Chapter Overview
Some Basic Concepts of Chemistry is the foundational chapter of CBSE Class 11 Chemistry. It introduces you to the language of chemistry — how we measure, quantify, and predict chemical reactions. From the SI unit system and significant figures to the mole concept and stoichiometric calculations, every concept in this chapter feeds directly into later chapters on atomic structure, chemical bonding, and thermodynamics.
This guide covers every exam-facing topic: the laws of chemical combination, Dalton's atomic theory, the mole concept and Avogadro's number, molar mass calculations, percentage composition, empirical and molecular formulae, stoichiometry, limiting reagent problems, and significant figures. Each section includes definitions, tables, worked numericals, and common mistakes to avoid.
1. Importance of Chemistry and Nature of Matter
Chemistry is the branch of science that studies the composition, structure, properties, and interactions of matter. Matter is anything that has mass and occupies space.
Classification of matter:
| Type | Definition | Example |
|---|---|---|
| Solid | Definite shape and volume | Iron, ice |
| Liquid | Definite volume, no definite shape | Water, oil |
| Gas | No definite shape or volume | Oxygen, CO₂ |
| Element | Single type of atom | Gold (Au), Oxygen (O₂) |
| Compound | Two or more elements chemically combined | H₂O, NaCl |
| Mixture | Two or more substances not chemically combined | Air, salt water |
Mixtures can be homogeneous (uniform throughout, e.g., salt in water) or heterogeneous (non-uniform, e.g., sand in water).
2. The International System of Units (SI)
Chemistry requires precise measurement. The SI system provides seven base units:
| Quantity | SI Unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Temperature | kelvin | K |
| Amount of substance | mole | mol |
| Electric current | ampere | A |
| Luminous intensity | candela | cd |
Important derived units:
| Quantity | Unit | Symbol | Expression |
|---|---|---|---|
| Force | newton | N | kg m s⁻² |
| Pressure | pascal | Pa | N m⁻² = kg m⁻¹ s⁻² |
| Energy | joule | J | kg m² s⁻² |
| Volume | cubic metre | m³ | — |
| Density | kilogram per cubic metre | kg m⁻³ | — |
Common SI prefixes:
| Prefix | Symbol | Factor |
|---|---|---|
| giga | G | 10⁹ |
| mega | M | 10⁶ |
| kilo | k | 10³ |
| deci | d | 10⁻¹ |
| centi | c | 10⁻² |
| milli | m | 10⁻³ |
| micro | μ | 10⁻⁶ |
| nano | n | 10⁻⁹ |
| pico | p | 10⁻¹² |
Temperature conversions:
- K = °C + 273.15
- F = (9/5)°C + 32
3. Significant Figures
Significant figures communicate the precision of a measurement.
Rules for counting significant figures:
1. All non-zero digits are significant. — 285 has 3 significant figures. 2. Zeros between non-zero digits are significant. — 2005 has 4 significant figures. 3. Leading zeros are NOT significant. — 0.025 has 2 significant figures. 4. Trailing zeros after a decimal point ARE significant. — 2.50 has 3 significant figures. 5. Trailing zeros in a whole number without a decimal are ambiguous — use scientific notation. — 2500 could be 2, 3, or 4 significant figures; write 2.5 × 10³ for 2 significant figures.
Rules for arithmetic:
- Addition/Subtraction: The result has the same number of decimal places as the number with the fewest decimal places.
- 12.11 + 3.4 = 15.5 (one decimal place)
- Multiplication/Division: The result has the same number of significant figures as the number with the fewest significant figures.
- 2.5 × 3.42 = 8.6 (2 significant figures)
Rounding rules: If the digit to be dropped is ≥ 5, increase the preceding digit by 1; if < 5, leave it unchanged. For exactly 5, round to make the preceding digit even (banker's rounding).
4. Laws of Chemical Combination
4.1 Law of Conservation of Mass (Lavoisier, 1789)
In a chemical reaction, the total mass of the reactants equals the total mass of the products. Mass is neither created nor destroyed.
This law is the basis of balancing chemical equations. It holds exactly for closed systems and is modified in nuclear reactions where mass-energy equivalence (E = mc²) applies.
4.2 Law of Definite Proportions (Proust, 1799)
A given chemical compound always contains the same elements in the same proportion by mass, regardless of the source or method of preparation.
Example: Water (H₂O) always contains hydrogen and oxygen in a 1:8 ratio by mass, whether obtained from a river, a laboratory, or melting ice.
