Why Simple Distillation Struggles with Similar Boiling Points

Imagine a flask containing two miscible liquids. Pure liquid A boils at 78°C, while pure liquid B boils at 82°C.

You heat the mixture in a simple distillation apparatus. You might expect A to evaporate first at 78°C, leaving B behind until the temperature reaches 82°C.

That prediction sounds reasonable, but it is wrong.

Both liquids evaporate at the same time. Because their boiling points are close, both contribute significantly to the vapour. When that vapour passes through the condenser, both liquids condense together, producing a mixture rather than pure A.

This is why simple distillation struggles with liquids of similar boiling points: one evaporation-condensation step does not produce enough separation between their components.

Before looking at the explanation, consider one prediction. If the original liquid contains 50% A and 50% B, will the first vapour be 100% A?

No. It will contain both liquids, although it will usually contain a greater proportion of the more volatile component, A.

The difference between those two compositions is the key to understanding distillation.

01The first design: what actually happens during simple distillation?

To understand the problem, follow the mixture through the apparatus.

A simple distillation setup has four important parts:

• A heated flask containing the original liquid mixture.
• A thermometer positioned near the outlet of the flask to measure the temperature of the escaping vapour.
• A condenser that cools the vapour.
• A receiving flask that collects the condensed liquid, called the distillate.

As the mixture heats up, particles gain kinetic energy. Some particles escape from the liquid surface and enter the gas phase.

Eventually, the total vapour pressure reaches the external pressure, and the mixture boils.

Here is where the separation problem begins.

Why both liquids evaporate

A common misconception is that a liquid with a lower boiling point evaporates while the higher-boiling liquid stays behind.

In reality, a higher boiling point does not mean a substance cannot evaporate at lower temperatures.

Both liquids have a vapour pressure at the mixture’s boiling temperature. The difference is how much each contributes to the vapour.

A liquid with a lower boiling point generally has a higher vapour pressure at the same temperature. It is therefore more volatile, meaning it has a greater tendency to enter the gas phase.

For our mixture:

• A has the lower boiling point, so it is more volatile.
• B has the higher boiling point, so it is less volatile.
• Both enter the vapour, but A is preferentially enriched.

The important point is that preferential evaporation is not exclusive evaporation.

A more precise model: partial vapour pressures

We can describe this behaviour mathematically using Raoult’s law for an ideal liquid mixture.

For two components, A and B:

\[ p_A=x_A P_A^\ast \]

\[ p_B=x_B P_B^\ast \]

The symbols mean:

• \(p_A\) and \(p_B\): partial vapour pressures of A and B, measured in kPa.
• \(x_A\) and \(x_B\): mole fractions of A and B in the liquid.
• \(P_A^\ast\) and \(P_B^\ast\): vapour pressures of the pure liquids at the mixture’s temperature.

The total vapour pressure is:

\[ P_{\mathrm{total}}=p_A+p_B \]

The mixture boils when its total vapour pressure equals the external pressure.

We can also calculate the proportion of A in the vapour:

\[ y_A=\frac{p_A}{P_{\mathrm{total}}} \]

Here, \(y_A\) is the mole fraction of A in the vapour.

Notice the distinction:

• \(x_A\) describes how much A is in the original liquid.
• \(y_A\) describes how much A is in the vapour.

If A is more volatile than B, we generally expect:

\[ y_A>x_A \]

This means the vapour is enriched in A compared with the original liquid.

However, it does not mean the vapour is pure A.

The equations above use an ideal-mixture model. Real mixtures can deviate from Raoult’s law, so actual separation depends on their measured vapour-liquid behaviour.

02What happens to each component in the apparatus?

It helps to distinguish what happens in the flask from what happens in the condenser.

There are two important observations here.

1. The condenser does not separate the liquids

A condenser has one main job: remove heat from the vapour so that it returns to the liquid state.

It does not automatically separate the different substances in that vapour.

If both A and B enter the condenser, both can condense and flow into the receiving flask.

Even replacing a short condenser with a much longer one does not solve the fundamental problem. It may improve condensation and recovery, but it does not provide the repeated separation needed to obtain purer fractions.

