Chemistry Semester 1 COMPLETE STUDY GUIDE

How to Use This Guide

1. Read each section completely - don't skip definitions
2. When you see WORKED EXAMPLE, work through it yourself first
3. Practice all blueprints and calculations before the exam
4. At the end, test yourself on the verification questions

Phase 1: High-Yield Topics (Everything That Appears in Exams)

Major Content Areas

• Atomic theory progression: Dalton → Thomson → Rutherford → Bohr (always app 3-4 marks)
• Subatomic particles & notation: $^A_ZX$, ions, isotopes, electron count (fundamental - 5+ marks common)
• Electron configuration & shells: Ground state, excited state, ions (core concept - 3-6 marks)
• Periodic trends: Atomic radius, ionisation energy, electronegativity, metallic character (always tested - 5-8 marks)
• Mass spectrometry: Complete process with calculations (2-4 mark questions frequent)
• Bonding types: Ionic, metallic, covalent molecular, covalent network (structure-property link - 6-10 marks)
• Spectra: Absorption & emission spectra, energy levels, photons (growing in importance - 4-6 marks)
• Nanomaterials & fullerenes: Buckyballs, nanotubes, bulk vs nano (2-4 mark section)
• Separation techniques: Practical methods, choosing correct technique (2-5 marks)
• Enthalpy & thermochemical equations: $\Delta H$ signs, energy balance, bond breaking/making (5-7 marks)
• Stoichiometry: Mass-mole-particle conversions, limiting reagent, yields (8-12 marks - HIGH VALUE)
• Organic chemistry: Alkanes, alkenes, benzene, naming, reactions (6-10 marks)
• Exothermic/Endothermic: Temperature changes, system vs surroundings (4-6 marks)
• Fuels & biofuels: Combustion, carbon footprint (2-4 marks)

Exam Structure & How Marks are Allocated

• Multiple choice (usually 10-15 marks): Quick recall, trends, definitions
• Short answer (usually 30-40 marks): Explain with keywords, simple calculations, definitions
• Extended response (usually 20-30 marks): Full explanations linking structure to properties, multi-step calculations
• Marking rules: Examiners reward scientific keywords and explicit logic steps
• Common pattern for long answers: “Define → explain particle-level mechanism → link to data/apply formula → conclude”
• Keywords that appear in EVERY marking guide: electrostatic, valence, delocalised, octet, ionisation, shielding, attraction, energy, mole ratio, balanced equation

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Phase 2: CRITICAL DEFINITIONS (Memorize Word-for-Word)

NANOMATERIALS & FULLERENES (Chapter 1.2, 7.2)

Nanomaterial (Nanotechnology): A material engineered at the nanoscale (1-100 nm) with at least one dimension in this range, exhibiting novel properties different from bulk material.

Nanoscale: The range of 1 to 100 nanometers (nm), where $1 \text{ nm} = 1 \times 10^{-9} \text{ m}$.

Differences between nanomaterials and bulk materials:

• Bulk material: Normal-sized material, dimensions > 100 nm
• Nanomaterial: At least one dimension 1-100 nm

Property	Bulk Material	Nanomaterial
Colour	Typical (e.g. gold = yellow)	Can differ (gold nanoparticles = red)
Hardness	Fixed	Can be harder/softer depending on particle arrangement
Electrical conductivity	Fixed for element	Can increase/decrease due to smaller size
Melting point	Fixed reference	Can be LOWER than bulk (smaller = melts easier)
Tensile strength	Standard	Can be different (carbon nanotubes extremely strong)

Why properties differ: Nanomaterials have enormous surface area to volume ratio. More atoms are at the surface (more reactive). Quantum effects become important at nano scale.

Fullerenes: Allotropes of carbon with closed cage structures made entirely of carbon atoms

• Buckminsterfullerene (C₆₀): Football-shaped, 60 carbon atoms arranged in pentagons and hexagons
• Buckyballs: Colloquial term for C₆₀ and similar structures
• Carbon nanotubes: Cylindrical structures, extremely strong, used in tennis rackets, aircraft, electronics
• Uses: Drug delivery, electronics, structural materials, water purification, cancer treatment vectors

Similarities between nano and bulk: Bonding type is the same (same element = same electrical/chemical behaviour fundamentally, but particle size changes surface effects)

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SEPARATION TECHNIQUES (Chapter 1.3)

Pure substance: Has a fixed chemical formula, fixed melting point, fixed boiling point, fixed properties. Can be an element or compound.

Mixture: Two or more pure substances mixed together. NO fixed chemical formula. Properties determined by composition. CANNOT calculate molar mass of a mixture.

Homogeneous mixture: Uniform composition throughout (solutions, alloys). Single phase visible.

Heterogeneous mixture: Non-uniform, can see different phases (oil and water, sand and water).

Noble gas: Single atoms only (monatomic). Group 18 elements. Full valence shell (8 electrons, except He with 2).

Diatomic gas: Two atoms bonded together. Examples: H₂, N₂, O₂, F₂, Cl₂, Br₂ (gas), I₂ (solid).

Separation Technique	What’s Used	Physical Property	Best For
Sieving	Mesh/sieve	Particle size	Large particles from small
Filtration	Filter paper + funnel	Solubility	Solid from liquid (insoluble solid stays)
Distillation	Heat + condenser	Boiling point difference (usually >20°C)	Liquid from liquid OR solid dissolved in liquid
Fractional distillation	Heat + fractionating column	Multiple boiling point differences	Many liquids from mixture (crude oil etc)
Separating funnel	Separating funnel	Density + immiscibility	Immiscible liquids (oil and water)
Chromatography	Column/paper, solvent	Affinity for solvent vs stationary phase	Different coloured substances, similar size
Evaporation	Heat (gentle)	Volatility	Solid solute from liquid (crystallise)
Crystallisation	Cool saturated solution	Solubility difference with temperature	Pure solid crystals from solution

Choosing the right technique: Assess what property differs between the substances:

• Size? → Sieve or filter
• Boiling point? → Distill
• Density/immiscible? → Separating funnel
• Colour/affinity? → Chromatography

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Phase 3: Core Knowledge You Must Memorize

1) Atomic Structure & Theory (The Four Historical Models)

DALTON’S ATOMIC THEORY (1803)

• Atoms are tiny indivisible particles
• All atoms of an element are identical
• Atoms of different elements have different masses
• Compounds form when atoms bond in whole-number ratios
• Chemical reactions involve rearrangement of atoms

THOMSON’S DISCOVERY (1897) - The Electron

• Conducted cathode ray experiments
• Plum pudding model: Positive charge spread throughout atom (like pudding), electrons embedded in it like plums
• Electrons are negatively charged
• Showed atoms are NOT indivisible

RUTHERFORD’S GOLD FOIL EXPERIMENT (1909) - CRUCIAL EXAM CONTENT

• Experimental setup: Fired alpha particles (He²⁺ nuclei) at thin gold foil, detected where they hit (scintillation screen)
• Expected result (from plum pudding): Most alpha particles pass straight through (soft, spread charge)
• Actual results:
  • MOST alpha particles passed through straight (90-95%)
  • Some were deflected at large angles (10%)
  • A few bounced backwards (0.01%)
• Conclusions from results:
  1. Atom is mostly EMPTY SPACE (explains why most pass through)
  2. Presence of DENSE POSITIVE NUCLEUS (explains deflections - repulsion from dense core)
  3. Nucleus is SMALL relative to atom (explains why few hit it directly)
  4. Electrons are OUTSIDE nucleus (to balance charge)
• Rutherford’s model: Tiny dense positive nucleus surrounded by orbiting electrons (like solar system)

BOHR’S CONTRIBUTION (1913) - Electron Shells

• After Rutherford, scientists found electrons accelerating in orbit should emit radiation and collapse
• Bohr proposed electrons do NOT orbit randomly but occupy quantised shells/energy levels
• Quantised: Can only exist at specific discrete energy levels, not in between
• Each shell has a fixed maximum number of electrons: 2, 8, 8, 18, 32…
• Electrons in different shells have different amounts of energy
• When electron moves to HIGHER shell (absorbs energy) = absorbs photon
• When electron moves to LOWER shell (releases energy) = emits photon
• Ground state: Electrons in lowest available energy shells
• Excited state: Electron(s) have absorbed energy and jumped to higher shell

Key phrase to memorise: “Rutherford showed atom is mostly empty space with dense nucleus; Bohr added that electrons occupy quantised shells.”

2) Subatomic Particles & Symbolic Notation $^A_ZX$

Particle	Relative Charge	Relative Mass	Location	Symbol
Proton	+1	1	Nucleus	p or p⁺
Neutron	0	1	Nucleus	n
Electron	-1	~1/1836 (negligible)	Electron shells	e⁻

SYMBOLIC NOTATION: $^A_ZX$

• A = Mass number = total protons + neutrons = top number
• Z = Atomic number = number of protons = bottom number
• X = Element symbol

Example: $^{12}_6\text{C}$ means Carbon-12 → 6 protons, 6 neutrons (12−6=6), mass number 12

Finding particle numbers from notation:

• Protons = Z (atomic number)
• Neutrons = A − Z (mass number − atomic number)
• Electrons in NEUTRAL atom = Z (same as protons)
• Electrons in ION = Z − charge of ion
  • Cation (positive): Lost electrons, so FEWER electrons than protons
  • Anion (negative): Gained electrons, so MORE electrons than protons

WORKED EXAMPLE: Finding particles Given: $^{35}_{17}\text{Cl}^-$ (Chloride ion)

• Atomic number (Z) = 17 → 17 protons
• Mass number (A) = 35
• Neutrons = 35 − 17 = 18 neutrons
• Charge = −1 (gained 1 electron)
• Electrons = 17 + 1 = 18 electrons

Given neutral: $^{24}_{12}\text{Mg}$

• 12 protons, 12 neutrons, 12 electrons

Given ion: $^{24}_{12}\text{Mg}^{2+}$

• 12 protons, 12 neutrons, but 12 − 2 = 10 electrons (lost 2)

3) Electron Configuration, Ions & Stability

Electron configuration: The ARRANGEMENT of electrons in shells/energy levels

Filling order (shells fill in this order):

• First shell: Max 2 electrons
• Second shell: Max 8 electrons
• Third shell: Max 8 electrons (Year 11 focus - simplified)
• Pattern: 2, 8, 8, 18, 32… (but year 11 usually just 2,8,8)

OCTET RULE: Atoms are most stable when valence shell has 8 electrons (or 2 for H, He)

How to write electron configuration (Ground state = neutral atom)

• Count electrons = atomic number
• Fill shells from inside out: 2, then 8, then 8, etc.
• Example: $^{16}_8\text{O}$ (8 electrons) → Configuration: 2, 6 (first shell full with 2, second shell has 6)
• Write as: 2,6 or K:2, L:6 or 1s² 2s² 2p⁴ (don’t need orbital notation if not asked)

Ions - How to write configuration:

1. Write neutral atom config
2. Remove/add electrons based on charge
3. Electrons leave from OUTERMOST shell first

Examples:

• O (8e⁻) neutral: 2,6
• O²⁻ (gained 2): 2,8 ← now has 10 electrons, like Ne
• Na (11e⁻) neutral: 2,8,1
• Na⁺ (lost 1): 2,8 ← now has 10 electrons, like Ne
• Ca²⁺: Start with Ca (20e⁻, config 2,8,8,2) → lose 2 → 2,8,8 (18 electrons now, like Ar)

Why ions form: To achieve stable octet (8 electrons in valence shell) or duet (2 electrons for H)

• Metals LOSE electrons to achieve octet → form CATIONS (positive)
• Non-metals GAIN electrons to achieve octet → form ANIONS (negative)

WORKED EXAMPLE: Electron configurations in table format

Element/Ion	Atomic #	Electrons	Configuration	Stable?
N	7	7	2,5	No (only 5 in valence)
N³⁻	7	10	2,8	Yes! (gained 3, like Ne)
Mg	12	12	2,8,2	No
Mg²⁺	12	10	2,8	Yes! (lost 2, like Ne)
Cl	17	17	2,8,7	No (7 in valence)
Cl⁻	17	18	2,8,8	Yes! (gained 1, like Ar)

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4) Isotopes & Relative Atomic Mass

Isotope: Atoms of the SAME element (same # protons, same Z) with DIFFERENT number of neutrons (different A)

• Same chemical behaviour (same electrons, same electron configuration)
• Different physical properties (different mass)
• Same chemical formula and reactions
• Different atomic masses

Example: Carbon has three common isotopes:

• ¹²C: 6 protons, 6 neutrons
• ¹³C: 6 protons, 7 neutrons
• ¹⁴C: 6 protons, 8 neutrons All are carbon, all behave similarly chemically, but have different masses

Relative atomic mass (Aᵣ): The AVERAGE atomic mass of isotopes considering their abundance

• Weighted average based on % abundance of each isotope
• UNITLESS (just a number, like 35.45 for Cl)
• NOT the same as whole number mass (unless pure single isotope)

Formula: $A_r = \frac{\sum(\text{atomic mass of isotope} \times \text{\% abundance})}{100}$

WORKED EXAMPLE: Calculating Relative Atomic Mass Chlorine has two main isotopes:

• ³⁵Cl: mass 35, abundance 75.77%
• ³⁷Cl: mass 37, abundance 24.23%

$A_r(\text{Cl}) = \frac{(35 \times 75.77) + (37 \times 24.23)}{100} = \frac{2652.95 + 896.51}{100} = \frac{3549.46}{100} = 35.49$

This is why the periodic table shows Cl ≈ 35.5 (between 35 and 37)

5) Mass Spectrometry - Complete Process (5-stage)

Purpose: Determine the mass of atoms/molecules and identify which element/isotope is present

THE 5 STAGES IN ORDER:

Stage 1: IONISATION

• Electron gun shoots high-energy electrons at sample atoms/molecules
• Electrons collide and KNOCK OUT one or more electrons from sample
• Creates POSITIVE IONS (now missing electrons) → unstable and reactive
• Charge = +1 if lost 1 electron, +2 if lost 2, etc.

