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Some Basic Concepts in Chemistry Class 11 Notes Chemistry

What is the Matter?

Anything that has mass and occupies space is called matter.

The following are the characteristics of the matter:

  • Particles have space between them
  • Particles are continuously moving
  • Particles attract each other

Classification of Matter

Based on the chemical composition of various substances matter is classified as Pure Substances & Mixtures.

Classification of MatterClassification of Matter

What is Pure Substance?

Pure substances are substances that are made up of only one kind of particles and has a fixed or constant structure.

Types of Pure Substances

Pure substances are further classified as elements and compounds.

1. Elements

  • It is the simplest form of the matter. 
  • The smallest unit of an element is known as an atom.
  • The total number of known elements is 118 out of which 98 elements occur naturally and 20 are formed by artificial transmutation.
  • Examples: Na, K, Mg. Al, Si, P, C, F, Br etc.

2. Compound

  • It is a non-elemental pure compound.
  • Formed by the chemical combination of two or more atoms of different elements in a fixed ratio.
  • Examples: H2O, CO2, C6H12O6 etc.

Characteristics & Properties Of Pure Substances

  • Pure substances are mostly homogeneous in nature containing only one type of atoms or molecules.
  • These substances mainly have a constant or uniform composition throughout.
  • The substances have fixed boiling and melting points.
  • A pure substance usually participates in a chemical reaction to form predictable products.

What is a Mixture?

In chemistry, when two or more substances mix with each other without participating in a chemical change, the resulting substance is called a Mixture.

Types of Mixtures

Type of MixturesType of Mixtures

1. Homogeneous mixtures are those whose composition for each part remains constant.
Example: Aqueous and gaseous solution.

2. Heterogeneous mixtures are those whose composition may vary for each and every part.

Example: Soil and concrete mixtures.

MULTIPLE CHOICE QUESTION

Try yourself: A ________ is made up of two or more pure substances which may be in any ratio.

A

Mixture

B

Element

C

Molecule

D

Atom

Physical Quantities & their Measurement

1. Fundamental Units

  • These units can neither be derived from one another nor can be further resolved into any other units. Seven fundamental units of the S.I. system
    1. Fundamental Units

2. Derived Units

  • These units are the function of more than one fundamental unit
    2. Derived Units

Measurement of Temperature

Three scales of temperature

  • Kelvin scale (K)
  • Degree Celsius scale (oC)
  • Degree Fahrenheit  scale (oF)

 Relations between the scales:

  • oF = 9/5(oC) + 32
  • K = oC + 273

0 K temperatures are called absolute zero.

Precision and Accuracy 

  • Precision: Closeness of outcomes of different measurements taken for the same quantity. 
  • Accuracy: Agreement  of experimental value to the true value

Significant Figures: Rules

  • All non-zero digits are significant.
  • Zeroes preceding the first non-zero digit are not significant.
  • Zeroes between two non-zero digits are significant.
  • Zeroes at the end of a number are significant when they are on the right side of the decimal point.
  • Counting numbers of objects have infinite significant figures.  

MULTIPLE CHOICE QUESTION

Try yourself: How many significant figures does 63180 have?

A

5

B

4

C

1

D

2


Scientific Notation

Numbers are represented in N × 10n form.
Where,

  • N = Digit term 
  • n = exponent having positive or negative value.

Examples:
12540000 = 1.254 × 107
0.00456 = 4.56 ×10-3

Mathematical Operations of Scientific Notation

 Multiplication and Division

  • Follow the same rules which are for exponential numbers.
    Example: (7.0×103)×(8.0×10-7) =  (7.0×8.0)×(10[3 + (-7)]) = 56.0×10-4
  • Result cannot have more digits to the rite of decimal point than either of the original numbers.
    (7.0×103)/(8.0×10-7) = (7.0/8.0)×(10[3 - (-7)]) =  0.875×1010  =  0.9×1010 

Addition and Subtraction

  • Numbers are written in such a way that they have the same exponent and after that coefficients are added or subtracted.
    (5×103) + (8×105) = (5×103) + (800×103) = (5+800)×103  = 805×103
  • The result must be reported with no more significant figures as there in the original number with few significant figures.

