How to Calculate Repeat Units in a Polymer Chain

A polystyrene sample has a number-average molar mass of \(52\,075\ \mathrm{g\,mol^{-1}}\). Each repeating unit has the formula \(\ce{C8H8}\), with a molar mass of \(104.15\ \mathrm{g\,mol^{-1}}\).

How many repeating units are in an average chain?

Simply divide:

\[ N_r=\frac{52\,075}{104.15}=500 \]

The average polystyrene chain contains approximately 500 repeating units.

The calculation is straightforward. The tricky part is identifying the correct repeating unit before calculating its molar mass. This becomes particularly important when comparing addition polymers, such as polystyrene, with condensation polymers, such as nylon-6,6.

Predict what would happen if you used the molar mass of just one monomer to calculate the chain length of nylon-6,6. Would you get the correct answer? No. Its repeating unit contains contributions from two different monomers, with water removed during polymer formation.

Let’s work through how to identify the correct structure and calculate the number of repeating units.

01The calculation: number of repeating units

The number of repeating units in a polymer chain can be estimated by dividing the molar mass of the polymer chain by the molar mass of one repeating unit.

\[ \boxed{N_r\approx\frac{M_{\mathrm{chain}}}{M_r}} \]

Where:

• \(N_r\) = number of repeating units in the polymer chain (dimensionless).
• \(M_{\mathrm{chain}}\) = molar mass of the polymer chain, in \(\mathrm{g\,mol^{-1}}\).
• \(M_r\) = molar mass of one repeating unit, in \(\mathrm{g\,mol^{-1}}\).

For a polymer sample containing chains of different lengths, use the number-average molar mass, \(M_n\):

\[ \boxed{\overline{N_r}\approx\frac{M_n}{M_r}} \]

The ratio has no units because the units in the numerator and denominator cancel.

For example, if the average polymer chain has a molar mass of \(40\,000\ \mathrm{g\,mol^{-1}}\), and each repeating unit has a molar mass of \(100\ \mathrm{g\,mol^{-1}}\), the average chain contains approximately 400 repeating units.

There are two important assumptions:

1. The polymer has a known, consistent repeating structure.
2. The mass of the terminal groups is negligible compared with the mass of the entire chain.

These assumptions are generally reasonable for long polymer chains.

02Identify the repeating unit before calculating

This is where most of the reasoning happens.

A repeating unit is the smallest structural section that repeats along the polymer chain. Its composition depends on how the polymer was formed.

The distinction between addition and condensation polymers is particularly important.

Addition polymers: retain all the atoms

Consider polystyrene, which is made from styrene monomers.

Styrene has the molecular formula \(\ce{C8H8}\). During addition polymerisation, its carbon-carbon double bond opens, allowing the molecules to join into a longer chain.

The repeating structure is:

\[ \left[-\mathrm{CH_2-CH(C_6H_5)-}\right]_n \]

The phenyl group, \(\ce{C6H5}\), remains attached to the carbon backbone.

No atoms are removed during the addition process, so the repeating unit has the same molecular formula as the original monomer:

\[ \text{Repeating unit}=\ce{C8H8} \]

Its molar mass is:

\[ M_r=8(12.011)+8(1.008) \]

\[ \boxed{M_r=104.15\ \mathrm{g\,mol^{-1}}} \]

This means that every repeating unit contributes approximately \(104.15\ \mathrm{g\,mol^{-1}}\) to the polymer chain.

The same principle applies to other familiar addition polymers.

Notice something about PVC.

Its repeating unit contains chlorine, even though the chlorine atom is attached to the carbon backbone rather than forming part of the backbone itself.

You must include every atom in the repeating unit when calculating its molar mass. Ignoring side groups will give you the wrong answer.

Condensation polymers: subtract the eliminated molecules

Now consider nylon-6,6.

Unlike polystyrene, nylon-6,6 is produced from two different monomers:

• Hexane-1,6-diamine: \(\ce{C6H16N2}\)
• Adipic acid: \(\ce{C6H10O4}\)

These monomers join through amide linkages, releasing water.