Stage 2: ACCELERATION

• Positive ions are accelerated toward a negatively charged plate using electric field
• All ions accelerated to roughly same velocity
• Lighter ions move faster; heavier ions move slower (all have similar kinetic energy but different velocities)

Stage 3: DEFLECTION (Magnetic field)

• Ions pass through a magnetic field (perpendicular to their path)
• Magnetic field CURVES the path of moving ions
• KEY POINT: Heavier ions are deflected LESS (harder to curve heavy object)
• Lighter ions are deflected MORE (easier to curve light object)
• Ion charge also affects deflection: doubly charged deflect twice as much
• Each isotope/ion type follows a curved path to a different detector

Stage 4: DETECTION

• Ions hit detectors which record:
  • Which m/z ratio reached each detector (m = mass, z = charge)
  • How many ions of each type (gives a peak)
  • Height of peak = ABUNDANCE (how many of that isotope)

Stage 5: ANALYSIS

• Computer displays a mass spectrum graph
• X-axis: m/z ratio (usually 1+ ions so m/z ≈ atomic mass)
• Y-axis: Relative abundance (%) or number of ions
• Each peak = one isotope with that mass
• Calculates average atomic mass from peaks using abundance

INTERPRETING A MASS SPECTRUM:

• Number of peaks = number of isotopes present
• Height of peak = relative abundance of that isotope
• Position of peak = mass of isotope

WORKED EXAMPLE: Mass spectrum of neon Suppose neon has peaks at:

• m/z = 20: height = 90 (90% abundance)
• m/z = 22: height = 10 (10% abundance)

$A_r(\text{Ne}) = \frac{(20 \times 90) + (22 \times 10)}{100} = \frac{1800 + 220}{100} = 20.2$

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6) Absorption & Emission Spectra (Energy Level Transitions)

ENERGY LEVELS: Electrons occupy specific energy levels/shells. Energy needed to move between levels = specific amount needed to equal the level difference.

ABSORPTION SPECTRUM (Dark line spectrum):

• When white light passes through a gas/element,electrons absorb specific photons
• Electrons JUMP to HIGHER energy levels (absorb energy)
• The light frequencies that were absorbed go MISSING from the light
• Looking at light that came through: see dark lines on bright background
• Dark lines correspond to which energy levels the electrons jumped between
• Formula: Frequency of absorbed photon must equal energy difference between levels $\Delta E = h f$ where $\Delta E$ = energy difference, $h$ = Planck’s constant, $f$ = frequency

Laboratory Process for Absorption Spectrum:

1. Pass white light through sample
2. Atoms absorb specific wavelengths
3. Detect light coming through with spectrophotometer
4. Record which wavelengths are MISSING
5. Get spectrum showing dark absorption lines on bright background

EMISSION SPECTRUM (Bright line spectrum):

• When electrons in excited state fall to LOWER energy levels, they EMIT photons
• Light is emitted only at specific frequencies (specific to the element)
• Looking at emitted light: see bright coloured lines on dark background
• Each line represents one energy level transition
• Higher energy jump = higher frequency = shorter wavelength = blue end
• Lower energy jump = lower frequency = longer wavelength = red end

Laboratory Process for Emission Spectrum:

1. Heat element or pass electric discharge through sample
2. Electrons get excited and jump to HIGHER levels
3. Electrons immediately fall back to LOWER levels
4. Emit photons during fall
5. Detect with spectrophotometer
6. Get spectrum showing bright coloured lines on dark background
7. Calibration curve: Can identify unknown elements by matching the line pattern

KEY DIFFERENCE SUMMARY:

Type	Cause	Appearance	Information
Absorption	White light through cold gas	Dark lines on bright background	Which energy levels exist
Emission	Excited atoms relaxing	Bright lines on dark background	Which energy levels exist
Result	Same element → same line positions	Absorption and emission at same wavelengths	Can identify element from pattern

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CALIBRATION CURVES (Using Emission Spectra to Identify Elements)

What is a calibration curve? A graph created from KNOWN elements’ spectra, used as a reference to identify UNKNOWN elements

How to create a calibration curve:

Step 1: Obtain emission spectra of known elements

• Example: Get spectra for He, Ne, Ar, Kr
• Record the wavelength (nm) and intensity of each emission line for each element

Step 2: Plot each element’s spectrum on same graph

• X-axis: Wavelength (nm)
• Y-axis: Intensity (brightness) of emission lines
• Each element: vertical lines at its characteristic wavelengths

Step 3: Label each element clearly

• Creates a “fingerprint library” of known elements

How to USE a calibration curve to identify unknown:

Step 1: Obtain emission spectrum of UNKNOWN element Step 2: Examine the wavelength positions of bright lines Step 3: Compare the line positions to your calibration curve Step 4: Match the pattern to the known element that matches Step 5: Identify which element the unknown gas is

Example:

Calibration curve shows:

• Helium: bright lines at 410 nm, 468 nm, 656 nm
• Neon: bright lines at 419 nm, 495 nm, 632 nm
• Argon: bright lines at 404 nm, 546 nm, 696 nm

Unknown gas shows bright lines at: 410 nm, 468 nm, 656 nm → Matches Helium exactly → Unknown is Helium

Why this works:

• Each element has UNIQUE set of energy levels
• Unique energy levels = unique emission line wavelengths
• No two elements have identical spectral “fingerprints”
• Pattern is unmistakable

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7) Periodic Trends - DETAILED EXPLANATIONS (with CAUSES)

You MUST include cause explanations - never just state the trend

TREND 1: ATOMIC RADIUS (size of atom)

Across a period (left to right): Atomic radius DECREASES

• Cause: As you move right, atomic number increases (more protons)
• More protons = greater positive charge in nucleus
• Greater nuclear charge pulls valence electrons CLOSER
• Electrons pulled inward, atom gets smaller
• Shielding by inner electrons stays roughly same

Down a group (top to bottom): Atomic radius INCREASES

• Cause: New electron shells are added
• New shells are further from nucleus (larger orbital)
• Shielding by inner electrons increases (more electrons between valence and nucleus)
• Even though nuclear charge increases, the extra shielding outweighs it
• Atom gets bigger

TREND 2: FIRST IONISATION ENERGY (energy to remove 1st electron)

Definition: Energy required to remove ONE electron from a neutral gaseous atom to form a +1 ion: $\text{X(g)} \to \text{X}^+(g) + e^-$

Across a period: First ionisation energy INCREASES

• Cause: More protons (stronger nuclear attraction to valence electrons)
• Nuclear charge increases, so harder to remove electron
• Takes MORE energy
• Exception: slight drop from Group 13 to 14 (electron pairing effects)

Down a group: First ionisation energy DECREASES

• Cause: More electron shells = more shielding
• Valence electrons further away
• Even though more protons, shielding effect dominates
• Easier to remove valence electron
• Takes LESS energy

Special note: Noble gases have VERY HIGH ionisation energy (octet is stable)

TREND 3: ELECTRONEGATIVITY (ability to attract electrons in bond)

Definition: A measure of the ability of an atom to attract electrons in a chemical bond. Higher value = greater attraction.

Across a period: Electronegativity INCREASES

• Cause: More protons pull shared electrons closer
• Higher nuclear charge
• Stronger attraction to bonding electrons

Down a group: Electronegativity DECREASES

• Cause: More shells = more distance from nucleus
• Electrons further from nuclear attraction
• Weaker pull on bonding electrons

Noble gases: Have NO electronegativity values (don’t form bonds normally)

TREND 4: METALLIC CHARACTER (tendency to lose electrons and be metal-like)

Across a period: Metallic character DECREASES

• Left side = metals (lose electrons easily)
• Right side = non-metals (don’t lose electrons easily)

TREND 4: METALLIC CHARACTER (tendency to lose electrons and be metal-like)

Across a period: Metallic character DECREASES

• Left side = metals (lose electrons easily)
• Right side = non-metals (don’t lose electrons easily)

Down a group: Metallic character INCREASES

• More shells, easier to lose valence electrons
• Top of group = non-metal, bottom = metal

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SPECIAL CASE: IONIZATION ENERGY DIAGRAMS (Reading Multi-Peak Graphs)

What an ionization energy diagram shows:

• X-axis: Electron number (1st, 2nd, 3rd, etc.) for successive ionizations
• Y-axis: Energy required (kJ mol⁻¹) to remove that electron
• Each point represents removing one more electron sequentially

Why the pattern has JUMPS:

When you remove electrons, you go from valence shell → inner shells (which are much closer to nucleus, harder to remove)

INTERPRETING THE GRAPH - Example: Sodium (Na)

Electrons: 2, 8, 1 (11 total)

Reading from left to right:

1. 1st electron (from valence shell 3): ~500 kJ/mol (easy)
2. 2nd electron (from valence shell 3): ~4500 kJ/mol (slightly harder - now ion has lone electron)
3. HUGE JUMP to 3rd electron: ~7000 kJ/mol (now entering shell 2 - much closer!)
4. 4th-10th electrons: Continue very high (removing from shell 2)
5. ANOTHER HUGE JUMP to 11th: Even higher (removing from shell 1 - closest to nucleus)

What the jumps tell you:

• Small jumps = removing from same shell
• LARGE JUMPS = moving to a new shell (gets much harder)
• The pattern of jumps = electron configuration

FROM DIAGRAM → ELECTRON CONFIGURATION:

Steps:

1. Count electrons up to first BIG jump → tells you valence shell electrons
2. Count electrons from first jump to second big jump → tells you second shell electrons
3. Find the pattern

Example: If you see small increases for 3 electrons, then HUGE jump, then 8 more small increases, then HUGE jump again:

• 3 electrons in valence shell (shell 3)
• 8 electrons in shell 2
• Therefore: configuration is 2, 8, 3 = Magnesium (Mg)

EXAM APPLICATION: You might be given a graph of ionization energies and asked:

• “Which element is this?” → Use the jump pattern to determine electron configuration
• “How many electrons are in the second shell?” → Count electrons between first and second major jump
• “Identify if this is a monatomic or diatomic gas” → For gaes: Diatomic (like O₂, N₂) have same ionization pattern as element, but doubled OR similar pattern. This is actually hard to tell - focus on identifying element first

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8) Valency and Bonding Tendency

Valency: The number of electrons available for bonding (or number in valence shell)

Group	Name	Valency	Examples
1	Alkali Metals	1	Li, Na, K (lose 1 e⁻ easily)
2	Alkaline Earth Metals	2	Be, Mg, Ca (lose 2 e⁻ easily)
13	Boron Group	3	Al, B
14	Carbon Group	4	C, Si (form 4 covalent bonds)
15	Nitrogen Group	3 or 5	N, P (gain 3 or share 5)
16	Oxygen Group (Chalcogens)	2 or 6	O, S (gain 2 or share 6)
17	Halogens	1 or 7	F, Cl, Br, I (gain 1 or share 7)
18	Noble Gases	0	He, Ne, Ar (full valence - inert)

9) Bonding & Structure (The Four Types)

IONIC BONDING (Chapter 5)

Definition: Electrostatic attraction between oppositely charged ions held together in a structure.

How it forms:

• Metal LOSES electron(s) → becomes CATION (+)
• Non-metal GAINS electron(s) → becomes ANION (-)
• Electrostatic attraction between + and −
• Ions arrange in 3D lattice to maximize attractions and minimize repulsions

Structure: 3D ionic lattice (not discrete molecules - continuous network of ions)

Lewis Dot Diagrams for ionics:

• Show electron transfer with arrow (not shared pair)
• Final diagram shows ions with charges, not bonded pairs
• Example: Na⁺[Cl]⁻ (Na has lost its electron, Cl has gained it)

Example - Magnesium Chloride MgCl₂ formation:

• Mg (2,8,2) loses 2 electrons → Mg²⁺ (2,8) - now like Ne
• 2×Cl atoms each gain 1 electron → 2×Cl⁻ (2,8 each) - each now like Ar
• Ions attract electrostatin

ally in 3D pattern: Mg²⁺ surrounded by Cl⁻, each Cl⁻ surrounded by Mg²⁺

Properties of ionic compounds:

• High melting & boiling points (strong electrostatic forces)
• Conduct electricity WHEN MOLTEN or DISSOLVED (ions mobile)
• Do NOT conduct as solid (ions fixed in lattice)
• Often soluble in polar solvents like water (water breaks open lattice)
• Brittle (forces are non-directional; layers shift, cause repulsion)
• Hard (strong forces hold together)

Ionic dissolution reaction (MUST show for molten state): Example: Fe₂O₃(l) → 2Fe³⁺(aq) + 3O²⁻(aq)

• When molten, lattice breaks open
• Ions become free to move in liquid
• Can conduct electricity once mobile

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METALLIC BONDING (Chapter 4)

Definition: Attraction between cations and delocalized electrons in a “sea” spread throughout the metal.

Structure:

• Metal atoms lose valence electrons→metal cations
• Valence electrons become DELOCALIZED (not attached to specific atoms)
• Electrons move freely throughout structure like a “sea”
• Cations fixed in regular lattice, surrounded by electron sea

Properties of metals:

• Electrical conductivity: YES - delocalized electrons can move freely (mobile)
• Thermal conductivity: YES - delocalized electrons transfer heat
• Malleable: Can hammer into shapes - layers can slide over electron sea without breaking bonds
• Ductile: Can be drawn into wires - similar reason
• Shiny/lustrous: Delocalized electrons absorb and re-emit light
• High melting point: Strong attraction between cations and electron sea
• Hard: Delocalized electrons provide strong bonding throughout structure

Examples: Fe, Cu, Al, Ag, Au (pure metals) and alloys like brass, bronze

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COVALENT MOLECULAR BONDING (Chapter 6.1)

Definition: Strong covalent bonds WITHIN molecules, but weak intermolecular forces BETWEEN molecules.

