Some Basic Concepts of Chemistry — NEET notes

Matter & Its Classification

Everything around us is matter — it has mass and occupies space. Matter is classified as pure substances (elements & compounds) or mixtures (homogeneous & heterogeneous). Physical changes don't alter identity; chemical changes create new substances. NEET loves classification-based MCQs.

Matter classify cases Pure Substances Elements, Compounds Mixtures Homogeneous, Heterogeneous cases

Matter & Its States

Matter exists as solid, liquid, or gas. The state depends on the balance between intermolecular forces (attraction) and thermal energy (motion).

Matter is anything that has mass and occupies space (volume). A rock, water, air, even the smell of perfume — all matter. Light and heat are NOT matter (they are energy).

• Solid: definite shape AND volume (strong intermolecular forces, particles vibrate in place)
• Liquid: definite volume but NO definite shape (particles slide past each other)
• Gas: NEITHER definite shape NOR volume (particles move freely, fill entire container)
• Plasma = ionized gas (4th state) — not in NEET syllabus but good to know
• Interconversion: solid heat liquid heat gas (and reverse on cooling)

Pure Substances

A pure substance has a definite, constant composition. Elements contain only one type of atom; compounds contain two or more elements in a fixed ratio by mass.

A pure substance has a fixed composition that doesn't vary from sample to sample. There are two types: elements (cannot be broken down further by chemical means) and compounds (two or more elements chemically combined in a fixed ratio).

• Elements: 118 known (94 natural + 24 synthetic). Metals (Fe, Cu), non-metals (O, N), metalloids (Si, Ge)
• Compounds: fixed ratio by mass (law of definite proportions). Water is ALWAYS 1:8 by mass (H:O)
• Properties of a compound differ COMPLETELY from its elements — NaCl (edible salt) vs Na (explosive metal) + Cl _2 (poisonous gas)
• A molecule is the smallest particle of an element or compound that can exist independently

Mixtures

Mixtures contain two or more substances in any proportion. Homogeneous mixtures look uniform throughout; heterogeneous mixtures have visible boundaries.

A mixture is a combination of two or more substances where each retains its own chemical identity. Unlike compounds, mixtures have variable composition — you can make tea with 1 spoon or 3 spoons of sugar.

• Mixtures can be separated by physical methods (filtration, distillation, chromatography, sublimation)
• Compounds can only be separated by chemical methods (electrolysis, thermal decomposition)
• No energy change during mixture formation (ideally); compound formation involves energy change
• Mixture: properties are average of components. Compound: properties are entirely new

Physical vs Chemical Changes

Physical changes alter appearance but not chemical identity. Chemical changes create entirely new substances with new properties.

A physical change is reversible and does not create new substances — only the form, shape, or state changes. A chemical change creates new substances with entirely different properties — the original substance is gone.

• Signs of chemical change: colour change, gas evolution, precipitate formation, energy (heat/light) release, change in smell
• Dissolving is tricky — sugar in water = physical change; HCl in water = chemical change (ionization)
• Allotropic conversions (diamond → graphite) = chemical change (bond rearrangement)
• Crystallization is a physical change — the dissolved substance returns unchanged

Quick recall

Is dissolving NaCl in water a physical or chemical change?

Technically, it involves both. NaCl dissociates into Na ^+ and Cl ^- ions (chemical aspect), but the ions recombine on evaporation to give back NaCl (reversible = physical aspect). NCERT treats it as a physical change for this chapter.

Laws of Chemical Combination

Five experimental laws governed chemistry before atomic theory explained them. Lavoisier's conservation of mass, Proust's definite proportions, Dalton's multiple proportions, Gay-Lussac's gaseous volumes, and Avogadro's law — these laid the foundation for quantitative chemistry. NEET asks direct application problems.

Mass of reactants = Mass of products (Law of Conservation of Mass)

Law of Conservation of Mass

In any chemical reaction, the total mass of reactants equals the total mass of products. Proposed by Antoine Lavoisier (1789), the 'father of modern chemistry.'

