Chemistry Notes

Some Basic Concepts of Chemistry — NEET Notes

This is the chapter that decides whether the rest of physical chemistry sits firmly in place or constantly slips. Every calculation in equilibrium, kinetics, electrochemistry, and stoichiometry of every reaction you will ever balance leans on three ideas built here — the mole, atomic and molecular masses, and the laws of chemical combination. NEET tests this chapter every year, and the patterns are remarkably stable: a stoichiometry calculation with percentage purity, a moles-in-one-gram comparison, an empirical-formula problem, a partial-pressure or mole-fraction setup. By the end of this chapter you should be able to convert mass to moles to molecules in a single line and identify the limiting reagent without writing a table.

Nature of matter

Matter is anything that has mass and occupies space. NCERT classifies it on two parallel axes. At the physical level, matter exists in three states — solid, liquid, gas — distinguished by how tightly the constituent particles are held. Solids have a definite shape and volume, liquids have a definite volume but take the shape of the container, and gases occupy the entire space available. These three states are interconvertible by changing temperature or pressure: heating a solid melts it to a liquid and then vaporises it to a gas; cooling reverses the sequence.

At the chemical level, matter is either a pure substance or a mixture. A pure substance has constituent particles of fixed chemical composition — copper, silver, water, glucose. A mixture contains two or more pure substances in any ratio; its composition is variable. Mixtures are homogeneous (sugar solution, air) when the components are uniformly distributed, and heterogeneous (salt + sugar, grain + dirt) when they are visibly distinct. Components of a mixture can be separated by physical methods such as filtration, distillation, or crystallisation; components of a compound cannot.

Pure substances are further divided into elements (all atoms of one kind — sodium, oxygen, copper) and compounds (atoms of two or more elements combined in a fixed ratio — water, ammonia, carbon dioxide). The properties of a compound are completely different from those of its constituent elements. Hydrogen burns and oxygen supports combustion, but their compound — water — is used to extinguish fire. This independence of compound properties from elemental properties is one of the central observations that the atomic theory had to explain.

Properties of matter & measurement

Every substance has characteristic properties, sorted into two categories. Physical properties — colour, odour, melting point, boiling point, density — can be measured or observed without altering the identity of the substance. Chemical properties — combustibility, acidity, reactivity with acids and bases — require a chemical change to be observed. Quantitative measurement is the engine of chemistry, and every measurement is reported as a number followed by a unit.

The International System of Units (SI), established by the 11th CGPM in 1960, defines seven base units that cover every measurable quantity. Each unit was redefined in 2019 in terms of fixed numerical values of fundamental physical constants — making the SI a system grounded entirely in nature rather than in artefacts.

Length

metre (m)

speed of light is fixed

Defined by fixing the speed of light in vacuum at 299 792 458 m s⁻¹.

Mass

kilogram (kg)

Planck constant is fixed

Defined by fixing Planck's constant h at 6.62607015 × 10⁻³⁴ J s.

Time

second (s)

caesium frequency is fixed

Defined by the unperturbed Cs-133 hyperfine transition at 9 192 631 770 Hz.

Amount of substance

mole (mol)

Avogadro's number is fixed

One mole contains exactly 6.02214076 × 10²³ elementary entities.

Temperature

kelvin (K)

Boltzmann constant is fixed

K = °C + 273.15. Negative kelvin temperatures are impossible.

Electric current

ampere (A)

elementary charge is fixed

Defined by fixing the elementary charge e at 1.602176634 × 10⁻¹⁹ C.

The chemistry laboratory uses derived units of everyday convenience — gram for mass (1 kg = 1000 g), cm³ or dm³ for volume, °C for temperature, g cm⁻³ for density. The litre (L) — a non-SI unit equal to 1 dm³ or 1000 cm³ — remains in widespread use for liquid volume. Mass is the amount of matter in a body and is invariant; weight is the gravitational force on that mass and varies from place to place. The two are routinely confused and NCERT is unusually firm about distinguishing them.

Two temperature conversions cover every NEET problem. The Fahrenheit–Celsius relation is °F = (9/5) × °C + 32. The Celsius–Kelvin relation is K = °C + 273.15. Density is mass per unit volume — the SI unit is kg m⁻³, but chemists use g cm⁻³. Higher density means more tightly packed particles.

Uncertainty & significant figures

Every experimental measurement carries some uncertainty — limited by the instrument and the experimenter. A platform balance might read 9.4 g where an analytical balance reads 9.4213 g; both are valid measurements, with different uncertainties. Significant figures are the meaningful digits in a measurement: all certain digits, plus one final uncertain digit. A reading of 11.2 mL means 11 is certain and 2 is the estimated tenth.