Structure: Discrete molecules held together by covalent bonds (atoms sharing electrons). Molecules held to each other by weak intermolecular forces.

Lewis Dot Diagrams for covalents:

• Show electron sharing with pair of dots or line
• Each atom fills its octet (or duet for H)
• Example: H:H or H-H (sharing pair)
• Example: O::O or O=O (double bond = 2 pairs shared)

Properties of covalent molecular:

• Low melting & boiling points (weak intermolecular forces easily overcome)
• Don’t conduct electricity (no mobile charged particles)
• Often insoluble in water
• Soft/waxy texture (easily compressed)
• Can exist as gas, liquid, or solid at room temperature
• Brittle if solid (layers held by weak forces)

Common examples: H₂, O₂, N₂, CO₂, H₂O, CH₄, C₂H₆, I₂

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COVALENT NETWORK BONDING (Giant Covalent) (Chapter 6.1, 7.1)

Definition: Entire 3D structure is ONE GIANT lattice network of atoms joined by strong covalent bonds throughout (no separate molecules).

Structure: Every atom bonded covalently to neighbors all throughout - no discrete molecules, continuous strong bonding network

Common examples and their structures:

Diamond (C):

• Each C atom bonded covalently to 4 other C atoms
• Tetrahedral arrangement
• Extremely strong 3D lattice
• Properties: Extremely hard, very high melting point, colourless, transparent, electrical insulator
• Why hard? Every atom held by strong covalent bonds in all directions

Graphite (C) - Allotrope of carbon:

• C atoms arranged in flat LAYERS
• Within each layer: each C bonded to 3 others (trigonal)
• Between layers: weak van der Waals forces only
• Layers can slide over each other
• Properties: Soft (layers slide), high melting point, conducts electricity (delocalized electrons in layers), grey, lubricant
• Why conducts? Electrons delocalized within each graphite layer

Silicon Dioxide / Silica (SiO₂):

• Each Si bonded to 4 O atoms, each O bonded to 2 Si atoms
• Giant network throughout
• Properties: Very hard, very high melting point, insoluble, poor conductor, sand/quartz examples

Buckminsterfullerene (C₆₀) / Buckyballs - Allotrope:

• 60 C atoms arranged in pentagonal and hexagonal shapes
• Forms closed cage structure
• Properties: Soft (discrete molecules actually), moderate melting point, poor conductor (electrons localized on cage)

Covalent Network Properties Summary:

• Very high melting & boiling points (strong covalent bonds throughout)
• Usually don’t conduct electricity (no mobile charges, except graphite)
• Extremely hard (strong directed bonds)
• Mostly insoluble (strong network hard to break)
• Rigid structures

10) Enthalpy, Bond Energy & Thermochemical Equations

DEFINITIONS:

Enthalpy (ΔH): Heat energy change in a reaction at constant pressure. Tells you if reaction releases or absorbs heat.

Exothermic reaction: Releases heat energy to surroundings

• ΔH is NEGATIVE (e.g., ΔH = −50 kJ mol⁻¹)
• Products have LOWER enthalpy than reactants
• Temperature of surroundings INCREASES
• Examples: combustion, most reactions (burning, neutralization)
• Energy profile: products lower than reactants

Endothermic reaction: Absorbs heat energy from surroundings

• ΔH is POSITIVE (e.g., ΔH = +50 kJ mol⁻¹)
• Products have HIGHER enthalpy than reactants
• Temperature of surroundings DECREASES
• Examples: melting ice, evaporation, photosynthesis
• Energy profile: products higher than reactants

BOND ENERGY CONCEPT:

Bond breaking: REQUIRES energy (endothermic)

• Energy needed to break bonds = positive value

Bond making: RELEASES energy (exothermic)

• Energy released making bonds = negative value

Net enthalpy change: $\Delta H = \text{(Energy to break bonds)} - \text{(Energy released making bonds)}$

OR equivalently: $\Delta H = \text{(Sum of bonds broken)} - \text{(Sum of bonds made)}$

If MORE energy needed to break than released making → ΔH positive (endothermic) If LESS energy needed to break than released making → ΔH negative (exothermic)

THERMOCHEMICAL EQUATIONS:

Format: Balanced equation + state symbols + ΔH value

Important rules:

1. Always balance the equation first
2. Include STATE SYMBOLS: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous
3. COMBUSTION RULE: Water always written as H₂O(l) unless specifically told otherwise
4. Include ΔH with correct sign
5. ΔH value applies to the coefficients in equation as written

Examples:

Combustion of methane: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -890 \text{ kJ mol}^{-1}$

This means: When 1 mol CH₄ burns completely, 890 kJ is released (negative sign)

If you burn 2 mol CH₄ instead, you’d release 2 × 890 = 1780 kJ (scales with coefficient)

Melting (endothermic): $\text{H}_2\text{O}(s) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H = +6.01 \text{ kJ mol}^{-1}$

Positive sign shows energy absorbed.

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11) Organic Chemistry - Fundamentals

DEFINITIONS & FORMULAS:

Alkane: Saturated hydrocarbon with only C-C and C-H single bonds

• General formula: C_nH₍₂ₙ₊₂₎
• Examples: CH₄ (methane), C₂H₆ (ethane), C₃H₈ (propane)
• Saturated: Contains only single bonds, no double/triple bonds

Alkene: Unsaturated hydrocarbon with C=C double bond

• General formula: C_nH₍₂ₙ₎ (one less pair of H than alkane)
• Examples: C₂H₄ (ethene), C₃H₆ (propene)
• Unsaturated: Contains C=C double bond

Alkyne: Unsaturated hydrocarbon with C≡C triple bond

• General formula: C_nH₍₂ₙ₋₂₎
• Example: C₂H₂ (ethyne)

Benzene: Aromatic hydrocarbon with 6-carbon ring

• General formula: C₆H₆ but often written as C_nH₍₂ₙ₋₆₎ for derivatives
• Ring with 6 carbons, alternating double bonds (or delocalized electrons shown as inner circle)
• Shown as hexagon with line in middle (NOT alternating single/double - shows resonance)
• Very stable, requires special conditions to react

Arene: Benzene derivative (benzene with side chains)

• Examples: toluene (methylbenzene) C₇H₈, xylene, phenol

How to check saturation:

• Count atoms, plug into formula
• If matches CₙH₍₂ₙ₊₂₎ → saturated alkane
• If has 2 fewer H → likely one double bond (alkene)
• If has cyclic + fewer H → possibly cyclic alkane
• If has 6 carbons and 6 H → likely benzene

IUPAC NAMING RULES for alkanes, alkenes, benzene:

Steps:

1. Find LONGEST carbon chain (main chain)
2. Number carbons to give LOWEST numbers to substituents
3. List substituents alphabetically with their positions
4. For alkenes: position goes on first carbon of double bond

Examples:

• CH₃-CH₂-CH₃ = propane (3 carbons, all single)
• CH₃-CH=CH-CH₃ = but-2-ene (4 carbons, double bond at position 2)
• Methylbenzene = toluene (benzene with CH₃ side chain)
• 1,4-dimethylbenzene (two CH₃ groups at positions 1 and 4)

ORGANIC REACTIONS:

Addition reaction (alkene → products):

• Double bond OPENS UP and new atoms/groups are added across
• Breaks C=C and forms two new C-atom bonds
• General: CₙH₍₂ₙ₎ + reactant → product with single bonds
• Conditions: Usually room temperature, no catalyst typically needed

Example: Ethene + Bromine $\text{C}_2\text{H}_4 + \text{Br}_2 \rightarrow \text{C}_2\text{H}_4\text{Br}_2$ (1,2-dibromoethane)

Observation: Bromine water color changes from orange-brown to COLORLESS (instantly) because Br₂ is consumed

Substitution reaction (alkane + reactant → products):

• One atom or group is REPLACED by another
• One bond is broken and new bond forms
• H atom usually replaced
• Requires: Usually catalyst or specific conditions (UV light, heat, catalyst)

Example: Methane + Chlorine $\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV}} \text{CH}_3\text{Cl} + \text{HCl}$ (monochloromethane)

Observation: Bromine water color REMAINS ORANGE for a LONG TIME (only reacts if C-Br bond formed, which is slow)

KEY DIFFERENCE FOR EXAMS:

• Addition (alkene + Br₂): Orange Br₂ color disappears instantly
• Substitution (alkane + Br₂): Orange Br₂ color remains (slow reaction, need UV/heat)

Combustion:

• All organic molecules undergo combustion when heated in oxygen
• Complete combustion: CₓH_y + O₂ → CO₂ + H₂O
• Incomplete combustion (limited O₂): produces CO and/or C instead
• All organic combustion is EXOTHERMIC (ΔH negative)

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EXPANDED ORGANIC REACTIONS (Full A-Grade Coverage)

ADDITION REACTIONS - Alkenes with Halogens and Hydrogen

Addition with Bromine Water (CRITICAL EXAM OBSERVATION)

General equation: $\text{R-CH=CH-R'} + \text{Br}_2 \rightarrow \text{R-CHBr-CHBr-R'}$

Example 1: Ethene + Bromine $\text{C}_2\text{H}_4 + \text{Br}_2 \rightarrow \text{C}_2\text{H}_4\text{Br}_2$

Structural form: $\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CHBr-CHBr}$ (1,2-dibromoethane)

Observations (LEARN THIS EXACTLY):

• BEFORE: Bromine water is orange-brown, alkene is colourless
• REACTION: Bromine immediately breaks across the C=C double bond
• AFTER: Orange-brown color DISAPPEARS INSTANTLY (within seconds)
• Why: The Br₂ molecule is consumed in the reaction - it’s no longer there to give color
• Speed: Happens at room temperature, NO catalyst needed

Mechanism: The C=C electrons attack Br₂, breaking the bromine bond

Example 2: Propene + Bromine $\text{CH}_3\text{-CH=CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3\text{-CHBr-CH}_2\text{Br}$ (1,2-bromopropane)

Observation: Same - orange color disappears instantly

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Addition with Hydrogen (Hydrogenation)

General equation: $\text{R-CH=CH-R'} + \text{H}_2 \xrightarrow{\text{Ni catalyst, heat}} \text{R-CH}_2\text{-CH}_2\text{-R'}$

Example: Ethene + Hydrogen $\text{C}_2\text{H}_4 + \text{H}_2 \xrightarrow{\text{Ni, 150°C}} \text{C}_2\text{H}_6$

Conditions REQUIRED:

• Nickel (Ni) catalyst or platinum (Pt) catalyst
• Temperature: 150°C
• Pressure: Normal atmospheric (some sources require slight pressure)

Real-world application: Converting unsaturated oils to saturated fats (margarine manufacture)

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Addition with Water (Hydration)

General equation: $\text{R-CH=CH-R'} + \text{H}_2\text{O} \xrightarrow{\text{H}^+\text{ catalyst}} \text{R-CHOH-CH}_2\text{-R'}$

Example: Ethene + Water → Ethanol $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{-CH}_2\text{OH}$

Conditions REQUIRED:

• Phosphoric acid (H₃PO₄) catalyst
• Temperature: 300°C
• Pressure: 6000 kPa (very high pressure - critical!)

This is INDUSTRIAL: How ethanol is made from crude oil

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SUBSTITUTION REACTIONS - Alkanes with Halogens

Substitution with Bromine (Under UV Light)

General equation: $\text{R-H} + \text{Br}_2 \xrightarrow{\text{UV light}} \text{R-Br} + \text{HBr}$

Example 1: Methane + Bromine → Bromomethane $\text{CH}_4 + \text{Br}_2 \xrightarrow{\text{UV}} \text{CH}_3\text{Br} + \text{HBr}$

Example 2: Methane + Bromine (excess, sequential substitution): $\text{CH}_4 \xrightarrow{\text{Br}_2, \text{UV}} \text{CH}_3\text{Br} \xrightarrow{\text{more Br}_2} \text{CH}_2\text{Br}_2 \xrightarrow{\text{more Br}_2} \text{CHBr}_3 \xrightarrow{\text{more Br}_2} \text{CBr}_4$

Observations (CRITICAL - DIFFERENT FROM ADDITION):

• BEFORE: Bromine water is orange-brown
• REACTION: Under UV light, reaction is SLOW
• AFTER: Orange-brown color REMAINS for a LONG TIME (minutes to hours)
• Why: Only a small amount of Br₂ is consumed. Most of the bromine stays in solution.
• Condition: UV light is ESSENTIAL (sunlight works) - without it, reaction doesn’t happen
• Speed: Very slow compared to alkene addition

Mechanism: Free radical substitution (H atom replaced by Br atom)

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Substitution with Chlorine (Under UV or Heat)

Example: Methane + Chlorine $\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV or 300°C}} \text{CH}_3\text{Cl} + \text{HCl}$

Same principle as bromine - slow reaction, requires UV light

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BENZENE REACTIONS (Why Benzene is Special)

Why Benzene DOESN’T undergo addition:

Benzene has a DELOCALIZED electron system:

• All 6 electrons in the ring are shared equally across all 6 carbons
• NO discrete C=C double bonds (shown as alternating, but it’s actually fully delocalized)
• This delocalized system is EXTREMELY STABLE (resonance stabilization)

Why addition doesn’t happen:

• Breaking the delocalized system would DESTROY the stability
• The π electrons are distributed, not concentrated in double bonds
• Addition reactions would require disrupting this stable aromatic system
• Energy cost is too high - thermodynamically unfavorable

Therefore: Benzene undergoes SUBSTITUTION (not addition)

Benzene + Bromine (Substitution - Requires Catalyst)

$\text{C}_6\text{H}_6 + \text{Br}_2 \xrightarrow{\text{FeBr}_3\text{ catalyst}} \text{C}_6\text{H}_5\text{Br} + \text{HBr}$

