Chapter 1: Some Basic Concepts of Chemistry

Introduction

Definition and Scope

• Chemistry studies preparation, properties, structures, and reactions of substances.
• Focuses on composition, structure, properties, and interactions of matter.
• Atoms and molecules are the basic units studied.
• Known as the science of atoms and molecules.

Solved Examples

• Ex 1: Identify the basic units studied in chemistry.
  Answer: Atoms and molecules.

Practice Set

• Q1: What is the primary focus of chemistry?
  A. Study of living organisms
  B. Study of atoms and molecules
  C. Study of celestial bodies
  D. Study of physical forces
• Q2: Chemistry is also called the science of:
  A. Cells and tissues
  B. Atoms and molecules
  C. Planets and stars
  D. Energy and work
• Q3: Which of the following is NOT a focus of chemistry?
  A. Properties of substances
  B. Reactions of substances
  C. Structure of substances
  D. Study of human behavior

Answer Key

• Q1: B - Chemistry focuses on atoms and molecules, the basic units of matter.
• Q2: B - Chemistry is known as the science of atoms and molecules.
• Q3: D - Chemistry does not study human behavior; it studies matter.

Memory Aids

• Chemistry = Science of Atoms + Molecules.
• Focus on composition, structure, properties, and reactions.

Importance of Chemistry

Applications and Role

• Studies properties, structures, and changes of substances.
• Central to science; linked with other branches.
• Explains weather, brain function, computer operation.
• Used in chemical industries: fertilizers, dyes, polymers, drugs, soaps, metals, alloys.
• Enables isolation and synthesis of lifesaving drugs (e.g., cisplatin, taxol, AZT).
• Develops safer alternatives to harmful refrigerants like CFCs.

Solved Examples

• Ex 1: Name a drug synthesized through chemistry used in cancer therapy.
  Answer: Cisplatin or Taxol.

Practice Set

• Q1: Which branch of science studies changes in substances?
  A. Physics
  B. Chemistry
  C. Biology
  D. Astronomy
• Q2: Which of the following is a product of chemical industry?
  A. Fertilizers
  B. Computers
  C. Planets
  D. Stars
• Q3: Which drug is used to help AIDS patients?
  A. Cisplatin
  B. Taxol
  C. AZT
  D. Penicillin

Answer Key

• Q1: B - Chemistry studies properties and changes of substances.
• Q2: A - Fertilizers are produced using chemical principles.
• Q3: C - AZT is used for AIDS treatment.

Memory Aids

• Chemistry = Central science linking various fields.
• Drugs: Cisplatin & Taxol (cancer), AZT (AIDS).
• CFCs replaced by safer refrigerants.

Nature of Matter

Physical States of Matter

• Matter occupies space and has mass.
• Exists in three states: solid, liquid, gas.
• Solid: Particles tightly packed, strong attraction, fixed positions, definite shape and volume.
• Liquid: Particles close but weak attraction, move past each other, definite volume, shape of container.
• Gas: Particles far apart, random motion, no definite shape or volume, fills container.

Interconversion of States

• States change by temperature and pressure.
• Heating solid → liquid (melting), liquid → gas (evaporation/boiling).
• Cooling gas → liquid (condensation), liquid → solid (freezing).
• Heating increases particle movement; cooling decreases it.

Classification of Matter

• Pure substances: Elements or compounds with fixed composition, e.g., copper, water, glucose.
• Cannot be separated by physical methods.
• Mixtures: Two or more substances, homogeneous or heterogeneous.
• Homogeneous mixture: Uniform composition, e.g., sugar solution, air.
• Heterogeneous mixture: Physically distinct phases, non-uniform, e.g., salt and sugar mix.

Solved Examples

• Ex 1: Identify the state of matter for a substance with fixed shape and volume.
  Answer: Solid, due to fixed particle positions and strong attraction.
• Ex 2: What happens when a liquid is heated?
  Answer: It evaporates or boils to form gas.
• Ex 3: Classify a mixture of salt and sugar.
  Answer: Heterogeneous mixture, as components are physically distinct.

