Class 11Chemistry Chapter 1 SOME BASIC CONCEPTS OF CHEMISTRY

Development of Chemistry

1. Origin and Early Aims

In ancient times, chemistry was not studied as a pure science; it grew from practical needs.
Two major goals of early practitioners:
1. Philosopher’s Stone (Paras) – a mythical substance believed to convert base metals like iron and copper into gold.
2. Elixir of Life – a substance believed to grant immortality.

2. Chemistry in Ancient India

Known as Rasayan Shastra, Ras Kriya, Rasvidya.
Involved metallurgy, medicine, cosmetics, glass, dyes, etc.
Evidence from Indus Valley Civilisation:
- Use of baked bricks in construction.
- Production of glazed pottery.
- Use of gypsum cement (lime + sand + CaCO₃).
- Manufacture of faience (a type of glass) for ornaments.
- Metal objects from lead, silver, gold, copper.
- Strengthening of copper with tin and arsenic.
- Glass objects found in Maski, Hastinapur, Taxila.

3. Metallurgy and Craftsmanship

Copper metallurgy began in the Chalcolithic Age in India.
Techniques for extraction of copper and iron developed indigenously.
Tanning of leather, dyeing of cotton practiced as per Rigveda.
Mastery over kiln temperatures in pottery making.
Kautilya’s Arthashastra: describes production of salt from sea.

4. Chemical Knowledge in Ancient Texts

Sushruta Samhita: Importance of alkalies in medicine.
Charaka Samhita: Preparation of acids, metal oxides, and salts.
Rasopanishad: Preparation of gunpowder.
Tamil texts: Fireworks using sulphur, charcoal, saltpetre, mercury, camphor.

5. Notable Scientists

Nagarjuna:
- Expert in chemistry, alchemy, and metallurgy.
- Wrote Rasratnakar – mercury compounds, metal extraction.
Chakrapani:
- Discovered mercury sulphide.
- Invented soap using mustard oil and alkalies.

6. Other Contributions

Ajanta & Ellora paintings – still fresh due to advanced natural adhesives.
Varahamihira’s Brihat Samhita: adhesives from plant extracts.
Ancient dyes: turmeric, madder, sunflower, lac, orpiment.
Ancient perfumes, cosmetics, hair dyes from plants and minerals.
Ink known in India since the 4th century CE.
Fermentation for liquors described in Vedas and Arthashastra.

7. Ancient Atomic Theory

Acharya Kanad (600 BCE):
- First proponent of atomic theory.
- Introduced the term Paramāṇu for indivisible particles.
- Atoms are eternal, indestructible, spherical, and in constant motion.
- Different atoms combine in fixed ratios to form matter.
Charaka Samhita also mentions reduction of particle size (early nanotechnology).

8. Decline and Revival

After medieval iatrochemistry, traditional Indian chemistry declined with Western medicine’s introduction.
Modern chemistry in India grew after the arrival of European scientists in the 19th century.

✅ Key Idea: Chemistry evolved from practical needs into a scientific discipline. Ancient India made significant contributions in metallurgy, medicine, dyes, and early atomic theory, many of which align with modern chemical principles.

Importance of Chemistry

1. Central Role in Science

Chemistry is interconnected with other branches like physics, biology, geology, environmental science, and materials science.
Principles of chemistry help explain natural phenomena — from weather patterns to brain function and computer operations.

2. Applications in Industry

Chemistry is used in manufacturing:
- Fertilisers (e.g., urea, ammonium nitrate)
- Acids & Bases (e.g., sulphuric acid, sodium hydroxide)
- Salts, dyes, polymers, metals, alloys
- Soaps and detergents
- Pharmaceutical products
These industries boost the national economy and create employment.

3. Role in Healthcare

Chemistry enables isolation and synthesis of life-saving drugs.
- Cisplatin, Taxol – effective in cancer therapy.
- AZT (Azidothymidine) – used in AIDS treatment.
Provides improved pesticides and insecticides to protect crops and ensure food supply.

