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K223 / K225 / 2027
Experimental Chemistry

Full chapter

Experimental Chemistry

Select apparatus, separate mixtures and judge what measurements actually show.

SEC G2 Science K223 / K225 (2027)

01

Choose a measurement that answers the question

Apparatus choice depends on range, precision and what changes.

Match apparatus to a task
Quantity or taskApparatusReason
TimeStopwatchMeasure an interval; define exactly when timing starts and ends.
TemperatureThermometer or temperature probeMeasure the reaction mixture, not the hot container wall.
MassBalanceMeasure mass directly; subtract the container mass.
Approximate liquid volumeMeasuring cylinderQuick measurement where modest precision is sufficient.
One accurate fixed volumeVolumetric pipette and fillerDelivers its calibrated volume, such as 25.0 cm3.
Accurate variable volumeBuretteVolume delivered = final minus initial reading.
Gas volumeGas syringeCollect gas directly and read the change in volume.

Read a liquid meniscus at eye level to avoid parallax; for a clear aqueous solution read its bottom. Record units and precision consistently. A larger container is not automatically more accurate. A pipette supplies one fixed volume; a burette is appropriate when the required volume is found during a titration.

Check your understanding

A burette reads 1.20 cm3 at the start and 23.65 cm3 at the end. What volume was delivered?

  1. 23.65 cm3, the reading at the end of the titration
  2. 24.85 cm3, the two burette readings added together
  3. 22.45 cm3, the final reading minus the initial reading
Show the answer

22.45 cm3, the final reading minus the initial reading

23.65 - 1.20 = 22.45 cm3. The final reading is the delivered volume only if the burette started at 0.00 cm3.

02

Collect a gas using its properties

Solubility and density determine a suitable setup.

A gas syringe is convenient for measuring gas volume. The bung and connections must be airtight, the plunger must move freely, and the capacity must exceed the expected volume. Begin collection promptly: gas lost before sealing makes measured volume too small.

Choose the collection method
MethodSuitable whenExample or limitation
Displacement of waterGas is insoluble or only slightly soluble in waterHydrogen or oxygen; unsuitable for very soluble ammonia.
Upright gas jar with gas entering near the bottomGas is denser than air and would dissolve in waterCarbon dioxide can displace air upwards.
Inverted gas jar with gas entering near the topGas is less dense than airAmmonia displaces air downwards.
Gas syringeGas can be collected without significant leakage or reaction with apparatusUseful for volume-time measurements; very soluble gases need an appropriate dry setup.

Check your understanding

Why is collecting ammonia over water unsuitable?

  1. Ammonia is denser than air, so it sinks out of the gas jar before it is sealed.
  2. Ammonia reacts with the glass of the gas jar when the jar is wet with water.
  3. Ammonia is very soluble in water, so much of it dissolves instead of collecting.
Show the answer

Ammonia is very soluble in water, so much of it dissolves instead of collecting.

A very soluble gas is lost into the water. Ammonia is less dense than air, so it is collected by upward delivery into an inverted dry container.

03

Separate by solubility and particle size

Recover the intended substance, not just any solid.

Filtration separates an insoluble solid from a liquid. The residue remains on the filter paper; the filtrate passes through. Dissolved salt particles pass through with water, so filtering salt solution does not remove the salt. Wash a residue with a small amount of suitable solvent to remove soluble impurities, then dry it.

Separate a sand-and-salt mixture
  1. Add water and stir

    Salt dissolves; sand remains insoluble.

  2. Filter

    Sand is the residue; salt solution is the filtrate.

  3. Wash and dry the sand

    Washing removes remaining salt solution.

  4. Recover salt from the filtrate

    Evaporate water for a heat-stable salt, or concentrate and crystallise.

For crystallisation, heat a solution gently to remove some solvent until it is nearly saturated, then let it cool. Crystals form as solubility decreases. Filter, wash with a little cold solvent and dry. Evaporation to dryness removes all solvent and suits a stable dissolved solid; strong heating may decompose some compounds. Crystallisation often leaves soluble impurities in the remaining solution.