Practice Set

• Q1: Which state of matter has definite volume but no definite shape?
  A. Solid
  B. Liquid
  C. Gas
  D. Plasma
• Q2: What process converts gas to liquid?
  A. Melting
  B. Freezing
  C. Condensation
  D. Evaporation
• Q3: Which of the following is a homogeneous mixture?
  A. Salt and sugar
  B. Air
  C. Oil and water
  D. Sand and water

Answer Key

• Q1: B - Liquids have definite volume but take the shape of the container.
• Q2: C - Condensation is gas to liquid transition on cooling.
• Q3: B - Air is a homogeneous mixture with uniform composition.

Memory Aids

• Solid = fixed shape and volume (S for Stable).
• Liquid = definite volume, shape varies (L for Loose shape).
• Gas = no fixed shape or volume (G for Go anywhere).
• Heating: Solid → Liquid → Gas (S to L to G).
• Cooling: Gas → Liquid → Solid (reverse order).
• Pure substances = fixed composition, no physical separation.
• Mixtures = physical combination, can be separated.

Properties of Matter and their Measurement

Physical Properties

• Associated with physical existence of matter.
• Determined without changing composition.
• Examples: states of matter, colour, taste, odour, melting point, boiling point, density.

Chemical Properties

• Ability to form new substances by decomposition or reaction.
• Examples: acidity, basicity, combustibility.

International System of Units (SI)

• Decimal system for measurement of physical quantities.
• Seven basic units: length (m), mass (kg), time (s), electric current (A), temperature (K), amount of substance (mol), luminous intensity (cd).
• Allows prefixes to indicate multiples or submultiples.

SI Prefixes

• Small prefixes: yocto (y, 10^-24), nano (n, 10^-9), micro (µ, 10^-6), etc.
• Common prefixes: milli (m, 10^-3), centi (c, 10^-2), deci (d, 10^-1), deca (da, 10¹), kilo (k, 10³), mega (M, 10⁶).
• Large prefixes: giga (G, 10⁹), tera (T, 10¹²), peta (P, 10¹⁵).

Mass and Weight

• Mass: amount of matter in a substance; SI unit is kilogram (kg).
• Measured accurately using an analytical balance.
• Weight: force exerted by gravity on an object.
• Mass is constant; weight varies with gravity.

Volume

• Amount of space occupied by a substance.
• SI unit: cubic meter (m³).
• Common lab units: cubic centimeter (cm³) or cubic decimeter (dm³).
• Conversions: 1 L = 1000 mL; 1000 cm³ = 1 dm³.

Density

• Mass per unit volume.
• SI unit: kg m^-3; commonly used unit: g cm^-3.

Temperature

• SI unit: Kelvin (K).
• Also measured in degree Celsius (°C) and degree Fahrenheit (°F).
• Conversions: °F = \( \frac{9}{5} \)(°C) + 32; K = °C + 273.15.

Solved Examples

• Ex 1: Convert 25 °C to Kelvin.
  Solution: K = 25 + 273.15 = 298.15 K.
• Ex 2: Calculate density if mass = 500 g and volume = 250 cm³.
  Solution: Density = mass/volume = 500/250 = 2 g cm^-3.

Practice Set

• Q1: Which of the following is a chemical property?
  A. Melting point
  B. Colour
  C. Combustibility
  D. Density
• Q2: What is the SI unit of electric current?
  A. Ampere (A)
  B. Kelvin (K)
  C. Mole (mol)
  D. Candela (cd)
• Q3: Convert 1000 mL to dm³.
  A. 0.1 dm³
  B. 1 dm³
  C. 10 dm³
  D. 100 dm³

Answer Key

• Q1: C. Combustibility is a chemical property involving reaction to form new substances.
• Q2: A. Ampere (A) is the SI unit of electric current.
• Q3: B. 1000 mL = 1 dm³ as 1000 cm³ = 1 dm³.

Memory Aids

• Physical properties: No change in composition (e.g., colour, melting point).
• Chemical properties: Involve new substance formation (e.g., combustibility).
• SI units: 7 basics - m, kg, s, A, K, mol, cd.
• Common prefixes: milli (10^-3), centi (10^-2), kilo (10³), mega (10⁶).
• Temperature conversions: K = °C + 273.15; °F = 9/5(°C) + 32.