4.3 Law of Multiple Proportions (Dalton, 1803)
When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.
Example: Carbon and oxygen form CO (12:16 = 3:4) and CO₂ (12:32 = 3:8). The ratio of oxygen masses combining with 12 g of carbon is 16:32 = 1:2, which is a small whole-number ratio.
4.4 Gay Lussac's Law of Gaseous Volumes (1808)
Gases combine in simple whole-number ratios by volume, measured at the same temperature and pressure.
Example: 2 volumes of hydrogen combine with 1 volume of oxygen to form 2 volumes of water vapour.
4.5 Avogadro's Law (1811)
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
This law explains Gay Lussac's observation and connects the microscopic (molecules) to the macroscopic (volumes).
5. Dalton's Atomic Theory
John Dalton proposed the first systematic atomic theory in 1808:
| Postulate | Statement |
|---|---|
| 1 | Matter is made up of tiny, indivisible particles called atoms. |
| 2 | Atoms of the same element are identical in mass and properties. |
| 3 | Atoms of different elements differ in mass and properties. |
| 4 | Atoms combine in simple whole-number ratios to form compounds. |
| 5 | Atoms can neither be created nor destroyed in a chemical reaction. |
Limitations of Dalton's theory:
- Atoms are divisible (subatomic particles: electrons, protons, neutrons).
- Isotopes are atoms of the same element with different masses — contradicting postulate 2.
- Isobars are atoms of different elements with the same mass — contradicting postulate 3.
6. Atomic and Molecular Masses
Atomic Mass
Atomic mass is the mass of an atom expressed in atomic mass units (u), where 1 u = 1/12 the mass of a carbon-12 atom = 1.66054 × 10⁻²⁴ g.
Average Atomic Mass
For elements with isotopes, the average atomic mass is a weighted average:
Example: Chlorine has two isotopes: ³⁵Cl (75.77%, mass = 34.9689 u) and ³⁷Cl (24.23%, mass = 36.9659 u).
Molecular Mass
Molecular mass = sum of atomic masses of all atoms in a molecule.
| Molecule | Calculation | Molecular Mass |
|---|---|---|
| H₂O | 2(1.008) + 16.00 | 18.016 u |
| CO₂ | 12.01 + 2(16.00) | 44.01 u |
| NH₃ | 14.01 + 3(1.008) | 17.034 u |
Formula Mass
Used for ionic compounds (which do not form discrete molecules). It is the sum of atomic masses of all atoms in the formula unit.
- NaCl: 22.99 + 35.45 = 58.44 u
- CaCl₂: 40.08 + 2(35.45) = 110.98 u
7. Mole Concept and Molar Mass
The mole is the SI unit for amount of substance.
One mole of any substance contains exactly 6.02214076 × 10²³ elementary entities. This number is called Avogadro's number (Nₐ).
| Substance | 1 Mole Contains | Mass of 1 Mole |
|---|---|---|
| Carbon-12 atoms | 6.022 × 10²³ atoms | 12 g |
| Water molecules | 6.022 × 10²³ molecules | 18.016 g |
| Sodium ions | 6.022 × 10²³ ions | 22.99 g |
Key relationships:
Molar volume at STP (0°C, 1 atm): 1 mole of any ideal gas occupies 22.4 L.
8. Percentage Composition
The percentage composition of a compound gives the mass percentage of each element.
Example: Percentage composition of H₂O:
9. Empirical and Molecular Formulae
Definitions
| Term | Definition |
|---|---|
| Empirical formula | Simplest whole-number ratio of atoms in a compound |
| Molecular formula | Actual number of atoms of each element in a molecule of the compound |
Relationship:
Comparison Table
| Property | Empirical Formula | Molecular Formula |
|---|---|---|
| Meaning | Simplest ratio | Actual composition |
| Example | CH₂O | C₆H₁₂O₆ (glucose) |
| For benzene | CH | C₆H₆ |
| For hydrogen peroxide | HO | H₂O₂ |
| Same as molecular? | Yes, when n = 1 | Always |
| Mass | Empirical formula mass | n × empirical formula mass |
Steps to Determine Empirical Formula from Percentage Data
1. Convert each element's percentage to grams (assume 100 g sample, so % = mass in grams). 2. Divide each mass by the atomic mass to get moles. 3. Divide all mole values by the smallest mole value. 4. If the ratios are not whole numbers, multiply by the smallest integer to make them whole.
