How to Choose Between Evaporation and Crystallisation

Two students are given the same clear salt solution. One heats it until all the water disappears. The other concentrates the solution, lets it cool, and filters out the crystals. Both recover a solid, but they won’t necessarily recover the same amount, and the products may have different levels of purity.

Choose evaporation to dryness when you want to recover the non-volatile dissolved solid and it can withstand heating. Choose crystallisation when you want to obtain solid crystals, often with improved purity, without removing all the solvent. If you also need to collect the solvent, use distillation instead.

There is one important detail: crystallisation often involves evaporation. The real difference is whether you remove all the solvent or create conditions in which the dissolved substance forms crystals that can be collected separately.

01The decision: what do you actually want to recover?

Start with the substance you want to collect. Then consider its physical properties and how it behaves when the solution is heated or cooled.

The choice between evaporation and crystallisation usually comes down to three considerations:

• The desired product: Do you need all the dissolved solid, or a crop of crystals?
• Solubility: Does the solute become significantly less soluble when the solution cools?
• Thermal stability: Can the solute withstand the heating required to remove the solvent?

This is a more specific decision than simply identifying the physical property that allows a mixture to be separated. For the broader method-selection process, see How to Choose Separation Techniques in HSC Chemistry.

02What happens inside each apparatus?

The easiest way to distinguish the methods is to follow the solvent and solute separately.

1. Evaporation to dryness

Imagine a student has sodium chloride dissolved in water and wants to recover the salt.

The student pours the solution into a shallow evaporating basin. The basin sits on wire gauze supported by a tripod, with a suitable heat source underneath.

As the solution heats up, water molecules gain enough energy to escape from the liquid surface as water vapour.

The sodium chloride behaves differently. Under ordinary laboratory heating, it is effectively non-volatile, so it remains in the evaporating basin.

As more water evaporates, the solution becomes increasingly concentrated. Eventually, the remaining solution can no longer keep all the sodium chloride dissolved. Solid sodium chloride forms.

Heating continues until the remaining liquid solvent has evaporated.

The final product is a dry solid residue.

Notice what happens to each component:

That last point matters.

Evaporation to dryness does not automatically produce pure salt. If the original solution contained two different dissolved, non-volatile substances, both would remain after the water disappeared.

There is also a limitation involving temperature. Some substances decompose or change their chemical form when heated strongly. Certain hydrated salts, for example, can lose their water of crystallisation.

If a substance is heat-sensitive, heating it until completely dry may damage the product.

2. Crystallisation

Now imagine that the student wants to obtain crystals of a dissolved substance rather than a completely dry residue.

The process is different because the student deliberately stops before all the solvent disappears.

A typical crystallisation procedure follows this sequence:

Step 1: Prepare a concentrated solution.

Place the solution in a beaker or evaporating basin and heat it gently.

Some of the solvent evaporates, increasing the concentration of the dissolved substance.

Step 2: Approach saturation.

Continue removing solvent until the solution is close to saturation.

A saturated solution contains the maximum amount of dissolved solute possible at that temperature.

Step 3: Allow the solution to cool.

Remove the heat source and leave the concentrated solution to cool, preferably slowly and undisturbed.

For many solids, solubility decreases as temperature decreases. The solution can therefore no longer keep as much solute dissolved.

The excess solute begins forming crystals.

Step 4: Separate the crystals.

Once crystals have formed, pour the mixture through a filter funnel containing filter paper.

The crystals remain on the filter paper, while the remaining liquid passes through.

Step 5: Wash and dry the crystals.

Wash the crystals with a small amount of cold solvent to remove liquid adhering to their surfaces. Dry them before weighing or further use.

The fate of each component is different from evaporation to dryness.

The liquid remaining after crystals have formed is called the mother liquor.

It still contains dissolved substances. This is important because crystallisation does not usually recover every particle of the original solute.

Also, crystallisation does not guarantee complete purity. If an impurity has similar solubility behaviour to the desired substance, it may crystallise alongside it.

03Why does temperature affect crystallisation?