2. The thermometer does not directly measure purity

The thermometer bulb should sit in the vapour path near the side arm of the distillation flask.

It measures the temperature of the vapour leaving the apparatus, not the temperature of the liquid at the bottom of the flask.

As distillation continues, A is gradually removed from the mixture. The remaining liquid becomes richer in B.

Consequently, the boiling temperature generally increases towards the boiling point of the higher-boiling component, assuming constant external pressure and ordinary mixture behaviour.

A steady thermometer reading does not, by itself, prove that the collected liquid is pure.

To establish purity, the composition of the collected fraction must be assessed using an appropriate analytical method.

03Why similar boiling points make separation difficult

The difficulty comes down to the difference between the vapour pressures of the two components.

If two liquids have similar boiling points, their pure-component vapour pressures are generally relatively close around the mixture’s boiling temperature.

As a result, both contribute substantially to the vapour.

Consider the following comparison.

The salt-water example is particularly important because it exposes a common misconception.

Contrasting example: Why simple distillation works for salt water

Imagine heating a solution of sodium chloride dissolved in water.

Water has a significant vapour pressure at its boiling temperature.

Dissolved sodium chloride, however, has negligible volatility under ordinary distillation conditions. It does not travel into the condenser as a vapour.

Therefore:

• Water evaporates and travels into the condenser.
• Water vapour cools and condenses into liquid water.
• Sodium chloride remains in the original flask.
• The remaining solution becomes increasingly concentrated in salt.

Simple distillation works well because the two components have dramatically different volatility.

It is not necessary to repeatedly separate two volatile vapours because the salt does not meaningfully enter the vapour in the first place.

You can explore this contrasting application in How to Recover Both Salt and Water by Simple Distillation.

Is there a specific boiling point difference that determines success?

Not a universal one.

You might hear that simple distillation works when boiling points differ by a particular number of degrees. These rules can be useful rough guides, but they are not absolute.

The effectiveness of separation depends on:

• The relative volatility of the components.
• The composition of the original mixture.
• The external pressure.
• The behaviour of the real liquid mixture.
• The purity required in the final distillate.

A difference of only a few degrees often makes simple distillation ineffective for obtaining high-purity components, but the actual vapour-liquid equilibrium is more informative than the boiling point difference alone.

04The solution: fractional distillation

If simple distillation provides only one effective evaporation-condensation step, how can we improve the separation?

We introduce a fractionating column between the heated flask and the condenser.

This changes the process significantly.

How the fractionating column improves separation

The column contains packing material, such as glass beads, which provides a large surface area for vapour-liquid contact.

As hot vapour rises through the column, it meets cooler surfaces and some of the vapour condenses.

This liquid then flows back down the column.

At the same time, rising hot vapour transfers energy to the descending liquid. Some of that liquid evaporates again.

Because A is more volatile, it preferentially enters the vapour during this process.

The result is a repeated pattern:

1. Vapour rises through the column.
2. Some vapour condenses on the cooler surfaces.
3. Liquid flows downwards.
4. More volatile A preferentially re-enters the vapour.
5. The vapour rising further up the column becomes increasingly enriched in A.

This repeated evaporation and condensation is what makes fractional distillation more effective.

The condensed liquid flowing back down the column is called reflux.

Reflux is important because it allows the rising vapour to interact repeatedly with liquid that is already becoming enriched in the less volatile component.

Why the column works better than a longer condenser

The distinction is the number of effective vapour-liquid equilibrium contacts.

A simple distillation apparatus allows the vapour to travel from the flask directly into the condenser.

A fractional distillation apparatus gives the vapour repeated opportunities to exchange components with descending liquid before reaching the condenser.

The fractionating column is responsible for improving the separation. The condenser is still needed, but its job remains condensation.

Heating rate also matters. If the mixture is heated too quickly, vapour can rush through the column without sufficient contact with the descending liquid. This reduces the effectiveness of the separation.

Controlled heating allows the column to maintain its temperature gradient and reflux.

05An important limitation: azeotropes

There is one further complication worth knowing as an extension beyond the core HSC treatment.

Some liquid mixtures form an azeotrope.

An azeotrope is a mixture that boils at a particular composition where the equilibrium vapour has the same composition as the liquid.

