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Define the objects
Chapters 1–2. Define the fraction before weighing
Build the argument
Chapters 3–5. Calculate count fraction and mass fraction
Compare and interpret
Chapters 6–8. Move from polymer to a filled specimen
Check and practise
Chapters 9–10. Identify what can invalidate the interpretation
Questions and next routes
Chapters 11–12. Questions readers ask
Full chapter index · Worked practice · Polymer Science subject root
A network containing 60% of the structural units does not necessarily contain 60% of the polymer mass. The answer depends on which units entered the network and how much each contributes to the dry polymer.
This guide separates a connectivity count, a mass fraction and an extraction result. It moves from a deliberately simple two-unit model to a filled specimen, showing why an apparently precise gel percentage can use the wrong denominator. All sample masses are invented for teaching.
Chapter index
Define the objects · Chapters 1–2
1. Define the fraction before weighing
2. Build an unequal-mass unit model
Build the argument · Chapters 3–5
3. Calculate count fraction and mass fraction
4. Derive when the two fractions agree
5. Change the composition of the network
Compare and interpret · Chapters 6–8
6. Move from polymer to a filled specimen
7. Close the extraction mass balance
8. Keep swelling separate from gel fraction
Check and practise · Chapters 9–10
9. Identify what can invalidate the interpretation
10. Practise with a different denominator
Questions and next routes · Chapters 11–12
11. Questions readers ask
12. Sources and connected learning routes
In a network-forming polymer system, the gel fraction concerns the mass associated with the network material. The sol fraction concerns finite molecules that can be dissolved or dispersed. These are mass-based quantities; counting constitutional units answers a different question.
For a closed, dry polymer-only accounting model, let mₚ be the initial polymer mass, mɢ the network mass and mₛ the sol mass. If the two categories exhaust that polymer without reaction or loss, mₚ = mɢ + mₛ. Define g = mɢ/mₚ and s = mₛ/mₚ, so g + s = 1.
Keep this ideal partition distinct from an operational solvent extraction. Whether an experiment isolates those categories depends on its chemistry, solvent, time, temperature and recovery procedure.
Consider polymer made from two types of incorporated units, A and B. Their dry contributions are 100 and 200 mass units per unit, respectively. These are contributions after incorporation; they are not automatically the starting monomer molar masses when polymerisation releases a small molecule.
There are 60 A units and 40 B units. The initial polymer mass is therefore 60×100 + 40×200 = 14,000 mass units. There are 100 units, but they do not all carry the same mass.
Suppose connectivity analysis assigns 50 A units and 10 B units to one spanning network. The other 10 A and 30 B units belong to finite molecules. This classification is supplied as part of the teaching model; unit composition alone would not reveal the network connections.
Sixty units belong to the network, giving a network unit-count fraction of 60/100 = 60%. Their mass is 50×100 + 10×200 = 7,000. The gel mass fraction is 7,000/14,000 = 50%.
The finite molecules contain 10×100 + 30×200 = 7,000 mass units. Their sol mass fraction is also 50%, even though they contain only 40% of the units. The count and mass ledgers both reconcile.
The discrepancy is not an instrument error. The gel is enriched in lighter A units while the sol is enriched in heavier B units. Changing the weights changes the fraction without changing the assigned connections.
| Component | A units | B units | Unit count | Mass |
|---|---|---|---|---|
| Gel | 50 | 10 | 60 | 7000 |
| Sol | 10 | 30 | 40 | 7000 |
| Total | 60 | 40 | 100 | 14000 |
Let nᵢ be the total number of incorporated units of type i, mᵢ their individual mass contribution and aᵢ the fraction of type i assigned to the gel. Then g = Σnᵢmᵢaᵢ / Σnᵢmᵢ. The unit-count fraction is c = Σnᵢaᵢ / Σnᵢ.
Using the unit-count distribution as weights, g − c = Cov(m,a)/E[m]. This is a bookkeeping identity for the specified unit partition. It is not a chemical gelation criterion.
Equal unit masses are sufficient for equality. Equal gel participation across all unit types is also sufficient. Other cancellations can produce equality, so matching percentages do not prove either sufficient condition holds.
Keep the total inventory at 60 A units and 40 B units. Now assign 20 A and all 40 B units to the network. It still contains 60 units, so its count fraction remains 60%.
Its mass becomes 20×100 + 40×200 = 10,000. The gel mass fraction is 10,000/14,000 = 5/7, approximately 71.43%. The remaining 40 A units supply the sol mass of 4,000.
Compare this with the first network’s 50% gel mass. A single count percentage conceals substantial compositional information. To predict mass from connectivity, retain which units belong to which component as well as how many there are.
A dry specimen weighs 1.000 g and contains 0.800 g polymer plus 0.200 g insoluble filler. After extraction and drying, the residue weighs 0.680 g. Assume all filler remains, all sol polymer is removed, no polymer degrades and no solvent remains.
