How to Count Carbon Atoms in Nylon Monomers

A nylon label reads nylon 6,6. You are given its two monomers and asked how many carbon atoms each contains. Should you count six, twelve, or something else?

The answer is six carbon atoms in each monomer. However, the combined repeating unit contains twelve carbon atoms. Understanding the difference is the key to counting carbon atoms in nylon monomers correctly.

The most common mistake is to count the \(\mathrm{CH_2}\) groups but forget the carbon atoms inside the carboxylic acid groups. Let’s work through why that happens and how to avoid it.

01Start by identifying the two monomers

Nylon 6,6 is formed from two monomers:

• Hexane-1,6-diamine, which contains two amine groups.
• Adipic acid (hexanedioic acid), which contains two carboxylic acid groups.

Their condensed structural formulas are:

Hexane-1,6-diamine:

\[ \ce{H2N-(CH2)6-NH2} \]

Adipic acid:

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

Before counting, make a prediction: does adipic acid contain four carbon atoms or six?

It contains six. The four in \((\mathrm{CH_2})_4\) tells you there are four methylene groups, but each \(\mathrm{COOH}\) group also contains one carbon atom.

Count the carbon atoms in each monomer

There are two important counting rules here.

Rule 1: Count every carbon atom, including carbonyl carbons.

A carboxylic acid group can be expanded as:

\[ \mathrm{-C(=O)-OH} \]

It contains one carbon atom, not zero or two. The carbon is bonded to two oxygen atoms, but it is still only one carbon.

Rule 2: Read the subscripts carefully.

In \((\mathrm{CH_2})_4\):

• The 2 means each group contains two hydrogen atoms.
• The 4 means there are four \(\mathrm{CH_2}\) groups.
• Therefore, there are four carbon atoms, not eight.

If a structural formula is drawn using line-angle notation, remember that each unlabeled vertex and line end represents a carbon atom unless another element is specified. A double bond also does not create an extra atom.

02Why does nylon 6,6 have twelve carbon atoms in its repeating unit?

Now that we know each monomer contains six carbon atoms, we can examine what happens when they join.

Nylon 6,6 is a polyamide. This means its polymer backbone contains repeating amide linkages:

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

The diamine and dicarboxylic acid react through condensation polymerisation. An amine group reacts with a carboxylic acid group, forming an amide linkage and eliminating water.

The repeating unit of nylon 6,6 is:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_6 -\mathrm{NH}-\mathrm{C(=O)} -(\mathrm{CH_2})_4-\mathrm{C(=O)}- \right]_n \]

Here, \(n\) represents the number of repeating units in the polymer chain.

Let’s count the carbon atoms in this structure.

The total is therefore:

\[ 6+4+1+1=12 \]

Each idealised repeating unit contains twelve carbon atoms.

Does condensation remove any carbon atoms?

No. This is an important distinction.

During condensation polymerisation, the OH from a carboxylic acid group and an H from an amine group combine to form water:

\[ \mathrm{-COOH + H_2N-} \]

\[ \longrightarrow \mathrm{-CONH- + H_2O} \]

The carbonyl carbon remains in the polymer. It becomes part of the amide linkage.

The reaction changes how atoms are connected, but it does not remove carbon atoms from the polymer backbone.

This is the same atom-accounting principle used when balancing ordinary chemical reactions. You can revisit that idea in Rearrangement of Atoms in Reactions: HSC Chemistry Guide.

For a polymer containing \(n\) idealised nylon 6,6 repeating units, the repeat-unit contribution is:

\[ N_{\mathrm C}=12n \]

If a question asks for the exact molecular formula of a finite chain, terminal groups may need to be considered separately. For ordinary repeating-unit carbon counts, use twelve carbon atoms per unit.

03Understanding the numbers in nylon names

The numbers in a nylon name are particularly useful when identifying its monomers.

For conventional nylons made from a diamine and a dicarboxylic acid:

• The first number represents the number of carbon atoms in the diamine.
• The second number represents the number of carbon atoms in the dicarboxylic acid.

This gives us:

Notice that the total number of carbon atoms in the repeating unit is not necessarily either number in the nylon name.

For example, nylon 6,10 contains sixteen carbon atoms per idealised repeating unit, not ten.

04A useful contrast: nylon 6

Not all nylons are made from two separate monomers.