Five rules cover every significant-figure question NEET asks. (i) All non-zero digits count — 285 cm has three significant figures. (ii) Leading zeros do not count — 0.0052 has two significant figures. (iii) Zeros between non-zero digits count — 2.005 has four. (iv) Trailing zeros count only if there is a decimal point — 0.200 has three, 100 has one (unless written as 100. or 1.00 × 10²). (v) Counted exact numbers (2 balls, 20 eggs) have infinite significant figures.

Scientific notation makes both extremely large and extremely small numbers tractable. Any number can be written as N × 10ⁿ where 1 ≤ N < 10. Thus 232.508 becomes 2.32508 × 10² and 0.00016 becomes 1.6 × 10⁻⁴. In addition and subtraction, the result is rounded to the fewest decimal places among the inputs; in multiplication and division, the result is rounded to the fewest significant figures.

Dimensional analysis — also called the factor-label method — converts a quantity from one unit to another by multiplying by unit factors equal to 1. To convert 3 in to centimetres, multiply by (2.54 cm / 1 in): 3 × 2.54 = 7.62 cm. The unit factor is always chosen so that the unwanted unit cancels. This single technique handles every NEET unit-conversion problem, from converting litres to cubic metres to converting days to seconds.

Laws of chemical combination

Five laws were stitched together over roughly forty years (1789–1811) to convert chemistry from an alchemical craft into a quantitative science. Each was derived from careful mass measurements, each was empirical, and together they forced the conclusion that matter is atomic. NEET tests them in two formats: identifying the law from a worked example, and applying a mass-ratio calculation to a multi-compound case.

1 · Conservation of mass

Lavoisier, 1789

mass in = mass out

Matter is neither created nor destroyed in any physical or chemical change.

2 · Definite proportions

Proust, 1799

fixed mass ratio

A given compound always contains the same elements in the same proportion by mass, regardless of source.

3 · Multiple proportions

Dalton, 1803

small whole-number ratios

When two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in ratios of small whole numbers.

4 · Gay-Lussac's law

Gay-Lussac, 1808

combining volumes

Reacting gases at the same T and P combine in volumes that bear a simple integer ratio.

5 · Avogadro's law

Avogadro, 1811

equal volumes, equal molecules

Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

Worked example

H + O

multiple proportions in action

Water: 2 g H + 16 g O. Hydrogen peroxide: 2 g H + 32 g O. Oxygen masses 16 and 32 — a 1:2 ratio.

The most fertile of these laws was Avogadro's. By distinguishing between atoms and molecules — a distinction Dalton resisted — Avogadro could explain Gay-Lussac's volume ratios. Hydrogen + oxygen combining in 2:1 by volume to give 2 volumes of water vapour only works if hydrogen and oxygen are diatomic (H₂ and O₂) and water is H₂O. Avogadro's idea was ignored for almost fifty years until Cannizzaro revived it at the 1860 Karlsruhe Congress.

Dalton's atomic theory

In 1808 John Dalton published A New System of Chemical Philosophy, in which he proposed that all matter is composed of indivisible atoms. His four postulates absorbed every empirical law that came before. The theory's power was that it explained — at a single stroke — conservation of mass, definite proportions, and multiple proportions, while predicting new chemistry that had not yet been observed.

The theory had two weaknesses. Dalton insisted that atoms of the same kind cannot combine, so he could not explain Gay-Lussac's gaseous volumes — which require diatomic molecules of elements (H₂, O₂, N₂). And he could not say why atoms combine — the answer came only with the electron and chemical bonding decades later. NCERT presents Dalton's theory as the first quantitatively useful model of matter, not as a complete one.

All matter — whether element, compound or mixture — is composed of small particles called atoms.— John Dalton, 1808

Atomic & molecular masses

Atoms are far too small to weigh individually — even the heaviest atoms come in at around 10⁻²² grams. Nineteenth-century chemists could only determine the mass of one atom relative to another. Hydrogen was initially taken as the standard with mass 1; in 1961 the international community switched to carbon-12, with one ¹²C atom defined as exactly 12 atomic mass units (amu, now written u).

Most elements occur naturally as mixtures of isotopes, so the value listed in the periodic table is an average atomic mass weighted by isotopic abundance. Carbon, for instance, exists as ¹²C (98.892%), ¹³C (1.108%), and trace ¹⁴C (≈10⁻¹⁰%). Plugging in:

(0.98892 × 12 u) + (0.01108 × 13.00335 u) + (2 × 10⁻¹² × 14.00317 u) = 12.011 uAverage atomic mass of natural carbon

Molecular mass is the sum of atomic masses of every atom in one molecule. Methane (CH₄) has molecular mass 12.011 + 4(1.008) = 16.043 u. Water has 2(1.008) + 16.00 = 18.02 u. Formula mass is used in place of molecular mass for ionic compounds — those that do not exist as discrete molecules. Sodium chloride is a three-dimensional lattice of Na⁺ and Cl⁻ ions in which one Na⁺ is surrounded by six Cl⁻ and vice versa; we therefore quote its formula mass as 23.0 + 35.5 = 58.5 u.

