Chapter 1: Some Basic Concepts of Chemistry

Part 1: Introduction, Development of Chemistry & Importance of Chemistry

Chapter Overview

Chemistry is the branch of science that studies matter—its composition, structure, properties, and the changes it undergoes. It explains how substances interact with one another and how these interactions can be used to improve our daily lives.

This chapter introduces the basic language of chemistry, including matter, atoms, molecules, measurement, chemical laws, the mole concept, and stoichiometry.

Learning Objectives

After studying this chapter, you should be able to:

Understand what chemistry is and why it is important.
Differentiate between different states and types of matter.
Measure physical quantities using SI units.
Use scientific notation and significant figures correctly.
Apply laws of chemical combination.
Understand atomic mass, molecular mass, and the mole concept.
Calculate percentage composition and determine empirical and molecular formulas.
Solve basic stoichiometric problems.

What is Chemistry?

Chemistry is the science that studies:

Composition of matter
Structure of substances
Physical and chemical properties
Chemical reactions
Energy changes during reactions

Unlike physics, which focuses on forces and energy, chemistry mainly deals with substances and how they transform into new substances.

Examples from Daily Life

Rusting of iron
Digestion of food
Burning of fuel
Cooking food
Formation of curd from milk
Ripening of fruits

All these processes involve chemical changes.

Development of Chemistry

Modern chemistry developed gradually over thousands of years.

Early Ideas

Ancient civilizations were interested in chemistry mainly for two purposes:

1. Philosopher’s Stone

People believed a mysterious substance could convert ordinary metals into gold.

2. Elixir of Life

Another belief was that a special substance could provide eternal life.

Although these ideas were incorrect, they encouraged experimentation, leading to the growth of chemistry.

Ancient Indian Contributions

India made remarkable contributions to chemistry long before modern laboratories existed.

Some important achievements include:

Metallurgy

Ancient Indians mastered:

Extraction of copper
Iron production
Gold and silver refining
Alloy preparation

The famous Iron Pillar of Delhi demonstrates advanced metallurgical knowledge because it has resisted rusting for centuries.

Pottery and Glass

People of the Indus Valley Civilization:

Produced baked bricks
Manufactured pottery
Made glazed ceramics
Prepared glass ornaments

These processes required controlled heating and chemical knowledge.

Medicines

Ayurveda included chemical preparation of medicines.

Texts such as:

Charaka Samhita
Sushruta Samhita

describe preparation of minerals, herbal medicines and metal-based compounds.

Cosmetics and Perfumes

Ancient Indians produced:

Perfumes
Hair dyes
Skin care products
Soaps

using natural plant extracts and minerals.

Dyes and Paints

Natural dyes were extracted from:

Turmeric
Indigo
Madder
Lac
Various flowers and roots

These dyes were used in textiles and paintings.

Fireworks

Ancient texts describe the use of:

Sulphur
Charcoal
Potassium nitrate

for producing fireworks.

Early Atomic Idea

Around 600 BCE, philosopher Acharya Kanada proposed that matter is made of extremely tiny indivisible particles called Paramanu.

This idea resembles the modern concept of atoms, although it was philosophical rather than experimental.

Important Indian Scientists

Scientist	Contribution
Acharya Kanada	Proposed the concept of Paramanu (atoms)
Nagarjuna	Worked on metallurgy and mercury compounds
Charaka	Developed Ayurvedic medicinal chemistry
Sushruta	Used chemical substances in medicine
Chakrapani	Worked on soap making and mercury sulphide

Evolution of Modern Chemistry

Chemistry passed through several stages:

Stage	Main Focus
Ancient Chemistry	Metals, medicines, dyes
Alchemy	Converting metals into gold
Iatrochemistry	Preparation of medicines
Modern Chemistry	Scientific study of matter

Modern chemistry began developing rapidly in Europe during the 18th century through careful experimentation and measurement.

Why is Chemistry Important?

Chemistry influences almost every aspect of modern life.

1. Agriculture

Chemistry helps produce:

Fertilizers
Insecticides
Pesticides
Plant growth regulators

These improve crop yield and food production.

2. Medicine

Many life-saving drugs are products of chemistry.

Examples include:

Antibiotics
Painkillers
Vaccines
Cancer medicines

Chemistry also helps discover new medicines.

3. Industry

Chemical industries manufacture:

Acids
Bases
Soaps
Detergents
Paints
Plastics
Polymers
Cement
Glass

These products support almost every manufacturing sector.

4. Environment

Chemistry helps:

Control pollution
Treat wastewater
Develop eco-friendly fuels
Reduce ozone depletion
Monitor greenhouse gases

Green chemistry focuses on reducing environmental damage.

5. Energy

Chemistry contributes to:

Batteries
Solar cells
Hydrogen fuel
Biofuels
Fuel cells

These technologies support cleaner energy production.

6. Everyday Life

Chemistry is involved in:

Cooking
Cleaning
Washing clothes
Preserving food
Cosmetics
Electronics
Mobile batteries
Packaging materials

Career Opportunities in Chemistry

Students studying chemistry can pursue careers in:

Chemical Engineering
Pharmacy
Medicine
Biotechnology
Environmental Science
Food Technology
Forensic Science
Nanotechnology
Research
Teaching

Key Terms

Term	Meaning
Chemistry	Study of matter and its changes
Matter	Anything having mass and occupying space
Alchemy	Ancient attempt to convert ordinary metals into gold
Iatrochemistry	Chemistry related to medicine
Metallurgy	Extraction and purification of metals

Quick Revision

✔ Chemistry studies matter and its transformations.

✔ Ancient chemistry developed through metallurgy, medicines, dyes and pottery.

✔ Acharya Kanada proposed the idea of atoms (Paramanu).

✔ Modern chemistry is based on experimentation and measurement.

✔ Chemistry plays an important role in agriculture, medicine, industries, environment and energy.

Part 2: Nature of Matter, States of Matter & Classification of Matter

1. Nature of Matter

Everything around us is made of matter.

Definition

Matter is anything that:

Has mass
Occupies space (volume)

Examples

Air
Water
Stone
Wood
Human body
Milk
Oxygen

Note: Heat, light and sound are forms of energy, not matter.

Characteristics of Matter

Every substance made of matter has certain characteristics:

Has mass
Occupies space
Made up of tiny particles
Exists in different physical states
Can undergo physical and chemical changes

2. States of Matter

Matter exists in three common physical states:

Solid
Liquid
Gas

The difference between these states depends mainly on:

Arrangement of particles
Distance between particles
Force of attraction
Movement of particles

A. Solid

In solids, particles are packed very closely together.

Characteristics

Definite shape
Definite volume
High density
Very small intermolecular space
Strong force of attraction
Least compressible
Particles only vibrate about fixed positions

Examples

Ice
Iron
Wood
Stone
Sugar

B. Liquid

Particles are close together but can move past one another.

Characteristics

Definite volume
No definite shape
Takes the shape of its container
Flows easily
Slightly compressible
Moderate intermolecular force

Examples

Water
Milk
Oil
Alcohol
Petrol

C. Gas

Particles are far apart and move freely in all directions.

