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State Board · Class 11 · Chemistry

Lesson 1: Some Basic Concepts of Chemistry

7 concepts · 56 practice questions with answers · Free to read

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Classification of matter: pure substances and mixtures

Link to Std 9–10 : In Std 9–10 you learned that matter occupies space and has mass; in Std 11 we sort matter by purity and composition.

From daily life

Milk, tea, water, salt are all matter; but tea contains sugar, milk and water mixed together while salt is a single substance.

Simple definition

A pure substance has the same composition throughout and fixed properties. An element is made of one kind of atom. A compound has two or more elements chemically combined in a fixed ratio. A mixture has two or more substances mixed in any proportion.

Classification of matterMatterPure substanceMixtureElementCompoundHomogeneousHeterogeneousCu, O₂, FeH₂O, NaClair, brinesand + water

In the figure : Matter → pure substance (element, compound) and mixture (homogeneous, heterogeneous).

Rule 1: Element or compound? — An element cannot be broken into simpler substances by chemical means; a compound can, and its properties differ from its elements.

  • Easy : Cu, O₂, Fe are elements; H₂O, NaCl are compounds.
  • Medium : Sodium (a metal) and chlorine (a poisonous gas) give NaCl (table salt), with entirely different properties.
  • Exam-type : Electrolysis of water gives H₂ and O₂, so water is a compound.

Rule 2: Types of mixtures — Homogeneous: uniform throughout, components not visible separately. Heterogeneous: components visible separately.

  • Easy : Sugar solution is homogeneous; sand + water is heterogeneous.
  • Medium : Air is a homogeneous mixture; brass (Cu + Zn) is a homogeneous alloy.
  • Exam-type : Milk looks uniform but fat droplets show under a microscope, so it is a colloid (heterogeneous).

Rule 3: Composition of a mixture (mass percent) — Mass % = (mass of the component ÷ total mass) × 100.

  • Easy : 20 g salt + 80 g water: salt = 20 ÷ 100 × 100 = 20 %.
  • Medium : In water, H = 2 ÷ 18 × 100 = 11.11 % and O = 16 ÷ 18 × 100 = 88.89 %.
  • Exam-type : In a compound the % is fixed (always 11.11 % H in water); in a mixture it varies.

Derivation: step method of classifying

  • Step 1: Is more than one substance present? Yes → mixture; no → pure substance.
  • Step 2: If pure, can it be broken into simpler substances chemically? Yes → compound; no → element.
  • Step 3: If a mixture, are the parts visibly separate? Yes → heterogeneous; no → homogeneous.

CET level (MCQ)

  • CET 1) Which is a compound? 1) O₃ 2) P₄ 3) H₂O₂ 4) S₈ — Answer: 3). Only H₂O₂ has more than one kind of atom.
  • CET 2) Which is not a homogeneous mixture? 1) air 2) brass 3) salt water 4) sand and water — Answer: 4).

Real-life use: Straining tea (filtration), salt from sea water (evaporation) and winnowing grain are all uses of separating mixtures.

Remember

Element = one kind of atom; compound = fixed ratio; mixture = any ratio.

Common mistake

O₃ (ozone) and P₄ are not compounds; they are molecules of elements. Air is a mixture, not a compound.

CET trap

Even with many atoms in the formula, if they are all one kind (O₂, S₈) it is an element; count the different letters, not the atoms.

Shortcut

Memory clue: "one kind of letter = element; two or more = compound".

Spot the difference

PointCompoundMixture
RatioFixedAny
PropertiesDifferent from partsParts keep theirs
SeparationChemicalPhysical

Practice questions

Easy
  1. 1What is an element?
  2. 2Give one example of a homogeneous mixture.
  3. 3Is water an element or a compound?
Medium
  1. 415 g sugar + 85 g water: mass % of sugar?
  2. 5From O₂, H₂O, CO₂, N₂ choose the elements.
  3. 6State two differences between a compound and a mixture.
Challenging
  1. 7Find the mass % of C in CO₂.
  2. 8A 40 g mixture is 12 % salt; find the mass of salt and of water.
Show answers (try them yourself first)
  1. 1A pure substance made of one kind of atom.
  2. 2Air / sugar solution.
  3. 3A compound.
  4. 415 ÷ 100 × 100 = 15 %.
  5. 5O₂, N₂.
  6. 6In a compound the ratio is fixed and properties change; in a mixture the ratio varies and properties persist.
  7. 712 ÷ 44 × 100 = 27.27 %.
  8. 8Salt = 0.12 × 40 = 4.8 g; water = 35.2 g.

