Chemistry is often called the central science because it bridges physics with biology, environmental science, and materials technology. But before mastering chemical bonding, thermodynamics, or organic reaction mechanisms, one must master the quantitative language of chemistry — atoms, molecules, moles, stoichiometry, and solution concentrations. Some Basic Concepts of Chemistry forms Chapter 1 of the CBSE Class 11 Chemistry syllabus and is a high-yield topic for JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers the five laws of chemical combination, Dalton's atomic theory, atomic mass unit (u), the mole concept, empirical and molecular formula calculations, stoichiometry, limiting reagent identification, solution concentration terms (molarity, molality, normality, mole fraction), and five step-by-step solved entrance exam numericals.
- 1. Five Fundamental Laws of Chemical Combination
- 2. Dalton's Atomic Theory, Atomic Mass Unit (u) & Molar Mass
- 3. The Mole Concept & Avogadro's Constant (NA)
- 4. Empirical Formula & Molecular Formula Determination
- 5. Stoichiometry & Limiting Reagent Algorithm
- 6. Concentration Terms of Solutions (Molarity, Molality, Normality)
- 7. Molarity vs Molality: Temperature Dependence Comparison
- 8. Solved Entrance Exam Numerical Problems (JEE / NEET)
- 9. Frequently Asked Questions (FAQ)
1. Five Fundamental Laws of Chemical Combination
Elements combine with one another according to five fundamental laws discovered during the 18th and 19th centuries:
| Law Name | Discovered By | Core Principle & Statement | Classic Example |
|---|---|---|---|
| 1. Law of Conservation of Mass | Antoine Lavoisier (1789) | In any physical change or chemical reaction, matter is neither created nor destroyed. Total mass of reactants equals total mass of products. | 2H₂ (4g) + O₂ (32g) → 2H₂O (36g) |
| 2. Law of Definite Proportions | Joseph Proust (1799) | A given chemical compound always contains exactly the same elements combined together in the same fixed proportion by mass, regardless of source. | Pure water (H₂O) always contains Hydrogen and Oxygen in 1 : 8 mass ratio (whether from rain, river, or lab). |
| 3. Law of Multiple Proportions | John Dalton (1803) | If 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 small whole-number ratios. | Carbon + Oxygen forms CO (12g C : 16g O) and CO₂ (12g C : 32g O). Ratio of Oxygen masses = 16 : 32 = 1 : 2. |
| 4. Gay Lussac's Law of Gaseous Volumes | Joseph Louis Gay-Lussac (1808) | When gases react together, they do so in volumes which bear a simple whole-number ratio to one another and to products (if gaseous) at constant T and P. | 1 vol H₂ + 1 vol Cl₂ → 2 vol HCl (Volume ratio 1 : 1 : 2). |
| 5. Avogadro's Law | Amedeo Avogadro (1811) | Equal volumes of all gases under identical conditions of temperature and pressure contain an equal number of molecules (V ∝ n). | 22.4 L of O₂, N₂, or CO₂ at STP all contain 6.022 × 10^23 molecules. |
2. Dalton's Atomic Theory, Atomic Mass Unit (u) & Molar Mass
In 1808, John Dalton published his Atomic Theory, proposing that matter consists of indivisible atoms.
1 u = 1.66056 × 10^-24 g = 1.66056 × 10^-27 kg
• Average Atomic Mass: Sum of (Isotopic Mass × Fractional Abundance) for elements existing as natural isotopes (e.g. Chlorine = 35.5 u).
• Molar Mass (M): The mass of one mole of a substance in grams (g/mol). Numerically equal to atomic/molecular mass in u.
3. The Mole Concept & Avogadro's Constant (NA)
The Mole (symbol: mol) is the SI base unit for amount of substance. One mole contains exactly 6.02214076 × 10^23 elementary entities (atoms, molecules, ions, or electrons). This number is called Avogadro's Constant (N_A).
n = Given Mass in grams (w) / Molar Mass (M)
2. In terms of Particles:
n = Number of Particles (N) / N_A (6.022 × 10^23)
3. In terms of Gas Volume at STP (0°C, 1 atm):
n = Volume of Gas in Liters (V) / 22.4 L
• STP (IUPAC modern standard): T = 273.15 K, P = 1 bar (10^5 Pa). Molar volume = 22.7 L/mol.
Note: For CBSE board & competitive exams, 22.4 L/mol is conventionally used unless specified otherwise.
4. Empirical Formula & Molecular Formula Determination
• Molecular Formula: Shows the actual number of atoms of each element present in one molecule of the compound (e.g., C₆H₁₂O₆ for Glucose).
Molecular Formula = n × (Empirical Formula)
where n = Molecular Mass / Empirical Formula Mass
Step-by-Step Procedure to Calculate Empirical Formula:
- Convert percentage composition of each element into grams (assume 100 g sample).
- Divide grams of each element by its atomic mass to get moles of atoms.
- Divide each mole value by the smallest mole value to get the simplest mole ratio.
- If ratios are not whole numbers, multiply all ratios by a small integer (2, 3, or 4) to obtain simplest whole numbers.
- Write the symbols of elements with respective whole-number subscripts.
5. Stoichiometry & Limiting Reagent Algorithm
Stoichiometry deals with the calculation of masses (and volumes) of reactants and products involved in a balanced chemical equation.
Algorithm to Identify Limiting Reagent:
1. Write the balanced chemical reaction: aA + bB → cC + dD.
2. Calculate initial moles of each reactant: nA and nB.
3. Compute the ratio: nA / a and nB / b (moles divided by stoichiometric coefficient).
4. The reactant with the SMALLER ratio is the Limiting Reagent! All product calculations must be based strictly on this limiting reagent.
