Reflux vs Distillation: How to Choose the Right Apparatus

Suppose you’re making an ester and need to heat the reactants for long enough to react. Heating speeds up the reaction, but the alcohol and ester can both escape as vapour. If you let that vapour leave the flask, you risk losing substances that you still need. If you cool it and send it back, the reaction mixture stays together.

So, which apparatus should you use?

Use reflux when you want to heat a reaction mixture without losing its volatile components. Use distillation when you want to separate and collect a volatile component from a mixture.

Both techniques use a condenser. Both involve vaporisation and condensation. The crucial difference is what happens to the condensed liquid: it either returns to the original flask or flows into a separate collecting vessel.

01The first decision: keep the mixture or separate it?

Imagine a flask containing a mixture of liquids. You heat it until vapour rises into a water-cooled condenser.

Predict what happens next. Does the liquid return to the flask, or does it travel into another container?

That single decision determines whether you’re using reflux or distillation.

A useful way to picture the difference is a loop versus a one-way journey:

• Reflux: Liquid evaporates, condenses, and returns to the same flask. The mixture stays together.
• Distillation: Liquid evaporates, condenses, and enters a separate receiver. The mixture is separated into a distillate and a residue.

The analogy describes the movement of material, not the chemical reactions occurring inside the flask. Those reactions depend on the substances and conditions involved.

02Reflux: heat the reaction without losing reactants

Reflux is used when a chemical reaction needs heating but the reactants or products are volatile.

A volatile substance evaporates relatively easily. If you heat a volatile organic mixture in an ordinary open flask, some of its components can escape before the reaction is complete.

That creates two problems:

1. You lose reactants that could otherwise continue reacting.
2. You may lose volatile products as they form.

Reflux solves this problem by returning condensed vapour to the reaction flask.

How the reflux apparatus works

The apparatus has three main parts:

• A round-bottomed flask containing the reaction mixture.
• A vertical water-cooled condenser fitted directly above the flask.
• A heat source beneath the flask, such as an electric heating mantle or suitable water bath.

The condenser has an inner tube surrounded by an outer water jacket.

Cooling water enters through the lower side arm and leaves through the upper side arm. This keeps the water jacket filled, providing effective cooling.

Here’s what happens during reflux:

Step 1: Heat is transferred into the reaction mixture. Its temperature rises until it begins boiling.

Step 2: Volatile components evaporate and travel upwards through the condenser.

Step 3: Cold water surrounding the inner tube removes heat from the vapour.

Step 4: The vapour condenses back into liquid.

Step 5: Gravity pulls the condensed liquid back into the original flask.

The cycle continues while heating is maintained.

The important result is that volatile reactants and products are returned rather than being collected elsewhere. Non-volatile substances generally remain in the flask as well.

This allows the reaction to be heated for an extended period without substantial loss of its volatile components.

A reflux condenser is normally left open at the top. Never seal a heated reflux apparatus: pressure can build up inside and cause the glassware to rupture. Organic liquids can also be flammable, so an appropriate electric heating source and laboratory safety procedures are essential.

Why does heating under reflux help a reaction?

Heating increases the rate of many chemical reactions.

According to collision theory, increasing temperature gives particles greater average kinetic energy. More particles can overcome the activation energy barrier, so successful collisions occur more frequently.

However, heating also increases evaporation.

Reflux allows you to benefit from faster reactions without continuously losing volatile substances.

There is an important distinction here: reflux does not automatically shift a chemical equilibrium towards products. It allows the mixture to be heated while retaining its components.

For a reversible reaction, the equilibrium position depends on the reaction conditions and composition. A catalyst increases the rates of the forward and reverse reactions but does not change the equilibrium constant.

03Distillation: separate and collect a liquid

Now consider a different problem.

You have a mixture containing water and dissolved copper sulfate. You want to collect the water separately, leaving the copper sulfate behind.

Would reflux help?

No. Reflux would simply return the condensed water to the original flask. You need the condensed liquid to travel somewhere else.

That’s the purpose of distillation.

Distillation separates substances using differences in volatility, which are closely related to their boiling points.

How the distillation apparatus works

A simple distillation setup contains:

• A heated round-bottomed flask containing the mixture.
• A still head connecting the flask to the condenser.
• A thermometer positioned in the vapour pathway.
• A sloping water-cooled condenser.
• A separate receiving flask to collect the distillate.

