Some Basic Concepts of ChemistryNEET Chemistry · Class 11 · NCERT Chapter 1

Introduction

Chemistry is the science of matter: what it is made of, how it changes, and how energy flows during those changes. Every drug you take, every food you digest, every material in your phone comes from chemical knowledge. For NEET 2027, this chapter is the mathematical foundation of all of physical and inorganic chemistry. Get these ideas right and numerical problems across the entire syllabus become easier.

You can expect 2 to 3 questions from this chapter in NEET each year. The most tested areas are the mole concept, stoichiometry (mole ratios in reactions), empirical vs molecular formula calculations, and occasionally the laws of chemical combination. Significant figures appear in some calculation-based questions.

Nature of Matter

Everything around you is matter: anything that has mass and occupies space. Chemistry classifies matter in two independent ways: by its physical state and by its composition.

Physical States of Matter

States can interconvert: melting (solid to liquid), vaporisation (liquid to gas), sublimation (solid directly to gas), and their reverses. These are physical changes: no new substance is formed.

Classification by Composition

Matter is first divided into pure substances (uniform composition throughout) and mixtures (two or more pure substances combined in variable proportions).

Pure substances are further divided into elements (cannot be broken into simpler substances by ordinary chemical means, e.g. hydrogen, iron, gold) and compounds(two or more elements combined in a fixed ratio by mass, e.g. water H₂O, salt NaCl).

Mixtures are divided into homogeneous mixtures (uniform composition throughout, like salt water or alloys; also called solutions) and heterogeneous mixtures(non-uniform; like sand and water, or milk).

Key Differences: Compound vs Mixture

Properties of Matter and Their Measurement

A property tells you something measurable or observable about a substance. Properties are classified as physical or chemical.

Physical Properties

Physical properties can be measured without changing the chemical identity of the substance. Examples: colour, density, melting point, boiling point, hardness, electrical conductivity, solubility, refractive index.

Chemical Properties

Chemical properties describe how a substance reacts with other substances or transforms into a different substance. Examples: flammability, acidity, basicity, reactivity with water. Measuring a chemical property always involves a chemical change (a new substance is formed).

Extensive vs Intensive Properties

An extensive property depends on the amount of matter present: mass, volume, heat content. An intensive property does not depend on amount: temperature, density, boiling point, colour. This distinction matters in thermodynamics questions.

SI Units and Prefixes

IUPAC recommends the International System of Units (SI) for all scientific measurements. It has seven base units; every other unit is derived from these.

Seven SI Base Units

Important Derived Units in Chemistry

In practice, chemists often use litre (L) for volume (1 L = 10⁻³ m³ = 1 dm³) and g mol⁻¹ for molar mass.

SI Prefixes

Prefixes scale a unit up or down. You must memorise the ones below; they appear in unit conversion questions.

Common conversions to know: 1 nm = 10⁻⁹ m = 10 Å (angstrom). 1 pm = 10⁻¹² m. 1 L = 1000 mL = 1000 cm³. 1 atm = 101325 Pa.

Temperature Scales

NCERT uses Celsius (°C) and Kelvin (K). The conversion is simple:

Kelvin is always used in gas law and thermodynamics calculations. Negative values are not possible in Kelvin (0 K is absolute zero).

Significant Figures

Every measurement has some uncertainty. Significant figures (sig figs) tell you how precisely a measurement was made. All certain digits plus the first uncertain digit are significant.

Rules for Counting Significant Figures

Arithmetic with Significant Figures

For addition and subtraction: the answer should have the same number of decimal places as the measurement with the fewest decimal places.

Example: 12.11 + 18.0 + 1.013 = 31.123, rounded to 31.1 (fewest decimal places = 1, from 18.0).

For multiplication and division: the answer should have the same number of significant figures as the measurement with the fewest sig figs.

Example: 2.5 × 1.25 = 3.125, rounded to 3.1 (fewest sig figs = 2, from 2.5).

Rounding Rules

If the digit to be dropped is less than 5, the preceding digit stays unchanged (round down). If it is greater than 5 or exactly 5 followed by non-zero digits, increase the preceding digit by 1 (round up). If it is exactly 5 with nothing after, round to the nearest even digit (round-half-to-even rule).

Laws of Chemical Combination

Before Dalton, experimentalists discovered five laws that describe how elements combine to form compounds. These laws guided Dalton to propose his atomic theory.

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

Matter can neither be created nor destroyed in a chemical reaction. The total mass of the reactants equals the total mass of the products.

Example: When 10 g of calcium carbonate decomposes, the total mass of calcium oxide and carbon dioxide produced is exactly 10 g. NEET questions sometimes ask you to identify the mass of an unidentified product using this law.

2. Law of Definite Proportions (Proust, 1799)

A pure chemical compound always contains the same elements combined in the same fixed ratio by mass, regardless of the source or method of preparation.

