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Nature of Matter: Elements, Compounds, and Mixtures - Compounds

Grade 8CBSE

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

🔑Concepts

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A compound is a pure substance composed of two or more different elements chemically combined in a fixed proportion by mass. For example, H2OH_2O (Water) and CO2CO_2 (Carbon Dioxide).

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The properties of a compound are entirely different from the properties of its constituent elements. For instance, Sodium (NaNa) is a highly reactive metal and Chlorine (Cl2Cl_2) is a poisonous gas, but together they form Sodium Chloride (NaClNaCl), which is edible table salt.

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Constituents of a compound cannot be separated by physical methods like filtration, evaporation, or magnetic separation. They can only be separated into their elements by chemical or electrochemical reactions, such as the electrolysis of water.

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A chemical formula represents the composition of a compound. It uses symbols of elements and subscripts to show the number of atoms of each element present in one molecule, such as C6H12O6C_6H_{12}O_6 for Glucose.

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Formation of a compound is always accompanied by energy changes. Energy is either absorbed or released in the form of heat, light, or sound during the chemical reaction.

📐Formulae

H2O (Water)H_2O \text{ (Water)}

CO2 (Carbon Dioxide)CO_2 \text{ (Carbon Dioxide)}

NaCl (Sodium Chloride)NaCl \text{ (Sodium Chloride)}

CaCO3 (Calcium Carbonate)CaCO_3 \text{ (Calcium Carbonate)}

Ratio of mass in H2O⇒H:O=1:8\text{Ratio of mass in } H_2O \Rightarrow H:O = 1:8

2H2O(l)→Electricity2H2(g)+O2(g)2H_2O(l) \xrightarrow{\text{Electricity}} 2H_2(g) + O_2(g)

💡Examples

Problem 1:

Determine the ratio by mass of the constituent elements in Ammonia (NH3NH_3). (Atomic masses: N=14uN = 14 u, H=1uH = 1 u)

Solution:

Mass of Nitrogen (NN) = 14×1=14u14 \times 1 = 14 u. Mass of Hydrogen (HH) = 1×3=3u1 \times 3 = 3 u. Therefore, the ratio N:HN:H by mass is 14:314:3.

Explanation:

According to the Law of Constant Proportions, the elements in a compound are always present in definite proportions by mass regardless of the source of the compound.

Problem 2:

Explain why Water (H2OH_2O) is considered a compound while a mixture of Hydrogen and Oxygen gas is not.

Solution:

In H2OH_2O, Hydrogen and Oxygen are chemically bonded in a fixed mass ratio of 1:81:8. In a mixture, they are simply physically blended in any proportion. Furthermore, Water has different properties (it extinguishes fire) than its constituents (Hydrogen is combustible and Oxygen supports combustion).

Explanation:

Compounds have fixed compositions and unique chemical properties, whereas mixtures retain the properties of their individual components and have variable compositions.

Problem 3:

Calculate the total relative molecular mass of Calcium Carbonate (CaCO3CaCO_3). Given: Ca=40uCa = 40 u, C=12uC = 12 u, O=16uO = 16 u.

Solution:

The molecular mass is calculated as: 40+12+(3×16)40 + 12 + (3 \times 16) =40+12+48= 40 + 12 + 48 =100u= 100 u

Explanation:

The molecular mass of a compound is the sum of the atomic masses of all the atoms present in its chemical formula.