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Chemistry: Matter, Mixtures, and Separation - Pure and Impure Substances

Grade 8IB

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

🔑Concepts

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Matter is anything that has mass and occupies space. It is classified into pure substances and mixtures.

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A Pure Substance consists of only one type of particle (atoms or molecules). These have a fixed chemical composition and distinct physical properties, such as a sharp melting point (TmT_m) and boiling point (TbT_b).

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Elements are pure substances that cannot be broken down into simpler substances by chemical means (e.g., O2O_{2}, FeFe, AuAu).

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Compounds are pure substances made of two or more elements chemically combined in a fixed ratio (e.g., H2OH_{2}O, CO2CO_{2}, NaClNaCl).

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An Impure Substance (Mixture) consists of two or more substances physically combined but not chemically bonded. They do not have fixed melting or boiling points; instead, they melt or boil over a range of temperatures.

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Homogeneous Mixtures (Solutions) have a uniform composition throughout, such as air or salt dissolved in water (NaCl(aq)NaCl(aq)).

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Heterogeneous Mixtures have a non-uniform composition where the different components can often be seen, such as oil and water or a mixture of sand and salt.

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Purity can be determined using Chromatography. A pure substance will produce only one spot on a chromatogram, while an impure substance will separate into multiple spots.

📐Formulae

Rf=Distance traveled by the substanceDistance traveled by the solvent frontR_{f} = \frac{\text{Distance traveled by the substance}}{\text{Distance traveled by the solvent front}}

Concentration=Mass of solute (g)Volume of solvent (dm3)\text{Concentration} = \frac{\text{Mass of solute (g)}}{\text{Volume of solvent (dm}^3)}

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\%

💡Examples

Problem 1:

A student tests a sample of liquid and finds that it starts boiling at 102∘C102^{\circ}C and finishes boiling at 105∘C105^{\circ}C. Determine if the liquid is pure water.

Solution:

The liquid is impure. Pure water has a fixed boiling point of exactly 100∘C100^{\circ}C at 11 atm pressure.

Explanation:

Pure substances melt and boil at specific, sharp temperatures. Because this liquid boils over a range (102∘C102^{\circ}C to 105∘C105^{\circ}C) and at a temperature higher than 100∘C100^{\circ}C, it contains dissolved impurities.

Problem 2:

In a paper chromatography experiment, a red dye travels 4.5 cm4.5\text{ cm} from the baseline, while the solvent front travels 9.0 cm9.0\text{ cm}. Calculate the RfR_{f} value of the dye.

Solution:

Rf=4.5 cm9.0 cm=0.5R_{f} = \frac{4.5\text{ cm}}{9.0\text{ cm}} = 0.5

Explanation:

The RfR_{f} (Retention Factor) is a ratio used to identify substances. It is calculated by dividing the distance moved by the solute by the distance moved by the solvent.

Problem 3:

A mixture contains 15 g15\text{ g} of salt and 85 g85\text{ g} of sand. Calculate the percentage of salt in this impure mixture.

Solution:

Percentage=1515+85×100=15%\text{Percentage} = \frac{15}{15 + 85} \times 100 = 15\%

Explanation:

To find the percentage of a component in a mixture, divide the mass of that component by the total mass of the mixture (solute + solvent/other components) and multiply by 100100.