How to Calculate Polymer Mass in Addition Polymerisation

Suppose a factory feeds 100.0 g of ethene into a reactor and converts every molecule into polyethylene (PE). How much polyethylene should it produce?

You might predict slightly less than 100.0 g because the carbon-carbon double bonds break open during polymerisation. But that prediction is wrong.

The answer is 100.0 g of polyethylene. In ideal addition polymerisation, every atom from the original monomer molecules becomes part of the polymer. No small molecule is released, so the mass of polymer produced equals the mass of monomer consumed.

The important word is consumed. If some monomer remains unreacted, you must exclude it from your calculation.

01The key rule: mass of polymer equals mass of monomer consumed

Think about connecting identical LEGO pieces into a long chain. You change how the pieces are connected, but you don’t throw any pieces away.

Addition polymerisation works similarly. The carbon-carbon double bond in an unsaturated monomer opens, allowing neighbouring molecules to form new carbon-carbon single bonds.

The atoms are rearranged into a polymer chain. They are not removed.

For example, ethene forms polyethene:

\[
\ce{nCH2=CH2 -> [-CH2-CH2-]_{n}}
\]

Each ethene molecule contributes one \(\ce{C2H4}\) unit to the polymer. The carbon and hydrogen atoms are all retained.

This gives us the central relationship:

\[
\boxed{
m_{\text{polymer formed}}
=
m_{\text{monomer consumed}}
}
\]

Here, \(m\) represents mass, measured in grams or kilograms.

If all the monomer reacts, the calculation becomes particularly simple:

\[
\boxed{
m_{\text{polymer}}=m_{\text{initial monomer}}
}
\]

You do not subtract anything for the double bonds opening. There is no water, hydrogen, or other small molecule released during ideal addition polymerisation.

What if not all the monomer reacts?

This is where HSC questions become slightly more interesting.

Imagine starting with 100.0 g of ethene, but only 80.0% of the ethene reacts.

The remaining 20.0% is still ethene. It has not become polyethylene.

Therefore:

\[
m_{\text{polymer}}
=
m_{\text{initial monomer}}\times X
\]

where \(X\) is the fractional conversion of the monomer.

For 80.0% conversion:

\[
m_{\text{polymer}}
=
100.0\times 0.800
\]

\[
\boxed{m_{\text{polymer}}=80.0\text{ g}}
\]

The remaining 20.0 g is unreacted ethene.

Notice that we have not lost any mass. We have simply divided the original monomer into two groups:

• 80.0 g has become polyethylene.
• 20.0 g remains as ethene.

Three masses you must distinguish

A common source of incorrect answers is treating the mass of polymer formed as identical to the mass of polymer collected.

These are not necessarily the same.

Two separate processes can affect the final result:

1. Conversion determines how much monomer reacts.
2. Recovery determines how much of the polymer produced is actually collected.

If the question supplies a polymer recovery percentage, use:

\[
m_{\text{recovered}}
=
m_{\text{formed}}\times R
\]

where \(R\) is the fractional recovery.

For example, if 80.0 g of polymer forms but only 90.0% is recovered:

\[
m_{\text{recovered}}
=
80.0\times0.900
\]

\[
m_{\text{recovered}}=72.0\text{ g}
\]

The reaction produced 80.0 g of polymer, even though the final collected sample weighs only 72.0 g.

02Apply the rule to different addition polymers

The same mass relationship applies to polyethylene, PVC, polystyrene, and PTFE.

However, you need to identify the correct monomer and its molecular formula before using mole calculations.

The important detail is that the repeating unit has the same molecular formula as its original monomer.

The molar masses are approximate and use standard atomic masses.

Let’s look at why the different structures matter when calculating polymer mass.

Polyethene (PE)

Ethene has the molecular formula \(\ce{C2H4}\).

Its repeating unit is:

\[
\left[-\mathrm{CH_2-CH_2-}\right]
\]

The repeating unit also has the formula \(\ce{C2H4}\).

Therefore, 1 mole of ethene produces 1 mole of ethene-derived repeating units, each with a mass of 28.05 g.

