HPLC and GC Chromatography Simulator

What does a chromatography simulator calculate?

A chromatography simulator predicts the chromatogram a column will give: when each compound in a mixture reaches the detector, how wide its peak is, and whether neighbouring peaks are separated. Each analyte leaves the column at its retention time, tR = tM(1 + k), where tM is the dead time and k is the analyte’s retention factor. Two neighbours are judged by their resolution, Rs = 2(tR2 − tR1)/(w1 + w2), the gap between them over their average width at the base. At Rs = 1.5 or more the trace returns to the baseline between them, called baseline resolution.

In reversed-phase HPLC the percentage of acetonitrile in the mobile phase sets every retention factor, and in gas chromatography the oven temperature does. In both, the column’s length and its particle size or internal diameter set its plate number through the van Deemter equation, and the plate number sets every peak’s width. Chromatography separates by partition between two phases; the gel electrophoresis simulator separates by size in an electric field instead, and a compound collected from a column is often identified by its NMR splitting pattern.

Retention time and the retention factor

A molecule spends part of its time held by the stationary phase and part carried along by the mobile phase. The retention factor k is time held over time moving, so its band travels at u/(1 + k) and leaves a column of length L at tR = (L/u)(1 + k). Rearranged, k = (tR − tM)/tM, which is how k is read from a chromatogram: the time past the dead time, in units of it.

A small k elutes close to the dead time, where peaks crowd together among anything unretained, and a large one wastes time on low, broad peaks. Isocratic methods usually aim for k between about 1 and 10. For reversed-phase HPLC the linear solvent strength model of Snyder, Dolan and Gant (1979) gives log k = log kw − Sφ, where φ is the volume fraction of acetonitrile, kw the retention factor extrapolated to pure water and S a constant for each compound, about 4 for a small molecule. With S = 4, ten points less acetonitrile multiplies k by 10^0.4 = 2.51, close to the common rule of thumb that k roughly triples.

Using the simulator

Choose HPLC or GC and how many analytes to inject. For HPLC the sliders set the acetonitrile, the column length, the particle size and the velocity; for GC, the oven temperature, the column length, its internal diameter and the helium velocity. Play runs one injection: the bands leave the inlet together, pull apart and spread as they travel, and draw their peaks at the detector, each shaded in its analyte’s colour under the summed trace. The scrubber steps through the run, and with the scene selected the arrow keys change the acetonitrile or the oven temperature.

The readouts give the critical pair, the two neighbouring peaks closest together, with its resolution and selectivity α = k2/k1, and the column’s plate count, plate height, best velocity, dead time and run time. The first plot is a window diagram, the critical pair’s resolution across the whole range of acetonitrile or oven temperature with a dashed line at Rs = 1.5; the second is the column’s van Deemter curve, with a dot at the set velocity.

A real HPLC detector is usually a UV absorbance cell, so a peak’s area turns into a concentration through the Beer-Lambert law, against standards made up with a dilution calculator. Here every analyte is injected in the same amount, so every peak has the same area and later, broader peaks are lower.

Worked example: four analytes at 50 percent acetonitrile

The simulator opens on a 150 mm HPLC column packed with 5 µm particles, run at 1.0 mm/s with 50 percent acetonitrile. The diffusion coefficient of a small molecule in the mobile phase is taken as 1 × 10⁻⁹ m²/s.

• Dead time: tM = L/u = 150 mm ÷ 1.0 mm/s = 150 s = 2.50 min.
• Reduced velocity: ν = u dp/Dm = 1.0 × 10⁻³ × 5 × 10⁻⁶ ÷ 10⁻⁹ = 5.
• The van Deemter terms: A = 1 × 5 µm = 5 µm, B/u = 2Dm/u = 2 µm and Cu = 0.125 × 5 µm × 5 = 3.125 µm, so H = 5 + 2 + 3.125 = 10.125 µm.
• Plate count: N = L/H = 150 mm ÷ 0.010125 mm = 14,815, which the readout rounds to 14,810, and √N = 121.72.
• The close pair: kB = 10^(2.5 − 4.0 × 0.5) = 3.16228 and kC = 10^(2.72 − 4.4 × 0.5) = 3.31131.
• Their retention times: tR = 2.5 × 4.16228 = 10.406 min for B and 2.5 × 4.31131 = 10.778 min for C, which the table rounds to 10.41 and 10.78. They are 2.5 × 0.14903 = 0.3726 min apart.
• Base widths, w = 4tR/√N: 4 × 10.406 ÷ 121.72 = 0.3420 min for B and 4 × 10.778 ÷ 121.72 = 0.3542 min for C.
• Resolution: Rs = 2 × 0.3726 ÷ (0.3420 + 0.3542) = 1.070.