Characteristics

No fixed shape
No fixed volume
Completely fills the container
Highly compressible
Lowest density
Very weak intermolecular force
Rapid movement of particles

Examples

Oxygen
Nitrogen
Hydrogen
Carbon dioxide

Comparison of the Three States

Property	Solid	Liquid	Gas
Shape	Fixed	Not fixed	Not fixed
Volume	Fixed	Fixed	Not fixed
Particle Arrangement	Very close	Close	Far apart
Intermolecular Force	Strong	Moderate	Weak
Compressibility	Negligible	Very low	Very high
Particle Movement	Vibrations only	Sliding movement	Free movement
Density	Highest	Moderate	Lowest

Interconversion of States

Matter can change from one state to another by changing:

Temperature
Pressure

Changes on Heating

Solid
   │
   ▼
Liquid
   │
   ▼
Gas

Changes on Cooling

Gas
   │
   ▼
Liquid
   │
   ▼
Solid

Important Processes

Process	Change
Melting	Solid → Liquid
Freezing	Liquid → Solid
Vaporisation	Liquid → Gas
Condensation	Gas → Liquid

Key Points

✔ Heating increases particle movement.

✔ Cooling decreases particle movement.

✔ Increasing pressure can convert gases into liquids.

3. Classification of Matter

Matter is broadly classified into:

Matter
│
├── Pure Substance
│     ├── Element
│     └── Compound
│
└── Mixture
      ├── Homogeneous
      └── Heterogeneous

Pure Substance

A pure substance consists of only one type of particle and has a fixed composition throughout.

Characteristics

Uniform composition
Fixed properties
Cannot be separated by physical methods
Sharp melting and boiling points

Examples

Gold
Silver
Copper
Water
Carbon dioxide
Sodium chloride

Element

An element is a pure substance made up of only one type of atom.

It cannot be broken down into simpler substances by ordinary chemical methods.

Characteristics

Contains only one kind of atom
Simplest form of matter
Has unique physical and chemical properties

Examples

Hydrogen (H)
Oxygen (O₂)
Nitrogen (N₂)
Iron (Fe)
Copper (Cu)
Gold (Au)

Note: Some elements exist as single atoms (e.g., Na, Fe), while others exist as molecules (e.g., H₂, O₂, N₂).

Compound

A compound is formed when two or more different elements combine chemically in a fixed ratio.

Characteristics

Fixed composition
Properties differ from constituent elements
Components cannot be separated by physical methods
Can be decomposed only by chemical methods

Examples

Compound	Elements Present
Water (H₂O)	Hydrogen + Oxygen
Carbon dioxide (CO₂)	Carbon + Oxygen
Ammonia (NH₃)	Nitrogen + Hydrogen
Sodium chloride (NaCl)	Sodium + Chlorine

Why are Compound Properties Different?

The properties of compounds are usually very different from those of the elements that form them.

Example

Hydrogen	Oxygen	Water
Burns easily	Supports burning	Used to extinguish fire

This shows that a chemical combination creates a substance with entirely new properties.

Mixture

A mixture contains two or more substances physically mixed together.

The substances retain their individual properties.

Characteristics

Variable composition
Components are not chemically combined
Can be separated by physical methods
Properties of components remain unchanged

Examples

Air
Tea
Salt solution
Soil
Milk

Types of Mixtures

A. Homogeneous Mixture

A homogeneous mixture has a uniform composition throughout.

The different components cannot be seen separately.

Examples

Air
Sugar solution
Salt solution
Vinegar
Brass

B. Heterogeneous Mixture

A heterogeneous mixture has a non-uniform composition.

The different components are visible.

Examples

Sand and water
Oil and water
Soil
Granite
Mixture of pulses

Homogeneous vs Heterogeneous Mixture

Homogeneous	Heterogeneous
Uniform composition	Non-uniform composition
Single phase	Two or more phases
Components not visible	Components visible
Same composition throughout	Composition varies

Separation of Mixtures

Mixtures can be separated using physical methods because no chemical bonds are formed between their components.

Common methods include:

Hand picking
Filtration
Evaporation
Crystallisation
Distillation
Sublimation
Magnetic separation

Pure Substance vs Mixture

Pure Substance	Mixture
Fixed composition	Variable composition
One kind of particle	Two or more kinds of particles
Fixed melting point	Melting occurs over a range
Cannot be separated physically	Can be separated physically
Uniform properties	Properties depend on composition

4. Physical and Chemical Properties

Every substance has characteristic properties that help identify it.

These properties are classified into:

Physical properties
Chemical properties

Physical Properties

Physical properties can be observed or measured without changing the chemical identity of a substance.

Examples

Colour
Odour
Density
Melting point
Boiling point
State
Solubility
Hardness

Chemical Properties

Chemical properties describe how a substance behaves during a chemical reaction.

They can only be observed when the substance undergoes a chemical change.

Examples

Combustibility
Reactivity with oxygen
Reactivity with acids
Reactivity with bases
Corrosion
Rusting tendency

Physical vs Chemical Properties

Physical Property	Chemical Property
No new substance formed	New substance formed
Easily observed	Requires chemical reaction
Identity remains same	Identity changes
Examples: Colour, density	Examples: Burning, rusting

Chapter Snapshot

Matter
│
├── States
│     ├── Solid
│     ├── Liquid
│     └── Gas
│
├── Classification
│     ├── Pure Substance
│     │      ├── Element
│     │      └── Compound
│     │
│     └── Mixture
│            ├── Homogeneous
│            └── Heterogeneous
│
└── Properties
      ├── Physical
      └── Chemical

Quick Revision

✔ Matter has mass and occupies space.

✔ Matter exists as solids, liquids and gases.

✔ Solids have fixed shape and volume.

✔ Liquids have fixed volume but no fixed shape.

✔ Gases have neither fixed shape nor fixed volume.

✔ Matter is classified into pure substances and mixtures.

✔ Pure substances include elements and compounds.

✔ Mixtures may be homogeneous or heterogeneous.

✔ Physical properties are observed without changing the substance.

✔ Chemical properties involve chemical reactions.

Part 3: Measurement of Physical Quantities, SI Units, Density & Temperature

1. Measurement in Chemistry

Chemistry is an experimental science. Most chemical studies require accurate measurement of physical quantities.

Examples of measurable quantities:

Mass of a substance
Volume of a liquid
Temperature
Density
Length
Amount of substance

A measurement always contains:

Numerical value + Unit

Example:

A bottle contains 2 L of water.

Here:

2 → numerical value
L → unit (litre)

Without a unit, a measurement is incomplete.

2. Systems of Measurement

Earlier, different countries used different measurement systems.

The two common systems were:

1. English System

Examples:

Inch
Foot
Pound

2. Metric System

Examples:

Metre
Gram
Litre

The metric system became popular because it follows a decimal system and is easier to use.

3. International System of Units (SI System)

To create a common measurement system worldwide, scientists introduced the International System of Units (SI).

SI units were established in 1960 by the General Conference on Weights and Measures.

The SI system contains seven fundamental base units.

