A copolymer can contain forty percent B by monomer-derived unit count and nearly sixty-seven percent B by mass. Both descriptions can be correct. The apparent disagreement disappears when we state what has been counted and what sits in the denominator.
We will convert composition bases, track unequal conversion in a feed and test what an overall composition cannot establish about individual chains or sequences. Unit masses and reaction ledgers are invented educational values. The examples do not specify a synthesis or characterise a commercial polymer.
Find the question you need
Choose a route below or read in order. All teaching chapters remain open; the index groups let you jump directly to a worked example or check.
Route 1 · Name the composition basis
Chapters 1–2: Name the composition basis.
Route 2 · Convert units into mass
Chapters 3–4: Convert units into mass.
Route 3 · Follow feed into incorporated units
Chapters 5–6: Follow feed into incorporated units.
Route 4 · Test averages and structural claims
Chapters 7–8: Test averages and structural claims.
Route 5 · Practise, check and continue
Chapters 9–12: Practise, check and continue.
Full chapter index · Try the transfer checks · Polymer Science subject root
Chapter contents
Name the composition basis · Chapters 1–2
Convert units into mass · Chapters 3–4
Follow feed into incorporated units · Chapters 5–6
Test averages and structural claims · Chapters 7–8
Practise, check and continue · Chapters 9–12
CHAPTER 1 OF 12 · Name the composition basis
1. Say which entities the fraction counts
Let nA and nB be amounts of incorporated A-derived and B-derived units in a binary copolymer population. Define xB = nB/(nA+nB). This is a unit amount fraction. Counting individual units gives the same ratio because the conversion between entities and amount is common.
It is not the fraction of polymer molecules that contain B. One long chain may carry thousands of B-derived units, while many short chains may carry none. A fraction of units, a fraction of chains and a fraction of specimen mass answer different questions.
Write “B-derived units: 40 mol%” when that is the actual basis. “Forty percent B” without a basis leaves the reader to guess. Also say whether the sample is the isolated polymer, one separated fraction, a whole formulation or the reaction feed.
CHAPTER 2 OF 12 · Name the composition basis
2. Define the mass assigned to each incorporated unit
Use MA and MB for the molar masses assigned to the incorporated units on a consistent chemical basis. Their contributions are nAMA and nBMB. In our binary idealisation, total polymer mass is their sum.
For a long addition copolymer, an appropriate monomer-derived unit may retain the monomer’s atoms. Condensation, chemical modification and other reactions require their own atom balance. Do not automatically substitute a starting reagent’s molar mass for the incorporated structural contribution.
Our initial ledger neglects end groups, additives, residual monomer and by-products. Those terms must be included when significant. Penn State’s introduction to polymer size and molar mass provides a route through repeat-unit distinctions and composition-weighted unit masses.
CHAPTER 3 OF 12 · Convert units into mass
3. Work from forty mole percent to mass fraction
Take nA = 60 mol of incorporated units, nB = 40 mol, MA = 50 g mol⁻¹ and MB = 150 g mol⁻¹. The stipulated masses are 3,000 g from A-derived units and 6,000 g from B-derived units.
Thus xB = 40/100 = 0.40, but wB = 6,000/9,000 = 2/3. B supplies fewer units than A yet twice its mass because each B-derived unit has three times the molar mass.
The mean molar mass per mole of incorporated units is 9,000/100 = 90 g mol⁻¹. It is a composition average over units. It is not the molar mass of an average polymer molecule.
| Contribution | A-derived | B-derived |
|---|---|---|
| Incorporated amount | 60 mol | 40 mol |
| Unit molar mass | 50 g mol⁻¹ | 150 g mol⁻¹ |
| Assigned mass | 3,000 g | 6,000 g |
| Unit fraction | 60% | 40% |
| Mass fraction | 1/3 | 2/3 |
Substitute nB = xB(nA+nB) and nA = (1−xB)(nA+nB) into the mass ratio. The common total amount cancels, giving wB = xBMB/[(1−xB)MA+xBMB].
Starting with mass fractions instead gives xB = (wB/MB)/[(1−wB)/MA+wB/MB]. Divide each mass contribution by its own unit molar mass before normalising the amounts.