Worked Example: A compound contains 40.0% C, 6.7% H, and 53.3% O. Find its empirical formula.
| Element | % (assume = g) | Atomic mass | Moles | Divide by smallest | Simplest ratio |
|---|---|---|---|---|---|
| C | 40.0 | 12 | 3.33 | 3.33/3.33 = 1 | 1 |
| H | 6.7 | 1 | 6.7 | 6.7/3.33 = 2.01 | 2 |
| O | 53.3 | 16 | 3.33 | 3.33/3.33 = 1 | 1 |
Empirical formula = CH₂O
If the molar mass is given as 180 g mol⁻¹:
10. Stoichiometry and Stoichiometric Calculations
Stoichiometry is the calculation of reactants and products in a chemical reaction using balanced equations.
Steps for stoichiometric calculations:
1. Write the balanced chemical equation. 2. Identify what is given and what is asked. 3. Convert given quantity to moles. 4. Use the mole ratio from the balanced equation. 5. Convert the result to the desired unit.
10.1 Limiting Reagent
In most reactions, one reactant is present in lesser amount than required by the balanced equation. This reactant limits the amount of product formed and is called the limiting reagent.
How to identify the limiting reagent:
1. Calculate the moles of each reactant. 2. Divide each by its stoichiometric coefficient. 3. The reactant with the smaller value is the limiting reagent.
Worked Example: In the reaction 2H₂ + O₂ → 2H₂O, if 5 g of H₂ and 24 g of O₂ are available, which is the limiting reagent?
10.2 Reactions in Solutions
Molarity (M): Moles of solute per litre of solution.
Dilution formula:
11. Solved Numericals
Numerical 1: Mole Calculations
Q: Calculate the number of atoms in 0.5 moles of carbon.
Numerical 2: Molar Mass and Number of Molecules
Q: How many molecules are present in 4.4 g of CO₂?
Numerical 3: Percentage Composition
Q: Calculate the percentage composition of CaCO₃.
Numerical 4: Empirical Formula
Q: A compound has 69.9% Fe and 30.1% O. Find its empirical formula.
| Element | % | Atomic mass | Moles | Ratio |
|---|---|---|---|---|
| Fe | 69.9 | 55.85 | 1.25 | 1.25/1.25 = 1 |
| O | 30.1 | 16 | 1.88 | 1.88/1.25 = 1.5 |
Since 1.5 is not a whole number, multiply by 2:
Fe = 2, O = 3
Empirical formula = Fe₂O₃
Numerical 5: Limiting Reagent
Q: 3.0 g of H₂ reacts with 29.0 g of O₂ to form water. Identify the limiting reagent and calculate the amount of water produced.
12. Important Definitions
| Term | Definition |
|---|---|
| Mole | Amount of substance containing 6.022 × 10²³ elementary entities |
| Molar mass | Mass of 1 mole of a substance in grams |
| Molarity | Number of moles of solute per litre of solution (mol L⁻¹) |
| Avogadro's number | 6.022 × 10²³ — number of particles in 1 mole |
| Empirical formula | Simplest whole-number ratio of atoms in a compound |
| Molecular formula | Actual number of atoms of each element in a molecule |
| Limiting reagent | Reactant completely consumed first, limiting product formation |
| Stoichiometry | Quantitative relationship between reactants and products in a balanced equation |
| Significant figures | Digits in a measurement that are known with certainty plus one uncertain digit |
| STP | Standard Temperature and Pressure — 0°C (273.15 K) and 1 atm pressure |
13. Common Mistakes to Avoid
1. Confusing empirical and molecular formulae — The molecular formula is always a whole-number multiple of the empirical formula. CH₂O (empirical) and C₆H₁₂O₆ (molecular) are not the same. 2. Forgetting to convert grams to moles before using mole ratios from balanced equations. Always convert to moles first. 3. Not balancing the equation first — All stoichiometric calculations require a balanced equation. 4. Significant figure errors — Remember: leading zeros are not significant, trailing zeros after a decimal point are significant. 5. Confusing molar mass with molecular mass — Molar mass is in g mol⁻¹ (for a mole of substance), molecular mass is in u (for a single molecule). They are numerically equal but have different units. 6. Assuming the reactant with lesser mass is the limiting reagent — The limiting reagent is determined by comparing mole-to-coefficient ratios, not by comparing masses directly. 7. Not converting temperature to Kelvin — Always use Kelvin for gas law calculations; 0°C = 273.15 K.