Polyvinyl chloride (PVC)

PVC is made from chloroethene, \(\ce{C2H3Cl}\).

Its repeating unit is:

\[
\left[-\mathrm{CH_2-CHCl-}\right]
\]

Pay attention to the chlorine atom.

It remains attached to the polymer chain, even though it is not part of the carbon backbone.

You must include chlorine when calculating the molar mass:

\[
M(\ce{C2H3Cl})=62.50\text{ g mol}^{-1}
\]

Ignoring chlorine would give you an incorrect result.

Polystyrene (PS)

Polystyrene forms from styrene, \(\ce{C8H8}\).

Its repeating unit is:

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

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

All eight carbon atoms and eight hydrogen atoms are retained.

Polytetrafluoroethene (PTFE)

PTFE is made from tetrafluoroethene, \(\ce{C2F4}\).

Its repeating unit is:

\[
\left[-\mathrm{CF_2-CF_2-}\right]
\]

All four fluorine atoms are retained in each repeating unit.

This is why the molar mass of the original monomer can be used directly when calculating the mass of PTFE produced.

For a more detailed explanation of these structures and their properties, see Addition polymers explained: PE, PVC, PS and PTFE for HSC Chemistry.

03A reliable method for calculating polymer mass

When you encounter a question asking for the mass of an addition polymer, work through these steps.

Step 1: Identify the monomer.

Determine which small molecule forms the polymer. Check its molecular formula, particularly if it contains atoms in side groups.

Step 2: Determine how much monomer reacts.

If the question gives 100% conversion, all the monomer reacts.

If it gives a conversion percentage, multiply the initial amount by the fractional conversion.

Step 3: Apply conservation of mass.

For ideal addition polymerisation:

\[
m_{\text{polymer formed}}
=
m_{\text{monomer consumed}}
\]

If the question gives the amount in moles, use:

\[
m=nM
\]

where:

• \(m\) is mass in grams,
• \(n\) is amount in moles,
• \(M\) is molar mass in g mol⁻¹.

The molar mass of the repeating unit is the same as the molar mass of the monomer.

Step 4: Account for recovery if necessary.

Only multiply by a recovery percentage if the question asks for the mass actually collected and provides information about polymer recovery.

Do not apply a recovery percentage to the mass already described as polymer formed.

04Worked examples

05When this rule does not apply: condensation polymerisation

There is one important contrast worth understanding.

Addition polymerisation retains all the atoms from the monomers. Condensation polymerisation, however, involves the elimination of small molecules when monomers join.

Consider nylon-6,6.

It forms from two monomers:

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

When these monomers join, water is eliminated as amide links form.

For an idealised repeating pair, the mass relationship is:

\[
\ce{C6H16N2 + C6H10O4 -> C12H22N2O2 + 2H2O}
\]

The polymer repeating unit therefore has a lower molar mass than the combined starting monomers.

Using the approximate molar masses:

The mass of one idealised nylon-6,6 repeating unit is:

\[
\begin{aligned}
M_{\text{repeat}}
&=116.21+146.14-2(18.015)\\
&=226.32\text{ g mol}^{-1}
\end{aligned}
\]

The two water molecules have been subtracted because they are released rather than incorporated into the polymer.

This is the key distinction:

For this article, our calculations focus on ideal addition polymerisation. Do not subtract water or any other by-product unless the reaction actually produces one.

Scope note: The named polymers connect to the Polymers content of the Chemistry Stage 6 (2017) syllabus. The quantitative mass-balance method in this article is an extension of the structure-and-use work, rather than a separate polymer-mass calculation dot point.

06Questions and solutions

The following questions move from straightforward mass calculations to more demanding applications involving conversion, recovery, condensation polymerisation, and copolymer composition.

Question 1

A chemist completely polymerises 0.250 mol of tetrafluoroethene to produce PTFE.

Tetrafluoroethene has the molecular formula \(\ce{C2F4}\) and a molar mass of \(100.02\text{ g mol}^{-1}\).

Calculate the mass of PTFE produced.

Solution 1

The mass of PTFE produced is 25.0 g.

