Some Basic Concepts of Chemistry

1. Matter & Its States

• Matter : Anything which has mass and occupies space.
• Non-Matter : Things that do not occupy space, do not have mass and are not made of particles.
• Three common states of matter : Solid, Liquid, and Gas.
• Two other states of matter :
  • Plasma : Exists at extremely high temperature. It is a hot ionized gas (e.g., lightning, solar wind, Aurora).
  • Bose-Einstein Condensate (B.E.C) : Exists at very low temperature, almost 0 K or −273°C (typically a few billionths of a degree above absolute zero). At such low temperature, atoms lose almost all their thermal energy and start behaving in a unique manner.
• Phase Changes :
  • Solid → Gas : Sublimation  |  Gas → Solid : Deposition
  • Solid → Liquid : Melting  |  Liquid → Solid : Freezing
  • Liquid → Gas : Vaporisation  |  Gas → Liquid : Condensation
  • Gas → Plasma : Ionization  |  Plasma → Gas : Deionization

2. Classification of Matter

• Matter is broadly classified into Pure Substances and Mixtures.

Pure Substance :

• A material containing only one type of substance. It cannot be separated into simpler substances by physical methods.
• Element : A pure substance containing one type of atom. (e.g., Fe, Na, Cs, Mg, O₂, S₈, P₄, O₃, F₂, Cl₂, N₂)
• Compound : A pure substance containing more than one kind of element which combine together in a fixed mass ratio. They can be separated into simpler substances by chemical methods. The property of the compound is entirely different from those of their constituent elements. (e.g., HCl, H₂O, NH₃, CaCO₃)

Mixture :

• A material which consists of more than one type of substance mixed in any ratio by weight. The property of the mixture is the property of its components. This can be separated into its substances by physical methods.
• Homogeneous : A mixture having the same composition throughout. (e.g., Salt + Water, Sugar + Water, Alcohol + Water, Urea + Water, Alloys)
• Heterogeneous : The mixture in which components are present non-uniformly. (e.g., Water + Sand, Water + Oil, Water + Fe-powder, Blood, Milk)

3. Atoms, Molecules & Atomicity

• Atom : Smallest particle of an element which may or may not have independent existence.
• Molecule : Smallest particle of an element or compound which always has independent existence. In a molecule, two or more atoms are connected by a chemical bond.
• Atomicity : Number of atoms present in a molecule or ion. (e.g. H₂O = 3, KMnO₄ = 6, H₂S₂O₈ = 12).

4. Atomic Representation & Subatomic Particles

• X = symbol of an element.
• A = mass number = total number of protons (p) and neutrons (n).
• Z = atomic number = number of protons (or electrons in a neutral atom).
• Nucleons : Particles that are part of a nucleus (i.e. protons and neutrons).

5. Ions & Salt Formulas

• Cation : M → Mⁿ⁺ + ne⁻ (loss of electrons).
• Anion : A + ne⁻ → Aⁿ⁻ (gain of electrons).

Common Cations :

Common Anions :

• Formula of a Salt : Obtained by criss-crossing the valencies of the cation and anion.
  • Calcium phosphate : Ca⁺² and PO₄⁻³ ⇒ Ca₃(PO₄)₂
  • Aluminium phosphide : Al⁺³ and P⁻³ ⇒ Al₃P₃ ⇒ AlP
  • Ferrous oxalate : Fe⁺² and C₂O₄⁻² ⇒ FeC₂O₄

6. Atomic Species

• Isotopes : Atoms of the same element having the same number of protons but different mass numbers (different number of neutrons).
  e.g. ¹²₆C and ¹⁴₆C (n = 6 and 8) ; ³⁵₁₇Cl and ³⁷₁₇Cl (n = 18 and 20).
• Isobars : Atoms of different elements having different number of protons and neutrons but the same number of nucleons (mass number).
  e.g. ⁴⁰₁₉K, ⁴⁰₂₀Ca, and ⁴⁰₁₈Ar (p+n = 40).
• Isotones : Atoms having different number of protons and mass numbers but the same number of neutrons.
  e.g. ³⁷₁₇Cl (n = 20) and ⁴⁰₂₀Ca (n = 20).
• Isodiaphers : Atoms having the same difference between protons and neutrons (i.e. |p − n| = same).
  e.g. ²³⁸₉₂U (p=92, n=146 ⇒ |p−n|=54) and ²³⁴₉₀Th (p=90, n=144 ⇒ |p−n|=54).