The repeating structure contains one diamine-derived section and one adipic-acid-derived section:

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

The repeating unit therefore has the molecular formula:

\[ \ce{C12H22N2O2} \]

Why isn’t its formula simply the sum of the two monomer formulas?

Because two water molecules are removed for each idealised repeating pair.

The corresponding formula balance is:

\[ \ce{C6H16N2 + C6H10O4} \]

\[ \ce{-> C12H22N2O2 + 2H2O} \]

This is a repeating-unit mass balance. A finite polymer chain also has terminal groups, which are neglected in this approximation.

We can now calculate the molar mass of the nylon-6,6 repeating unit:

\[ \begin{aligned} M_r &= M_{\mathrm{diamine}}+M_{\mathrm{diacid}}\\ &\quad -2M_{\ce{H2O}}\\ &=116.21+146.14-2(18.015)\\ &=226.32\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The correct molar mass is therefore:

\[ \boxed{M_r=226.32\ \mathrm{g\,mol^{-1}}} \]

A common mistake is to add the molar masses of the two monomers without subtracting the water molecules.

That would give \(262.35\ \mathrm{g\,mol^{-1}}\), which is too high.

Another mistake is to use only the molar mass of the diamine. That ignores the adipic-acid-derived section entirely.

The same approach applies to poly(ethylene terephthalate), commonly called PET.

PET is formed from ethane-1,2-diol and terephthalic acid. Its repeating unit has the formula \(\ce{C10H8O4}\).

Its molar mass can be calculated as:

\[ M_r=62.07+166.13-2(18.015) \]

\[ \boxed{M_r=192.17\ \mathrm{g\,mol^{-1}}} \]

The key decision is simple:

For the examples above, the atomic masses are rounded using standard values. In an examination, use the atomic masses supplied in the question or on the periodic table.

Repeat units versus monomer units: an important distinction

There is a small terminology trap worth understanding.

The constitutional repeating unit is the smallest structural section that repeats along a regular polymer chain. A monomeric unit is the section contributed by one monomer molecule.

For an addition polymer such as polystyrene, one repeating unit corresponds to one monomer-derived unit.

For nylon-6,6, however, one repeating unit contains contributions from two different monomer molecules.

Consequently:

• 500 polystyrene repeating units correspond to 500 monomer-derived units.
• 250 nylon-6,6 repeating units correspond to 500 monomer-derived units.

IUPAC formally defines degree of polymerisation in terms of the number of monomeric units. Some school-level material uses the term more loosely to mean the number of repeating units. To avoid confusion, this article uses \(N_r\) specifically for the number of repeating units. IUPAC: constitutional repeating unit, monomeric unit, and degree of polymerisation.

03How do chemists determine the molar mass of a polymer chain?

So far, we’ve assumed that the molar mass of the polymer is already known.

But how would a chemist actually measure it?

A single polymer chain is far too small to weigh on an ordinary laboratory balance. Instead, chemists can determine an average molar mass by measuring the behaviour of a solution containing enormous numbers of polymer molecules.

One method is called membrane osmometry.

The experimental setup

Imagine two compartments separated by a semipermeable membrane.

• One compartment contains pure solvent.
• The other contains a dilute solution of the polymer dissolved in the same solvent.

The membrane allows small solvent molecules to pass through, but prevents the much larger polymer chains from crossing.

Initially, solvent moves towards the polymer solution. As more solvent enters, pressure builds on the solution side.

A pressure sensor or liquid column measures the pressure difference. The pressure needed to stop further net solvent movement is the osmotic pressure, \(\Pi\).

What should the experiment show?

Suppose two polymer solutions have the same mass concentration.

One contains relatively short polymer chains, while the other contains much longer chains.

Which solution should produce the greater osmotic pressure?

The solution containing shorter chains.

Here’s why.

At the same mass concentration, shorter chains mean more individual polymer molecules are present in each litre of solution.

