GATE Chromatographic Methods — Plates, Resolution and van Deemter
Chromatography looks descriptive until you meet it in an exam, and then it turns out to be one of the most numerical topics in analytical chemistry. Five quantities — retention factor, selectivity, plate number, plate height and resolution — carry almost every question, and they are linked by one master equation that tells you which of them to change when two peaks overlap. This guide works each of them in full, then adds the van Deemter treatment of flow rate and a compact comparison of the techniques.
A separate article in this knowledge base covers the broader separation toolkit — solvent extraction, precipitation, ion exchange and distillation. Here we stay inside the column and quantify what happens on it.
The five working quantities
| Symbol | Name | What it measures | Unit |
|---|---|---|---|
| tM | Dead / void time | Time for an unretained species to pass through | min or s |
| tR | Retention time | Time to the peak maximum | min or s |
| k | Retention factor (capacity factor) | How much longer the solute stays than the mobile phase | dimensionless |
| α | Selectivity factor | Whether the column can tell two solutes apart at all | dimensionless, ≥ 1 by convention |
| N | Plate number | Column efficiency — how narrow the peaks are | dimensionless |
| H | Plate height (HETP) | Efficiency per unit length; smaller is better | cm or µm |
| Rs | Resolution | How well two peaks are actually separated | dimensionless |
Read Rs like this: about 1.0 gives roughly 94 % separation with visible overlap; 1.5 is the accepted baseline-resolution target; above 2 you are wasting analysis time you could spend on a harder pair.
Worked example 1 — k, α, N and H from one chromatogram
On a 25.0 cm column, an unretained marker elutes at tM = 1.20 min. Peak A appears at 6.60 min with a baseline width of 0.30 min; peak B at 8.40 min with a baseline width of 0.36 min. Find k for both, α, N and H.
kA = (6.60 − 1.20) / 1.20 = 5.40 / 1.20 = 4.50
kB = (8.40 − 1.20) / 1.20 = 7.20 / 1.20 = 6.00
α = 6.00 / 4.50 = 1.333 (always the larger k on top, so α ≥ 1)
N from peak A: N = 16 × (6.60 / 0.30)2 = 16 × (22.0)2 = 16 × 484 = 7 744 plates
H = L / N = 25.0 / 7 744 = 3.23 × 10−3 cm = 32.3 µm
Sanity check on k: solute A spends 4.50 times as long in the stationary phase as in the mobile phase, so it must elute at tM(1 + k) = 1.20 × 5.50 = 6.60 min ✓ — the relation runs both ways and is worth using as a free verification.
Worked example 2 — resolution
Using the same two peaks, calculate Rs.
Rs = 2(8.40 − 6.60) / (0.30 + 0.36) = 2 × 1.80 / 0.66 = 3.60 / 0.66 = 5.45
That is far above 1.5, so the pair is fully baseline separated with room to spare. In practice you would now shorten the run — raise the flow rate or use a stronger mobile phase — and trade some of that surplus resolution for speed.
The master resolution equation — which knob to turn
This factorises resolution into three independent terms: efficiency, selectivity and retention. The practical lesson is that they are not equally useful. Because N enters as a square root, doubling resolution by efficiency alone needs four times the plates — a four-fold longer column, and four times the analysis time and back-pressure. Selectivity is far cheaper to improve: change the stationary phase, the mobile-phase composition, the pH or the temperature.
Two solutes have α = 1.05 and k2 = 4.0. How many plates are needed for Rs = 1.5? Then, how many if a different stationary phase raises α to 1.10?
Rearrange: √N = 4Rs ÷ [ ((α − 1)/α) × (k2/(1 + k2)) ]
Case 1, α = 1.05:
(α − 1)/α = 0.05 / 1.05 = 0.047619
k2/(1 + k2) = 4.0 / 5.0 = 0.800
product = 0.047619 × 0.800 = 0.038095
√N = (4 × 1.5) / 0.038095 = 6 / 0.038095 = 157.5
N = 157.52 = 2.48 × 104 plates
Case 2, α = 1.10:
(α − 1)/α = 0.10 / 1.10 = 0.090909
product = 0.090909 × 0.800 = 0.072727
√N = 6 / 0.072727 = 82.5
N = 82.52 = 6.81 × 103 plates
A selectivity change from 1.05 to 1.10 — chemically a modest change — cuts the required efficiency by a factor of about 3.6. That single comparison is the most useful thing on this page, and it is a standard conceptual question in its own right.
Note also what happens as k2 grows: k/(1 + k) approaches 1 and stops helping. Past about k = 10 you are only adding run time. The useful working window is roughly k = 2–10.