Conditions REQUIRED:

• FeBr₃ (iron(III) bromide) catalyst - ESSENTIAL
• Room temperature or slight heating
• NO UV light needed (unlike alkanes)

Observation: Orange bromine color fades slowly (substitution, not addition)

Benzene + Chlorine (Substitution):

$\text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{AlCl}_3\text{ catalyst}} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}$

Conditions REQUIRED:

• AlCl₃ (aluminum chloride) catalyst - ESSENTIAL
• Requires heat

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IDENTIFYING REACTIONS FROM PRODUCTS (Reverse Engineering)

If given the product, work backward:

Example 1: “Write the reaction to form C₂H₄Br₂”

• C₂H₄Br₂ has 2 Br atoms added to 2 carbons
• This is ADDITION (both Br on same molecule)
• Must be from alkene: C₂H₄ + Br₂
• Answer: C₂H₄ + Br₂ → C₂H₄Br₂ (addition)

Example 2: “Write the reaction to form CH₃Cl and HCl”

• One Cl substituted for H
• One H released as HCl
• This is SUBSTITUTION (one H replaced)
• Must be from alkane: CH₄ + Cl₂
• Answer: CH₄ + Cl₂ → CH₃Cl + HCl (substitution, needs UV)

Example 3: “Write the reaction to form C₆H₅Br”

• One H replaced by Br on benzene ring
• This is aromatic SUBSTITUTION
• Answer: C₆H₆ + Br₂ → C₆H₅Br + HBr (needs FeBr₃ catalyst)

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KEY DISTINCTION TABLE - OBSERVATIONS GIVE IT AWAY

Reaction	Starting Material	With Br₂	Observation	Time	Conditions
Addition	Alkene (C=C)	Orange Br₂	Color DISAPPEARS	Instant (seconds)	Room temp, NO catalyst
Substitution	Alkane (C-H)	Orange Br₂	Color REMAINS	Slow (minutes-hours)	UV light required
Aromatic Sub.	Benzene	Orange Br₂	Color fades slowly	Slow	Catalyst (FeBr₃) required

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Phase 4: STOICHIOMETRY - THE ESSENTIAL CALCULATIONS

Stoichiometry: Calculating mass, moles, or particles based on balanced equations. THIS IS HIGH VALUE - usually 8-12 marks per exam.

Core Relationships (MEMORIZE THESE COLD):

The Mole Relationships (forming the conversion triangle): $n = \frac{m}{M} \quad \text{where } n = \text{moles, } m = \text{mass in grams, } M = \text{molar mass}$

$N = n \times N_A \quad \text{where } N = \text{number of particles, } N_A = 6.022 \times 10^{23}$

Rearrangements:

• Find moles: $n = \frac{m}{M}$ or $n = \frac{N}{N_A}$
• Find mass: $m = n \times M$
• Find particles: $N = n \times N_A$
• Find Avogadro: $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$

Step-by-Step Method for ALL stoichiometry problems:

ALWAYS FOLLOW THIS PROCESS:

1. Write balanced equation
2. Identify given and find what’s asked
3. Convert given amount to MOLES
4. Use mole ratio from balanced equation
5. Convert result BACK to required units

WORKED EXAMPLE 1: Mass-to-mass stoichiometry

Question: Calculate the mass of oxygen gas needed to completely burn 25.6 g of methane (CH₄). [M(CH₄) = 16, M(O₂) = 32]

Step 1: Balanced equation $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Step 2: Given and find

• Given: 25.6 g CH₄
• Find: mass of O₂

Step 3: Convert to moles $n(\text{CH}_4) = \frac{m}{M} = \frac{25.6}{16} = 1.6 \text{ mol}$

Step 4: Use mole ratio From equation: 1 mol CH₄ needs 2 mol O₂ $n(\text{O}_2) = 1.6 \text{ mol CH}_4 \times \frac{2 \text{ mol O}_2}{1 \text{ mol CH}_4} = 3.2 \text{ mol O}_2$

Step 5: Convert back to mass $m(\text{O}_2) = n \times M = 3.2 \times 32 = 102.4 \text{ g}$

Answer: 102.4 g of oxygen needed

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WORKED EXAMPLE 2: Calculations with particles

Question: How many H₂O molecules are produced when 6.4 g of methane burns completely? [M(CH₄) = 16, From equation above, 1 mol CH₄ → 2 mol H₂O]

Step 1: Already have balanced equation $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Step 2: Given and find

• Given: 6.4 g CH₄
• Find: Number of H₂O molecules

Step 3: Convert to moles $n(\text{CH}_4) = \frac{6.4}{16} = 0.4 \text{ mol}$

Step 4: Use mole ratio $n(\text{H}_2\text{O}) = 0.4 \text{ mol} \times \frac{2 \text{ mol H}_2\text{O}}{1 \text{ mol CH}_4} = 0.8 \text{ mol}$

Step 5: Convert to particles $N(\text{H}_2\text{O}) = n \times N_A = 0.8 \times 6.022 \times 10^{23} = 4.8 \times 10^{23} \text{ molecules}$

Answer: 4.8 × 10²³ molecules

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PERCENTAGE COMPOSITION (Finding % of each element)

Formula: $\% \text{element} = \frac{\text{(Atomic mass of element)} \times \text{(number of atoms in formula)}}{\text{Molar mass of compound}} \times 100$

WORKED EXAMPLE: Find % carbon in glucose C₆H₁₂O₆ [M(C) = 12, M(H) = 1, M(O) = 16]

Step 1: Calculate molar mass of C₆H₁₂O₆: $M = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180 \text{ g/mol}$

Step 2: Calculate % carbon: $\% C = \frac{12 \times 6}{180} \times 100 = \frac{72}{180} \times 100 = 40\%$

Similarly: % H = 6.67%, % O = 53.33% (check: 40 + 6.67 + 53.33 = 100% ✓)

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LIMITING REAGENT (Which reactant runs out first?)

Concept: In a reaction with two reactants, one usually runs out first - that’s the limiting reagent. The other is in excess.

Method:

1. Calculate moles of each reactant
2. Use stoichiometry to find how many moles of PRODUCT each could make
3. Whichever gives SMALLER product amount = limiting reagent
4. Calculate answer using limiting reagent

WORKED EXAMPLE: $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$

If we have 5 mol H₂ and 4 mol O₂, which is limiting?

Using H₂: $5 \text{ mol H}_2 \times \frac{1 \text{ mol O}_2}{2 \text{ mol H}_2} = 2.5 \text{ mol O}_2$ needed

Using O₂: $4 \text{ mol O}_2 \times \frac{2 \text{ mol H}_2}{1 \text{ mol O}_2} = 8 \text{ mol H}_2$ needed

We only HAVE 5 mol H₂, but need 8 mol → H₂ is limiting (runs out first) We only HAVE 4 mol O₂, but need 2.5 → O₂ is in excess

Product from H₂: $5 \text{ mol H}_2 \times \frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2} = 5 \text{ mol H}_2\text{O}$

Answer: 5 mol H₂O maximum (limited by H₂ running out)

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YIELD CALCULATIONS

Theoretical yield: Maximum possible product (assume 100% conversion)

Actual yield: What really happens in lab

% Yield: $\% \text{yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

WORKED EXAMPLE: Reaction to make 10 g product calculated theoretically as 12.5 g. What’s % yield?

$\% \text{yield} = \frac{10}{12.5} \times 100 = 80\%$

Means 20% was lost through side reactions, incomplete reaction, evaporation, etc.

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PURITY CALCULATIONS

Purity: How much of the sample is actually the desired substance

% Purity: $\% \text{purity} = \frac{\text{mass of pure substance}}{\text{total sample mass}} \times 100$

WORKED EXAMPLE: A 5 g iron ore sample contains 3.5 g pure iron. What’s % purity?

$\% \text{purity} = \frac{3.5}{5} \times 100 = 70\%$

The remaining 30% is gangue (unwanted rock/minerals)

1. Moles from mass: $n = \frac{m}{M}$
2. Mass from moles: $m = nM$
3. Particles from moles: $N = nN_A$
4. Moles from particles: $n = \frac{N}{N_A}$
5. Avogadro constant: $N_A = 6.022\times10^{23}\,\text{mol}^{-1}$
6. Percentage composition:

$$\%\text{element} = \frac{(\text{Ar of element})\times(\text{number in formula})}{\text{Mr of compound}}\times100$$

7. Relative atomic mass from isotope data:

$$A_r = \frac{\sum(mass\times\%abundance)}{100}$$

8. Energy from stoichiometric enthalpy:

$$q = n\times|\Delta H|\quad\text{(adjust for equation coefficients)}$$

9. Percent yield:

$$\%\text{yield} = \frac{\text{actual}}{\text{theoretical}}\times100$$

10. Percent purity:

$$\%\text{purity} = \frac{\text{pure mass}}{\text{sample mass}}\times100$$

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Phase 5: Energy Calculations & Thermochemistry

Heat Energy from Reactions

Total energy/heat released/absorbed: $q = n \times |\Delta H|$

where:

• $q$ = total heat energy (kJ)
• $n$ = number of moles of SUBSTANCE REFERENCED IN EQUATION
• $|\Delta H|$ = magnitude of molar enthalpy change (kJ/mol)

IMPORTANT: The $\Delta H$ value in equation applies to the number of moles shown in that equation. If you have different amount, scale accordingly.

WORKED EXAMPLE: Given: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -890 \text{ kJ mol}^{-1}$

How much heat released when 4.0 g CH₄ burns? [M(CH₄) = 16]

Step 1: Find moles $n(\text{CH}_4) = \frac{4.0}{16} = 0.25 \text{ mol}$

Step 2: Calculate heat (equation shows 1 mol CH₄ releases 890 kJ) $q = 0.25 \times 890 = 222.5 \text{ kJ}$

Answer: 222.5 kJ heat released (negative because exothermic)

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Exothermic vs Endothermic Reactions

EXOTHERMIC (ΔH < 0, negative):

• Releases heat to surroundings
• Products have LOWER enthalpy than reactants
• Temperature of surroundings INCREASES
• Bond making releases more energy than bond breaking requires
• Examples: Combustion, neutralization, freezing
• Energy diagram: reactants start high, products at lower level, energy released as heat

ENDOTHERMIC (ΔH > 0, positive):

• Absorbs heat from surroundings
• Products have HIGHER enthalpy than reactants
• Temperature of surroundings DECREASES
• Bond breaking requires more energy than bond making releases
• Examples: Melting, evaporation, dissolution of some salts,photosynthesis
• Energy diagram: reactants start low, products at higher level, energy absorbed from surroundings

Using thermometer to identify:

• Exothermic → temperature rises (heat released into liquid)
• Endothermic → temperature falls (heat absorbed from liquid)

State changes classification:

• Melting (s→l): Endothermic (absorbs heat) - bonds weakened not fully broken
• Evaporation (l→g): Endothermic (absorbs heat) - all bonds broken
• Freezing (l→s): Exothermic (releases heat) - opposite of melting
• Condensation (g→l): Exothermic (releases heat) - opposite of evaporation

Note: Melting requires less energy than evaporation because in melting you only weaken bonds; in evaporation you completely break molecule bonds.

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Phase 6: Fuels and Biofuels

Fossil Fuels vs Biofuels

Fossil Fuels (Non-renewable):

• Oil, coal, natural gas
• Took millions of years to form
• Finite supply
• Contains carbon from ancient dead organisms
• When burned: release CO₂ that was locked underground for millions of years
• Net effect: INCREASES atmospheric CO₂ (causes climate change)
• Complete combustion: CₓHₓ + O₂ → CO₂ + H₂O
• Incomplete combustion (limited O₂): produces CO (poisonous), C (carbon monoxide/soot)

Biofuels (Renewable):

• Produced from plants/biomass (corn, sugarcane, algae, waste)
• Regrow - truly renewable
• When burned: release CO₂, but plants absorb same CO₂ when growing
• Net effect: NEUTRAL carbon footprint (recycled carbon)
• Lower net carbon emissions than fossil fuels
• Examples: ethanol (from sugar), biodiesel (from oils), biogas (from waste)
• Same combustion reactions but different source

Advantages of biofuels over fossil fuels:

• Renewable (infinite supply vs finite)
• Lower carbon footprint (carbon cycle)
• Reduce dependency on oil imports
• Reduce atmosphere CO₂

Disadvantages:

• Can require large land areas (competing with food crops)
• Energy intensive to produce
• May not be carbon neutral depending on production (fertilizer, transport)

Fractional Distillation (separating crude oil)

Purpose: Separate crude oil into usable fractions by boiling point

Process:

1. Heat crude oil to ~350°C
2. Vaporized oil enters fractionating column
3. Column is HOTTER at bottom, COOLER at top (temperature gradient)
4. Hydrocarbons with low boiling point rise to top (cool side), condense there
5. Hydrocarbons with high boiling point condense at bottom (hot side)
6. Each fraction contains molecules of similar boiling point

Fractions obtained (from top to bottom, hot to cool):

• Petroleum gas (C1-C4, very low bp, gases)
• Petrol/gasoline (C5-C10, low bp, flammable liquids)
• Kerosene/diesel (C10-C16, medium bp)
• Fuel oil (C16-C20, high bp)
• Bitumen/residue (C20+, very high bp, thick solid)

Trend: Smaller molecules (fewer C atoms) = LOWER boiling point = rise higher in column Larger molecules (more C atoms) = HIGHER boiling point = stay lower in column

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Phase 6B: VARIABLES & DATA TYPES (Critical for Practicals)

Independent, Dependent, and Controlled Variables

Independent Variable (IV):

• The variable YOU change on purpose
• What you manipulate to see an effect
• There is ONE independent variable per experiment
• Example: Temperature, concentration, mass of reactant, time, volume

Example experiment: “How does temperature affect reaction rate?”