SI Base Quantities and Units

Physical Quantity	SI Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

Importance of SI Units

SI units provide:

Uniform measurement worldwide
Accurate scientific communication
Easy conversion between units
Standard reference values

4. SI Prefixes

Very large or very small quantities are difficult to write using ordinary numbers.

SI prefixes help represent such quantities easily.

Common Prefixes

Prefix	Symbol	Value
kilo	k	10³
mega	M	10⁶
giga	G	10⁹
centi	c	10⁻²
milli	m	10⁻³
micro	μ	10⁻⁶
nano	n	10⁻⁹
pico	p	10⁻¹²

Examples

1 kilometre: $1 km = 1000 m = 10^3 m$ 1km=1000m=103m

1 milligram: $1 mg = 10^{-3} g$ 1mg=10−3g

5. Mass and Weight

Although mass and weight are often used interchangeably, they are different.

Mass

Mass is the amount of matter present in an object.

Characteristics

Remains constant everywhere
Independent of gravity
SI unit: kilogram (kg)

Measurement

In laboratories, mass is measured using balances such as:

Analytical balance
Electronic balance

Weight

Weight is the force with which gravity attracts an object.

Formula: $\text{Weight} = \text{Mass} \times \text{Acceleration due to gravity}$ Weight=Mass×Acceleration due to gravity $W = mg$ W=mg

Characteristics

Changes from place to place
Depends on gravity
SI unit: newton (N)

Difference Between Mass and Weight

Mass	Weight
Amount of matter	Gravitational force on matter
Constant	Changes with gravity
SI unit: kg	SI unit: N
Measured using balance	Measured using spring balance

6. Volume

Volume is the amount of space occupied by matter.

SI Unit

$m^3$ m3

However, chemistry laboratories commonly use:

cm³
dm³
litre (L)
millilitre (mL)

Important Volume Conversions

$1 L = 1000 mL$ 1L=1000mL $1 L = 1000 cm^3$ 1L=1000cm3 $1 dm^3 = 1 L$ 1dm3=1L $1 m^3 = 1000 L$ 1m3=1000L

Measuring Volume in Laboratory

Common instruments:

1. Measuring Cylinder

Used for approximate measurement of liquids.

2. Burette

Used for accurate delivery of liquids during titration.

3. Pipette

Used to transfer a fixed volume of liquid accurately.

4. Volumetric Flask

Used for preparing solutions of accurate concentration.

7. Density

Density tells us how much mass is present in a given volume.

Formula:

$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$ Density=VolumeMass

or $d=\frac{m}{V}$ d=Vm

Units of Density

SI unit: $kg\,m^{-3}$ kgm−3

Common chemistry unit: $g\,cm^{-3}$ gcm−3

Understanding Density

Higher density means:

Particles are packed more closely
More mass is present in the same volume

Lower density means:

Particles are farther apart
Less mass occupies the same volume

Example

Iron sinks in water because its density is higher than water.

Wood floats because its density is lower than water.

8. Temperature Measurement

Temperature indicates the degree of hotness or coldness of a substance.

Common temperature scales:

Celsius scale (°C)
Fahrenheit scale (°F)
Kelvin scale (K)

Celsius Scale

Reference points:

Freezing point of water = 0°C
Boiling point of water = 100°C

Fahrenheit Scale

Reference points:

Freezing point = 32°F
Boiling point = 212°F

Kelvin Scale

Kelvin is the SI unit of temperature.

Relationship: $K = °C + 273.15$ K=°C+273.15

Example:

If temperature = 25°C $K = 25 + 273.15$ K=25+273.15 $K = 298.15K$ K=298.15K

Celsius-Fahrenheit Conversion

$\frac{C}{5}=\frac{F-32}{9}$ 5C=9F−32

Important Points About Kelvin Scale

✔ Kelvin temperature cannot be negative.

✔ It is called the absolute temperature scale.

✔ 0 K represents absolute zero.

9. National Standards of Measurement

Accurate measurements require standard references.

Every country maintains measurement standards through national laboratories.

In India, this responsibility is handled by:

National Physical Laboratory (NPL), New Delhi

It maintains standards for:

Length
Mass
Time
Temperature
Other physical quantities

Quick Revision Table

Quantity	SI Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Temperature	kelvin	K
Amount of substance	mole	mol
Volume	cubic metre	m³
Density	kg/m³	kg m⁻³

Important Formulas

Density

$d=\frac{m}{V}$ d=Vm

Weight

$W=mg$ W=mg

Temperature Conversion

$K=°C+273.15$ K=°C+273.15 $\frac{C}{5}=\frac{F-32}{9}$ 5C=9F−32

Quick Revision Points

✔ Every measurement needs a number and unit.

✔ SI system provides internationally accepted units.

✔ Mass is constant but weight depends on gravity.

✔ Volume represents space occupied by matter.

✔ Density is mass per unit volume.

✔ Kelvin is the SI unit of temperature.

✔ Chemistry commonly uses g cm⁻³ for density and litre/mL for volume.

Part 4: Uncertainty in Measurement, Scientific Notation, Significant Figures & Dimensional Analysis

1. Accuracy and Uncertainty in Measurement

In chemistry, measurements are never perfectly exact because every measuring instrument has some limitation.

For example:

A simple balance may measure mass up to 0.1 g.
An analytical balance may measure mass up to 0.0001 g.

Therefore, every experimental measurement contains some degree of uncertainty.

Why Does Uncertainty Occur?

Uncertainty may arise due to:

1. Limitation of Instruments

No instrument can measure an infinite number of decimal places.

Example:

A balance may show: $9.4g$ 9.4g

while a more advanced balance may show: $9.4213g$ 9.4213g

The extra digits depend on the sensitivity of the instrument.

2. Human Errors

Errors may occur due to:

Incorrect observation
Reading scale incorrectly
Experimental conditions

3. Environmental Conditions

Factors such as:

Temperature
Pressure
Humidity

may affect measurements.

2. Scientific Notation

Chemistry often deals with extremely large or extremely small numbers.

Writing all zeros is inconvenient, so scientific notation is used.

General Form

$N \times 10^n$ N×10n

where:

N is a number between 1 and 10
n is an integer

Examples

Large Number

602200000000000000000000

can be written as: $6.022 \times 10^{23}$ 6.022×1023

Small Number

0.000000001

can be written as: $1 \times 10^{-9}$ 1×10−9

Rules for Scientific Notation

Positive exponent

Used for numbers greater than 1.

Example: $5000 = 5 \times 10^3$ 5000=5×103

Negative exponent

Used for numbers between 0 and 1.

Example: $0.005 = 5 \times 10^{-3}$ 0.005=5×10−3

Mathematical Operations with Scientific Notation

Addition and Subtraction

The powers of 10 must be made equal first.

Example: $(6.65 \times 10^4)+(8.95\times10^3)$ (6.65×104)+(8.95×103)

Convert: $8.95\times10^3=0.895\times10^4$ 8.95×103=0.895×104

Now: $(6.65+0.895)\times10^4$ (6.65+0.895)×104 $=7.545\times10^4$ =7.545×104

Multiplication

Multiply numerical parts and add powers.

