How to Recover Both Salt and Water by Simple Distillation

You have a beaker of salt water, and you need to recover both substances. Heating the solution in an evaporating basin gives you the salt, but the water escapes into the air. Filtering the solution doesn’t work either: the dissolved salt passes through the filter paper with the water.

So how do you collect both?

Use simple distillation. Heat the salt solution so that the water evaporates, pass the water vapour through a cooled condenser, and collect the condensed liquid in a separate container. The salt stays behind in the original flask.

This method works because water is volatile under laboratory conditions, while sodium chloride is effectively non-volatile. The two substances therefore follow different paths through the apparatus.

01Why simple distillation separates salt and water

Start by thinking about what happens when salt dissolves in water.

Solid sodium chloride, \(\ce{NaCl}\), separates into positively charged sodium ions and negatively charged chloride ions:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

These ions become surrounded by water molecules. They are distributed throughout the solution rather than sitting at the bottom as separate grains of salt.

Now imagine heating this solution.

Which particles can escape into the gas phase?

Water molecules can gain enough energy to escape from the liquid and become water vapour. Sodium and chloride ions do not evaporate with them under normal laboratory conditions.

This is the important difference:

• Water is volatile: it can readily enter the gas phase when heated.
• Sodium chloride is non-volatile under these conditions: it remains in the liquid or solid residue.

As the water evaporates, the remaining solution becomes more concentrated in salt.

Eventually, if enough water is removed, the remaining salt forms solid crystals.

There is a useful distinction here. When you pour salt water, the water carries dissolved ions with it because they are part of the liquid solution. That does not mean the ions travel with water vapour.

A filter works a little like a sieve: it separates suspended particles from a liquid. Dissolved ions are far too small to be trapped by ordinary filter paper, so filtration cannot separate dissolved salt from water.

Distillation uses a different physical property: the difference in volatility between the components.

Why not just evaporate the water?

You might predict that boiling the solution in an open container should be enough.

It is enough if you only want the salt. However, the water vapour escapes into the surroundings, so you cannot collect it.

The problem with the original arrangement is not that the separation fails. It is that the apparatus does not capture the substance you also want to recover.

Simple distillation solves this by providing a pathway for the escaping water vapour to become liquid again.

02How to set up the distillation apparatus

A simple distillation apparatus has five important components:

1. A distillation flask containing the salt solution.
2. A heat source to boil the solution.
3. A thermometer to monitor the vapour temperature.
4. A condenser to cool the water vapour.
5. A receiving flask or beaker to collect the distilled water.

Let’s follow the substances through the apparatus in the order they move.

Step 1: Heat the salt solution

Place the salt solution in the distillation flask.

Add a few anti-bumping granules before heating. These help the liquid boil smoothly rather than suddenly erupting in large bubbles.

As the solution heats, its temperature increases until it reaches its boiling point.

At atmospheric pressure, pure water boils at approximately \(100^\circ\mathrm{C}\). A salt solution has a slightly higher boiling point, which increases as the solution becomes more concentrated.

Water molecules escape from the solution as water vapour.

The sodium and chloride ions remain behind.

Step 2: Move the water vapour into the condenser

The water vapour travels from the distillation flask through the connecting tube and into the condenser.

A Liebig condenser contains two separate pathways:

• An inner tube carrying the hot water vapour.
• An outer water jacket carrying cold cooling water.

The cooling water removes thermal energy from the vapour.

Importantly, the cooling water never mixes with the water being distilled. Heat passes through the glass wall separating the two pathways.

Step 3: Condense the water vapour

As the water vapour loses thermal energy, it changes back into liquid water.

This physical change is called condensation.

The liquid water flows out of the condenser and into the receiving flask.

The collected liquid is called the distillate.

One detail about the condenser is particularly important in practical questions.

Cooling water must enter through the lower connection and leave through the upper connection.

Why?

This arrangement keeps the water jacket filled with cooling water and reduces the chance of air pockets forming. It allows efficient heat transfer from the hot vapour to the cooling water.

If you reverse the connections, the condenser may not fill properly, reducing its cooling efficiency.

Step 4: Collect the salt residue

While water travels into the receiving flask, sodium chloride remains in the distillation flask.

As more water is removed, the remaining solution becomes increasingly concentrated.

Eventually, solid sodium chloride can form.

The remaining material is called the residue.

If you need dry salt crystals, allow the apparatus and residue to cool, then evaporate the small amount of remaining water gently in an evaporating basin.

Avoid heating the distillation flask aggressively until completely dry. This can cause bumping, splashing, and loss of solid product.