4. Development of New Materials

Knowledge of chemical principles allows scientists to design substances with special properties:
- Superconducting ceramics
- Conducting polymers
- Optical fibres
- Magnetic and electronic materials

5. Environmental Protection

Chemistry contributes to solutions for environmental degradation:
- Developing eco-friendly refrigerants as alternatives to CFCs (which damage ozone).
- Research on controlling greenhouse gases like CO₂ and CH₄.
- Development of biodegradable materials.

6. Future Challenges

Understanding complex biochemical processes.
Large-scale production of chemicals using enzymes.
Synthesis of exotic materials for advanced technologies.
India needs talented chemists to address these global and national challenges.

✅ Key Idea: Chemistry not only helps us understand the natural world but also improves human life through industry, healthcare, technology, and environmental conservation. It is essential for sustainable development.

Here are Class 11 Chemistry notes for the next topic from your file — “Nature of Matter” — in a clear, exam-oriented format:

Nature of Matter

1. Definition

Matter: Anything that has mass and occupies space.
Examples: book, pen, air, water, all living beings.

2. States of Matter

Matter exists in three physical states: Solid, Liquid, Gas.

Property	Solid	Liquid	Gas
Volume	Definite	Definite	No definite volume
Shape	Definite	No definite shape (takes container’s shape)	No definite shape
Particle Arrangement	Very close, fixed positions	Close, but can move	Far apart, move freely
Movement of particles	Very little	Moderate	Free & fast

Interconversion:
Solid ⇄ Liquid ⇄ Gas (by changing temperature & pressure).

3. Classification of Matter

At the macroscopic level, matter is classified into:

(A) Mixtures

Composed of two or more pure substances mixed physically.
Variable composition.
Components can be separated by physical methods (filtration, distillation, etc.).
Types:
1. Homogeneous mixture – uniform composition (e.g., sugar solution, air).
2. Heterogeneous mixture – non-uniform composition (e.g., sand in water, soil).

(B) Pure Substances

Fixed composition, properties same throughout.
Cannot be separated into simpler substances by physical methods.
Two types:
1. Elements – composed of only one type of atom.
  Examples:
  - Atoms: Na, Cu, Ag
  - Molecules: H₂, N₂, O₂
2. Compounds – atoms of different elements combined in a fixed ratio.
  - Properties differ from constituent elements.
    Example: Water (H₂O) – liquid, even though H₂ is flammable and O₂ supports combustion.
  - Can be separated only by chemical methods.

✅ Key Idea: Matter can be classified into mixtures and pure substances. Pure substances are either elements or compounds, while mixtures may be homogeneous or heterogeneous.

Here are Class 11 Chemistry notes for the next topic — “Properties of Matter and Their Measurement” — from your file, in a clear, exam-focused format:

Properties of Matter and Their Measurement

1. Types of Properties

(A) Physical Properties

Can be observed without changing the composition of the substance.
Examples: colour, odour, melting point, boiling point, density.

(B) Chemical Properties

Observed only when a chemical change occurs.
Examples: combustibility, reactivity with acids/bases, chemical composition.

2. Measurement of Physical Properties

Any quantitative measurement = number + unit
Example: Length of a rod = 5 m.

3. International System of Units (SI)

Adopted in 1960 by CGPM.
Seven base quantities & units:

Quantity	Symbol	SI Unit	Symbol
Length	l	metre	m
Mass	m	kilogram	kg
Time	t	second	s
Electric Current	I	ampere	A
Temperature	T	kelvin	K
Amount of Substance	n	mole	mol
Luminous Intensity	Iv	candela	cd

Prefixes used for multiples/submultiples (e.g., kilo = 10³, milli = 10⁻³).

4. Mass and Weight

Mass: Amount of matter in an object (constant).
Weight: Force due to gravity on mass (varies with location).
SI Unit of mass = kilogram (kg). In lab work, gram (g) is common.

5. Volume

Space occupied by matter.
SI unit: m³
Common lab units: cm³, dm³, litre (L).
1 L = 1000 mL = 1000 cm³ = 1 dm³.