The thermometer bulb should sit level with the side arm leading towards the condenser. This allows it to measure the temperature of the vapour entering the condenser.

The condenser slopes downwards towards the receiving flask.

Just as in reflux, cooling water enters at the lower end of the water jacket and leaves at the upper end.

The process works as follows.

Step 1: Heat the mixture in the flask.

Step 2: The more volatile component preferentially enters the vapour phase.

Step 3: The vapour travels through the still head and into the sloping condenser.

Step 4: Cooling water removes heat, causing the vapour to condense.

Step 5: The condensed liquid flows into the receiving flask.

The collected liquid is called the distillate.

The less volatile components tend to remain in the original flask, becoming more concentrated as distillation continues.

Why does boiling point matter?

At a given external pressure, a substance with a lower boiling point is generally more volatile than one with a higher boiling point.

This means it can enter the vapour phase more readily.

Consider this simplified comparison:

There is one important limitation.

It is tempting to think that only the lowest-boiling substance evaporates. That’s not quite right.

In a mixture of liquids, both substances can contribute to the vapour. The vapour is generally enriched in the more volatile component, but it may still contain some of the less volatile component.

This is why simple distillation does not always produce a pure substance.

Simple distillation versus fractional distillation

If two liquids have sufficiently different boiling points, simple distillation may provide a useful separation.

However, what if their boiling points are close?

A simple condenser may not provide enough separation.

Fractional distillation improves the separation by adding a fractionating column between the flask and condenser.

The column allows repeated vaporisation and condensation. This enriches the rising vapour in the more volatile component before it reaches the condenser.

The distinction is useful:

• Simple distillation: suitable when separating a volatile liquid from a non-volatile solute or when liquid boiling points are sufficiently different.
• Fractional distillation: preferred when separating liquids with relatively close boiling points.
• Reflux: not a separation technique; it keeps a heated reaction mixture together.

For a separate explanation of refinery-scale fractional distillation and why it differs from chemical cracking, see Fractional Distillation vs Cracking: From Crude Oil to PVC.

04Reflux vs distillation: the comparison

Here’s the comparison to remember when selecting apparatus in an HSC question.

A quick decision process

When you encounter an unfamiliar experiment, work through these questions.

1. Is the main aim to carry out a reaction?

If yes, consider reflux.

2. Does the reaction mixture contain volatile substances that need to remain in the flask?

If yes, reflux is usually appropriate.

3. Is the aim to collect a liquid separately from the original mixture?

If yes, consider distillation.

4. Are you separating two liquids with close boiling points?

Consider fractional distillation rather than simple distillation.

5. Is a volatile intermediate being produced that could react further?

Distillation may be necessary to remove it before it undergoes another reaction.

One final exam tip: don’t just write that reflux is used because it prevents evaporation. Explain that vapour is cooled, condensed, and returned to the reaction flask. Similarly, for distillation, explain that vapour is condensed and collected in a separate receiver.

The destination of the condensate is the physical difference your explanation needs.

05Questions and solutions

Question 1

A student has 100.0 g of an aqueous copper(II) sulfate solution containing 12.0 g of dissolved copper(II) sulfate.

The student wants to recover the water as a separate liquid. Simple distillation is performed, producing 79.0 g of distillate.

Assuming the distillate is pure water:

a) Identify the appropriate apparatus.

b) Calculate the percentage recovery of the original water.

c) Explain where the copper(II) sulfate ends up.

Solution 1

The correct apparatus is simple distillation, and the percentage recovery of water is 89.8%. Copper(II) sulfate remains in the original flask because it is non-volatile under these conditions.

Step 1: Calculate the original mass of water.

The total solution contains 100.0 g, of which 12.0 g is copper(II) sulfate.

\[ m_{\text{water}}=100.0-12.0 \]

\[ m_{\text{water}}=88.0\text{ g} \]

Step 2: Identify the separation method.

Water is volatile and can evaporate when heated.

Copper(II) sulfate is non-volatile under the conditions used, so it remains in the flask.

The water vapour enters the condenser, becomes liquid, and collects in the receiver.

Step 3: Calculate percentage recovery.

