How to Identify a Polymer from a Three-Unit Drawing

An HSC Chemistry question shows three consecutive sections of a polymer chain. Each section contains a \(\ce{CH2}\) group joined to a \(\ce{CCl2}\) group, and a single bond continues beyond both ends of the drawing. Does that mean the polymer contains exactly three repeating units?

No. The drawing shows three units, but it doesn’t tell you how many exist in the complete chain. To identify the polymer, find the smallest pattern that repeats, check that every atom has the correct number of bonds, and work backwards to a possible monomer.

In this case, the repeating unit is \(\ce{-CH2-CCl2-}\). Assuming that pattern continues throughout the chain, the polymer is poly(1,1-dichloroethene), commonly called PVDC. Its repeat-unit representation is:

\[ \ce{[-CH2-CCl2-]_{n}} \]

If the polymer formed by direct addition polymerisation, its monomer was 1,1-dichloroethene, \(\ce{CH2=CCl2}\).

The important point is that the three-unit drawing provides evidence about the local structure, not necessarily the full chain length or the reaction used to manufacture it.

01What three repeating units actually tell you

Before identifying a monomer, decide whether you’re looking at a complete molecule or a small section of something larger.

Imagine that the question shows three copies of \(\ce{-CH2-CCl2-}\), connected through carbon-carbon single bonds.

The full drawing contains six backbone carbon atoms:

• Three \(\ce{CH2}\) groups.
• Three \(\ce{CCl2}\) groups.

Each \(\ce{CCl2}\) carbon has two chlorine atoms attached. These chlorine atoms are side groups, not part of the continuous carbon backbone.

The smallest pattern that recreates this arrangement is:

\[ \ce{-CH2-CCl2-} \]

You don’t need all six backbone carbons inside the polymer brackets. One two-carbon section is enough to represent the regular structure.

Open bonds are not necessarily the ends of a molecule

Look at the short single bonds extending from either side of a polymer fragment.

When the question identifies the drawing as part of a continuing polymer chain, those bonds connect to atoms outside the section being shown. They aren’t unused bonds, missing hydrogens, or evidence that the polymer stops there.

Compare two situations.

A continuing chain: Three \(\ce{-CH2-CCl2-}\) sections are shown, with bonds extending beyond both ends. The rest of the chain is omitted.

A finite molecule: Three identical sections are joined together, and the two outer ends are explicitly capped with hydrogen atoms. No further atoms are attached.

Only the second drawing identifies the displayed molecule as having exactly three of those structural sections.

The symbol \(n\) in a formula such as \(\ce{[-CH2-CCl2-]_{n}}\) represents the number of repeating units in the complete chain.

It is not automatically equal to the number of units printed in the question.

This gives you a useful evidence rule:

For the separate problem of choosing the correct positions of brackets and continuing bonds, see How to Draw Polymer Brackets and Continuing Bonds Correctly.

02How to identify the polymer by working backwards

The most reliable approach is to reconstruct the bonding rather than trying to remember what a familiar polymer looks like.

Step 1: Trace the continuous backbone

Start at the left-hand continuation bond and follow the connected atoms towards the right.

For our opening example, the backbone passes through alternating \(\ce{CH2}\) and \(\ce{CCl2}\) groups.

The chlorine atoms branch away from that path.

This matters because a branch is not automatically part of the repeating backbone. A larger substituent could contain several carbon atoms of its own.

Step 2: Find a pattern that reproduces the whole section

Try joining two copies of your proposed repeat unit.

If you choose only \(\ce{-CCl2-}\), copying it would place chlorine-bearing carbons next to each other throughout the chain. That doesn’t match the original alternating arrangement.

If you choose \(\ce{-CH2-CCl2-}\), copying it reproduces the correct sequence.

That’s the better repeat unit.

You could also write the unit in the reverse direction, \(\ce{-CCl2-CH2-}\). For this regular polymer, it represents the same connectivity when repeated.

Step 3: Check carbon valence

A neutral carbon atom in these ordinary organic structures normally has a total bond order of four.

Consider the \(\ce{CCl2}\) carbon in the middle of a polymer chain:

The neighbouring \(\ce{CH2}\) carbon also has four bonds: two to hydrogen and one to each adjacent backbone carbon.

An open bond at the edge of the drawing must be counted as one of those carbon-carbon bonds.

Step 4: Decide what reaction is consistent with the structure

For a simple addition polymer made directly from an alkene, two carbon atoms from the original \(\ce{C=C}\) bond become part of the polymer backbone.