6. Density

Density=MassVolume\text{Density} = \frac{\text{Mass}}{\text{Volume}}

SI unit: kg m⁻³
Common in chemistry: g cm⁻³.
High density = particles closely packed.

7. Temperature Scales

Celsius (°C), Fahrenheit (°F), Kelvin (K).
Conversions:
- °F=95(°C)+32°F = \frac{9}{5} (°C) + 32
- K=°C+273.15K = °C + 273.15
Kelvin scale has no negative values.

✅ Key Idea: Physical and chemical properties describe matter. Accurate measurement requires SI units, proper tools, and understanding of mass, volume, density, and temperature scales.

Here are Class 11 Chemistry notes for the next topic — “Uncertainty in Measurement” — in a clear, student-friendly style:

Uncertainty in Measurement

1. Why Uncertainty Occurs

No measurement is perfectly exact due to:
- Instrument limitations (e.g., scale precision).
- Human error in reading.
Example:
Platform balance → 9.4 g
Analytical balance → 9.4213 g
→ The extra digits from the analytical balance are more precise.

2. Scientific Notation

Used for very large or very small numbers.
General form: N × 10ⁿ
- NN = number between 1 and 10
- nn = positive (large numbers) or negative (small numbers) integer.

Examples:

232.508 → 2.32508 × 10²
0.00016 → 1.6 × 10⁻⁴

Rules for operations:

Multiplication/Division → multiply/divide N values, add/subtract exponents.
Addition/Subtraction → make exponents equal first.

3. Significant Figures (Sig. Figs.)

Definition: Meaningful digits in a measurement (all certain + one uncertain digit).
Rules:
1. All non-zero digits are significant.
  Example: 285 (3 sig. figs.)
2. Zeros before first non-zero digit → not significant.
  Example: 0.0052 (2 sig. figs.)
3. Zeros between non-zero digits → significant.
  Example: 2.005 (4 sig. figs.)
4. Zeros after decimal and a non-zero digit → significant.
  Example: 0.200 (3 sig. figs.)
5. Exact counting numbers have infinite significant figures.
  Example: 2 balls, 20 eggs.

4. Precision vs Accuracy

Precision: closeness of repeated measurements to each other.
Accuracy: closeness of measurement to the true value.

Example:
True value = 2.00 g

Precise, not accurate: 1.95 g, 1.93 g
Neither: 1.94 g, 2.05 g
Precise & accurate: 2.01 g, 1.99 g

5. Rounding Off Rules

Last digit removed > 5 → increase preceding digit by 1.
(1.386 → 1.39)
Last digit removed < 5 → keep preceding digit same.
(4.334 → 4.33)
Last digit removed = 5 →
- Preceding digit even → keep same (6.25 → 6.2)
- Preceding digit odd → increase by 1 (6.35 → 6.4)

6. Dimensional Analysis / Unit Factor Method

Used to convert from one unit to another.
Multiply by a unit factor = 1, arranged so unwanted units cancel.

Example:
Convert 3 inches to cm:
1 inch = 2.54 cm 3 in×2.54 cm1 in=7.62 cm3\ \text{in} \times \frac{2.54\ \text{cm}}{1\ \text{in}} = 7.62\ \text{cm}

✅ Key Idea: Scientific notation, significant figures, and dimensional analysis help express measurements clearly and convert between units while maintaining accuracy and precision.

Here are Class 11 Chemistry notes for the next topic — “Laws of Chemical Combinations” — in a simple, exam-oriented format:

Laws of Chemical Combinations

These laws describe how elements combine to form compounds.

1. Law of Conservation of Mass

Proposed by: Antoine Lavoisier (1789)
Statement: Matter can neither be created nor destroyed in a chemical reaction.
Implication: Mass of reactants = Mass of products.
Example: CaCO3→ΔCaO+CO2\text{CaCO}_3 \xrightarrow{\Delta} \text{CaO} + \text{CO}_2 Total mass before heating = Total mass after heating.

