No matter where liquid chromatography is used, it is never immune to problems.
One of the privileges of teaching liquid chromatography (LC) classes for a living is that I get to meet chromatographers from around the world during my travels. As I write this installment, I'm just finishing six weeks of classes, where I have had a chance to discuss problems that attendees were having with a variety of LC applications. Although the topics in this column originate in Poland, Turkey, and Argentina, they are no different than those that arise in Singapore, Galway, Charlottetown, or Huntsville. Let's look at a couple of problems that students had related to both conventional LC and ultrahigh-pressure LC (UHPLC). These problems apply equally to both techniques.
How Much Can I Inject?One of my students from Poland, who was just starting out using UHPLC, asked if she could inject 5 µL of sample onto her UHPLC column without problems. Her conditions comprised a 75 mm × 2.1 mm column packed with 1.8-µm diameter particles operated at a flow rate of 0.8 mL/min. It is fairly easy to perform a few simple calculations and answer her question.
A good rule of thumb (1) is that you can inject ≈15% of the volume of the first peak of interest without causing a more than ≈5% increase in peak broadening, as long as you use mobile phase for the injection solvent. We need to determine the volume of the first peak of interest to see how this applies in the present case. We will assume a retention factor, k, of 1 for the first peak; you can make similar calculations for other retention factors.
First, we need to estimate a column plate number (efficiency), N, which can be done as
N = 300 L/d p 
where L is the column length in millimeters and d p is the particle diameter in micrometers. A well-packed column tested with an ideal solute will give larger values of N, but equation 1 is a reasonable estimate for real samples. With our L = 75 mm, d p = 1.8 µm column, we get N = 12,500.
Next, it is necessary to estimate the column volume, V M, in milliliters, using the equation
V M ≈ 5 × 10-4 Ld c 2 
where dc is the column diameter in millimeters. For the 75 mm × 2.1 mm column we get 0.165 mL. From this we can get the column dead time, t 0, by dividing by the flow rate, 0.8 mL/min: 0.21 min. (Note: If you are trying to repeat these calculations, carry along several extra decimal places of precision — I'm rounding the numbers for simplicity of presentation).
With the knowledge of the column dead time, we can calculate the retention time, t R, for our k = 1 peak by rearranging the equation for k:
t R = t 0 (1 + k) 
This gives t R = 0.41 min.
The next to last step is to estimate the peak volume of the first peak. We estimated N with equation 1, but if we were measuring N from an isocratic chromatogram, we would use
N = 16 (t R/w)2 
where w is the baseline width of the peak measured between tangents drawn along the sides of the peak. Solving equation 4 for the peak width, we get
w = 4 t R/N 0.5 
For t R = 0.41 min and N = 12,500, this gives w = 0.015 min, or at 0.8 mL/min, w = 11.8 µL.
Using our 15% rule of thumb from the beginning of our discussion, an injection of (0.15 × 11.8 µL) = 1.8 µL, or ≈ 2 µL would be allowed in the mobile phase. This is less than half the desired 5-µL injection, so it is expected that excess peak broadening would be observed with a 5-µL injection. Armed with this knowledge, I would perform a series of injections of 1, 2, 3, 4, and 5 µL and observe the effect on the peak width of the first peak of interest. I agree wholeheartedly with Izaak Kolthoff, the father of analytical chemistry, who is credited with the quote, "Theory guides, experiment decides." So, use the results of the injection-volume experiment for the final decision.