To choose crystallisation confidently, you need to understand solubility.

Solubility is the maximum amount of a substance that can dissolve in a given amount of solvent at a particular temperature.

For a solid dissolved in water, solubility is commonly expressed in grams of solute per 100 g of water.

Consider the following illustrative data for a hypothetical substance, solute A.

These are illustrative values, not reference measurements for a particular chemical.

Notice the pattern: as temperature increases, solubility increases.

This means that a hot saturated solution can contain much more dissolved solute than a cold saturated solution.

When the solution cools, the excess solute can no longer remain dissolved and forms crystals.

There is a useful prediction to make here.

If a hot saturated solution cools, will every dissolved particle become a crystal?

No. At the lower temperature, the solvent can still dissolve some of the solute. That amount remains in the mother liquor.

The difference between the original dissolved amount and the amount that can remain dissolved determines the mass of crystals formed.

What if solubility barely changes with temperature?

This is where a common assumption causes problems.

Students sometimes assume that cooling a hot solution will always produce a large quantity of crystals. That is not true.

Consider sodium chloride. Its solubility changes only modestly across ordinary laboratory temperatures.

Cooling a saturated sodium chloride solution therefore produces relatively few crystals compared with a substance whose solubility changes substantially with temperature.

What should you do instead?

You can use controlled evaporation to remove more solvent. As the amount of water decreases, the solution becomes more concentrated until further sodium chloride forms crystals.

This is called evaporative crystallisation.

If your only objective is to recover all the sodium chloride, evaporation to dryness is also suitable.

The important lesson is to examine the solubility data rather than automatically choosing cooling.

Does slower cooling produce more crystals?

Not necessarily.

If two solutions have the same starting composition, reach the same final temperature, and reach equilibrium without losing solvent, the equilibrium amount of crystallised solute is determined by solubility.

Slower cooling can influence crystal growth and the size of the crystals. It does not automatically increase the equilibrium mass recovered.

This is why the amount of solute remaining in the mother liquor matters more than simply how long you leave the solution sitting.

04A fair experimental comparison

Suppose you want to investigate whether evaporation or crystallisation is more suitable for recovering a particular dissolved solid.

A fair comparison needs to account for both the solid collected and the solute left in the liquid.

Start with one homogeneous solution and divide it into two equal, measured portions.

• Place the first portion in an evaporating basin. Heat it until the solvent has been removed and record the mass of dry residue.
• Place the second portion in a suitable vessel. Concentrate it, allow it to cool, filter the crystals, and dry them before recording their mass.

What should you expect?

The first portion should leave a dry residue containing the non-volatile dissolved substances.

The second should produce a crop of crystals and a remaining mother liquor.

A student might compare the two masses and conclude that crystallisation has failed because fewer crystals were collected.

That conclusion is premature.

Some solute is still dissolved in the mother liquor. The student has compared the dry residue with only the first crop of crystals, not with the total amount of solute recovered.

A better investigation includes a mass balance.

If the initial dissolved solute mass is known, the amount remaining in the mother liquor can be estimated using:

\[
m_{\text{remaining}}
=
m_{\text{initial}}-m_{\text{crystals}}
\]

Alternatively, a measured portion of the mother liquor can be evaporated to determine how much dissolved solid remains.

There is another experimental limitation: the appearance of crystals does not prove that they are pure. A suitable purity test is needed to support that conclusion.

The distinction between mass recovered and purity is important. A method can produce a smaller quantity of solid while giving a more useful product.

07A practical decision process for HSC questions

When a question asks you to choose between evaporation and crystallisation, work through these decisions in order.

In your written explanation, don’t just name the technique.

Explain:

• Which substance you want to recover.
• Which physical property makes the separation possible.
• Which apparatus is appropriate.
• What happens to the solvent, desired solute, and impurities.
• Why the alternative method is less suitable.

That final point often separates a basic answer from a complete chemical explanation.

08Questions and solutions

The following questions progress from straightforward method selection to problems involving solubility, impurities, experimental design, and competing constraints.

