Learning Goals #
After this first class of the course you will be able to
- evaluate fundamental properties of separation systems, such as retention times, retention factors, selectivity, plate height and resolution.
- distinguish and relate retention, selectivity and efficiency in method development.
- understand how chromatographic separations are governed by fundamental thermodynamic and kinetic concepts such as distribution coefficients, diffusion, and flow.
- calculate the various properties to characterize your separation.
READ SECTION 1.1.1
Chromatographic Separation
1. Chromatography #
In chromatography a mixture of analytes can be separated based on differential partitioning of the analytes between two (or more) phases that move at different velocities.
1.1. Dynamic equilibria #
It relies on the distribution of analytes between a mobile and a stationary phase. This may be through absorption, where the volume of the stationary phase (e.g. a liquid or soft polymer) matters, or through adsorption, where the surface area of a solid adsorbent matters.
Absorption is a process in which one substance is taken up into the bulk (volume) of another, such as is illustrated in Figure 1 below. In contrast, adsorption is the process in which molecules adhere to the surface of a solid or a liquid.
Figure 1. Schematic illustration of a partitioning process based on absorption. See discussion below for clarification on symbols.
The process is a dynamic equilibrium and is characterized by a distribution coefficient (K_\text{d}) in terms of concentrations in the two phases
Here, c_{i,\text{s}} and c_{i,\text{m}} represent the concentration of the analyte in the stationary and mobile phase, respectively. This can be related to the retention factor:
Equation 1.2: k=\frac{q_{i,\text{s}}} {q_{i,\text{m}}}=\frac{c_{i,\text{s}}} {c_{i,\text{m}}} \cdot \frac{V_{\text{s}}}{V_{\text{m}}}=K_{\text{d}} \cdot \Phi
Now, q_{i,\text{s}} and q_{i,\text{m}} represent the quantities of the analyte in the stationary and mobile phase, respectively. \Phi is the phase ratio, which represents the ratio between the volume of the stationary V_{\text{s}} and mobile phase V_{\text{m}}.
Analyte molecules can be present at certain numbers, or quantities, and both the mobile- and stationary phase can have certain volumes. In reality, we typically do not know these specific numbers and therefore it is more useful to consider the fractions of analytes present in the stationary (f_{\text{s}}) and mobile (f_{\text{m}}) phase. We’ll see soon why.
Looking back at the definition of chromatography, we have extensively defined the partitioning of analytes in both phases, but the component of velocity is still missing.
1.2. Linear velocity #
As the analytes travel through the chromatographic column, velocity is typically expressed as the linear velocity (u) in mm∙s-1. To determine the velocity of analyte i, we can use
Equation 1.4: u_i=f_{i,\text{m}} \cdot u_{i,\text{m}} + f_{i,\text{s}} \cdot {u_{i,\text{s}}} = f_{i,\text{m}} \cdot u_0
Inside the pores, the velocity of the mobile phase is infinitely low (say, zero), which cancels out a chunk of Equation 1.4. This means that the average velocity of the analytes is completely determined by the velocity of the mobile phase (which can be computed from the flow rate), and the fraction of the time that the analytes spend in the mobile phase.
If we combine Equations 1.2 and 1.4 we obtain:
which yields the average velocity of analyte i
Equation 1.5: u_i = \frac{1}{1+k_i} \cdot u_0
Equation 1.5 specifies that the average velocity of each analyte depends on its retention by the stationary phase. This also implies that separation can be achieved as different analytes will exhibit different retention factors.
READ SECTION 1.1.3
COLUMN CHROMATOGRAPHY
1.3. Column chromatography #
For an analyte to be more retained, it must spend more time in the stationary phase. Before we continue it is useful to realise that the stationary phase (and chromatography) exist in different formants. In thin-layer chromatography (TLC), thin plates are used (discussed in Section 1.1.2 of the book). However, in this course we focus on analytical separations in chromatographic columns (e.g. liquid chromatography). Regardless of what is used, Equation 1.5 still applies, underlining that all of these techniques are guided by the same fundamental principle.
For column chromatography typically cylindrically shaped channels are employed that contain the stationary phase. This is also shown in Figure 2. Eluent (a solvent mixture) is introduced as mobile phase along with the analyte mixture in the injector. We speak of elution as the process of mobile phase transporting the analyte through the column (“The analyte eluted from the column.”). Once the mobile phase emerges from the column, we speak of effluent instead of fluent.
Figure 2. Schematic overview of a chromatographic system with enlarged depictions of the stationary phase and their formats.