Lavoisier (1789) performed careful experiments with sealed vessels and found that the total mass before and after a reaction is exactly the same. Matter is neither created nor destroyed — it just rearranges.

m_ reactants = m_ products

• Also called the Law of Indestructibility of Matter
• Verified by Landolt (1908) with sealed tube reactions
• Applies to chemical reactions. In nuclear reactions, mass converts to energy ( E = mc^2 ) — conservation of mass alone doesn't hold
• This law is the basis for balancing chemical equations — atoms in = atoms out

Quick recall

If 12 g of carbon reacts with 32 g of oxygen, how much CO _2 is formed?

By conservation of mass: 12 + 32 = 44 g of CO _2 . Simple addition — total mass in = total mass out.

Law of Definite Proportions

A pure compound always contains the same elements in the same fixed ratio by mass, regardless of source or method of preparation. Proposed by Joseph Proust (1799).

Proust (1799) showed that a given compound always contains the same elements combined in the same proportion by mass. Water from a river, from a lab, or from Mars — it's always 1 part H : 8 parts O by mass (or 11.1% H and 88.9% O).

• Also called the Law of Constant Composition
• CO _2 from any source: always 12:32 (= 3:8) by mass of C:O
• H _2 O from any source: always 1:8 by mass of H:O (or 2:16 = 1:8)
• This law does NOT apply to non-stoichiometric compounds (e.g., Fe _ 0.93 O, wustite) — called Berthollides
• Isotopic variations can cause slight deviations (heavy water D _2 O has different mass ratio than H _2 O)

Law of Multiple Proportions

When two elements form more than one compound, the different masses of one element that combine with a fixed mass of the other are in simple whole-number ratios. Proposed by Dalton (1803).

Dalton (1803): If two elements form multiple compounds, fix the mass of one element — the masses of the other element in different compounds will be in a simple whole-number ratio (1:2, 2:3, etc.).

• This law provides evidence for atoms — matter comes in discrete units, so ratios must be whole numbers
• The ratio must be small whole numbers: 1:2, 2:3, 1:1:2 etc.
• This law applies ONLY when two elements form MORE than one compound
• Dalton's atomic theory was built to explain this law (and the law of definite proportions)

Gay-Lussac's Law & Avogadro's Law

Gay-Lussac found that gas volumes in reactions are in simple ratios. Avogadro explained this by proposing that equal volumes of gases at same T and P contain equal numbers of molecules.

Gay-Lussac (1808): When gases react, their volumes (at same T and P) are in simple whole-number ratios. Example: 1 volume H _2 + 1 volume Cl _2 → 2 volumes HCl.

H_2(g) + Cl_2(g) arrow 2HCl(g) Volume ratio: 1:1:2

Avogadro (1811) proposed the explanation: equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. This also means gases are made of molecules (not just atoms) — 1 molecule of H _2 has 2 atoms.

V n (at constant T, P) V_1 n_1 = V_2 n_2

• Gay-Lussac's law applies ONLY to gases (not solids or liquids)
• Avogadro's law: at STP (0°C, 1 atm), 1 mole of any gas occupies 22.4 L (molar volume)
• At STP: 22.4 L of O _2 , H _2 , CO _2 , or ANY gas = 1 mole = 6.022 10^ 23 molecules
• Avogadro's law resolved the H _2 + Cl _2 puzzle that Dalton couldn't explain

Atomic & Molecular Mass

Atoms are incredibly tiny — we need a relative scale to compare their masses. The atomic mass unit (u) is defined as 1/12 the mass of a C-12 atom. Molecular mass is the sum of atomic masses of all atoms in a molecule. Formula mass is used for ionic compounds. These are the building blocks for the mole concept.

1 u = 1.66054 10^ -24 g = 1 12 mass of one ^ 12 C atom

Atomic Mass Unit (u)

Since atoms are too small to weigh on any balance, we use a relative scale. One atomic mass unit (u) is exactly 1/12 the mass of a carbon-12 atom.

An atom of hydrogen has a mass of about 1.67 10^ -24 g — an absurdly small number. Instead of working with such tiny values, chemists use a relative scale where the mass of one ^ 12 C atom is defined as exactly 12 u.