Question 1

A student has 80.0 mL of aqueous sodium chloride solution. They want to recover dry sodium chloride, not the water.

The sodium chloride is stable under gentle laboratory heating.

Which separation technique should the student use? Identify the appropriate apparatus and explain what happens to the water and sodium chloride.

Solution 1

Evaporation to dryness using an evaporating basin is the appropriate technique.

The objective is to recover the entire amount of dissolved sodium chloride, rather than produce a selected crop of crystals.

Step 1: Place the solution in a shallow evaporating basin.

Step 2: Heat the solution gently using a suitable heat source.

The water evaporates and escapes as water vapour.

Step 3: Continue heating until the solvent has been removed.

The non-volatile sodium chloride remains in the evaporating basin as a solid residue.

There is no need to collect crystals separately because the student wants all the dissolved salt.

Question 2

A saturated aqueous solution of hypothetical salt R is prepared at 70°C.

It contains 60.0 g of dissolved salt R in 150.0 g of water.

At 20°C, the solubility of R is 15.0 g per 100 g of water.

The solution is cooled to 20°C without losing any water.

Calculate:

(a) The mass of R that crystallises.

(b) The mass of R remaining in the mother liquor.

(c) The percentage recovery of R as crystals.

Assume that the solution reaches equilibrium and no solute is lost during filtration.

Solution 2

37.5 g of R crystallises, 22.5 g remains dissolved, and the percentage recovery is 62.5%.

The key is to calculate how much solute can remain dissolved at the final temperature.

Step 1: Calculate the mass of R remaining dissolved at 20°C.

The solubility is 15.0 g per 100 g of water.

There are 150.0 g of water, so:

\[
m_{\text{remaining}}
=
\frac{15.0}{100}\times150.0
\]

\[
=22.5\text{ g}
\]

Step 2: Calculate the mass of crystals formed.

The original solution contained 60.0 g of dissolved R.

\[
m_{\text{crystals}}
=
60.0-22.5
\]

\[
\boxed{m_{\text{crystals}}=37.5\text{ g}}
\]

Step 3: Calculate the percentage recovery.

\[
\text{Recovery}
=
\frac{37.5}{60.0}\times100
\]

\[
\boxed{\text{Recovery}=62.5\%}
\]

Step 4: Interpret the result.

The solution produces 37.5 g of solid crystals, but 22.5 g remains dissolved.

This demonstrates why crystallisation does not necessarily recover all the original solute. The final solubility determines how much remains in the mother liquor.

Question 3

A student is given a hot mixture containing:

• 40.0 g of dissolved substance P.
• 5.0 g of dissolved substance Q, which is a soluble impurity.
• 2.0 g of suspended charcoal.
• 100.0 g of water.

At 75°C, both P and Q are completely dissolved.

At 20°C, their solubilities are:

The student wants to obtain the purest possible crystals of P.

(a) Describe the correct sequence of separation techniques.

(b) Calculate the mass of P crystals formed.

(c) Explain what happens to Q and the charcoal.

Assume that Q does not enter the crystal structure of P, and no water evaporates during the separation.

Solution 3

The correct sequence is hot filtration followed by crystallisation, producing 28.0 g of P crystals.

The student must remove the insoluble charcoal before allowing P to crystallise.

Step 1: Remove the insoluble charcoal.

Filter the hot mixture through filter paper.

The charcoal is trapped on the filter paper because it does not dissolve in the water.

The dissolved P and Q pass through with the water as the filtrate.

Filtering while the solution is hot helps prevent P from crystallising prematurely in the filter funnel.

Step 2: Allow the filtrate to cool.

The filtrate contains:

• 40.0 g of P.
• 5.0 g of Q.
• 100.0 g of water.

As the temperature decreases to 20°C, the solubility of P falls.

The solution can now retain only 12.0 g of P per 100 g of water.