The network polymer mass is 0.680 − 0.200 = 0.480 g. The polymer-basis gel fraction is 0.480/0.800 = 60%. The residue fraction on the original specimen is 68%, a different quantity.
Simply calling 68% the polymer gel fraction credits the filler to the polymer network. The correct subtraction depends on independently justified filler mass and retention; those conditions cannot be inferred from the two weighings.
| Quantity | Mass or fraction | Meaning |
|---|---|---|
| Initial specimen | 1.000 g | Polymer plus filler |
| Initial polymer | 0.800 g | Polymer denominator |
| Dry residue | 0.680 g | Gel polymer plus filler |
| Gel polymer | 0.480 g | Corrected numerator |
| Residue fraction | 68% | Specimen basis |
| Gel fraction | 60% | Polymer basis |
In this specimen model, sol polymer mass is 0.800 − 0.480 = 0.320 g. The residue and recovered sol sum to 0.680 + 0.320 = 1.000 g. Polymer-only fractions are 60% gel and 40% sol.
If recovered dissolved material weighs 0.350 g instead, the result needs investigation. It may contain extractable additives or retained solvent, or the initial composition may be wrong. Do not force it into the polymer sol category merely because it came from the solvent.
A mass balance is a diagnostic, not a guarantee of chemical identification. Two compensating errors can close a balance. Identify the material in each stream and document which components the calculation includes.
CHAPTER 8 OF 12 · Compare and interpret
8. Keep swelling separate from gel fraction
Suppose the dried 0.680 g residue later takes up solvent and weighs 2.080 g in its swollen state. The added solvent mass is 1.400 g under the stated assumptions. A swollen-to-dry-residue ratio would be 2.080/0.680, approximately 3.06.
That ratio is not the gel fraction. It compares a solvent-containing state with a dry state. A polymer-basis swelling ratio would also require the filler correction and a precise definition of the reported numerator.
Swelling may help investigate network structure under an appropriate model. Converting it into crosslink density requires further thermodynamic and structural assumptions. One extraction mass fraction alone does not determine the network’s elastic response.
CHAPTER 9 OF 12 · Check and practise
9. Identify what can invalidate the interpretation
An insoluble crystalline or physically associated polymer may resist a particular extraction without constituting the permanent chemical network intended by the analysis. Conversely, harsh extraction can break network bonds and remove material that originally belonged to the gel.
Incomplete extraction inflates the dry residue; incomplete drying can do the same. Mechanical loss of network fragments can reduce it. Fillers, pigments and extractable additives can affect different sides of the balance.
Specify the polymer system, solvent, extraction protocol, drying endpoint and component corrections. Use a validated material-specific method when an experimental or quality decision requires it. The arithmetic here explains denominators; it does not establish a universal laboratory protocol.
A dry 2.000 g specimen contains 0.500 g retained insoluble filler. Its final dry residue is 1.400 g. Under the same ideal assumptions as before, calculate the polymer gel and sol fractions.
Initial polymer mass is 1.500 g. Gel polymer mass is 1.400 − 0.500 = 0.900 g. Therefore g = 0.900/1.500 = 60%, and sol polymer mass is 0.600 g, giving s = 40%.
The specimen residue fraction is 70%. Dividing the corrected gel polymer mass by the whole specimen would give 45%, which is gel-polymer mass per specimen mass. That can be a useful composition quantity, but it must carry that label.
Is gel fraction the proportion of reacted functional groups? No. Conversion counts reaction progress, whereas gel fraction assigns polymer mass to a network. They require different information.
Does a large gel fraction prove high crosslink density? No. Network mass and the density of elastically relevant connections describe different features. Loops, dangling chains, chemistry and swelling conditions matter.
Can a unit-count probability equal a mass fraction? Yes, under suitable weighting conditions. The first worked model shows why equality cannot be assumed with unequal unit masses and selective network participation.
Is every insoluble residue gel? No. The measured residue can include filler, trapped material or polymer that is insoluble for another reason. The interpretation requires chemical and procedural evidence.
CHAPTER 12 OF 12 · Questions and next routes
12. Sources and connected learning routes
The IUPAC terminology links establish the mass basis of gel and sol fractions. The two-unit models, covariance identity and specimen corrections are original bookkeeping examples with their assumptions stated explicitly.
Read the network-formation article for connectivity and the gel point, then the cluster-average article for why different molecular averages behave differently near network formation. This guide adds the composition and specimen denominators needed to interpret a fraction.
For a transfer check, change the unit masses, the network assignment and the filler loading one at a time. Explain which changes affect connectivity, which affect weighting and which affect measurement interpretation. Return to the Polymer Science root to choose the next mechanism.
References checked 22 October 2026
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