How to Count Water Molecules in a Short Polyester Chain

Two molecules of ethane-1,2-diol and two molecules of terephthalic acid can form a short, open polyester chain. How many water molecules are released?

Three, not four.

Although there are four monomer molecules, only three ester linkages are needed to connect them into one continuous, open chain. One hydroxyl group (\(\ce{-OH}\)) and one carboxyl group (\(\ce{-COOH}\)) remain unreacted at the two ends.

This is the detail that makes counting water molecules in short polyester chains slightly different from simply counting repeat units.

01Polyester is a polymer family; PET is one example

Before counting anything, distinguish between two terms.

A polyester is a family of polymers containing repeating ester linkages in their main molecular chain. PET, or poly(ethylene terephthalate), is one particular polyester.

In the direct-condensation model, PET forms from two monomers:

Terephthalic acid is also called benzene-1,4-dicarboxylic acid.

Both monomers have two reactive ends. This allows them to connect to neighbouring molecules and form an extended chain.

Other polyesters can be made from different monomers. The counting method in this article applies to direct condensation between carboxylic acid and hydroxyl groups.

For the broader explanation of how condensation polymerisation produces nylon and polyester, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

02One ester linkage produces one water molecule

The important reaction occurs when a carboxylic acid group reacts with an alcohol group.

The general reaction is:

\[ \ce{RCOOH + R'OH -> RCOOR' + H2O} \]

Here, \(R\) and \(R'\) represent the remaining parts of the molecules.

Two things happen:

• The carboxylic acid group loses an \(\ce{-OH}\) group.
• The alcohol group loses an \(\ce{H}\) atom.

These combine to form one water molecule, \(\ce{H2O}\).

The remaining atoms form an ester linkage:

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

Notice that the oxygen from the alcohol remains in the ester linkage.

This gives us the most important counting rule:

One ester linkage formed by direct acid-alcohol condensation produces one water molecule.

Therefore:

\[ N(\ce{H2O})=N(\text{ester linkages}) \]

Here, \(N\) represents the number of molecules.

For bulk quantities, the equivalent relationship is:

\[ n(\ce{H2O})=n(\text{ester linkages}) \]

where \(n\) represents the amount in moles.

The reaction route matters. If an ester is formed from an acyl chloride and an alcohol instead, the by-product is hydrogen chloride, not water. Always check the starting functional groups before applying the water-counting rule.

03Count the ester linkages in the actual chain

Let’s return to the short PET chain containing two molecules of each monomer.

Its structure can be represented as:

\[ \begin{aligned} \ce{HO-CH2-CH2-O-C(=O)}\\ \ce{-C6H4-C(=O)-O-CH2-CH2-O-C(=O)}\\ \ce{-C6H4-COOH} \end{aligned} \]

Read the structure from left to right.

You can identify three ester linkages:

• The first ethane-1,2-diol molecule connects to the first terephthalic acid molecule.
• The first terephthalic acid molecule connects to the second ethane-1,2-diol molecule.
• The second ethane-1,2-diol molecule connects to the second terephthalic acid molecule.

Each connection releases one water molecule.

Why aren’t there four ester linkages?

Because the chain is open. Its two ends have not reacted with each other.

A useful way to picture this is four train carriages connected in a straight line. You need three couplings to connect four carriages. You would need a fourth coupling only if you connected the last carriage back to the first.

The analogy helps count connections, but the chemistry determines what each connection produces: in this case, one water molecule.

04The counting rule for different chain structures

The train analogy leads to a useful mathematical shortcut.

For a single, open, linear chain containing \(N\) bifunctional monomer molecules:

\[ \boxed{N(\ce{H2O})=N-1} \]

This assumes every monomer is incorporated into the same chain and every connection forms through direct esterification.

If the monomers form multiple separate chains, adjust the calculation:

The distinction between an open chain and a cyclic chain is particularly important.

The repeat-unit trap in PET

You might encounter the PET repeating unit written as:

\[ \left[ \ce{-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-} \right]_n \]

Each idealised repeat unit contains one ethane-1,2-diol residue and one terephthalate residue.

The repeating pattern contains two ester-link positions per pair of residues. However, the bracket notation does not explicitly show the terminal groups of a finite polymer molecule.

For a finite, open PET chain containing \(n\) molecules of each monomer:

• Total monomer molecules = \(2n\).
• Actual ester linkages = \(2n-1\).
• Water molecules released = \(2n-1\).