2. Law of Definite Proportions (Law of Constant Composition)

Proposed by: Joseph Proust
Statement: A given compound always contains the same elements combined in the same proportion by mass.
Example: Water (H₂O) from any source contains H and O in the mass ratio 1:8.

3. Law of Multiple Proportions

Proposed by: John Dalton (1803)
Statement: If two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole-number ratios.
Example:
- Water (H₂O): 2 g H + 16 g O
- Hydrogen Peroxide (H₂O₂): 2 g H + 32 g O
  Ratio of oxygen masses: 16:32 = 1:2

4. Gay-Lussac’s Law of Gaseous Volumes

Proposed by: Joseph Louis Gay-Lussac (1808)
Statement: When gases combine or are produced, they do so in simple whole-number ratios by volume (measured at same temperature and pressure).
Example:
2 volumes H₂ + 1 volume O₂ → 2 volumes H₂O vapour
Volume ratio = 2:1:2

5. Avogadro’s Law

Proposed by: Amedeo Avogadro (1811)
Statement: Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Example:
1 L O₂, 1 L N₂, and 1 L H₂ (at same T & P) → each has the same number of molecules.

✅ Key Idea: These five laws form the foundation of chemical formulas and equations, explaining the fixed and simple ratios in which substances combine.

Here are Class 11 Chemistry notes for the next topic — “Dalton’s Atomic Theory” — in a clear, exam-oriented format:

Dalton’s Atomic Theory

1. Background

The idea that matter is made of tiny indivisible particles dates back to Democritus (Greek philosopher, 460–370 BCE).
John Dalton (1808) gave the first modern scientific theory of the atom in his book A New System of Chemical Philosophy.
His theory was based on the experimental evidence provided by the laws of chemical combinations.

2. Postulates of Dalton’s Atomic Theory

Matter consists of indivisible particles called atoms.
All atoms of a given element are identical in mass and properties; atoms of different elements differ in mass and properties.
Compounds are formed when atoms of different elements combine in a fixed whole-number ratio.
Chemical reactions involve rearrangement of atoms.
- Atoms are neither created nor destroyed in a chemical reaction.

3. Merits

Successfully explained:
- Law of Conservation of Mass
- Law of Definite Proportions
- Law of Multiple Proportions

4. Limitations

Could not explain:
- Gay-Lussac’s Law of Gaseous Volumes.
- Nature of forces between atoms.
- Divisibility of atoms into subatomic particles (discovered later: electrons, protons, neutrons).
- Isotopes (atoms of same element with different masses).
- Isobars (atoms of different elements with same mass).

✅ Key Idea: Dalton’s theory laid the foundation for modern atomic structure, even though later discoveries modified some of its postulates.

Here are Class 11 Chemistry notes for the next topic — “Atomic and Molecular Masses” — from your file, in a clear and concise format:

Atomic and Molecular Masses

1. Atomic Mass

Definition: Mass of a single atom, expressed in atomic mass units (u).
Standard: Carbon-12 isotope (¹²C) assigned exactly 12 u.
1 atomic mass unit (1 u) = 112\frac{1}{12} of the mass of one carbon-12 atom.
= 1.66056×10−241.66056 \times 10^{-24} g.
Example:
- Mass of H atom = 1.6736×10−241.6736 \times 10^{-24} g ≈ 1.008 u.
- Mass of O atom = 15.995 u.
Note: ‘amu’ is now replaced by u (unified mass unit).

2. Average Atomic Mass

Most elements occur as a mixture of isotopes.
Average atomic mass = weighted average of isotope masses based on their abundance.

Example – Carbon:

Isotope	Abundance (%)	Atomic Mass (u)
¹²C	98.892	12.00000
¹³C	1.108	13.00335
¹⁴C	~0	14.00317

Calculation: (0.98892×12)+(0.01108×13.00335)=12.011 u(0.98892 \times 12) + (0.01108 \times 13.00335) = 12.011\ \text{u}

This value is shown in the periodic table.