Nylon 6 is commonly produced from a single cyclic monomer called caprolactam.

Caprolactam contains:

• Five \(\mathrm{CH_2}\) groups.
• One carbonyl carbon.
• One nitrogen atom.

Therefore, its carbon count is:

\[ N_{\mathrm C}=5+1=6 \]

Caprolactam undergoes ring-opening polymerisation to form nylon 6.

Its repeating unit is:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_5-\mathrm{C(=O)}- \right]_n \]

Count the carbon atoms:

\[ 5+1=6 \]

So nylon 6 has six carbon atoms per repeating unit.

Notice the difference:

There is another interesting detail.

The nylon 6 repeating unit can also be produced through condensation polymerisation of 6-aminohexanoic acid:

\[ \ce{H2N-(CH2)5-COOH} \]

This molecule also contains six carbon atoms.

Both routes can produce the same idealised nylon 6 repeating-unit structure. However, their polymerisation mechanisms differ. Ring-opening polymerisation does not eliminate water as a condensation product, whereas the amino acid route does.

This distinction becomes important when a question asks you to identify the starting monomer from experimental evidence, rather than simply counting atoms in the polymer.

05Why the historical nylon experiments matter

The development of nylon 6,6 provides a useful example of how chemists connected molecular structure with polymer formation.

In 1935, Gerard Berchet, working in Wallace Carothers’ research group, prepared nylon 6,6 using hexamethylenediamine and adipic acid.

The problem was to produce long polymer chains from relatively small molecules. The reaction formed amide linkages, but it also produced water.

Why was that a problem?

Condensation polymerisation is reversible. If water remains in the reaction mixture, it can limit further chain formation. Removing water helps favour continued condensation.

Following the experimental arrangement

The reported preparation involved heating the reactants in meta-cresol.

A simplified picture of the apparatus helps explain the reasoning:

1. A heated reaction vessel contains the diamine and adipic acid in the solvent.
2. As condensation occurs, water is produced.
3. Water vapour leaves the reaction mixture through a distillation system and is collected.
4. The polymer becomes increasingly viscous and remains in the reaction vessel.
5. After the water has distilled off, the temperature is raised to approximately \(255\text{-}260\,^\circ\mathrm C\), and the cresol is subsequently removed under vacuum.

What did chemists predict, and what did they observe?

The prediction was that repeated reactions between amine and carboxylic acid groups would produce long chains containing amide linkages. Removing water would help the reaction continue.

The resulting polymer could be formed into fibres, supporting the idea that small bifunctional molecules could be joined into long-chain synthetic materials.

There was also an important limitation to the earlier approach. Polyester fibres had been produced, but their relatively low melting points and sensitivity to some organic solvents restricted their usefulness.

The research team investigated diamines instead of diols. This changed the linkage from an ester to an amide, introducing additional opportunities for hydrogen bonding between neighbouring polymer chains.

The historical experiment demonstrated polymer formation. It did not provide a direct measurement of the number of carbon atoms in each monomer. That count still comes from the molecular structures.

A small calculation: estimating the water produced

Consider an idealised, very long nylon 6,6 polymer chain.

Each diamine-diacid pair contributes two amide linkages to the repeating-unit structure. Each linkage forms one water molecule.

Therefore, approximately two moles of water are produced per mole of idealised repeating units, assuming chain-end effects are negligible.

For \(1.00\) mol of repeating units:

\[ n(\ce{H2O})\approx 2.00\ \mathrm{mol} \]

Using the molar mass of water, \(18.0\ \mathrm{g\,mol^{-1}}\):

\[ \begin{aligned} m(\ce{H2O}) &=nM\\ &=2.00\times18.0\\ &=36.0\ \mathrm g \end{aligned} \]

Approximately \(36.0\) g of water would be removed.

For a finite linear chain containing \(N\) diamine molecules and \(N\) diacid molecules, the exact number of condensation events is \(2N-1\), because the chain has two ends. The approximation of two water molecules per repeating unit becomes increasingly accurate for long chains.

For the fuller history of nylon development and the relationship between amide linkages and material properties, see Polyamides: How Nylon Forms, Its Structure and Properties.

06How carbon counting connects to nylon properties

Counting carbon atoms is useful for more than identifying a polymer’s monomers. It also helps you compare the structures of different nylons.

Consider nylon 6,6 and nylon 6,10.

