Separation of Mixtures: Choose a Sequence and Track Every Component

A mixture-separation question is easier when you follow each component through every stage. Naming a magnet, a filter and an evaporating dish is not enough. You need to explain why a step separates particular substances, where each substance goes, and whether the final products meet the question’s goal.

The cases below are fictional paper models. Their properties and ideal recoveries are supplied, and no experiment is required. They do not give operating instructions for heating, handling chemicals or constructing apparatus. Real work would need appropriate supervision and a separate safety assessment.

Begin with properties and the recovery goal

Consider a dry mixture of sand, iron and salt T. In this model iron is attracted by a magnet; sand and salt T are not. Salt T dissolves in the available water, while sand and iron remain insoluble. Salt T is nonvolatile and remains unchanged under the modelled recovery conditions. None of the three reacts with the water.

The first target is to recover all three original solids separately; the water does not need to be collected. A useful sequence therefore exploits magnetism, then solubility, then the different volatility of water and salt. Those are three different reasons for separation, not three interchangeable apparatus names.

Always notice the word “all”. Recovering a little clean sand would not satisfy this target if the iron and salt were discarded. Likewise, obtaining one dry solid mixture at the end would conserve the material without separating it. State the identity of each final fraction.

Worked analysis 1: track a three-solid mixture

1. Magnetic separation gives an iron fraction containing the model’s 7 g of iron. The nonmagnetic fraction contains 13 g sand and 10 g salt T.
2. In the paper model, contact with sufficient water transfers the salt into solution. Sand remains as solid particles. The salt has changed its physical distribution, not disappeared.
3. Filtration separates the insoluble sand residue from the salt solution. Salt T and water both pass into the filtrate.
4. Modelled removal of the water leaves the salt fraction. Under the stated complete-recovery assumptions, this contains 10 g salt T.

The recovered original solids total 7 + 13 + 10 = 30 g, matching the initial mixture. This is a check on the ideal bookkeeping. It is not a guarantee that a real separation has no loss. Some liquid may remain with a residue, some material may adhere to equipment, and an incompletely separated magnetic fraction may contain other particles.

A wet residue cannot be compared directly with an initial dry solid mass. If a purported sand fraction weighs 16 g while containing 3 g of retained water, its sand mass is still 13 g. The extra water is not evidence that sand was produced. Distinguish the identity of the fraction from the total mass placed on a balance.

Why the filtrate contains the salt

Ordinary filter paper retains sufficiently large suspended particles; it does not isolate a dissolved solute simply because the solute began as a solid. In this model the dissolved salt passes with the water. “Residue” names the material retained by the filter; “filtrate” names the liquid fraction that passes through it.

This distinction is more reliable than saying “solids stay and liquids pass”. Dissolved material is present in the filtrate even though the pure substance may be solid at room temperature. Conversely, the residue can be wet. The separation concerns the state of the mixture at that particular stage.

For each filter symbol in a flow chart, write both exits. A chart showing only a residue can hide the loss of a wanted dissolved component. If the problem asks for salt, the filtrate is the fraction to follow after sand is removed.

Worked analysis 2: change the goal to include water

A second fictional sample contains 12 g nonvolatile salt U fully dissolved in 80 g water, plus 8 g insoluble grit. Recover grit, salt U and the water separately. Salt U is stipulated to be stable during the theoretical process, and the model ignores transfer losses.

First separate the grit by filtration. The dry grit fraction is 8 g, and the filtrate contains 92 g of solution: 12 g salt U plus 80 g water. Simply allowing the water to escape would recover the salt but fail the water-recovery requirement.

A distillation-based paper model instead transfers water to a vapour fraction and then to a condensed liquid fraction. Salt U stays in the nonvolatile fraction. Complete ideal separation is represented as 80 g collected water and 12 g recovered salt. This is an endpoint mass model, not an instruction to distil a container to dryness.

The distinction between evaporation and distillation is therefore tied to the requested product. Both involve vaporisation. Distillation additionally collects the condensed volatile material. A correct answer says which component is collected where, rather than assuming the longer technique name is always better.

Worked analysis 3: find the error in a proposed sequence

A learner writes: “Add water to mixture A, filter out sand and salt, then use a magnet to remove iron from the filtrate.” Identify two mistakes and give a corrected account.

The first mistake puts dissolved salt in the residue. Under the supplied properties, the residue contains sand and iron; salt T is in the filtrate. The second mistake sends iron through the ordinary filter even though the model leaves it as insoluble particles retained with the sand.

A corrected and simpler sequence removes iron magnetically from the original dry mixture, then separates sand from salt solution, then recovers salt from that solution. The key repair is not merely swapping words. It is rebuilding the component record so each operation matches the supplied physical state.

Could a different order work? In principle, a question might allow an alternative route that first separates the salt solution and later separates the retained iron and sand. The answer would need to explain that extra handling and any changed conditions. An alternative is not wrong because it differs from a memorised order; it is wrong if a component is lost, mixed incorrectly or assigned to an impossible fraction.

Test the selectivity before choosing a method

A method is selective when the components respond differently enough for the required separation. A magnet does not distinguish two components that are both nonmagnetic. Ordinary filtration does not separate two fully dissolved salts from each other. Evaporating their shared solvent would leave them together.

When the supplied properties do not provide a usable difference, say what further information is needed. Do not silently assume a second salt is insoluble or a liquid volatile. A careful limitation can be the most scientifically useful part of an answer.

Three practice tasks with solved answers

Practice 1: identify both filter fractions

A fictional mixture has 5 g magnetic iron, 9 g insoluble sand and 6 g soluble salt V. Iron is removed first. All salt V then dissolves in 40 g water. What are the dry residue mass, filtrate composition and ideal final salt mass?

Answer: Dry residue is 9 g sand. Filtrate contains 6 g salt V and 40 g water, so its mass is 46 g. Complete ideal solvent removal leaves 6 g salt V. The original solid masses still sum to 5 + 9 + 6 = 20 g. The 40 g water was introduced for separation and is accounted for separately.

Practice 2: recover a specified liquid

A sample contains 4 g insoluble grit, 5 g fully dissolved nonvolatile salt and 45 g water. The only required product is collected water. Which two separation principles are appropriate, and what happens to the other components?

Answer: Filtration first separates grit from the salt solution. A theoretical distillation then collects water as condensed vapour while the nonvolatile salt remains behind. The ideal water amount is 45 g if recovery is complete. Evaporation without collection does not meet the target, even if the salt left behind is pure.

Practice 3: recognise an insufficient difference

Two nonmagnetic salts are fully dissolved in water. Both remain stable and nonvolatile in the stated conditions. A learner proposes filtration followed by complete evaporation to obtain the salts separately. Is the proposal supported?

Answer: No. Both dissolved salts pass through the filter and both remain when the water is removed. The proposal recovers a mixed solid fraction. Further properties, such as sufficiently different solubilities under specified conditions, would be needed before proposing a selective route. No quantitative crystallisation calculation is required to diagnose this failure.

Sources and further reading

The references below support the scientific background. The cases, calculations and illustrations in this lesson are original. External teaching guidance does not establish current KCSE marking rules or grade placement.

• OpenStax Chemistry 2e: classification of matter
• College of Saint Benedict and Saint John’s University: separation methods

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