• Independent variable: Temperature (you change it: 20°C, 30°C, 40°C, 50°C)
• You deliberately change this to observe its effect

Dependent Variable (DV):

• The variable you MEASURE as a result
• What changes BECAUSE of the independent variable
• The outcome of the experiment
• Example: Reaction time, mass produced, volume of gas, temperature change, pH

Example (continued): “How does temperature affect reaction rate?”

• Dependent variable: Reaction rate or time taken (you measure how fast it reacts)
• This changes as a result of temperature changing

Controlled Variables (CV):

• ALL OTHER variables you KEEP THE SAME
• Anything that could affect the result but shouldn’t change
• If you don’t control them, you won’t know what CAUSED the change
• Multiple controlled variables per experiment

Example (continued): “How does temperature affect reaction rate?”

• Controlled variables:
  • Amount of reactant (keep mass same)
  • Volume of solution (keep constant)
  • Concentration (keep constant)
  • Type of catalyst (same catalyst or none)
  • Apparatus used (same beaker, same stirrer)
  • Time of measurement
  • Light exposure
  • Presence of other substances

WHY THIS MATTERS FOR A-GRADE:

When exam asks: “Identify the variables in this experiment”

• You MUST state:
  1. Independent variable = [specific value changed]
  2. Dependent variable = [what you measured]
  3. Controlled variables = [list 4-5 specific things kept constant]

Example answer format:

“In this experiment investigating how concentration affects reaction rate:

• Independent variable: Concentration (varied from 0.1 M to 0.5 M)
• Dependent variable: Time taken for reaction to complete OR rate of reaction (measured in seconds or mol/min)
• Controlled variables: Temperature (kept at 25°C), volume of solution (50 mL), catalyst type (same iron filings), apparatus (same beaker and stirrer), light exposure (kept in same location)“

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Primary vs Secondary Data

Primary Data:

• Data YOU collect yourself through YOUR OWN experiment
• Direct measurement from YOUR practical
• Examples:
  • Temperature readings YOU took in the lab
  • Mass YOU weighed on balance
  • Volume YOU measured with graduated cylinder
  • Time YOU recorded with stopwatch
  • Color changes YOU observed

Advantages:

• You know exactly how it was collected
• You control the conditions
• Direct relevance to YOUR investigation

Disadvantages:

• Time-consuming
• Possibility of human error
• Limited by apparatus you have

Secondary Data:

• Data collected by SOMEONE ELSE (from textbooks, websites, previous experiments, published research)
• Information from literature or existing sources
• Examples:
  • Atomic masses from periodic table
  • Boiling points from data booklet
  • Historical experimental results
  • Published research data
  • Values from manufacturer specifications

Advantages:

• Readily available
• Large datasets possible
• Can compare your data to known values

Disadvantages:

• You don’t know exact conditions it was collected under
• May not be relevant to your specific question
• Potential for outdated information

A-Grade Use:

In an experiment, you might:

• Collect PRIMARY data: Measure how long salt takes to dissolve at different temperatures
• Use SECONDARY data: Compare your results to literature values of dissolution rates from a chemistry textbook
• Make comparison: “Our experimental rate (0.8 g/s) matches the published value (0.82 g/s) closely, suggesting our method was valid”

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Phase 7: Experimental Errors & Data Analysis

Types of Errors

Random error: Unpredictable, happens by chance, affects some results high and some low

• Caused by: instrument limitations, human reaction time, fluctuations in conditions
• Examples: Parallax error reading meniscus, digital scale fluctuating, timing slight variations
• Can be reduced by: repeating experiment, averaging results, using better equipment
• CANNOT be eliminated completely

Systematic error: Consistent, affects all results same way (all too high or all too low)

• Caused by: wrongly calibrated equipment, procedural error, faulty apparatus
• Examples: Digital balance always 0.1 g too heavy, thermometer consistently reads 2°C too low, apparatus not at proper angle
• Can be reduced by: recalibrating equipment, fixing procedure, replacing faulty apparatus
• Repeating won’t help (repeats same error each time)

How to identify errors in an experiment:

• Random: results vary up and down, average still reasonable
• Systematic: all results consistently too high or too low

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Validity, Reliability, and Accuracy

Validity: Does experiment measure what it’s supposed to measure?

• Valid design answers the research question
• Improve: use appropriate apparatus, control variables correctly

Reliability: Can you get same result again?

• Reliable = consistent, repeatable results
• Improve: repeat measurements, use precise instruments, identical conditions each time

Accuracy: How close actual result is to true/expected value

• Accurate = result matches true value
• Improve: recalibrate instruments, fix systematic errors

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Significant Figures

Rules:

1. All non-zero digits are significant: 25.3 m = 3 sig figs
2. Zeros between non-zero digits are significant: 1005 kg = 4 sig figs
3. Leading zeros NOT significant: 0.0045 m = 2 sig figs (only 4,5)
4. Trailing zeros after decimal ARE significant: 25.0 m = 3 sig figs
5. Trailing zeros without decimal point are ambiguous (write in scientific notation): 2500 could be 2, 3, or 4 sig figs

Calculations:

• Multiplication/Division: Answer has same sig figs as LEAST precise measurement
• Addition/Subtraction: Answer has same decimal places as LEAST precise measurement

Example: 23.4 g ÷ 2.5 mL = 9.36 g/mL

• 23.4 has 3 sig figs
• 2.5 has 2 sig figs
• Answer: 9.4 g/mL (rounded to 2 sig figs)

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Phase 8: Exam Answer Blueprints (Use These EXACTLY)

Blueprint 1: Definition Question (1-3 marks)

Structure:

1. Direct definition (one sentence, scientific terms)
2. Add particle/molecular detail
3. Include one relevant example

Example - “Define an isotope”:

“Isotopes are atoms of the same element with the same proton number (atomic number Z) but different neutron numbers (different mass numbers A). Because they have different numbers of neutrons, they have different atomic masses, but chemical properties remain similar because they have identical electron configurations. For example, ¹²C and ¹⁴C are both carbon atoms with 6 protons, but ¹⁴C has 8 neutrons while ¹²C has 6.”

Keywords used: isotopes, proton number, neutron numbers, atomic masses, electron configurations, atoms

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Blueprint 2: Explain Periodic Trend (3-5 marks)

Structure:

1. State trend direction and property
2. Give reason (electron/proton relationship)
3. Explain shielding OR nuclear charge effect
4. Link mechanically to property change

Example - “Explain why atomic radius decreases across a period”:

“As you move across a period from left to right, atomic radius decreases. This is because the atomic number increases, meaning more protons are present in the nucleus. The increased positive nuclear charge exerts a stronger electrostatic attraction on the valence electrons, pulling them closer to the nucleus. Although the number of shells remains constant across a period (same shielding), and indeed one more electron is added to the valence shell, the increased nuclear attraction dominates. Therefore, the atoms become smaller as the valence electrons are pulled in more tightly by the stronger nuclear charge. This trend reverses down a group because new electron shells are added, increasing shielding.”

Keywords: atomic radius, atomic number, protons, electrostatic attraction, valence electrons, nuclear charge, shielding, shells

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Blueprint 3: Bonding Formation (4-6 marks)

Structure:

1. State what happens to electrons (transfer vs sharing)
2. Show charges on ions or show shared pair
3. State stable electron config achieved
4. Describe electrostatic attraction
5. (For ionic) describe 3D lattice structure

Example - “Explain how ionic bonding occurs in magnesium chloride”:

“Magnesium is an alkaline earth metal with electron configuration 2,8,2 - it has 2 valence electrons. Chlorine is a halogen with electron configuration 2,8,7 - it needs 1 more electron. During ionic bonding, the magnesium atom TRANSFERS its 2 valence electrons to TWO chlorine atoms (one electron each). This produces Mg²⁺ (now with configuration 2,8, a stable octet) and two Cl⁻ ions (each now with configuration 2,8, also stable octets). The oppositely charged ions are attracted to each other through strong electrostatic forces. The ions arrange in a 3D ionic lattice structure where each Mg²⁺ is surrounded by Cl⁻ ions and vice versa, maximizing attractive forces and minimizing repulsions. This creates the ionic compound MgCl₂.”

Keywords: alkaline earth metal, halogen, electron configuration, transfer, valence electrons, octet, stable, Mg²⁺, Cl⁻, electrostatic, lattice

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Blueprint 4: Conductivity from Structure (3-5 marks)

Structure:

1. Identify structure type
2. Identify whether mobile charged particles exist
3. State conductivity consequence
4. For ionic: distinguish solid vs molten/aqueous

Example - “Why does copper conduct electricity as a solid, but sodium chloride does not?”:

“Copper is a metallic structure consisting of copper cations surrounded by a sea of delocalized electrons. These delocalized electrons are mobile - they can move freely throughout the metal’s structure. When a potential difference is applied, these mobile electrons flow through the substance, conducting electricity.

In contrast, sodium chloride has an ionic structure with Na⁺ and Cl⁻ ions fixed in a rigid 3D lattice. Although charged ions are present, they are FIXED in position and cannot move. Without mobile charged particles, no electric current can flow, and the solid does not conduct.

However, if sodium chloride is melted, the ionic lattice breaks apart and ions become free to move in the liquid. In this molten state, the mobile ions can conduct electricity. Similarly, if dissolved in water, the ions separate and become mobile, and the solution conducts.”

Keywords: metallic, delocalized electrons, mobile, ionic, lattice, fixed position, molten, dissolved, conduct

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Blueprint 5: Energy Diagram & Enthalpy (4-6 marks)

Structure:

1. Identify exo/endothermic from ΔH sign
2. Describe energy level positions
3. Describe flow of energy with surroundings
4. Link to bond breaking/making if asked

Example - “Draw an energy diagram for combustion of methane (ΔH = -890 kJ/mol) and explain what the diagram shows”:

Draw diagram with:

• Vertical Y-axis labeled “Enthalpy/Energy”
• Horizontal X-axis labeled “Reaction progress”
• “Reactants” starting at high level
• “Products” at lower level (lower than reactants)
• Arrow downward from reactants → products labeled “ΔH = -890 kJ/mol”
• “Heat released” or “exothermic” noted

Then explain:

“The energy diagram shows an exothermic reaction. The reactants (CH₄ and O₂) start at a higher enthalpy level. As the reaction proceeds, bonds break in the reactants. The energy released when new bonds form in the products (CO₂ and H₂O) is GREATER than the energy required to break the original bonds. Therefore, the products have a lower enthalpy than the reactants, and the difference (890 kJ per mole of CH₄) is released as heat energy to the surroundings. The negative ΔH value confirms this is exothermic. The system loses energy; the surroundings gain it, increasing temperature.”

Keywords: exothermic, enthalpy, reactants, products, bonds break, bonds form, ΔH, negative, heat released, surroundings

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WORKED SOLUTION 5: Bonding Type Identification & Property Explanation (A-Grade)

Question: Three substances are: Diamond (C), Copper (Cu), and Carbon Dioxide (CO₂)

a) Identify the bonding type in each substance b) Explain how the bonding type determines THREE different properties c) Predict which substance has the highest melting point and justify your answer

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SOLUTION a): Bonding Type Identification

Substance	Bonding Type	Evidence
Diamond	Covalent Network (Giant Covalent)	Each C bonded to 4 other C atoms in continuous 3D network
Copper	Metallic	Metal element; cations surrounded by delocalized electron sea
CO₂	Covalent Molecular	Discrete molecules held together by weak intermolecular forces

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SOLUTION b): Bonding Type Determines Properties

Property 1: Melting Point

Substance	Bonding	Melting Point	Reason
Diamond	Covalent Network	~3823°C (very high)	Entire structure is ONE giant network of strong C-C covalent bonds throughout. ALL bonds must break simultaneously to melt. Enormous energy required.
Copper	Metallic	1084°C (high)	Strong electrostatic attraction between cations and electron sea, but delocalized electrons more mobile than covalent bonds. Energy required is less than diamond.
CO₂	Covalent Molecular	-78°C (sublimes - very low)	Within each molecule, C=O bonds are strong. BUT between molecules only weak van der Waals forces exist. Very little energy needed to overcome these weak forces and separate molecules.

Explanation: More bonds to break = higher melting point. Continuous strong bonds (diamond) > metallic bonds (copper) > weak intermolecular forces (CO₂)

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Property 2: Electrical Conductivity

Substance	Bonding	Conducts?	State	Reason
Diamond	Covalent Network	NO	Solid or molten	Electrons are localized in C-C bonds. No mobile charged particles. Even when melten, carbon forms covalent bonds, not ions.
Copper	Metallic	YES	Solid AND molten	Delocalized electrons are mobile in both solid and liquid states. These electrons can move freely when potential difference applied.
CO₂	Covalent Molecular	NO	Solid, liquid, or gas	Electrons are localized in C=O bonds. No mobile charged particles. Molecules are neutral.

Explanation: Conductivity requires mobile charged particles (ions or delocalized electrons). Only copper has these in all states.

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Property 3: Hardness & Brittleness

Substance	Bonding	Hardness	Why
Diamond	Covalent Network	Extremely hard	3D network of strong covalent bonds in all directions. Cannot be compressed or scratched. Bonds won’t break under pressure.
Copper	Metallic	Hm, malleable (not hard)	Delocalized electrons act like “glue” allowing layers to slide over each other without breaking bonds. Can be hammered into shapes.
CO₂	Covalent Molecular	Soft (waxy if solid)	Molecules held only by weak van der Waals forces between them. Easily compressed or broken apart.

Explanation: Hard materials have strong, directed bonds throughout. Malleable materials have bonding that allows layer movement. Soft materials have weak forces between particles.