Step 3: Calculate the mass of P remaining dissolved.

\[
m_{\text{remaining P}}
=
\frac{12.0}{100}\times100.0
\]

\[
=12.0\text{ g}
\]

Step 4: Calculate the mass of P crystals.

\[
m_{\text{crystals}}
=
40.0-12.0
\]

\[
\boxed{m_{\text{crystals}}=28.0\text{ g}}
\]

Step 5: Explain the fate of the impurities.

The 5.0 g of Q remains dissolved because its solubility at 20°C is 40.0 g per 100 g of water.

The charcoal was removed during the initial filtration.

The crystals of P can therefore be collected separately from the mother liquor, which contains the remaining 12.0 g of P and all 5.0 g of Q.

Conclusion: Hot filtration removes the insoluble impurity, while crystallisation separates P from the more soluble impurity Q. Using the techniques in the correct order is essential for obtaining a relatively pure product.

Question 4

A student is investigating the crystallisation of hypothetical salt M.

The following illustrative solubility data are provided:

A saturated solution at 80°C contains 100.0 g of water and 37.5 g of dissolved M.

The student cools the solution to 20°C and expects to obtain at least 20.0 g of crystals.

(a) Calculate the actual mass of M that crystallises.

(b) Explain why leaving the solution to cool for longer will not produce the expected 20.0 g.

(c) Recommend a different approach to obtain a larger quantity of crystals without heating the solution to dryness.

Solution 4

Only 1.5 g of M crystallises during cooling. The student should use controlled evaporation to increase the concentration if a larger crop of crystals is required.

The important feature is that the solubility of M changes very little with temperature.

Step 1: Calculate the amount of M that remains dissolved.

At 20°C, the solubility is 36.0 g per 100 g of water.

The solution contains 100.0 g of water, so 36.0 g of M can remain dissolved.

Step 2: Calculate the mass of crystals.

\[
m_{\text{crystals}}
=
37.5-36.0
\]

\[
\boxed{m_{\text{crystals}}=1.5\text{ g}}
\]

Step 3: Explain why additional cooling time does not solve the problem.

Once the solution reaches equilibrium at 20°C, 36.0 g of M can remain dissolved in the 100.0 g of water.

Waiting longer does not change this equilibrium solubility.

The student cannot obtain another 18.5 g of crystals simply by leaving the solution undisturbed for longer.

Step 4: Recommend an alternative.

The student should use controlled evaporation to remove some of the water.

As the amount of water decreases, the solution becomes more concentrated. Eventually, additional M must form crystals.

The student can then filter and dry the crystals without necessarily removing all the solvent.

Conclusion: Cooling is ineffective as the main crystallisation method when the solubility changes only slightly with temperature. Controlled evaporation provides a better way to increase the amount of solid recovered.

Question 5

A chemist needs to recover a heat-sensitive solid P from an aqueous solution containing a soluble impurity Q.

The initial solution is at 40°C and contains:

• 30.0 g of P.
• 4.0 g of Q.
• 150.0 g of water.

All the substances are initially dissolved.

P decomposes at temperatures above 45°C.

The following illustrative solubility data are available:

The chemist wants to obtain the greatest possible mass of pure P crystals while keeping Q dissolved in the mother liquor.

Assume:

• P crystals do not incorporate Q into their crystal structure.
• Both substances reach their equilibrium solubilities.
• No chemical reactions occur.
• No solute is lost during transfer or filtration.
• Only water is evaporated.
• The final solution is at 20°C.

Answer the following:

(a) Explain why simply cooling the original solution from 40°C to 20°C is not the best approach for maximising crystal recovery.

(b) Determine the minimum mass of water that must remain at 20°C to keep all 4.0 g of Q dissolved.

(c) Calculate the mass of P that remains dissolved at this final water content.

(d) Calculate the maximum theoretical mass of pure P crystals that can be recovered.

(e) Explain why evaporating the solution completely to dryness would be a poor decision, even though it would leave more solid in the apparatus.

Solution 5

The maximum theoretical mass of pure P crystals is 28.8 g. The chemist must concentrate the solution without exceeding 40°C, then cool it to 20°C while retaining enough water to keep Q dissolved.