In other words:

\[ y_A=x_A \]

At this composition, evaporation does not preferentially enrich either component.

Even if you introduce a fractionating column, repeated equilibrium contacts cannot separate the mixture beyond this point using ordinary distillation at that pressure.

This is a useful reminder that pure-component boiling points alone do not determine how effectively a real mixture can be separated. The actual vapour-liquid equilibrium matters.

06Questions and solutions

Try these questions to check whether you can apply the reasoning to unfamiliar mixtures and experimental arrangements.

Question 1

A student heats an aqueous sodium chloride solution using simple distillation.

Assuming no liquid droplets are mechanically carried into the condenser, identify:

• The substance collected in the receiving flask.
• The substance remaining in the heated flask.

Explain why simple distillation is effective for this separation.

Solution 1

The receiving flask collects water, while sodium chloride remains in the original flask.

Water is volatile and has a significant vapour pressure at its boiling temperature. It evaporates, travels through the condenser, and returns to the liquid state.

Sodium chloride is essentially non-volatile under these conditions. It does not enter the vapour in a meaningful amount.

As water is removed, the remaining solution becomes more concentrated in sodium chloride.

This is different from separating two volatile liquids with similar boiling points. In salt water, there is no significant competition between two components entering the vapour.

Question 2

An ideal liquid mixture contains 40% A and 60% B by mole.

At its boiling temperature, the pure-component vapour pressures are:

\[ P_A^\ast=125\ \mathrm{kPa} \]

\[ P_B^\ast=85\ \mathrm{kPa} \]

The external pressure is \(101\ \mathrm{kPa}\).

Calculate:

(a) The partial vapour pressure of each component.

(b) The mole percentage of A in the first equilibrium vapour.

(c) Whether the first distillate is pure A.

Solution 2

The first equilibrium vapour contains approximately 49.5% A and 50.5% B. It is enriched in A, but it is not pure.

(a) Calculate the partial pressures.

The liquid mole fractions are:

\[ x_A=0.40,\qquad x_B=0.60 \]

Using Raoult’s law:

\[ p_A=x_A P_A^\ast \]

\[ p_A=0.40\times125 \]

\[ p_A=50.0\ \mathrm{kPa} \]

For B:

\[ p_B=x_B P_B^\ast \]

\[ p_B=0.60\times85 \]

\[ p_B=51.0\ \mathrm{kPa} \]

(b) Calculate the vapour composition.

First, calculate the total pressure:

\[ P_{\mathrm{total}}=50.0+51.0 \]

\[ P_{\mathrm{total}}=101.0\ \mathrm{kPa} \]

Now calculate the mole fraction of A:

\[ y_A=\frac{p_A}{P_{\mathrm{total}}} \]

\[ y_A=\frac{50.0}{101.0} \]

\[ y_A=0.4950 \]

Therefore:

\[ y_A=49.5\% \]

The vapour contains approximately 50.5% B.

(c) Assess the purity of the distillate.

The original liquid contained 40% A, whereas the vapour contains 49.5% A.

A has been enriched, confirming that it is the more volatile component.

However, B still makes up slightly more than half of the vapour.

When both components condense, they enter the receiving flask together. Therefore, the first distillate is still a mixture.

Notice something interesting: A is more volatile, but B remains the majority component in the vapour. Greater volatility does not automatically mean greater abundance.

Question 3

A student performs batch simple distillation on an ideal mixture of A and B.

A has the lower boiling point.

The following illustrative equilibrium data are obtained from a model at constant external pressure.

Calculate the mole fraction of A in the equilibrium vapour at both stages.

Explain how the composition of the distillate and the thermometer reading are expected to change as distillation continues.

Solution 3

The initial vapour contains approximately 54.5% A, while the later vapour contains approximately 25.7% A.

Although A remains preferentially enriched in the vapour, its proportion in the collected distillate decreases as the original liquid becomes depleted in A.

Step 1: Calculate the initial vapour composition.