Osmotic pressure depends on the number of dissolved particles. More polymer molecules therefore produce a greater osmotic pressure.

This gives chemists a way to estimate the number of polymer molecules in a solution and, consequently, their average molar mass.

For a sufficiently dilute polymer solution, the relationship is:

\[ \Pi\approx\frac{cRT}{M_n} \]

Where:

• \(\Pi\) = osmotic pressure, in pascals (Pa).
• \(c\) = polymer mass concentration, in \(\mathrm{kg\,m^{-3}}\).
• \(R\) = ideal gas constant, \(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
• \(T\) = absolute temperature, in kelvin (K).
• \(M_n\) = number-average molar mass, in \(\mathrm{kg\,mol^{-1}}\).

Rearranging:

\[ \boxed{M_n\approx\frac{cRT}{\Pi}} \]

Why isn’t one measurement enough?

The simple equation assumes a sufficiently dilute solution.

At higher concentrations, polymer chains can interact with one another. These interactions affect the measured osmotic pressure, so a single measurement may produce an inaccurate molar mass.

Chemists address this by measuring osmotic pressure at several dilute concentrations.

They then plot \(\Pi/c\) against concentration and extrapolate towards zero concentration. At this limit, the effect of interactions between polymer chains becomes negligible.

The intercept is related to \(RT/M_n\), allowing the number-average molar mass to be determined.

This is an important improvement over simply applying the ideal dilute-solution equation to a concentrated sample. Membrane osmometry and number-average molar mass. Further technical discussion.

A small measurement calculation

Suppose a dilute polystyrene solution has:

• Mass concentration: \(1.50\ \mathrm{g\,L^{-1}}\)
• Temperature: \(298\ \mathrm{K}\)
• Osmotic pressure: \(18.6\ \mathrm{Pa}\)

Assume the solution is sufficiently dilute for the ideal relationship to apply.

First, convert the concentration:

\[ 1.50\ \mathrm{g\,L^{-1}} =1.50\ \mathrm{kg\,m^{-3}} \]

Substitute into the equation:

\[ \begin{aligned} M_n&=\frac{cRT}{\Pi}\\ &=\frac{1.50(8.314)(298)}{18.6}\\ &=199.8\ \mathrm{kg\,mol^{-1}} \end{aligned} \]

Therefore:

\[ M_n=1.998\times10^5\ \mathrm{g\,mol^{-1}} \]

Using the polystyrene repeating-unit molar mass:

\[ N_r=\frac{1.998\times10^5}{104.15} \]

\[ \boxed{N_r\approx1.92\times10^3} \]

The average chain contains approximately 1,920 repeating units.

As a scale check, an osmotic pressure of \(18.6\ \mathrm{Pa}\) corresponds to a water-column height of approximately:

\[ h=\frac{\Pi}{\rho g} =\frac{18.6}{1000(9.81)} \]

\[ h\approx1.9\times10^{-3}\ \mathrm{m} =1.9\ \mathrm{mm} \]

This is a very small pressure difference, which helps explain why sensitive measurement equipment is needed.

The numerical measurements here are illustrative, rather than results from an actual experiment.

04Worked examples

Now let’s apply the method to the types of calculations you should be comfortable completing.

05Common mistakes and important limitations

Before attempting examination questions, check these four points.

Why the type of average matters

A real polymer sample generally contains chains of different lengths.

For example, a sample might contain some chains with 100 repeating units and others with 900. It would be misleading to assume every chain has the same length.

Two important averages are:

• Number-average molar mass, \(M_n\): each polymer chain contributes equally to the average.
• Weight-average molar mass, \(M_w\): heavier polymer chains contribute more strongly to the average.

To calculate the average number of repeating units per chain, use:

\[ \boxed{\overline{N_r}=\frac{M_n}{M_r}} \]

Do not automatically substitute \(M_w\).

The distinction becomes important when a sample has a broad distribution of chain lengths. IUPAC technical guide to polymer terminology.