7. Relative Atomic Mass (RAM)

• The mass of any atom is because of protons (p) and neutrons (n).

Total nucleons (Mass number) = p + n

• Examples of RAM :
  • H — 1
  • C — 12 (6p + 6n)
  • N — 14 (7p + 7n)
  • O — 16 (8p + 8n)
  • Na — 23 (11p + 12n)

8. Atomic Mass Unit (amu, u)

• m_p ≈ m_n
• 1 u or 1 amu = 1.66 × 10⁻²⁴ g

9. Atomic Mass and Molecular Mass

• Atomic or Molecular mass = mass of total nucleons present
• Mass of total nucleons present = Total nucleons × 1 u

10. Calculation of Entities (Moles, Atoms, Nucleons)

Note: 11.66 × 10⁻²⁴ = 6.022 × 10²³ = N_A

• No. of atoms in a molecule = No. of molecules × atomicity
• No. of protons/neutrons/electrons/nucleons = Total entities × (p/n/e/nucleons present in one entity)

11. Mole & Molar Mass

• Definition : A mole is the amount of the substance that contains as many entities (atom, molecule, ion or other particle) as there are atoms present in exactly 0.012 Kg or 12 g of C-12 sample.
• 1 u = 1.66 × 10⁻²⁴ g.
• Number of C atoms in 12 g of C-12 sample = 12 g12 u = 12 g12 × 1.66 × 10⁻²⁴ g = 6.022 × 10²³ = N_A.
• N_A entities means one mole.
  • Moles × N_A = Entities
  • Entities ÷ N_A = Moles
• Gram Atomic or Gram Molecular Weight (Molar mass) : Weight of one mole of atoms or molecules in grams.
• e.g. Molar mass of Na = 23 × 1.66 × 10⁻²⁴ × 6.022 × 10²³ g = 23 g
  

  (since 1.66 × 10⁻²⁴ × 6.022 × 10²³ ≈ 1).

• Note : In modern practice, gm-atom, gm-molecule, and gm-ion all mean Mole.
  • 2 gm-atom of sodium = 2 moles of sodium atoms.
  • 2 gm-molecule of water = 2 moles of water molecules.
  • 2 gm-ion of oxalate = 2 moles of oxalate ions.

12. Mass & Entities Conversion

• If the weight of the sample is given in 'u', to calculate the number of entities we divide that weight by the weight of one entity (in u).
• If the weight of the sample is given in gm, to calculate the number of entities we divide that weight by the molar mass (in gm) to get moles, then multiply it by N_A.

13. Avogadro's Hypothesis & Molar Volume

• Avogadro's Hypothesis (PV = nRT) : Equal volume of all gases have equal number of molecules (not atoms) at same temperature and pressure conditions.
• Volume of one mole gas depends on conditions :

  Temperature	Pressure	Volume of 1 mole gas
  NTP (0°C or 273 K)	1 atm	22.4 L
  STP (0°C or 273 K)	1 bar	22.7 L

• To convert between Moles and Volume of gas :
  • Volume at NTP = Moles × 22.4 L
  • Volume at STP = Moles × 22.7 L

14. Mole Conversions & Atomicity

• For a generic molecule A₄B₃ (atomicity = 7) in an a mole sample :
  • Total moles of atoms = a × atomicity of molecule = 7a moles
  • Moles of A atoms = a × atomicity of A = 4a moles
  • Moles of B atoms = a × atomicity of B = 3a moles

Conversion Flowchart :

• Forward : Sample weight → (÷ M. wt.) → Sample mole → (× atomicity of atom) → Atom mole → (× Molar mass) → Atom weight
• Reverse : Atom weight → (÷ Molar mass) → Atom mole → (÷ atomicity of atom) → Sample mole → (× M. wt.) → Sample weight

15. Percentage Composition

16. Percentage Composition & Minimum Molecular Weight

• To find the minimum molecular weight of a compound given the mass percentage of an element, assume the molecule contains at least 1 atom of that element.

• All valid molecular weights for the compound will be integral multiples of this minimum molecular weight.

17. Sample Mass Fraction & Percentage Purity

• For an unknown atomic weight (M), the mass fraction of an element in a compound can be used to solve for M.

e.g. In X₄O₆, molar mass = 4M + 96. Mass of X in sample = Weight of sample × 4M4M + 96.

• Weight loss on heating hydrated salts (e.g., Glauber's salt, Na₂SO₄·10H₂O) is due to the loss of water of crystallization.