SI units, density and temperature scales

From daily life

Vegetables are weighed in kg, milk in litres and fever in °F; to avoid confusion the world uses one system, the SI.

Simple definition

The SI has 7 base units. Density = mass ÷ volume. The kelvin is the SI unit of temperature.

Temperature scales°C1000K373.15273.15°F21232K = °C + 273.15°F = (9 ÷ 5) × °C + 32

In the figure : °C, K and °F: freezing and boiling points of water.

Rule 1: Base units and prefixes — Length m, mass kg, time s, current A, temperature K, amount of substance mol, luminous intensity cd. Prefixes: kilo 10³, centi 10⁻², milli 10⁻³, micro 10⁻⁶, nano 10⁻⁹.

  • Easy : 1 km = 1000 m; 1 mg = 10⁻³ g.
  • Medium : 1 L = 1000 cm³ = 1 dm³; 1 mL = 1 cm³.
  • Exam-type : 2.5 nm = 2.5 × 10⁻⁹ m = 2.5 × 10⁻⁷ cm.

Rule 2: Density — ρ = m ÷ V (ρ = density, m = mass, V = volume). 1 g/cm³ = 1000 kg/m³.

  • Easy : 54 g aluminium, 20 cm³: ρ = 54 ÷ 20 = 2.7 g/cm³.
  • Medium : A 500 g body of 250 cm³: ρ = 2 g/cm³ = 2000 kg/m³.
  • Exam-type : 50 mL of a liquid with ρ = 0.8 g/cm³ has mass 0.8 × 50 = 40 g.

Rule 3: Temperature scales — K = °C + 273.15 and °F = (9 ÷ 5) × °C + 32.

  • Easy : 25 °C = 298.15 K = 77 °F.
  • Medium : 98.6 °F → °C = (98.6 − 32) × 5 ÷ 9 = 37 °C.
  • Exam-type : −40 °C = −40 °F (the two scales agree).

Derivation: °F = (9 ÷ 5) °C + 32

  • Step 1: Between freezing and boiling there are 100 divisions on °C and 212 − 32 = 180 on °F.
  • Step 2: So 1 °C = 180 ÷ 100 = 9 ÷ 5 °F.
  • Step 3: At 0 °C the °F reading is 32, so °F = (9 ÷ 5) × °C + 32.

Formulas, units and dimensional formulae

QuantityFormulaSI unit
Densityρ = m ÷ Vkg m⁻³
TemperatureK = °C + 273.15K
Massm = ρ Vkg
VolumeV = m ÷ ρm³ (1 L = 10⁻³ m³)

CET level (MCQ)

  • CET 1) 300 K equals how many °C? 1) 27 2) 573 3) 30 4) −27 — Answer: 1) 27. 300 − 273.15 = 26.85 ≈ 27.
  • CET 2) 1 g/cm³ equals? 1) 1 kg/m³ 2) 10 kg/m³ 3) 100 kg/m³ 4) 1000 kg/m³ — Answer: 4).

Real-life use: Medicine mass in mg, weather in K or °C and fuel density in kg/L — SI prefixes keep these simple.

Remember

K = °C + 273.15; °F = 9/5 °C + 32; ρ = m ÷ V; 1 L = 1 dm³ = 1000 cm³.

Common mistake

Do not write the degree sign with K (300 K, not 300 °K). Convert cm³ to m³ by dividing by 10⁶, not 100.

CET trap

Unit conversion is the biggest trap; wrong options are off by factors of 1000 (g ↔ kg, mL ↔ L).

Shortcut

Memory clue: "freezing 0 – 273 – 32; boiling 100 – 373 – 212".