Example: $(2\times10^3)(3\times10^4)$ (2×103)(3×104) $=6\times10^7$ =6×107

Division

Divide numerical parts and subtract powers.

Example: $\frac{6\times10^5}{2\times10^2}$ 2×1026×105 $=3\times10^3$ =3×103

3. Significant Figures

Meaning

Significant figures are the meaningful digits in a measurement.

They include:

All certain digits
One uncertain (estimated) digit

They indicate the precision of a measurement.

Example

A measurement: $11.2\,mL$ 11.2mL

contains:

11 → certain digits
2 → uncertain digit

Therefore, it has three significant figures.

Rules for Counting Significant Figures

Rule 1: All Non-Zero Digits Are Significant

Examples:

Number	Significant Figures
285	3
25.6	3
0.25	2

Rule 2: Zeros Before First Non-Zero Digit Are Not Significant

These zeros only locate the decimal point.

Examples:

Number	Significant Figures
0.03	1
0.0052	2

Rule 3: Zeros Between Non-Zero Digits Are Significant

Examples:

Number	Significant Figures
2.005	4
1002	4

Rule 4: Terminal Zeros After Decimal Are Significant

Examples:

Number	Significant Figures
0.200	3
5.00	3

Rule 5: Terminal Zeros Without Decimal Are Uncertain

Example:

100 may represent: $1\times10^2$ 1×102

(1 significant figure)

or $1.00\times10^2$ 1.00×102

(3 significant figures)

Scientific notation removes confusion.

Exact Numbers

Numbers obtained by counting objects have unlimited significant figures.

Examples:

5 students
20 books
12 eggs

These are exact values.

Significant Figures in Scientific Notation

All digits written in scientific notation are significant.

Examples: $4.01\times10^2$ 4.01×102

has:

3 significant figures $8.256\times10^{-3}$ 8.256×10−3

has:

4 significant figures

4. Precision and Accuracy

These two terms describe the quality of measurements.

Precision

Precision means the closeness of repeated measurements to each other.

It shows reproducibility.

Example:

Measurements:

1.95 g
1.93 g

These values are close to each other.

Therefore, they are precise.

Accuracy

Accuracy means how close a measured value is to the true value.

Example:

True value = 2.00 g

Measured values:

2.01 g
1.99 g

These are accurate.

Difference Between Accuracy and Precision

Accuracy	Precision
Closeness to true value	Closeness among repeated values
Indicates correctness	Indicates reproducibility
Depends on errors	Depends on consistency

Relationship Between Accuracy and Precision

A good measurement should be:

Accurate
Precise

The best results are both close to the true value and close to each other.

5. Rounding Off Rules

When a calculated answer contains more digits than required, it must be rounded.

Rule 1: If the Removed Digit is Greater Than 5

Increase the previous digit by one.

Example:

1.386 rounded to three digits:

Remove 6: $1.39$ 1.39

Rule 2: If the Removed Digit is Less Than 5

Previous digit remains unchanged.

Example:

4.334

Remove 4: $4.33$ 4.33

Rule 3: If the Removed Digit is Exactly 5

If the previous digit is even → leave unchanged
If the previous digit is odd → increase by one

Examples: $6.25 \rightarrow 6.2$ 6.25→6.2

(2 is even) $6.35 \rightarrow 6.4$ 6.35→6.4

(3 is odd)

6. Mathematical Operations Using Significant Figures

Addition and Subtraction Rule

The answer should contain the same number of decimal places as the measurement having the least decimal places.

Example: $12.11+18.0+1.012$ 12.11+18.0+1.012

Actual answer: $31.122$ 31.122

Since 18.0 has only one decimal place:

Final answer: $31.1$ 31.1

Multiplication and Division Rule

The final answer should contain the same number of significant figures as the value having the fewest significant figures.

Example: $2.5 \times 1.25$ 2.5×1.25 $=3.125$ =3.125

2.5 has only 2 significant figures.

Therefore: $3.1$ 3.1

is the final answer.

7. Dimensional Analysis

Dimensional analysis is a method used to convert one unit into another.

It is also called:

Factor label method
Unit factor method

Basic Principle

A conversion factor is a ratio equal to 1.

Example: $1\,inch = 2.54\,cm$ 1inch=2.54cm

Therefore: $\frac{2.54\,cm}{1\,inch}=1$ 1inch2.54cm=1

and $\frac{1\,inch}{2.54\,cm}=1$ 2.54cm1inch=1

Both are unit factors.

Example: Convert 3 inches into centimetres

Given: $1\,inch=2.54\,cm$ 1inch=2.54cm

Using unit factor: $3\,inch\times\frac{2.54\,cm}{1\,inch}$ 3inch×1inch2.54cm

The inch units cancel. $=3\times2.54$ =3×2.54 $=7.62\,cm$ =7.62cm

Advantages of Dimensional Analysis

✔ Helps convert units easily.

✔ Prevents calculation mistakes.

✔ Units can be cancelled like algebraic quantities.

✔ Useful in chemical calculations.

Important Conversions

$1L=1000cm^3$ 1L=1000cm3 $1m=100cm$ 1m=100cm $1m^3=10^6cm^3$ 1m3=106cm3

Quick Revision

✔ Every measurement has some uncertainty.

✔ Scientific notation simplifies very large or very small numbers.

✔ Significant figures show measurement reliability.

✔ Non-zero digits are always significant.

✔ Leading zeros are not significant.

✔ Precision means closeness between measurements.

✔ Accuracy means closeness to true value.

✔ Rounding rules are important while reporting answers.

✔ Dimensional analysis converts units using conversion factors.

Part 5: Laws of Chemical Combination & Dalton’s Atomic Theory

1. Laws of Chemical Combination

When elements combine to form compounds, they follow certain fixed rules. These rules are known as laws of chemical combination.

The major laws are:

Law of Conservation of Mass
Law of Definite Proportions
Law of Multiple Proportions
Gay-Lussac’s Law of Gaseous Volumes
Avogadro’s Law

These laws formed the foundation for understanding atoms and molecules.

1.1 Law of Conservation of Mass

Statement

The law states:

Mass can neither be created nor destroyed during a physical or chemical change. The total mass before and after a reaction remains the same.

This law was proposed by Antoine Lavoisier (1789).

Explanation

In a chemical reaction: $\text{Mass of Reactants}=\text{Mass of Products}$ Mass of Reactants=Mass of Products

Atoms are only rearranged during a reaction; they are not created or destroyed.

Example

When carbon burns in oxygen: $C + O_2 \rightarrow CO_2$ C+O2→CO2

If:

Carbon mass = 12 g
Oxygen mass = 32 g

Then: $\text{Mass of CO}_2=12+32=44g$ Mass of CO2=12+32=44g

The total mass remains constant.

Importance of the Law

This law helped scientists:

Perform accurate chemical calculations
Understand chemical reactions
Develop atomic theory

1.2 Law of Definite Proportions

Statement

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

This law was proposed by Joseph Proust.

It is also called:

Law of Constant Composition

Example: Water

Water always contains hydrogen and oxygen in a fixed mass ratio.