% H₂O = Weight of H₂O in formulaMolecular weight × 100 = 180322 × 100 = 55.9%

18. Empirical & Molecular Formula

• Molecular formula (M.F): The formula which gives the exact number of atoms of elements present in one molecule of the substance.
• Empirical formula (E.F): The formula which gives the simplest ratio of atoms of different elements present in one molecule of the substance.
• e.g. Benzene : Molecular formula = C₆H₆. Simplest ratio of C : H = 1 : 1. Empirical formula = CH.
• e.g. Glucose : Molecular formula = C₆H₁₂O₆. Simplest ratio of C : H : O = 1 : 2 : 1. Empirical formula = CH₂O.

• For Glucose : n = 180 / 30 = 6 ⇒ M.F = (CH₂O)₆ = C₆H₁₂O₆.
• Steps to find E.F from % composition : Set up a table with Element, % composition, and Atomic Weight. Find the number of atoms (Moles = % / At. wt), then divide by the smallest value to get the simplest atomic ratio, and convert to a whole number ratio.

19. Average Atomic Mass & Molecular Weight

Average Atomic Mass (A.A.W)

• Used for elements with multiple isotopes.

Average Molecular Weight (A.M.W)

• The average molar mass of the different substances present in a container.

where n = moles, M = molar mass, and x = mole fraction.

20. Density of Gases & Vapour Density

1. Absolute Density

• Absolute density has units (e.g., g/L, g/mL).
• For ideal gases : PV = nRT ⇒ PV = wM RT

2. Relative Density & Specific Gravity

• Relative Density : Density of anything with respect to another (unitless number).
• Specific Gravity (sp. gr.) = density of the substancedensity of H₂O at 4°C
• Density of H₂O at 4°C ≈ 1 g/mL or 1 g/cc (where 1 c.c = 1 mL).
• e.g. If density of a solution = 1.1 g/mL, then its sp. gr = 1.1 g/mL1 g/mL = 1.1.

3. Density of Gas 1 wrt Gas 2

• Under similar conditions (same Pressure P and Temperature T) :

4. Vapour Density (V.D)

• Density of a substance in the vapour phase with respect to H₂ gas under similar conditions.

21. Vapour Density of a Mixture

Where M_Mix = Σ M_iX_i = M_AX_A + M_BX_B + ... (X = mole fraction)

22. Combustion of Hydrocarbons

• Propane (C₃H₈) : 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
  

  (1 mole of propane requires 5 moles of O₂)

• Butane (C₄H₁₀) : 1 C₄H₁₀ + 132 O₂ → 4 CO₂ + 5 H₂O
  

  (1 mole of butane requires 13/2 moles of O₂)

• For gases (same T, P) : V mL of hydrocarbon needs (x + y4)V mL of O₂ and gives xV mL of CO₂.

23. Stoichiometry & Important Reactions

Steps for Stoichiometry :

• Write down the chemical reaction.
• Balance the chemical reaction.
• Read the chemical reaction (relate moles).
• Now see the data given in the question.

Key Reactions :

• Metal Displacement: Zn + 2 HCl → ZnCl₂ + H₂
• Base Neutralisation (Metal Hydroxides):
  MOH + HCl → MCl + H₂O
  2 MOH + H₂SO₄ → M₂SO₄ + 2 H₂O

24. Stoichiometry & Limiting Reagent

• Limiting Reagent (LR) : The reactant that is completely consumed. It is identified by the minimum value of the ratio:

25. Percentage Yield

% Yield = (Jitna bana) / (Jitna banana chahiye) × 100

• Actual yield is either obtained experimentally or given in the question. Theoretical yield is calculated assuming 100% conversion via stoichiometry.
• For a sequential reaction sequence, the overall % yield is the product of the fractional yields of the individual steps (e.g., a 60% yield followed by a 50% yield gives an overall yield of 0.6 × 0.5 = 0.3 or 30%).

26. Principle of Atomic Conservation (POAC)

• The mass and number of atoms of an individual element are conserved during a chemical reaction.
• e.g. Decomposition of Potassium chlorate : 1 KClO₃ →^Δ KCl + 32 O₂
• Applying POAC on Oxygen atoms :
  Number of O-atoms in KClO₃ = Number of O-atoms in O₂
  Moles of O-atoms in KClO₃ = Moles of O-atoms in O₂

27. Concentration Terms & Molarity

• Concentration terms define the composition of solute and solvent in the solution.
• A solution is a homogeneous mixture of solute and solvent. The concentration of a solution coming from the same tank/source is always the same.
• Molarity (M) : Number of moles of solute dissolved in 1 L of solution.