Band broadening — the van Deemter equation
| Term | Physical origin | Behaviour with flow rate u | How to reduce it |
|---|---|---|---|
| A — eddy diffusion | Different flow paths through the packing | Independent of u | Smaller, more uniform particles; open tubular column (A = 0) |
| B/u — longitudinal diffusion | Solute diffuses along the column while it sits there | Dominates at low u | Run faster; use a denser mobile phase (matters most in GC) |
| Cu — mass-transfer resistance | Solute takes time to move between phases | Dominates at high u | Thinner stationary film; smaller particles; higher temperature |
Because one term falls with u and another rises with it, H passes through a minimum. That is why "faster flow" is not automatically better: past uopt, peaks broaden again and resolution falls.
A column has A = 0.10 cm, B = 0.30 cm2 s−1 and C = 0.020 s. Find the optimum linear velocity and the minimum plate height.
uopt = √(B/C) = √(0.30 / 0.020) = √15 = 3.87 cm s−1
√(BC) = √(0.30 × 0.020) = √0.0060 = 0.07746
Hmin = 0.10 + 2 × 0.07746 = 0.10 + 0.15492 = 0.255 cm
Verify by substituting back into H = A + B/u + Cu:
0.10 + 0.30/3.873 + 0.020 × 3.873 = 0.10 + 0.07746 + 0.07746 = 0.2549 ✓
The two variable terms come out equal at the minimum, which is a quick way to confirm you have the right u without redoing the derivative.
Planar chromatography — Rf and its link to k
On a TLC plate a spot moves 3.6 cm while the solvent front moves 9.0 cm. Find Rf and the equivalent retention factor.
Rf = 3.6 / 9.0 = 0.40
k = (1 − 0.40) / 0.40 = 0.60 / 0.40 = 1.50
Rf lies strictly between 0 and 1: a value of 0 means the spot never left the baseline, and 1 means it ran with the front and was not retained at all. A reported Rf above 1 is an arithmetic error, not a result.
Choosing the technique
| Technique | Mobile phase | Separation is based on | Suited to |
|---|---|---|---|
| Gas chromatography (GC) | Inert gas | Volatility and partitioning into a liquid film | Volatile, thermally stable compounds |
| Normal-phase HPLC | Non-polar solvent | Adsorption on a polar phase (silica) | Non-polar elutes first; polar retained |
| Reversed-phase HPLC | Water / methanol / acetonitrile | Partition into a non-polar C18 phase | Polar elutes first; most common mode today |
| Ion-exchange | Buffered aqueous | Charge and ion affinity for the resin | Ions, amino acids, proteins |
| Size-exclusion (gel permeation) | Aqueous or organic | Hydrodynamic size — large molecules elute first | Polymer molar-mass distributions |
| TLC / paper | Solvent rising by capillary action | Adsorption or partition | Fast qualitative checks, reaction monitoring |
The elution-order reversals are examiner favourites. In normal phase, increasing solvent polarity elutes solutes faster; in reversed phase, increasing the water content retains them longer. In size exclusion the biggest molecule leaves first because it cannot enter the pores — the opposite of the intuition every other technique builds.
Common mistakes
- Forgetting to subtract tM. k uses the adjusted retention time (tR − tM), not tR.
- Mixing the two N formulas. Use 16(tR/wb)2 with the baseline width and 5.54(tR/w½)2 with the width at half height. Swapping them changes N by roughly a factor of three.
- Reporting α < 1. By convention the later-eluting peak goes on top.
- Mixing units of tR and w. Both are times; both must be in the same unit before the ratio is squared.
- Assuming higher flow always means better separation. Beyond uopt the Cu term takes over and resolution deteriorates.
- Confusing efficiency with selectivity. If α = 1 the peaks are chemically indistinguishable on that column, and no number of plates will separate them.
- Chasing resolution with a longer column first. Rs ∝ √N, so it is the least efficient of the three routes.
- Quoting Rf without the solvent system. Rf is only reproducible for a stated plate, solvent and temperature.
Where this appears in GATE Chemistry
| Question type | What is tested | Usual form |
|---|---|---|
| Compute k and α from a chromatogram | Adjusted retention time | Numerical answer |
| Compute N and H | Correct width formula and L/N | Numerical answer |
| Compute Rs | Peak separation versus average width | Numerical answer |
| Plates required for a target Rs | Rearranging the master equation | Numerical answer |
| van Deemter reasoning | uopt, Hmin, term origins | Numerical or conceptual |
| Elution order | Normal versus reversed phase; size exclusion | Multiple choice |
The balance between analytical and other topics is revised from time to time, so confirm the current syllabus and question pattern from the official GATE notification for your exam year rather than from any secondary summary, this one included.
These are short calculations that punish careless arithmetic. Squaring a ratio, taking a square root of B/C, and dividing by a small number like 0.038 are exactly the steps where a dropped decimal turns correct method into a wrong option. There is no dedicated chromatography tool in the suite, so this button opens the calculator home view honestly — use the Scientific Calculator for the roots and squares.
Open the ABC Chemistry Calculator Suite →Preparing for GATE, IIT-JAM, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated chemical-sciences batches — coaching centre and online classes across India. Details at abcchemistry.in.