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Identify the calibration coordinate
Chapters 1–2. Separate the measured position from the assigned mass
Convert and weight the fractions
Chapters 3–5. Derive the mass conversion before calculating averages
Recalculate the distribution
Chapters 6–8. Recalculate after converting every fraction
Test limits and a new sample
Chapters 9–10. Keep experimental limitations attached to the model
Questions and measurement routes
Chapters 11–12. Questions readers ask
Full chapter index · Worked practice · Polymer Science subject root
Two polymer fractions can elute at the same column volume without having the same molar mass. A reference calibration translates an elution position using the reference polymer’s behaviour. If the sample behaves differently, both its mass scale and its apparent distribution can change.
This guide builds a two-fraction example from a dimensionless universal-calibration model. It calculates the resulting number and mass averages rather than assuming calibration adds one constant correction. The parameters are invented for teaching, not values for any named polymer or solvent.
Chapter index
Identify the calibration coordinate · Chapters 1–2
1. Separate the measured position from the assigned mass
2. Define a dimensionless calibration coordinate
Convert and weight the fractions · Chapters 3–5
3. Derive the mass conversion before calculating averages
4. Specify what the detector weights mean
5. Calculate the reference-equivalent averages
Recalculate the distribution · Chapters 6–8
6. Recalculate after converting every fraction
7. Explain why correcting the final average fails
8. Identify the special case of a constant scale factor
Test limits and a new sample · Chapters 9–10
9. Keep experimental limitations attached to the model
10. Practise with new masses and unequal weights
Questions and measurement routes · Chapters 11–12
11. Questions readers ask
12. Continue from separation to defensible interpretation
CHAPTER 1 OF 12 · Identify the calibration coordinate
1. Separate the measured position from the assigned mass
Size-exclusion chromatography, or SEC, produces a separation in elution volume. A conventional calibration associates that position with the molar mass of a reference material. Calling the assigned result a reference-equivalent mass keeps the interpretation visible.
The existing SEC guide explains pore access, hydrodynamic size and detectors. Here the narrower question is quantitative: if the reference and sample have different size–mass relations, how can the reported distribution change?
For the model, assume ideal size-exclusion behaviour in the working range, dilute noninteracting species, valid universal calibration and sufficiently narrow fractions. Exclude adsorption, aggregation and unresolved mixtures within a fraction.
These assumptions make the conversion identifiable. If several architectures with different masses co-elute and their proportions are unknown, one calibration assignment cannot recover every constituent mass from elution position alone.
CHAPTER 2 OF 12 · Identify the calibration coordinate
2. Define a dimensionless calibration coordinate
Universal calibration uses the product of intrinsic viscosity [η] and molar mass M as a size-related coordinate under suitable conditions. Let M* and η* be fixed reference units, and define m = M/M* and y = [η]/η*.
The dimensionless product is h = ym = [η]M/(η*M*). Equal elution volumes are assigned equal h in this ideal model. Normalisation avoids taking fractional powers of an unlabelled dimensional mass.
Suppose the reference polymer A obeys yA = mA^0.5, whereas sample B obeys yB = mB. These are invented relations over the limited range used here. They are not a material database or an assertion that these powers apply universally.
The resulting coordinates are hA = mA^1.5 and hB = mB². The same value of h can therefore correspond to different m values.
CHAPTER 3 OF 12 · Convert and weight the fractions
3. Derive the mass conversion before calculating averages
At a shared elution position, mA^1.5 = mB². Taking the positive square root gives mB = mA^0.75. This converts each reference-equivalent slice mass into the model’s sample mass.
An apparent mA of 16 becomes mB = 16^0.75 = 8. An apparent mA of 81 becomes mB = 81^0.75 = 27. Check through the product: 16^1.5 = 8² = 64, while 81^1.5 = 27² = 729.
If M* is 10,000 g mol⁻¹, those positions are assigned reference masses of 160,000 and 810,000 g mol⁻¹. Their sample masses are instead 80,000 and 270,000 g mol⁻¹ under our model.
