Chapter 1: Some Basic Concepts of Chemistry

1.1 Importance and Scope of Chemistry

Chemistry is the branch of science that deals with the composition, structure, properties, and transformations of matter. It is often called the central science because it connects physics with other natural sciences such as biology, geology, and environmental science.

Applications of Chemistry

Chemistry plays a vital role in nearly every aspect of our daily lives:

• Medicine & Health: Development of drugs, anaesthetics, and antibiotics
• Agriculture: Fertilizers, pesticides, and herbicides
• Materials: Polymers, alloys, ceramics, and nanomaterials
• Energy: Fuels, batteries, and solar cells
• Environment: Water purification, pollution control, and green chemistry
• Food: Preservatives, artificial sweeteners, and food processing

1.2 Nature of Matter

Matter is anything that has mass and occupies space. Matter can be classified in two ways:

Classification Based on Physical State

Classification Based on Chemical Composition

1.3 Properties of Matter and Their Measurement

Physical Properties

Physical properties can be measured or observed without changing the identity of the substance. Examples include colour, odour, melting point, boiling point, and density.

SI Units (International System of Units)

Important Derived Quantities

Temperature Conversions

\[ T(\text{K}) = T(°\text{C}) + 273.15 \]

\[ T(°\text{F}) = \frac{9}{5},T(°\text{C}) + 32 \]

Significant Figures

Rules for determining significant figures:

1. All non-zero digits are significant. (e.g., 285 has 3 significant figures)
2. Zeros between non-zero digits are significant. (e.g., 2.005 has 4)
3. Leading zeros are not significant. (e.g., 0.0025 has 2)
4. Trailing zeros in a number with a decimal point are significant. (e.g., 2.500 has 4)
5. Trailing zeros in a number without a decimal point may or may not be significant.

1.4 Laws of Chemical Combination

1. Law of Conservation of Mass (Lavoisier, 1789)

“In all physical and chemical changes, the total mass of the reactants is equal to the total mass of the products.”

\[ \text{Mass of reactants} = \text{Mass of products} \]

2. Law of Definite Proportions (Proust, 1799)

“A given compound always contains exactly the same proportion of elements by weight, regardless of the source or method of preparation.”

Example: Water (H₂O) always contains hydrogen and oxygen in the mass ratio 1:8.

3. Law of Multiple Proportions (Dalton, 1803)

“If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.”

Example: Carbon forms two oxides with oxygen:

Ratio of oxygen = 16 : 32 = 1 : 2 (simple whole numbers)

4. Gay-Lussac’s Law of Gaseous Volumes (1808)

“When gases react together, they do so in volumes which bear a simple whole number ratio to each other and to the volumes of the products, if gaseous — all volumes measured at same temperature and pressure.”

\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2,\text{HCl}(g) \] \[ 1;\text{vol} : 1;\text{vol} : 2;\text{vol} \]

5. Avogadro’s Law (1811)

“Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.”

1.5 Dalton’s Atomic Theory

John Dalton proposed the atomic theory in 1808:

1. Matter consists of indivisible atoms.
2. Atoms of a given element are identical in mass and properties.
3. Atoms of different elements differ in mass and properties.
4. Compounds are formed when atoms of different elements combine in fixed ratios.
5. Chemical reactions involve reorganisation of atoms — atoms are neither created nor destroyed.

Limitations:

• Atoms can be further divided into subatomic particles (electrons, protons, neutrons).
• Atoms of the same element can have different masses (isotopes).
• Atoms of different elements can have the same mass (isobars).

1.6 Atomic and Molecular Masses

Atomic Mass Unit (amu or u)

\[ 1;\text{amu} = \frac{1}{12} \times \text{mass of one } {}^{12}\text{C atom} = 1.66054 \times 10^{-24};\text{g} \]

Average Atomic Mass

For elements with isotopes, the average atomic mass is calculated as:

\[ \text{Average atomic mass} = \sum_{i} f_i \times m_i \]

where \( f_i \) is the fractional abundance and \( m_i \) is the mass of isotope \( i \).

