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Choose a route or begin at Chapter 1. Teaching chapters stay open; the chapter index expands by section.
Define the objects
Chapters 1–2. Separate the specimen into named components
Build the argument
Chapters 3–5. Construct a complete mass ledger
Compare and interpret
Chapters 6–8. Correct the extract without double-counting losses
Check and practise
Chapters 9–10. Bound the answer when retention is only bounded
Questions and next routes
Chapters 11–12. Questions readers ask
Full chapter index · Worked practice · Polymer Science subject root
An extraction can remove polymer and additives together. A dry residue can retain network polymer, filler and some additive. Even when every balance reading is accurate, those mixed streams do not identify the polymer gel fraction by themselves.
This guide builds a component balance for a formulated specimen and shows which additional measurements resolve it. The purpose is to interpret mass accounting, not to prescribe an extraction solvent or laboratory method. All compositions and recoveries are original teaching cases.
Chapter index
Define the objects · Chapters 1–2
1. Separate the specimen into named components
2. State the idealised extraction assumptions
Build the argument · Chapters 3–5
3. Construct a complete mass ledger
4. Identify three different naive answers
5. Correct the residue with identified components
Compare and interpret · Chapters 6–8
6. Correct the extract without double-counting losses
7. Use the mass deficit as a diagnostic
8. Demonstrate that the weighings do not identify gel
Check and practise · Chapters 9–10
9. Bound the answer when retention is only bounded
10. Practise a new component balance
Questions and next routes · Chapters 11–12
11. Questions readers ask
12. Sources and connected learning routes
CHAPTER 1 OF 12 · Define the objects
1. Separate the specimen into named components
Let the initial dry specimen contain polymer mass P, insoluble filler mass F and additive mass D. Its total mass is M₀ = P + F + D. This is a specified three-component bookkeeping model; real formulations may need more components.
Partition the polymer into network material G and finite soluble molecules S, so P = G + S under the ideal chemical classification used here. The polymer-basis gel fraction is G/P. It is not the fraction of the entire specimen remaining in a container.
An additive is distinguished here by its role as a non-polymer formulation component. Whether it extracts, reacts, adsorbs or remains trapped is a separate question. Do not identify the chemical category of a recovered mass solely from which vessel contains it.
Assume all finite soluble polymer leaves the specimen, all network polymer remains chemically intact, and all filler stays with the residue. Let Dᵣ be additive retained in the dry residue, leaving D − Dᵣ to enter the extract.
The dry residue mass is R = G + F + Dᵣ. The material transferred out of the specimen has mass S + D − Dᵣ. Some of that transferred material may be lost during subsequent collection or handling.
Keep extraction completeness separate from recovery completeness. Complete removal from the specimen does not imply complete recovery in the collection vessel. Likewise, high collection recovery does not establish that the specimen was fully extracted.
Take a 1.000 g specimen containing P = 0.800 g polymer, F = 0.100 g filler and D = 0.100 g additive. Its polymer consists of G = 0.480 g gel and S = 0.320 g sol, so the true polymer gel fraction is 60%.
Suppose 0.020 g additive stays with the residue and 0.080 g enters the extract. The residue is 0.480 + 0.100 + 0.020 = 0.600 g. The transferred material is 0.320 + 0.080 = 0.400 g.
Now suppose only 90% of the transferred sol polymer is recovered, while all transferred additive is recovered. The collection vessel contains 0.288 g polymer and 0.080 g additive, totalling 0.368 g. The missing 0.032 g is polymer lost after extraction in this constructed case.
| Component | Initially present (g) | Dry residue (g) | Recovered extract (g) | Unrecovered (g) |
|---|---|---|---|---|
| Gel polymer | 0.480 | 0.480 | 0 | 0 |
| Sol polymer | 0.320 | 0 | 0.288 | 0.032 |
| Filler | 0.100 | 0.100 | 0 | 0 |
| Additive | 0.100 | 0.020 | 0.080 | 0 |
| Total | 1.000 | 0.600 | 0.368 | 0.032 |
Calling R/M₀ the gel fraction gives 0.600/1.000 = 60%. That happens to equal the true polymer gel fraction here, but only by numerical coincidence. Its numerator contains filler and additive, and its denominator contains the whole formulation.
Subtracting filler but ignoring retained additive gives (R − F)/P = 0.500/0.800 = 62.5%. Treating all recovered extract as sol polymer gives 1 − 0.368/0.800 = 54%.
The three reported percentages disagree because they mix different components and recovery assumptions. An answer that coincides with the truth in one specimen can still come from an invalid method. Change the formulation and the coincidence may disappear.
CHAPTER 5 OF 12 · Build the argument
5. Correct the residue with identified components
If P, F and Dᵣ are independently established and the retention assumptions hold, then G = R − F − Dᵣ. In the worked case, G = 0.600 − 0.100 − 0.020 = 0.480 g, and G/P = 0.60.
Notice what is not needed for this residue-based estimate: the fraction of extracted sol later recovered in a collection vessel. The corrected residue can determine G even when extract recovery is incomplete, provided extraction itself was complete and the network was preserved.
The word independently matters. Assuming Dᵣ equals zero because the procedure was called an extraction does not measure it. A validated separation or component-specific analysis is needed when retained additive materially affects the interpretation.