The correction factors are 0.5 and one third. A single multiplier cannot perform both conversions. The change is nonlinear because the two size–mass exponents differ.
| Fraction | h | Reference mA | Sample mB | mB/mA |
|---|---|---|---|---|
| 1 | 64 | 16 | 8 | 0.5 |
| 2 | 729 | 81 | 27 | 0.333… |
CHAPTER 4 OF 12 · Convert and weight the fractions
4. Specify what the detector weights mean
Assume the two fractions contain equal polymer mass: mass fractions w₁ = w₂ = 0.5. These fractions are established by a correctly calibrated concentration measurement and remain fixed while we change only the mass assignment.
Equal mass does not mean equal molecule count. At the same total mass, the lower-mass fraction contains more molecules. That distinction determines which average formula to use.
For mass fractions wi and slice molar masses Mi, mass-average molar mass is Mw = ΣwiMi. Number-average molar mass is Mn = 1/(Σwi/Mi). The second expression follows because molecule count is proportional to mass divided by molecular molar mass.
Here each fraction is idealised as having one mass. Real detector slices can contain unresolved heterogeneity. Detector response, concentration calibration and slice resolution must be checked before treating the observed trace as this two-fraction ledger.
CHAPTER 5 OF 12 · Convert and weight the fractions
5. Calculate the reference-equivalent averages
Use dimensionless apparent masses 16 and 81 with equal mass fractions. The apparent mass average is mw,A = 0.5×16 + 0.5×81 = 48.5.
The apparent number average is mn,A = 1/(0.5/16 + 0.5/81) = 2592/97, approximately 26.722. Thus apparent dispersity is 48.5/26.722, approximately 1.815.
Multiplying the two averages by M* converts them to g mol⁻¹. With the chosen M*, apparent Mn is approximately 267,216 g mol⁻¹ and apparent Mw is 485,000 g mol⁻¹. Dispersity remains dimensionless.
These are internally consistent calculations on the reference-equivalent scale. The arithmetic is not the failure. The error would be labelling them as the sample’s actual masses while leaving the calibration mismatch undisclosed.
CHAPTER 6 OF 12 · Recalculate the distribution
6. Recalculate after converting every fraction
Now use the corrected dimensionless masses 8 and 27 while keeping the same equal mass fractions. The mass average is mw,B = 17.5.
The number average is mn,B = 1/(0.5/8 + 0.5/27) = 432/35, approximately 12.343. Corrected dispersity is 17.5/12.343, approximately 1.418.
For M* = 10,000 g mol⁻¹, the corrected averages are approximately 123,429 and 175,000 g mol⁻¹. The distribution is narrower on this relative summary than the apparent result suggested.
This does not mean the physical sample changed during recalculation. Its two component masses and mass fractions were held fixed. We changed the mapping from measured elution coordinate to assigned molar mass. Calibration can alter apparent breadth as well as the central mass scale.
| Quantity | Reference-equivalent scale | Corrected sample scale |
|---|---|---|
| Mn/M* | 26.722 | 12.343 |
| Mw/M* | 48.500 | 17.500 |
| Dispersity | 1.815 | 1.418 |
CHAPTER 7 OF 12 · Recalculate the distribution
7. Explain why correcting the final average fails
A tempting shortcut applies mB = mA^0.75 to the already calculated Mw. That gives 48.5^0.75, approximately 18.38, rather than the correct 17.5.
The conversion is nonlinear, so converting the average differs from averaging the converted fractions. In symbols, f(Σwi mi) generally differs from Σwi f(mi). The distribution has to survive long enough for the conversion to be applied to each fraction.
Applying the same shortcut to Mn also fails, because its reciprocal weighting adds another nonlinear operation. There is no justification for treating a reported mean as if it were a single narrow slice.
When only aggregate Mn and Mw are available, an exact distribution-dependent correction may be impossible. Request the trace, slice weights and calibration information instead of presenting an invented whole-sample correction as a recovered measurement.