3. Molecular Mass

Definition: Sum of atomic masses of all atoms present in a molecule.
Formula:

Molecular Mass=∑(Atomic Mass of each element×Number of atoms)\text{Molecular Mass} = \sum (\text{Atomic Mass of each element} \times \text{Number of atoms})

Examples:
- Methane (CH₄): 12.011+4(1.008)=16.043 u12.011 + 4(1.008) = 16.043\ \text{u}
- Water (H₂O): 2(1.008)+16.00=18.02 u2(1.008) + 16.00 = 18.02\ \text{u}

4. Formula Mass

Used for ionic compounds which do not have discrete molecules (e.g., NaCl).
Definition: Sum of atomic masses of all atoms in the formula unit.
Example:
NaCl = 23.0+35.5=58.5 u23.0 + 35.5 = 58.5\ \text{u}
(Here, Na⁺ is surrounded by Cl⁻ ions in a crystal lattice, not separate molecules.)

✅ Key Idea:

Atomic mass is for single atoms.
Average atomic mass accounts for isotopes.
Molecular mass is for covalent compounds.
Formula mass is for ionic compounds.

Here are Class 11 Chemistry notes for the next topic — “Atomic and Molecular Masses” — from your file, in a clear and concise format:

Atomic and Molecular Masses

1. Atomic Mass

Definition: Mass of a single atom, expressed in atomic mass units (u).
Standard: Carbon-12 isotope (¹²C) assigned exactly 12 u.
1 atomic mass unit (1 u) = 112\frac{1}{12} of the mass of one carbon-12 atom.
= 1.66056×10−241.66056 \times 10^{-24} g.
Example:
- Mass of H atom = 1.6736×10−241.6736 \times 10^{-24} g ≈ 1.008 u.
- Mass of O atom = 15.995 u.
Note: ‘amu’ is now replaced by u (unified mass unit).

2. Average Atomic Mass

Most elements occur as a mixture of isotopes.
Average atomic mass = weighted average of isotope masses based on their abundance.

Example – Carbon:

Isotope	Abundance (%)	Atomic Mass (u)
¹²C	98.892	12.00000
¹³C	1.108	13.00335
¹⁴C	~0	14.00317

Calculation: (0.98892×12)+(0.01108×13.00335)=12.011 u(0.98892 \times 12) + (0.01108 \times 13.00335) = 12.011\ \text{u}

This value is shown in the periodic table.

3. Molecular Mass

Definition: Sum of atomic masses of all atoms present in a molecule.
Formula:

Molecular Mass=∑(Atomic Mass of each element×Number of atoms)\text{Molecular Mass} = \sum (\text{Atomic Mass of each element} \times \text{Number of atoms})

Examples:
- Methane (CH₄): 12.011+4(1.008)=16.043 u12.011 + 4(1.008) = 16.043\ \text{u}
- Water (H₂O): 2(1.008)+16.00=18.02 u2(1.008) + 16.00 = 18.02\ \text{u}

4. Formula Mass

Used for ionic compounds which do not have discrete molecules (e.g., NaCl).
Definition: Sum of atomic masses of all atoms in the formula unit.
Example:
NaCl = 23.0+35.5=58.5 u23.0 + 35.5 = 58.5\ \text{u}
(Here, Na⁺ is surrounded by Cl⁻ ions in a crystal lattice, not separate molecules.)

✅ Key Idea:

Atomic mass is for single atoms.
Average atomic mass accounts for isotopes.
Molecular mass is for covalent compounds.
Formula mass is for ionic compounds.

Here are Class 11 Chemistry notes for the next topic — “Mole Concept and Molar Masses” — in a clear, exam-focused style:

Mole Concept and Molar Masses

1. Mole – Definition

SI unit of amount of substance (symbol: mol).
1 mole = 6.02214076×10236.02214076 \times 10^{23} elementary entities
(atoms, molecules, ions, electrons, formula units, etc.).
This constant is called the Avogadro Number (NA_A).