1 u (or amu) = Mass of one ^ 12 C atom 12 = 1.66054 10^ -24 g

• u (unified atomic mass unit) = amu (atomic mass unit) = Da (dalton) — all three names mean the same thing
• Carbon-12 was chosen as the standard because it gives nearly whole-number masses for most elements
• Atomic mass of H = 1.008 u, C = 12.011 u, N = 14.007 u, O = 15.999 u
• These are average atomic masses (weighted average over all natural isotopes)

Average Atomic Mass

Most elements have multiple isotopes in nature. The atomic mass listed in the periodic table is the weighted average based on natural abundance of each isotope.

Chlorine has two isotopes: ^ 35 Cl (75.77%) and ^ 37 Cl (24.23%). The atomic mass you see in the periodic table is NOT 35 or 37 — it's the weighted average.

Average atomic mass = (isotopic mass fractional abundance)

• The average atomic mass is always between the lightest and heaviest isotope masses
• It is closer to the mass of the MORE abundant isotope
• Carbon: ^ 12 C (98.9%) + ^ 13 C (1.1%) → average = 12.011 u (very close to 12)
• This is why no individual atom of chlorine has mass 35.5 — it's a statistical average

Quick recall

Boron has two isotopes: ^ 10 B (19.9%) and ^ 11 B (80.1%). Calculate average atomic mass.

Average = 10 0.199 + 11 0.801 = 1.99 + 8.811 = 10.81 u. This matches the periodic table value for boron.

Molecular Mass & Formula Mass

Molecular mass is the sum of atomic masses of all atoms in a molecule. For ionic compounds (which don't have discrete molecules), we use 'formula mass' instead.

The molecular mass of a substance is the sum of the atomic masses of all atoms in one molecule. For H_2O : molecular mass = 2(1.008) + 16.00 = 18.02 u.

Molecular mass of H_2SO_4 = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 u

• Molecular mass → for covalent/molecular compounds (H _2 O, CO _2 , glucose)
• Formula mass → for ionic compounds (NaCl, CaCO _3 ) since they don't form discrete molecules
• Formula mass of NaCl = 58.5 u (not 'molecular mass' because NaCl exists as a lattice, not molecules)
• Both molecular mass and formula mass have the same unit: u (or amu)

Mole Concept

The mole is chemistry's counting unit — just like a 'dozen' means 12, a 'mole' means 6.022 × 10²³. One mole of any substance contains Avogadro's number of particles and has mass equal to its atomic/molecular mass in grams. The mole bridges the atomic world (invisible) to the lab world (weighable). This is the MOST tested topic from this chapter in NEET.

n = m M = N N_A where N_A = 6.022 10^ 23 mol^ -1

What is a Mole?

The mole is the SI unit of 'amount of substance.' One mole = exactly 6.02214076 × 10²³ entities. It bridges the gap between atoms (too small to count) and grams (what you weigh in a lab).

A mole (symbol: mol) is defined as exactly 6.02214076 10^ 23 entities (atoms, molecules, ions, electrons — whatever you're counting). This number is called Avogadro's number ( N_A ).

1 mole of X = cases 6.022 10^ 23 particles of X M_X grams (molar mass) 22.4 L at STP (if gas) cases

• 1 mole of H atoms = 6.022 10^ 23 H atoms = 1 g
• 1 mole of H _2 molecules = 6.022 10^ 23 H _2 molecules = 2 g
• 1 mole of e ^- = 6.022 10^ 23 electrons (mass ≈ 0.00055 g — negligible)
• Specify what you're counting! '1 mole of oxygen' is ambiguous — 1 mol O atoms (16 g) vs 1 mol O _2 molecules (32 g)

Molar Mass

Molar mass is the mass of one mole of a substance. Numerically equal to the atomic/molecular mass but with units of g/mol instead of u.

The molar mass ( M ) of a substance is the mass of one mole of that substance, expressed in g/mol. The beautiful connection: the molar mass in g/mol is numerically equal to the atomic/molecular mass in u.