There is a practical trade-off here: the final traces of water removed during drying are not collected as distillate. You can recover most of the water through distillation, but perfect quantitative recovery is not guaranteed.

03What happens to each component?

It helps to track the salt and water separately rather than thinking of the solution as one substance.

For a solution containing only sodium chloride and water, the expected result is:

• Distillate: predominantly water.
• Residue: concentrated sodium chloride solution, which can be dried to obtain solid salt.

This distinction also explains an important limitation.

Distillation does not automatically produce pure water from every solution. If the original solution contains another volatile substance, that substance may also enter the vapour and contaminate the distillate.

For example, ethanol and water are both volatile. Simple distillation of an ethanol-water mixture does not automatically produce pure water.

The method works particularly well for salt water because sodium chloride is effectively non-volatile at the temperatures used.

How can you check whether the separation worked?

A useful test is to compare the electrical conductivity of the original solution and the collected distillate.

Salt water conducts electricity because it contains mobile ions.

If the distillation works properly, the collected water should have much lower electrical conductivity than the original solution because most of the dissolved ions have remained in the distillation flask.

This supports the conclusion that the salt and water have been separated.

However, low conductivity does not prove absolute purity. Other dissolved substances, contamination, and experimental limitations may still affect the result.

04How does distillation compare with other separation methods?

Students sometimes confuse filtration, evaporation, crystallisation, and distillation because all four can be used when working with mixtures.

The easiest way to distinguish them is to identify what you actually need to collect.

Consider two different questions.

Situation A: You only want the salt.

Evaporation to dryness or crystallisation may be suitable, depending on the desired product and the properties of the solute.

Situation B: You want both the salt and the water.

Simple distillation is the better choice because it captures the water vapour rather than allowing it to escape.

For a closer comparison of the first two methods, see How to Choose Between Evaporation and Crystallisation.

There is also a useful distinction between distillation and reflux. Both use a condenser, but during distillation the condensed liquid travels into a separate receiving container. During reflux, it returns to the original reaction flask. See Reflux vs Distillation: How to Choose the Right Apparatus.

05Worked examples

06Questions and solutions

Question 1

A student is given \(50.0\,\mathrm{g}\) of sodium chloride solution containing \(6.00\%\) salt by mass.

They want to collect both the salt and the water.

a) Identify the most appropriate separation technique.

b) Calculate the initial mass of salt.

c) Calculate the initial mass of water.

Solution 1

Simple distillation is the appropriate method. It allows the student to collect the water as distillate while retaining the non-volatile sodium chloride in the distillation flask.

Step 1: Calculate the mass of salt.

\[ m(\ce{NaCl})= \frac{6.00}{100}\times50.0 \]

\[ =3.00\,\mathrm{g} \]

Step 2: Calculate the mass of water.

\[ m(\ce{H2O})=50.0-3.00 \]

\[ =47.0\,\mathrm{g} \]

The initial solution contains \(3.00\,\mathrm{g}\) of sodium chloride and \(47.0\,\mathrm{g}\) of water.

In an ideal distillation, these would be recovered separately as solid salt and liquid water.

Question 2

A student assembles a simple distillation apparatus to recover water from salt solution.

They connect the cooling-water supply to the upper connection of the Liebig condenser and the drainage tube to the lower connection.

The condenser becomes cool, but the student notices that its cooling performance is inconsistent.

a) Explain why the student’s arrangement is not ideal.

b) State how the cooling-water connections should be arranged.

c) Explain whether the cooling water comes into direct contact with the water vapour.

Solution 2

The cooling-water connections should be reversed: cold water should enter through the lower connection and leave through the upper connection. This helps keep the condenser jacket completely filled, improving heat transfer.

Step 1: Identify the problem.

When cooling water enters from the top and exits from the bottom, the jacket may not fill completely. Air pockets can remain, reducing the effective cooling surface.

The condenser may still cool the vapour, but its efficiency is less reliable.

Step 2: Correct the connections.

The correct arrangement is:

• Lower connection: cooling-water inlet.
• Upper connection: cooling-water outlet.

Water entering from below pushes air upwards and helps keep the jacket full.

Step 3: Explain the separation between the two pathways.

Cooling water flows through the outer jacket, while water vapour travels through the inner glass tube.

They never directly mix.

Thermal energy passes through the glass wall, cooling the vapour until it condenses into liquid water.

Question 3

A student has a mixture containing:

• Insoluble sand.
• Dissolved sodium chloride.
• Water.

The student wants to recover all three components separately.

They suggest placing the entire mixture directly into a distillation flask and collecting the distillate.

a) Explain why direct distillation is not the best first step.

b) Describe a suitable sequence of separation techniques.

c) Explain what happens to each component during the process.