Uncertainty in Measurement

Scientific Notation

• Express numbers as N × 10ⁿ, where 1 ≤ N < 10 and n is an integer.
• Example: 0.0000298 = 2.98 × 10^-5, 130596 = 1.30596 × 10⁵.
• Multiplication: (N₁ × 10ⁿ¹) × (N₂ × 10ⁿ²) = (N₁ × N₂) × 10ⁿ¹⁺ⁿ².
• Division: (N₁ × 10ⁿ¹) / (N₂ × 10ⁿ²) = (N₁ / N₂) × 10^n1-n2.
• Addition/Subtraction: Convert to same exponent, then add/subtract mantissas.

Precision and Accuracy

• Precision: Closeness of repeated measurements.
• Accuracy: Closeness to true value.
• Uncertainty shown by significant figures.
• Significant figures include all certain digits plus one uncertain digit.

Rules for Significant Figures

• All non-zero digits are significant.
• Exact numbers have infinite significant figures.
• Leading zeros are not significant.
• Zeros between non-zero digits or after non-zero digits (right of decimal) are significant.
• Result cannot have more decimal digits than original numbers.

Rounding Off Rules

• Addition/Subtraction: Round to least decimal places among terms.
• Multiplication/Division: Round to least significant figures among terms.
• If digit to remove > 5, increase preceding digit by 1.
• If digit to remove < 5, leave preceding digit unchanged.
• If digit to remove = 5, increase preceding digit if odd, else leave unchanged.

Dimensional Analysis

• Convert quantities using fundamental units.
• Use factor label or unit factor method.
• Example: 1 inch = 2.54 cm, so 1 inch / 2.54 cm = 1 = 2.54 cm / 1 inch.
• Unit conversion factors equal 1 and convert units without changing quantity.
• Quantity required = Quantity given × Conversion factor.

Solved Examples

• Ex 1: Multiply (5.6 × 10⁵) and (6.9 × 10⁸):
  Step 1: Multiply mantissas: 5.6 × 6.9 = 38.64
  Step 2: Add exponents: 5 + 8 = 13
  Step 3: Result = 38.64 × 10¹³ = 3.864 × 10¹⁴.
• Ex 2: Add 6.65 × 10⁴ and 8.95 × 10³:
  Step 1: Convert 8.95 × 10³ to 0.895 × 10⁴
  Step 2: Add mantissas: 6.65 + 0.895 = 7.545
  Step 3: Result = 7.545 × 10⁴.
• Ex 3: Convert 10 inches to cm:
  Step 1: Use conversion factor 1 inch = 2.54 cm
  Step 2: Multiply: 10 × 2.54 = 25.4 cm.

Practice Set

• Q1: What is the product of (3.2 × 10³) and (4.5 × 10²)?
  A. 1.44 × 10⁵
  B. 7.2 × 10⁵
  C. 1.44 × 10⁶
  D. 7.2 × 10⁶
• Q2: Which digit is significant in 0.00450?
  A. Only 4 and 5
  B. 4, 5 and trailing zero
  C. Only 4
  D. None
• Q3: Round off 0.05678 to three significant figures.
  A. 0.056
  B. 0.057
  C. 0.0567
  D. 0.0570

Answer Key

• Q1: B. 7.2 × 10⁵ - Multiply mantissas (3.2 × 4.5 = 14.4), add exponents (3 + 2 = 5), adjust to 7.2 × 10⁵.
• Q2: B. 4, 5 and trailing zero - Leading zeros are not significant; trailing zeros after decimal are significant.
• Q3: B. 0.057 - Third digit is 6 (>5), so second digit increases from 5 to 7.

Memory Aids

• Scientific notation: Multiply/divide mantissas, add/subtract exponents.
• Significant figures: Non-zero digits always count; leading zeros never count.
• Rounding 5 rule: Increase if preceding digit odd, else keep same.
• Dimensional analysis: Multiply by conversion factor = 1 to change units.

Law of Conservation of Mass

Explanation

• Matter cannot be created or destroyed in a chemical reaction.
• Total mass of reactants equals total mass of products.
• Mass remains constant during physical or chemical changes.