At the beginning:

\[ x_A=0.50,\qquad x_B=0.50 \]

The partial pressures are:

\[ p_A=0.50\times110=55\ \mathrm{kPa} \]

\[ p_B=0.50\times92=46\ \mathrm{kPa} \]

The total pressure is:

\[ P_{\mathrm{total}}=55+46 =101\ \mathrm{kPa} \]

Therefore:

\[ y_A=\frac{55}{101} \]

\[ y_A=0.545 \]

The initial vapour contains 54.5% A.

Step 2: Calculate the later vapour composition.

Later in the experiment:

\[ x_A=0.20,\qquad x_B=0.80 \]

The partial pressures are:

\[ p_A=0.20\times130 =26.0\ \mathrm{kPa} \]

\[ p_B=0.80\times94 =75.2\ \mathrm{kPa} \]

The total pressure is:

\[ P_{\mathrm{total}}=26.0+75.2 \]

\[ P_{\mathrm{total}}=101.2\ \mathrm{kPa} \]

Therefore:

\[ y_A=\frac{26.0}{101.2} \]

\[ y_A=0.257 \]

The later vapour contains approximately 25.7% A.

Step 3: Explain the changing distillate composition.

Initially, the flask contains equal amounts of A and B, so the vapour is relatively rich in A.

As distillation continues, A is preferentially removed. The remaining liquid becomes richer in B.

Consequently, the vapour leaving the flask also becomes richer in B.

The early distillate is therefore more enriched in A than the later distillate.

Step 4: Predict the thermometer reading.

At constant external pressure, the boiling temperature generally increases as the remaining liquid becomes richer in the higher-boiling component B.

The thermometer at the vapour outlet should therefore show a gradual increase towards B’s higher boiling temperature.

The important point is that simple distillation does not maintain a constant distillate composition throughout the experiment.

Question 4

Two miscible liquids, A and B, have pure boiling points of 78°C and 82°C, respectively.

A student wants to improve the separation and considers three modifications:

• Option I: Replace the existing condenser with a much longer condenser.
• Option II: Install a packed fractionating column between the heated flask and the condenser, using controlled heating.
• Option III: Install a packed fractionating column but increase the heating rate so that vapour travels through it as quickly as possible.

Which option is most likely to improve the separation?

Explain why the other two options are less effective. In your answer, refer to the role of the condenser, the fractionating column, and reflux.

Solution 4

Option II provides the best separation because it introduces repeated vapour-liquid equilibrium contacts while allowing sufficient time for reflux.

The reason is the difference between condensation and separation.

Why Option I is less effective

A longer condenser provides more surface area for heat transfer.

This can improve the efficiency of condensation, helping ensure that vapour becomes liquid before reaching the receiving flask.

However, both A and B still condense into the same receiver.

The condenser does not provide the repeated vapour-liquid contacts needed to enrich the more volatile component significantly.

Increasing condenser length therefore improves product recovery rather than substantially improving separation.

Why Option II is more effective

The fractionating column provides a large surface area for condensation and repeated evaporation.

As vapour rises, some of it condenses and flows back down the column as reflux.

The less volatile component B becomes relatively enriched in the descending liquid, while the more volatile component A becomes increasingly enriched in the rising vapour.

Controlled heating allows these repeated interactions to occur effectively.

The vapour reaching the condenser is therefore richer in A than the vapour leaving a simple distillation flask.

Why Option III is less effective than Option II

Heating too quickly can force vapour through the column before sufficient vapour-liquid interaction occurs.

This reduces the effectiveness of the equilibrium contacts and can disrupt reflux. Excessive vapour flow can also cause liquid to be carried upwards or flood the column.

A fractionating column only improves separation when it operates under suitable conditions.

The correct approach is to provide sufficient heating to maintain boiling while allowing effective contact between rising vapour and descending liquid.

Question 5

Extension beyond the core HSC treatment:

A mixture contains two miscible liquids, C and D.

Their pure boiling points are:

• C: 76°C
• D: 84°C

A flask initially contains 100.0 mol of the mixture:

• 60.0 mol C
• 40.0 mol D

At the mixture’s boiling temperature and the stated pressure, an equilibrium vapour sample is found to contain:

• 60.0 mol% C
• 40.0 mol% D

A student removes 10.0 mol of vapour and condenses it.

The student argues that a sufficiently tall fractionating column must eventually produce pure C because C has a lower pure-component boiling point.