Both have two amide linkages per idealised repeating unit, but their hydrocarbon sections have different lengths.

Nylon 6,10 has four additional methylene groups per repeating unit. These additional carbon atoms do not create additional amide linkages.

Why does that matter?

An amide linkage contains a polar N-H bond and a carbonyl oxygen. The hydrogen attached to nitrogen can form a hydrogen bond with a carbonyl oxygen on a neighbouring polymer chain.

This produces attractions between chains:

\[ \mathrm{N-H\cdots O=C} \]

Nylon 6,6 has a greater concentration of amide linkages relative to its carbon backbone than nylon 6,10.

This can influence the relative importance of hydrogen bonding and hydrocarbon-chain interactions.

However, don’t conclude that a nylon with fewer carbon atoms must always be stronger. Actual mechanical properties also depend on chain length, crystallinity, molecular orientation, and processing conditions.

The useful structural principle is that the number and arrangement of functional groups matter, not simply the total number of carbon atoms.

07Questions and solutions

Question 1

Nylon 6,6 is produced from hexane-1,6-diamine and adipic acid.

Their structural formulas are:

\[ \ce{H2N-(CH2)6-NH2} \]

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

Determine:

a. The number of carbon atoms in each monomer.

b. The total number of carbon atoms in one idealised repeating unit.

Solution 1

Both monomers contain six carbon atoms, and the repeating unit contains twelve carbon atoms.

Step 1: Count the carbon atoms in hexane-1,6-diamine.

There are six \(\mathrm{CH_2}\) groups:

\[ N_{\mathrm C}=6 \]

Step 2: Count the carbon atoms in adipic acid.

There are four methylene carbons and two carboxyl carbon atoms:

\[ N_{\mathrm C}=4+2=6 \]

Step 3: Calculate the total number of carbon atoms in the repeating unit.

\[ N_{\mathrm C}=6+6=12 \]

The polymer is called nylon 6,6 because each monomer contains six carbon atoms. Its idealised repeating unit contains twelve.

Question 2

A straight-chain saturated dicarboxylic acid has the molecular formula:

\[ \ce{C10H18O4} \]

It reacts with hexane-1,6-diamine to produce a polyamide.

A student proposes that the acid contains ten methylene groups and that the resulting polymer is nylon 6,12.

Assuming the acid has the structure \(\ce{HOOC-(CH2)_m-COOH}\), determine:

a. The correct number of methylene groups in the acid.

b. The correct nylon designation.

c. The total number of carbon atoms in one idealised repeating unit.

Explain the student’s error.

Solution 2

The acid contains eight methylene groups, the polymer is nylon 6,10, and each repeating unit contains sixteen carbon atoms.

Step 1: Establish how the carbon atoms are distributed.

The general structure is:

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

There are two carbon atoms in the terminal carboxylic acid groups.

Therefore:

\[ m+2=10 \]

Rearranging:

\[ m=8 \]

The acid contains eight methylene groups and two carboxyl carbon atoms.

Step 2: Check the hydrogen count.

Eight methylene groups contribute sixteen hydrogen atoms, while the two carboxylic acid groups contribute two more:

\[ H=2(8)+2=18 \]

This agrees with the supplied molecular formula, \(\ce{C10H18O4}\).

Step 3: Determine the nylon designation.

Hexane-1,6-diamine contains six carbon atoms.

The dicarboxylic acid contains ten.

Therefore, the polymer is nylon 6,10.

Step 4: Calculate the repeating-unit carbon count.

\[ N_{\mathrm C}=6+10=16 \]

Each idealised repeating unit contains sixteen carbon atoms.

The student’s mistake was treating the ten carbon atoms in the acid as though they were all methylene carbons. The two carboxyl carbon atoms must also be included, leaving eight methylene groups.

Question 3

An unknown polyamide has the following idealised repeating unit:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_5 -\mathrm{NH}-\mathrm{C(=O)} -(\mathrm{CH_2})_7-\mathrm{C(=O)}- \right]_n \]

The polymer was produced by condensation polymerisation of a diamine and a dicarboxylic acid.

Determine:

a. The structural formula of each monomer.

b. The nylon designation.

c. The total number of carbon atoms in one repeating unit.

Solution 3

The monomers are pentane-1,5-diamine and nonanedioic acid. The polymer is nylon 5,9, with fourteen carbon atoms per repeating unit.