Solved Examples

• Ex 1: If 10 g of reactants react, mass of products = 10 g (mass conserved).

Practice Set

• Q1: In a chemical reaction, the total mass of reactants is 20 g. What is the total mass of products?
  • A. 10 g
  • B. 20 g
  • C. 30 g
  • D. Cannot be determined
• Q2: Which law states that mass is neither created nor destroyed?
  • A. Law of definite proportions
  • B. Law of conservation of mass
  • C. Law of multiple proportions
  • D. Avogadro's law
• Q3: When 5 g of hydrogen reacts with 40 g of oxygen, the mass of water formed is?
  • A. 35 g
  • B. 45 g
  • C. 50 g
  • D. 55 g

Answer Key

• Q1: B - Mass of products equals mass of reactants as per the law.
• Q2: B - Law of conservation of mass states mass is conserved.
• Q3: C - Total mass = 5 g + 40 g = 45 g water formed.

Memory Aids

• Mass before = Mass after reaction.
• Think: "Mass is conserved like money in a closed system."

Law of Definite Proportions

Explanation

• Ratio of masses of elements in a compound is constant.
• Same compound always contains elements in fixed mass ratio.
• Also called Law of Definite Composition.

Solved Examples

• Ex 1: Water always contains hydrogen and oxygen in 1:8 mass ratio.

Practice Set

• Q1: What does the law of definite proportions state?
  • A. Mass is conserved
  • B. Elements combine in fixed mass ratio
  • C. Gases combine in volume ratio
  • D. Equal volumes have equal molecules
• Q2: If 18 g of water contains 2 g hydrogen, mass of oxygen is?
  • A. 16 g
  • B. 20 g
  • C. 18 g
  • D. 2 g
• Q3: Which law is also called law of definite composition?
  • A. Law of conservation of mass
  • B. Law of multiple proportions
  • C. Law of definite proportions
  • D. Avogadro's law

Answer Key

• Q1: B - Elements combine in fixed mass ratio in a compound.
• Q2: A - Oxygen mass = 18 g - 2 g = 16 g.
• Q3: C - Law of definite proportions is also called law of definite composition.

Memory Aids

• Fixed mass ratio = fixed compound composition.
• Water always H:O = 1:8 by mass.

Law of Multiple Proportions

Explanation

• Two elements form more than one compound.
• Masses of one element combining with fixed mass of other are in small whole number ratios.
• Example: CO and CO₂ have C:O mass ratios in simple multiples.

Solved Examples

• Ex 1: Carbon monoxide and carbon dioxide show mass ratios 1:2 for oxygen combining with fixed carbon mass.

Practice Set

• Q1: Law of multiple proportions applies when?
  • A. One element forms one compound
  • B. Two elements form multiple compounds
  • C. Gases combine in volume ratio
  • D. Equal volumes have equal molecules
• Q2: If 1 g of element A combines with 2 g and 4 g of element B in two compounds, ratio is?
  • A. 1:2
  • B. 1:3
  • C. 1:4
  • D. 2:1
• Q3: Which pair illustrates law of multiple proportions?
  • A. H₂O and H₂O₂
  • B. CO and CO₂
  • C. NaCl and KCl
  • D. NH₃ and CH₄

Answer Key

• Q1: B - Two elements forming multiple compounds.
• Q2: A - Ratio of masses 2:4 = 1:2 small whole numbers.
• Q3: B - CO and CO₂ show law of multiple proportions.

Memory Aids

• Multiple compounds = simple mass ratios.
• CO and CO₂ classic example.

Gay Lussac's Law of Gaseous Volumes

Explanation

• Gases combine or form in simple volume ratios at same temperature and pressure.
• Volume ratios are small whole numbers.
• Example: 2 volumes H₂ + 1 volume O₂ → 2 volumes H₂O (gas).

Solved Examples

• Ex 1: 2 L H₂ reacts with 1 L O₂ to produce 2 L H₂O vapor.