Spot the difference

Point°CK°F
Freezing0273.1532
Boiling100373.15212
Divisions100100180

Practice questions

Easy
  1. 1What is the SI unit of temperature?
  2. 2How many mm in 1 km?
  3. 3Write the formula for density.
Medium
  1. 4Convert 37 °C to K.
  2. 5A 60 g body of 24 cm³: find the density.
  3. 6Convert 104 °F to °C.
Challenging
  1. 7Find the mass in kg of 250 cm³ of mercury of density 13.6 g/cm³.
  2. 8The difference between K and °C readings is always 273.15; at what temperature do °F and °C coincide?
Show answers (try them yourself first)
  1. 1Kelvin (K).
  2. 21 km = 10⁶ mm.
  3. 3ρ = m ÷ V.
  4. 4310.15 K.
  5. 560 ÷ 24 = 2.5 g/cm³.
  6. 6(104 − 32) × 5 ÷ 9 = 40 °C.
  7. 713.6 × 250 = 3400 g = 3.4 kg.
  8. 8Always; at −40 °C and °F coincide.

Laws of chemical combination

From daily life

When making dough the ratio of flour and water is fixed; change it and it fails. In reactions too, substances combine in fixed ratios.

Simple definition

Law of conservation of mass (Lavoisier): total mass does not change in a reaction. Law of constant proportions (Proust): a compound always has the same mass ratio of elements. Law of multiple proportions (Dalton): when two elements form several compounds, for a fixed mass of one the masses of the other are in a simple whole-number ratio.

Laws of chemical combinationConservation of mass: before = afterC + O₂12 g + 32 g = 44 gCO₂44 gConstant proportions: any water has H : O = 1 : 8 by mass9 g water = 1 g H + 8 g O36 g water = 4 g H + 32 g OMultiple proportions: 12 g C with O = 16 g (CO) and 32 g (CO₂) → 1 : 2

In the figure : Three laws: 12 + 32 = 44; water has H : O = 1 : 8; oxygen in CO and CO₂ is in the ratio 1 : 2.

Rule 1: Conservation of mass — Total mass of reactants = total mass of products.

  • Easy : 12 g C + 32 g O₂ → 44 g CO₂.
  • Medium : CaCO₃ → CaO + CO₂: 100 g limestone gives 56 g CaO and 44 g CO₂ (56 + 44 = 100).
  • Exam-type : If 50 g CaCO₃ gives 22 g CO₂ then CaO = 50 − 22 = 28 g.

Rule 2: Constant proportions — A pure compound, wherever obtained, has the same mass ratio of elements.

  • Easy : 9 g water = 1 g H + 8 g O.
  • Medium : 36 g water = 4 g H + 32 g O (ratio 1 : 8).
  • Exam-type : In CO₂ C : O = 12 : 32 = 3 : 8; 11 g CO₂ contains 3 g C.

Rule 3: Multiple proportions — Different compounds of the same two elements show simple whole-number ratios.

  • Easy : CO: 16 g O with 12 g C; CO₂: 32 g O with 12 g C → 16 : 32 = 1 : 2.
  • Medium : N₂O, NO, NO₂: with 28 g N the O is 16, 32, 64 g → 1 : 2 : 4.
  • Exam-type : H₂O (2 g H : 16 g O) and H₂O₂ (2 g H : 32 g O): O in the ratio 1 : 2.

Derivation: how much C in 11 g CO₂?

  • Step 1: Molar mass of CO₂ = 12 + 2 × 16 = 44 g.
  • Step 2: 44 g CO₂ has 12 g C, so the C fraction = 12 ÷ 44 = 3 ÷ 11.
  • Step 3: In 11 g, C = 11 × 3 ÷ 11 = 3 g.

CET level (MCQ)

  • CET 1) 50 g CaCO₃ decomposes giving 22 g CO₂; how much CaO? 1) 28 g 2) 22 g 3) 50 g 4) 72 g — Answer: 1). 50 − 22 = 28.
  • CET 2) Two oxides of carbon have 16 g and 32 g O with 12 g C; which law? 1) constant proportions 2) multiple proportions 3) conservation 4) Gay-Lussac — Answer: 2).

Real-life use: Pharmaceutical and fertiliser plants add reactants in fixed ratios, so these laws let them predict the product.

Remember

Conservation: same mass; constant: same ratio; multiple: simple whole-number ratio.

Common mistake

In multiple proportions fix the mass of one element first; do not compare unequal samples directly.

CET trap

CET options swap constant and multiple proportions: one compound → constant; several compounds → multiple.

Shortcut

Memory clue: "one compound = constant; many compounds = multiple; before-after mass = conservation".