For columns packed with porous particles, such as is the case in LC, analytes are considered to sample the entire volume of the column that is also occupied by the mobile phase (V_{\text{m}}): the pore volume inside the pores of the stationary phase (V_{\text{pores}}) and the interstitial volume between the stationary-phase particles (V_{\text{int}}).
The detector ultimately records the response signal in the form of a chromatogram (see Figure 3) that contains the different signal intensities of each analyte band (“chromatographic peak“).
1.4. Void volume and Porosity #
One useful relation at this stage is the computation of the column void volume, or column dead volume (V_{\text{0}}), and corresponding void time (t_{\text{0}}), or dead time, which are defined as
Equation 1.11: t_{\text{0}}=\frac{V_0}{F}=\frac{\epsilon \cdot 0.25\pi \cdot d^{2}_{\text{c}} \cdot L}{F}=\frac{L}{u_0}
Here, \epsilon depicts the column porosity, which is the fraction of mobile phase in the cylinder column, F the volumetric flow rate and d_\text{c} the column internal diameter.
EXERCISE 1 #
Estimate the dead time (t_0) in minutes of a 100 mm long column (L) with an internal diameter (d_{\text{c}}) of 2.1 mm that is running at a flow rate (F) of 0.6 mL min-1. Assume that the porosity (\epsilon) is 0.65. Round your answer two decimals.
This is correct! Well done!
Unfortunately, this is not correct. Make sure that you convert the units correctly. Hint: Convert all metric values to mm, so that you can apply 1 mm3 = 0.001 mL. Your estimated dead volume should around 225 mm3.
READ MODULES 1.2 AND 1.3
RETENTION & SELECTIVITY
1.5. Retention #
As different analytes will spend on average different amounts of time in the stationary phase, they will also elute at different retention times (t_{\text{R}}). If the column has length L then the average linear velocity is given by u_i=L/t_{\text{R}}. In other words, the retention time is given by
Equation 1.12: t_{\text{R},i}=(1+k_i) \cdot t_0
Here, t_{\text{0}} is referred to as the dead time, or void time. It is the time that the mobile-phase molecules take to travel through the column, and therefore also often referred to as (t_{\text{m}}). From Equation 1.11 in the book it follows that the mobile-phase molecules will travel so at average linear velocity u_0=L/t_{\text{0}}.
The retention time is the sum of t_{\text{m}} and the time that the analytes spent in the stationary phase (t_{\text{s}}). The latter is sometimes referred to as the net retention time (indicated with pink in Figure 3).
For molecules that travel unretained through the column (t_{\text{s}}=0), the retention time will be equal to (t_{\text{0}}).
Figure 3. Schematic overview of a chromatographic system with enlarged depictions of the stationary phase and their formats.
Ultimately, the retention time is useful, but one intrinsic flaw is that does not allow for easy comparisons. For example, a different column length (with the same stationary-phase chemistry) will yield different retention times, even though retention process is the same. It is therefore more useful to consider the retention factor (k), given also by
Equation 1.13: k_i=\frac{t_{\text{R},i}} {t_\text{0}} -1
The retention factor corrects the observed retained peak for the dead time and thus also for changes in column length, or flow rate, allowing the values to be compared from different columns (as long as their stationary-phase chemistry is the same!).
Of course, it is important to realize that the analytes must also travel through the entire system (tubing, injector, etc), which technically do not contribute to retention. This is known as the extra column volume of extra-column time (t_{\text{ec}}. In practice we often assume that it is neglible, but when accuracy matters (e.g. for fundamental applications, or library searches) it is important to correct for it through
Equation 1.14: k_i=\frac{t_{\text{R,obs},i}-t_{\text{ec}}} {t_{\text{0,obs}}-t_{\text{ec}}} -1
EXERCISE 2 #
This is the results table obtained by your UHPLC instrument for a simple separation of two standard compounds on a 100mm long column. The first peak is unretained.
| PEAK # | t_{\text{R}} (min) | W_{\text{0.5}} (min) |
|---|---|---|
|
1 |
0.39 |
0.008 |
|
2 |
2.67 |
0.051 |
|
3 |
2.77 |
0.053 |
Calculate the selectivity \alpha_{3,2} of peaks 2 and 3. Round to two decimals. Hint: You will first need to calculate the retention factors of peaks 2 and 3.
This is the correct answer! As retention factors (l for peaks 2 and 3 we found 6.12 and 6.38, respectively.
Unfortunately this is not correct! Let’s check the retention factors. As retention factors (l for peaks 2 and 3 you should find 6.12 and 6.38, respectively.
