends adsorbed on the stationary phase, and that this is a function of the chemical properties of the analyte, mobile phase, and stationary phase.

References

1. Skoog, D.A.; West, D.M.; and Holler, E.J.; Analytical Chemistry: An Introduction, 6th Edition, Saunders College Publishing: New York, NY, 1994.
2. Robinson, J.W.; Undergraduate Instrumental Analysis 5th Edition, Marcel Dekker, New York, 1995, Boca Raton, 2011.
3. Smith, B.C.; Principles of Chromatography, Part I: Theory, Cannabis Science and Technology®, 2022, 5 (9), 8-11.
4. Giese, M.W.; Lewis, M.A.; Giese, L.; and Smith, K.M.; Journal of AOAC International 2015, 98 (6), 1503.
5. Ruppel, T.; Kuffel, M.; Cannabis Analysis: Potency Testing Identification and Quantification of THC and CBD by GC/FID and GC/MS, PerkinElmer Application Note, 2013.
6. Smith, B,C.; Infrared Spectral Interpretation: A Systematic Approach, CRC Press, Boca Raton, 1999.

About the Columnist

Brian C. Smith, PhD, is Founder, CEO, and Chief Technical Officer of Big Sur Scientific. He is the inventor of the BSS series of patented mid-infrared based cannabis analyzers. Dr. Smith has done pioneering research and published numerous peer-reviewed papers on the application of mid-infrared spectroscopy to cannabis analysis, and sits on the editorial board of Cannabis Science and Technology. He has worked as a laboratory director for a cannabis extractor, as an analytical chemist for Waters Associates and PerkinElmer, and as an analytical instrument salesperson. He has more than 30 years of experience in chemical analysis and has written three books on the subject. Dr. Smith earned his PhD on physical chemistry from Dartmouth College.
Direct correspondence to: brian@bigsurscientific.com

How to Cite this Article:

Smith, B., Chromatography Theory, Part II: How Separations Occur Illustrated and Quantitated, Cannabis Science and Technology® 2023, 6(1), 7-12.