During addition polymerisation, the carbon-carbon double bond is converted into a single bond as new carbon-carbon bonds form between neighbouring monomer-derived sections.

Working backwards, you remove the two external chain connections of one repeat unit and restore a double bond between its two backbone carbons.

For our example:

\[ \ce{CH2=CCl2} \]

The two chlorine atoms stay on the same carbon. You must not move one chlorine to the neighbouring carbon just to make the structure look more familiar.

The result is 1,1-dichloroethene, a plausible monomer for the displayed addition-polymer structure.

But this reverse method has a condition: you are assuming direct addition polymerisation of the alkene. A finished polymer structure does not always reveal every reaction used to make it.

03Why chemists had to establish that polymers really were long chains

That distinction between a short drawing and a complete molecule has an interesting experimental history.

In the early twentieth century, chemists disagreed about what made rubber and other polymeric substances behave so differently from ordinary small molecules.

One explanation treated these materials as aggregates of relatively small molecules held together by weaker interactions. Another proposed that they contained extremely large molecules whose atoms were joined by covalent bonds.

Hermann Staudinger argued for the second model.

The experimental question: what holds rubber together?

Some early aggregate models proposed that interactions associated with carbon-carbon double bonds helped smaller rubber molecules associate into larger structures.

That produced a testable prediction.

If those double bonds were chemically removed, would the supposedly large structures fall apart?

Staudinger investigated this using hydrogenation of rubber.

Hydrogenation reacts hydrogen with carbon-carbon double bonds, converting them into single bonds under suitable catalytic conditions.

To picture the experiment’s logic, consider a schematic laboratory arrangement rather than a reconstruction of his exact historical glassware.

A vessel contains rubber undergoing a hydrogenation reaction in the presence of a suitable catalyst. Hydrogen reaches the unsaturated sections of the rubber molecules. After the reaction, the chemically modified material is isolated and its behaviour compared with that of the starting rubber.

Under the particular aggregate model being tested, removing the double bonds should disrupt the interactions holding the smaller molecules together.

Instead, hydrogenated rubber retained broadly similar high-molecular-material behaviour.

That challenged the proposed aggregate explanation and supported the idea that rubber consisted of long, covalently connected molecules.

It didn’t, by itself, establish the exact number of repeating units in every molecule. Hydrogenation could also change physical properties for reasons other than a change in molecular size.

A second line of evidence: measuring solution viscosity

Staudinger also developed evidence using viscosity measurements during the chemical modification of polymers.

Viscosity describes a fluid’s resistance to flow. Long polymer molecules can strongly affect how a solution flows, although viscosity also depends on concentration, temperature, solvent, and molecular structure.

One way to study viscosity is with a capillary viscometer.

The sample solution flows through a narrow glass tube. An investigator measures the time taken for the liquid meniscus to move between two marked positions.

The procedure is repeated for a reference liquid under comparable conditions.

Suppose an illustrative comparison gives these results:

These are invented teaching values, not Staudinger’s experimental data.

Let \(t_0\) be the pure solvent’s flow time and \(t\) the polymer solution’s flow time. Both are measured in seconds, s.

For the same viscometer at the same temperature, and assuming the two liquid densities are approximately equal, the relative viscosity can be estimated by:

\[ \begin{aligned} \eta_{\mathrm{rel}}&\approx\frac{t}{t_0}\\ &=\frac{64.0\ \mathrm{s}}{40.0\ \mathrm{s}}\\ &=1.60 \end{aligned} \]

Here, \(\eta_{\mathrm{rel}}\) is the relative viscosity, a dimensionless ratio.

The solution has an estimated viscosity 1.60 times that of the solvent.

Does that establish that the polymer molecule is 1.60 times as long as some reference molecule?

No. Flow time is not directly proportional to chain length in such a simple way.

The value of this measurement comes from comparing samples under controlled conditions, examining how the results change when the polymer is chemically modified, and testing predictions from competing molecular models.

Staudinger’s hydrogenation and viscosity evidence helped establish the macromolecular interpretation of polymers. It explains why modern polymer drawings use continuing covalent chains rather than treating every short printed section as a complete molecule.

The historical account is described by the American Chemical Society’s Foundation of Polymer Science landmark.