6. Concentration Terms of Solutions (Molarity, Molality, Normality)
A solution consists of a Solute (B) dissolved in a Solvent (A).
| Concentration Term | Mathematical Formula | SI / Conventional Unit | Temperature Dependence |
|---|---|---|---|
| Mass Percentage (% w/w) | % w/w = (Mass of Solute wB / Total Mass of Solution) × 100 | Unitless (%) | Independent of Temperature |
| Molarity (M) | M = Moles Solute nB / Volume Solution in Liters = (wB × 1000) / (MB × V_mL) | mol/L or M | Temperature DEPENDENT (Volume expands with temperature!) |
| Molality (m) | m = Moles Solute nB / Mass Solvent in kg = (wB × 1000) / (MB × wA_g) | mol/kg or m | Temperature INDEPENDENT (Mass does not change with temp) |
| Normality (N) | N = Gram Equivalents of Solute / Volume Solution in Liters = M × n-factor | eq/L or N | Temperature DEPENDENT |
| Mole Fraction (χ) | χA = nA / (nA + nB), χB = nB / (nA + nB), (χA + χB = 1) | Unitless | Temperature INDEPENDENT |
7. Molarity vs Molality: Temperature Dependence Comparison
• Molality (m) depends on the mass of solvent. Since mass is invariant to temperature, molality remains strictly constant across all temperatures.
Hence, molality is preferred when studying colligative properties (boiling point elevation, freezing point depression)!
• Acids (Basicity): n-factor = number of replaceable H+ ions (e.g. HCl = 1, H₂SO₄ = 2, H₃PO₄ = 3).
• Bases (Acidity): n-factor = number of replaceable OH¯ ions (e.g. NaOH = 1, Ca(OH)₂ = 2).
• Salts: n-factor = total positive charge on cations (e.g. Na₂CO₃ = 2, Al₂(SO₄)₃ = 6).
8. Solved Entrance Exam Numerical Problems (JEE / NEET)
Moles of NaOH nB = 4 / 40 = 0.1 mol.
Molarity M = (nB × 1000) / V_mL = (0.1 × 1000) / 250 = 0.4 M.
Moles of N2 = 50,000 g / 28 g/mol = 1785.7 mol. Ratio n/a = 1785.7 / 1 = 1785.7.
Moles of H2 = 10,000 g / 2.016 g/mol = 4960.3 mol. Ratio n/b = 4960.3 / 3 = 1653.4.
Since 1653.4 < 1785.7, H2 is the Limiting Reagent!
From stoichiometry: 3 moles H2 produce 2 moles NH3.
Moles of NH3 formed = (2/3) × 4960.3 = 3306.87 mol.
Mass of NH3 = 3306.87 mol × 17.03 g/mol = 56,316 g = 56.3 kg.
2. Moles of H = 6.7 / 1 = 6.70. Ratio = 6.70 / 3.33 = 2.
3. Moles of O = 53.3 / 16 = 3.33. Ratio = 3.33 / 3.33 = 1.
• Empirical Formula: CH₂O (Empirical Mass = 12 + 2(1) + 16 = 30 g/mol).
• n = Molar Mass / Empirical Mass = 180 / 30 = 6.
• Molecular Formula: (CH₂O)6 = C_6H_12O_6 (Glucose).
Moles of ethanol nB = 2.5 mol. Mass of ethanol wB = 2.5 × 46 = 115 g.
Mass of 1000 mL solution = V × d = 1000 × 1.02 = 1020 g.
Mass of solvent (water) wA = Mass solution - wB = 1020 - 115 = 905 g = 0.905 kg.
Molality m = nB / wA_kg = 2.5 / 0.905 = 2.76 m.
• For H2SO4: n-factor = 2 ⇒ N1 = M1 × 2 = 0.1 × 2 = 0.2 N.
• For NaOH: n-factor = 1 ⇒ N2 = M2 × 1 = 0.2 × 1 = 0.2 N.
N1 V1 = N2 V2 ⇒ 0.2 × V1 = 0.2 × 50 ⇒ V1 = 50 mL.
Explore Related CBSE Chemistry & Lab Instrument Guides
9. Frequently Asked Questions (FAQ)
1 mole is the amount of substance containing 6.02214076 × 10^23 elementary entities (Avogadro Constant NA). It equals the molar mass of the substance in grams, or 22.4 L of an ideal gas at STP (0°C, 1 atm).
Molarity (M) depends on liquid volume, which expands or contracts with temperature. Molality (m) depends on the mass of solvent, which is temperature-independent and remains constant regardless of temperature fluctuations.
Divide the initial moles of each reactant by its respective stoichiometric coefficient in the balanced equation (n / coefficient). The reactant with the smallest ratio is the Limiting Reagent!
The Empirical Formula gives the simplest whole-number ratio of atoms in a compound (e.g. CH₂O), while the Molecular Formula shows the actual number of atoms in a molecule (e.g. C₆H₁₂O₆). Molecular Formula = n × (Empirical Formula).
Normality (N) = Molarity (M) × n-factor. For acids, n-factor is basicity (H+ count); for bases, n-factor is acidity (OH- count); for salts, n-factor is total cation charge.
Equip Your Class 11 Chemistry Lab with Ambala Volumetric Glassware & Balances
AJKANT Overseas manufactures and supplies complete CBSE Class 11 chemistry lab equipment including Class A volumetric flasks, burettes, pipettes, electronic analytical balances, magnetic stirrers, and pure borosilicate 3.3 glassware. Factory-direct from Ambala, India. Trusted by educational institutions across 28 states and 25+ countries.
Request Chemistry Lab Equipment Quote →