Spot the difference

LawByMain idea
ConservationLavoisierTotal mass same
Constant proportionsProustFixed ratio in a compound
Multiple proportionsDaltonSimple whole-number ratio

Practice questions

Easy
  1. 1State the law of conservation of mass.
  2. 2What is the H : O mass ratio in water?
  3. 3Who proposed the law of multiple proportions?
Medium
  1. 4How much H and O are in 18 g of water?
  2. 5How much CO₂ forms when 24 g C burns completely?
  3. 6Two samples of CuO have the same Cu : O ratio — which law?
Challenging
  1. 7Find the ratio of O combining with 28 g N in N₂O₃ and N₂O₅.
  2. 8How much CaO and CO₂ form from 80 g CaCO₃?
Show answers (try them yourself first)
  1. 1Total mass is unchanged in a reaction.
  2. 21 : 8.
  3. 3Dalton.
  4. 42 g H and 16 g O.
  5. 524 g C = 2 mol → 88 g CO₂.
  6. 6Law of constant proportions.
  7. 7N₂O₃: 48 g O; N₂O₅: 80 g O → 48 : 80 = 3 : 5.
  8. 8CaO = 80 × 56 ÷ 100 = 44.8 g; CO₂ = 80 × 44 ÷ 100 = 35.2 g.

Gay-Lussac, Avogadro and Dalton's theory

From daily life

Identical balloons filled at the same temperature and pressure hold the same amount of gas, so volume tells the number of particles.

Simple definition

Gay-Lussac's law: gases combine in simple whole-number volume ratios at the same T and P. Avogadro's law: equal volumes of gases at the same T and P hold equal numbers of molecules. Dalton's atomic theory: matter is made of indivisible atoms; atoms of one element are alike; compounds form by atoms joining in fixed ratios.

Gay-Lussac's law2 H₂ + O₂ → 2 H₂O (g)H₂H₂2 vol+O₂1 volH₂OH₂O2 volVolume ratio 2 : 1 : 2 — simple whole numbers(same temperature and pressure)

In the figure : 2 volumes H₂ + 1 volume O₂ → 2 volumes steam.

Rule 1: Volume ratio (Gay-Lussac) — Coefficients of a balanced equation give the gas volume ratio.

  • Easy : N₂ + 3H₂ → 2NH₃: 10 mL N₂ + 30 mL H₂ → 20 mL NH₃.
  • Medium : CH₄ + 2O₂ → CO₂ + 2H₂O: 50 mL CH₄ needs 100 mL O₂ and gives 50 mL CO₂.
  • Exam-type : C₃H₈ + 5O₂ → 3CO₂ + 4H₂O: 10 mL propane needs 50 mL O₂ and gives 30 mL CO₂.

Rule 2: Avogadro's law — Volume ∝ number of molecules (T, P fixed), so volume ratio = molecule ratio = mole ratio.

  • Easy : 1 L of H₂ and 1 L of O₂ hold equal numbers of molecules.
  • Medium : H₂ + Cl₂ → 2HCl: 100 mL H₂ + 100 mL Cl₂ → 200 mL HCl.
  • Exam-type : He is monatomic, O₂ diatomic: equal volumes hold equal molecules but different numbers of atoms.

Rule 3: Dalton's atomic theory — Limitations: atoms are divisible (electrons, protons exist); isotopes differ in mass.

  • Easy : It explained conservation of mass (atoms are not destroyed).
  • Medium : Constant proportions: H₂O always has 2 H and 1 O.
  • Exam-type : Multiple proportions: CO has 1 O and CO₂ 2 O per C — 1 : 2.

Derivation: how Avogadro showed H₂ is diatomic

  • Step 1: Experiment: 1 volume H₂ + 1 volume Cl₂ → 2 volumes HCl.
  • Step 2: Avogadro: n molecules H₂ + n molecules Cl₂ → 2n molecules HCl.
  • Step 3: Each HCl has 1 H and 1 Cl; 2n HCl contain 2n H atoms, which came from n H₂ molecules.
  • Step 4: So each H₂ molecule has 2 H atoms — H₂ is diatomic.

CET level (MCQ)

  • CET 1) Cl₂ needed to react completely with 100 mL H₂? 1) 50 mL 2) 100 mL 3) 200 mL 4) 25 mL — Answer: 2) 100 mL. Ratio 1 : 1.
  • CET 2) CO₂ formed on burning 10 mL propane? 1) 10 mL 2) 20 mL 3) 30 mL 4) 40 mL — Answer: 3). C₃H₈ + 5O₂ → 3CO₂ + 4H₂O; ratio 1 : 3.