• Moles of solute = Molarity × Volume of solution (L).
• Molarity is temperature dependent.
• Unit of molarity = mol/L or Molar (M).
• e.g. 0.2 M NaOH solution means 0.2 mol of NaOH is dissolved in 1 L of solution.

Resultant Molarity after Dilution :

• Key Concept : The moles of solute remain constant upon adding water.

Resultant Molarity during Mixing (same solute) :

• Key Concept : (Moles of solute)_Final = (Moles of solute)₁ + (Moles of solute)₂

28. Molality & Mole Fraction

• Molality (m) : Number of moles of solute dissolved in 1 kg of solvent.

• Moles of solute = Molality × Weight of solvent (kg).
• Molality is temperature independent.
• e.g. 0.4 m NaCl solution means 0.4 moles of NaCl are dissolved in 1 kg of solvent.

Mole Fraction (X) :

• Mole fraction is temperature independent.

29. Percentage Concentration Terms

• % w/w (Percent by mass) : Temperature independent.

wt of solution = wt of solute + wt of solvent

• e.g. 10% w/w aq urea solution means 10 g urea is dissolved in 100 g solution.
• % w/V (Mass by volume percent) : Temperature dependent.

• e.g. 11% w/V aq urea solution means 11 g urea is dissolved in 100 mL solution.

• % V/V (Volume by volume percent) : Used when both solute and solvent are liquids. Temperature dependent.

30. Parts Concentration Terms & Strength

• ppm (Parts per million) :

W_solvent ≈ W_solution ∵ W_solute is negligible wrt solvent.

• ppb (Parts per billion) :

• Strength of solution : Temperature dependent.

• e.g. 1.2 g/L strength glucose (aq) solⁿ means 1.2 g glucose is present in 1 L solution.

31. Concentration Conversions & Relations

• Molarity to Molality (using density) : Find the mass of the solution from its volume and density, then subtract the mass of the solute to get the mass of the solvent.
• % (V/V) to Molality : Use volume percentages with respective densities to find the mass of solute and solvent.
• Mole fraction (X) relations : e.g. If X_solute = 0.1 = 110 = 11 + 9, it implies 1 mole of solute is present for every 9 moles of solvent.
• If solute molar mass is p and solvent molar mass is q, weight of solution = p + 9q. This can be used with solution density to find the volume and equate Molarity with Molality.

32. Temperature Dependence of Molarity

• Molarity (M) depends on the volume of the solution : M = moles of solutevol of solⁿ (L).
• If an aqueous solution is heated from 1°C to 25°C (assuming solubility remains unchanged), the molarity increases to a peak at 4°C and then continuously decreases.
• This happens because the density of water is maximum at 4°C, meaning the volume of the solution is minimum at this temperature, resulting in the maximum molarity.

33. Standard Concentrations & Pure Solvents

• Molarity of pure water : Assuming density d = 1 g/mL, 1 L of water has a mass of 1000 g.
• Moles of water = 100018 = 55.55 mol.

34. Ion Concentration in Solutions

• For a strong electrolyte, the concentration of constituent ions depends on the stoichiometry of its dissociation.
• e.g. For an aqueous solution of MgF₂ with concentration C : MgF₂ → Mg²⁺ + 2F⁻. Therefore, [Mg²⁺] = C and [F⁻] = 2C.
• If [F⁻] = 0.4 M, then 2C = 0.4 M ⇒ C = 0.2 M (concentration of MgF₂).
• e.g. 100 mL of Ba₃(PO₄)₂ solution containing 13.7 g of Barium (Ba = 137). Moles of Ba²⁺ = 13.7137 = 0.1 mol. Since 1 molecule of Ba₃(PO₄)₂ gives 3 Ba²⁺ ions, moles of Ba₃(PO₄)₂ = 0.13 mol. Molarity = 0.1 / 3100 / 1000 = 0.33 M.

35. Mixing Solutions

• When mixing non-reacting solutions (e.g., strong acids together), total concentration of an ion is the sum of moles of that ion divided by total volume.

36. Laws of Chemical Combination

1. Law of Conservation of Mass :

• "In all physical and chemical changes the total mass of the reactants is equal to that of products."
• Limitation: Not applicable in the case of nuclear reactions, where mass is converted into energy (E = Δmc²).