CHAPTER 8 OF 12 · Recalculate the distribution
8. Identify the special case of a constant scale factor
Suppose both materials instead share the same exponent a in y = k m^a, but have different positive coefficients kA and kB. Equal h gives kA mA^(a+1) = kB mB^(a+1).
Provided a + 1 is nonzero, mB = (kA/kB)^(1/(a+1)) mA. The conversion is a constant multiplier c across the applicable range.
With fixed mass fractions, both Mn and Mw then multiply by c, so their ratio is unchanged. Calibration mismatch can shift a mass scale without changing dispersity in this special case.
This separates two possibilities that a general warning might blur. Different prefactors with the same exponent give a constant rescaling in the model. Different exponents can reshape the assigned distribution. Determine which relation is supported before choosing the correction.
CHAPTER 9 OF 12 · Test limits and a new sample
9. Keep experimental limitations attached to the model
Universal calibration is conditional on the separation and the solution behaviour. The linked NIST material describes the physical coordinate and the need to preserve the relevant column, solvent, temperature and flow conditions.
Our assumed power laws need evidence over the analysed range. A conformation change, chemical heterogeneity or architecture-dependent relation can invalidate a single fitted power. An excellent curve fit does not identify all co-eluting species.
The concentration detector needs its own response basis. If response differs between components, equal trace areas need not mean equal mass. Correcting masses while keeping wrong weights would leave another error intact.
Additional viscometry or light-scattering measurements can add useful information. Their interpretation still requires detector alignment, concentration information and method-specific checks. The present example isolates one source of bias rather than certifying an analytical pipeline.
CHAPTER 10 OF 12 · Test limits and a new sample
10. Practise with new masses and unequal weights
Keep mB = mA^0.75, but take apparent masses 1 and 16 with mass fractions 0.25 and 0.75. Convert the fractions first: corrected masses are 1 and 8.
The apparent mass average is 12.25. Apparent number average is 1/(0.25/1 + 0.75/16) = 64/19, approximately 3.368. Their ratio is approximately 3.637.
The corrected mass average is 6.25. Corrected number average is 1/(0.25/1 + 0.75/8) = 32/11, approximately 2.909. Corrected dispersity is approximately 2.148.
Check that the number average does not exceed the mass average in either case. Then replace the nonlinear conversion with a constant factor of two: both averages double and dispersity stays fixed. The changed example tests whether the source of distortion has been understood.
Does the same elution volume prove equal molar mass? No. Under our ideal universal-calibration model it identifies equal [η]M, which can correspond to different masses.
Does a reference-equivalent result have no value? It can be useful when accurately labelled and used for a suitable comparison. Its meaning depends on the calibrant and conditions.
Can I correct dispersity by dividing it by a mass correction factor? No. Under a constant mass rescaling it is already unchanged. Under a nonlinear mapping, recalculate both averages from converted fractions.
Does a narrower corrected dispersity prove better material performance? No. Dispersity is a summary of assigned masses. Structure, processing and the relevant property measurements remain separate questions.
CHAPTER 12 OF 12 · Questions and measurement routes
12. Continue from separation to defensible interpretation
The two-fraction distributions and calculations are original teaching examples. NIST’s SEC chapter and viscometric-detection reference provide the scientific context for universal calibration and its experimental scope.
Use the existing SEC mechanism guide before this calculation if pore access and detector signals are unfamiliar. Follow the molar-mass-average guide when the averaging basis is the problem, and the matched-averages guide when distribution shape remains uncertain after two means agree.
For a practical reading check, ask for the measured coordinate, calibration relation, concentration weights, validity range and exact quantity being reported. An appropriately limited result is more useful than an apparently precise value whose mass scale is unclear.
The transferable lesson is to apply a justified nonlinear transformation to the underlying components before aggregating them. It appears in many measurement systems, but each application needs its own physical relation and weights.
References checked 22 October 2026
NIST: Size-exclusion chromatography, universal calibration and coupled detectors
NIST: Viscometric detection in size-exclusion chromatography
Continue through Polymer Science
SEC pore access, hydrodynamic size and detector evidence
Matching molar-mass averages with different chain populations
Chain scission, molecule counts and number-average molar mass
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