Real-life use: Fuel-air ratios in engines and gas cylinder capacities use volume ratios.

Remember

Volume ratio = coefficient ratio (for gases); same T and P.

Common mistake

The volume ratio does not apply to liquids and solids; only to gases.

CET trap

Balance the equation first; using an unbalanced one is the commonest CET trap. Steam counts as a gas, liquid water does not.

Shortcut

Memory clue: "coefficient = volume = mole" (for gases).

Spot the difference

LawAboutIdea
Gay-LussacVolume ratiosSimple whole numbers
AvogadroNumber of moleculesV ∝ n
DaltonAtomsIndivisible atoms

Practice questions

Easy
  1. 1State Gay-Lussac's law.
  2. 2State Avogadro's law.
  3. 3State one limitation of Dalton's theory.
Medium
  1. 4How much H₂ reacts with 20 mL N₂, and how much NH₃ forms?
  2. 5O₂ needed to burn 25 mL CH₄?
  3. 6How much steam forms from 40 mL H₂ in 2H₂ + O₂ → 2H₂O?
Challenging
  1. 7100 mL C₂H₄ burns: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O; find O₂ needed and CO₂ formed.
  2. 812 mL H₂ and 8 mL O₂ form water; which gas is left and how much?
Show answers (try them yourself first)
  1. 1Gases combine in simple whole-number volume ratios.
  2. 2Equal volumes at the same T and P hold equal molecules.
  3. 3Atoms are divisible.
  4. 4H₂ = 60 mL; NH₃ = 40 mL.
  5. 550 mL.
  6. 640 mL.
  7. 7O₂ = 300 mL; CO₂ = 200 mL.
  8. 812 mL H₂ needs 6 mL O₂; 2 mL O₂ is left.

Atomic mass, molecular mass and formula mass

From daily life

Weighing one grain of sand is hard, so we use kilograms. Atoms are far smaller, so we compare them with carbon-12.

Simple definition

The atomic mass unit (u) = 1 ÷ 12 of the mass of a carbon-12 atom = 1.66 × 10⁻²⁷ kg. Average atomic mass = abundance-weighted mean of isotope masses. Molecular mass = sum of atomic masses in a molecule; formula mass for ionic compounds.

Average atomic mass75 %25 %³⁵Cl³⁷Cl%(35 × 75 + 37 × 25) ÷ 100 = 35.5 uAverage atomic mass = sum of (mass × abundance)

In the figure : Chlorine: ³⁵Cl 75 %, ³⁷Cl 25 % ⇒ average 35.5 u.

Rule 1: Average atomic mass — Average = (m₁ × %₁ + m₂ × %₂ + …) ÷ 100.

  • Easy : Cl: (35 × 75 + 37 × 25) ÷ 100 = 35.5 u.
  • Medium : C: 98.9 % ¹²C and 1.1 % ¹³C ⇒ 12.011 u.
  • Exam-type : B: 20 % ¹⁰B, 80 % ¹¹B ⇒ (10 × 20 + 11 × 80) ÷ 100 = 10.8 u.

Rule 2: Molecular mass — Add atomic masses (H = 1, C = 12, N = 14, O = 16, S = 32).

  • Easy : H₂O = 2 + 16 = 18 u; CO₂ = 12 + 32 = 44 u.
  • Medium : H₂SO₄ = 2 + 32 + 64 = 98 u; glucose C₆H₁₂O₆ = 72 + 12 + 96 = 180 u.
  • Exam-type : Include water of crystallisation: CuSO₄·5H₂O = 63.5 + 32 + 64 + 90 = 249.5 u.

Rule 3: Formula mass — Ionic substances like NaCl have no separate molecules, so we use 'formula mass'.

  • Easy : NaCl = 23 + 35.5 = 58.5 u.
  • Medium : CaCO₃ = 40 + 12 + 48 = 100 u.
  • Exam-type : Al₂(SO₄)₃ = 54 + 3 × 96 = 342 u.

Derivation: 1 u in kg

  • Step 1: 12 g of carbon-12 has 6.022 × 10²³ atoms.
  • Step 2: Mass of one atom = 12 ÷ 6.022 × 10²³ = 1.993 × 10⁻²³ g.
  • Step 3: 1 u = 1.993 × 10⁻²³ ÷ 12 = 1.66 × 10⁻²⁴ g = 1.66 × 10⁻²⁷ kg.