2. Significance

1 mole of:
- Atoms → NAN_A atoms
- Molecules → NAN_A molecules
- Formula units → NAN_A units
Examples:
- 1 mol H atoms = 6.022×10236.022 \times 10^{23} atoms H
- 1 mol H₂O molecules = 6.022×10236.022 \times 10^{23} molecules H₂O

3. Molar Mass

Definition: Mass of 1 mole of a substance in grams.
Numerically equal to its atomic/molecular/formula mass in u.
Examples:
- H₂O: Molecular mass = 18.02 u → Molar mass = 18.02 g/mol
- NaCl: Formula mass = 58.5 u → Molar mass = 58.5 g/mol

4. Relationship Between Moles, Mass, and Number of Particles

Moles (n)=Given mass (g)Molar mass (g/mol)\text{Moles (n)} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}} Number of particles=n×NA\text{Number of particles} = n \times N_A

Example:
How many molecules are in 36 g of H₂O?

Molar mass of H₂O = 18 g/mol
n=3618=2 moln = \frac{36}{18} = 2\ \text{mol}
Number of molecules = 2×6.022×1023=1.2044×10242 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} molecules.

✅ Key Idea: The mole concept links mass, number of particles, and molar mass, allowing easy conversion between microscopic (atoms/molecules) and macroscopic (grams/litres) quantities.

Here are Class 11 Chemistry notes for the next topic — “Percentage Composition and Empirical/Molecular Formulas” — in a clear, exam-oriented style:

Percentage Composition and Empirical/Molecular Formulas

1. Percentage Composition

Definition: The percentage by mass of each element present in a compound.
Formula:

% Element=Mass of element in 1 mol of compoundMolar mass of compound×100\% \ \text{Element} = \frac{\text{Mass of element in 1 mol of compound}}{\text{Molar mass of compound}} \times 100

Example – Water (H₂O):
- Molar mass = 18 g/mol
- H: 2.01618×100=11.19%\frac{2.016}{18} \times 100 = 11.19\%
- O: 1618×100=88.81%\frac{16}{18} \times 100 = 88.81\%

2. Empirical Formula

Definition: Simplest whole-number ratio of atoms of each element in a compound.
Steps to Calculate:
1. Find mass % of each element.
2. Convert % into moles (divide by atomic mass).
3. Divide each mole value by the smallest one to get a whole-number ratio.
Example:
- Compound has 40% C, 6.7% H, 53.3% O.
- C: 4012=3.33\frac{40}{12} = 3.33
  H: 6.71=6.7\frac{6.7}{1} = 6.7
  O: 53.316=3.33\frac{53.3}{16} = 3.33
- Ratio: C : H : O = 1 : 2 : 1 → Empirical formula = CH₂O

3. Molecular Formula

Definition: Actual number of atoms of each element in one molecule of the compound.
Relationship:

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

where n=Molar mass of compoundEmpirical formula massn = \frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}

Example:
- Empirical formula mass of CH₂O = 30 g/mol
- Molar mass = 60 g/mol
- n=6030=2n = \frac{60}{30} = 2
- Molecular formula = C₂H₄O₂

✅ Key Idea:

Percentage composition helps identify how much of each element is present.
Empirical formula gives the simplest ratio; molecular formula gives the actual count of atoms.

Here are Class 11 Chemistry notes for the next topic — “Stoichiometry and Stoichiometric Calculations” — in a simple, exam-oriented format:

Stoichiometry and Stoichiometric Calculations

1. Stoichiometry – Definition

Stoichiometry: Branch of chemistry dealing with the quantitative relationship between reactants and products in a chemical reaction.
Based on the balanced chemical equation.

2. Balanced Chemical Equation

Represents the correct mole ratio of reactants and products.
Example:

2 H2+O2→2 H2O2\ \text{H}_2 + \text{O}_2 \rightarrow 2\ \text{H}_2\text{O}

Interpretation:
- Mole ratio: 2 mol H₂ : 1 mol O₂ → 2 mol H₂O
- Mass ratio: (2 × 2 g H₂) : (32 g O₂) → (2 × 18 g H₂O)
  i.e., 4 g H₂ reacts with 32 g O₂ to produce 36 g H₂O.