CHAPTER 6 OF 12 · Compare and interpret
6. Correct the extract without double-counting losses
Let Eₚ be the recovered sol-polymer mass and rₛ its recovery fraction after removal from the specimen. Under the stated model, Eₚ = rₛS, so S = Eₚ/rₛ when rₛ is known and nonzero.
Here the total recovered extract E is 0.368 g. Subtract its independently identified additive mass of 0.080 g to obtain Eₚ = 0.288 g. Dividing by rₛ = 0.90 gives S = 0.320 g. Then G = P − S = 0.480 g.
Do not divide the entire mixed extract by the polymer recovery fraction unless every component shares that recovery. In this example, additive recovery is complete while polymer recovery is 90%. One correction factor cannot represent both streams.
The measured dry residue and collected dry extract sum to 0.600 + 0.368 = 0.968 g. Relative to the initial specimen, the deficit is 0.032 g, or 3.2%.
In our constructed ledger, we know its cause: unrecovered sol polymer. In an experiment, the same deficit could have other causes, including incomplete transfer, loss of particles, volatile material or an incorrect initial dry-mass basis. The deficit alone does not identify its composition.
Do not add the entire deficit to recovered sol polymer by default. That repair assumes the conclusion being investigated. First account for possible missing components, measurement uncertainty and transformations before assigning the missing mass to one category.
CHAPTER 8 OF 12 · Compare and interpret
8. Demonstrate that the weighings do not identify gel
Keep the same known initial P = 0.800 g, F = 0.100 g and D = 0.100 g, and the same observed R = 0.600 g and E = 0.368 g. A second possible world has G = 0.460 g, S = 0.340 g and Dᵣ = 0.040 g.
Its residue is again 0.460 + 0.100 + 0.040 = 0.600 g. If 0.308 g sol polymer and 0.060 g additive are recovered, the extract is again 0.368 g. Polymer recovery is now 0.308/0.340, approximately 90.59%.
Both worlds match every total weighing, yet their gel fractions are 60% and 57.5%. These totals are therefore insufficient to identify G. A component-specific measurement can distinguish the worlds; another repetition of the same two totals cannot remove this structural ambiguity.
| Quantity | World 1 | World 2 |
|---|---|---|
| Dry residue (g) | 0.600 | 0.600 |
| Recovered extract (g) | 0.368 | 0.368 |
| Retained additive (g) | 0.020 | 0.040 |
| Gel polymer (g) | 0.480 | 0.460 |
| Polymer gel fraction | 60% | 57.5% |
CHAPTER 9 OF 12 · Check and practise
9. Bound the answer when retention is only bounded
If all you know is 0 ≤ Dᵣ ≤ D, the residue equation gives R − F − D ≤ G ≤ R − F, also constrained by 0 ≤ G ≤ P.
For the example, 0.400 ≤ G ≤ 0.500 g, so the polymer gel fraction lies between 50% and 62.5%. This is an algebraic identification range under the model, not a statistical confidence interval.
If an independent analysis narrows retained additive to 0.015–0.025 g, the range becomes G = 0.475–0.485 g, or 59.375–60.625% of polymer. Retention uncertainty alone creates this range; balance and composition uncertainty would require additional treatment.
A 2.000 g dry specimen contains 1.600 g polymer, 0.250 g retained filler and 0.150 g additive. Its dry residue is 1.260 g, including 0.050 g retained additive. Determine gel and sol polymer masses.
Gel mass is 1.260 − 0.250 − 0.050 = 0.960 g. Sol mass is 1.600 − 0.960 = 0.640 g. Thus the polymer gel fraction is 60% and sol fraction is 40%.
If 0.576 g sol polymer and 0.100 g additive are collected, total extract is 0.676 g. The final deficit is 2.000 − 1.260 − 0.676 = 0.064 g. It equals 10% of the sol polymer only because the exercise specifies all other retention and recovery terms. Explain that condition before applying the interpretation elsewhere.
Is recovered extract the same as sol polymer? Only when its composition and relevant recovery conditions justify that identification. Additives can share the extract.
Does a closing mass balance prove the gel fraction is correct? No. Component misclassification or compensating errors can leave a total balance closed. Closure checks accounting, not chemical identity.
Can I reuse a recovery factor from another formulation? Not without evidence that the relevant separation and collection behaviour is comparable. A recovery factor is attached to a component and procedure.
Does dry residue always equal a permanent chemical network? No. Insolubility can have other causes, and the procedure can retain filler or other material. This guide assumes the chemical partition; it does not infer network topology from weighing alone.
CHAPTER 12 OF 12 · Questions and next routes
12. Sources and connected learning routes
IUPAC’s gel and sol terminology supplies the polymer mass categories. The mixed-component ledgers, two indistinguishable weighing scenarios and algebraic bounds are original derivations under the assumptions stated here.
Continue from the sol-and-gel denominator guide when unit masses or filler change the reference mass. Use the network-formation guide for connectivity and the gel point. These solve different parts of the interpretation problem.
For transfer, ask which observations distinguish the two worlds in Chapter 8. An independent retained-additive measurement or a resolved extract composition adds information; extra decimal places on the same total masses do not. Return to Polymer Science for the wider route through structure, measurement and properties.
References checked 22 October 2026
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Sol and gel fractions with unequal unit masses
Branching, gel points and network formation
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