CET level (MCQ)

  • CET 1) Boron: ¹⁰B 20 %, ¹¹B 80 % ⇒ average? 1) 10.2 2) 10.8 3) 10.5 4) 11 — Answer: 2) 10.8. Closer to the more abundant isotope.
  • CET 2) Mass of CuSO₄·5H₂O? 1) 159.5 2) 249.5 3) 177.5 4) 223.5 — Answer: 2) 249.5.

Real-life use: Drug dosages, fertiliser purity and water analysis all need molecular masses.

Remember

1 u = 1/12 × m(¹²C); the average lies near the more abundant isotope.

Common mistake

Forgetting to divide by 100; forgetting that a number outside brackets multiplies every atom inside.

CET trap

Forgetting water of crystallisation; use Cl = 35.5 and Cu = 63.5, not rounded values.

Shortcut

Memory clue: "more abundant = average sits near it".

Spot the difference

PointAtomic massMolecular massFormula mass
ForAtomMoleculeIonic compound
ExampleCl = 35.5H₂O = 18NaCl = 58.5

Practice questions

Easy
  1. 1What is 1 u?
  2. 2Molecular mass of H₂SO₄?
  3. 3Why 'formula mass' for NaCl?
Medium
  1. 4Mass of Mg(OH)₂ (Mg = 24).
  2. 5Average mass of Cl (³⁵Cl 75 %, ³⁷Cl 25 %).
  3. 6Find the mass of glucose.
Challenging
  1. 7Formula mass of Al₂(SO₄)₃ (Al = 27).
  2. 8Mass of one ¹²C atom in g (Nₐ = 6.022 × 10²³).
Show answers (try them yourself first)
  1. 11/12 of the mass of a ¹²C atom.
  2. 298 u.
  3. 3Ionic solids have no discrete molecules.
  4. 424 + 2 × 17 = 58 u.
  5. 535.5 u.
  6. 6180 u.
  7. 7342 u.
  8. 81.993 × 10⁻²³ g.

Mole concept, Avogadro constant and molar mass

From daily life

A dozen is 12 bananas and a ream is 500 sheets; a mole is 6.022 × 10²³ particles. Atoms are so tiny that we need this huge 'dozen'.

Simple definition

1 mole is the amount containing as many particles (6.022 × 10²³) as atoms in 12 g of carbon-12. This is the Avogadro constant (Nₐ). Molar mass (M) is the mass of 1 mole in g mol⁻¹, numerically equal to the atomic/molecular mass.

Mole concept mapMolenMass m (g)Particles NVolume V at STP÷ M× Nₐ× Vₘn = m ÷ MN = n × Nₐ

In the figure : The mole is the centre: ÷ M gives mass, × Nₐ gives particles, × Vₘ gives volume at STP.

Live figure

Mole concept, Avogadro constant and molar mass

In the app you can move this figure with your finger.

Open in the app

Rule 1: Moles and mass — n = m ÷ M (n = moles, m = mass in g, M = molar mass in g/mol).

  • Easy : 18 g water ⇒ n = 18 ÷ 18 = 1 mol.
  • Medium : 90 g water ⇒ 5 mol; 4.4 g CO₂ ⇒ 0.1 mol.
  • Exam-type : 9.8 g H₂SO₄ ⇒ n = 9.8 ÷ 98 = 0.1 mol.

Rule 2: Moles and number of particles — N = n × Nₐ (N = number of particles).

  • Easy : 1 mol water = 6.022 × 10²³ molecules.
  • Medium : 0.5 mol NH₃: molecules = 3.011 × 10²³; atoms (4 each) = 1.204 × 10²⁴.
  • Exam-type : 1.7 g NH₃ = 0.1 mol ⇒ atoms = 0.4 × 6.022 × 10²³ = 2.41 × 10²³.

Rule 3: Mass of one particle — Mass of one particle = M ÷ Nₐ.

  • Easy : One water molecule = 18 ÷ 6.022 × 10²³ = 2.99 × 10⁻²³ g.
  • Medium : One carbon atom = 1.99 × 10⁻²³ g.
  • Exam-type : Molecules in 22 g CO₂ = 0.5 × 6.022 × 10²³ = 3.011 × 10²³.