M (g/mol) Atomic/Molecular mass (u) numerically

• Molar mass links mass (grams) to moles: n = m M
• SI unit of molar mass: kg/mol, but g/mol is universally used in practice
• 1 u = 1.66054 10^ -24 g, and N_A = 6.022 10^ 23 , so 1 u N_A = 1 g/mol (this is the connection!)
• For elements, use the periodic table value directly as molar mass

Moles ↔ Mass ↔ Particles ↔ Volume

The central skill of chemistry: converting between moles, mass (grams), number of particles, and volume of gases. Master this triangle and you can solve 90% of NEET stoichiometry problems.

The mole sits at the centre of all conversions. From moles, you can go to mass (multiply by M ), to particles (multiply by N_A ), or to gas volume at STP (multiply by 22.4 L).

n = m M = N N_A = V_ STP 22.4 L

• Grams → Moles: n = m / M
• Moles → Particles: N = n N_A
• Moles → Volume (gas at STP): V = n 22.4 L
• Grams → Particles (shortcut): N = m N_A M
• 22.4 L/mol at STP applies ONLY to gases — not liquids or solids!

Quick recall

How many molecules are in 9 g of water?

Moles of H _2 O = 9 18 = 0.5 mol. Molecules = 0.5 6.022 10^ 23 = 3.011 10^ 23 molecules. Bonus: total atoms = 3.011 10^ 23 3 = 9.033 10^ 23 (each H _2 O has 3 atoms).

Quick recall

What volume does 11 g of CO _2 occupy at STP?

Moles of CO _2 = 11 44 = 0.25 mol. Volume at STP = 0.25 22.4 = 5.6 L.

Percentage Composition & Empirical Formula

Percentage composition tells you what fraction of a compound's mass comes from each element. From percentage data, you can work backwards to find the empirical formula (simplest ratio) and then the molecular formula (actual formula). This is a classic NEET numerical topic.

Mass % of element = Mass of element in 1 mol of compound Molar mass of compound 100

Percentage Composition

Percentage composition tells you the mass percentage of each element in a compound. It's calculated from the molecular formula and atomic masses.

The mass percent of an element in a compound = (total mass of that element in one formula unit ÷ molar mass of compound) × 100.

Mass % of X = n_X A_X M_ compound 100

where n_X = number of atoms of element X in the formula, A_X = atomic mass of X , and M = molar mass of the compound.

• All percentages in a compound must add up to 100% (useful error check)
• % composition is a property of the compound, not the sample size — 1 g or 1 kg of H _2 O both have 11.1% H
• If a question gives mass of compound and asks mass of an element, use: mass of element = (mass %) × (total mass) / 100

Empirical Formula

The empirical formula shows the simplest whole-number ratio of atoms in a compound. It's derived from percentage composition data by converting mass % to moles and then simplifying.

The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. Glucose has molecular formula C _6 H _ 12 O _6 but empirical formula CH _2 O (ratio 1:2:1).

• Step 1: Assume 100 g of sample → mass of each element = its percentage
• Step 2: Convert mass to moles → divide each by its atomic mass
• Step 3: Divide all mole values by the smallest mole value
• Step 4: If needed, multiply all by a small integer to get whole numbers
• Step 5: Write the empirical formula using these whole-number ratios

Molecular Formula from Empirical Formula

The molecular formula is a whole-number multiple of the empirical formula. You need the molar mass (from experiment) to find this multiple.

The molecular formula = n (empirical formula), where n is a whole number found by comparing the actual molar mass with the empirical formula mass.

n = Molar mass (experimental) Empirical formula mass

• Molecular formula is ALWAYS a whole-number multiple of empirical formula
• n must be a positive integer (1, 2, 3, ...). If n = 1 , molecular formula = empirical formula
• You NEED the molar mass from an experiment (e.g., vapour density, colligative properties) to find n
• Vapour density shortcut: Molar mass = 2 vapour density (VD)

Quick recall

A compound has empirical formula NO _2 and molar mass 92. Find the molecular formula.