Formula: $H_2O$ H2O

Mass contribution:

Hydrogen: $2 \times 1 = 2$ 2×1=2

Oxygen: $16$ 16

Ratio: $H:O=2:16$ H:O=2:16

or $1:8$ 1:8

So, every sample of pure water contains hydrogen and oxygen in the ratio 1:8 by mass.

Importance

This law shows that:

Compounds have fixed composition.
Chemical formulas represent definite ratios of atoms.

1.3 Law of Multiple Proportions

Statement

When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are always in the ratio of small whole numbers.

Proposed by:

John Dalton (1803)

Example: Carbon and Oxygen

Carbon forms two compounds:

Carbon monoxide (CO)

12 g carbon combines with 16 g oxygen.

Carbon dioxide (CO₂)

12 g carbon combines with 32 g oxygen.

Oxygen masses: $16:32$ 16:32

Simplifying: $1:2$ 1:2

This is a simple whole-number ratio.

Importance

This law supported the idea that atoms combine in fixed numbers.

1.4 Gay-Lussac’s Law of Gaseous Volumes

Statement

When gases combine or are produced in a chemical reaction, their volumes are in simple whole-number ratios, provided all gases are measured at the same temperature and pressure.

Proposed by:

Joseph Louis Gay-Lussac (1808)

Example: Formation of Water Vapour

Reaction: $2H_2 + O_2 \rightarrow 2H_2O$ 2H2+O2→2H2O

Volume relationship:

Gas	Volume
Hydrogen	2 volumes
Oxygen	1 volume
Water vapour	2 volumes

Ratio: $2:1:2$ 2:1:2

Importance

This law helped in understanding the relationship between gas molecules and led to Avogadro’s hypothesis.

1.5 Avogadro’s Law

Statement

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

Proposed by:

Amedeo Avogadro (1811)

Explanation

According to Avogadro:

Gas volume depends on the number of molecules present.
Equal volumes under identical conditions contain equal molecular numbers.

Example

At the same temperature and pressure:

1 litre hydrogen gas
1 litre oxygen gas

contain the same number of molecules.

Importance of Avogadro’s Law

It helped explain:

Difference between atoms and molecules
Molecular formulas of gases
Mole concept

Comparison of Chemical Combination Laws

Law	Scientist	Main Idea
Conservation of Mass	Lavoisier	Mass remains constant
Definite Proportions	Proust	Fixed mass ratio in compounds
Multiple Proportions	Dalton	Simple whole number ratios
Gaseous Volumes	Gay-Lussac	Gas volume ratios are simple
Avogadro’s Law	Avogadro	Equal gas volumes contain equal molecules

2. Dalton’s Atomic Theory

After studying laws of chemical combination, John Dalton proposed an atomic theory in 1808.

His theory explained how atoms participate in chemical reactions.

Main Postulates of Dalton’s Theory

1. Matter is Made of Atoms

All matter is composed of extremely small particles called atoms.

Atoms were considered indivisible.

2. Atoms of the Same Element Are Identical

According to Dalton:

All atoms of an element have the same mass and properties.

Example:

All oxygen atoms were considered identical.

3. Atoms of Different Elements Are Different

Atoms of different elements have:

Different masses
Different properties

Example:

Hydrogen atoms differ from oxygen atoms.

4. Compounds Form by Combination of Atoms

Atoms combine in simple whole-number ratios to form compounds.

Example:

Water: $H_2O$ H2O

Two hydrogen atoms combine with one oxygen atom.

5. Chemical Reactions Involve Rearrangement of Atoms

During chemical reactions:

Atoms are rearranged.
Atoms are neither created nor destroyed.

This explains conservation of mass.

Limitations of Dalton’s Theory

Later discoveries showed that some ideas were incomplete.

1. Atoms Are Divisible

Dalton considered atoms indivisible.

However, atoms contain smaller particles:

Electrons
Protons
Neutrons

2. Atoms of Same Element May Differ

The discovery of isotopes showed that atoms of the same element can have different masses.

Example:

Hydrogen exists as:

Protium
Deuterium
Tritium

3. Atoms of Different Elements Can Have Similar Masses

Some atoms of different elements may have similar atomic masses.

Contribution of Dalton’s Theory

Despite limitations, Dalton’s theory was important because it:

✔ Explained laws of chemical combination.

✔ Introduced the concept of atoms scientifically.

✔ Provided the foundation of modern chemistry.

Quick Revision

✔ Chemical laws explain how elements combine.

✔ Conservation of mass states that total mass remains unchanged.

✔ Definite proportions states that compounds have fixed composition.

✔ Multiple proportions explains simple whole-number ratios.

✔ Gay-Lussac explained gas volume relationships.

✔ Avogadro proposed equal gas volumes contain equal numbers of molecules.

✔ Dalton introduced atomic theory in 1808.

✔ Atoms are not indivisible; they contain subatomic particles.

Part 6: Atomic Mass, Molecular Mass, Formula Mass & Mole Concept

1. Atomic Mass

Atoms are extremely small, so their actual masses cannot be expressed conveniently in grams.

For example, the mass of a hydrogen atom is approximately: $1.67 \times 10^{-24} g$ 1.67×10−24g

Such small values are difficult to use in calculations. Therefore, scientists use a special unit called atomic mass unit (u).

2. Atomic Mass Unit (u)

Definition

One atomic mass unit (1 u) is defined as:

One-twelfth of the mass of one carbon-12 atom.

Carbon-12 is used as the standard reference for measuring atomic masses. $1u=\frac{1}{12}\text{ mass of one carbon-12 atom}$ 1u=121 mass of one carbon-12 atom

Value of 1 u

$1u = 1.66056 \times 10^{-24}g$ 1u=1.66056×10−24g

Why Carbon-12 is Used as Standard?

Carbon-12 was selected because:

It is stable.
It is easily available.
Its mass can be accurately measured.
It provides a convenient reference for all elements.

Relative Atomic Mass

Atomic masses are compared with the mass of carbon-12.

Example:

Hydrogen has an atomic mass close to: $1.008u$ 1.008u

Oxygen has an atomic mass close to: $16u$ 16u

3. Average Atomic Mass

Many elements exist naturally as mixtures of isotopes.

Isotopes

Atoms of the same element having:

Same atomic number
Different mass numbers

are called isotopes.

Example:

Carbon exists as:

Carbon-12
Carbon-13
Carbon-14

Calculation of Average Atomic Mass

Average atomic mass depends on:

Mass of each isotope
Percentage abundance of each isotope

Formula: $\text{Average atomic mass} = \frac{ (\text{mass of isotope 1 × abundance}) + (\text{mass of isotope 2 × abundance}) } {100}$ Average atomic mass=100(mass of isotope 1 × abundance)+(mass of isotope 2 × abundance)

Example: Carbon

Carbon has:

Carbon-12 (major isotope)
Carbon-13
Carbon-14

After considering their natural abundance, the average atomic mass of carbon is approximately: $12.011u$ 12.011u

Important Point

The atomic masses shown in the periodic table are generally average atomic masses, not the mass of a single atom.