Example: Chlorine has two isotopes: \({}^{35}\text{Cl}\) (75.77%) and \({}^{37}\text{Cl}\) (24.23%)

\[ \text{Average atomic mass} = (0.7577 \times 35) + (0.2423 \times 37) = 35.48;\text{u} \]

Molecular Mass

The molecular mass is the sum of atomic masses of all atoms in a molecule.

Example: Molecular mass of H₂SO₄:

\[ = 2(1.008) + 32.06 + 4(16.00) = 98.08;\text{u} \]

Formula Mass

Used for ionic compounds which do not exist as discrete molecules.

Example: Formula mass of NaCl = 23.0 + 35.5 = 58.5 u

1.7 Mole Concept and Molar Mass

The Mole

One mole of any substance contains exactly \( 6.022 \times 10^{23} \) elementary entities (atoms, molecules, ions, etc.). This number is called Avogadro’s number (\( N_A \)).

\[ 1;\text{mol} = 6.022 \times 10^{23};\text{entities} \]

Molar Mass

The molar mass of a substance is the mass of one mole, expressed in g/mol. It is numerically equal to the atomic/molecular mass in u.

Relationship Between Moles, Mass, and Number

Key Formulas:

\[ n = \frac{\text{Given mass (w)}}{\text{Molar mass (M)}} \]

\[ n = \frac{N}{N_A} \]

\[ n = \frac{V(\text{at STP})}{22.4;\text{L}} \quad \text{(for gases)} \]

1.8 Percentage Composition

The mass percentage of an element in a compound:

\[ \text{Mass %} = \frac{\text{Mass of element in 1 mol of compound}}{\text{Molar mass of compound}} \times 100 \]

Example: Percentage composition of H₂O:

\[ \text{Mass % of H} = \frac{2 \times 1.008}{18.02} \times 100 = 11.19% \]

\[ \text{Mass % of O} = \frac{16.00}{18.02} \times 100 = 88.81% \]

1.9 Empirical and Molecular Formula

\[ \text{Molecular formula} = n \times \text{Empirical formula} \]

where \( n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} \)

Example: Glucose has empirical formula CH₂O (empirical formula mass = 30 u) and molar mass = 180 u.

\[ n = \frac{180}{30} = 6 \]

Therefore, molecular formula = \( \text{C}6\text{H}{12}\text{O}_6 \)

Steps to Determine Empirical Formula

1. Convert percentage composition to grams (assume 100 g of compound).
2. Convert mass to moles by dividing by atomic mass.
3. Divide each mole value by the smallest mole value.
4. If ratios are not whole numbers, multiply by a suitable integer.

1.10 Chemical Reactions and Stoichiometry

Balanced Chemical Equations

A balanced chemical equation has equal numbers of atoms of each element on both sides.

Example:

\[ 4,\text{Fe}(s) + 3,\text{O}_2(g) \rightarrow 2,\text{Fe}_2\text{O}_3(s) \]

This equation tells us:

• 4 atoms of Fe react with 3 molecules of O₂
• 4 mol Fe reacts with 3 mol O₂ to form 2 mol Fe₂O₃
• \( 4 \times 55.85 = 223.4 \) g Fe reacts with \( 3 \times 32 = 96 \) g O₂

Limiting Reagent

The limiting reagent is the reactant that is completely consumed in a reaction, determining the maximum amount of product formed.

Worked Example:

Calculate the mass of water formed when 4 g of H₂ reacts with 16 g of O₂.

\[ 2,\text{H}_2 + \text{O}_2 \rightarrow 2,\text{H}_2\text{O} \]

Moles of H₂ = \( \frac{4}{2} = 2 \) mol

Moles of O₂ = \( \frac{16}{32} = 0.5 \) mol

From stoichiometry: 2 mol H₂ requires 1 mol O₂.

Available ratio: \( \frac{2}{2} = 1 \) for H₂ and \( \frac{0.5}{1} = 0.5 \) for O₂.