06Questions and solutions

Question 1

A polymer chain is drawn with three consecutive \(\ce{-CH2-CH(F)-}\) sections. All backbone carbon-carbon bonds shown are single bonds, and the chain continues beyond both ends.

(a) Write the repeating unit in bracket notation.

(b) Identify the likely alkene monomer if the polymer formed by direct addition.

(c) Explain whether the drawing proves that \(n=3\).

Solution 1

The repeating unit is \(\ce{-CH2-CH(F)-}\), the likely monomer is fluoroethene, \(\ce{CH2=CHF}\), and the number of units in the full chain is unknown.

(a) The bracketed structure is:

\[ \ce{[-CH2-CH(F)-]_{n}} \]

The backbone alternates between \(\ce{CH2}\) and \(\ce{CH(F)}\), so both carbon environments must be included.

(b) Under the direct-addition assumption, restore the carbon-carbon double bond between the two backbone carbons while retaining the fluorine substituent.

That gives fluoroethene, \(\ce{CH2=CHF}\). The corresponding polymer is poly(fluoroethene), also called poly(vinyl fluoride).

Carbon valence is satisfied: the fluorine-bearing carbon in the monomer has a double bond to carbon, a single bond to hydrogen, and a single bond to fluorine.

(c) The three visible sections are not proof that \(n=3\). The continuation bonds show that the drawing omits additional chain structure.

The tempting mistake is to count printed sections as the complete degree of polymerisation. Those are different quantities.

Question 2

Two drawings contain the same internal pattern: three consecutive \(\ce{-CH2-CH(Cl)-}\) sections.

Drawing A has continuing bonds beyond both ends.

Drawing B represents a complete finite molecule. Its six backbone carbons have the same internal arrangement, but each outer continuation bond has been replaced by a bond to one terminal hydrogen atom. No additional atoms are attached.

(a) Are A and B representations of the same complete molecule?

(b) Calculate the molecular formula of B.

(c) A student measures a molecular mass consistent with the formula calculated in part (b). Does that result prove that the polymer represented by A contains exactly three repeat units? Explain.

Solution 2

The drawings are not equivalent complete-molecule representations. B is a finite hydrogen-capped molecule with formula \(\ce{C6H11Cl3}\), while A is an incomplete view of a continuing chain.

(a) Drawing A does not specify the complete molecule. Its outer bonds continue beyond the region being shown.

Drawing B specifies actual terminal atoms. Because hydrogen atoms complete the two outer carbon valencies, there is no continuation beyond them.

(b) Each internal \(\ce{-CH2-CH(Cl)-}\) section contains:

• Two carbon atoms.
• Three hydrogen atoms.
• One chlorine atom.

Three sections therefore contribute:

\[ \ce{C6H9Cl3} \]

The two terminal hydrogen atoms add another \(\ce{H2}\).

Counting each element gives:

\[ \begin{aligned} N_{\mathrm C}&=3(2)=6\\ N_{\mathrm H}&=3(3)+2=11\\ N_{\mathrm{Cl}}&=3(1)=3 \end{aligned} \]

Here, \(N_{\mathrm C}\), \(N_{\mathrm H}\), and \(N_{\mathrm{Cl}}\) represent the numbers of carbon, hydrogen, and chlorine atoms, respectively.

The molecular formula is therefore:

\[ \ce{C6H11Cl3} \]

This formula satisfies the valence requirements of the specified saturated, hydrogen-capped structure.

(c) No. A molecular mass consistent with \(\ce{C6H11Cl3}\) would support that molecular formula for the measured species, but it would not establish that the continuing polymer in A has three units.

There are two separate constraints.

First, A has unspecified connections beyond both ends, whereas B does not. You cannot simply treat A’s continuation bonds as terminal hydrogens.

Second, a molecular formula alone does not uniquely establish connectivity. Different structures can have the same molecular formula.

A complete structural assignment needs evidence about both composition and bonding.

The tempting route is to count six backbone carbons, divide by two, and conclude that the whole polymer must contain three units. That reasoning ignores the distinction between an observed fragment and an actual complete molecule.

Question 3

A polyamide is formed from these two monomers:

\[ \ce{H2N-(CH2)4-NH2} \]

and

\[ \ce{HOOC-(CH2)4-COOH} \]

The first molecule is butane-1,4-diamine. The second is hexanedioic acid.

Exactly three molecules of each monomer react to form one linear oligomer. Assume there are no side reactions, no cyclic structures, and no leftover monomers.

(a) Identify the type of polymerisation and write one suitable repeating unit.