4. Molecular Mass

Definition

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

Formula:

$\text{Molecular mass} = \text{Sum of atomic masses of all atoms}$ Molecular mass=Sum of atomic masses of all atoms

Unit: $u$ u

Example 1: Water (H₂O)

Water contains:

2 hydrogen atoms
1 oxygen atom

Atomic masses:

H = 1.008 u

O = 16.00 u

Calculation: $=2(1.008)+16.00$ =2(1.008)+16.00 $=18.016u$ =18.016u

Approximately: $18.02u$ 18.02u

Example 2: Methane (CH₄)

Methane contains:

1 carbon atom
4 hydrogen atoms

Calculation: $=12.011+4(1.008)$ =12.011+4(1.008) $=16.043u$ =16.043u

5. Formula Mass

Some compounds do not exist as individual molecules.

Example:

Sodium chloride (NaCl)

In solid sodium chloride, sodium and chloride ions form a large crystal structure.

Therefore, instead of molecular mass, we use formula mass.

Definition

Formula mass is the sum of atomic masses of all atoms present in the formula unit of an ionic compound.

Unit: $u$ u

Example: Sodium Chloride (NaCl)

Atomic mass:

Na = 23.0 u

Cl = 35.5 u

Formula mass: $=23.0+35.5$ =23.0+35.5 $=58.5u$ =58.5u

Molecular Mass vs Formula Mass

Molecular Mass	Formula Mass
Used for molecular compounds	Used for ionic compounds
Based on molecules	Based on formula units
Example: H₂O, CO₂	Example: NaCl, CaCl₂

6. Mole Concept

Atoms and molecules are extremely small, so even a small amount of substance contains a very large number of particles.

To count these particles, chemists use a special counting unit called the mole.

Definition of Mole

A mole is the amount of substance containing: $6.022 \times 10^{23}$ 6.022×1023

particles.

This number is called:

Avogadro Number or Avogadro Constant

Symbol: $N_A$ NA

Value: $N_A=6.022\times10^{23}mol^{-1}$ NA=6.022×1023mol−1

Understanding Mole

Just as:

1 dozen = 12 objects
1 pair = 2 objects

Similarly:

1 mole = $6.022\times10^{23}$ 6.022×1023 particles

One Mole Represents

Substance	One Mole Contains
Hydrogen atoms	$6.022\times10^{23}$ 6.022×1023 atoms
Water molecules	$6.022\times10^{23}$ 6.022×1023 molecules
Sodium chloride	$6.022\times10^{23}$ 6.022×1023 formula units

Importance of Mole Concept

The mole connects:

Atomic scale
Laboratory scale

It helps convert between:

Number of particles
Mass
Amount of substance

7. Molar Mass

Definition

The mass of one mole of a substance is called its molar mass.

Unit: $g\,mol^{-1}$ gmol−1

Relationship Between Atomic Mass and Molar Mass

Numerically: $\text{Atomic mass in u} = \text{Molar mass in }g\,mol^{-1}$ Atomic mass in u=Molar mass in gmol−1

Examples

Hydrogen

Atomic mass: $1.008u$ 1.008u

Molar mass: $1.008g\,mol^{-1}$ 1.008gmol−1

Water

Molecular mass: $18.02u$ 18.02u

Molar mass: $18.02g\,mol^{-1}$ 18.02gmol−1

Sodium chloride

Formula mass: $58.5u$ 58.5u

Molar mass: $58.5g\,mol^{-1}$ 58.5gmol−1

8. Important Mole Relationships

Relationship 1: Mole and Number of Particles

$\text{Number of particles} = \text{Number of moles}\times N_A$ Number of particles=Number of moles×NA

Relationship 2: Mole and Mass

$\text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}}$ Number of moles=Molar massGiven mass

Relationship 3: Mass from Mole

$\text{Mass} = \text{Number of moles}\times\text{Molar mass}$ Mass=Number of moles×Molar mass

Example

Calculate number of moles in 36 g of water.

Given:

Mass = 36 g

Molar mass of water = 18 g/mol

Formula: $n=\frac{mass}{molar\,mass}$ n=molarmassmass $n=\frac{36}{18}$ n=1836 $n=2mol$ n=2mol

Therefore:

36 g water contains 2 moles of water molecules.

Mole Conversion Map

             Number of particles
                    ▲
                    │
                    │ × 6.022×10²³
                    │
Mass  ◄──────────►  Moles
      divide by       multiply by
      molar mass      molar mass

Quick Revision

✔ Atomic masses are expressed in atomic mass unit (u).

✔ 1 u is one-twelfth mass of carbon-12 atom.

✔ Periodic table values represent average atomic masses.

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

✔ Formula mass is used for ionic compounds.

✔ One mole contains $6.022\times10^{23}$ 6.022×1023 particles.

✔ Molar mass is mass of one mole of a substance.

✔ Mole concept connects microscopic particles with measurable quantities.

Part 7: Percentage Composition, Empirical Formula, Molecular Formula & Stoichiometry

1. Percentage Composition

A chemical compound contains different elements in a fixed ratio.

Percentage composition tells us the percentage by mass of each element present in a compound.

It is useful for:

Identifying unknown compounds
Checking purity of substances
Calculating formulas of compounds

Formula for Mass Percentage

$\text{Mass percentage of element} = \frac{\text{Mass of element in one mole of compound}} {\text{Molar mass of compound}} \times100$ Mass percentage of element=Molar mass of compoundMass of element in one mole of compound×100

Example: Percentage Composition of Water (H₂O)

Molar mass of water: $=2(1.008)+16.00$ =2(1.008)+16.00 $=18.016g/mol$ =18.016g/mol

Percentage of Hydrogen

Mass of hydrogen: $=2.016g$ =2.016g $\%H=\frac{2.016}{18.016}\times100$ %H=18.0162.016×100 $=11.18\%$ =11.18%

Percentage of Oxygen

$\%O=\frac{16.00}{18.016}\times100$ %O=18.01616.00×100 $=88.82\%$ =88.82%

Therefore, water contains approximately:

Hydrogen = 11.18%
Oxygen = 88.82%

2. Empirical Formula

Definition

The empirical formula represents the simplest whole-number ratio of atoms of different elements present in a compound.

It does not show the actual number of atoms in a molecule.

Examples

Molecular Formula	Empirical Formula
H₂O	H₂O
C₆H₁₂O₆	CH₂O
N₂O₄	NO₂
C₂H₄	CH₂

Steps to Find Empirical Formula

Step 1: Convert Percentage into Grams

Assume the compound sample is 100 g.