For equal mass contributions, wB = 0.5. With the same unit masses, xB = (0.5/150)/(0.5/50+0.5/150) = 0.25. A material called “50:50” can therefore mean 25 mol% B on this basis.
Check the endpoints: xB = 0 gives wB = 0, and xB = 1 gives wB = 1. If MA = MB, the two fractions coincide. These checks explain when the tempting equality is valid.
CHAPTER 5 OF 12 · Follow feed into incorporated units
5. Keep feed composition separate from polymer composition
A feed records what enters the reaction system; incorporated-unit composition records what has entered the polymer population being measured. Equal feed and product fractions are not an accounting identity.
Suppose the initial feed contains 60 mol A and 40 mol B. At a chosen stopping point, 30 mol A and 30 mol B have become polymer units, with no side reactions or material losses in this constructed addition model.
A conversion is 30/60 = 50%, while B conversion is 30/40 = 75%. The polymer’s B unit fraction is 30/60 = 50%, even though the starting feed’s B fraction was 40%. The remaining monomer contains 30 mol A and 10 mol B: only 25 mol% B.
This is a stipulated conversion ledger, not a kinetic prediction. To predict conversion or instantaneous incorporation, we would need suitable reaction chemistry, rate information and a justified model.
CHAPTER 6 OF 12 · Follow feed into incorporated units
6. Reconcile unit conversion with mass conversion
Under the same retained-atom model, the polymer contains 30×50 + 30×150 = 6,000 g. The initial feed mass was 9,000 g, so the polymer contains two thirds of that starting monomer mass.
By unit count, 60 of 100 mol have been incorporated, giving 60% conversion on the total-unit basis. Sixty percent and sixty-six-and-two-thirds percent describe different totals, not conflicting laboratory results.
The polymer’s B mass fraction is 4,500/6,000 = 75%. One ledger now contains feed B unit fraction 40%, polymer B unit fraction 50%, polymer B mass fraction 75%, total-unit conversion 60% and converted-mass fraction 66.67%. Every number needs its noun.
| Quantity | Result | Denominator |
|---|---|---|
| Initial B unit fraction | 40% | 100 mol initial units |
| Polymer B unit fraction | 50% | 60 mol incorporated units |
| Polymer B mass fraction | 75% | 6,000 g polymer |
| Total-unit conversion | 60% | 100 mol starting units |
| Converted-mass fraction | 66.67% | 9,000 g starting monomer mass |
CHAPTER 7 OF 12 · Test averages and structural claims
7. Distinguish an increment from the accumulated product
Continue the hypothetical reaction by incorporating another 20 mol A and 5 mol B. The new increment is 20 mol% B. The accumulated polymer now contains 50 mol A and 35 mol B, so its B unit fraction is 35/85 = 7/17 ≈ 41.18%.
The increment and the accumulated product do not share the same composition. An analysis of the full isolated polymer averages material incorporated across the history. It need not reveal the composition of the most recent growth.
The cumulative B mass fraction is 35×150/(50×50+35×150) = 21/31 ≈ 67.74%. The residual feed is 10 mol A and 5 mol B. Initial feed, accumulated product, latest increment and residual feed can all have different compositions.
Reconcile each species: 60 = 50+10 for A and 40 = 35+5 for B. The ledger closes without assuming that composition stayed constant during growth.
CHAPTER 8 OF 12 · Test averages and structural claims
8. Weight combined samples by the quantity being counted
Lot 1 contains 100 mol of incorporated units at xB = 0.20. Lot 2 contains 300 mol of units at xB = 0.80. Mixing all the material gives 20+240 = 260 mol B units out of 400, so overall xB = 0.65.
The unweighted mean of 0.20 and 0.80 is 0.50. It fails because the lots contain different amounts of units. The correct weights are 100/400 and 300/400.
With our unit masses, lot masses are 7,000 and 39,000 g. B-derived mass totals 39,000 g out of 46,000, giving wB = 39/46 ≈ 84.78%. Mass-weighting the two lot mass fractions gives the same answer.
Use unit amounts to combine unit fractions and masses to combine mass fractions. The fact that both sets of weights sum to one does not make the weights interchangeable.