What about the ends of the polymer chain?

Our calculation treats a chain as a collection of complete repeating units.

In reality, a finite chain has terminal groups that may not be part of the repeating structure.

A more complete expression is:

\[ M_{\mathrm{chain}} =N_rM_r+M_{\mathrm{ends}} \]

For a long polymer chain, the terminal-group contribution is usually small compared with the total chain mass.

For a short oligomer, however, the terminal groups can make a noticeable difference. If a question supplies their composition, include their mass rather than ignoring it.

Also remember that a polymer sample’s total mass alone cannot tell you how many repeating units are in each chain. You need information about the molar mass or number of polymer molecules as well.

06Questions and solutions

The following questions progress from standard calculations to more demanding applications. Pay particular attention to the information provided about the polymer structure and the type of average molar mass.

Question 1

A polystyrene sample has a number-average molar mass of \(41\,660\ \mathrm{g\,mol^{-1}}\).

The repeating unit has the molecular formula \(\ce{C8H8}\).

Calculate the average number of repeating units per chain.

Solution 1

The average chain contains 400 repeating units.

Polystyrene is an addition polymer, so the repeating unit has the same molecular formula as styrene.

First, calculate its molar mass:

\[ \begin{aligned} M_r&=8(12.011)+8(1.008)\\ &=104.15\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Now divide the number-average molar mass by the repeating-unit molar mass:

\[ \begin{aligned} N_r&=\frac{41\,660}{104.15}\\ &=400 \end{aligned} \]

Therefore, the average polystyrene chain contains approximately 400 repeating units.

The key is recognising that no atoms are lost during addition polymerisation.

Question 2

Polyvinyl chloride (PVC) is formed from chloroethene, \(\ce{C2H3Cl}\).

A PVC sample has a number-average molar mass of \(250\,000\ \mathrm{g\,mol^{-1}}\).

Calculate the average number of repeating units per chain.

A student argues that chlorine should not be included when calculating the repeating-unit molar mass because it is attached to the carbon backbone rather than forming part of it.

Explain why this reasoning is incorrect.

Solution 2

The PVC sample contains approximately 4,000 repeating units per chain.

The repeating structure is:

\[ \left[-\mathrm{CH_2-CH(Cl)-}\right]_n \]

The chlorine atom is part of the repeating unit, even though it is a side group.

Therefore, the complete molecular formula must be used:

\[ \ce{C2H3Cl} \]

Calculate its molar mass:

\[ \begin{aligned} M_r&=2(12.011)+3(1.008)+35.45\\ &=62.50\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Calculate the number of repeating units:

\[ \begin{aligned} N_r&=\frac{250\,000}{62.50}\\ &=4\,000 \end{aligned} \]

The student’s reasoning is incorrect because the repeating unit includes both backbone atoms and attached side groups.

Ignoring chlorine would underestimate the mass of one repeating unit. This would produce an artificially large estimate of the number of repeating units.

Always calculate the molar mass using the entire repeating-unit formula, not just the atoms forming the main chain.

Question 3

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

Their molecular formulas and molar masses are:

A nylon-6,6 sample has a number-average molar mass of \(79\,212\ \mathrm{g\,mol^{-1}}\).

(a) Calculate the average number of constitutional repeating units per chain.

(b) Calculate the corresponding number of monomer-derived units per chain using the formal IUPAC meaning of degree of polymerisation.

Assume terminal-group contributions are negligible.

Solution 3

The average nylon-6,6 chain contains 350 repeating units, corresponding to 700 monomer-derived units.

Step 1: Calculate the repeating-unit molar mass.

Nylon-6,6 is a condensation polymer. One repeating unit contains contributions from both monomers, with two water molecules eliminated.

\[ \begin{aligned} M_r&=116.21+146.14-2(18.015)\\ &=226.32\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Step 2: Calculate the number of repeating units.

\[ \begin{aligned} N_r&=\frac{79\,212}{226.32}\\ &=350 \end{aligned} \]

Therefore, there are approximately 350 constitutional repeating units per chain.