2. Law of Constant or Definite Proportion :

• A chemical compound is always made up of the same elements combined together in the same fixed proportion by mass.
• Percentage of an element in a compound by mass remains the same even if the source of the compound is different. (e.g. in H₂O, % H = 218 × 100%, % O = 1618 × 100%).
• Limitations :
  • Not applicable if the element exists in different isotopes which form the compound. (e.g. % C in ¹²CO₂ is different from % C in ¹⁴CO₂).
  • The reverse of this law is not true: elements may combine in the same mass ratio but the compounds may be different (Isomers). e.g. C₃H₆O can be CH₃−C(=O)−CH₃ or CH₃−CH₂−C(=O)−H ; C₂H₆O can be C₂H₅OH or CH₃OCH₃.
  • In non-stoichiometric compounds, the ratio of elements is not fixed and varies (e.g. Wustite Fe₀.₉₃O, which contains both Fe²⁺ and Fe³⁺).

3. Law of Multiple Proportion :

• When two elements combine with each other to form two or more chemical compounds, then different amounts of one element which react with a definite amount of the second element are in proportion of simple whole numbers.
• Does not hold well when an element exists in the form of different isotopes.
• e.g. Compounds of Carbon and Oxygen: CO (12g C : 16g O) and CO₂ (12g C : 32g O). For 12g C, the ratio of O is 16 : 32 = 1 : 2.
• e.g. Compounds of Nitrogen and Oxygen: N₂O (28g N : 16g O), NO (14g : 16g ⇒ 28g : 32g), NO₂ (14g : 32g ⇒ 28g : 64g), N₂O₃ (28g : 48g). The ratio of O for a fixed 28g of N is 16 : 32 : 64 : 48 = 1 : 2 : 4 : 3.

4. Law of Reciprocal Proportion :

• The ratio of masses of two elements A and B which combine separately with the fixed mass of a third element C is either the same or some simple multiple of the ratio of the masses in which A and B combine directly with each other.
• e.g. Elements C, H, O. Carbon forms CH₄ with H (12g C : 4g H) and CO₂ with O (12g C : 32g O). The mass ratio of H to O combining with a fixed mass of C (12g) is 4 : 32 = 1 : 8.
  When H and O combine directly to form H₂O, the mass ratio is 2g H : 16g O = 1 : 8, which matches exactly.

37. Gay Lussac's & Avogadro's Laws

• Gay Lussac's Law of combining volume : When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume provided that all the gases are at the same temperature and pressure.
  e.g. 1 H₂ + 1 Cl₂ → 2 HCl (1 vol : 1 vol : 2 vol). (Note: reaction must be balanced).
• Avogadro's Law : At constant temperature and pressure, volume is directly proportional to the moles of gas (V ∝ n, from PV = nRT).

38. Eudiometry (Gas Analysis)

• Gaseous reactions are often studied in a closed graduated tube (Eudiometer tube) provided with platinum terminals for the passage of electricity.
• Purpose : Useful for determining volume-volume relationships amongst gases, composition of gaseous mixtures, and molecular formulae of gases/hydrocarbons.
• Reagents used for absorbing different gases :

• Key rules for problems :
  • It is assumed that sparking occurs at room temperature.
  • Water formed would be in the liquid state.
  • Nitrogen gas (N₂) is inert towards oxidation.
  • If a gas is added externally, it will be assumed to be in excess unless otherwise stated.
  • "Without air spark" means decomposition of gas ; "With air spark" means combustion.

39. Volume Strength of H₂O₂ Solution

• Definition : The volume of O₂ formed at STP on decomposition of 1 ml of H₂O₂ solution is called the Volume Strength (V.S.) of H₂O₂.
• Decomposition reaction: 1 H₂O₂ → H₂O + ½ O₂
• Relationship with Molarity (M): moles of H₂O₂ in 1 ml of its solution = M × 11000.

40. Percentage Labelling of Oleum

• Oleum is a mixture of H₂SO₄ and free SO₃.
• Reaction on adding water: SO₃ + H₂O → H₂SO₄.
• A labelling of "109% Oleum" means that when 100 g of the oleum sample is treated with water, a total of 109 g of H₂SO₄ is formed.
  (This implies exactly 9 g of H₂O was added to react completely with the free SO₃).

• Minimum labelling is just above 100% (almost pure H₂SO₄). Maximum labelling is 122.5%, when the 100 g sample is pure SO₃ (100 g SO₃ → 10080 × 98 = 122.5 g H₂SO₄). So 100% < % labelling ≤ 122.5%.

41. Quick Formula Revision