CBSE Class 11 Chemistry Revision Notes Chapter 1 Some Basic Concepts of Chemistry

CBSE Class 11 Chemistry Revision Notes Chapter 1 explain Some Basic Concepts of Chemistry through matter, measurement, mole concept, formulas and stoichiometry.

For CBSE 2026 Chemistry, Some Basic Concepts of Chemistry builds the calculation base for laws, molar mass, concentration and balanced equations.

How can 2 g of hydrogen gas contain about 6.022 × 10²³ molecules, while one hydrogen atom has a mass close to 1.6736 × 10⁻²⁴ g? The NCERT Class 11 Chemistry chapter Some Basic Concepts of Chemistry starts from this scale problem and builds the language needed for chemical calculations. Atoms and molecules are too small to count one by one, so chemistry uses units, scientific notation, significant figures and the mole concept.

The chapter connects daily substances with measurable quantities. Matter is classified into elements, compounds and mixtures, physical properties are measured with SI units, and chemical reactions follow fixed laws of combination. The same base helps students calculate molecular mass, percentage composition, empirical formula, limiting reagent, molarity and molality in CBSE 2026 Chemistry.

Key Takeaways

• Avogadro constant: One mole contains exactly 6.02214076 × 10²³ elementary entities.
• Carbon-12 standard: One atomic mass unit equals one-twelfth of the mass of one carbon-12 atom.
• Mass conservation: Antoine Lavoisier stated in 1789 that matter is neither created nor destroyed.
• Molarity relation: Dilution calculations use M1 × V1 = M2 × V2 when concentration and volume change.

CBSE Class 11 Chemistry Revision Notes Chapter 1 Structure 2026

Chemistry, Matter and Measurement

Chemistry studies the composition, structure, properties and reactions of material substances. The NCERT Class 11 Chemistry chapter Some Basic Concepts of Chemistry also calls chemistry the science of atoms and molecules.

The chapter begins with matter because every chemical calculation uses mass, volume, particles or formula units. These ideas prepare students for mole-based numericals and balanced equations.

Importance of chemistry

Chemistry supports industries that produce fertilisers, acids, alkalis, dyes, polymers, drugs, soaps, detergents, metals and alloys. It also helps in healthcare, food production and environmental protection.

Examples from the NCERT chapter include cisplatin and taxol in cancer therapy. AZT is mentioned as a drug used for helping AIDS patients.

Nature of matter

Matter is anything that has mass and occupies space. Books, air, water, living beings and laboratory chemicals are made of matter.

Matter exists as solid, liquid or gas. These states differ because particles have different arrangements and freedom of movement.

Classification of matter

Matter is classified into mixtures and pure substances. Mixtures contain two or more pure substances in any ratio.

Pure substances are further divided into elements and compounds. Elements contain one type of atom, while compounds contain atoms of different elements in a fixed ratio.

Properties of Matter and SI Units

Physical properties can be measured without changing the identity of a substance. Chemical properties require a chemical change to be observed.

Class 11 chemistry chapter 1 some basic concepts of chemistry notes use SI units because chemical measurements need common standards. Length, mass, time, temperature and amount of substance become important in calculations.

Physical and chemical properties

Physical properties include colour, odour, melting point, boiling point and density. These can be measured without changing the substance.

Chemical properties include composition, combustibility and reactivity with acids or bases. These properties are studied through chemical reactions.

Seven SI base units

The SI system has seven base units. These units represent fundamental scientific quantities.

Other SI base units include second, ampere, kelvin and candela. Derived units such as density and volume are built from base units.

Mass, volume and density

Mass is the amount of matter present in a substance. Weight is the force exerted by gravity.

Volume is the space occupied by a substance. Density relates mass and volume.

Formula:

Density = Mass / Volume

SI unit of density = kg m⁻³

Common chemistry unit = g cm⁻³

Temperature conversion

Temperature is commonly measured in Celsius, Fahrenheit and Kelvin. Kelvin is the SI unit of thermodynamic temperature.

Formula:

K = °C + 273.15

Negative temperatures are possible on the Celsius scale. Negative temperatures are not possible on the Kelvin scale.

Scientific Notation and Significant Figures

Scientific notation helps write very large and very small numbers used in chemistry. A number is written as N × 10ⁿ, where N lies between 1 and 9.999.

This is important because atoms and molecules have extremely small masses. Their counts in ordinary samples are extremely large.

Scientific notation

Scientific notation writes numbers using powers of 10.