Example: Water (H₂O) always contains hydrogen and oxygen in the mass ratio 1:8. Whether you obtain water from rain, a river, or the laboratory, the ratio is always 1:8.

3. Law of Multiple Proportions (Dalton, 1803)

When two elements A and B form more than one compound, the masses of B that combine with a fixed mass of A are in simple whole-number ratios.

Example: In carbon monoxide (CO), 12 g of carbon combines with 16 g of oxygen. In carbon dioxide (CO₂), 12 g of carbon combines with 32 g of oxygen. The ratio of oxygen masses = 16:32 = 1:2. A simple whole-number ratio.

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

When gases react and produce gaseous products at the same temperature and pressure, the volumes of the reacting gases and the products are in simple whole-number ratios.

Example: 1 volume of hydrogen + 1 volume of chlorine gives 2 volumes of hydrogen chloride (1:1:2 ratio). 2 volumes of hydrogen + 1 volume of oxygen gives 2 volumes of steam (2:1:2 ratio). This law applies only to gases measured at the same conditions.

5. Avogadro's Law (Avogadro, 1811)

Equal volumes of all gases, at the same temperature and pressure, contain equal numbers of molecules.

This law explained Gay-Lussac's observations. It told chemists that molecules (not just atoms) are the fundamental units of gases. It also confirmed that hydrogen, oxygen, and nitrogen exist as diatomic molecules (H₂, O₂, N₂).

Dalton's Atomic Theory (1808)

John Dalton proposed the first scientific atomic theory to explain the laws above. The main postulates are:

1. All matter is made of indivisible, indestructible particles called atoms.
2. All atoms of a given element are identical in mass and chemical properties. Atoms of different elements differ in mass and properties.
3. Atoms cannot be created or destroyed in a chemical reaction (explains conservation of mass).
4. Atoms combine in simple whole-number ratios to form compound atoms (what we now call molecules). This explains the law of definite proportions.
5. When two elements form more than one compound, the different numbers of atoms combine in simple whole-number ratios (explains the law of multiple proportions).

Limitations of Dalton's Atomic Theory

Later discoveries showed that Dalton's theory was not completely correct. The key limitations are:

• Atoms are not indivisible. They contain sub-atomic particles: electrons, protons, and neutrons.
• Atoms of the same element can have different masses (isotopes, e.g. ¹H and ²H). So not all atoms of an element are identical.
• Atoms can be created and destroyed in nuclear reactions (though not in ordinary chemical reactions).
• The law of definite proportions does not hold perfectly for non-stoichiometric compounds (defect structures in crystals).

Atomic and Molecular Masses

Atomic Mass Unit (amu or u)

Atoms are too small to weigh directly. Chemists use a relative scale. One atomic mass unit (u or amu) is defined as exactly 1/12 of the mass of one carbon-12 atom.

The atomic mass of an element on this scale tells you how many times heavier one atom of that element is compared to 1/12 of a carbon-12 atom. For example, the atomic mass of oxygen is 16 u, meaning one oxygen atom is 16 times heavier than 1/12 of a carbon-12 atom.

Average Atomic Mass

Most elements exist as a mixture of isotopes. The atomic mass you see in the periodic table is the weighted average of all naturally occurring isotopes.

Example: Chlorine has two isotopes. ³⁵Cl (75.77% abundance, mass 34.97 u) and ³⁷Cl (24.23% abundance, mass 36.97 u).

Average atomic mass = (0.7577 × 34.97) + (0.2423 × 36.97) = 26.50 + 8.96 = 35.46 u.

Molecular Mass

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

Example: Molecular mass of H₂SO₄ = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 u.

Formula Mass

Ionic compounds do not exist as discrete molecules; they form lattice structures. For these, we use formula mass instead of molecular mass. The formula mass is the sum of atomic masses of all ions in one formula unit.

Example: Formula mass of NaCl = 23 + 35.5 = 58.5 u. Formula mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 u.

Mole Concept

A mole is the SI unit for amount of substance. It is defined as the amount of a substance that contains exactly 6.022 × 10²³ elementary entities (atoms, molecules, ions, electrons, formula units, or any specified particle). This number is called Avogadro's number($N_A$).

Molar Mass

The molar mass of a substance is the mass of one mole of that substance, expressed in g mol⁻¹. Numerically, it equals the atomic mass (for elements) or molecular mass (for compounds) in grams.

Examples:

• Molar mass of Na = 23 g mol⁻¹
• Molar mass of O₂ = 32 g mol⁻¹
• Molar mass of H₂O = 18 g mol⁻¹
• Molar mass of NaCl = 58.5 g mol⁻¹
• Molar mass of H₂SO₄ = 98 g mol⁻¹
• Molar mass of glucose (C₆H₁₂O₆) = 180 g mol⁻¹

The Mole Triangle

Three quantities connect through the mole: mass (in grams), number of moles, and number of particles.