Therefore:

\[ \boxed{N(\ce{H2O})=2n-1} \]

If the chain closes into a ring, the final connection forms another ester linkage:

\[ N(\ce{H2O})=2n \]

You may see an idealised polymerisation equation that writes \(2n\ce{H2O}\) for \(n\) PET repeat units. That notation suppresses the finite chain-end details. For an explicitly drawn short, open molecule, count its actual ester linkages instead.

This one-molecule correction becomes much more noticeable when the polymer is short.

07How could an experiment check the water count?

Counting ester linkages from a molecular structure is usually the most direct method. However, measuring the water produced can provide an independent check.

Imagine a simplified experiment designed to collect the water released during polyester formation.

The reaction mixture containing the diol and dicarboxylic acid is placed in a dry reaction flask and heated under controlled conditions.

A vapour outlet connects the flask to a water-cooled condenser. The condenser directs the condensed liquid into a dry, pre-weighed collection vessel.

The water is therefore transported from the reaction flask, through the condenser, and into the receiver, where its mass can be measured.

What should the apparatus show?

Suppose the reaction forms \(0.0020\) mol of the five-monomer chains from Worked example 2.

We predicted that the reaction should produce:

\[ n(\ce{H2O})=0.0080\text{ mol} \]

The expected water mass is:

\[ m=0.0080\times18.0=0.144\text{ g} \]

If the reaction proceeds as assumed and all the water is collected, the receiver should gain \(0.144\) g.

What if the measured mass is different?

Suppose the receiver gains only \(0.126\) g. This is illustrative measurement data, not a result from an actual experiment.

The measured amount of water is:

\[ \begin{aligned} n(\ce{H2O}) &=\frac{m}{M}\\ &=\frac{0.126}{18.0}\\ &=0.00700\text{ mol} \end{aligned} \]

This is less than the predicted \(0.0080\) mol.

It would be tempting to conclude that only 87.5% of the expected esterification occurred:

\[ \frac{0.00700}{0.00800}\times100 =87.5\% \]

However, that conclusion is only justified if the collection is quantitative and no other sources of water are present.

Some water might remain in the reaction mixture, escape through a poorly sealed connection, or condense elsewhere in the apparatus. Moisture already present in the reactants or glassware could also increase the measured mass.

A better design uses dried reactants and apparatus, sealed connections, and a blank measurement to account for background moisture. The collection vessel should also be checked to ensure that all condensate is captured.

There is one further limitation: measuring total water does not tell you whether the polymer molecules are linear or cyclic. Both structures can release water. To distinguish them, you need information about the remaining terminal functional groups as well.

08Questions and solutions

Question 1

An open polyester molecule contains the following monomer sequence:

\[ \mathrm{D-A-D} \]

Here, D represents a diol residue and A represents a dicarboxylic acid residue.

Assuming direct esterification, calculate:

a) The number of ester linkages formed.

b) The number of water molecules released.

Solution 1

The molecule contains two ester linkages and releases two water molecules.

There are three monomer residues in the sequence:

\[ \mathrm{D-A-D} \]

Each neighbouring pair must be connected by an ester linkage.

The first diol connects to the acid, and the acid connects to the second diol.

Therefore:

\[ N(\text{ester links})=3-1=2 \]

Each ester linkage produces one water molecule.

\[ N(\ce{H2O})=2 \]

The two terminal diol residues each retain one hydroxyl group. The central dicarboxylic acid residue has used both of its carboxyl groups.

The common mistake is to count the two carboxyl groups on the acid and assume that each produces a separate water molecule without checking the number of connections actually formed. Here, both are used, giving two waters.

Question 2

An illustrative batch contains four molecules of a diol and four molecules of a dicarboxylic acid.

They form two separate open polyester chains:

• Chain X: \(\mathrm{D-A-D-A-D}\)
• Chain Y: \(\mathrm{A-D-A}\)

Calculate the total number of water molecules released.

Also determine how many hydroxyl groups and carboxyl groups remain unreacted across both chains.

Solution 2

The reaction releases six water molecules, leaving two hydroxyl groups and two carboxyl groups unreacted.

Start by calculating the number of ester linkages in each chain.

Chain X contains five monomer molecules:

\[ N_X=5-1=4 \]

Therefore, Chain X produces four water molecules.

Chain Y contains three monomer molecules:

\[ N_Y=3-1=2 \]

Therefore, Chain Y produces two water molecules.