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SOLUTION c): Highest Melting Point Prediction

Prediction: Diamond has highest melting point

Justification (A-Grade Required):

“Diamond is a covalent network structure where each carbon atom is bonded covalently to four other carbon atoms in a tetrahedral arrangement, forming a continuous 3D lattice extending throughout the entire structure. Unlike molecular substances like CO₂, which have weak intermolecular van der Waals forces between discrete molecules, diamond has strong covalent bonds between all atoms. Unlike metallic substances like copper, which have metallic bonding allowing some electron delocalization and ion movement, diamond’s electrons are fully localized in C-C bonds.

To melt diamond, every single C-C covalent bond in the entire structure must be broken simultaneously. This requires enormous energy - approximately 3823°C. This is exceptionally high compared to copper (1084°C) and CO₂ (-78°C).

The fundamental principle is: melting point ∝ strength of bonds holding structure together

In this case:

• Diamond: Strong, directed covalent bonds throughout → highest m.p.
• Copper: Metallic bonds (strong, but delocalized) → medium m.p.
• CO₂: Weak van der Waals forces between molecules → lowest m.p.

Therefore, diamond > copper > CO₂ in melting point order.”

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WORKED SOLUTION 6: Organic Addition Reaction (A-Grade Working)

Question: a) When 2.8 g of ethene (C₂H₄) undergoes addition with bromine water, write the balanced equation and describe the observations b) Calculate how many moles of Br₂ are needed to completely react c) Is this addition or substitution? How can you tell?

[M(C₂H₄) = 28, M(Br₂) = 160]

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SOLUTION a): Balanced Equation & Observations

Balanced Equation: $\textsf{C}_2\textsf{H}_4 + \textsf{Br}_2 \rightarrow \textsf{C}_2\textsf{H}_4\textsf{Br}_2$

(1,2-dibromoethane formed)

Structural representation: $\textsf{CH}_2=\textsf{CH}_2 + \textsf{Br-Br} \rightarrow \textsf{CHBr-CHBr}$

Observations (A-Grade Description):

Before reaction:

• Ethene: Colorless liquid or gas
• Bromine water: Orange-brown color

During reaction:

• The C=C double bond electrons attack the Br-Br molecule, breaking it
• The bromine atom adds across the double bond (one Br to each carbon)
• The double bond becomes a single bond

After reaction:

• Orange-brown color DISAPPEARS INSTANTLY (within seconds)
• Product formed: Colorless (or pale) 1,2-dibromoethane
• Observation: The rapid decolorization indicates: (1) fast reaction, (2) bromine is consumed (no longer in solution to give color)

Why this observation matters: This is how you distinguish addition (instant color loss) from substitution (slow color change)

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SOLUTION b): Moles of Br₂ Required

Step 1: Convert mass of ethene to moles $n(\text{C}_2\text{H}_4) = \frac{m}{M} = \frac{2.8 \text{ g}}{28 \text{ g/mol}} = 0.1 \text{ mol}$

Step 2: Use balanced equation to find mole ratio

From equation: 1 mol C₂H₄ reacts with 1 mol Br₂

Therefore: $n(\text{Br}_2) = 0.1 \text{ mol C}_2\text{H}_4 \times \frac{1 \text{ mol Br}_2}{1 \text{ mol C}_2\text{H}_4} = 0.1 \text{ mol Br}_2$

Answer b): 0.1 mol of Br₂ (or 16 g of Br₂) is needed

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SOLUTION c): Addition vs Substitution Identification

This is ADDITION because:

1. Molecular formula changes in specific way: C₂H₄ + Br₂ → C₂H₄Br₂

   • Atoms are ADDED to the molecule
   • No atoms are REPLACED or REMOVED
   • Total number of atoms increases

2. How to identify from structural formula:

   • Starting: C₂H₄ has C=C double bond
   • Ending: C₂H₄Br₂ has C-C single bond with Br atoms attached
   • The double bond BROKE and became single bond
   • Br atoms ADDED across where double bond was

3. Observation confirms it:

   • Orange bromine water color disappears INSTANTLY
   • This speed indicates addition reaction (not substitution)
   • If this were substitution, the color would remain for minutes/hours

Compare to substitution (hypothetical):

• Substitution would be: C₂H₆ + Br₂ → C₂H₅Br + HBr
• Color would remain orange (alkane, slow reaction, requires UV)
• Would need UV light to proceed

**Answer c): This is ADDITION because:

• Double bond breaks (C=C becomes C-C)
• Atoms are added (2 Br atoms added)
• Observation shows instant color loss (characteristic of addition)**

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WORKED SOLUTION 3: Thermochemistry - Heat Energy Calculation (A-Grade Working)

Question: Given the thermochemical equation: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -890 \text{ kJ mol}^{-1}$

Calculate the total heat energy released when: a) 4.0 g of methane is burned completely b) 2 moles of methane are burned

[M(CH₄) = 16 g/mol]

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SOLUTION a): Heat released from mass of fuel (4.0 g CH₄)

Step 1: Understand what ΔH value means

The equation shows ΔH = -890 kJ mol⁻¹

This means: When 1 mole of CH₄ burns completely, 890 kJ of heat is released (negative sign means exothermic)

The ΔH value ONLY applies to the amounts shown in the balanced equation

Step 2: Convert mass to moles

Given: mass of CH₄ = 4.0 g

$n(\text{CH}_4) = \frac{m}{M} = \frac{4.0 \text{ g}}{16 \text{ g/mol}} = 0.25 \text{ mol}$

Step 3: Calculate total heat using q = n × ΔH

The formula is: $q = n(\text{substance referenced in equation}) \times |\Delta H|$

The ΔH of -890 kJ refers to the coefficient of CH₄ (which is 1), so:

$q = 0.25 \text{ mol} \times 890 \text{ kJ/mol} = 222.5 \text{ kJ}$

Step 4: Include the sign and state what it means

Since the equation has ΔH = negative, the reaction is exothermic (releases heat)

$q = -222.5 \text{ kJ}$

(Negative indicates heat released TO the surroundings)

Step 5: Report with appropriate significant figures

Given mass has 2 sig figs (4.0), so: 220 kJ released or -222.5 kJ

Answer a): 222.5 kJ (or 220 kJ considering sig figs) of heat is released

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SOLUTION b): Heat released from moles of fuel (2 moles CH₄)

Step 1: Understand the scaling principle

The balanced equation shows 1 mol CH₄ → 890 kJ released

If you burn 2 moles of CH₄, you get 2 × 890 kJ

Step 2: Apply the scaling

$\text{For 1 mol CH}_4: q = 890 \text{ kJ}$

$\text{For 2 mol CH}_4: q = 2 \times 890 = 1780 \text{ kJ}$

Alternative method using formula: $q = n \times |\Delta H| = 2 \text{ mol} \times 890 \text{ kJ/mol} = 1780 \text{ kJ}$

Answer b): 1780 kJ of heat is released

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CRITICAL POINT FOR A-GRADE:

If the equation were doubled: $2\text{CH}_4(g) + 4\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \quad \Delta H = -1780 \text{ kJ mol}^{-1}$

Then burning 2 mol CH₄ would release 1780 kJ (straight from ΔH), OR you’d calculate: $q = 2 \text{ mol} \times 1780 \text{ kJ} / 2 \text{ mol} = 1780 \text{ kJ}$

The answer is the same - you get 1780 kJ either way.

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WORKED SOLUTION 4: Percentage Composition Calculation (A-Grade Working)

Question: Calculate the percentage composition of each element in glucose (C₆H₁₂O₆). Then use it to verify the formula is correct by summing to 100%.

[M(C) = 12, M(H) = 1, M(O) = 16]

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SOLUTION: Complete A-Grade Working

Step 1: Calculate molar mass of glucose

$M(\text{C}_6\text{H}_{12}\text{O}_6) = (6 \times 12) + (12 \times 1) + (6 \times 16)$

$M = 72 + 12 + 96 = 180 \text{ g/mol}$

Step 2: Recall percentage composition formula

$\% \text{ element} = \frac{(\text{Aᵣ of element}) \times (\text{number in formula})}{M_r \text{ of compound}} \times 100$

Step 3: Calculate %Carbon

$\% \text{C} = \frac{12 \times 6}{180} \times 100 = \frac{72}{180} \times 100$

$= 0.4 \times 100 = 40\%$

Step 4: Calculate %Hydrogen

$\% \text{H} = \frac{1 \times 12}{180} \times 100 = \frac{12}{180} \times 100$

$= 0.0667 \times 100 = 6.67\%$

Step 5: Calculate %Oxygen

$\% \text{O} = \frac{16 \times 6}{180} \times 100 = \frac{96}{180} \times 100$

$= 0.5333 \times 100 = 53.33\%$

Step 6: Verify - sum should equal 100%

$40\% + 6.67\% + 53.33\% = 100\%$ ✓

Answer: C = 40%, H = 6.67%, O = 53.33%

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COMMENTARY (A-Grade Required):

“Three key observations:

1. Verification method: Always add percentages - must sum to exactly 100% (allow ±0.01% for rounding). If not, you’ve made a calculation error.

2. Which element is most abundant? Oxygen at 53.33% - this makes sense because:

   • Oxygen atoms have highest atomic mass (16)
   • Glucose has most oxygen atoms (6 out of 20 total atoms)
   • Combined effect makes O dominate the mass

3. Practical application: If you’re given an unknown compound is 40% C, 6.67% H, 53.33% O, you could deduce it could be C₆H₁₂O₆ (glucose) without being told.”

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WORKED SOLUTION 1: Full Stoichiometry Problem (A-Grade Complete Working)

Question: Iron oxide can be reduced by carbon: $2\text{Fe}_2\text{O}_3(s) + 3\text{C}(s) \rightarrow 4\text{Fe}(s) + 3\text{CO}_2(g)$

a) Calculate the mass of iron produced when 3.2 g of carbon reacts with excess iron oxide [M(C) = 12, M(Fe) = 56]

b) If 15.68 g of iron was actually produced, calculate the percentage yield. Comment on the yield.

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SOLUTION a): Complete A-Grade Working

Step 1: Check equation is balanced and identify given information

Counting atoms:

• Left: Fe (2×2=4), O (3×3=9), C (3)
• Right: Fe (4), O (6+6=12) ❌ Not balanced in O

Recheck: Fe₂O₃ has 3 oxygen atoms per molecule, 2 molecules = 6 O

• Left: 2 Fe₂O₃ = 4 Fe, 6 O; 3 C
• Right: 4 Fe, 3 CO₂ = 3 C, 6 O ✓ Now balanced

Given data:

• Mass of carbon = 3.2 g
• Molar mass of C = 12 g/mol
• Molar mass of Fe = 56 g/mol
• Excess Fe₂O₃ (so carbon is limiting reagent)

What to find: Mass of Fe produced

Step 2: Convert mass of carbon to moles using n = m/M

$n(\text{C}) = \frac{m(\text{C})}{M(\text{C})} = \frac{3.2 \text{ g}}{12 \text{ g/mol}} = 0.2667 \text{ mol}$

(Keep to 4 significant figures during calculation)

Step 3: Use balanced equation to find mole ratio

From the equation: 3 mol C → 4 mol Fe

This means: For every 3 moles of carbon that react, 4 moles of iron are produced

Step 4: Calculate moles of Fe using mole ratio

$n(\text{Fe}) = n(\text{C}) \times \frac{\text{mol Fe}}{\text{mol C}} = 0.2667 \text{ mol} \times \frac{4}{3}$

$n(\text{Fe}) = 0.2667 \times 1.333... = 0.3556 \text{ mol Fe}$

Step 5: Convert moles of Fe back to mass using m = nM

$m(\text{Fe}) = n(\text{Fe}) \times M(\text{Fe}) = 0.3556 \text{ mol} \times 56 \text{ g/mol}$

$m(\text{Fe}) = 19.91 \text{ g}$

Step 6: Round to appropriate significant figures

Given data has 2 sig figs (3.2), so answer = 19.9 g or 20 g (2 sig figs)

But better to report as 19.91 g (4 sig figs) for part a, use for part b

Answer a): 19.91 g or 20 g of Fe produced (depending on sig fig instruction)

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SOLUTION b): Percentage Yield with Commentary

Step 1: Recall percentage yield formula

$\% \text{yield} = \frac{\text{actual yield (experimental)}}{\text{theoretical yield (calculated)}} \times 100$

Step 2: Identify the two values

• Theoretical yield (calculated in part a) = 19.91 g Fe
• Actual yield (given) = 15.68 g Fe

Step 3: Calculate percentage yield

$\% \text{yield} = \frac{15.68}{19.91} \times 100$

$\% \text{yield} = 0.7873 \times 100 = 78.73\%$

Step 4: Round to appropriate significant figures

Given values have 4 sig figs, so report as: % yield = 78.7% (4 sig figs)

Answer b): 78.7%

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COMMENTARY (A-Grade Required):

“The percentage yield of 78.7% indicates that the reaction did not proceed to completion (100% yield). This is typical in real laboratory conditions. The loss of 21.3% can be attributed to several factors:

1. Side reactions: Other reactions may have occurred, consuming some reactants to form undesired products
2. Incomplete reaction: Not all carbon reacted with the iron oxide (despite being stated as excess)
3. Evaporation/spillage: Some of the product may have been lost during handling, transfer, or heating
4. Equilibrium limitations: The reaction may not reach equilibrium completely in the time given
5. Apparatus limitations: Mass may be retained in pipettes, beakers, or lost as small particles

A yield of 78.7% is considered reasonable for most organic synthetic reactions, which typically range from 70-90%.”

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WORKED SOLUTION 2: Relative Atomic Mass Calculation (A-Grade Working)

Question: Chlorine exists as two main isotopes:

• ³⁵Cl with mass 35 and abundance 75.77%
• ³⁷Cl with mass 37 and abundance 24.23%

Calculate the relative atomic mass of chlorine. Comment on how this relates to the periodic table.

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SOLUTION: Complete A-Grade Working

Step 1: Understand what relative atomic mass is

Relative atomic mass (Aᵣ) is the weighted average of all isotopes’ masses, taking into account the percentage abundance of each isotope. It is calculated per 12 carbon atoms (C-12 = 12 exactly).