Examples:

232.508 = 2.32508 × 10²

0.00016 = 1.6 × 10⁻⁴

This makes multiplication, division, addition and subtraction easier in chemistry numericals.

Significant figures

Significant figures are meaningful digits in a measured value. They include certain digits and one uncertain digit.

Examples:

285 cm has 3 significant figures.

0.0052 has 2 significant figures.

2.005 has 4 significant figures.

0.200 g has 3 significant figures.

Precision and accuracy

Precision means repeated measurements are close to each other. Accuracy means a measured value is close to the true value.

Example:

If the true value is 2.00 g, values 2.01 g and 1.99 g are both precise and accurate. Values 1.95 g and 1.93 g are precise but not accurate.

Dimensional Analysis and Unit Conversion

Dimensional analysis converts a quantity from one unit to another using unit factors. It is also called the factor label method or unit factor method.

This method helps students avoid unit errors in stoichiometry class 11 numericals. Units are cancelled like numbers during calculation.

Unit factor method

A unit factor is formed from two equivalent quantities.

Example:

1 inch = 2.54 cm

So, 2.54 cm / 1 inch = 1

If length = 3 inch,

3 inch × 2.54 cm / 1 inch = 7.62 cm

Time conversion example

Convert 2 days into seconds.

1 day = 24 h

1 h = 60 min

1 min = 60 s

2 days = 2 × 24 × 60 × 60 s

2 days = 172800 s

Laws of Chemical Combination

Chemical combination follows five basic laws. These laws explain how elements combine to form compounds in fixed mass or volume relationships.

In CBSE Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry, law-based questions need exact statements and examples. These laws also support Dalton’s atomic theory.

Law of conservation of mass

The law of conservation of mass states that matter can neither be created nor destroyed. Lavoisier proposed this law in 1789.

In a chemical reaction, the total mass of reactants equals the total mass of products. This law supports balanced chemical equations.

Law of definite proportions

The law of definite proportions states that a given compound always contains the same elements in the same proportion by mass.

Example:

Water always contains hydrogen and oxygen in a fixed mass ratio. The source of water does not change this composition.

Law of multiple proportions

The law of multiple proportions states that when two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in small whole-number ratios.

Example:

Hydrogen and oxygen form water and hydrogen peroxide.

In water:

2 g hydrogen combines with 16 g oxygen.

In hydrogen peroxide:

2 g hydrogen combines with 32 g oxygen.

Oxygen mass ratio = 16 : 32 = 1 : 2

Gay Lussac’s law of gaseous volumes

Gay Lussac’s law states that gases combine or form products in simple whole-number volume ratios at the same temperature and pressure.

Example:

Hydrogen + Oxygen → Water vapour

100 mL hydrogen combines with 50 mL oxygen to form 100 mL water vapour.

Volume ratio = 2 : 1 : 2

Avogadro’s law

Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

This law explained Gay Lussac’s volume relationship. It also helped distinguish atoms from molecules.

Dalton’s Atomic Theory

Dalton’s atomic theory explained the laws of chemical combination. It was published in 1808 in A New System of Chemical Philosophy.

The theory treated atoms as the basic particles involved in chemical reactions. It supported fixed ratios in compounds and conservation of atoms.

Main postulates

Dalton’s atomic theory states:

1. Matter consists of indivisible atoms.
2. Atoms of the same element have identical properties and mass.
3. Atoms of different elements differ in mass.
4. Compounds form when atoms combine in fixed ratios.
5. Chemical reactions involve reorganisation of atoms.

Limitation of Dalton’s theory

Dalton’s theory could explain laws of chemical combination. It could not explain Gay Lussac’s law of gaseous volumes.

It also did not explain why atoms combine. Later theories explained bonding and molecular structure.

Atomic Mass, Molecular Mass and Formula Mass

Atomic and molecular masses help compare tiny particles using a practical scale. The present atomic mass scale uses carbon-12 as the standard.

Some basic concepts of chemistry class 11 revision notes often test these ideas through formula mass, molecular mass and molar mass calculations.

Atomic mass

Atomic mass is the mass of an atom expressed relative to carbon-12. One atomic mass unit is exactly one-twelfth of the mass of one carbon-12 atom.

Formula:

1 amu = 1.66056 × 10⁻²⁴ g

At present, amu is replaced by u, called unified mass.

Average atomic mass

Average atomic mass considers isotopes and their relative abundance. The atomic masses in the periodic table are average atomic masses.