Example:

A compound contains:

Carbon = 40%
Hydrogen = 6.67%
Oxygen = 53.33%

Then:

Carbon = 40 g
Hydrogen = 6.67 g
Oxygen = 53.33 g

Step 2: Convert Mass into Moles

Formula: $\text{Moles}=\frac{\text{Given mass}}{\text{Atomic mass}}$ Moles=Atomic massGiven mass

Example:

Carbon: $\frac{40}{12}=3.33$ 1240=3.33

Hydrogen: $\frac{6.67}{1}=6.67$ 16.67=6.67

Oxygen: $\frac{53.33}{16}=3.33$ 1653.33=3.33

Step 3: Divide by the Smallest Mole Value

Smallest value = 3.33

Carbon: $\frac{3.33}{3.33}=1$ 3.333.33=1

Hydrogen: $\frac{6.67}{3.33}=2$ 3.336.67=2

Oxygen: $\frac{3.33}{3.33}=1$ 3.333.33=1

Ratio: $C:H:O=1:2:1$ C:H:O=1:2:1

Step 4: Write the Empirical Formula

$\boxed{CH_2O}$ CH2O

3. Molecular Formula

Definition

The molecular formula shows the actual number of atoms of each element present in one molecule of a compound.

Relationship Between Empirical and Molecular Formula

$\text{Molecular Formula} = (\text{Empirical Formula})_n$ Molecular Formula=(Empirical Formula)n

where: $n= \frac{\text{Molar Mass}} {\text{Empirical Formula Mass}}$ n=Empirical Formula MassMolar Mass

Example

Suppose:

Empirical formula: $CH_2O$ CH2O

Empirical formula mass: $=12+2(1)+16$ =12+2(1)+16 $=30g/mol$ =30g/mol

Given molar mass: $180g/mol$ 180g/mol

Then: $n=\frac{180}{30}$ n=30180 $n=6$ n=6

Therefore: $(CH_2O)_6$ (CH2O)6

Molecular formula: $\boxed{C_6H_{12}O_6}$ C6H12O6

Difference Between Empirical and Molecular Formula

Empirical Formula	Molecular Formula
Simplest ratio of atoms	Actual number of atoms
May not represent actual molecule	Represents one molecule
Example: CH₂O	Example: C₆H₁₂O₆

4. Chemical Equations

A chemical reaction is represented using a chemical equation.

It shows:

Reactants
Products
Relative amounts involved

General Form

$\text{Reactants} \rightarrow \text{Products}$ Reactants→Products

Example: $CH_4+2O_2\rightarrow CO_2+2H_2O$ CH4+2O2→CO2+2H2O

Parts of a Chemical Equation

Reactants

Substances that take part in the reaction.

Example: $CH_4,\ O_2$ CH4, O2

Products

New substances formed during the reaction.

Example: $CO_2,\ H_2O$ CO2, H2O

Coefficients

Numbers written before formulas are called coefficients.

Example: $2O_2$ 2O2

The coefficient indicates the number of molecules or moles.

Balanced Chemical Equation

A balanced equation has equal numbers of atoms of each element on both sides.

Example:

Unbalanced: $H_2+O_2\rightarrow H_2O$ H2+O2→H2O

Balanced: $2H_2+O_2\rightarrow2H_2O$ 2H2+O2→2H2O

5. Stoichiometry

Meaning

The word stoichiometry comes from Greek words:

Stoicheion = element
Metron = measure

Stoichiometry deals with calculations involving:

Reactant quantities
Product quantities

Importance of Stoichiometry

It helps calculate:

Amount of reactants required
Amount of products formed
Limiting reactant
Percentage yield

Stoichiometric Relationships

Consider: $CH_4+2O_2\rightarrow CO_2+2H_2O$ CH4+2O2→CO2+2H2O

This tells us:

Mole Relationship

1 mole CH₄ reacts with:

2 moles O₂

to produce:

1 mole CO₂ and 2 moles H₂O

Mass Relationship

Molar masses:

CH₄ = 16 g

O₂ = 32 g

CO₂ = 44 g

H₂O = 18 g

Therefore: $16g CH_4+64g O_2$ 16gCH4+64gO2

produces: $44g CO_2+36g H_2O$ 44gCO2+36gH2O

Steps in Stoichiometric Calculations

Step 1

Write a balanced chemical equation.

Step 2

Convert given quantity into moles. $n=\frac{mass}{molar\,mass}$ n=molarmassmass

Step 3

Use mole ratio from balanced equation.

Step 4

Convert required moles into desired units.

6. Limiting Reagent

Definition

The reactant that gets completely consumed first during a reaction is called the limiting reagent.

It limits the amount of product formed.

Excess Reagent

The reactant left behind after the reaction is completed is called the excess reagent.

Example

Reaction: $2H_2+O_2\rightarrow2H_2O$ 2H2+O2→2H2O

Requirement:

2 moles H₂ need 1 mole O₂.

If we have:

5 moles H₂
1 mole O₂

Only 2 moles H₂ react with 1 mole O₂.

Therefore:

O₂ is completely consumed.
O₂ is the limiting reagent.

7. Percentage Yield

In real experiments, the actual amount of product obtained is usually less than the theoretical amount.

Formula

$\%Yield= \frac{\text{Actual yield}} {\text{Theoretical yield}} \times100$ %Yield=Theoretical yieldActual yield×100

Theoretical Yield

The maximum amount of product predicted by calculation.

Actual Yield

The amount of product actually obtained experimentally.

Reasons for Lower Yield

Incomplete reaction
Side reactions
Loss during separation
Experimental errors

Important Formula Sheet

Mass Percentage

$\%= \frac{\text{Mass of element}} {\text{Molar mass of compound}} \times100$ %=Molar mass of compoundMass of element×100

Number of Moles

$n=\frac{\text{Given mass}}{\text{Molar mass}}$ n=Molar massGiven mass

Particles from Moles

$N=n\times6.022\times10^{23}$ N=n×6.022×1023

Molecular Formula

$= (\text{Empirical Formula})_n$ =(Empirical Formula)n $n= \frac{\text{Molar mass}} {\text{Empirical formula mass}}$ n=Empirical formula massMolar mass

Percentage Yield

$\%Yield= \frac{\text{Actual yield}} {\text{Theoretical yield}} \times100$ %Yield=Theoretical yieldActual yield×100

Quick Revision

✔ Percentage composition gives the mass percentage of elements in a compound.

✔ Empirical formula gives the simplest atomic ratio.

✔ Molecular formula gives actual atoms present.

✔ Balanced equations follow conservation of mass.

✔ Stoichiometry calculates quantities involved in reactions.

✔ Limiting reagent determines maximum product formation.

✔ Actual yield is usually lower than theoretical yield.

Part 8: Complete Chapter Revision, Formula Sheet & Exam Preparation

Chapter Complete Revision

1. Chemistry and Matter

Chemistry

Chemistry is the branch of science that studies:

Composition of matter
Structure of substances
Properties of substances
Chemical transformations

Matter

Matter is anything that:

✔ Has mass
✔ Occupies space

Examples:

Air
Water
Metals
Food

2. Classification of Matter

Matter
│
├── Pure Substance
│     │
│     ├── Element
│     │
│     └── Compound
│
└── Mixture
      │
      ├── Homogeneous
      │
      └── Heterogeneous

Element

Contains only one type of atom.
Cannot be broken into simpler substances by chemical methods.

Examples:

Fe
Cu
O₂
H₂

Compound

Formed by chemical combination of elements.
Elements combine in fixed ratios.

Examples:

H₂O
CO₂
NaCl

Mixture

Physical combination of substances.
Composition can vary.
Components retain their properties.