3. Types of Stoichiometric Calculations

(A) Mass–Mass Calculations

Write the balanced equation.
Convert given mass → moles.
Use mole ratio from equation.
Convert moles of required substance → mass.

(B) Mass–Volume Calculations

Use molar volume of a gas at STP = 22.4 L/mol.

(C) Volume–Volume Calculations (for gases)

Apply Gay-Lussac’s law (simple ratio of volumes).

4. Limiting Reagent

Definition: The reactant that is completely consumed first, limiting the amount of product formed.
Steps to identify:
1. Convert all reactant masses to moles.
2. Divide each mole value by its coefficient in the balanced equation.
3. The smallest value → limiting reagent.

Example:
N₂ + 3 H₂ → 2 NH₃
Given: 5 mol N₂ and 12 mol H₂

Required H₂ for 5 mol N₂ = 5×3=155 \times 3 = 15 mol (not enough available)
Hence, H₂ is limiting reagent.

5. Percentage Yield

Formula:

% Yield=Actual yieldTheoretical yield×100\% \ \text{Yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100

Actual yield: amount actually obtained in lab.
Theoretical yield: amount predicted by stoichiometry.

✅ Key Idea:
Stoichiometry connects the macroscopic quantities of reactants and products using the mole concept, and helps in predicting yields, identifying limiting reagents, and performing mass–volume conversions.

Reactions in Solutions

1. Concentration of Solutions

Concentration expresses the amount of solute present in a given quantity of solution or solvent.

(A) Mass Percentage (% w/w)

% by mass=Mass of soluteMass of solution×100\% \ \text{by mass} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100

Example: 10 g NaCl in 90 g water → solution mass = 100 g
% by mass = 10100×100=10%\frac{10}{100} \times 100 = 10\%

(B) Volume Percentage (% v/v)

% by volume=Volume of soluteVolume of solution×100\% \ \text{by volume} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100

Example: 25 mL ethanol in 75 mL water → solution volume = 100 mL
% v/v = 25100×100=25%\frac{25}{100} \times 100 = 25\%

(C) Mass by Volume Percentage (% w/v)

% w/v=Mass of solute (g)Volume of solution (mL)×100\% \ \text{w/v} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100

Common in medical preparations.

(D) Molarity (M)

Definition: Moles of solute per litre of solution.

M=Moles of soluteVolume of solution in litresM = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}

Example: 0.5 mol NaOH in 1 L solution → 0.5 M NaOH.

(E) Molality (m)

Definition: Moles of solute per kg of solvent.

m=Moles of soluteMass of solvent in kgm = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}

Independent of temperature (unlike molarity).

(F) Mole Fraction (X)

Definition: Ratio of moles of one component to total moles of all components.

XA=Moles of ATotal moles of all componentsX_A = \frac{\text{Moles of A}}{\text{Total moles of all components}}

For a binary solution: XA+XB=1X_A + X_B = 1.

(G) Parts Per Million (ppm)

Definition: Number of parts of solute per million parts of solution.

ppm=Mass of soluteMass of solution×106\text{ppm} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6

Useful for very dilute solutions.

2. Preparing Solutions for Reactions

Required steps:
1. Weigh the exact amount of solute.
2. Dissolve in small volume of solvent.
3. Transfer to a volumetric flask and make up to the mark.

3. Dilution Formula

When diluting a stock solution: M1V1=M2V2M_1 V_1 = M_2 V_2

M1M_1, V1V_1 → molarity and volume of concentrated solution
M2M_2, V2V_2 → molarity and volume of diluted solution.

✅ Key Idea: Different units (molarity, molality, mole fraction, ppm) are used to express concentration depending on the type of reaction and required accuracy. Molarity changes with temperature; molality does not.