PTFE is an addition polymer, so every atom from the tetrafluoroethene molecules is retained in the polymer.

The amount of repeating units formed is therefore 0.250 mol.

Using:

\[
m=nM
\]

Substitute:

\[
\begin{aligned}
m&=0.250\times100.02\\
&=25.005\text{ g}
\end{aligned}
\]

To three significant figures:

\[
\boxed{m_{\text{PTFE}}=25.0\text{ g}}
\]

The mass of PTFE equals the mass of tetrafluoroethene consumed because no small molecule is released.

Question 2

A manufacturer uses 52.0 g of styrene to produce polystyrene.

At the end of the reaction, 80.0% of the styrene has undergone addition polymerisation.

Calculate:

(a) the mass of polystyrene produced,

(b) the mass of unreacted styrene remaining.

Use \(M(\ce{C8H8})=104.15\text{ g mol}^{-1}\).

Solution 2

The reaction produces 41.6 g of polystyrene, with 10.4 g of styrene remaining unreacted.

The conversion is 80.0%, meaning 80.0% of the original styrene has reacted.

(a) Mass of polystyrene produced

Because this is ideal addition polymerisation:

\[
m_{\text{polymer}}
=
m_{\text{monomer consumed}}
\]

Therefore:

\[
\begin{aligned}
m_{\text{polystyrene}}
&=52.0\times0.800\\
&=41.6\text{ g}
\end{aligned}
\]

(b) Mass of unreacted styrene

The unreacted fraction is:

\[
100.0\%-80.0\%=20.0\%
\]

Therefore:

\[
\begin{aligned}
m_{\text{unreacted}}
&=52.0\times0.200\\
&=10.4\text{ g}
\end{aligned}
\]

Check the mass balance:

\[
41.6+10.4=52.0\text{ g}
\]

The result is consistent with conservation of mass.

A tempting mistake is to use the 52.0 g as the polymer mass simply because addition polymerisation retains all atoms. However, only the styrene that actually reacts becomes polystyrene.

Question 3

A chemist uses 1.50 mol of chloroethene to produce PVC.

The reaction has a monomer conversion of 68.0%. During processing, 92.0% of the PVC formed is successfully recovered.

The molar mass of chloroethene is \(62.50\text{ g mol}^{-1}\).

Calculate:

(a) the mass of PVC formed,

(b) the mass of unreacted chloroethene,

(c) the mass of PVC recovered, and

(d) the mass of polymer not recovered.

Solution 3

The reaction produces 63.8 g of PVC, of which 58.7 g is recovered. There are 30.0 g of unreacted chloroethene and 5.10 g of unrecovered polymer.

(a) Mass of PVC formed

First, calculate the amount of chloroethene that reacts:

\[
\begin{aligned}
n_{\text{reacted}}
&=1.50\times0.680\\
&=1.02\text{ mol}
\end{aligned}
\]

Each mole of chloroethene contributes one mole of PVC repeating units.

Therefore:

\[
\begin{aligned}
m_{\text{PVC}}
&=nM\\
&=1.02\times62.50\\
&=63.75\text{ g}
\end{aligned}
\]

To three significant figures:

\[
\boxed{m_{\text{PVC formed}}=63.8\text{ g}}
\]

(b) Mass of unreacted chloroethene

The unreacted amount is:

\[
\begin{aligned}
n_{\text{unreacted}}
&=1.50-1.02\\
&=0.480\text{ mol}
\end{aligned}
\]

Calculate its mass:

\[
\begin{aligned}
m_{\text{unreacted}}
&=0.480\times62.50\\
&=30.0\text{ g}
\end{aligned}
\]

(c) Mass of PVC recovered

The recovery is 92.0%, so:

\[
\begin{aligned}
m_{\text{recovered}}
&=63.75\times0.920\\
&=58.65\text{ g}
\end{aligned}
\]

Therefore:

\[
\boxed{m_{\text{recovered}}=58.7\text{ g}}
\]

(d) Mass of polymer not recovered

Subtract the collected mass from the mass actually formed:

\[
\begin{aligned}
m_{\text{not recovered}}
&=63.75-58.65\\
&=5.10\text{ g}
\end{aligned}
\]

The final distinction is important: the reaction produced 63.75 g of PVC, but only 58.65 g was collected.

