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Experimental Techniques - Methods of purification (Filtration, Crystallisation, Distillation)

Grade 11IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Filtration is used to separate an insoluble solid from a liquid. The solid trapped on the filter paper is called the residue, while the liquid that passes through is the filtrate.

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Crystallisation is employed to obtain a pure solid (solute) from its solution. The solution is heated to evaporate the solvent until a saturated solution is reached (the crystallization point), followed by slow cooling to form crystals like CuSO4⋅5H2OCuSO_4 \cdot 5H_2O.

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Simple Distillation separates a volatile solvent from a non-volatile solute (e.g., obtaining pure H2OH_2O from sea water). It relies on the process of boiling followed by condensation in a Liebig condenser.

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Fractional Distillation separates two or more miscible liquids with different boiling points (e.g., ethanol at 78∘C78 ^\circ C and water at 100∘C100 ^\circ C). A fractionating column filled with glass beads provides a large surface area for repeated evaporation and condensation.

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Sublimation is a technique used for substances that transition directly from solid to gas without becoming liquid, such as I2(s)I_2(s) or NH4Cl(s)NH_4Cl(s).

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Purity can be assessed using melting and boiling points. A pure substance has a fixed, sharp melting point. Impurities cause the melting point to decrease and the boiling point to increase over a range of temperatures.

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Paper Chromatography separates mixtures of solutes (like dyes) based on their relative solubilities in a mobile phase (solvent) and their affinity for the stationary phase (paper).

📐Formulae

Rf=distance moved by the substancedistance moved by the solvent frontR_f = \frac{\text{distance moved by the substance}}{\text{distance moved by the solvent front}}

Percentage Purity=Mass of pure substanceTotal mass of sample×100%\text{Percentage Purity} = \frac{\text{Mass of pure substance}}{\text{Total mass of sample}} \times 100\%

Percentage Yield=Actual YieldTheoretical Yield×100%\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

💡Examples

Problem 1:

A student needs to obtain pure water from a mixture of NaClNaCl and H2OH_2O. Which technique should be used, and what is the function of the condenser?

Solution:

Simple Distillation.

Explanation:

The mixture is heated until H2OH_2O reaches its boiling point of 100∘C100 ^\circ C. The water vapor enters the Liebig condenser, where cold water circulating in the outer jacket cools the vapor, causing it to condense back into liquid H2OH_2O (distillate). The non-volatile NaClNaCl remains in the distillation flask.

Problem 2:

Calculate the RfR_f value for a red dye that travelled 4.5 cm4.5 \text{ cm} from the baseline, while the solvent front travelled 6.0 cm6.0 \text{ cm}.

Solution:

Rf=4.5 cm6.0 cm=0.75R_f = \frac{4.5 \text{ cm}}{6.0 \text{ cm}} = 0.75

Explanation:

The RfR_f value is a ratio and therefore has no units. It indicates the relative solubility of the dye in the chosen solvent.

Problem 3:

How would you separate a mixture of sand and KNO3KNO_3 solid to obtain pure KNO3KNO_3 crystals?

Solution:

  1. Add water and stir to dissolve KNO3KNO_3. 2. Filter to remove insoluble sand. 3. Heat the filtrate (KNO3KNO_3 solution) to the point of crystallisation. 4. Allow to cool and filter the crystals.

Explanation:

This uses the difference in solubility. Sand is insoluble in H2OH_2O, while KNO3KNO_3 is soluble. Filtration removes the sand, and crystallisation recovers the KNO3KNO_3 from the liquid phase.

Methods of purification (Filtration, Crystallisation, Distillation) Revision - Grade 11 Chemistry…