Solution 3

Filter the mixture first, then use simple distillation on the filtrate. Filtration separates the insoluble sand, while distillation separates dissolved salt from water.

Step 1: Explain why direct distillation is unsuitable.

If the entire mixture is heated in a distillation flask, the water will evaporate and condense in the receiver.

However, both sand and sodium chloride will remain in the original flask.

The final solid mixture will contain two substances, so the student will not have recovered the sand and salt separately.

Step 2: Filter the original mixture.

Pour the mixture through filter paper in a filter funnel.

Two fractions form:

• Residue: insoluble sand remains on the filter paper.
• Filtrate: water containing dissolved sodium chloride passes through.

The dissolved ions pass through the filter because they are part of the liquid solution.

Step 3: Wash and dry the sand.

A small amount of distilled water can be used to wash the sand, removing salt solution trapped between the particles.

Collect the washings with the original filtrate.

Dry the sand to recover the solid.

Step 4: Distil the combined filtrate and washings.

Place the liquid in a distillation flask.

Heat it so that water evaporates, travels through the condenser, and is collected in the receiving flask.

The sodium chloride remains in the distillation flask.

Step 5: Recover the salt.

Allow the remaining concentrated solution to cool, then evaporate the remaining water gently to obtain solid sodium chloride.

The final products are:

The important idea is that separation techniques can be combined. Each technique should be chosen according to the physical properties of the component being separated.

Question 4

An \(80.0\,\mathrm{g}\) sodium chloride solution contains \(12.5\%\) salt by mass.

A student distils the solution and obtains:

• \(65.0\,\mathrm{g}\) of water distillate.
• \(9.60\,\mathrm{g}\) of dry sodium chloride.

Assume the distillate contains only water and the dried solid contains only sodium chloride.

a) Calculate the initial mass of salt and water.

b) Calculate the percentage recovery of water.

c) Calculate the percentage recovery of salt.

d) Determine the total mass that was not recovered.

e) Explain why the missing mass does not necessarily indicate a chemical reaction.

Solution 4

The water recovery is \(92.9\%\), the salt recovery is \(96.0\%\), and the total unrecovered mass is \(5.40\,\mathrm{g}\). The missing mass can be explained by incomplete collection and transfer losses.

Step 1: Calculate the initial mass of sodium chloride.

\[ m(\ce{NaCl})= \frac{12.5}{100}\times80.0 \]

\[ =10.0\,\mathrm{g} \]

Step 2: Calculate the initial mass of water.

\[ m(\ce{H2O})=80.0-10.0 \]

\[ =70.0\,\mathrm{g} \]

Step 3: Calculate the water recovery.

\[ \text{Water recovery}= \frac{65.0}{70.0}\times100 \]

\[ =92.9\% \]

Step 4: Calculate the salt recovery.

\[ \text{Salt recovery}= \frac{9.60}{10.0}\times100 \]

\[ =96.0\% \]

Step 5: Calculate the total unrecovered mass.

The combined mass of the collected products is:

\[ 65.0+9.60=74.60\,\mathrm{g} \]

Therefore:

\[ m_{\text{unrecovered}}=80.0-74.60 \]

\[ =5.40\,\mathrm{g} \]

We can also separate the losses:

• Water not recovered: \(70.0-65.0=5.0\,\mathrm{g}\).
• Salt not recovered: \(10.0-9.60=0.40\,\mathrm{g}\).

The total is \(5.40\,\mathrm{g}\), as expected.

Step 6: Explain the missing mass.

The missing mass does not mean that the salt or water must have reacted chemically.

Some water may have remained in the apparatus or escaped without being condensed. Some salt may have remained on the glassware during transfer.

These are physical losses associated with the experimental procedure.

Question 5

A student performs a simple distillation on a solution known to contain only sodium chloride and water.

The original solution contains \(12.0\,\mathrm{g}\) of sodium chloride.

The apparatus is clean, and the thermometer is positioned correctly.

The student collects three fractions under different conditions:

The thermometer remains between \(101\) and \(102^\circ\mathrm{C}\) during fractions B and C.

The student concludes:

“The solution became more concentrated, so the salt must have started evaporating and entering the condenser.”

Evaluate this explanation.

In your answer:

a) Calculate the mass percentage of sodium chloride in fraction B.

b) Compare the two possible explanations for the salt appearing in fraction B: evaporation of sodium chloride or mechanical carryover of liquid droplets.

c) Identify the explanation best supported by the observations.

d) Explain how the experimental procedure could be improved.

e) Explain one limitation of using a clear distillate as evidence that the collected water is pure.