Recovery affects the amount of product obtained, not the amount of polymerisation that occurred.

Question 4

Nylon-6,6 is produced by condensation polymerisation between hexane-1,6-diamine and adipic acid.

A reaction contains:

• 0.400 mol of hexane-1,6-diamine.
• 0.400 mol of adipic acid.

Assume both monomers react completely in the correct 1:1 ratio, and terminal-group contributions are negligible.

Use the following molar masses:

• Hexane-1,6-diamine: \(116.21\text{ g mol}^{-1}\).
• Adipic acid: \(146.14\text{ g mol}^{-1}\).
• Water: \(18.015\text{ g mol}^{-1}\).

Calculate:

(a) the mass of nylon-6,6 produced,

(b) the mass of water released.

Solution 4

The reaction produces 90.5 g of nylon-6,6 and 14.4 g of water.

Unlike addition polymerisation, condensation polymerisation releases small molecules as the polymer chains form.

For nylon-6,6, two water molecules are eliminated per idealised repeating pair.

(a) Mass of nylon-6,6 produced

First, establish the repeating-unit mass.

The idealised reaction is:

\[
\ce{C6H16N2 + C6H10O4 -> C12H22N2O2 + 2H2O}
\]

Calculate the molar mass of the nylon-6,6 repeating unit:

\[
\begin{aligned}
M_{\text{repeat}}
&=116.21+146.14\\
&\quad-2(18.015)\\
&=226.32\text{ g mol}^{-1}
\end{aligned}
\]

Because the monomers react in a 1:1 ratio, 0.400 mol of each produces 0.400 mol of idealised repeating pairs.

Therefore:

\[
\begin{aligned}
m_{\text{nylon}}
&=0.400\times226.32\\
&=90.528\text{ g}
\end{aligned}
\]

\[
\boxed{m_{\text{nylon}}=90.5\text{ g}}
\]

(b) Mass of water released

There are two moles of water released for every mole of idealised repeating pairs.

Therefore:

\[
\begin{aligned}
n_{\ce{H2O}}
&=2\times0.400\\
&=0.800\text{ mol}
\end{aligned}
\]

Calculate the mass:

\[
\begin{aligned}
m_{\ce{H2O}}
&=0.800\times18.015\\
&=14.412\text{ g}
\end{aligned}
\]

\[
\boxed{m_{\ce{H2O}}=14.4\text{ g}}
\]

Check the mass balance using the original monomers:

\[
\begin{aligned}
m_{\text{reactants}}
&=0.400(116.21+146.14)\\
&=104.94\text{ g}
\end{aligned}
\]

The products have a combined mass of:

\[
90.528+14.412=104.94\text{ g}
\]

The mass is conserved.

The tempting approach would be to add the masses of both monomers and call the result the mass of nylon. That would incorrectly include the water released during condensation.

This is precisely why you must identify the type of polymerisation before calculating.

Question 5

A chemist prepares an idealised copolymer using two different addition-polymer monomers:

• 60.0 g of ethene, \(\ce{C2H4}\).
• 104.0 g of styrene, \(\ce{C8H8}\).
• 20.0 g of inert solvent.

The ethene and styrene react together to form a copolymer.

After polymerisation, analysis shows that the reaction mixture contains:

• 12.0 g of unreacted ethene.
• 26.0 g of unreacted styrene.

The mixture is then processed. All the solvent is removed, 120.0 g of dry copolymer is collected, and a further 6.0 g of dry copolymer is found remaining on the filter and reaction vessel.

Assume:

• All consumed monomer is incorporated into the copolymer.
• No monomer is lost through evaporation or other processes.
• No other products form.
• The 6.0 g remaining on the apparatus is entirely polymer.
• Initiator and terminal-group contributions are negligible.

Calculate:

(a) the total mass of copolymer formed.