Practice Set

• Q1: Gay Lussac's law relates to?
  • A. Mass ratios
  • B. Volume ratios of gases
  • C. Mole ratios
  • D. Energy changes
• Q2: If 3 L of N₂ reacts with 9 L of H₂, volume of NH₃ formed is?
  • A. 6 L
  • B. 9 L
  • C. 12 L
  • D. 18 L
• Q3: In reaction 2H₂ + O₂ → 2H₂O, volume ratio of reactants is?
  • A. 1:1
  • B. 2:1
  • C. 1:2
  • D. 3:2

Answer Key

• Q1: B - Law deals with volume ratios of gases.
• Q2: A - N₂:H₂:NH₃ = 1:3:2 volume ratio.
• Q3: B - Volume ratio of H₂ to O₂ is 2:1.

Memory Aids

• Gas volumes combine in simple ratios.
• Remember: 2 volumes H₂ + 1 volume O₂ → 2 volumes H₂O.

Avogadro Law

Explanation

• Equal volumes of gases at same temperature and pressure contain equal number of molecules.
• Volume proportional to number of molecules.
• Basis for molar volume concept (22.4 L at STP).

Solved Examples

• Ex 1: 22.4 L of any gas at STP contains 6.022 × 10²³ molecules.

Practice Set

• Q1: Avogadro's law states equal volumes of gases have?
  • A. Equal mass
  • B. Equal molecules
  • C. Equal density
  • D. Equal pressure
• Q2: Volume of 1 mole gas at STP is?
  • A. 11.2 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 1 L
• Q3: If 2 volumes of gas A and 3 volumes of gas B contain equal molecules, which is true?
  • A. Gas A is heavier
  • B. Gas B is heavier
  • C. Both have same mass
  • D. Cannot determine

Answer Key

• Q1: B - Equal volumes have equal molecules at same T and P.
• Q2: B - Molar volume at STP is 22.4 L.
• Q3: B - More volume with same molecules means gas B is lighter.

Memory Aids

• Equal volume = equal molecules (Avogadro).
• Molar volume = 22.4 L at STP.

Law of Reciprocal Proportions

Explanation

• When two elements separately combine with fixed mass of third element, their mass ratio is same or simple multiple of ratio when they combine with each other.
• Helps in determining atomic masses.
• Example: If A combines with fixed mass of C, and B combines with same mass of C, ratio of A to B masses is simple multiple of A-B combination ratio.

Solved Examples

• Ex 1: If A and B combine with fixed mass of C in ratio 1:2, and A and B combine with each other in ratio 1:1 or 1:2.

Practice Set

• Q1: Law of reciprocal proportions relates to?
  • A. Mass conservation
  • B. Mass ratios of elements combining with third element
  • C. Volume ratios of gases
  • D. Mole ratios
• Q2: If A combines with fixed mass of C in 1:3 ratio and B combines with same mass of C in 1:6 ratio, ratio of A to B is?
  • A. 1:2
  • B. 2:1
  • C. 1:1
  • D. 3:6
• Q3: Which is true for law of reciprocal proportions?
  • A. Ratios are always equal
  • B. Ratios are simple multiples
  • C. Ratios are unrelated
  • D. Ratios depend on temperature

Answer Key

• Q1: B - It relates to mass ratios with a third element.
• Q2: A - 1:3 and 1:6 gives 1:2 ratio between A and B.
• Q3: B - Ratios are simple multiples or equal.

Memory Aids

• Mass ratios with third element guide element ratios.
• Simple multiples rule applies.

Dalton's Atomic Theory

Main Points

• Matter is made of indivisible atoms.
• Atoms of the same element have identical properties and mass; atoms of different elements differ in mass.
• Compounds form when atoms of different elements combine in fixed ratios.
• Chemical reactions rearrange atoms; atoms are neither created nor destroyed.
• Explains laws of chemical combination but not gaseous volume laws or reasons for atom combination.

Solved Examples

• Ex 1: Identify if the statement aligns with Dalton's theory: "Atoms can be divided into smaller particles."
  Answer: No, Dalton's theory states atoms are indivisible.
• Ex 2: If 2 atoms of element A combine with 3 atoms of element B to form a compound, what does this illustrate?
  Answer: Fixed ratio combination of atoms forming compounds.