In SHORT

1. Importance of Chemistry

Central Role in Science
- Chemistry connects physical sciences (physics, geology) and life sciences (biology, medicine).
- Explains the composition, structure, properties, and transformations of matter.
Applications in Industry
- Fertilizers: Urea, ammonium sulphate.
- Polymers: Plastics, synthetic fibres.
- Metals, alloys, soaps, detergents.
Role in Healthcare
- Life-saving drugs: antibiotics, anticancer agents.
- Pesticides, insecticides, antiseptics.
Development of New Materials
- Superconductors, optical fibres, advanced alloys.
Environmental Protection
- Eco-friendly refrigerants, biodegradable materials, greenhouse gas control.
Future Challenges
- Enzyme-based production, exotic materials, biochemical research.

2. Properties of Matter and Their Measurement

Physical Properties: Measured without changing composition (colour, boiling point).
Chemical Properties: Observed during chemical change (reactivity, flammability).

SI Units:
Metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), candela (cd).

Mass: constant; Weight: changes with gravity.
Volume: m³, L, cm³.
Density: mass/volume.
Temperature:
- K = °C + 273.15
- °F = (9/5 × °C) + 32

3. Uncertainty in Measurement

Causes: instrument precision limits, human error.
Scientific Notation: N × 10ⁿ, where 1 ≤ N < 10.
Significant Figures:
1. Non-zero digits significant.
2. Leading zeros not significant.
3. Zeros between non-zero digits significant.
4. Trailing zeros after decimal significant.
Precision: closeness of repeated results.
Accuracy: closeness to true value.
Dimensional Analysis: unit conversions using factor method.

4. Laws of Chemical Combinations

Law of Conservation of Mass – Mass is neither created nor destroyed.
Law of Definite Proportions – Same compound has same ratio by mass.
Law of Multiple Proportions – Ratios of masses are small whole numbers.
Gay-Lussac’s Law – Gases combine in simple volume ratios.
Avogadro’s Law – Equal volumes of gases (same T & P) contain equal molecules.

5. Dalton’s Atomic Theory

Postulates:

Matter is made of indivisible atoms.
Atoms of same element are identical; atoms of different elements differ.
Compounds form by fixed whole-number ratio of atoms.
Chemical reactions rearrange atoms without creation/destruction.

Limitations: Could not explain isotopes, isobars, subatomic particles, gas volume laws.

6. Atomic and Molecular Masses

Atomic Mass: Mass of one atom in u, based on C-12 isotope.
Average Atomic Mass: Weighted average considering isotopes.
Molecular Mass: Sum of atomic masses in a molecule.
Formula Mass: For ionic compounds (no discrete molecules).

7. Mole Concept and Molar Masses

1 mole = 6.022×10236.022 × 10^{23} entities (Avogadro’s number).
Molar Mass = mass of 1 mole (in g), numerically equal to molecular/formula mass in u.

Formulas: Moles=MassMolar Mass,Particles=Moles×NA\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}, \quad \text{Particles} = \text{Moles} × N_A

8. Percentage Composition & Empirical/Molecular Formulas

% Composition:

%Element=Mass of element in 1 moleMolar mass×100\% \text{Element} = \frac{\text{Mass of element in 1 mole}}{\text{Molar mass}} × 100

Empirical Formula: Simplest atom ratio. Steps: % → mass → moles → ratio.
Molecular Formula: n×Empirical formula\text{n} × \text{Empirical formula}, where
n=Molar massEmpirical massn = \frac{\text{Molar mass}}{\text{Empirical mass}}.

9. Stoichiometry and Stoichiometric Calculations

Based on balanced chemical equations.
Types: Mass–Mass, Mass–Volume, Volume–Volume.
Limiting Reagent: Consumed first.
% Yield:

%Yield=Actual yieldTheoretical yield×100\% \text{Yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} × 100

10. Reactions in Solutions

Concentration Units:
- Mass %, Volume %, Mass by Volume %, Molarity (mol/L), Molality (mol/kg), Mole Fraction, ppm.
Dilution Formula: M1V1=M2V2M_1V_1 = M_2V_2
Note: Molarity varies with temperature; molality does not.