Solution 5

The salt in fraction B most likely entered the condenser through mechanical carryover of tiny liquid droplets, not because sodium chloride evaporated. The strongest evidence is that salt contamination occurs during violent bumping but disappears when smooth boiling resumes, even though the thermometer reading remains similar.

This question requires you to distinguish two different ways material can travel through a distillation apparatus.

Step 1: Calculate the mass percentage of sodium chloride in fraction B.

The entire \(10.0\,\mathrm{g}\) fraction leaves \(0.150\,\mathrm{g}\) of solid after evaporation.

Therefore:

\[ \text{Mass percentage}= \frac{0.150}{10.0}\times100 \]

\[ =1.50\% \]

Fraction B contains \(1.50\%\) sodium chloride by mass.

This confirms that the collected liquid was contaminated with a measurable amount of salt.

Step 2: Consider the student’s explanation.

The student suggests that the increasingly concentrated salt solution caused sodium chloride to evaporate.

This explanation might initially seem reasonable because the solution becomes hotter and more concentrated as water is removed.

However, sodium chloride is effectively non-volatile at the temperatures involved.

At approximately \(101\)-\(102^\circ\mathrm{C}\), water can readily enter the gas phase, but sodium and chloride ions do not evaporate as gaseous particles in any significant amount.

Increasing the concentration of the solution does not suddenly make sodium chloride volatile.

Step 3: Consider mechanical carryover.

There is another way for salt to reach the condenser.

During violent boiling, bubbles can burst at the liquid surface and throw tiny droplets of the original salt solution upwards.

These droplets contain dissolved sodium and chloride ions.

If they enter the connecting tube, they travel into the condenser along with the water vapour.

When the droplets cool, they become part of the collected liquid. The salt remains dissolved in that liquid until the fraction is evaporated.

This process is called mechanical entrainment.

Step 4: Use the experimental evidence to choose between the explanations.

Look closely at the three fractions.

Fraction A contains no detectable solid residue.

Fraction B contains salt, but it is collected during violent bumping.

Fraction C returns to having no detectable solid residue when the heating is reduced and smooth boiling resumes.

The thermometer remains between \(101\) and \(102^\circ\mathrm{C}\) during fractions B and C.

This is important.

If the contamination resulted from sodium chloride becoming volatile at a particular temperature, reducing the heating would not explain why the salt contamination disappears while the measured vapour temperature remains similar.

Instead, the change in boiling behaviour provides a convincing explanation.

Violent bumping causes liquid droplets to enter the condenser. Smooth boiling reduces this mechanical carryover.

The evidence therefore supports mechanical entrainment rather than sodium chloride evaporation.

Step 5: Calculate the proportion of the original salt carried over.

The original solution contained \(12.0\,\mathrm{g}\) of sodium chloride.

Fraction B contains \(0.150\,\mathrm{g}\).

Therefore:

\[ \text{Salt carryover}= \frac{0.150}{12.0}\times100 \]

\[ =1.25\% \]

Approximately \(1.25\%\) of the original salt mass has entered the collected fraction.

This is a measurable loss from the salt remaining in the distillation flask.

Step 6: Improve the experimental procedure.

The student should:

• Add anti-bumping granules before heating.
• Use a lower heating rate to maintain steady boiling.
• Avoid overfilling the distillation flask.
• Avoid allowing the solution to boil violently as it becomes concentrated.

If fraction B has already been collected, it can be redistilled to separate the dissolved salt from the water.

Simply filtering fraction B will not remove the salt because it is dissolved, not suspended.

Step 7: Identify the limitation of the visual observation.

A clear, colourless distillate is not necessarily pure water.

Dissolved sodium chloride is colourless, and other dissolved substances may also be colourless. Some volatile impurities can travel into the condenser without producing any visible change.

Therefore, appearance alone cannot establish purity.

The experiment provides evidence about the separation of sodium chloride and water, but further testing would be needed to establish the purity of the collected water.

This is the deeper lesson: distillation separates substances according to their volatility, but the actual purity of the products also depends on how effectively the apparatus prevents contamination and material loss.

06The key decision to remember

When a question asks how to recover both salt and water from a salt solution, identify the physical property that makes their separation possible.

• Water is volatile, so it evaporates and can be condensed.
• Sodium chloride is non-volatile under the conditions used, so it remains in the distillation flask.
• The condenser cools the water vapour and allows the liquid water to be collected separately.

The correct technique is simple distillation.

The next useful decision is whether your mixture contains one volatile component or multiple volatile components. If both components can evaporate, you may need to consider fractional distillation rather than simple distillation.