This question requires balancing two competing objectives:

• Remove enough water to maximise the amount of P that crystallises.
• Retain enough water to prevent Q from crystallising and contaminating the product.

The second condition changes the answer.

Step 1: Explain why cooling alone gives a poor recovery.

The initial solution contains 150.0 g of water.

At 20°C, P has a solubility of 18.0 g per 100 g of water.

Therefore, the amount of P that can remain dissolved after cooling is:

\[
m_{\text{P remaining}}
=
\frac{18.0}{100}\times150.0
\]

\[
=27.0\text{ g}
\]

Only:

\[
30.0-27.0=3.0\text{ g}
\]

of P crystallises.

Cooling alone therefore recovers only 3.0 g of P.

The chemist needs to remove some water before cooling to increase the concentration of P.

Step 2: Calculate the minimum amount of water needed to keep Q dissolved.

At 20°C, Q has a solubility of 60.0 g per 100 g of water.

The chemist needs to keep all 4.0 g of Q dissolved.

Let \(m_w\) represent the mass of water remaining, in grams.

Using the solubility relationship:

\[
60.0\text{ g Q}\longrightarrow100\text{ g water}
\]

\[
4.0\text{ g Q}\longrightarrow m_w
\]

Therefore:

\[
m_w=\frac{4.0}{60.0}\times100
\]

\[
\boxed{m_w=6.67\text{ g}}
\]

The solution must retain at least 6.67 g of water at 20°C to keep all the Q dissolved.

Step 3: Calculate the mass of P that remains dissolved.

At 20°C, P has a solubility of 18.0 g per 100 g of water.

With 6.67 g of water remaining:

\[
m_{\text{P remaining}}
=
\frac{18.0}{100}\times6.67
\]

\[
\boxed{m_{\text{P remaining}}=1.20\text{ g}}
\]

Only 1.20 g of P can remain dissolved at this final condition.

Step 4: Calculate the maximum mass of pure P crystals.

The initial solution contained 30.0 g of P.

Therefore:

\[
m_{\text{crystals}}
=
m_{\text{initial}}-m_{\text{remaining}}
\]

\[
=30.0-1.20
\]

\[
\boxed{m_{\text{crystals}}=28.8\text{ g}}
\]

The chemist can theoretically recover 28.8 g of pure P crystals.

Step 5: Explain the required experimental procedure.

The chemist should use a shallow evaporating basin and a controlled water bath maintained at 40°C.

This allows water to evaporate without exceeding the temperature at which P decomposes.

As water is removed, P gradually reaches saturation and begins forming crystals.

The solution is then cooled to 20°C.

At the final theoretical limit, 6.67 g of water remains. This is enough to keep all 4.0 g of Q dissolved, while only 1.20 g of P remains dissolved.

The P crystals can then be collected by filtration.

In practice, the chemist should retain slightly more than 6.67 g of water to allow for measurement uncertainty and avoid accidentally crystallising Q.

Step 6: Explain why evaporation to dryness is unsuitable.

Evaporating to dryness would leave both P and Q in the apparatus.

The chemist would lose the separation advantage provided by their different solubilities.

There is also the risk of decomposing P if excessive heating is used to remove the final solvent.

The objective is not simply to obtain the greatest mass of solid. It is to obtain the greatest mass of the desired solid while keeping the impurity separate.

Conclusion: The correct approach is controlled evaporation below the decomposition temperature, followed by cooling and filtration. The minimum remaining water content is determined by the solubility of the impurity Q, not just the solubility of the desired product P. This gives a theoretical maximum recovery of 28.8 g of pure P.

09What to consider next

Once you can choose between evaporation and crystallisation, the next important decision is whether you need to recover the solvent as well.

If the solvent is a valuable product, allowing it to escape as vapour is wasteful. Distillation provides a way to collect that vapour as a separate liquid.

For a closer comparison of the apparatus and the movement of volatile substances, see Reflux vs Distillation: How to Choose the Right Apparatus.

The key is to choose the technique based on the product you need, the physical properties of the substances, and what happens to every component during the separation.