The total water released is:

\[ \begin{aligned} N(\ce{H2O})&=4+2\\ &=6 \end{aligned} \]

Now examine the chain ends.

Chain X starts and finishes with a diol residue. Each terminal diol retains one hydroxyl group.

Therefore, Chain X has two unreacted \(\ce{-OH}\) groups.

Chain Y starts and finishes with a dicarboxylic acid residue. Each terminal acid retains one carboxyl group.

Therefore, Chain Y has two unreacted \(\ce{-COOH}\) groups.

The final count is:

A tempting answer is eight water molecules because four diol molecules and four acid molecules provide eight possible ester-forming connections on each side of the reaction.

However, the monomers form two separate open chains. Each chain has one fewer connection than its number of monomer molecules.

This is why you must account for the number of separate chains, not just the total number of monomers.

Question 3

A finite PET molecule contains four ethylene terephthalate repeat pairs. It consists of four ethane-1,2-diol molecules and four terephthalic acid molecules, connected in one open linear chain.

A student claims that eight water molecules must have been released because there are two ester-link positions per repeat pair.

Determine the exact number of water molecules released.

Explain why the student’s answer is incorrect and describe the structural condition under which eight water molecules would be produced.

Solution 3

The exact number is seven water molecules, not eight. Eight would be produced if the chain formed a closed ring.

There are four diol molecules and four dicarboxylic acid molecules.

Therefore:

\[ N=4+4=8 \]

Because all eight monomers form one open linear chain:

\[ \begin{aligned} N(\text{ester links})&=N-1\\ &=8-1\\ &=7 \end{aligned} \]

Each ester linkage releases one water molecule.

\[ \boxed{N(\ce{H2O})=7} \]

The student’s reasoning comes from counting two ester-link positions per repeat pair.

That is a useful way to recognise the repeating pattern in an idealised PET structure. However, it ignores the terminal groups of the finite molecule.

The open chain has one unreacted hydroxyl group and one unreacted carboxyl group. These represent the two ends that have not formed another ester linkage.

For eight water molecules to be released, the terminal hydroxyl and carboxyl groups would need to react with each other, closing the molecule into a ring.

That final ring-closing reaction would produce one additional water molecule.

Therefore:

• Open chain: 7 ester linkages and 7 water molecules.
• Closed ring: 8 ester linkages and 8 water molecules.

The difference is not the number of repeat pairs. It is whether the final connection has formed.

Question 4

An illustrative reaction uses:

• \(0.0100\) mol of ethane-1,2-diol.
• \(0.0100\) mol of terephthalic acid.

The reaction produces \(0.0180\) mol of water.

Assume all monomer molecules are incorporated into open linear polyester chains. There are no cyclic products, branches, or unreacted monomers, and all the water produced is collected.

Calculate:

a) The amount of ester linkages formed, in moles.

b) The percentage of the original carboxyl groups that reacted.

c) The average number of original monomer molecules in each polymer chain.

Solution 4

The reaction forms \(0.0180\) mol of ester linkages, achieves 90.0% carboxyl-group conversion, and produces chains containing an average of ten monomer molecules each.

Step 1: Calculate the amount of ester linkages formed.

One mole of ester linkages formed by direct condensation produces one mole of water.

Therefore:

\[ n(\text{ester links})=0.0180\text{ mol} \]

Step 2: Calculate the initial amount of carboxyl groups.

Each terephthalic acid molecule contains two carboxyl groups.

Therefore:

\[ \begin{aligned} n(\ce{-COOH})_{\text{initial}} &=2\times0.0100\\ &=0.0200\text{ mol} \end{aligned} \]

The amount that reacted is \(0.0180\) mol.

The percentage conversion is:

\[ \begin{aligned} \text{Conversion} &=\frac{0.0180}{0.0200}\times100\\ &=90.0\% \end{aligned} \]

Step 3: Determine the remaining functional groups.

The unreacted carboxyl groups are:

\[ 0.0200-0.0180=0.0020\text{ mol} \]

The initial amount of hydroxyl groups is also:

\[ 2\times0.0100=0.0200\text{ mol} \]

After the same number of esterification reactions, the remaining hydroxyl groups are:

\[ 0.0200-0.0180=0.0020\text{ mol} \]

Therefore, the total amount of terminal functional groups is:

\[ 0.0020+0.0020=0.0040\text{ mol} \]

Step 4: Calculate the amount of polymer chains.

Each open linear chain has two ends.