Empirical formula mass of NO _2 = 14 + 32 = 46 . n = 92/46 = 2 . Molecular formula = (NO _2 ) _2 = N _2 O _4 (dinitrogen tetroxide).

Stoichiometry & Limiting Reagent

Stoichiometry uses balanced equations to calculate how much reactant is needed or product is formed. The limiting reagent is the one that runs out first and determines the maximum product. Percent yield compares actual vs theoretical product. High-weightage NEET numericals come from here.

% Yield = Actual yield Theoretical yield 100

Balanced Chemical Equations

A balanced equation has equal numbers of each type of atom on both sides. The coefficients tell you the mole ratios for calculations.

A balanced chemical equation obeys the law of conservation of mass — atoms are neither created nor destroyed. The stoichiometric coefficients (numbers before formulae) tell you the exact mole ratios in which reactants combine and products form.

2H_2(g) + O_2(g) arrow 2H_2O(l)

This equation tells us: 2 molecules (or moles) of H _2 react with 1 molecule (or mole) of O _2 to form 2 molecules (or moles) of H _2 O. In terms of mass: 4 g H _2 + 32 g O _2 → 36 g H _2 O.

• Coefficients give mole ratios, NOT mass ratios. 2 mol H _2 : 1 mol O _2 ≠ 2 g : 1 g
• For gas reactions at same T, P: coefficients also give volume ratios (Avogadro's law)
• State symbols: (g) gas, (l) liquid, (s) solid, (aq) aqueous solution
• An equation MUST be balanced before doing any stoichiometric calculation

Mole-Ratio Method

Given the amount of one substance, use the balanced equation's mole ratio to find the amount of another substance. This is the workhorse method for all NEET stoichiometry numericals.

The mole-ratio method is a 3-step process: (1) convert given quantity to moles, (2) use the mole ratio from the balanced equation, (3) convert moles to the desired quantity.

• Step 1: Convert given quantity (g, L, particles) to moles
• Step 2: Use the mole ratio from the balanced equation to find moles of desired substance
• Step 3: Convert moles of desired substance to required quantity (g, L, particles)

Quick recall

How many grams of NH _3 are produced from 14 g of N _2 ? ( N_2 + 3H_2 arrow 2NH_3 )

mol N _2 = 14/28 = 0.5 mol. From equation: 1 mol N _2 → 2 mol NH _3 , so 0.5 mol N _2 → 1 mol NH _3 . Mass = 1 × 17 = 17 g.

Limiting Reagent

When reactants are not in the exact stoichiometric ratio, one runs out first (limiting reagent) and the other is left over (excess reagent). The limiting reagent determines the maximum amount of product.

The limiting reagent (or limiting reactant) is the reactant that is completely consumed first in a reaction. It limits the amount of product that can form. The other reactant is in excess — some of it remains unreacted.

• Method 1: Calculate moles of each reactant. Divide each by its coefficient. The smallest value → limiting reagent
• Method 2: Assume each reactant is limiting, calculate product from each. The one giving LESS product is the limiting reagent
• Product is ALWAYS calculated from the limiting reagent
• Excess reagent remaining = initial moles − moles consumed

Percent Yield

Theoretical yield is the maximum product from stoichiometry. Actual yield is what you really get (always less due to side reactions, losses, incomplete reaction). Percent yield measures efficiency.

The theoretical yield is the maximum amount of product calculated from stoichiometry (assuming 100% conversion). The actual yield is what you actually obtain in the lab — always less than theoretical.

% Yield = Actual yield Theoretical yield 100

• % yield is always 100 % (if you get > 100%, your product is impure or you made an error)
• Reasons for < 100% yield: incomplete reaction, side reactions, loss during purification, reversible reactions
• Theoretical yield is calculated from the limiting reagent only
• If % yield is given, actual yield = theoretical yield × (% yield / 100)

Quick recall

If 10 g of CaCO _3 is heated and 3.36 g of CO _2 is collected, what is the % yield? ( CaCO_3 arrow CaO + CO_2 )

Theoretical: mol CaCO _3 = 10/100 = 0.1 mol → 0.1 mol CO _2 → 0.1 × 44 = 4.4 g. % yield = 3.36 4.4 100 = 76.4 % .