Use:

\[ \%\text{ recovery}= \frac{\text{mass collected}} {\text{initial mass}}\times100 \]

Substitute:

\[ \%\text{ recovery}= \frac{79.0}{88.0}\times100 \]

\[ \boxed{89.8\%} \]

Step 4: Explain the final separation.

The receiving flask contains 79.0 g of water.

The original flask retains the copper(II) sulfate and approximately 9.0 g of water.

The total mass balance is:

\[ 100.0=79.0+12.0+9.0 \]

This accounts for the original solution.

The student has recovered most, but not all, of the available water.

Question 2

A student prepares an ester by heating ethanol and ethanoic acid under reflux with an acid catalyst.

After 30 minutes, the reaction has reached equilibrium.

A second student suggests running the same experiment for 60 minutes instead, arguing that twice the heating time must produce twice as much ester.

Evaluate this claim. In your answer, explain the separate roles of heating, reflux, and the acid catalyst.

Solution 2

The claim is incorrect: doubling the heating time does not double the equilibrium yield. Once equilibrium has been reached under unchanged conditions, extending the heating period does not change the equilibrium composition.

Step 1: Consider the effect of heating.

Heating increases the rate of the esterification reaction.

More particles have sufficient energy to overcome the activation energy barrier, so the reaction approaches equilibrium more quickly.

However, heating does not mean the reaction will continue producing ester indefinitely.

Step 2: Consider the role of reflux.

The reaction mixture contains volatile substances, including ethanol and ester.

Reflux allows these substances to evaporate, condense, and return to the flask.

This prevents substantial losses while the reaction is heated.

Step 3: Consider the role of the acid catalyst.

The acid catalyst provides an alternative reaction pathway with a lower activation energy.

It increases the rates of both the forward and reverse reactions.

It does not change the equilibrium constant or independently shift the equilibrium position.

Step 4: Evaluate the additional heating time.

If equilibrium has already been reached after 30 minutes, the forward and reverse reaction rates are equal.

Heating for another 30 minutes under the same conditions does not double the amount of ester.

The reaction continues dynamically, but the overall concentrations remain approximately constant.

If equilibrium had not yet been reached, additional heating could allow the reaction to move closer to equilibrium.

Conclusion: Heating increases reaction rate, reflux prevents volatile material from being lost, and the catalyst speeds up the reaction. None of these effects means that twice the heating time produces twice the equilibrium yield.

Question 3

A student wants to produce propanal by oxidising propan-1-ol.

Instead, the student heats propan-1-ol with excess acidified dichromate under reflux for 20 minutes.

After cooling, the mixture is distilled at atmospheric pressure.

The first collected fraction appears at approximately 97°C. No fraction is collected near 49°C.

The approximate boiling points are:

a) Explain why no propanal is collected.

b) Identify the most likely substance in the first fraction.

c) Does the absence of a fraction at 49°C prove that propanal was never produced? Explain.

Solution 3

The most likely explanation is that propanal was formed but then oxidised further to propanoic acid during reflux. The first fraction near 97°C is consistent with unreacted propan-1-ol.

Step 1: Consider the oxidation pathway.

Propan-1-ol is oxidised in two stages:

\[ \ce{CH3CH2CH2OH -> CH3CH2CHO} \]

\[ \ce{CH3CH2CHO -> CH3CH2COOH} \]

The first reaction forms propanal, while the second forms propanoic acid.

Step 2: Consider what reflux does to propanal.

Propanal is volatile, with a boiling point of approximately 49°C.

However, the vertical condenser returns its vapour to the reaction flask.

It therefore remains in contact with the excess acidified dichromate and can undergo further oxidation.

Step 3: Explain the missing fraction.

By the time distillation begins, much or all of the propanal may have been converted into propanoic acid.

Consequently, there may be no propanal left to produce a fraction near 49°C.

Step 4: Identify the first fraction.

Propan-1-ol has a boiling point of approximately 97°C.

The first collected fraction near this temperature is therefore consistent with unreacted propan-1-ol.

Propanoic acid is less volatile and is more likely to remain in the original flask.

Step 5: Evaluate what the observation proves.

The absence of propanal in the final distillate does not establish that propanal was never formed.

It could have existed as an intermediate during the reaction and subsequently been consumed.

This is an important distinction between observing the final products and reconstructing the reaction pathway.