Since 0.5 < 1, O₂ is the limiting reagent.

Moles of H₂O formed = \( 2 \times 0.5 = 1 \) mol

Mass of H₂O = \( 1 \times 18 = 18 \) g

Reactions in Solutions — Concentration Terms

Molarity (M):

\[ M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}} \]

Molality (m):

\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \]

Worked Examples

Example 1: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its molar mass is 180 g/mol. Find the molecular formula.

Solution:

Empirical formula = CH₂O (empirical formula mass = 30 u)

\[ n = \frac{180}{30} = 6 \]

Molecular formula = C₆H₁₂O₆ (Glucose)

Example 2: How many molecules of methane are present in 0.80 g of CH₄?

Molar mass of CH₄ = 16 g/mol

\[ n = \frac{0.80}{16} = 0.05;\text{mol} \]

\[ N = n \times N_A = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22};\text{molecules} \]

Practice Questions

Multiple Choice Questions (MCQs)

1. The number of atoms in 4.25 g of NH₃ is approximately:

 (a) \( 6.022 \times 10^{23} \)

 (b) \( 1.505 \times 10^{23} \)

 (c) \( 4 \times 6.022 \times 10^{23} \)

 (d) \( 6.022 \times 10^{22} \)

2. The empirical formula of a compound with 40% sulphur and 60% oxygen is:

 (a) SO₂

 (b) SO₃

 (c) SO

 (d) S₂O₃

3. The number of moles of CO₂ containing 8.0 g of oxygen is:

 (a) 0.25

 (b) 0.50

 (c) 1.0

 (d) 2.0

4. If 1.0 g of a metal oxide contains 0.5 g of oxygen, the equivalent weight of the metal is:

 (a) 4

 (b) 8

 (c) 16

 (d) 32

5. Which has the maximum number of molecules?

 (a) 1 g of CO₂

 (b) 1 g of N₂

 (c) 1 g of H₂

 (d) 1 g of CH₄

Short Answer Questions (2–3 Marks)

6. Define the law of conservation of mass. Give an example.

7. Calculate the molarity of a solution prepared by dissolving 5.85 g of NaCl in water to make 500 mL of solution.

8. What is the difference between empirical formula and molecular formula? Explain with an example.

9. A compound has the following composition: Na = 29.11%, S = 40.51%, O = 30.38%. Determine its empirical formula.

10. Calculate the mass percentage of each element in calcium carbonate (CaCO₃).

Long Answer Questions (5 Marks)

11. (a) State and explain the law of multiple proportions with a suitable example.

 (b) Calculate the amount of carbon dioxide that could be produced when:

  (i) 1 mole of carbon is burnt in air

  (ii) 1 mole of carbon is burnt in 16 g of dioxygen

12. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).

 (a) Express this in percent by mass.

 (b) Determine the molality of chloroform in the water sample.

13. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g CO₂ and 0.690 g H₂O and no other products. A volume of 10.0 L (measured at STP) of this welding gas weighs 11.6 g.

 (a) Find the empirical formula.

 (b) Find the molar mass and molecular formula.

Numerical Problems

14. Calculate the number of atoms of hydrogen present in 52 g of methane (CH₄). What would be the mass of an individual atom of hydrogen?

15. How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

\[ \text{CuSO}_4 + \text{Fe} \rightarrow \text{FeSO}_4 + \text{Cu} \]

16. Determine the molecular formula of an oxide of iron in which the mass percentage of iron and oxygen are 69.9% and 30.1% respectively. (Given: molar mass ≈ 160 g/mol)

Assertion-Reason Questions

In each of the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is NOT the correct explanation of A.

(c) A is true but R is false.

(d) A is false but R is true.

17. Assertion (A): One mole of different gases at STP have different volumes.

Reason (R): At STP, the molar volume of any ideal gas is 22.4 L.

18. Assertion (A): The empirical formula and molecular formula of glucose are the same.

Reason (R): The empirical formula gives the simplest whole number ratio of atoms.

Answer Key