Therefore:

\[ \begin{aligned} n(\text{chains}) &=\frac{0.0040}{2}\\ &=0.0020\text{ mol} \end{aligned} \]

Step 5: Calculate the average number of monomer molecules per chain.

The total amount of original monomer molecules is:

\[ 0.0100+0.0100=0.0200\text{ mol} \]

Divide this by the amount of polymer chains:

\[ \begin{aligned} \text{Average monomers per chain} &=\frac{0.0200}{0.0020}\\ &=10 \end{aligned} \]

Each chain therefore contains an average of ten original monomer molecules.

This is an average, not a claim that every individual chain has exactly ten monomer molecules.

The useful cross-check is that ten monomer molecules in an open linear chain require nine ester linkages. The calculated average is also:

\[ \frac{0.0180}{0.0020}=9 \]

The two methods agree.

This question demonstrates why water production can tell you more than just the number of ester linkages. With suitable assumptions about the products, it can also help determine the average chain length.

Question 5

A condensation reaction begins with six molecules of a diol and six molecules of a dicarboxylic acid.

All twelve monomer molecules are incorporated into exactly three separate polyester molecules.

The following information is obtained:

• Exactly two terminal hydroxyl groups remain across all products.
• Exactly two terminal carboxyl groups remain across all products.
• Every monomer is bifunctional.
• There are no branches, crosslinks, or unreacted monomers.

Determine:

a) How many of the three polyester molecules are open linear chains?

b) How many are cyclic?

c) How many water molecules were released in total?

Explain why simply subtracting the number of polymer molecules from the total number of monomers gives the wrong answer.

Solution 5

The products consist of two open linear chains and one cyclic polyester molecule. A total of ten water molecules are released.

This question requires you to combine information about the number of molecules, their terminal groups, and their connectivity.

Start by examining what the terminal groups tell us.

An open linear polyester chain has two ends. These ends contain unreacted functional groups.

A cyclic polyester has no terminal ends because the last monomer connects back to the first.

The reaction mixture contains:

• Two terminal hydroxyl groups.
• Two terminal carboxyl groups.

That is four terminal functional groups altogether.

Since each open linear chain has two ends:

\[ N(\text{linear chains})=\frac{4}{2}=2 \]

There are three separate polyester molecules in total.

Therefore:

\[ N(\text{cyclic molecules})=3-2=1 \]

The three products must consist of two open chains and one ring.

Now calculate the number of water molecules.

Step 1: Count the original carboxyl groups.

There are six dicarboxylic acid molecules, each containing two carboxyl groups.

\[ N(\ce{-COOH})_{\text{initial}}=6\times2=12 \]

Step 2: Determine how many carboxyl groups reacted.

Two terminal carboxyl groups remain unreacted.

Therefore:

\[ N(\text{ester links})=12-2=10 \]

Step 3: Convert ester linkages into water molecules.

Each ester linkage formed by direct condensation releases one water molecule.

\[ \boxed{N(\ce{H2O})=10} \]

We can check this result using the hydroxyl groups.

There are six diol molecules, each containing two hydroxyl groups:

\[ N(\ce{-OH})_{\text{initial}}=6\times2=12 \]

Two hydroxyl groups remain unreacted.

Therefore:

\[ 12-2=10 \]

This independently confirms that ten ester linkages have formed.

Why is subtracting three from twelve incorrect?

A tempting approach is:

\[ 12-3=9 \]

This assumes that all three polyester molecules are open linear chains, each requiring one fewer connection than its number of monomer molecules.

However, the terminal-group information contradicts that assumption.

Only two molecules are open chains. The third is cyclic and has no unreacted ends. It contains an additional ester linkage compared with an open chain containing the same number of monomer residues.

The alternative shortcut of treating all twelve monomers as one continuous chain would also be incorrect because the monomers form three separate molecules.

The key is to use both pieces of information:

• The number of separate polymer molecules tells you how many products exist.
• The terminal functional groups tell you which products are open chains and which are cyclic.

Only after establishing the structure can you count the ester linkages correctly.

09Where to go next

Once you can count ester linkages in a finite polyester, the next useful step is calculating its average chain length and molar mass.

The same distinction between repeat units and terminal groups becomes important when determining the degree of polymerisation. In a short molecule, ignoring the ends can produce a noticeably incorrect result.

For broader practice with polymer structures, ester linkages, and condensation reactions, return to Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

The essential habit to carry into your HSC Chemistry questions is simple: identify the actual ester linkages in the structure before counting the water molecules.