Concentration Terms

Concentration tells you how much solute is present in a given amount of solution or solvent. Molarity (M), molality (m), normality (N), mole fraction (χ), and ppm are the key terms. NEET loves interconversion numericals — especially molarity ↔ molality using density, and normality using n-factor.

M = n_ solute V_ solution (L) m = n_ solute W_ solvent (kg) N = n-factor M

Molarity (M)

The most commonly used concentration term. Molarity = moles of solute per litre of solution. It changes with temperature because volume changes with temperature.

Molarity ( M ) is the number of moles of solute dissolved in 1 litre (1000 mL) of solution (not solvent!).

M = n_ solute V_ solution (L) = w_ solute M_ solute V_ solution (L)

• Unit: mol/L or mol L ^ -1 (also written as M)
• Temperature-dependent because volume expands/contracts with temperature
• Dilution formula: M_1 V_1 = M_2 V_2 (moles of solute are conserved on dilution)
• For mixing solutions of same solute: M_1V_1 + M_2V_2 = M_ final V_ final
• 0.1 M = decimolar, 0.5 M = semimolar, 0.01 M = centimolar

Quick recall

What is the molarity of a solution containing 4 g of NaOH in 250 mL?

Moles of NaOH = 4/40 = 0.1 mol. Volume = 250 mL = 0.25 L. M = 0.1/0.25 = 0.4 M.

Molality (m)

Molality uses mass of solvent (not volume of solution), making it temperature-independent. Preferred for colligative property calculations.

Molality ( m ) is the number of moles of solute per kilogram of solvent (not solution).

m = n_ solute W_ solvent (kg) = w_ solute 1000 M_ solute W_ solvent (g)

• Unit: mol/kg or mol kg ^ -1 (also written as m)
• For dilute aqueous solutions at room temperature: M m (since density ≈ 1 g/mL and solvent mass ≈ solution volume)
• Temperature-independent because mass doesn't change with temperature (unlike volume)

Normality (N) & Equivalent Weight

Normality = number of gram-equivalents per litre. It depends on the reaction context through the n-factor. N = n-factor × M. Though removed from some syllabi, NEET still tests it indirectly.

Normality ( N ) is the number of gram-equivalents of solute per litre of solution. The gram-equivalent weight depends on the reaction — the same substance can have different equivalent weights in different reactions.

N = w_ solute E V(L) where E = M_ solute n-factor

N = n-factor M N_1V_1 = N_2V_2 (at equivalence point)

• n-factor for acids = number of replaceable H ^+ ions (basicity)
• n-factor for bases = number of replaceable OH ^- ions (acidity)
• n-factor for salts = total positive (or negative) charge
• n-factor for redox = change in oxidation state per formula unit
• At equivalence point: N_1V_1 = N_2V_2 (milliequivalents of acid = milliequivalents of base)

Mole Fraction & ppm

Mole fraction is the ratio of moles of one component to total moles. ppm (parts per million) is used for very dilute solutions. Both are dimensionless and temperature-independent.

Mole fraction ( or x ) of a component = moles of that component ÷ total moles of all components. It's always between 0 and 1, and the sum of all mole fractions = 1.

_A = n_A n_A + n_B , _B = n_B n_A + n_B , _A + _B = 1

ppm = mass of solute mass of solution 10^6

• Mole fraction is dimensionless and temperature-independent (mass-based)
• Used extensively in Raoult's law: P_A = _A P_A^
• ppm = parts per million. 1 ppm = 1 mg solute per 1 kg solution (for dilute aqueous solutions, 1 kg ≈ 1 L)
• Used for trace quantities: pollutant levels, dissolved gases, impurities
• Relation: _ solute = m M_ solvent 1000 + m M_ solvent where m = molality

Quick recall

A solution has 2 mol ethanol and 8 mol water. Find the mole fraction of ethanol.

_ ethanol = 2 2+8 = 2 10 = 0.2 . Mole fraction of water = 1 - 0.2 = 0.8 .