krit.club logo

Nature of Matter: Elements, Compounds, and Mixtures - How Do We Use Elements, Compounds, and Mixtures?

Grade 8CBSE

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

🔑Concepts

•

Matter is broadly classified into pure substances (Elements and Compounds) and mixtures (Homogeneous and Heterogeneous).

•

Elements are pure substances consisting of only one type of atom. They are used based on their properties, such as metals like Aluminum (AlAl) for making utensils and non-metals like Oxygen (O2O_2) for respiration.

•

Compounds are substances formed when two or more elements chemically combine in a fixed ratio by mass. For example, Water (H2OH_2O) is essential for life, and Common Salt (NaClNaCl) is used in food.

•

Mixtures consist of two or more substances physically combined in any proportion. Examples include Air (a mixture of gases like N2N_2, O2O_2, and CO2CO_2) and Alloys like Brass (a mixture of CuCu and ZnZn).

•

Uses of Compounds: Baking Soda (NaHCO3NaHCO_3) is used in baking, and Carbon Dioxide (CO2CO_2) is used in fire extinguishers because it does not support combustion.

•

Uses of Alloys: Stainless Steel (a mixture of FeFe, CrCr, and NiNi) is used to make surgical instruments and kitchenware because it is resistant to corrosion.

•

The Law of Constant Proportions states that in a chemical substance, the elements are always present in definite proportions by mass.

📐Formulae

Mass Ratio of H2O→H:O=1:8\text{Mass Ratio of } H_2O \rightarrow H:O = 1:8

Mass Ratio of NH3→N:H=14:3\text{Mass Ratio of } NH_3 \rightarrow N:H = 14:3

Mass Ratio of CO2→C:O=3:8\text{Mass Ratio of } CO_2 \rightarrow C:O = 3:8

Mass Percentage=Mass of SoluteMass of Solution×100\text{Mass Percentage} = \frac{\text{Mass of Solute}}{\text{Mass of Solution}} \times 100

💡Examples

Problem 1:

Calculate the mass of Oxygen required to react completely with 3 g3\text{ g} of Hydrogen to form Water (H2OH_2O), given that the mass ratio of H:OH:O in water is 1:81:8.

Solution:

Given the ratio H:O=1:8H:O = 1:8. If 1 g1\text{ g} of Hydrogen requires 8 g8\text{ g} of Oxygen, then 3 g3\text{ g} of Hydrogen requires: 3×8=24 g of Oxygen3 \times 8 = 24\text{ g of Oxygen}

Explanation:

According to the Law of Constant Proportions, elements in a compound are always in a fixed ratio by mass regardless of the source.

Problem 2:

A student mixes 20 g20\text{ g} of salt in 80 g80\text{ g} of water. Calculate the total mass of the resulting mixture using vertical addition.

Solution:

20+80100\begin{array}{r} 20 \\ + 80 \\ \hline 100 \end{array} The total mass is 100 g100\text{ g}.

Explanation:

In a mixture, the total mass is the sum of the masses of the individual components (solute + solvent).

Problem 3:

Identify the type of matter: (a) 24 Karat Gold24\text{ Karat Gold}, (b) 22 Karat Gold22\text{ Karat Gold}, (c) CO2CO_2.

Solution:

(a) Element (Pure Gold), (b) Mixture (Gold + Copper/Silver), (c) Compound.

Explanation:

Pure Gold is an element (AuAu). 22K22\text{K} gold is an alloy (mixture) used in jewelry to provide strength. CO2CO_2 is a chemical combination of Carbon and Oxygen in a fixed 3:83:8 mass ratio.

How Do We Use Elements, Compounds, and Mixtures? Class 8 Notes & Examples