Conclusion: Reflux was unsuitable for collecting propanal because it returned the aldehyde to the oxidising mixture. Distillation should have been used from the beginning to remove propanal as it formed.

Question 4

A student is given a mixture of two liquids, X and Y.

Their normal boiling points at atmospheric pressure are:

The student uses simple distillation. During the collection of the first 15 mL, the thermometer remains close to 65.5°C.

Analysis of the collected fraction shows that it contains both X and Y.

The student argues that the nearly constant thermometer reading proves that pure X has been collected.

Evaluate the student’s reasoning and recommend a more suitable separation method.

Solution 4

The student’s conclusion is incorrect. Fractional distillation is more suitable because X and Y have close boiling points, and the collected fraction demonstrably contains both liquids.

Step 1: Compare the boiling points.

The difference is:

\[ \Delta T=67.0-64.0 \]

\[ \Delta T=3.0^\circ\mathrm{C} \]

The boiling points are very close.

This means both liquids can contribute significantly to the vapour during simple distillation.

Step 2: Interpret the thermometer reading.

The thermometer measures the temperature of the vapour entering the condenser.

It does not directly measure the purity of the collected liquid.

A relatively stable temperature can occur while a mixture of substances is being collected.

The fact that the fraction contains both X and Y confirms that the separation has not produced pure X.

Step 3: Explain why simple distillation is insufficient.

The two substances have similar volatilities.

As a result, the vapour leaving the flask is not sufficiently enriched in X to achieve a good separation in a single vaporisation-condensation stage.

Step 4: Select fractional distillation.

A fractionating column provides repeated opportunities for vapour to condense and liquid to vaporise again.

The rising vapour becomes increasingly enriched in the more volatile component, X.

This improves the separation compared with simple distillation.

Step 5: State the limitation.

Fractional distillation improves separation, but it does not guarantee complete purity in every mixture.

The actual separation depends on the substances’ vapour-liquid behaviour.

Conclusion: Fractional distillation is the better choice because the boiling points are close. A stable thermometer reading alone does not establish that a collected fraction is pure.

Question 5

A student is investigating the reversible esterification reaction:

\[ \ce{CH3COOH + CH3CH2OH <=> CH3COOCH2CH3 + H2O} \]

The student first heats the mixture under reflux until equilibrium is reached.

At equilibrium, the reaction flask contains:

For this idealised model, the equilibrium constant is \(K_c=0.250\).

The student then switches to distillation, arguing:

“If I remove ethyl ethanoate from the mixture, equilibrium must shift to the right, producing more ester.”

During collection, the receiver accumulates:

• 0.20 mol of ethyl ethanoate
• 0.50 mol of ethanol

Assume:

• No ethanoic acid or water enters the receiver.
• The reaction is effectively paused during collection.
• The liquid mixture is homogeneous and behaves ideally.
• Concentrations can be calculated using the remaining amounts divided by their common solution volume.

At atmospheric pressure, the approximate normal boiling points of the pure liquids are 77.1°C for ethyl ethanoate and 78.4°C for ethanol.

a) Calculate the reaction quotient, \(Q_c\), immediately after distillation.

b) Predict the direction in which the reaction will shift when equilibrium is re-established.

c) Explain why the student’s argument is incomplete, even though ethyl ethanoate has been removed.

d) Recommend which apparatus should be used during the reaction if the aim is to retain the reaction mixture while it approaches equilibrium.

Solution 5

The reaction shifts to the left, towards ethanoic acid and ethanol. The reaction quotient increases to approximately 0.267, which is greater than the equilibrium constant of 0.250.

This is the important complication: removing a product tends to favour the forward reaction, but removing a reactant tends to favour the reverse reaction. Both substances have been removed, so we need to consider their combined effect.

Step 1: Write the equilibrium expression.

For the reaction:

\[ \ce{A + B <=> C + D} \]

the equilibrium expression is:

\[ K_c=\frac{[C][D]}{[A][B]} \]

For esterification:

\[ K_c= \frac{ [\text{ester}][\text{water}] }{ [\text{acid}][\text{ethanol}] } \]

Step 2: Calculate the amounts remaining after distillation.