Derivation: the g ↔ particles chain

  • Step 1: M grams contain Nₐ particles (definition of mole).
  • Step 2: m grams contain m ÷ M moles.
  • Step 3: Particles = (m ÷ M) × Nₐ, i.e. N = (m ÷ M) × Nₐ.

Formulas, units and dimensional formulae

QuantityFormulaUnit
Amountn = m ÷ Mmol
Molar massM = m ÷ ng mol⁻¹
ParticlesN = n × Nₐnumber
Avogadro constantNₐ = 6.022 × 10²³mol⁻¹

CET level (MCQ)

  • CET 1) Moles in 90 g water? 1) 1 2) 5 3) 18 4) 0.5 — Answer: 2) 5. 90 ÷ 18.
  • CET 2) Molecules in 22 g CO₂? 1) 6.022 × 10²³ 2) 3.011 × 10²³ 3) 1.2 × 10²⁴ 4) 12.04 × 10²² — Answer: 2). 22 ÷ 44 = 0.5 mol.

Real-life use: Counting ingredients in a tablet, fertiliser dosing and pollutants in water all use the mole as a common measure.

Remember

n = m ÷ M; N = n × Nₐ; Nₐ = 6.022 × 10²³ mol⁻¹; M is numerically the molecular mass.

Common mistake

Giving molecules when atoms are asked; atoms in NH₃ = molecules × 4.

CET trap

Options swap atoms and molecules; read whether the question gives grams or moles.

Shortcut

Memory clue: "grams ÷ M = moles; moles × Nₐ = particles".

Spot the difference

PointMoleMolar mass
MeaningAmount of particlesMass of 1 mole
Unitmolg mol⁻¹

Practice questions

Easy
  1. 1How many particles in 1 mole?
  2. 2Give the meaning of each letter in n = m ÷ M.
  3. 3Molar mass of NH₃?
Medium
  1. 4How many molecules are in 4.4 g CO₂?
  2. 5Mass of 0.25 mol H₂O?
  3. 6Mass of one H₂O molecule in g.
Challenging
  1. 7Total atoms in 1.7 g NH₃?
  2. 8How many carbon atoms in 0.1 mol C₆H₁₂O₆?
Show answers (try them yourself first)
  1. 16.022 × 10²³.
  2. 2n = moles, m = mass, M = molar mass.
  3. 317 g/mol.
  4. 40.1 mol × 6.022 × 10²³ = 6.022 × 10²².
  5. 50.25 × 18 = 4.5 g.
  6. 62.99 × 10⁻²³ g.
  7. 70.1 × 4 × 6.022 × 10²³ = 2.41 × 10²³.
  8. 80.6 mol C = 0.6 × 6.022 × 10²³ = 3.61 × 10²³.

Molar volume and mole calculations

From daily life

Pressure cookers, gas cylinders and balloons: the volume of a gas depends on the number of particles, so moles of a gas follow from volume.

Simple definition

Molar volume (Vₘ) is the volume of 1 mole of a gas: 22.4 L at STP (273 K, 1 atm); 22.7 L at 1 bar. n = V ÷ Vₘ. Percent composition, empirical formula and limiting reagent problems all run on moles.

Molar volumeH₂2 gO₂32 gCO₂44 g1 mole of any gas: same volume at STP (273 K)22.4 L (1 atm)  ≈  22.7 L (1 bar)each holds 6.022 × 10²³ molecules

In the figure : 1 mole of any gas — H₂, O₂, CO₂ — occupies the same volume at STP.

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Molar volume and mole calculations

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Rule 1: Volume of a gas — V (STP) = n × 22.4 L. If the question says 1 bar use 22.7 L.

  • Easy : 5.6 L of gas = 5.6 ÷ 22.4 = 0.25 mol.
  • Medium : 11.2 L O₂ = 0.5 mol = 16 g.
  • Exam-type : 2.4 g Mg + HCl → H₂: 0.1 mol Mg ⇒ 0.1 mol H₂ = 2.24 L.

Rule 2: Reaction problems and the limiting reagent — Coefficients of a balanced equation are mole ratios. The reagent that runs out first is limiting and fixes the yield.

  • Easy : CaCO₃ → CaO + CO₂: 25 g CaCO₃ = 0.25 mol ⇒ CO₂ = 0.25 mol = 5.6 L.
  • Medium : 2H₂ + O₂ → 2H₂O: 4 g H₂ (2 mol) and 16 g O₂ (0.5 mol) ⇒ O₂ is limiting (0.5 mol O₂ needs only 1 mol H₂).
  • Exam-type : Water formed = 1 mol = 18 g.