Examples:

Air
Salt solution
Soil

3. States of Matter

State	Main Features
Solid	Fixed shape and volume
Liquid	Fixed volume but no fixed shape
Gas	No fixed shape or volume

4. Measurement in Chemistry

Every measurement contains: $\text{Numerical value + Unit}$ Numerical value + Unit

Example:

5 g

5 → value
g → unit

SI Base Units

Quantity	SI Unit
Length	metre (m)
Mass	kilogram (kg)
Time	second (s)
Temperature	kelvin (K)
Amount of substance	mole (mol)

Important Conversions

$1L=1000mL$ 1L=1000mL $1L=1000cm^3$ 1L=1000cm3 $1kg=1000g$ 1kg=1000g

5. Density

Density represents mass per unit volume.

Formula: $d=\frac{m}{V}$ d=Vm

Common unit: $g\,cm^{-3}$ gcm−3

6. Temperature Conversion

Kelvin scale: $K=^{\circ}C+273.15$ K=∘C+273.15

Celsius-Fahrenheit: $\frac{C}{5}=\frac{F-32}{9}$ 5C=9F−32

7. Scientific Notation

General form: $N\times10^n$ N×10n

where:

N is between 1 and 10
n is an integer

Example: $602000000000000000000000$ 602000000000000000000000

can be written as: $6.02\times10^{23}$ 6.02×1023

8. Significant Figures

Significant figures show the accuracy of a measurement.

Rules

Significant:

✔ Non-zero digits

Example:

245 → 3 significant figures

✔ Zeros between non-zero digits

Example:

1005 → 4 significant figures

✔ Zeros after decimal

Example:

5.00 → 3 significant figures

Not Significant:

✘ Zeros before the first non-zero digit

Example:

0.0052 → 2 significant figures

9. Accuracy and Precision

Accuracy

Closeness to the actual value.

Precision

Closeness among repeated measurements.

A good measurement should have:

✔ High accuracy
✔ High precision

10. Laws of Chemical Combination

Law of Conservation of Mass

Given by:

Lavoisier

Statement:

Mass is neither created nor destroyed during a chemical reaction. $\text{Mass of reactants}= \text{Mass of products}$ Mass of reactants=Mass of products

Law of Definite Proportions

Given by:

Proust

A compound always contains elements in a fixed ratio by mass.

Example:

Water: $H:O=1:8$ H:O=1:8

Law of Multiple Proportions

Given by:

Dalton

When elements form different compounds, masses combine in simple whole-number ratios.

Example:

CO and CO₂

Gay-Lussac’s Law

Gas volumes combine in simple whole-number ratios.

Avogadro’s Law

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

11. Dalton’s Atomic Theory

Main ideas:

✔ Matter consists of atoms.

✔ Atoms combine in whole-number ratios.

✔ Chemical reactions involve rearrangement of atoms.

✔ Atoms are not created or destroyed.

Limitations

Atoms are divisible into:

Electrons
Protons
Neutrons

Isotopes show atoms of the same element may have different masses.

12. Atomic Mass

Atomic mass is expressed in: $u$ u

1 atomic mass unit: $1u=\frac{1}{12}$ 1u=121

mass of carbon-12 atom.

13. Molecular Mass

Sum of atomic masses of atoms present in a molecule.

Example:

Water: $H_2O$ H2O $=2(1)+16$ =2(1)+16 $=18u$ =18u

14. Formula Mass

Used for ionic compounds.

Example:

NaCl: $23+35.5$ 23+35.5 $=58.5u$ =58.5u

15. Mole Concept

One mole contains: $6.022\times10^{23}$ 6.022×1023

particles.

This number is called:

Avogadro Constant

Important Mole Formulas

Number of moles

$n=\frac{\text{Given mass}}{\text{Molar mass}}$ n=Molar massGiven mass

Number of particles

$N=n\times N_A$ N=n×NA

where: $N_A=6.022\times10^{23}$ NA=6.022×1023

Mass

$\text{Mass}=n\times\text{Molar mass}$ Mass=n×Molar mass

16. Percentage Composition

Formula: $\%\,\text{element} = \frac{\text{Mass of element in one mole}} {\text{Molar mass of compound}} \times100$ %element=Molar mass of compoundMass of element in one mole×100

17. Empirical Formula

Represents:

Simplest whole-number ratio of atoms.

Steps:

Assume 100 g sample.
Convert percentage into grams.
Convert grams into moles.
Divide by smallest mole value.
Obtain simplest ratio.

18. Molecular Formula

Formula: $\text{Molecular Formula} = (\text{Empirical Formula})_n$ Molecular Formula=(Empirical Formula)n

where: $n= \frac{\text{Molar Mass}} {\text{Empirical Formula Mass}}$ n=Empirical Formula MassMolar Mass

19. Stoichiometry

Stoichiometry deals with quantitative relationships in chemical reactions.

Steps:

Write balanced equation.
Convert given data into moles.
Apply mole ratio.
Convert into required quantity.

20. Limiting Reagent

The reactant that gets consumed first is called the limiting reagent.

It decides:

Maximum product formed

21. Percentage Yield

Formula: $\%Yield= \frac{\text{Actual Yield}} {\text{Theoretical Yield}} \times100$ %Yield=Theoretical YieldActual Yield×100

Most Important Formula Sheet (One Page Revision)

Concept	Formula
Density	$d=\frac{m}{V}$ d=Vm
Moles	$n=\frac{mass}{molar\,mass}$ n=molarmassmass
Particles	$N=n\times6.022\times10^{23}$ N=n×6.022×1023
Mass	$m=n\times molar\,mass$ m=n×molarmass
Percentage composition	$\frac{Mass\ of\ element}{Molar\ mass}\times100$ Molar massMass of element×100
Molecular formula	(Empirical formula)ₙ
Value of n	$\frac{Molar\ mass}{Empirical\ formula\ mass}$ Empirical formula massMolar mass
Percentage yield	$\frac{Actual}{Theoretical}\times100$ TheoreticalActual×100
Kelvin conversion	°C + 273.15

Common Mistakes Students Make

❌ Confusing mass with weight
✅ Mass is constant; weight depends on gravity.

❌ Forgetting to balance chemical equations
✅ Always balance before stoichiometric calculations.

❌ Treating empirical formula as molecular formula
✅ Molecular formula is a multiple of empirical formula.

❌ Incorrect significant figures
✅ Apply rules carefully during calculations.

❌ Using wrong molar mass
✅ Always calculate using correct atomic masses.

Important Exam Questions

Very Short Answer Questions

What is one mole?

Answer: Amount of substance containing $6.022\times10^{23}$ 6.022×1023 particles.

Define molar mass.

Answer: Mass of one mole of a substance.

State law of conservation of mass.

Answer: Mass remains constant during a chemical reaction.

What is an empirical formula?

Answer: Simplest whole-number ratio of atoms in a compound.

Numerical Practice Areas

Students should practice:

✔ Mole calculations
✔ Mass-mole conversions
✔ Particle calculations
✔ Percentage composition
✔ Empirical formula problems
✔ Stoichiometry
✔ Limiting reagent problems