Check your understanding

A clear sugar solution passes through filter paper unchanged. Why?

  1. Sugar melts in water, so it becomes a liquid that soaks through the paper.
  2. The sugar is dissolved, so its particles pass through the paper with the water.
  3. The filter paper is too coarse; a finer paper would hold back the sugar.
Show the answer

The sugar is dissolved, so its particles pass through the paper with the water.

Filtration separates insoluble solids from a liquid. Dissolved sugar particles are small and spread through the water, so evaporation or crystallisation is needed to recover the sugar.

04

Distillation and liquid layers

Use boiling-point differences or immiscibility.

Make a guess. You are not marked.

A mixture of ethanol (boiling point 78 degrees C) and water (boiling point 100 degrees C) is heated slowly in fractional distillation. Which liquid collects first?

  1. Both together, in the same ratio as the mixture
  2. Mostly ethanol, as it boils at a lower temperature
  3. Mostly water, as there is more of it in the mixture
Show the answer

Mostly ethanol, as it boils at a lower temperature

Ethanol has the lower boiling point, so its vapour reaches the top of the column first while most water condenses and drips back.

Simple distillation recovers a solvent from a solution containing a non-volatile solute. Heat salt water: water vaporises, enters a condenser, cools and becomes liquid distillate. Salt remains in the flask. Place the thermometer bulb near the side-arm to measure the vapour entering the condenser. Cooling water enters the condenser at the lower end and exits at the upper end to keep the jacket full.

Fractional distillation separates miscible liquids with different boiling points, especially when those boiling points are close. A fractionating column provides repeated evaporation and condensation. The vapour reaching its top is richer in the more volatile, lower-boiling component. Collect fractions over suitable temperature ranges. Crude oil is separated into groups of hydrocarbons this way; a fraction is generally a mixture, not one pure compound.

Check your understanding

You need pure water from seawater. Why choose distillation rather than evaporation to dryness?

  1. Distillation condenses and collects the water vapour; evaporation lets it escape.
  2. Evaporation leaves salt dissolved in the water, so the water that remains is impure.
  3. Distillation removes the salt faster, because it heats the seawater to a higher temperature.
Show the answer

Distillation condenses and collects the water vapour; evaporation lets it escape.

Both methods boil off the water, but only distillation cools the vapour back into liquid and collects it. Evaporation keeps the salt and loses the water.

05

Read the evidence in a chromatogram

Compare spots under the same conditions.

Place small sample spots on a pencil baseline above the solvent level. Ink would dissolve and add unwanted spots. As solvent rises through the paper, substances repeatedly distribute between the solvent and paper; different substances move different distances. Keep the vessel covered and mark the solvent front before it evaporates.

Compare an unknown with standards

Unknown U has two spots at the same heights as one spot in A and one in B.

A and B were run alongside U using the same paper and solvent.

Several spots indicate a mixture of separated components. A single spot suggests purity under these conditions, but two substances might move together. A spot matching a known sample supports identification; comparison is meaningful only with the same solvent and conditions.

Check your understanding

U gives two spots that match the spots of A and B. Can you be certain that U contains A and B and nothing else?

  1. No. U cannot contain A or B, because a mixture gives spots in new positions.
  2. Yes. Two spots means two substances, and both spots match a known reference.
  3. No. A and B are probably present, but another substance could hide under a spot.
Show the answer

No. A and B are probably present, but another substance could hide under a spot.

Matching spots support A and B being present. Another substance could have the same Rf in this solvent, or not show up under these conditions, so the conclusion is not certain.

06

Use melting and boiling data

A physical measurement supports identity; a single match is not proof.

Make a guess. You are not marked.

Pure aspirin melts sharply at a fixed temperature. How does a sample of aspirin with a little impurity melt?