So B and C are partly resolved: a clear valley, but no return to the baseline. Their selectivity is α = 3.31131 ÷ 3.16228 = 1.047, which is 10^0.02. A elutes at 5.305 min and D at 18.27 min, each far clear of its neighbour, so the whole separation stands or falls on B and C.

Resolution in chromatography

Resolution weighs the gap between two peaks against their widths. The base width of a Gaussian peak, between the tangents through its inflection points, is four standard deviations, so at Rs = 1.5 two equal peaks have their apexes six standard deviations apart, and only about 0.13 percent of each peak’s area lies past the midpoint between them. At Rs = 1.0 that share is about 2.3 percent, and the trace dips only 73 percent of the way to the baseline between the apexes. At 0.5 or less two equal peaks merge into a single lump with no dip at all, so a badly overlapped pair can pass for one compound.

Purnell’s equation shows what Rs depends on: Rs = (√N/4) × ((α − 1)/α) × (k2/(1 + k2)), a product of efficiency, selectivity and retention. For the opening pair it gives 30.43 × 0.04501 × 0.7681 = 1.052, against 1.070 from the widths, because it gives both peaks the width of the later one. Selectivity is the strongest lever, since α − 1 is a small number that can easily double. Efficiency enters only as a square root. Retention helps only while k is small: k2/(1 + k2) is 0.5 at k = 1 but already 0.91 at k = 10.

Three ways to resolve a close pair

• Selectivity. Drop the acetonitrile from 50 to 48 percent. B and C have different S values, so their retention factors rise at different rates: α grows from 1.047 to 1.067 and Rs from 1.07 to 1.563, while D takes 22.54 min to come out instead of 18.27. Go the other way and the pair closes until, at 55 percent, B and C co-elute, and above that C comes out first.
• Efficiency, by length. A 300 mm column has twice the plates, 29,630, but resolution grows only as √N, to 1.514, and the run doubles to 36.55 min.
• Efficiency, by particle size. Pack the 150 mm column with 2 µm particles and run it at 2.0 mm/s: the plate height falls to 4 µm, N rises to 37,500 and Rs to 1.703, and the run halves to 9.137 min. The catch is back pressure, which scales as u/dp², so this column needs 12.5 times the pressure of the opening one, which is why sub-2 µm particles need UHPLC pumps.

On the window diagram, the closest pair is baseline resolved at 48 percent acetonitrile or less, at a growing cost in time. Between the two crossings, at 55 and 75 percent, the order changes to A, C, B, D, and the closest pair never reaches 1.5: the best is about 1.34, near 70 percent, in a run under 4 minutes.

The van Deemter equation

The van Deemter equation, published by van Deemter, Zuiderweg and Klinkenberg in 1956, gives the plate height H, the length of column per theoretical plate, against the mobile-phase velocity: H = A + B/u + Cu. Each term is a different way a band spreads.

• A, eddy diffusion. Paths of different lengths through the packing, in proportion to the particle size and whatever the velocity.
• B/u, longitudinal diffusion. A band diffuses along the column for as long as it is inside, so this term dominates at low velocity. B goes as Dm.
• Cu, resistance to mass transfer. Molecules take time to move in and out of the stationary phase, so the faster the flow, the further the moving ones get ahead. C goes as dp²/Dm.

The curve has a minimum at u = √(B/C), where H = A + 2√(BC). The packed column here uses the reduced coefficients 1, 2 and 0.125, which put the smallest plate height at twice the particle size, the usual benchmark for a well-packed column, at a reduced velocity of 4. For the opening column that is 0.8 mm/s and 10 µm. The opening velocity, 1.0 mm/s, costs only 1.25 percent in plate height, and slowing to the optimum raises Rs only from 1.070 to 1.077 while making the run 25 percent longer.