(b) Determine how many water molecules are eliminated.

(c) A student argues that three repeating units should produce six water molecules because the ideal repeat contains two amide linkages. Explain the error.

Solution 3

This is condensation polymerisation producing a polyamide. Five water molecules are eliminated when the six starting molecules form one linear chain.

(a) The diamine contains two amino groups, while the dicarboxylic acid contains two carboxyl groups.

An amino group can react with a carboxylic acid group to form an amide linkage:

\[ \ce{-C(=O)-NH-} \]

One suitable repeating unit is:

\[ \ce{[-NH-(CH2)4-NH-} \]

\[ \ce{-C(=O)-(CH2)4-C(=O)-]_{n}} \]

The two lines together represent one continuous repeat unit. The polymer formed is a polyamide.

(b) Initially, there are six separate molecules.

To join six separate molecules into one linear chain, five connections must be formed.

Every new amide linkage formed by this direct condensation reaction eliminates one water molecule.

Therefore:

\[ \begin{aligned} N_{\mathrm{links}}&=6-1\\ &=5\\ N_{\ce{H2O}}&=5 \end{aligned} \]

Here, \(N_{\mathrm{links}}\) is the number of new amide connections, and \(N_{\ce{H2O}}\) is the number of water molecules released.

Thus, five water molecules are formed.

(c) The student’s error is treating an ideal repeating-unit representation as though it includes every connection of a finite chain.

An indefinitely continuing polyamide chain can be represented with two amide connections associated with each complete ideal repeat.

But this particular molecule has actual ends.

For an alternating linear arrangement of three diamine and three diacid molecules, one terminal amino group and one terminal carboxylic acid group remain unreacted.

Forming a sixth connection would require joining those two ends, creating a ring, or connecting to another molecule. Both possibilities violate the specified final arrangement.

This is why counting water molecules from a finite condensation product requires counting actual bonds formed, not simply multiplying the number of ideal repeating units by a memorised number.

Question 4

A polymer was made solely by addition polymerisation of propene, \(\ce{CH2=CHCH3}\), without subsequent chemical modification.

An unusual section contains six consecutive backbone carbon atoms. Reading from left to right, they are:

\(\ce{CH2}\), \(\ce{CH(CH3)}\), \(\ce{CH(CH3)}\), \(\ce{CH2}\), \(\ce{CH2}\), and \(\ce{CH(CH3)}\).

All backbone bonds are single bonds, and the structure continues beyond both ends.

A student claims:

“Because this polymer was made from propene, those six carbon atoms must be three identical copies of \(\ce{-CH2-CH(CH3)-}\).”

(a) Is the displayed structure consistent with having been made from three propene molecules? Justify your answer.

(b) Does repeating \(\ce{-CH2-CH(CH3)-}\) three times reproduce the exact displayed arrangement?

(c) Explain how a polymer made from only one type of monomer can contain a section that does not match its usual idealised repeating pattern.

Solution 4

Yes, the displayed section can come from three propene monomers, but no, it is not three identical copies of the usual idealised propene repeat unit.

(a) Each propene monomer has two carbon atoms involved in its double bond:

\[ \ce{CH2=CH-CH3} \]

After addition, those two carbon atoms become part of the backbone. One is a \(\ce{CH2}\) group, and the other is a \(\ce{CH(CH3)}\) group.

The displayed six-carbon sequence can be divided into three possible monomer-derived pairs:

First pair: \(\ce{CH2-CH(CH3)}\)

Second pair: \(\ce{CH(CH3)-CH2}\)

Third pair: \(\ce{CH2-CH(CH3)}\)

Every pair has the two carbon environments contributed by propene.

The middle pair is simply oriented in the opposite direction.

(b) No.

Repeating the usual unit \(\ce{-CH2-CH(CH3)-}\) would produce alternating \(\ce{CH2}\) and \(\ce{CH(CH3)}\) groups throughout the displayed region.

But the supplied sequence contains:

• Two adjacent \(\ce{CH(CH3)}\) carbons.
• Two adjacent \(\ce{CH2}\) carbons.

Those adjacent identical carbon environments do not appear in the standard alternating representation.

(c) Propene is an unsymmetrical alkene. Its two double-bonded carbons have different substituents.

During polymerisation, monomer units can sometimes connect in different orientations.

The usual idealised structure is associated with head-to-tail addition, where a methyl-bearing carbon on one unit connects to a \(\ce{CH2}\) carbon on the next.