Step 3: Determine the number of monomer-derived units.

Each repeating unit contains contributions from two monomer molecules:

• One hexane-1,6-diamine-derived section.
• One adipic-acid-derived section.

Therefore:

\[ N_{\mathrm{monomeric}}=2(350) \]

\[ \boxed{N_{\mathrm{monomeric}}=700} \]

This question tests an important distinction.

The answer to part (a) is 350 because it asks for repeating units. The answer to part (b) is 700 because it asks for monomer-derived units.

Question 4

Two separate polymer samples are prepared:

• Sample A contains 2.00 g of polystyrene.
• Sample B contains 2.00 g of PVC.

Both samples have the same number-average molar mass:

\[ M_n=62\,500\ \mathrm{g\,mol^{-1}} \]

The repeating-unit molar masses are:

• Polystyrene: \(104.15\ \mathrm{g\,mol^{-1}}\).
• PVC: \(62.50\ \mathrm{g\,mol^{-1}}\).

(a) Calculate the average number of repeating units per chain in each sample.

(b) Calculate the amount, in moles, of polymer chains in each sample.

(c) Calculate the amount, in moles, of repeating units in each sample.

(d) Explain why the two samples contain the same amount of polymer chains but different amounts of repeating units.

Solution 4

The polystyrene chains contain approximately 600 repeating units, while the PVC chains contain 1,000 repeating units.

Despite this difference, both samples contain the same amount of polymer chains because they have equal masses and equal number-average molar masses.

Step 1: Calculate the average number of polystyrene repeating units.

\[ \begin{aligned} N_{r,\mathrm{PS}}&=\frac{62\,500}{104.15}\\ &\approx600 \end{aligned} \]

Step 2: Calculate the average number of PVC repeating units.

\[ \begin{aligned} N_{r,\mathrm{PVC}}&=\frac{62\,500}{62.50}\\ &=1\,000 \end{aligned} \]

PVC has more repeating units per chain because each PVC repeating unit has a smaller molar mass.

Step 3: Calculate the amount of polymer chains in each sample.

Use:

\[ n_{\mathrm{chains}}=\frac{m}{M_n} \]

For both samples:

\[ \begin{aligned} n_{\mathrm{chains}}&=\frac{2.00}{62\,500}\\ &=3.20\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Both samples therefore contain \(3.20\times10^{-5}\) mol of polymer chains.

Step 4: Calculate the amount of repeating units.

The relationship is:

\[ n_{\mathrm{repeat}}= N_r\times n_{\mathrm{chains}} \]

For polystyrene:

\[ \begin{aligned} n_{\mathrm{repeat,PS}} &=600(3.20\times10^{-5})\\ &=0.0192\ \mathrm{mol} \end{aligned} \]

For PVC:

\[ \begin{aligned} n_{\mathrm{repeat,PVC}} &=1\,000(3.20\times10^{-5})\\ &=0.0320\ \mathrm{mol} \end{aligned} \]

Step 5: Explain the difference.

Both samples contain the same amount of polymer chains because their total masses and number-average molar masses are identical.

However, a PVC repeating unit is lighter than a polystyrene repeating unit.

Consequently, each PVC chain contains more repeating units on average, and the PVC sample contains a greater total amount of repeating units.

The important distinction is that the number of polymer molecules and the number of repeating units are different quantities.

Question 5

A polystyrene sample contains only two populations of polymer chains:

• Population A contains chains with exactly 100 repeating units.
• Population B contains chains with exactly 900 repeating units.

Exactly 50.0% of the total polymer mass comes from Population A, and the remaining 50.0% comes from Population B.

Assume all chains are linear, terminal-group masses are negligible, and the molar mass of one polystyrene repeating unit is \(104.15\ \mathrm{g\,mol^{-1}}\).

A supplier reports the mass-average molar mass, \(M_w\), of the complete sample.