Practice Set

• Q1: According to Dalton's theory, atoms are:
  A) Divisible
  B) Indivisible
  C) Created in reactions
  D) Destroyed in reactions
• Q2: Atoms of the same element have:
  A) Different masses
  B) Identical properties
  C) Different chemical behavior
  D) Different atomic numbers
• Q3: Dalton's theory could NOT explain:
  A) Law of conservation of mass
  B) Law of definite proportions
  C) Laws of gaseous volumes
  D) Fixed ratio in compounds

Answer Key

• Q1: B - Dalton stated atoms are indivisible.
• Q2: B - Atoms of the same element have identical properties including mass.
• Q3: C - Dalton's theory could not explain gaseous volume laws.

Memory Aids

• Atoms are "A"tomic: indivisible units.
• Same element = Same mass and properties.
• Compounds = Fixed ratio combinations.
• Atoms rearranged, not created/destroyed in reactions.

Atomic and Molecular Masses

Concepts and Definitions

• Atomic mass (atomic weight) is the average relative mass of atoms compared to carbon-12, which is exactly 12 u.
• 1 unified mass unit (u) = 1.66056 × 10^-24 g; 1 u = 1/12 mass of carbon-12 atom.
• Average atomic mass accounts for isotopes and their relative abundances.
• Formula: Atomic mass = (Average mass of atom) / (1/12 × mass of C-12 atom).
• Average atomic mass = \( \frac{(RA_1 \times At.mass_1) + (RA_2 \times At.mass_2)}{RA_1 + RA_2} \), where RA = relative abundance.
• Molecular mass = sum of atomic masses in a molecule; Formula mass applies to ionic compounds.
• Molecular mass = (Average mass of one molecule) / (1/12 × mass of C-12 atom).
• Key relations:
  • Atomic mass = Equivalent mass × Valency
  • Atomic mass = Molecular mass / Atomicity
  • Actual mass of one molecule = Molecular mass × 1.66 × 10^-24 g
  • Atomic mass = 6.4 / Specific heat (cal)
  • Molecular mass = 2 × Vapour density

Isotopes and Average Atomic Mass

• Isotopes: atoms of same element with different neutrons.
• Example: Carbon isotopes - C-12 (98.892%), C-13 (1.108%), C-14 (trace).
• Average atomic mass of carbon = (0.98892 × 12 u) + (0.01108 × 13.00335 u) + (2 × 10^-12 × 14.00317 u) = 12.011 u.
• Isotopic masses affect dating and chemical studies.

Solved Examples

• Ex 1: Calculate average atomic mass of carbon given isotopes and abundances.
  • Use formula: Average atomic mass = Σ (RA × atomic mass).
  • Substitute values: (0.98892 × 12) + (0.01108 × 13.00335) + (2 × 10^-12 × 14.00317) = 12.011 u.
• Ex 2: Find molecular mass of H₂O.
  • Atomic masses: H = 1 u, O = 16 u.
  • Molecular mass = 2 × 1 + 16 = 18 u.

Practice Set

• Q1: What is the atomic mass unit (u) defined as?
  • A) Mass of one proton
  • B) One-twelfth mass of carbon-12 atom
  • C) Mass of one neutron
  • D) Mass of one electron
• Q2: Which isotope of carbon is most abundant?
  • A) Carbon-14
  • B) Carbon-13
  • C) Carbon-12
  • D) Carbon-11
• Q3: Calculate molecular mass of CO₂ (C=12 u, O=16 u).
  • A) 28 u
  • B) 44 u
  • C) 32 u
  • D) 16 u

Answer Key

• Q1: B - Atomic mass unit is defined as one-twelfth the mass of carbon-12 atom.
• Q2: C - Carbon-12 is the most abundant isotope (~98.9%).
• Q3: B - Molecular mass of CO₂ = 12 + 2×16 = 44 u.

Memory Aids

• Atomic mass unit (u) = 1/12 mass of C-12.
• Average atomic mass = weighted average of isotopes.
• Molecular mass = sum of atomic masses in molecule.
• Use vapour density × 2 for molecular mass quick calculation.

