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Chromatography peak resolution and plate count calculator

Chromatographic peak spacing alone does not describe separation: the same retention-time gap is less useful when the peaks are broad.

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Chromatographic peak spacing alone does not describe separation: the same retention-time gap is less useful when the peaks are broad. Resolution combines spacing with widths, while apparent plate counts relate individual peak width to retention time.

Worked example comparison
ExampleOutputWhat to notice
Worked exampleEntered peak pair resolution2For retention times five and six with baseline widths 0.5 each, resolution is two. The first apparent plate count is sixteen hundred and the second is twenty-three hundred four. With dead time one, the adjusted retention factors are four and five, so selectivity is 1.25. Replacing widths with valid half-height measurements requires changing the convention and the measured widths together.
Compare a changed assumptionEntered peak pair resolution1Bring the second peak closer while retaining widths; resolution falls even though both individual peaks still have finite plate estimates.

Check your understanding

Why is selectivity not simply six divided by five?

It uses adjusted retention factors, with the unretained-marker dead time removed before taking the ratio.

Predict the comparison before running it

Bring the second peak closer while retaining widths; resolution falls even though both individual peaks still have finite plate estimates.

Behind the numbers

How this calculation works

Chromatographic peak spacing alone does not describe separation: the same retention-time gap is less useful when the peaks are broad. Resolution combines spacing with widths, while apparent plate counts relate individual peak width to retention time. This calculator keeps baseline and half-height width conventions separate. It also calculates retention-factor selectivity using an entered unretained-marker time rather than using the raw retention-time ratio.

THE FORMULABaseline: Rs=2(t₂−t₁)/(w₁+w₂), N=16(tR/w)²; half-height coefficients 1.18 and 5.54

A worked example

For retention times five and six with baseline widths 0.5 each, resolution is two. The first apparent plate count is sixteen hundred and the second is twenty-three hundred four. With dead time one, the adjusted retention factors are four and five, so selectivity is 1.25. Replacing widths with valid half-height measurements requires changing the convention and the measured widths together.

Good to know

The page does not detect peaks from an image or instrument file. Tailing factors, extra-column broadening, gradient corrections and column-length calculations are outside this model. Identical numbers entered under different width conventions are a mathematical comparison, not two legitimate measurements of the same ideal peak. Retain the actual measurement definition when comparing columns or analytical runs.

What this calculation answers

Chromatographic peak spacing alone does not describe separation: the same retention-time gap is less useful when the peaks are broad. Resolution combines spacing with widths, while apparent plate counts relate individual peak width to retention time.

This calculator keeps baseline and half-height width conventions separate. It also calculates retention-factor selectivity using an entered unretained-marker time rather than using the raw retention-time ratio.

The Simple linear regression calculator fits an observation line. Peak resolution instead depends on retention-time spacing and widths measured with a consistent baseline or half-height convention.

Use the right input conventions

All retention times, dead time and widths must use the same time unit. Enter the earlier peak first, followed by the later peak, and place dead time below the first peak retention time.

Choose the convention that matches how both widths were measured. Baseline widths use the coefficients two for resolution and sixteen for plates; half-height widths use 1.18 and 5.54. Mixing conventions gives inconsistent results.

Interpret the result

Resolution is dimensionless. Each peak receives its own apparent plate count because its retention time and width may differ. Selectivity uses adjusted retention factors after subtracting dead time from both peak times.

These quantities describe the entered pair; they do not automatically classify a method as suitable. Peak asymmetry, integration choices and an unresolved shoulder can make manually entered widths poor representations of the actual chromatogram.

Work through the example

For retention times five and six with baseline widths 0.5 each, resolution is two. The first apparent plate count is sixteen hundred and the second is twenty-three hundred four.

With dead time one, the adjusted retention factors are four and five, so selectivity is 1.25. Replacing widths with valid half-height measurements requires changing the convention and the measured widths together.

Where the model stops

The page does not detect peaks from an image or instrument file. Tailing factors, extra-column broadening, gradient corrections and column-length calculations are outside this model.

Identical numbers entered under different width conventions are a mathematical comparison, not two legitimate measurements of the same ideal peak. Retain the actual measurement definition when comparing columns or analytical runs.

A few questions, answered

Why is selectivity not simply six divided by five?

It uses adjusted retention factors, with the unretained-marker dead time removed before taking the ratio.

What changes in the comparison?

Bring the second peak closer while retaining widths; resolution falls even though both individual peaks still have finite plate estimates.

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