Rules for limiting the result of mathematical operations:

  • If the rightmost digit to be removed is more than 5, the preceding number is increased by one.
  • If the rightmost digit to be removed is less than 5, the preceding number is not changed.
  • If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but is increased by one if it is an odd number.

Dimensional Analysis

  • This is based on the fact that the ratio of each fundamental quantity in one unit with their equivalent quantity in another unit is equal to one. 
  • Derived unit first expressed in dimension and each fundamental quantities like mass length time are converted in another system of the desired unit to work out the conversion factor 
  • Original Quantity × Conversion factor = Equivalent Quantity
    (In former unit)                                  (In final Unit)
    Example: (1 kg/2.205 pound) = 1=(1kg/1000gm)
    So 1 kg = 2.205 pound = 1000 gm

Laws of Chemical Combination

1. Law of conservation of mass

"For any chemical change total mass of active reactants are always equal to the mass of the product  formed"

Law of Conservation of MassLaw of Conservation of Mass

2. Law of constant proportions

A chemical compound always contains same elements in definite proportion by mass and it does not depend on the source of compound.

Law of Constant ProportionsLaw of Constant Proportions

3. Law of multiple proportions

When two elements combine to form two or more than two different compounds then the different masses  of one element B which combine with fixed mass of the other element bear a simple ratio to one another.

Law of Multiple ProportionsLaw of Multiple Proportions

MULTIPLE CHOICE QUESTION

Try yourself: What did Dalton propose?

A

Law of Multiple Proportions

B

Avogadro’s Law

C

Law of Definite Composition

D

Law of Conservation of Mass

4. Law of reciprocal proportion

If two elements B and C react with the same mass of a third element (A), the ratio in which they do so will be the same or simple multiple if B and C reacts with each other.

Law of Reciprocal ProportionLaw of Reciprocal Proportion

5. Gay Lussac's law of combining volumes

At given temperature and pressure the volumes of all gaseous reactants and products bear a simple whole number ratio to each other.

Gay Lussac`s LawGay Lussac's Law

What is Dalton's Atomic Theory?

Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

  • Every matter consists of indivisible atoms.
  • Atoms can neither be created nor destroyed.
  • Atoms of a given element are identical in  properties
  • Atoms of different elements differ in properties.
  • Atoms of different elements combine in a fixed ratio to form a molecule of a compound.

MULTIPLE CHOICE QUESTION

Try yourself: According to Dalton’s Atomic Theory, matter consists of indivisible _______

A

Molecules

B

Atoms

C

Ions

D

Mixtures

Atomic and Molecular Masses

Atomic and Molecular Masses

Atomic Mass

  • Mass of an atom.
  • Reported in the atomic mass unit "amu" or unified mass "u"
  • One atomic mass unit i.e. amu, is the mass exactly equal to one-twelfth the mass of one carbon-12 atom.

Molecular Mass

  • Mass of a molecule of the covalent compound.
  • It is equal to the sum of the atomic masses of all the elements present in the molecule.

 Formula Unit Mass

  • Mass of a molecule of an ionic compound
  • It is also equal to the sum of the atomic masses of all the elements present in the molecule

Mole Concept

Mole

  • Unit of the amount of substance.
  • One mole amount of substance that contains as many particles or entities as there are atoms in exactly 12 g of the 12C isotope.

Molar mass

  • Mass of one mole of a substance in gram
  • Molar mass in gram in numerically equal to atomic/molecular/formula mass in amu or u.

Percentage composition

Mass percentage of an element in a compound = (Mass of that element in the compound/Molecular mass of the compound)×100 

Percentage yield

  • It is the ratio of the actual yield of the reaction to the theoretical yield multiplied by 100.
  • % yield = (Actual yield /Theoretical yield)×100

Empirical Formula and Molecular Formula

Molecular Formula

Represents the actual number of each individual atom in any molecule is known as the molecular formula.

Empirical Formula

Expresses the smallest whole number ratio of the constituent atom within the molecule.