Step 1: Identify the diamine section.

The nitrogen-containing section is:

\[ \mathrm{-NH-(CH_2)_5-NH-} \]

This corresponds to the diamine:

\[ \ce{H2N-(CH2)5-NH2} \]

It contains five carbon atoms.

Step 2: Identify the dicarboxylic acid section.

The carbonyl-containing section is:

\[ \mathrm{-C(=O)-(CH_2)_7-C(=O)-} \]

The original dicarboxylic acid must therefore be:

\[ \ce{HOOC-(CH2)7-COOH} \]

Count its carbon atoms:

\[ N_{\mathrm C}=7+2=9 \]

The acid is nonanedioic acid.

Step 3: Determine the nylon designation.

The diamine contains five carbon atoms, and the diacid contains nine.

The polymer is therefore nylon 5,9.

Step 4: Count the carbon atoms in the repeating unit.

\[ N_{\mathrm C}=5+7+2=14 \]

The repeating unit contains fourteen carbon atoms.

The two carbonyl carbons are essential to this calculation. Counting only the methylene groups would give twelve and incorrectly identify the acid as an eight-carbon molecule.

Question 4

Two polymers are made from different monomer combinations:

• Nylon 6,6 has twelve carbon atoms and two amide linkages per idealised repeating unit.
• Nylon 6,10 has sixteen carbon atoms and two amide linkages per idealised repeating unit.

A student argues:

“Nylon 6,10 must have stronger hydrogen bonding because it contains more carbon atoms, so its polymer chains should attract each other more strongly.”

Evaluate this statement.

In your answer:

a. Calculate the ratio of amide linkages to carbon atoms for each polymer.

b. Identify which polymer has the greater concentration of amide linkages relative to its carbon backbone.

c. Explain why the student’s conclusion about mechanical strength is not justified by carbon counting alone.

Solution 4

Nylon 6,6 has the greater amide-linkage concentration relative to its carbon backbone. However, this alone does not prove that it has greater tensile strength.

Step 1: Calculate the ratio for nylon 6,6.

There are two amide linkages for every twelve carbon atoms:

\[ \frac{2}{12}=0.167 \]

This corresponds to approximately 0.167 amide linkages per carbon atom.

Step 2: Calculate the ratio for nylon 6,10.

There are two amide linkages for every sixteen carbon atoms:

\[ \frac{2}{16}=0.125 \]

This corresponds to 0.125 amide linkages per carbon atom.

Step 3: Compare the results.

\[ 0.167>0.125 \]

Nylon 6,6 therefore has a greater concentration of amide linkages relative to its carbon backbone.

Both polymers contain the same number of amide linkages per repeating unit. Nylon 6,10 simply contains four additional methylene groups.

Step 4: Evaluate the proposed strength comparison.

The amide linkages provide N-H groups that can donate hydrogen bonds and carbonyl oxygen atoms that can accept them.

Nylon 6,6 has more amide linkages relative to its hydrocarbon section. This suggests a greater relative contribution from amide-related interactions.

However, tensile strength also depends on factors such as:

• The degree of crystallinity.
• How well polymer chains are aligned.
• The average chain length.
• The conditions used to manufacture and draw the fibre.

Therefore, the student’s explanation is incorrect. More carbon atoms do not automatically produce stronger hydrogen bonding or greater tensile strength.

Question 5

Nylon 6 provides an interesting challenge because the same repeating-unit structure can be produced through two different polymerisation routes.

Consider the repeating unit:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_5-\mathrm{C(=O)}- \right]_n \]

Two possible starting monomers are proposed.

Route A: Ring-opening polymerisation

Caprolactam has the molecular formula:

\[ \ce{C6H11NO} \]

It undergoes ring-opening polymerisation. In the idealised process, no small molecule is eliminated.

Route B: Condensation polymerisation

6-aminohexanoic acid has the structural formula:

\[ \ce{H2N-(CH2)5-COOH} \]

It undergoes condensation polymerisation, with each amide linkage formed releasing one water molecule.

An idealised model reactor contains exactly 100 monomer molecules. They form one linear polymer chain containing 100 repeating units.

Assume:

• No water is initially present.
• No side reactions occur.
• The polymer chain is linear and unbranched.
• All condensation reactions proceed to completion.
• The ring-opening route produces no small-molecule by-product.

The reactor collects exactly 99 water molecules.

