# Chromatography Theory, Part III: Calculating Elution Times and Capacity Factors

Published on:
Cannabis Science and Technology, March 2023, Volume 6, Issue 2
Pages: 8-11

Part III of this series will delve into understanding how to optimize methods for chromatographic separations, Smith will also introduce the concepts of capacity factor and selectivity.

In our ongoing quest to understand how to optimize methods for chromatographic separations, we will see how to calculate retention times and introduce the concept of the capacity factor. The capacity factor normalizes retention time so that these quantities can be compared to each other even if separations are performed under different conditions. The capacity factor will help us predict how well analytes will be separated from each other, and we will find that this depends upon an important concept from the last column, the fraction of time the analyte spends on the stationary phase, Fs.

Following up on our discussion of quantitating the separation of mixtures from the last column, here we will discuss further how to calculate the expected retention time of an analyte, and then define what are called capacity factors. The capacity factor is a way of normalizing retention times so we can compare chromatograms that are measured, for example, at different flow rates.

### How to Calculate Elution Times

Recall that in the last column (1), we derived an equation that allows us to calculate the elution time for an analyte, as seen in Equation 1:

where te is the retention time, t0 is the void time in minutes, and Fs is the fraction of time spent adsorbed on the stationary phase.

Note that the quantity (1-Fs) is in the denominator. So as Fs goes up, that is, the higher the fraction of time an analyte spends adsorbed and immobile on the stationary phase, (1-Fs) goes down and hence te goes up. If you imagine our different types of analyte molecules are in a horse race as in the last column (1), Equation 1 says, for example, that a horse that spends more time standing around grazing, as horses are wont to do, and not moving will get to the finish line later than a horse that spends all its time running.

Equation 1 is, in my opinion, the most important equation in chromatography because it relates the quantity we care about as measured by a chromatograph, the elution time of an analyte, to a property of the analyte­—the fraction of time it spends on the stationary phase. We will be seeing Equation 1 a lot going forward because it will provide us with the theoretical foundation for understanding how different chromatography separation methods work. As I have mentioned (1), and will probably mention again, the tricky part in achieving a separation is getting Fs to be different for all the analytes in a mixture you wish to separate.

Figure 1 shows a chromatogram from last time (1). We assumed the sample was tetrahydrocannabinol (THC) dissolved in ethanol, and that the ethanol molecules had no affinity for the stationary phase, did not adsorb, and hence had a Fs value of 0. We also established last time (1) that the t0 for ethanol is 2 min. The elution time of ethanol then using Equation 1 is:

which shows that for unadsorbed molecules the elution time is the void time, which here is 2 min.

Now Equations 2-4 were an easy calculation because we were not dealing with an analyte. Last column we assumed (1) that our THC molecule spent ½ of its time on the stationary phase, or in other words its Fs is 0.5. How then would we calculate the retention time of this analyte?

Using Equation 1 as such:

The reason THC has an elution time of 4 min in Figure 1 is that 2 x 2 min = 4 min. Note that the elution time for THC is twice that of ethanol because it spent half its time stationary whereas the ethanol spent no time stationary.

The beauty of Equation 1 is that it allows us to calculate the retention time of an analyte as long as we know Fs. The practical problem here is that we rarely know Fs exactly because it depends upon several factors including the chemical natures of the analyte, mobile phase, and stationary phase. Thus, our lack of knowledge of the exact value of Fs means we must experiment with chromatographic conditions to optimize separations. However, Equation 1 is still true and can help guide our experiments. Given knowledge of the chemical structures of the analytes to be separated, one can deduce based on Equation 1 what types of experiments to perform to obtain different values of Fs for the analytes and hence achieve a separation. More on that in future columns.

### The Capacity Factor: k

As with all things chromatography, I refer you to publications (2,3) and references therein for a more detailed discussion of what is presented here.