Example:

Carbon has isotopes such as ¹²C, ¹³C and ¹⁴C. Their relative abundances are used to calculate average atomic mass.

Molecular mass

Molecular mass is the sum of atomic masses of atoms present in a molecule.

Example:

Molecular mass of methane, CH₄

= 12.011 u + 4 × 1.008 u

= 16.043 u

Molecular mass of water, H₂O

= 2 × 1.008 u + 16.00 u

= 18.02 u

Formula mass

Formula mass is used for ionic compounds that do not exist as discrete molecules.

Example:

Formula mass of NaCl

= Atomic mass of Na + Atomic mass of Cl

= 23.0 u + 35.5 u

= 58.5 u

Mole Concept and Molar Mass

The mole is the SI unit for amount of substance. In CBSE Class 11 Chemistry Chapter 1, the mole concept connects particle count with measurable mass through Avogadro constant and molar mass.

Mole concept class 11 chemistry is central to numericals because it connects mass, particles and molar mass. It allows chemists to count atoms and molecules through measurable mass.

Definition of mole

One mole contains exactly 6.02214076 × 10²³ elementary entities. These entities may be atoms, molecules, ions, electrons or formula units.

Examples:

1 mol hydrogen atoms = 6.022 × 10²³ hydrogen atoms

1 mol water molecules = 6.022 × 10²³ water molecules

1 mol NaCl = 6.022 × 10²³ formula units of NaCl

Molar mass

Molar mass is the mass of one mole of a substance in grams. It is numerically equal to atomic, molecular or formula mass in u.

Examples:

Molar mass of water = 18.02 g mol⁻¹

Molar mass of sodium chloride = 58.5 g mol⁻¹

Mole calculation formulas

Useful formulas:

Number of moles = Given mass / Molar mass

Given mass = Number of moles × Molar mass

Number of particles = Number of moles × Avogadro constant

Number of moles = Number of particles / Avogadro constant

Percentage Composition and Chemical Formula

Percentage composition gives the mass percentage of each element in a compound. It helps check purity and find empirical formula.

Empirical and molecular formula class 11 questions usually begin with percentage composition. The steps depend on converting mass percentage into mole ratio.

Mass percentage

Formula:

Mass % of an element = Mass of that element in compound × 100 / Molar mass of compound

Example:

Water, H₂O, has molar mass 18.02 g mol⁻¹.

Mass % of hydrogen = 2.016 × 100 / 18.02

= 11.18%

Mass % of oxygen = 16.00 × 100 / 18.02

= 88.79%

Empirical formula

Empirical formula shows the simplest whole-number ratio of atoms in a compound.

Steps:

1. Write mass percentage of each element.
2. Divide each percentage by atomic mass.
3. Divide all mole values by the smallest mole value.
4. Convert ratios into whole numbers.
5. Write the empirical formula.

Molecular formula

Molecular formula shows the actual number of atoms of each element in a molecule.

Formula:

n = Molar mass / Empirical formula mass

Molecular formula = Empirical formula × n

Example:

If empirical formula = CH₂Cl and n = 2,

Molecular formula = C₂H₄Cl₂

Stoichiometry and Balanced Equations

Stoichiometry uses balanced chemical equations to calculate reactant and product amounts. In Some Basic Concepts of Chemistry, stoichiometric calculations connect mole ratio, limiting reagent and mass conversion.

A balanced equation gives mole ratios between reactants and products. These ratios are used to calculate mass, moles or volume involved in a reaction.

Information from a balanced equation

A balanced equation shows the relative number of moles of reactants and products.

Example:

N₂(g) + 3H₂(g) → 2NH₃(g)

This equation means:

1 mol N₂ reacts with 3 mol H₂.

2 mol NH₃ are formed.

Limiting reagent

Limiting reagent is the reactant that gets consumed first. It limits the amount of product formed.

Example:

In N₂(g) + 3H₂(g) → 2NH₃(g), if hydrogen is less than required, hydrogen becomes the limiting reagent. Product formation depends on available hydrogen.

Stoichiometric calculation steps

Step 1: Write the balanced chemical equation.

Step 2: Convert given mass into moles.

Step 3: Use mole ratio from the equation.

Step 4: Convert required moles into mass if needed.

Step 5: Check limiting reagent when two reactants are given.

Concentration of Solutions

Concentration tells how much solute is present in a given amount of solution or solvent. In CBSE Class 11 Chemistry Chapter 1, NCERT explains mass per cent, mole fraction, molarity and molality as four common ways to express concentration.