Molecular formula = (Empirical formula)n
Molecular weight = n × Empirical weight
also,
Molecular weight = 2 × Vapour density

Limiting Reagent

  • The reactant which is totally consumed during the course of reaction and when it is consumed reaction stops.
  • For a balanced reaction reaction:
    A +B →  C + D
    B would be a limiting reagent if n/ nB>nB/nA
    Similarly, A is a limiting reagent if nA / nB<nB/nA

Concentration of the Solutions

Concentration of the Solutions

Mass by Mass Percentage

  • Amount of solute in gram present per 100 gm of the solution.
  • Mass percentage of solute = [(Mass of solute)/(Mass of solution)] x100
  • Mass by Volume Percentage
  • Amount solute in gram present per 100 mL of the solution.

Volume by Volume Percentage

  • The volume of solute per 100 mL of the solution
  • Volume by volume percentage of solute = [(Volume of solute)/(volume of solution)] x100

Parts per million (ppm)

  • The amount of solute in grams per million (106) grams of the solution.
  • ppm = [(mass of solute/mass of solution)]x 106

Mole fraction

  • Ratio of the moles of one component of the solution to the total number of moles of solution.
  • Total mole fraction of all the components of a solution is equal to 1.
  • For binary solutions having two components A and B

Mole fraction of A   
XA =  (nA)/(nA+nB)]
Mole fraction of B
X=  (nB)/(nA+nB)]
or  XB = 1- XA

Molarity (M)

  • Number of moles of solute per 1000 mL of the solution.
  • M = (Number of moles of solute)/(Volume of solution in L)  

Molality (m)

  • The number of moles of solute per 1000 grams of the solvent.
  • m = (Number of moles of solute)/(Weight of solvent in kg)

MULTIPLE CHOICE QUESTION

Try yourself: Which of the following  be a molecular formula for an empirical formula HO?

A

H2O

B

H2O2

C

HO

D

HO2

The document Revision Notes: Some Basic Concepts in Chemistry is a part of the JEE Course Mock Tests for JEE Main and Advanced 2026.
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FAQs on Revision Notes: Some Basic Concepts in Chemistry

1. What's the difference between molar mass and molecular weight in chemistry?
Ans. Molar mass and molecular weight are essentially the same thing-both express the mass of one mole of a substance in grams per mole (g/mol). Molecular weight refers to the sum of atomic masses of all atoms in a molecule, while molar mass applies this concept to one mole of any substance. For JEE Main problems, they're used interchangeably, though molar mass is the more precise term in formal chemistry contexts.
2. How do I calculate the number of moles from mass and molar mass for JEE problems?
Ans. Use the fundamental formula: number of moles = mass (in grams) ÷ molar mass (in g/mol). First, identify the compound and calculate its molar mass by adding atomic masses of all constituent elements. Then divide the given mass by this value to find moles. This stoichiometric conversion is essential for solving chemical equations and percentage composition problems on JEE Advanced and Main exams.
3. Why do we need to understand atomic mass units and Avogadro's number for basic chemistry concepts?
Ans. Atomic mass units (amu) and Avogadro's number (6.022 × 10²³) form the bridge between microscopic atomic scales and macroscopic laboratory measurements. One mole contains exactly Avogadro's number of particles, making these constants crucial for converting between individual atom masses and measurable quantities. Mastering this relationship unlocks solutions to molarity calculations, limiting reagent problems, and empirical formula determination.
4. What are the most common mistakes students make with stoichiometry and mole calculations in JEE?
Ans. Students frequently forget to balance chemical equations before using stoichiometric ratios, leading to incorrect mole relationships. Others confuse molar mass with molarity or skip unit conversions between grams and moles. Many also miscalculate atomic masses by forgetting to count all atoms in polyatomic compounds. Referring to revision notes and MCQ tests helps identify and correct these recurring errors before exams.
5. How do percentage composition and empirical formulas relate to molar mass calculations?
Ans. Percentage composition shows the mass fraction of each element in a compound, calculated using: (mass of element ÷ molar mass) × 100. Empirical formula reveals the simplest whole-number ratio of atoms; converting it to molecular formula requires knowing the compound's molar mass. Together, these concepts connect elemental percentages to actual molecular structure-critical for JEE problem-solving in quantitative analysis.
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