(a) Calculate \(M_w\).

(b) A student divides this value by the repeating-unit molar mass and concludes that the average chain contains 500 repeating units.

Is this conclusion correct?

(c) Calculate the actual number-average number of repeating units per chain.

Explain why the two calculated averages are different.

Solution 5

The actual number-average chain contains 180 repeating units, not 500.

The value of 500 is the mass-weighted average chain length. The distinction arises because the two populations contribute equal masses, but they do not contain equal numbers of polymer molecules.

Step 1: Calculate the molar mass of each chain population.

For Population A:

\[ \begin{aligned} M_A&=100(104.15)\\ &=10\,415\ \mathrm{g\,mol^{-1}} \end{aligned} \]

For Population B:

\[ \begin{aligned} M_B&=900(104.15)\\ &=93\,735\ \mathrm{g\,mol^{-1}} \end{aligned} \]

A chain in Population B has nine times the molar mass of a chain in Population A.

Step 2: Calculate the mass-average molar mass.

The two populations each contribute 50.0% of the total mass.

Therefore:

\[ \begin{aligned} M_w&=0.500M_A+0.500M_B\\ &=0.500(10\,415)+0.500(93\,735)\\ &=52\,075\ \mathrm{g\,mol^{-1}} \end{aligned} \]

So:

\[ \boxed{M_w=52\,075\ \mathrm{g\,mol^{-1}}} \]

Step 3: Evaluate the student’s calculation.

The student calculates:

\[ \begin{aligned} \frac{M_w}{M_r} &=\frac{52\,075}{104.15}\\ &=500 \end{aligned} \]

The arithmetic is correct, but the interpretation is wrong.

The student has calculated a mass-weighted average number of repeating units, not the average number of repeating units per polymer molecule.

Step 4: Determine the relative number of chains in each population.

The two populations contribute equal total masses.

Therefore:

\[ N_A M_A=N_B M_B \]

Rearranging:

\[ \frac{N_A}{N_B}=\frac{M_B}{M_A} \]

Substituting:

\[ \begin{aligned} \frac{N_A}{N_B} &=\frac{93\,735}{10\,415}\\ &=9 \end{aligned} \]

There are nine times as many chains in Population A as in Population B.

This is the crucial step.

Although the longer chains contain nine times as many repeating units, there are nine times as many shorter chains.

Step 5: Calculate the number-average number of repeating units.

We can represent the chain population as nine short chains for every one long chain.

Therefore:

\[ \begin{aligned} \overline{N_r} &=\frac{9(100)+1(900)}{9+1}\\ &=\frac{900+900}{10}\\ &=180 \end{aligned} \]

Hence:

\[ \boxed{\overline{N_r}=180} \]

We can check this result by calculating the number-average molar mass:

\[ \begin{aligned} M_n&=\frac{9M_A+M_B}{10}\\ &=\frac{9(10\,415)+93\,735}{10}\\ &=18\,747\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Then:

\[ \begin{aligned} \overline{N_r} &=\frac{18\,747}{104.15}\\ &=180 \end{aligned} \]

The result is consistent.

The tempting approach was to treat the equal mass contributions as equal numbers of chains. But a 900-unit chain contains nine times as much polymer mass as a 100-unit chain.

Consequently, the shorter-chain population contains nine times as many molecules.

This is why the number-average chain length is 180 rather than 500.

The distinction between number-average and weight-average molar mass becomes increasingly important when polymer samples contain a wide range of chain lengths.

07Where to go next

Once you can calculate the number of repeating units in a polymer chain, the next useful step is understanding how polymer structure influences physical properties.

Chain length is only part of the picture. Side groups, branching, intermolecular forces, and the ability of chains to pack closely together also influence properties such as flexibility, strength, and softening behaviour.

For example, the phenyl side groups in polystyrene and the chlorine substituents in PVC affect how their chains interact and move.

Being able to connect the repeating-unit structure to the properties and uses of a named polymer is the natural next step.