So far, we have only considered a single chromatogram measured under controlled conditions. A problem with chromatography is that variations in the experimental variables affect retention time. For example, when we first discussed Figure 1 (1), we stated that the flow rate was 1 mL/min which gave t0 = 2 min. Now, if we slow down the flow rate for this separation to 0.5 mL/min, an unadsorbed analyte molecule will take twice as long to elute as before and give a t0 of 4 min. Note that in the calculation for the retention time of THC in Equations 5-7 they all depend on t0. Specifically, Equation 6 says that the retention time for THC equals 2t0. By slowing down our flow rate to 0.5 mL/min and increasing t0 to 4 min the elution time for THC is now 8 min, compared to 4 min at a flow rate of 1 mL/min. These results mean that separations performed at different flow rates have different retention times, which ultimately means retention times in chromatograms run at different flow rates cannot be compared to each other.

A practical problem with chromatography, particularly with the pumps used in liquid chromatography, is that a given pump may not give the same exact flow rate over time. Worse yet, pumps on different instruments set to the same flow rate may give measurably different flow rates. This means that elution times for the same sample can vary from instrument to instrument, which is a real problem when comparing data across instruments and developing methods.

Fortunately, there is a solution to this problem. We calculate for each peak in a chromatogram what is called its capacity factor, k. The capacity factor gives us a way of normalizing retention times so we can compare them across separations on different chromatographs. For example, capacity factors allow us to compare retention times when separations are performed at differing flow rates. The equation for calculating a capacity factor for a given chromatographic peak is seen in Equation 8:

where k is the capacity factor, te is the elution time, and t0 is the void time.

Let’s calculate the capacity factors for the two peaks in Figure 1. For ethanol te = t0, which means (te – t0) = 0 hence k = 0. This shows that for all void time peaks the capacity factor is 0. From Figure 1 te for THC is 4 min and t0 is 2 min, hence its capacity factor is:

Figure 2 shows the chromatogram from Figure 1 with its peaks labeled with their capacity factors.

We said above that if we slowed our flow rate down to 0.5 mL/min t0 for our separation becomes 4 min and the te for THC becomes 8 min. We have established that the capacity factor for all peaks at t0 is 0. What is the capacity factor for THC at this new flow rate?

Note that despite the change in flow rate the capacity factor for THC is still 1 as seen in both Equations 10 and 12. The reason for this is that in equation 8 the numerator is in units of time, the denominator is in units of time, and when we calculate k the two units cancel, meaning the capacity factor is a unitless quantity that does not depend upon flow rate. This means that by using chromatograms labeled with their capacity factors we can compare retention times to each other even if they are measured at different flow rates.

Figure 3 shows a chromatogram with a void volume peak and two analyte peaks. In this chromatogram t0 is 2 min and the analyte elution time for peak A is 4 min and the elution time for peak B is 8 min. What are the capacity factors for the peaks in this separation? Well, the first two peaks are the same as in Figure 1, so their capacity factors are still 0 and 1, respectively. The capacity factor for peak B, kb, is calculated as follows:

kb = 3

Peaks A and B are said to be baseline separated because there is no overlap between them and there is a flat baseline between them. Note that they also have quite different capacity factors, which indicates that the relative capacity factors of two peaks might be a measure of the quality of a separation. More on that next time.

### Conclusions

We presented the equation from last time for how to calculate elution time, emphasizing the importance of the fraction of time, Fs, an analyte spends adsorbed and immobile on the stationary phase. We then calculated some elution times based on assumed values of Fs. We found that elution times for the same separation measured at different flow rates are different, and that the capacity factor, k, is a flow rate independent quantity that normalizes retention time and allows us to compare retention times measured under different conditions. Finally, we saw that for baseline separated peaks their capacity factors are different.

References

1. Smith, B.C., Chromatography Theory, Part II: How Separations Occur Illustrated and Quantitated, Cannabis Science and Technology®, 2023, 6 (1), 7-12.
2. Skoog, D.A.; West, D.M.; and Holler, E.J.; Analytical Chemistry: An Introduction,Saunders College Publishing. 1994, 6.
3. Robinson, J.W.; Undergraduate Instrumental Analysis, Marcel Dekker, 1995, 5.