1.6. Selectivity #
To establish how well a chromatographic system can separate two analytes from a chemical perspective, the selectivity is useful, which is given by
Equation 1.19: \alpha_{j,i}=\frac{K_{\text{d},j}}{K_{\text{d},i}}=\frac{k_{j}} {k_{i}}
Relating to Equation 1.2, we can see from Equation 1.20 that the selectivity is purely determined by thermodynamic factors. Its value is by definition larger than one, and is largely determined by the chemistry of the mobile and stationary phases in relation to the analytes.
EXERCISE 3 #
Which of the following statements is correct?
This is correct! Column length and particle size have no influence on the selectivity. The retention factor normalizes the retention time using the dead time for the column length. Even if factors affect the retention factor, then – as long as they do so proportionally to all analytes – the selectivity will still stay the same.
This is not correct. Try again! Hint: the retention factor normalizes the retention time using the dead time. Also note that the selectivity is defined as a ratio of two retention factors, not retention times!
You now may wonder what the usefulness is of \alpha if it depends on k. Well, parameters that only affect the phase ratio, also affect the retention factor, but not the selectivity. Examples include the inner diameter of open-tubular columns and the surface area of stationary-phase packing materials.
READ MODULE 1.4
EFFICIENCY
2. Efficiency #
We have thus far assumed that identical analytes all have the same retention time, but the chromatogram shown in Figure 3 does not display peaks as spikes, but broader distributions. Indeed, ensembles of identical analytes do not arrive exactly at the same time at the other side of the column. Instead these analyte bands disperse due to different factors.
2.1. Diffusion #
The most important factor for band broadening is diffusion. In a nutshell, a high concentration of analytes in a large medium of mobile phase is thermodynamically not favourable: the system attempts to establish equilibrium (increase entropy) and the analytes diffuse randomly in all directions, but mainly in the direction where its concentration is lower.
Figure 4. A) Schematic illustration of diffusion, B) different quantities for peak width for Gaussian peak. See Module 1.4 in the book for extended schematics and explanations.
The fact that our separation takes place inside the column makes things both easier and more complicated at the same time. Diffusion technically takes place in all directions (Figure 4A). Fortunately for us, the column walls prevent radial diffusion (perpendicular to the flow direction) mostly. Not so fortunate is axial diffusion (in the direction of the length of the column), which causes our analyte bands to diffuse in both directions. This results in the band broadening shown in Figure 4B.
This band broadening is a major problem, because even if we do a good job in creating selectivity between two analyte bands, they may still not be separated because of the peaks being so broad that they overlap!
2.2. The plate number #
The last thing you may expect when doing this lesson is to suddenly see the word “plate” pop-up. Well, you are reading it correctly, and no, they do not exist in chromatographic columns.
The word plate is derived from distillation technology used in industry and there are a number of analogies, as is explained in more detail in Figure 5 and Module 1.4 of the book.
Figure 5. Similarties and differences between distillation towers (left) and chromatographic columns (right).
The efficiency is a measure of the quality of the separation system and typically expressed in the theoretical plate number. It is a comparison of the travelled distance (L) with band broadening (\sigma_{\text{t}}) and given by:
Equation 1.26: N=\left(\frac{t_{\text{R}}}{\sigma_{\text{t}}}\right)^2
Here, \sigma_{\text{t}} is the standard deviation of the chromatographic peak, which may be obtained using various methods (Figure 4B). Given that a longer column will automatically yield more theoretical plates, it is more useful to consider the height equivalent of a theoretical plate (HETP), abbreviated as plate height (H), which is given by:
Equation 1.29: H=L/N
A lower plate height reflects a more efficient the chromatographic column. Within one specific separation, N will be similar for all analytes, although t_0 deviates often.
EXERCISE 4 #
| PEAK # | t_{\text{R}} (min) | W_{\text{0.5}} (min) |
|---|---|---|
|
1 |
0.39 |
0.008 |
|
2 |
2.67 |
0.051 |
|
3 |
2.77 |
0.053 |
Calculate the plate heights (H) for peaks 2 and 3 in μm. You will first need to calculate the plate numbers (N) for each, which requires you to use the relation in Figure 4B to convert peak width into peak standard deviation. Round your plate height values to two decimals.
For peak 2:
Correct!
Unfortunately, this is not correct. Your plate number for this peak should be 15136, based on a \sigma_{\text{t},2} of 0.022 for peak 2, using the relation W_0.5 = 2.35 \sigma to convert the peak width into the peak standard deviation.
For peak 3:
Correct, it is roughly also 6.6 μm! Generally you expect plate numbers for similar compounds to also be very similar, because the plate number is (approximately) considered to be a system property.
Unfortunately, this is not correct. Your plate number for this peak should be 15085, based on a \sigma_{\text{t},3} of 0.023 for peak 3, using the relation W_0.5 = 2.35 \sigma to convert the peak width into the peak standard deviation.