In the unusual section, the reversed middle unit creates one connection between two methyl-bearing carbons and another between two \(\ce{CH2}\) carbons.

These are often described as head-to-head and tail-to-tail connections.

The bonding remains chemically valid. Every backbone carbon still has four bonds.

What fails is the assumption that every local section must match the idealised two-carbon pattern.

This is an example of a structural irregularity in a polymer chain.

The original claim feels reasonable because a homopolymer contains units derived from one type of monomer. But that does not require every monomer-derived unit to have exactly the same orientation.

Consequently, the familiar bracketed polypropene structure represents the regular idealised arrangement. It is not an exact description of this particular irregular six-carbon section.

Question 5

A chemist starts with a sample of poly(vinyl acetate), formed by addition polymerisation of vinyl acetate.

Its original repeat-unit structure is:

\[ \ce{[-CH2-CH(O-C(=O)CH3)-]_{n}} \]

The sample contains \(0.100\ \mathrm{mol}\) of ester-bearing repeating units. This quantity refers to repeating units across the sample, not to the number of polymer molecules.

The chemist treats the polymer with excess aqueous sodium hydroxide, causing alkaline hydrolysis of some ester side groups. Each hydrolysed side group becomes \(\ce{-OH}\) and produces one sodium ethanoate molecule.

After the reaction, \(0.0600\ \mathrm{mol}\) of sodium ethanoate is recovered.

A structural drawing taken from the product shows three consecutive \(\ce{-CH2-CH(OH)-}\) sections, with continuation bonds beyond both ends.

Assume the following:

• The sodium ethanoate is recovered quantitatively and comes only from the stated hydrolysis reaction.
• Every hydrolysed ester group produces one sodium ethanoate molecule.
• The carbon-carbon polymer backbone is not broken.
• There are no other reactions that change the polymer’s composition.
• Chain-end contributions to mass are negligible.

Use these illustrative molar masses:

(a) Calculate the percentage of the original ester side groups that have been hydrolysed.

(b) A student says the product must consist entirely of poly(vinyl alcohol), because the drawing contains three consecutive \(\ce{-CH2-CH(OH)-}\) units. Evaluate this conclusion.

(c) Another student says that producing sodium ethanoate proves condensation polymerisation occurred. Explain whether this is correct.

(d) Calculate the original and final dry polymer masses. Use the sodium hydroxide consumed and sodium ethanoate produced to check conservation of mass.

Solution 5

The sample is a partially hydrolysed addition polymer. Overall, 60.0% of its original acetate ester side groups have become hydroxyl groups, but the three-unit drawing does not establish complete conversion. The sodium ethanoate comes from hydrolysis, not from condensation polymerisation.

The critical distinction is between the original formation of the polymer backbone and the later modification of groups attached to that backbone.

(a) Percentage hydrolysed

Every original repeating unit contains one ester side group.

The sample initially contains:

\[ n_{\mathrm{ester}}=0.100\ \mathrm{mol} \]

Here, \(n_{\mathrm{ester}}\) is the amount of original ester-bearing repeat units.

Each hydrolysed ester produces one sodium ethanoate molecule.

Therefore, \(0.0600\ \mathrm{mol}\) of sodium ethanoate corresponds to \(0.0600\ \mathrm{mol}\) of hydrolysed ester groups.

The percentage hydrolysed is:

\[ \begin{aligned} \%\,\text{hydrolysed} &=\frac{0.0600}{0.100}\times100\\ &=60.0\% \end{aligned} \]

Thus, 60.0% of the ester sites have been hydrolysed, and 40.0% remain unchanged.

The amount of unchanged sites is:

\[ \begin{aligned} n_{\mathrm{unchanged}} &=0.100-0.0600\\ &=0.0400\ \mathrm{mol} \end{aligned} \]

(b) Can the three-unit drawing prove complete conversion?

No. Three consecutive hydroxyl-bearing repeat units can occur within a polymer sample that still contains unhydrolysed ester-bearing units.

The measured conversion tells us the overall proportions:

• 60.0% of the original repeat sites now carry \(\ce{-OH}\).
• 40.0% retain \(\ce{-O-C(=O)CH3}\).

It does not tell us where those sites occur along the polymer chains.

For example, some chains might contain mixed sequences of hydrolysed and unchanged units. Alternatively, conversion might differ between individual chains.

The evidence supplied does not establish a particular distribution.