Mole Concept and Molar Masses

Fundamentals

• One mole contains 6.023 × 10²³ particles (atoms, molecules, ions).
• Avogadro constant (N_a) = 6.023 × 10²³ entities/mol.
• Molar mass (g/mol) = mass of 1 mole of substance; numerically equals atomic/molecular/formula mass in atomic mass units (u).
• At STP, 1 mole of gas occupies 22.4 L.
• 1 gram atom = N_a atoms = gram atomic mass.
• 1 gram molecule = N_a molecules = gram molecular mass.

Solved Examples

• Ex 1: Calculate number of atoms in 12 g of carbon.
  • Given: 12 g of ¹²C = 1 mole.
  • Number of atoms = 6.023 × 10²³ atoms.
• Ex 2: Find molar mass of H₂O.
  • Atomic masses: H = 1 u, O = 16 u.
  • Molecular mass = 2 × 1 + 16 = 18 u.
  • Molar mass = 18 g/mol.

Practice Set

• Q1: Avogadro number is:
  • A) 6.023 × 10²²
  • B) 6.023 × 10²³
  • C) 6.023 × 10²⁴
  • D) 6.023 × 10²⁵
• Q2: Molar mass of CO₂ is:
  • A) 12 g/mol
  • B) 16 g/mol
  • C) 44 g/mol
  • D) 28 g/mol
• Q3: Volume occupied by 2 moles of gas at STP is:
  • A) 11.2 L
  • B) 22.4 L
  • C) 44.8 L
  • D) 56.0 L

Answer Key

• Q1: B) 6.023 × 10²³ - Avogadro number defines particles in 1 mole.
• Q2: C) 44 g/mol - CO₂ molar mass = 12 + 2 × 16 = 44 g/mol.
• Q3: C) 44.8 L - 1 mole gas at STP = 22.4 L, so 2 moles = 44.8 L.

Memory Aids

• Avogadro number = 6.023 × 10²³ (think: 6.02 is a common constant).
• Molar mass in grams = atomic/molecular mass in u.
• Gas volume at STP = 22.4 L per mole.

Percentage Composition

Concept and Formula

• Mass % of an element = (Mass of that element in the compound × 100) / Molar mass of the compound
• Used to find the percentage by mass of each element in a compound

Empirical and Molecular Formulas

• Empirical formula: simplest whole number ratio of atoms in a compound
• Molecular formula: exact number of atoms of each element in a molecule
• Different compounds can share the same empirical formula but not the same molecular formula
• Molecular formula = n × Empirical formula = (Empirical formula)_n
• n = Molecular mass / Empirical formula mass (n = 1, 2, 3, ...)

Solved Examples

• Ex 1: Calculate % of C in CO₂.
  • Molar mass CO₂ = 12 + 2×16 = 44 g/mol
  • %C = (12/44) × 100 = 27.27%
• Ex 2: Find molecular formula if empirical formula is CH₂ and molecular mass is 56 g/mol.
  • Empirical mass = 12 + 2×1 = 14 g/mol
  • n = 56 / 14 = 4
  • Molecular formula = (CH₂)₄ = C₄H₈

Practice Set

• Q1: What is the % of H in H₂O?
  • A) 11.11%
  • B) 22.22%
  • C) 66.67%
  • D) 88.89%
• Q2: The empirical formula of a compound is CH₂O. Its molecular mass is 180 g/mol. What is its molecular formula?
  • A) CH₂O
  • B) C₂H₄O₂
  • C) C₃H₆O₃
  • D) C₆H₁₂O₆
• Q3: Calculate the % of oxygen in H₂SO₄ (Molar mass = 98 g/mol).
  • A) 32.65%
  • B) 48.98%
  • C) 65.31%
  • D) 75.51%

Answer Key

• Q1: A) 11.11% - H mass = 2 g, molar mass H₂O = 18 g; (2/18)×100 = 11.11%
• Q2: D) C₆H₁₂O₆ - Empirical mass = 30 g; n = 180/30 = 6; molecular formula = (CH₂O)₆
• Q3: C) 65.31% - O mass = 4×16 = 64 g; (64/98)×100 = 65.31%