CHAPTER 9 OF 12 · Practise, check and continue
9. Test what one overall fraction cannot identify
Consider a ten-unit chain containing five A and five B units. ABABABABAB and AAAAABBBBB have identical overall unit fractions and identical mass fractions in the model. Their order along the chain differs.
A bulk 50 mol% B measurement therefore does not prove alternation, blocks or any particular statistical sequence. It also does not establish that every chain is 50 mol% B. A mixture of A-rich and B-rich chains can share the same bulk total.
Instrument response adds another layer. A measured signal fraction equals a unit fraction only under the appropriate response-per-unit model. The existing copolymer SEC detector-response guide treats signal-to-mass weighting; that correction answers a different question from converting a known mass fraction to a unit fraction.
State the structural claim first, then ask which additional measurement distinguishes the competing chain populations or sequences. A single total composition is a constraint, not a complete architecture.
CHAPTER 10 OF 12 · Practise, check and continue
10. Audit the specimen and the denominator
An isolated polymer analysis may exclude insoluble network material or preferentially lose one composition during extraction. Its reported fraction then describes the recovered population unless representative recovery is established.
Additives and residual monomer may contribute to whole-specimen mass or to a chemical signal. Decide whether “B mass fraction” means B-derived polymer-unit mass divided by polymer mass, or divided by the entire formulation. These differ whenever other material is present.
For short chains, include end-group mass explicitly. For modified or condensation-derived units, use an appropriate atom balance. The convenient binary denominator is a model that must be revised when the sample contains additional constituents.
Keep raw quantities, chemical assignments, conversion steps and recovery information together. A normalised total of 100% is a check on arithmetic, not proof that the specimen and method represent the intended material.
First replace the unit masses by MA = 80 and MB = 120 g mol⁻¹, keeping xB = 0.40. Predict whether B’s mass fraction remains above its unit fraction. Then calculate wB = 48/(48+48) = 0.50.
Reverse the calculation using wB = 0.50: equal masses correspond to amounts proportional to 1/80 and 1/120, giving xB = 0.40. The round-trip check returns the starting composition.
Second, take an idealised composition measurement with nonoverlapping quantitative signals. Each A unit contributes three signal units and each B unit two. Measured signals are six and four. Corrected unit amounts are 6/3 = 2 and 4/2 = 2, so xB = 0.50; the raw B signal share was 0.40.
This last exercise stipulates exact response and acquisition conditions. Actual spectroscopy requires suitable assignments, quantitative acquisition and validation. The algebra does not certify those experimental conditions.
CHAPTER 12 OF 12 · Practise, check and continue
12. Questions, sources and further reading
Can feed ratio stand in for measured polymer composition?
Only with evidence supporting that equivalence for the actual reaction and recovered population. The unequal-conversion ledger demonstrates why it is not automatic.
Is the heavier unit always the majority by mass?
No. Amount matters too. The heavier unit has a mass fraction above its own unit fraction in this binary model, but may still be below half the total mass if sufficiently rare.
Does one composition determine material behaviour?
No. Sequence, architecture, molar-mass distribution, morphology, formulation and testing conditions can matter. A composition measurement narrows the explanation without completing it.
Source checked 22 October 2026: Penn State’s linked polymer lesson supports the distinction between units, chain molar mass and composition-weighted unit mass. All masses, lot balances, reaction increments and signal calculations here are original constructed exercises.
Choose the next question
Signals, recovery and composition
- Detector Response Factors and Mass Fractions in Copolymer SEC
- Extractable Additives and Incomplete Mass Recovery
- Sol and Gel Fractions with Unequal Unit Masses
- Polymer Chemistry — Monomers, Chains, Architecture, Molecular Weight and Material Behaviour
Chain populations and molecular architecture
- Matching Molar-Mass Averages, Different Chain Populations
- Step-Growth Imbalance and the Ceiling on Average Chain Length
- Chain Scission, Molecule Counts and the Fall in Number-Average Molar Mass
- Branching and the Molar-Mass Ambiguity Behind One SEC Peak
Thermal interpretation needs its own evidence
- Fox Equation, Glass Transition and Blend Miscibility
- Scan Rate, Thermal Lag and the Reported Glass Transition
Return to the Polymer Science subject root · The Gold Standard of Education: learning transfers into life · Explore How X Works.