14. Exam Tips
- For numerical problems: Always write the formula, substitute values with units, and state the answer with correct significant figures and units.
- For empirical formula problems: Show the table with columns for element, %, atomic mass, moles, and simplest ratio.
- For limiting reagent problems: Calculate moles, divide by stoichiometric coefficient, and identify the smaller value.
- For definition questions: Write the definition as given in NCERT, then give one example.
- For law-based questions: State the law, give the mathematical form (if any), and provide one illustration/example.
- Common 1-mark questions: Define mole, state Avogadro's number, define molarity, write the value of 1 u in grams.
- Common 3-mark questions: Calculate percentage composition, determine empirical formula, explain a law of chemical combination with example.
- Common 5-mark questions: Limiting reagent problems, empirical + molecular formula determination, stoichiometric calculations with two reactants.
15. Quick Revision Notes
- 1 mole = 6.022 × 10²³ particles = gram atomic/molecular mass = 22.4 L of gas at STP
- Molar mass is numerically equal to molecular mass; units differ (g mol⁻¹ vs u)
- Percentage composition is independent of sample size
- Empirical formula ≤ Molecular formula; n = molecular formula mass / empirical formula mass
- Five laws: Conservation of Mass, Definite Proportions, Multiple Proportions, Gay Lussac's Law, Avogadro's Law
- Dalton's atomic theory has 5 postulates; limitations include isotopes and isobars
- Significant figures: non-zero digits count; sandwiched zeros count; leading zeros do not count; trailing zeros after decimal count
- In addition/subtraction, match decimal places; in multiplication/division, match significant figures
- Limiting reagent determines maximum product; calculate mole/coefficient ratios
- Molarity = moles of solute / volume (L); dilution: M₁V₁ = M₂V₂
- 1 u = 1/12 mass of C-12 atom = 1.66054 × 10⁻²⁴ g
- Always balance equations before stoichiometric calculations
16. Practice Questions
1. Calculate the number of atoms in 0.1 moles of sulphur (S₈). 2. What is the molar mass of (a) NaOH (b) CuSO₄·5H₂O (c) CH₃COOH? 3. A compound contains 4.07% hydrogen, 24.27% carbon, and 71.65% chlorine. Its molar mass is 98.96 g. Determine its molecular formula. 4. Determine the empirical formula of an oxide of iron with 69.9% Fe and 30.1% O. 5. How many significant figures are in: (a) 0.0025 (b) 208 (c) 5005 (d) 0.0070? 6. In the reaction N₂ + 3H₂ → 2NH₃, if 28 g of N₂ reacts with 6 g of H₂, which is the limiting reagent? How much NH₃ is produced? 7. Calculate the molarity of a solution containing 5.85 g of NaCl in 500 mL of solution. 8. State and illustrate the Law of Multiple Proportions. 9. What volume at STP will 0.5 moles of oxygen gas occupy? 10. Round off to 2 significant figures: (a) 2.546 (b) 0.00347 (c) 14.75
Quick Answers to Practice Questions
1. Number of atoms = 0.1 × 6.022 × 10²³ × 8 = 4.816 × 10²³ atoms 2. (a) NaOH = 40 g mol⁻¹ (b) CuSO₄·5H₂O = 249.5 g mol⁻¹ (c) CH₃COOH = 60 g mol⁻¹ 3. Empirical formula = CH₂Cl; n = 98.96/49.48 = 2; Molecular formula = C₂H₄Cl₂ 4. Empirical formula = Fe₂O₃ (see worked example in section 11) 5. (a) 2 (b) 3 (c) 4 (d) 2 6. Moles of N₂ = 1, Moles of H₂ = 3; H₂ is the limiting reagent; 2 mol NH₃ = 34 g 7. Molarity = 0.1 / 0.5 = 0.2 M 8. See section 4.3; use CO and CO₂ as illustration 9. 0.5 × 22.4 = 11.2 L 10. (a) 2.5 (b) 0.0035 (c) 15
17. Final Revision Checklist
Before the exam, make sure you can:
- Define mole, Avogadro's number, molarity, and empirical/molecular formula
- Calculate molar mass, number of moles, and number of particles
- Determine percentage composition from a formula
- Derive empirical and molecular formulae from percentage data
- Identify the limiting reagent in a reaction
- Apply significant figure rules to calculations
- State all five laws of chemical combination with examples
- List the postulates and limitations of Dalton's atomic theory
- Convert between mass, moles, and number of particles fluently