Rule 3: Empirical formula — Percent → grams → moles → divide by the smallest → whole numbers.

  • Easy : C 40 %, H 6.67 %, O 53.33 %: moles 3.33 : 6.67 : 3.33 ⇒ 1 : 2 : 1 ⇒ CH₂O.
  • Medium : Molar mass 180 ⇒ n = 180 ÷ 30 = 6 ⇒ molecular formula C₆H₁₂O₆.
  • Exam-type : Molecular formula = empirical formula × n.

Derivation: where 22.4 L comes from (ideal gas)

  • Step 1: Ideal gas equation PV = nRT (see the gaseous-state lesson).
  • Step 2: n = 1, R = 0.0821 L atm K⁻¹ mol⁻¹, T = 273 K, P = 1 atm.
  • Step 3: V = nRT ÷ P = 0.0821 × 273 ÷ 1 = 22.4 L.

Formulas, units and dimensional formulae

QuantityFormulaUnit
Moles of gasn = V ÷ Vₘmol
Volume at STPV = n × 22.4L
Molar volumeVₘ = 22.4 (1 atm) / 22.7 (1 bar)L mol⁻¹
Molecular formula(empirical) × n—

CET level (MCQ)

  • CET 1) 11.2 L of a gas at STP weighs 14 g; molar mass? 1) 14 2) 28 3) 56 4) 7 — Answer: 2) 28. n = 0.5; 14 ÷ 0.5.
  • CET 2) Volume at STP of 3.011 × 10²³ molecules? 1) 22.4 L 2) 11.2 L 3) 5.6 L 4) 44.8 L — Answer: 2) 11.2 L. n = 0.5.

Real-life use: Gas in balloons, welding oxygen and gas cylinder sizing all use the molar volume idea.

Remember

1 mol gas = 22.4 L (STP, 1 atm) = 22.7 L (1 bar); the limiting reagent fixes the yield.

Common mistake

Applying 22.4 L to liquids or solids; confusing STP with room temperature.

CET trap

Options include both 22.4 and 22.7 — pick by '1 atm' or '1 bar'. Do not skip finding the limiting reagent.

Shortcut

Memory clue: "22.4 for one atmosphere; 22.7 for one bar".

Spot the difference

PointEmpirical formulaMolecular formula
MeaningSimplest ratioTrue atoms in molecule
GlucoseCH₂OC₆H₁₂O₆

Practice questions

Easy
  1. 1What is STP?
  2. 2Volume of 1 mole of any gas at STP?
  3. 3What is a limiting reagent?
Medium
  1. 4Mass of 44.8 L CO₂ at STP.
  2. 5Volume of 8 g O₂ at STP.
  3. 6C 40 %, H 6.67 %, O 53.33 % — empirical formula.
Challenging
  1. 7How much water from 4 g H₂ and 16 g O₂ (2H₂ + O₂ → 2H₂O)?
  2. 8H₂ from 2.4 g Mg with HCl at STP?
Show answers (try them yourself first)
  1. 1273 K and 1 atm.
  2. 222.4 L.
  3. 3The reactant that is used up first.
  4. 42 mol × 44 = 88 g.
  5. 50.25 mol × 22.4 = 5.6 L.
  6. 6CH₂O.
  7. 7O₂ is limiting ⇒ 18 g.
  8. 80.1 mol ⇒ 2.24 L.

Quick revision of the whole lesson

ConceptKey formula / fact
MatterPure (element, compound) / mixture
TemperatureK = °C + 273.15; °F = 9/5 °C + 32
Densityρ = m ÷ V
LawsConservation, constant, multiple proportions
GasesVolume ratio = coefficient ratio
Average massΣ(m × %) ÷ 100
Molen = m ÷ M; N = n × Nₐ
Moles of gasn = V ÷ 22.4 L (STP)
Empirical formula% → g → mol → divide by smallest
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🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Identify element, compound, mixture.
  • Unit conversion and K, °C, °F.
  • Recognise the three laws; volumes from coefficients.
  • Average atomic mass and molecular mass.
  • Mole ↔ gram ↔ particles ↔ volume conversions.
  • Limiting reagent and empirical formula.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.