READ MODULE 1.5.1
RESOLUTION
3. Resolution #
Equation 1.35: R_{\text{S},j,i}=\frac{t_{\text{R},j}-t_{\text{R},i}}{2(\sigma_{\text{t},i}+\sigma_{\text{t},j})}\approx \frac{\Delta t_{\text{R},j,i}}{4 \sigma_{\text{t}}}
Figure 6. Schematic example of two peaks with different degrees of resolution.
EXERCISE 5 #
Using your answers from Exercise 2, calculate the resolution between peaks 2 and 3 using Equation 1.35. Round your answer to 2 decimals.
This is correct!
Unfortunately, this is not correct. Did you use retention factors of 6.12 and 6.38 for peaks 2 and 3, respectively (such as obtained in Exercise 2)?
Will the peaks be baseline separated based on your obtained resolution value?
This is correct! Baseline separation is reached when the resolution is 1.5. Only with a value of 0.5 will they appear to be a single peak. At a value of 1.13, they are visible as two peaks, but are not baseline separated. We can see examples in Figure 1.12 of the book.
This is not correct. Baseline separation is reached when the resolution is 1.5 or higher. See examples in Figure 1.12 of the book.
Equation 1.39: R_{\text{S},j,i}=\frac{\sqrt{N}}{2} \cdot \frac{\alpha_{j,i}-1}{\alpha{j,i}+1} \cdot \frac{\bar{k}}{1+\bar{k}}
Different forms of this equation exist, including the Purnell equation (Eq. 1.40, which strongly resembles this one; see the book), but they all share in common that they present three key options to chromatographers facing a separation problem.
Indeed, Equation 1.39 shows us that, to resolve co-elution, we have three options:
- We improve the efficiency of the chromatographic system (i.e. the plate number), through e.g. adjusting the column length or flow rate.
- We adjust the selectivity, for example by changing the mobile-phase or stationary-phase chemistry.
- We modify the retention through adjustment of the strength of the mobile phase, changing the temperature, etc.
Changing the selectivity is the most effective approach, but is also the most difficult to change as it is not always clear what – from a chemical perspective – should be done. Moreover, while it may improve resolution between two specific peaks, it could also undo the separation of other peak pairs. Changing the selectivity changes the landscape for the entire chromatogram.
Changing the efficiency is relatively easy, and will work equally well for all peaks, but it is limited by fundamental limits (e.g. back pressure, we see this in later lessons), but also resources. More advanced columns typically require more sophisticated instruments.
Finally, changing the retention factor is usually the first thing that we try, as we can achieve this through changing method parameters. We will see this in action in later lessons.
EXTENSIVE EXERCISE #
What plate number would be required to obtain a resolution of 1.5 (baseline separation)? You will need to rearrange Equation 1.39 for this exercise. Your answer should be an integer (no decimals). Assume that \bar{k} = 6.249 and \alpha = 1.044.
This is correct! We rearrange Equation 1.39 to \sqrt{N_{\text{req}}}=\frac{2 R_S \cdot (\alpha+1)(1+k)}{k(\alpha+1)}
Unfortunately, this is incorrect. Make sure you rearrange Equation 1.39 correctly to \sqrt{N_{\text{req}}}=\frac{2 R_S \cdot (\alpha+1)(1+k)}{k(\alpha+1)}
What is the corresponding column length (mm) that would be required to achieve this plate number? Assume that the plate height (H) is 6.6 μm. Round to an integer number (no decimals).
This is correct! You may have noticed that it will change slightly if you use the unrounded plate height numbers obtained from Exercise 4, but these changes are not significant.
This is incorrect. Did you use a plate number of 26634?
What resolution do you expect to obtain when you, instead of increasing the plate number, use a weaker eluent, that gives double the retention factor with the same selectivity (i.e. \alpha is still 1.044)? Assume that the plate number is the mean of those obtained for peaks 2 and 3: \bar{N} 15111. Round your answer to two decimals.
This is the correct answer!
This is unfortunately not correct. Did you use the values specified in the question? Your \bar{k} should now be 12.498 (double the value of 6.249).
Concluding remarks #
We have learned that chromatography considers differences in partitioning of analytes between two phases that move at different velocities. We also saw that three concepts govern the separation: retention, selectivity and efficiency. In the next two lessons we will zoom in further on the thermodynamic and kinetic processes that guide these concepts.
Please find here an Excel sheet that contains the calculates above. In order to facilitate your capability to really learn these calculations it is STRONGLY recommended that you first try them yourself without the sheet.
Download: (SS_01, .XLSX).