  1. Higher, and over a range of temperatures
  2. Lower, and over a range of temperatures
  3. At the same temperature, but more slowly
Show the answer

Lower, and over a range of temperatures

Impurities lower the melting point and spread the melting over a range. A sharp, correct melting point is evidence of purity.

At a specified pressure, a pure substance melts sharply and boils at a characteristic temperature. Soluble impurities usually lower and broaden the melting range and can raise the boiling point. Compare measured data with reference values taken under comparable conditions; a mixture may have a changing boiling temperature.

Worked exampleInterpret a melting range

Pure substance P melts at 82 degrees C. A sample melts over 76-80 degrees C.

Try it first. Then show the worked stepsHide the worked steps
  1. The lower, broader melting range is inconsistent with a pure sample of P.
  2. It is consistent with P containing impurities, although more evidence is needed to establish identity.
Answer

The sample is impure if it is P. A matching single temperature alone would not uniquely identify an unknown.

Worked exampleMatch two measurements to reference data

At the same pressure, pure P melts at 18 degrees C and boils at 126 degrees C; Q melts at 18 degrees C and boils at 154 degrees C. Unknown X melts sharply at 18 degrees C and boils steadily at 154 degrees C. A second sample, claimed to be Q, melts over 12-16 degrees C. Interpret both samples.

Try it first. Then show the worked stepsHide the worked steps
  1. The melting point alone cannot distinguish P and Q: both have the same value.
  2. The boiling point of X matches Q, not P. Within the supplied candidates, both measurements support pure Q.
  3. The second sample melts below the reference temperature over a range. If it is Q, the data indicate impurities.
Answer

X is consistent with pure Q. The second sample is impure Q if its claimed identity is correct; its melting range alone cannot establish identity.

Check your understanding

A liquid is clear and colourless. Is it pure?

  1. Not necessarily. A colourless substance could be dissolved in it.
  2. Yes. An impurity would make the liquid cloudy or coloured.
  3. Not necessarily. Pure liquids are slightly cloudy, so clear means it has been filtered.
Show the answer

Not necessarily. A colourless substance could be dissolved in it.

Many dissolved substances, such as salt or sugar, are colourless. Use a fixed boiling point, chromatography or another test, not appearance.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Pure substanceA single substance that is not mixed with anything else. It melts and boils at fixed (sharp) temperatures.
MixtureTwo or more substances that are not chemically combined. They can be separated by physical methods.
Residue and filtrateThe residue is the solid left on the filter paper; the filtrate is the liquid that passes through it.
Which property selects the method?
DifferenceMethod
Insoluble solid in liquidFiltration
Soluble solid wantedCrystallisation or suitable evaporation
Solvent wantedSimple distillation
Miscible liquids with different boiling pointsFractional distillation
Different movement with solvent and paperChromatography

Choose measurements with suitable precision. Distinguish observation from inference. A clear liquid is not automatically pure.

Scope

Learning outcomes

1. Experimental Chemistry (K223 / K225). Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map
  1. 1.1(a) Select measuring apparatus

    • Time, temperature, mass and volume
    • Burettes, pipettes, measuring cylinders and gas syringes

    Choose a measurement that answers the question

  2. 1.1(b) Select simple experimental setups

    • Gas collection
    • Suitable apparatus from given information

    Choose a measurement that answers the questionCollect a gas using its properties

  3. 1.2(a) Describe separation methods

    • Solvent, filtration, crystallisation, evaporation
    • Simple/fractional distillation
    • Paper chromatography

    Separate by solubility and particle sizeDistillation and liquid layersRead the evidence in a chromatogram

  4. 1.2(b) Choose methods from properties

    • Solid-solid
    • Solid-liquid
    • Miscible liquid-liquid

    Separate by solubility and particle sizeDistillation and liquid layers

  5. 1.2(c) Interpret chromatograms

    • Comparison with known samples
    • Rf not required

    Read the evidence in a chromatogram

  6. 1.2(d) Infer identity and purity

    • Melting and boiling data

    Use melting and boiling data