Ethyl ethanoate:

\[ n=1.00-0.20=0.80\text{ mol} \]

Ethanol:

\[ n=2.00-0.50=1.50\text{ mol} \]

Ethanoic acid remains at 2.00 mol, and water remains at 1.00 mol.

The new mixture is:

Step 3: Calculate the new reaction quotient.

Because there are two dissolved species on each side of the equation, the common solution volume cancels.

Therefore, we can use the mole amounts directly in this idealised calculation:

\[ Q_c= \frac{n_{\text{ester}}n_{\text{water}}} {n_{\text{acid}}n_{\text{ethanol}}} \]

Substitute:

\[ Q_c= \frac{(0.80)(1.00)} {(2.00)(1.50)} \]

\[ Q_c=\frac{0.80}{3.00} \]

\[ \boxed{Q_c=0.267} \]

Step 4: Compare \(Q_c\) with \(K_c\).

The equilibrium constant is:

\[ K_c=0.250 \]

The new reaction quotient is:

\[ Q_c=0.267 \]

Therefore:

\[ Q_c>K_c \]

The reaction quotient is too large compared with its equilibrium value.

To decrease the quotient, the system shifts towards the reactants.

The reverse reaction consumes ethyl ethanoate and water while producing ethanoic acid and ethanol.

Step 5: Explain why the student’s argument fails.

The student correctly identifies one effect of distillation: removing ethyl ethanoate reduces the amount of product and tends to drive the forward reaction.

However, the student ignores the simultaneous removal of ethanol.

Consider the percentage of each substance removed:

Ethyl ethanoate:

\[ \frac{0.20}{1.00}\times100=20\% \]

Ethanol:

\[ \frac{0.50}{2.00}\times100=25\% \]

A larger proportion of ethanol has been removed than ethyl ethanoate.

This reduces the denominator of the equilibrium expression proportionally more than the numerator, causing \(Q_c\) to increase.

The reverse shift is therefore the net result.

Step 6: Recommend the correct apparatus.

If the aim is to heat the esterification mixture while retaining its volatile components, reflux is the appropriate choice.

Distillation is not automatically a good way to improve ester yield. It may remove reactants along with products, depending on their volatility.

A more effective equilibrium strategy would involve selectively removing a product, such as water, or using an appropriate excess of a reactant.

Conclusion: Removing a product does not automatically guarantee a forward equilibrium shift when reactants are removed at the same time. You must consider the complete equilibrium expression.

This is why understanding what happens to each component is more useful than memorising that distillation collects the lowest-boiling substance.

06Extension: when boiling points aren’t enough

There is one further limitation worth knowing.

Some liquid mixtures form an azeotrope. At a particular composition, the vapour and liquid have the same composition, so ordinary fractional distillation cannot separate them beyond that point.

This is relevant to ethyl ethanoate and ethanol, which form a minimum-boiling azeotrope at atmospheric pressure.

The pure substances have boiling points of approximately 77.1°C and 78.4°C, respectively, but their mixture can behave differently from what you would predict using those two values alone.

This is extension material rather than a requirement for the basic HSC apparatus comparison. It demonstrates why boiling points are useful guides, not guarantees of purity.

For further detail, see the 2025 research published in the AIChE Journal on ethyl ethanoate production and separation.

07What to remember for the HSC

When an exam question asks you to select or compare reflux and distillation, organise your answer around the purpose of the experiment.

• Reflux: Heat a reaction mixture while returning volatile substances to the original flask.
• Distillation: Vaporise, condense, and collect a volatile component in a separate receiver.
• Fractional distillation: Improve the separation of liquids with relatively close boiling points using repeated vaporisation and condensation.

For reflux, explain why the reaction needs heating and how the condenser prevents substantial loss of volatile components.

For distillation, explain how differences in volatility allow one component to become enriched in the vapour and collected separately.

One final syllabus note: as at November 2026, NSW schools are still teaching the Chemistry Stage 6 Syllabus (2017). The Chemistry 11-12 Syllabus (2025) is scheduled for implementation from 2028, with the first HSC examination under the new syllabus in 2029. This article uses the 2017 course context; students in later cohorts should check the content applying to their year group. See NESA’s Chemistry syllabus information.

Once you’ve selected the correct apparatus, the next decision is whether simple distillation will give a sufficiently pure product or whether fractional distillation, or another purification method, is needed. That’s where boiling-point differences become especially important.