Answer the following:

a. Calculate the number of carbon atoms in one monomer for each proposed route.

b. Calculate the total number of carbon atoms in the final polymer chain.

c. Determine which polymerisation route is consistent with the collected water.

d. Explain why the reactor collects 99 water molecules rather than 100.

e. Can the repeating-unit structure alone establish which monomer was used? Justify your answer.

Solution 5

Route B, condensation polymerisation of 6-aminohexanoic acid, is consistent with the collection of 99 water molecules. The final chain contains 600 carbon atoms, and the repeating-unit structure alone cannot distinguish the two starting routes.

This question tests three connected ideas: carbon conservation, the difference between ring-opening and condensation polymerisation, and the number of links in a finite polymer chain.

Step 1: Count the carbon atoms in caprolactam.

Caprolactam contains five methylene groups and one carbonyl carbon.

\[ N_{\mathrm C}=5+1=6 \]

The caprolactam monomer contains six carbon atoms.

Step 2: Count the carbon atoms in 6-aminohexanoic acid.

Its structural formula is:

\[ \ce{H2N-(CH2)5-COOH} \]

There are five methylene carbon atoms and one carboxyl carbon atom.

\[ N_{\mathrm C}=5+1=6 \]

This monomer also contains six carbon atoms.

Notice that both proposed monomers have the same carbon count, even though their structures and polymerisation mechanisms differ.

Step 3: Calculate the total number of carbon atoms in the polymer.

There are 100 monomer molecules, each containing six carbon atoms.

Therefore:

\[ N_{\mathrm C}=100\times6 \]

\[ N_{\mathrm C}=600 \]

The final polymer chain contains 600 carbon atoms.

Neither polymerisation route removes carbon atoms. The carbon atoms are retained in the repeating units.

Step 4: Determine which route produces the collected water.

In Route A, caprolactam undergoes ring-opening polymerisation.

The ring opens and the monomer units connect to form the polymer backbone. Under the stated idealised conditions, no water is eliminated.

Route A therefore predicts zero water molecules.

In Route B, each reaction between an amine group and a carboxylic acid group forms an amide linkage and releases one water molecule.

The collected water is therefore consistent with Route B.

Step 5: Explain why 99 water molecules are produced.

This is the most important part of the question.

There are 100 monomer molecules, but they form one linear chain.

To connect 100 separate molecules into one continuous chain, only 99 connections are needed.

The pattern is:

• Joining two monomers creates one link.
• Joining three monomers creates two links.
• Joining four monomers creates three links.
• Joining 100 monomers creates 99 links.

Each condensation link releases one water molecule.

Therefore:

\[ N_{\ce{H2O}}=100-1=99 \]

The remaining reactive groups are located at the two ends of the chain.

A common mistake is to assume that every monomer must release one water molecule. However, the end groups do not both react with another monomer. Only the connections between neighbouring monomers produce water.

Step 6: Decide whether the repeating-unit structure identifies the monomer.

It does not.

Both caprolactam ring-opening polymerisation and 6-aminohexanoic acid condensation can produce the same idealised nylon 6 repeating unit:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_5-\mathrm{C(=O)}- \right]_n \]

The polymer’s repeating-unit structure establishes the arrangement of atoms in the chain, but it does not necessarily reveal the exact route used to produce it.

The water measurement provides the additional evidence needed to distinguish the two proposed routes.

This is a useful general lesson in polymer chemistry: the final polymer structure does not always contain enough information to uniquely identify its starting monomers or polymerisation mechanism.

08The method to remember

When counting carbon atoms in nylon monomers, follow this sequence:

1. Identify the monomer or monomers from the structural formula.
2. Count every \(\mathrm{CH_2}\) group.
3. Count the carbon atom in every carboxylic acid group.
4. Do not count nitrogen, oxygen, or hydrogen as carbon.
5. For nylon made from a diamine and a dicarboxylic acid, use the two separate monomer counts to identify the nylon name.
6. If asked for the repeating-unit carbon count, add the carbon atoms contributed by both monomers.
7. Remember that condensation removes water, not carbon atoms.

The key distinction is simple: count the carbon atoms in each monomer separately before calculating the total in the polymer repeating unit.

Once you can do this reliably, the next step is to track which atoms are removed during condensation polymerisation and explain how the resulting amide linkages influence the properties of polyamides such as nylon.