For gases at STP (Standard Temperature and Pressure: 0°C, 1 atm), 1 mole of any ideal gas occupies 22.4 L (molar volume). This is used frequently in numerical problems.

I know the

Percentage Composition

The percentage composition of a compound tells you what fraction of its total mass comes from each element. It is calculated from the molecular formula.

Example: Find the percentage of nitrogen in ammonia (NH₃). Molar mass of NH₃ = 14 + 3(1) = 17 g mol⁻¹. % N = (14/17) × 100 = 82.35%. % H = (3/17) × 100 = 17.65%. These add up to 100%.

Empirical and Molecular Formula

Empirical Formula

The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. It is derived from percentage composition or combustion data.

Steps to find empirical formula from percentage composition:

1. Divide the percentage of each element by its atomic mass. This gives the mole ratio (relative number of moles of each atom).
2. Divide all mole ratios by the smallest mole ratio to get a provisional ratio.
3. If the ratios are not whole numbers, multiply all by the smallest integer that makes them whole.
4. Write the empirical formula with subscripts as the whole-number ratios.

Molecular Formula

The molecular formula gives the actual number of atoms of each element in one molecule. It is a whole-number multiple of the empirical formula.

Example: A compound has empirical formula CH₂O (empirical mass = 12 + 2 + 16 = 30 g mol⁻¹). If the molar mass is 180 g mol⁻¹, then n = 180/30 = 6. Molecular formula = C₆H₁₂O₆(glucose).

Enter % composition

Stoichiometry

Stoichiometry is the study of quantitative relationships in chemical reactions. It uses the balanced chemical equation as a map of mole ratios between reactants and products.

Reading a Balanced Equation

Consider the combustion of methane:

The coefficients (1, 2, 1, 2) are the mole ratios. You read this as: 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O.

Types of Stoichiometry Problems

Mole-to-mole: Use the mole ratio directly from the equation.

Mass-to-mass: Convert given mass to moles, use mole ratio, convert product moles to mass.

Mass-to-volume: Same as above but convert product moles to volume using 22.4 L mol⁻¹ at STP.

General Steps for Stoichiometry

1. Write the balanced chemical equation.
2. Convert given quantity (mass/volume/number of particles) to moles.
3. Use the mole ratio from the equation to find moles of the required substance.
4. Convert moles of required substance to the desired unit (mass, volume, or number of particles).

Limiting Reagent

In real reactions, reactants are rarely present in exact stoichiometric proportions. The limiting reagent (or limiting reactant) is the reactant that is completely consumed first. It determines how much product can form. The other reactant is in excess.

How to Identify the Limiting Reagent

1. Convert the given masses of all reactants to moles.
2. Divide each reactant's moles by its stoichiometric coefficient in the balanced equation. This gives the "mole ratio value".
3. The reactant with the smallest mole ratio value is the limiting reagent.
4. Use the moles of the limiting reagent to calculate the amount of product formed (theoretical yield).

Example: Reaction is N₂ + 3H₂ → 2NH₃. You have 28 g of N₂(molar mass 28, so 1 mol) and 6 g of H₂ (molar mass 2, so 3 mol). Mole ratio values: N₂= 1/1 = 1, H₂ = 3/3 = 1. Both are exactly equal. So neither is limiting; they are in exact stoichiometric ratio and both will be fully consumed.

Change the problem: use 28 g N₂ and 3 g H₂ (= 1.5 mol). Mole ratio values: N₂ = 1/1 = 1, H₂ = 1.5/3 = 0.5. H₂ has the smaller value, so H₂ is the limiting reagent. Moles of NH₃ formed = 1.5 × (2/3) = 1 mol = 17 g.

Percentage Yield

In practice, the actual amount of product obtained is often less than the theoretical yield due to incomplete reactions, side reactions, or losses during collection.

Adjust reactant masses

Molarity and Solutions

Many chemical reactions are carried out in solution. The concentration of a solution tells you how much solute is dissolved in a given volume of solution.

Molarity (M)

Molarity is the most common concentration unit in NEET chemistry. It is defined as the number of moles of solute dissolved per litre of solution.

Or equivalently:

Example: 5.85 g of NaCl (molar mass 58.5 g mol⁻¹) is dissolved in 500 mL (0.5 L) of solution. Moles of NaCl = 5.85/58.5 = 0.1 mol. Molarity = 0.1/0.5 = 0.2 M.

Dilution

When you dilute a solution, the amount of solute stays the same but the volume increases. The useful dilution formula is:

where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume after dilution.

Mole Fraction

The mole fraction of component A in a mixture is the ratio of moles of A to the total moles of all components.

The sum of mole fractions of all components always equals 1.

Mass Percent (Weight Percent)

Parts per Million (ppm)

For very dilute solutions (trace contaminants, pollutants), concentration is expressed in ppm.

Worked NEET Problems

Summary Cheat Sheet