Most laboratory reactions happen in solutions. These concentration terms help calculate solute amount, solvent mass and solution volume.

Mass per cent

Mass per cent shows solute mass as a percentage of solution mass.

Formula:

Mass % = Mass of solute × 100 / Mass of solution

Example:

If 2 g of A is added to 18 g water,

Mass of solution = 20 g

Mass % of A = 2 × 100 / 20

= 10%

Mole fraction

Mole fraction is the ratio of moles of one component to total moles of solution.

For solute A and solvent B:

Mole fraction of A = nA / (nA + nB)

Mole fraction of B = nB / (nA + nB)

Molarity

Molarity is the number of moles of solute present in 1 litre of solution. It is denoted by M.

Formula:

Molarity = Number of moles of solute / Volume of solution in litres

Dilution formula:

M1 × V1 = M2 × V2

Molarity changes with temperature because volume changes with temperature.

Molality

Molality is the number of moles of solute present in 1 kg of solvent. It is denoted by m.

Formula:

Molality = Number of moles of solute / Mass of solvent in kg

Molality does not change with temperature because mass remains unchanged.

Important Formulas in Some Basic Concepts of Chemistry

The main formulas in CBSE Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry connect mass, moles, particles, concentration and chemical formula. These formulas support most numerical questions from mole concept, molarity, molality and stoichiometry.

More formulas:

1. Density = Mass / Volume
2. K = °C + 273.15
3. Mass % = Mass of element × 100 / Molar mass of compound
4. Number of particles = Moles × 6.022 × 10²³
5. Molecular formula = Empirical formula × n
6. M1 × V1 = M2 × V2
7. Molality = Moles of solute / Mass of solvent in kg

Important Terms in Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry uses fixed terms in measurement, mole concept, chemical laws and stoichiometry. These terms help students answer one-mark questions and calculation-based questions in CBSE Class 11 Chemistry Chapter 1.

Matter

Matter is anything that has mass and occupies space.

Pure substance

A pure substance has particles of the same chemical nature and fixed composition.

Mole

Mole is the SI unit of amount of substance and contains 6.02214076 × 10²³ entities.

Molar mass

Molar mass is the mass of one mole of a substance in grams.

Empirical formula

Empirical formula is the simplest whole-number ratio of atoms in a compound.

Molecular formula

Molecular formula shows the exact number of atoms in one molecule of a compound.

Limiting reagent

Limiting reagent is the reactant that is consumed first and limits product formation.

Molarity

Molarity is moles of solute per litre of solution.

Molality

Molality is moles of solute per kilogram of solvent.

NCERT-Style Questions from Some Basic Concepts of Chemistry

In CBSE Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry, NCERT-style questions usually test units, laws of chemical combination, mole calculations, empirical formula and stoichiometry. Strong answers begin with the correct formula, followed by substitution, units and the final value.

Q1. Calculate the molecular mass of glucose, C₆H₁₂O₆.

The molecular mass of glucose is 180.162 u.

Step 1:

C₆H₁₂O₆ has 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms.

Step 2:

Molecular mass = 6 × 12.011 + 12 × 1.008 + 6 × 16.00

Step 3:

= 72.066 + 12.096 + 96.00

= 180.162 u

Q2. What is the difference between empirical formula and molecular formula?

Empirical formula gives the simplest whole-number ratio of atoms, while molecular formula gives the actual number of atoms in a molecule.

Explanation:

CH₂O can be an empirical formula. C₆H₁₂O₆ is the molecular formula of glucose.

Formula:

Molecular formula = Empirical formula × n

Q3. Why is mole concept needed in chemistry?

Mole concept is needed because atoms and molecules are too small to count directly.

Explanation:

One mole gives a fixed count of 6.02214076 × 10²³ entities. This connects particle count with measurable mass.

Fact:

Molar mass in grams contains one mole of particles.

Q4. What is the limiting reagent in a reaction?

The limiting reagent is the reactant that is consumed first.

Explanation:

Once the limiting reagent is used up, the reaction stops even if other reactants remain. It decides the maximum amount of product.

Example:

In ammonia formation, shortage of H₂ can limit NH₃ production.

Q5. What is the difference between molarity and molality?

Molarity is moles of solute per litre of solution, while molality is moles of solute per kilogram of solvent.

Explanation:

Molarity depends on volume, so it changes with temperature. Molality depends on mass, so it does not change with temperature.