Assume the stated vapour-liquid equilibrium relationship persists at this composition, and there are no side reactions or mechanical losses.

(a) Calculate the number of moles of C and D remaining in the flask.

(b) Calculate the mole fractions of C and D in the remaining liquid.

(c) Determine whether adding more ideal equilibrium stages can separate the mixture from this starting composition. Explain why the difference between the pure boiling points does not settle the question.

Solution 5

The flask contains 54.0 mol C and 36.0 mol D after the removal of vapour. Its composition remains exactly 60.0 mol% C and 40.0 mol% D.

Adding more ideal equilibrium stages cannot enrich C from this starting composition under the stated conditions.

This is the important distinction: a lower pure-component boiling point does not guarantee that a component will be enriched in the vapour of every possible mixture.

(a) Calculate the moles remaining.

The initial mixture contains:

\[ n_C=60.0\ \mathrm{mol} \]

\[ n_D=40.0\ \mathrm{mol} \]

The removed vapour contains 60.0% C and 40.0% D.

For 10.0 mol of vapour:

\[ n_C(\mathrm{removed})=0.60\times10.0 \]

\[ =6.0\ \mathrm{mol} \]

For D:

\[ n_D(\mathrm{removed})=0.40\times10.0 \]

\[ =4.0\ \mathrm{mol} \]

Subtract these amounts from the original mixture:

\[ n_C(\mathrm{remaining})=60.0-6.0 \]

\[ =54.0\ \mathrm{mol} \]

Similarly:

\[ n_D(\mathrm{remaining})=40.0-4.0 \]

\[ =36.0\ \mathrm{mol} \]

(b) Calculate the remaining mole fractions.

The total number of moles remaining is:

\[ n_{\mathrm{total}}=54.0+36.0 \]

\[ n_{\mathrm{total}}=90.0\ \mathrm{mol} \]

The mole fraction of C is:

\[ x_C=\frac{54.0}{90.0} \]

\[ x_C=0.60 \]

The mole fraction of D is:

\[ x_D=\frac{36.0}{90.0} \]

\[ x_D=0.40 \]

The remaining liquid has exactly the same composition as the original mixture.

(c) Explain why fractional distillation cannot improve the separation.

The key information is that the equilibrium vapour and liquid have identical compositions:

\[ y_C=x_C=0.60 \]

Normally, a more volatile component becomes enriched in the vapour. That difference in composition is what allows distillation to separate the substances.

Here, no such enrichment occurs.

When 10.0 mol of vapour leaves the flask, it removes C and D in exactly the same proportions as those present in the liquid.

The remaining liquid therefore retains its original composition.

If the same equilibrium relationship continues to apply, removing more vapour produces the same result. The liquid composition does not move towards pure C or pure D.

This behaviour identifies an azeotropic mixture.

At an azeotropic composition, the liquid and equilibrium vapour have the same composition at a particular pressure. Repeated evaporation and condensation cannot enrich either component beyond that point.

The student’s mistake is assuming that the pure boiling points completely determine the behaviour of the mixture.

In reality, interactions between unlike molecules can change the vapour-liquid equilibrium. These non-ideal interactions can produce an azeotrope even when the pure liquids have different boiling points.

This is why actual mixture behaviour matters more than simply comparing the boiling points printed in a data table.

07Choosing the right separation method

When deciding between simple and fractional distillation, start by identifying the physical properties of the substances being separated.

• If one component is non-volatile, simple distillation may be sufficient.
• If two volatile liquids have substantially different volatilities, simple distillation may provide useful separation.
• If two volatile liquids have similar boiling points, fractional distillation is generally more effective because it provides repeated vapour-liquid contacts.
• If a mixture forms an azeotrope, ordinary fractional distillation may not achieve complete separation.

Also remember that not every liquid mixture requires distillation. If the substances have different solubilities or form separate liquid layers, another separation method may be more appropriate.

For example, separating an ester from an aqueous reaction mixture may involve differences in solubility and the use of a separating funnel. See Simple Ester Production: HSC Chemistry Lab Explained for that contrasting application.

The central decision is simple: match the separation apparatus to the physical properties of the substances, rather than assuming that heating and condensation will automatically produce pure components.