A three-unit drawing showing hydroxyl groups is therefore consistent with the experimental results, but it is not a complete structural description of the sample.

This is the same basic limitation as the opening problem: the drawing describes a local section, not necessarily every repeating unit in the material.

(c) Does sodium ethanoate production prove condensation polymerisation?

No. The sodium ethanoate is formed by alkaline hydrolysis of existing ester side groups.

The polymer was originally produced by addition polymerisation of vinyl acetate.

Its carbon-carbon backbone already existed before the sodium hydroxide treatment.

During hydrolysis, a side-group reaction occurs:

\[ \ce{R-O-C(=O)-CH3 + NaOH} \]

\[ \ce{-> R-OH + CH3COONa} \]

Here, \(R\) represents the part of the existing polymer chain attached to the ester oxygen.

The reaction replaces an acetate-containing side group with a hydroxyl group while leaving the stated carbon backbone intact.

It does not join separate monomers together to form new polymer backbone connections.

The presence of a small-molecule product is therefore not enough to identify a reaction as condensation polymerisation.

You must ask what bond-making process occurred.

(d) Original mass, final mass, and conservation of mass

First, calculate the original mass.

The initial amount of repeating units is \(0.100\ \mathrm{mol}\), and their molar mass is \(86.1\ \mathrm{g\,mol^{-1}}\).

Using \(m=nM\), where \(m\) is mass in grams, \(n\) is amount in moles, and \(M\) is molar mass:

\[ \begin{aligned} m_{\mathrm{initial}} &=n_{\mathrm{initial}}M_{\mathrm{initial}}\\ &=(0.100)(86.1)\ \mathrm{g}\\ &=8.61\ \mathrm{g} \end{aligned} \]

Next, calculate the final polymer mass.

The product contains \(0.0600\ \mathrm{mol}\) of hydroxyl-bearing repeat sites and \(0.0400\ \mathrm{mol}\) of unchanged ester-bearing sites.

The hydroxyl-bearing contribution is:

\[ \begin{aligned} m_{\mathrm{OH}} &=(0.0600)(44.1)\ \mathrm{g}\\ &=2.646\ \mathrm{g} \end{aligned} \]

The unchanged ester-bearing contribution is:

\[ \begin{aligned} m_{\mathrm{ester}} &=(0.0400)(86.1)\ \mathrm{g}\\ &=3.444\ \mathrm{g} \end{aligned} \]

Adding them gives:

\[ \begin{aligned} m_{\mathrm{final}} &=2.646+3.444\\ &=6.090\ \mathrm{g} \end{aligned} \]

The dry polymer loses:

\[ \begin{aligned} \Delta m&=8.61-6.09\\ &=2.52\ \mathrm{g} \end{aligned} \]

Why has the polymer become lighter?

The hydroxyl-bearing repeat units have a lower molar mass because the acetate-containing side groups have been replaced by hydroxyl groups.

But mass hasn’t disappeared from the overall reaction.

The sodium hydroxide consumed is:

\[ \begin{aligned} m_{\ce{NaOH}} &=(0.0600)(40.0)\ \mathrm{g}\\ &=2.40\ \mathrm{g} \end{aligned} \]

The sodium ethanoate produced has mass:

\[ \begin{aligned} m_{\mathrm{ethanoate}} &=(0.0600)(82.0)\ \mathrm{g}\\ &=4.92\ \mathrm{g} \end{aligned} \]

Now compare the masses on both sides of the simplified hydrolysis reaction.

Reactants:

\[ \begin{aligned} m_{\mathrm{reactants}} &=8.61+2.40\\ &=11.01\ \mathrm{g} \end{aligned} \]

Products:

\[ \begin{aligned} m_{\mathrm{products}} &=6.09+4.92\\ &=11.01\ \mathrm{g} \end{aligned} \]

The masses agree.

This calculation also rules out a tempting interpretation of the experiment: the missing mass from the dry polymer has not been lost through shortening the carbon backbone. It is accounted for by the changed side groups and the sodium ethanoate product.

The deeper lesson is that three identical-looking repeat units can tell you what bonds exist in one region of a chain, but not necessarily the composition of the entire sample or its reaction history.

When a question asks for the polymer’s structure, start with the visible backbone, substituents, and valence. When it asks how that polymer was actually manufactured or modified, you need an additional piece of evidence: the starting materials, reaction conditions, products, or chemical measurements that distinguish the possible pathways.