Memory Aids

• Empirical formula = simplest ratio; molecular formula = exact count
• Use n = Molecular mass / Empirical mass to find molecular formula
• Mass % = (mass of element / molar mass) × 100

Stoichiometry and Stoichiometric Calculations

Concepts

• Stoichiometry calculates masses or volumes of reactants and products in balanced chemical reactions.
• Limiting reagent: Reactant fully consumed, limits product formation.
• Concentration expressions:
  • Mass percent = (Mass of solute / Mass of solution) × 100
  • Volume percent = (Volume of solute / Volume of solution) × 100
  • Mole fraction = Moles of solute / (Moles of solute + Moles of solvent)
  • Molarity (M) = Moles of solute / Volume of solution (L)
  • Molality (m) = Moles of solute / Mass of solvent (kg)
• Effect of temperature: Molality and mole fraction are temperature-independent; molarity changes with temperature due to volume change.
• Normality (N) = Gram equivalent of solute / Volume of solution (L) = (W × 1000) / (GEM × V in mL), where GEM = gram equivalent mass.
• Equivalent masses:
  • Element = Atomic mass / Valency
  • Acid = Molecular mass / Basicity
  • Base = Molecular mass / Acidity
  • Salt = Formula mass / Total positive or negative charge
  • Oxidising agent = Molecular mass / Total change in oxidation number
• Relation: Molarity × GMM = Normality × GEM (GMM = gram molecular mass)
• Dilution equations:
  • N₁V₁ = N₂V₂
  • M₁V₁ = M₂V₂
  • M₁V₁ / n₁ = M₂V₂ / n₂ (n = moles in balanced reaction)
• Mixing solutions: M₃(V₁ + V₂) = M₁V₁ + M₂V₂ (for non-reacting solutions)
• Conversions:
  • Millimoles = Molarity × Volume (mL)
  • Equivalents = Normality × Volume (L)
  • Milliequivalents = Normality × Volume (mL)
  • Gram atoms or moles = Mass of element / Gram atomic mass
• Mole definitions:
  • 1 mole = 6.023 × 10²³ particles (atoms/molecules)
  • 1 mole atoms = Gram atomic mass = 6.023 × 10²³ atoms
  • 1 mole molecules = Gram molecular mass = 6.023 × 10²³ molecules = 22.4 L at STP
  • 1 mole ionic compound = Gram formula mass = 6.023 × 10²³ formula units
• Gram equivalents = Weight of solute / Equivalent weight
• Milliequivalents = (Weight of solute / Equivalent weight) × 1000
• Strength of solution = Weight of solute (g) / Volume of solution (L)
• Parts per million (ppm) = (Mass or Volume of A / Mass or Volume of solution) × 10⁶

Solved Examples

• Ex 1: Calculate molality if 2 moles of solute (M₂ = 50 g/mol) are dissolved in 1 kg solvent with solution density 1.2 g/mL.
  • Use formula: M = n₁ / (n₁M₁ + n₂M₂) / ρ
  • Calculate mass of solute and solvent, then molarity and molality.
• Ex 2: Find normality of 0.5 M H₂SO₄ (basicity = 2).
  • Normality = Molarity × Basicity = 0.5 × 2 = 1 N

Practice Set

• Q1: What is the limiting reagent?
  • A. Reactant in excess
  • B. Reactant completely consumed
  • C. Product formed first
  • D. Catalyst
• Q2: Molarity is defined as:
  • A. Moles of solute per kg solvent
  • B. Moles of solute per L solution
  • C. Mass of solute per mass of solution
  • D. Volume of solute per volume of solution
• Q3: Calculate normality of 1 M H₃PO₄ (basicity = 3).
  • A. 1 N
  • B. 2 N
  • C. 3 N
  • D. 0.33 N

Answer Key

• Q1: B - Limiting reagent is the reactant fully consumed, limiting product formation.
• Q2: B - Molarity is moles of solute per liter of solution.
• Q3: C - Normality = Molarity × Basicity = 1 × 3 = 3 N.

Memory Aids

• Limiting reagent = fully consumed reactant.
• Molarity changes with temperature; molality does not.
• Normality = Molarity × Basicity/Acidity.
• 1 mole = 6.023 × 10²³ particles.