(b) The percentage conversion of ethene and styrene separately.

(c) The overall monomer conversion by mass.

(d) The mass percentage contributed by each monomer to the copolymer.

(e) The mole percentage of ethene-derived and styrene-derived repeating units.

(f) The polymer recovery percentage.

Finally, evaluate these two statements:

• The overall monomer conversion is the arithmetic mean of the two individual conversion percentages.
• Since the overall monomer conversion is approximately 76.8%, 76.8% of the original reactor mixture became polymer.

Solution 5

The reaction produces 126.0 g of copolymer, with an overall monomer mass conversion of 76.8% and a polymer recovery of 95.2%.

The copolymer contains more ethene-derived repeating units by number, even though styrene contributes more to its mass.

This question tests several connected ideas. The important thing is to keep mass conversion, mole composition, and product recovery separate.

(a) Total mass of copolymer formed

Start by calculating how much of each monomer reacted.

For ethene:

\[
\begin{aligned}
m_{\text{ethene consumed}}
&=60.0-12.0\\
&=48.0\text{ g}
\end{aligned}
\]

For styrene:

\[
\begin{aligned}
m_{\text{styrene consumed}}
&=104.0-26.0\\
&=78.0\text{ g}
\end{aligned}
\]

Both monomers undergo ideal addition polymerisation, so all their consumed mass becomes part of the copolymer.

Therefore:

\[
\begin{aligned}
m_{\text{copolymer}}
&=48.0+78.0\\
&=126.0\text{ g}
\end{aligned}
\]

\[
\boxed{m_{\text{copolymer formed}}=126.0\text{ g}}
\]

Notice that the solvent is not included. It does not form part of the polymer.

(b) Percentage conversion of each monomer

Calculate conversion separately for each monomer.

For ethene:

\[
\begin{aligned}
X_{\text{ethene}}
&=\frac{48.0}{60.0}\times100\\
&=80.0\%
\end{aligned}
\]

For styrene:

\[
\begin{aligned}
X_{\text{styrene}}
&=\frac{78.0}{104.0}\times100\\
&=75.0\%
\end{aligned}
\]

Therefore:

• Ethene conversion = 80.0%.
• Styrene conversion = 75.0%.

The two monomers have different conversion percentages.

(c) Overall monomer conversion by mass

The overall conversion must account for the different initial masses of the monomers.

The total initial monomer mass is:

\[
m_{\text{monomers}}
=60.0+104.0
=164.0\text{ g}
\]

The total mass consumed is 126.0 g.

Therefore:

\[
\begin{aligned}
X_{\text{overall}}
&=\frac{126.0}{164.0}\times100\\
&=76.829\ldots\%
\end{aligned}
\]

\[
\boxed{X_{\text{overall}}=76.8\%}
\]

Why isn’t this simply the average of 80.0% and 75.0%?

The arithmetic mean would be:

\[
\frac{80.0+75.0}{2}=77.5\%
\]

But the two monomers were not initially present in equal masses.

Styrene contributed 104.0 g, while ethene contributed only 60.0 g. The overall conversion must therefore be weighted according to the initial monomer masses.

The correct calculation is:

\[
X_{\text{overall}}
=
\frac{60.0(0.800)+104.0(0.750)}
{60.0+104.0}
\]

This gives 76.8%.

(d) Mass percentage contributed by each monomer

The copolymer contains 48.0 g of ethene-derived material and 78.0 g of styrene-derived material.

Calculate the mass percentages using the total polymer mass of 126.0 g.

Ethene contribution:

\[
\begin{aligned}
\%\text{ ethene}
&=\frac{48.0}{126.0}\times100\\
&=38.1\%
\end{aligned}
\]

Styrene contribution:

\[
\begin{aligned}
\%\text{ styrene}
&=\frac{78.0}{126.0}\times100\\
&=61.9\%
\end{aligned}
\]

The copolymer therefore contains:

• 38.1% ethene-derived mass.
• 61.9% styrene-derived mass.

(e) Mole percentage of repeating units

Here is the more subtle part.