Step 2: Recall the formula

$A_r = \frac{(\text{mass of isotope 1} \times \% \text{ abundance 1}) + (\text{mass of isotope 2} \times \% \text{ abundance 2})}{100}$

OR more simply:

$A_r = \frac{\sum(\text{mass} \times \text{abundance})}{100}$

Step 3: Identify given data

• Isotope 1: ³⁵Cl, mass = 35, abundance = 75.77%
• Isotope 2: ³⁷Cl, mass = 37, abundance = 24.23%
• Check: 75.77% + 24.23% = 100% ✓

Step 4: Substitute into formula

$A_r(\text{Cl}) = \frac{(35 \times 75.77) + (37 \times 24.23)}{100}$

Step 5: Calculate each term in numerator

First term: $35 \times 75.77 = 2,652.95$

Second term: $37 \times 24.23 = 896.51$

Sum: $2,652.95 + 896.51 = 3,549.46$

Step 6: Divide by 100 to get average

$A_r(\text{Cl}) = \frac{3,549.46}{100} = 35.49 \text{ (or 35.5 to 1 decimal place)}$

Step 7: Report with appropriate sig figs

Given abundances to 4 sig figs, so report: $A_r(\text{Cl}) = 35.49$

Answer: 35.49 or 35.5 (both acceptable, 35.5 commonly seen)

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COMMENTARY (A-Grade Required):

“Notice that the relative atomic mass (35.49) is:

1. Between the two isotope masses (35 and 37) ✓ This must always be true
2. Closer to 35 than 37 because ³⁵Cl is MORE abundant (75.77% vs 24.23%)
3. NOT a whole number because it’s an average of two masses
4. Matches the periodic table value (Cl = 35.45 in most tables) - our minor difference is due to rounding

This calculation explains why most periodic table values for relative atomic mass are NOT whole numbers. For example:

• O ≈ 16 (appears whole because O-16 is 99.76% abundant)
• Cu ≈ 63.5 (mixture of Cu-63 and Cu-65 in roughly equal amounts)
• Br ≈ 79.9 (mixture of Br-79 and Br-81 nearly equally abundant)

Key insight for exams: If an element has relative atomic mass far from a whole number (like Cu at 63.5), this indicates the element has multiple isotopes present in significant amounts.”

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Phase 10: Common Mistakes to AVOID

1. Forgetting to BALANCE before using mole ratio

   • Incorrect: Using unbalanced equation to find mole ratio
   • Correct: Always balance first, then use coefficients

2. Wrong ΔH sign for classification

   • Incorrect: “Reaction is endothermic, ΔH = -50 kJ”
   • Correct: Endothermic means ΔH = +50 kJ (positive)

3. Not scaling heat when coefficients are doubled

   • If equation: CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ
   • If 2 mol CH₄ burns, then q = 2 × 890 = 1780 kJ (not still 890)

4. Confusing conductivity in different states

   • Incorrect: “Ionic compounds don’t conduct, period”
   • Correct: Ionic solids don’t conduct (fixed ions), but molten/dissolved ionic DO (mobile ions)

5. Treating relative atomic mass as having units

   • Incorrect: A_r(Cl) = 35.5 g/mol
   • Correct: A_r(Cl) = 35.5 (unitless number)

6. Skipping explanation keywords in extended responses

   • Incorrect: Just giving answer without “because,” “therefore,” “due to”
   • Correct: Every 3 words include a chemistry keyword

7. Ignoring significant figures and units

   • Incorrect: 3.22 g ÷ 5.6 mL = 0.574642857 g/mL
   • Correct: 0.57 g/mL (2 sig figs, limited by 5.6)

8. Writing water as H₂O(g) in combustion equations

   • Guideline SAYS: Water should be written as H₂O(l) unless told otherwise
   • This is an actual marking rule in your exams

9. Not showing state symbols

   • Incorrect: CH₄ + 2O₂ → CO₂ + 2H₂O
   • Correct: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

10. Ion charges in formulas

    • Incorrect: MgCl (wrong charge balance)
    • Correct: Mg²⁺ loses 2, needs 2 Cl⁻ → MgCl₂

11. Confusing addition vs substitution reactions with observations

    • Incorrect: “Bromine water addition → stays orange”
    • Correct: Addition (alkene) → orange disappears INSTANTLY; Substitution (alkane) → orange stays for long time

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Phase 11: A-GRADE Answer Language (Use These Exactly)

Use these sentence starters and phrases to sound like an A student:

For trend explanations:

• “This trend occurs because…”
• “As atomic number increases/down the group…”
• “The increased nuclear charge…”
• “Due to the addition of electron shells…”
• “The shielding effect of inner electrons…”

For particle-level explanations:

• “At the particle level…”
• “The electrostatic attraction between…”
• “The delocalized electrons…”
• “Mobile charged particles (ions/electrons)…”
• “Within the lattice structure…”

For energetics:

• “By conservation of energy…”
• “The energy released when bonds form…”
• “A greater amount of energy is required to…”
• “Therefore, the net energy change is…”

For mathematical working:

• “Using the balanced equation, the mole ratio is…”
• “Converting to moles: n = m/M =…”
• “This confirms that…”
• “Rounded to appropriate significant figures…”

Transition phrases:

• “Consequently…”
• “As a result of…”
• “Similarly, this explains why…”
• “This is because…”
• “In contrast…”

Conclusion phrases:

• “In summary, the structure determines the…”
• “Therefore, the bonding model explains why…”
• “This demonstrates that…”

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Phase 12: Night-Before Exam 90-Minute Cram Plan

• 0-15 min: Memorize all formulas (n = m/M, etc.)
• 15-30min: Review subatomic particles, electron config, isotopes
• 30-45 min: Practice one stoichiometry problem start to finish
• 45-60 min: Review periodic trends with cause language
• 60-75 min: Practice one bonding explanation (ionic, then metallic)
• 75-85 min: Review enthalpy signs and thermochemical equations
• 85-90 min: Skim through organic naming and combustion equation

Final check:

• Balanced equations with state symbols?
• H₂O written as H₂O(l)?
• ΔH signs correct?
• Significant figures ok?
• Keywords included in explanations?

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WORKED SOLUTION 7: Limiting Reagent Problem (A-Grade Working)

Question: Zinc reacts with copper sulfate solution: $\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)$

When 6.5 g of Zn is reacted with 12.8 g of CuSO₄: a) Identify the limiting reagent b) Calculate the maximum mass of Cu that can be produced c) Calculate the mass of the excess reagent remaining

[M(Zn) = 65, M(CuSO₄) = 160, M(Cu) = 64]

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SOLUTION a): Identify Limiting Reagent

Step 1: Convert both masses to moles

$n(\text{Zn}) = \frac{m}{M} = \frac{6.5 \text{ g}}{65 \text{ g/mol}} = 0.1 \text{ mol}$

$n(\text{CuSO}_4) = \frac{m}{M} = \frac{12.8 \text{ g}}{160 \text{ g/mol}} = 0.08 \text{ mol}$

Step 2: Use balanced equation to find mole ratio

From equation: 1 mol Zn reacts with 1 mol CuSO₄ (1:1 ratio)

Step 3: Determine which is limiting

Method: For each reactant, calculate how much product it COULD make

Using Zn as limiting: $n(\text{Cu from Zn}) = 0.1 \text{ mol Zn} \times \frac{1 \text{ mol Cu}}{1 \text{ mol Zn}} = 0.1 \text{ mol Cu}$

Using CuSO₄ as limiting: $n(\text{Cu from CuSO}_4) = 0.08 \text{ mol CuSO}_4 \times \frac{1 \text{ mol Cu}}{1 \text{ mol CuSO}_4} = 0.08 \text{ mol Cu}$

Comparison:

• If Zn is limiting: theoretical yield = 0.1 mol Cu
• If CuSO₄ is limiting: theoretical yield = 0.08 mol Cu
• The LOWER amount (0.08 mol) is what actually gets produced
• Therefore CuSO₄ is the limiting reagent (runs out first)

Answer a): CuSO₄ is the limiting reagent

Verification:

• We have 0.1 mol Zn but need only 0.08 mol to react with all the CuSO₄
• We have 0.08 mol CuSO₄ but all of it reacts
• CuSO₄ is consumed completely → it’s limiting
• Zn has 0.1 - 0.08 = 0.02 mol left over → it’s in excess

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SOLUTION b): Maximum Mass of Cu Produced

Step 1: Use the limiting reagent (CuSO₄) to calculate product

From balanced equation: 1 mol CuSO₄ produces 1 mol Cu

$n(\text{Cu}) = 0.08 \text{ mol CuSO}_4 \times \frac{1 \text{ mol Cu}}{1 \text{ mol CuSO}_4} = 0.08 \text{ mol Cu}$

Step 2: Convert moles to mass

$m(\text{Cu}) = n \times M = 0.08 \text{ mol} \times 64 \text{ g/mol} = 5.12 \text{ g}$

Step 3: Significant figures

Given data has 2-3 sig figs, report as: 5.1 g or 5.12 g

Answer b): Maximum 5.12 g of Cu produced

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SOLUTION c): Mass of Excess Reagent Remaining

Step 1: Identify which is in excess From part a: Zn is in excess (CuSO₄ was limiting)

Step 2: Calculate how much Zn was consumed

From balanced equation: 1 mol Zn reacts with 1 mol CuSO₄

Since all 0.08 mol of CuSO₄ reacted: $n(\text{Zn consumed}) = 0.08 \text{ mol CuSO}_4 \times \frac{1 \text{ mol Zn}}{1 \text{ mol CuSO}_4} = 0.08 \text{ mol Zn}$

Step 3: Calculate how much Zn remains

$n(\text{Zn remaining}) = n(\text{Zn initial}) - n(\text{Zn consumed})$

$n(\text{Zn remaining}) = 0.1 - 0.08 = 0.02 \text{ mol}$

Step 4: Convert remaining Zn to mass

$m(\text{Zn remaining}) = n \times M = 0.02 \text{ mol} \times 65 \text{ g/mol} = 1.3 \text{ g}$

Answer c): 1.3 g of Zn remains unreacted

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VERIFICATION OF ANSWER:

Check mass balance:

• Initial Zn consumed: 0.08 mol × 65 = 5.2 g
• Initial CuSO₄ consumed: all 12.8 g
• Cu produced: 5.12 g
• Zn remaining: 1.3 g

Total in: 6.5 g Zn + 12.8 g CuSO₄ = 19.3 g Total out: 5.12 g Cu + 1.3 g Zn remaining = 6.42 g out and…

Actually, this doesn’t balance because CuSO₄ → ZnSO₄ (ions go into solution, not counted as solid product)

Better check:

• Zn consumed = 5.2 g ✓ (0.08 mol)
• CuSO₄ consumed = 12.8 g ✓ (all of it)
• Cu produced = 0.08 mol ✓
• These ratios follow 1:1 from balanced equation ✓

Answer c): 1.3 g Zn remains

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Phase 13: A-GRADE ESSENTIALS - Final Checklist

What Makes an A-Grade Answer (vs Pass-Grade)

Pass-Grade (60-70%) includes:

• Correct formula substitution
• Basic calculations shown
• Definition of terms
• Correct answer with units

A-Grade (85-95%) ADDITIONALLY includes:

1. Complete working: EVERY step shown, not skipped

   • Converting units clearly
   • Showing fraction before calculating
   • Intermediate values with correct sig figs

2. Explanations of WHY:

   • “Because the nuclear charge increases…” (not just “it increases”)
   • “This is exothermic (ΔH negative) because more energy is released making bonds than required breaking bonds”
   • Particle-level reasoning for all macroscopic observations

3. Significant figures:

   • Identified limiting value
   • Rounded final answer appropriately
   • Justified rounding decision if asked

4. Connection to concepts:

   • “This explains why Na has lower ionization energy than Mg because…”
   • “Unlike ionic compounds which conduct when molten, covalent compounds do not because…”
   • Links observations to underlying structure

5. Commentary & Analysis:

   • “A yield of 78% is typical because…”
   • “The orange color disappears instantly, indicating this is addition (not substitution) because…”
   • “Comparing all three trends across the period shows…”

6. Verification:

   • Checking percentages add to 100%
   • Checking products make sense from balanced equation
   • Confirming theoretical > actual in yield calculations

7. Using correct terminology:

   • “Electrostatic attraction” not just “attraction”
   • “Delocalized electrons” not “free electrons”
   • “Lattice structure” not just “structure”
   • “Octet rule” not “full shell”

8. Addressing ALL parts of question:

   • Part a. (calculation) → Part b. (explanation) → Part c. (comparison)
   • Not leaving any part blank or rushed

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Phase 13: VERIFICATION - Can You Solve Exam Questions?

Test yourself on these exam-style questions using ONLY the cheatsheet

Question 1 (Definition - 2 marks): Define what an isotope is, including what particles differ and give an example.

Model answer: Isotopes are atoms with same atomic number (protons) but different mass number (neutrons). Same element, same chemical properties but different mass. Example: ¹²C and ¹⁴C carbon.

Question 2 (Calculation - 3 marks): Calculate A_r for chlorine given: Cl-35 (mass 35, abundance 75%), Cl-37 (mass 37, abundance 25%)

Calculation: $A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = 35.5$

Question 3 (Bonding - 5 marks): Explain how ionic bonding forms in magnesium oxide (MgO), including electron transfer and structure.

Model answer: Mg atom has electron configuration 2,8,2 and O has 2,6. Mg transfers 2 electrons to O. Mg becomes Mg²⁺ (stable 2,8) and O becomes O²⁻ (stable 2,8). Electrostatic attraction between opposite charges. 3D ionic lattice forms with Mg²⁺ surrounded by O²⁻ and vice versa.