Smaller particles lower the whole curve and move its minimum to faster flow, since the optimum velocity goes as 1/dp: for 1.7 µm particles it is 2.353 mm/s, with H = 3.4 µm. Their C term is also far flatter, going as dp², so small-particle columns can be run fast with little loss. Running too slowly is the opposite mistake: at 0.2 mm/s the opening column’s plate height is 15.63 µm, more than half as large again as at the optimum, because longitudinal diffusion takes over.

Gas chromatography: temperature in place of solvent

In gas chromatography the mobile phase is an inert gas that hardly interacts with the analytes, so the oven temperature sets retention. The retention factor follows the van ’t Hoff equation, ln k = ln k100 + (ΔH/R)(1/T − 1/T100), where ΔH is the enthalpy of moving the analyte out of the stationary phase into the gas and k100 is its retention factor at 100 °C. Retention falls exponentially as the oven warms: near 100 °C, B’s retention factor halves for every 21 °C, the same exponential dependence on 1/T that the Arrhenius equation gives a reaction rate.

A capillary GC column is an open tube with the stationary phase as a film on its wall, so it has no packing and no A term. Golay’s equation for an open tube gives H = B/u + Cu, with B = 2Dm and C set by the square of the internal diameter, and for a well-retained analyte its smallest plate height is 0.957 times the internal diameter. At the GC opening settings, a 30 m × 0.25 mm column with helium at 25 cm/s and the oven at 100 °C:

• tM = 30 m ÷ 0.25 m/s = 120 s = 2.00 min.
• H = B/u + Cu = 0.1200 + 0.1194 = 0.2394 mm, at the bottom of the curve, since the optimum is 25.07 cm/s.
• N = 30 m ÷ 0.2394 mm = 125,300, over eight times the HPLC column’s plates.
• B and C, with k = 3.00 and 3.05, come out at 8.00 and 8.10 min with Rs = 1.099.

With so many plates, a selectivity of only 1.017 nearly separates them. Cooling the oven to 94 °C raises the resolution to 1.522, with D out at 21.11 min rather than 17.00. Warming it closes the gap until B and C co-elute at about 120 °C, and above that they swap order. A 0.53 mm megabore column has far fewer plates, and the pair falls to 0.6639.

What this model leaves out

• Gradients and temperature programmes. Most real methods raise the acetonitrile or the oven temperature during the run, bringing late peaks out sooner and sharper; here both are constant.
• A plate height for each peak. Every peak shares its column’s plate height. In reality the C term depends on k, most of all in an open tube, where Golay’s factor runs from 1/96 for an unretained solute to 11/96 for a well-retained one, and the model takes the second.
• Pressure and gas compression. Back pressure is never computed, though it limits how long, fine-grained and fast an HPLC column can be. In GC the gas expands along the column, so its velocity rises towards the outlet; the velocity here is an average.
• Injection, tubing and peak shape. Injection is instantaneous, nothing outside the column adds broadening, and every peak is a Gaussian. Real peaks tail when a few strong sites hold an analyte, and distort when too much is injected.
• Fixed diffusion. Diffusion coefficients change with temperature, solvent and pressure; here they are fixed, and the HPLC column’s temperature plays no part.

Common mistakes

• Using half-height widths in the base-width formula. Rs = 2(tR2 − tR1)/(w1 + w2) needs widths at the base. With widths at half height it becomes Rs = 1.18(tR2 − tR1)/(w½,1 + w½,2); using them in the base-width formula overstates Rs by a factor of 1.70.
• Expecting twice the column to double the resolution. Doubling L doubles N, and Rs grows only as √N: 1.070 becomes 1.514, not 2.14, and the run takes twice as long.
• Working out k from the injection. k = (tR − tM)/tM, so peak B, out at 10.406 min with tM = 2.50 min, has k = 3.162, not 10.406 ÷ 2.5 = 4.162.
• Mixing up the plate count formulas. From a peak, N = 16(tR/w)² with the base width, or N = 5.545(tR/w½)² with the width at half height. From peak B, 16 × (10.406 ÷ 0.3420)² = 14,813, the same as L/H to within the rounding of the width.