A styrene-derived repeating unit has a greater molar mass than an ethene-derived repeating unit.

Therefore, a greater mass contribution does not necessarily mean a greater number of repeating units.

First, calculate the amount of ethene-derived repeating units:

\[
\begin{aligned}
n_{\text{ethene units}}
&=\frac{48.0}{28.05}\\
&=1.71\text{ mol}
\end{aligned}
\]

Now calculate the amount of styrene-derived repeating units:

\[
\begin{aligned}
n_{\text{styrene units}}
&=\frac{78.0}{104.15}\\
&=0.749\text{ mol}
\end{aligned}
\]

The total amount of monomer-derived units is:

\[
\begin{aligned}
n_{\text{total}}
&=1.71+0.749\\
&=2.46\text{ mol}
\end{aligned}
\]

Calculate the mole percentage of ethene-derived units:

\[
\begin{aligned}
\%\text{ ethene units}
&=\frac{1.71}{2.46}\times100\\
&\approx69.6\%
\end{aligned}
\]

For styrene-derived units:

\[
\begin{aligned}
\%\text{ styrene units}
&=\frac{0.749}{2.46}\times100\\
&\approx30.4\%
\end{aligned}
\]

This gives us an interesting result:

Styrene contributes more than half the polymer’s mass, but ethene contributes more than half the repeating units.

The reason is simple: each styrene-derived unit is much heavier than each ethene-derived unit.

You also cannot determine the order in which these units appear along the copolymer chain from this mass balance alone. That would require additional information about the polymerisation process.

(f) Polymer recovery percentage

The reaction formed 126.0 g of copolymer.

The chemist collected 120.0 g, with another 6.0 g remaining on the equipment.

The recovery is:

\[
\begin{aligned}
R
&=\frac{120.0}{126.0}\times100\\
&=95.238\ldots\%
\end{aligned}
\]

\[
\boxed{R=95.2\%}
\]

The difference between the polymer formed and the polymer recovered is:

\[
126.0-120.0=6.0\text{ g}
\]

The reaction itself did produce this additional 6.0 g. It simply wasn’t collected.

Evaluating the two statements

The first statement claims that overall conversion is the arithmetic mean of the individual conversion percentages.

This is incorrect.

The arithmetic mean gives 77.5%, but the correct mass-weighted overall conversion is 76.8%, because the initial monomer masses were different.

The second statement claims that 76.8% of the original reactor mixture became polymer.

This is also incorrect.

The original reactor mixture contained:

\[
60.0+104.0+20.0=184.0\text{ g}
\]

The 20.0 g of solvent must be included when calculating the fraction of the entire reactor charge that became polymer, but excluded when calculating monomer conversion.

Therefore:

\[
\begin{aligned}
\%\text{ of reactor charge becoming polymer}
&=\frac{126.0}{184.0}\times100\\
&=68.5\%
\end{aligned}
\]

Only 68.5% of the original reactor charge became polymer.

If we instead calculate the fraction of the initial reactor charge represented by the isolated product:

\[
\begin{aligned}
\%\text{ isolated product}
&=\frac{120.0}{184.0}\times100\\
&=65.2\%
\end{aligned}
\]

These percentages answer different questions:

• 76.8% is the conversion of the initial monomer mass.
• 95.2% is the recovery of polymer formed.
• 68.5% is the fraction of the entire initial reactor charge converted into polymer.
• 65.2% is the fraction of the entire initial reactor charge represented by the collected polymer.

The crucial lesson is that a percentage is meaningless until you identify its denominator.

07Where to go next

Once you can calculate the total mass of polymer produced, the next step is understanding how that mass is distributed across individual polymer chains.

For example, knowing that you have 52.1 g of polystyrene does not tell you how many repeating units are in each chain. You need the average molar mass of the polymer chains as well.

That is a different calculation, covered in How to Calculate Repeat Units in a Polymer Chain.

Keep the two questions separate:

• Polymer mass: How much monomer has been incorporated into the polymer?
• Number of repeating units: How many monomer-derived units make up an average polymer chain?

Getting the first mass balance right gives you the foundation for the second calculation.