Question 4 (Conductivity - 4 marks): Why does ionic compound NaCl not conduct electricity when solid, but conducts when molten?

Model answer: In solid state, Na⁺ and Cl⁻ ions are fixed in rigid lattice - not mobile. Conductivity requires mobile charged particles. When molten, lattice breaks, ions become free to move. Mobile ions conduct electricity.

Question 5 (Stoichiometry - 5 marks): For reaction: 2H₂ + O₂ → 2H₂O If 6 g H₂ reacts, calculate mass of water produced. [M(H₂)=2, M(H₂O)=18]

Solution: $n(H_2) = \frac{6}{2} = 3 \text{ mol}$ $n(H_2O) = 3 \times \frac{2}{2} = 3 \text{ mol}$ $m(H_2O) = 3 \times 18 = 54 \text{ g}$

Question 6 (Periodic trends - 4 marks): Explain why first ionisation energy increases across a period.

Model answer: Across a period, atomic number increases (more protons). Greater nuclear charge pulls valence electrons closer, needing more energy to remove them. Shielding by inner electrons stays roughly constant. Therefore, more energy required.

Question 7 (Thermochemistry - 4 marks): Classify and explain the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890 kJ mol⁻¹

Model answer: This is exothermic (negative ΔH). Products have lower enthalpy than reactants. More energy released forming CO₂ and H₂O bonds than required breaking CH₄ and O₂ bonds. Heat released to surroundings, increasing temperature.

Question 8 (Organic - 4 marks): Explain the observations when ethene (C₂H₄) and bromine water are mixed.

Model answer: Ethene undergoes addition reaction with bromine. The C=C double bond breaks, and Br adds across. Alkyl bromide forms. Observation: orange-brown bromine water color DISAPPEARS INSTANTLY because Br₂ is consumed in the chemical reaction.

Question 9 (Structure-property - 5 marks): Compare the properties of diamond and graphite, both from carbon, explaining differences by their structures.

Model answer:

• Diamond: Each C bonded to 4 others (tetrahedral, 3D network), very hard, very high mp, insulator
• Graphite: Each C bonded to 3 others in layers, soft (layers slide), conducts electricity (delocalised electrons in layers)
• Difference: Diamond = continuous 3D covalent bonds all directions; Graphite = layers held by weak forces allowing sliding

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APPENDIX: Common Ions (KNOW THESE!)

Group 1 & 2 Cations (always +)

• H⁺, Li⁺, Na⁺, K⁺, Be²⁺, Mg²⁺, Ca²⁺, Ba²⁺

Other Metal Cations

• Al³⁺, Fe²⁺, Fe³⁺, Cu²⁺, Pb²⁺, Sn²⁺

Non-metal Anions

• H⁻, F⁻, Cl⁻, Br⁻, I⁻, O²⁻, S²⁻, N³⁻

Polyatomic Ions

• SO₄²⁻ (sulfate), NO₃⁻ (nitrate), CO₃²⁻ (carbonate)
• HCO₃⁻ (bicarbonate), ClO⁻ (hypochlorite)
• OH⁻ (hydroxide), NH₄⁺ (ammonium)

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Question 3 (Bonding - 5 marks): Explain how ionic bonding forms in magnesium oxide (MgO), including electron transfer and structure.

Model answer: Mg atom has electron configuration 2,8,2 and O has 2,6. Mg transfers 2 electrons to O. Mg becomes Mg²⁺ (stable 2,8) and O becomes O²⁻ (stable 2,8). Electrostatic attraction between opposite charges. 3D ionic lattice forms with Mg²⁺ surrounded by O²⁻ and vice versa.

Question 4 (Conductivity - 4 marks): Why does ionic compound NaCl not conduct electricity when solid, but conducts when molten?

Model answer: In solid state, Na⁺ and Cl⁻ ions are fixed in rigid lattice - not mobile. Conductivity requires mobile charged particles. When molten, lattice breaks, ions become free to move. Mobile ions conduct electricity.

Question 5 (Stoichiometry - 5 marks): For reaction: 2H₂ + O₂ → 2H₂O If 6 g H₂ reacts, calculate mass of water produced. [M(H₂)=2, M(H₂O)=18]

Solution: $n(H_2) = \frac{6}{2} = 3 \text{ mol}$ $n(H_2O) = 3 \times \frac{2}{2} = 3 \text{ mol}$ $m(H_2O) = 3 \times 18 = 54 \text{ g}$

Question 6 (Periodic trends - 4 marks): Explain why first ionisation energy increases across a period.

Model answer: Across a period, atomic number increases (more protons). Greater nuclear charge pulls valence electrons closer, needing more energy to remove them. Shielding by inner electrons stays roughly constant. Therefore, more energy required.

Question 7 (Thermochemistry - 4 marks): Classify and explain the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890 kJ mol⁻¹

Model answer: This is exothermic (negative ΔH). Products have lower enthalpy than reactants. More energy released forming CO₂ and H₂O bonds than required breaking CH₄ and O₂ bonds. Heat released to surroundings, increasing temperature.

Question 8 (Organic - 4 marks): Explain the observations when ethene (C₂H₄) and bromine water are mixed.

Model answer: Ethene undergoes addition reaction with bromine. The C=C double bond breaks, and Br adds across. Alkyl bromide forms. Observation: orange-brown bromine water color DISAPPEARS INSTANTLY because Br₂ is consumed in the chemical reaction.

Question 9 (Structure-property - 5 marks): Compare the properties of diamond and graphite, both from carbon, explaining differences by their structures.

Model answer:

• Diamond: Each C bonded to 4 others (tetrahedral, 3D network), very hard, very high mp, insulator
• Graphite: Each C bonded to 3 others in layers, soft (layers slide), conducts electricity (delocalised electrons in layers)
• Difference: Diamond = continuous 3D covalent bonds all directions; Graphite = layers held by weak forces allowing sliding

NEW: Question 10 (Variables - 4 marks):

Question: “An experiment to investigate how changing the concentration of hydrochloric acid affects the rate of reaction with magnesium metal”

Identify: a) Independent variable b) Dependent variable c) Three controlled variables

Model answer: a) Independent variable: Concentration of hydrochloric acid (varied from 0.1 M to 0.5 M OR 0.5 M to 2.0 M)

b) Dependent variable: Rate of reaction (measured as time taken for magnesium to dissolve completely, or volume of hydrogen gas produced per minute, or mass of magnesium remaining over time)

c) Controlled variables (any three of):

• Temperature of acid solution (kept at 25°C in water bath)
• Mass of magnesium metal (kept at exactly 2.0 g)
• Size/surface area of magnesium (same sized pieces/powder)
• Volume of acid used (kept at 50 mL)
• Type of apparatus (same beaker, same stirring rod)
• Atmospheric pressure (same location)
• Light exposure (kept in same conditions)

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NEW: Question 11 (Organic Reaction Identification - 3 marks):

Question: “When a colourless organic compound is added to bromine water at room temperature, the orange-brown colour persists for several minutes. When 2-methylpropene is added to bromine water at room temperature, the color disappears instantly. Explain which reaction is addition and which is substitution.”

Model answer:

2-methylpropene + Br₂ = ADDITION

• Observation: Orange color disappears instantly (within seconds)
• Explanation: The C=C double bond in the alkene breaks open. Bromine atoms add directly across the double bond, consuming the Br₂. With Br₂ consumed, no orange color remains in solution.
• Equation: (CH₃)₂C=CH₂ + Br₂ → (CH₃)₂CBr-CH₂Br
• Mechanism: π electrons from C=C attack Br-Br, breaking the bromine molecule

Colourless compound (alkane) + Br₂ = SUBSTITUTION

• Observation: Orange color persists for minutes (slow reaction)
• Explanation: Only a hydrogen atom is replaced by a bromine atom. Most of the bromine water remains unconsumed because very little Br₂ is actually incorporated. The orange color remains because Br₂ is still in solution.
• Equation (example with methane): CH₄ + Br₂ → CH₃Br + HBr (requires UV light)
• Mechanism: Free radical substitution - requires UV light to break Br-Br bond into radicals first

Key Distinction:

• Addition: FAST, no catalyst needed, color disappears (Br₂ consumed)
• Substitution: SLOW, requires UV light, color remains (little Br₂ consumed)

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FINAL 60-SECOND PRE-EXAM CHECKLIST

✓ Balanced equations written? ✓ State symbols (s), (l), (g), (aq) included? ✓ Water written as H₂O(l) in combustion? ✓ ΔH has correct sign (+ endothermic, − exothermic)? ✓ Units on final answers (g, mol, J, kJ)? ✓ Significant figures rounded correctly? ✓ Mole ratio used from balanced equation? ✓ Calculations shown step by step? ✓ Chemistry keywords in explanations (electrostatic, octet, lattice, etc)? ✓ Periodic trends explained with CAUSE (nuclear charge or shielding)?

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If you can answer all 9 verification questions above with full working and explanation, you are ready for an A grade.

THIS SHEET COVERS EVERYTHING IN YOUR GUIDELINE SHEET AND PAST PAPERS.

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SECTION COMPLETENESS VERIFICATION

✅ FULLY COVERED (Original + Expanded):

• ✓ Atomic theory (Dalton, Thomson, Rutherford, Bohr) with detailed mechanisms
• ✓ Subatomic particles & $^A_ZX$ notation with worked examples
• ✓ Electron configuration & ions with table format
• ✓ Isotopes & relative atomic mass with detailed calculation and commentary
• ✓ Mass spectrometry (all 5 stages clear)
• ✓ Absorption & emission spectra with calibration curves (ADDED)
• ✓ Ionization energy diagrams with interpretation guide (ADDED)
• ✓ Periodic trends with ALL causes explained
• ✓ Bonding (ionic, metallic, covalent molecular, covalent network) with properties
• ✓ Separation techniques with decision-making
• ✓ Nanomaterials & fullerenes
• ✓ Enthalpy & thermochemical equations
• ✓ Exothermic/endothermic reactions
• ✓ Stoichiometry (mass-mole conversions, limiting reagent, yield, purity) - EXPANDED
• ✓ Heat energy calculations from reactions - EXPANDED WITH FULL WORKING
• ✓ Organic chemistry: alkanes, alkenes, benzene naming
• ✓ Organic reactions: ADDITION (alkene + Br₂, + H₂, + H₂O) FULLY ADDED WITH ALL CONDITIONS
• ✓ Organic reactions: SUBSTITUTION (alkane + Br₂, + Cl₂) FULLY ADDED WITH CONDITIONS
• ✓ Organic reactions: AROMATIC SUBSTITUTION (benzene + Br₂) FULLY ADDED
• ✓ Observation differences for ADD vs SUB (color changes, timing) CRITICAL TABLE ADDED
• ✓ Combustion & biofuels
• ✓ Errors (random, systematic), validity, reliability, accuracy
• ✓ Significant figures
• ✓ Variables: Independent, Dependent, Controlled (FULLY ADDED WITH EXAMPLES)
• ✓ Primary vs Secondary Data (FULLY ADDED WITH DEFINITIONS)
• ✓ Ionization energy graphs (FULLY ADDED WITH INTERPRETATION STEPS)
• ✓ Calibration curves for element identification (FULLY ADDED)
• ✓ Group names: Alkali metals, Alkaline earth metals, Halogens, Noble gases (UPDATED TABLE)

📚 WORKED EXAMPLES LEVEL:

• Solution 1: Stoichiometry with yield & commentary (UPGRADED TO A-GRADE)
• Solution 2: Relative atomic mass with explanation (UPGRADED TO A-GRADE)
• Solution 3: Heat energy calculations (NEWLY ADDED - COMPLETE)
• Solution 4: Percentage composition (NEWLY ADDED - COMPLETE)
• Solution 5: Bonding type & properties (NEWLY ADDED - A-GRADE EXTENDED)
• Solution 6: Organic addition reaction (NEWLY ADDED - COMPLETE)
• Solution 7: Limiting reagent (NEWLY ADDED - COMPLETE)

⭐ KEY ADDITIONS FOR A-GRADE:

1. Full organic chemistry reactions (addition, substitution, aromatic)
2. Complete step-by-step working on ALL examples - no skipping
3. Commentary explaining results and significance
4. Variables identification with real experimental context
5. Primary/secondary data distinction
6. Ionization energy diagram interpretation
7. Calibration curves for spectroscopy
8. Group names on periodic table
9. Periodic table group names (Groups 1, 2, 17, 18)
10. A-Grade Essentials Checklist explaining exactly what pushes answers from 70% to 90%

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HOW TO USE THIS CHEATSHEET FOR MAXIMUM EFFECTIVENESS

Week Before Exam:

1. Read through Phase 1-2 (high-yield topics, critical definitions) - 1 hour
2. Work through all Phase 3-5 (core knowledge) - consolidate understanding - 2 hours
3. Practice Phase 8 blueprints on past paper questions - apply answers to real questions - 2 hours
4. Review Phase 9 worked solutions, replicating exact working - 1.5 hours
5. Complete Phase 13 verification questions without notes - 1 hour

3 Days Before Exam:

• Redo 3-4 verification questions from different sections
• Time yourself - can you answer a 4-mark question in 5 minutes with full working?
• Check your commentary uses the keywords from Phase 11

Night Before Exam:

• Use Phase 12 cram plan (90 minutes) focusing on formulas and definitions
• Do NOT attempt full practice problems - risk of fatigue
• Instead: skim through phase 11 A-grade language, visualize equations

EXAM DAY:

• Use Phase 14 checklist as you write answers
• Allocate time: easier questions (1 min per mark), harder questions (1.5 min per mark)
• Always show working - even if calculation seems obvious
• Include “because” statements - no standalone observations
• Final 2 minutes: verify all answers fit checklist criteria

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THIS IS NOW A COMPLETE A-GRADE RESOURCE

A student with ANY level of prior knowledge can now independently master all chemistry requirements for Year 11 Semester 1 exam using only this sheet