Mole concept & molar mass

The mole is chemistry's bridge between the microscopic world of atoms and the macroscopic world of grams. A dozen counts twelve, a score counts twenty, a gross counts 144; a mole counts 6.02214076 × 10²³ — Avogadro's number. The mole was added in 1971 as the seventh SI base unit, and in 2019 was redefined so that this number itself is fixed by convention rather than derived from a mass measurement of carbon-12.

An "elementary entity" can be anything you choose to specify — an atom, a molecule, an ion, an electron, a formula unit, or any other particle. One mole of hydrogen atoms is 6.022 × 10²³ H atoms; one mole of hydrogen molecules is 6.022 × 10²³ H₂ molecules, which contains 2 × 6.022 × 10²³ atoms. Confusing these is the single most reliable NEET trap from this chapter.

The molar mass of a substance is the mass of one mole, expressed in grams per mole. Numerically, molar mass in g mol⁻¹ equals atomic, molecular, or formula mass in u — only the units differ. Molar mass of water is 18.02 g mol⁻¹; of sodium chloride 58.5 g mol⁻¹; of methane 16.043 g mol⁻¹.

Percentage composition & empirical formula

If you have an unknown compound, the first question chemistry asks is — what is in it, and in what ratio? Mass per cent of an element in a compound is the mass of that element divided by the molar mass of the compound, multiplied by 100. For water, mass per cent of hydrogen = (2 × 1.008 / 18.02) × 100 = 11.18%; mass per cent of oxygen = (16.00 / 18.02) × 100 = 88.79%. Mass per cent is independent of the sample size — a direct corollary of the law of definite proportions.

From per cent composition, two formulae can be derived. The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms in one molecule. For glucose, the empirical formula is CH₂O and the molecular formula is C₆H₁₂O₆ — the molecular formula is 6 × the empirical formula.

NEET 2021 tested this in compact form (Q.82): an organic compound is 78% C and 22% H by mass; find the empirical formula. Moles: C = 78/12 = 6.5; H = 22/1 = 22. Ratio: H/C = 22/6.5 = 3.38 ≈ 3. Empirical formula = CH₃. The arithmetic takes thirty seconds when the steps are memorised.

Stoichiometry & stoichiometric calculations

The word stoichiometry comes from Greek stoicheion (element) and metron (measure). It is the quantitative bookkeeping of a chemical reaction — calculating how much of each reactant is consumed and how much of each product is formed. The currency is the balanced chemical equation, which respects the law of conservation of mass by having the same number of atoms of each element on both sides.

Consider the combustion of methane:

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g)Stoichiometric coefficients: 1 : 2 : 1 : 2

The coefficients (1, 2, 1, 2) are called stoichiometric coefficients. They simultaneously represent four parallel ratios that NCERT spells out exactly:

In moles

1 : 2 : 1 : 2

moles, the universal scale

1 mol CH₄ + 2 mol O₂ → 1 mol CO₂ + 2 mol H₂O.

In molecules

1 : 2 : 1 : 2

microscopic count

1 molecule of CH₄ + 2 molecules of O₂ → 1 CO₂ + 2 H₂O molecules.

In mass

16 : 64 : 44 : 36

grams (conserves total)

16 g CH₄ + 64 g O₂ → 44 g CO₂ + 36 g H₂O. 80 in, 80 out.

In gas volume

1 : 2 : 1 : 2

at the same T & P

22.4 L CH₄ + 44.8 L O₂ → 22.4 L CO₂ + 44.8 L H₂O (g) at STP.

The standard stoichiometric calculation has three steps. First, write and balance the equation. Second, convert the given mass to moles using molar mass. Third, use the mole ratio in the balanced equation to find moles of the unknown, then convert back to mass, volume, or particles as the question demands.

NEET 2023 Q.57 is a textbook application. Heat 20 g of 20% pure CaCO₃; calculate the mass of CO₂ produced. CaCO₃ → CaO + CO₂. Pure CaCO₃ in the sample = 20 × 0.20 = 4 g. Mole ratio is 1:1: 100 g CaCO₃ gives 44 g CO₂, so 4 g gives (4 × 44 / 100) = 1.76 g of CO₂. Three short lines.

Limiting reagent

Real reactions are rarely run with exactly stoichiometric amounts. One reactant is usually in excess; the other is in short supply. The reactant in short supply runs out first and stops the reaction — it limits how much product can be formed and is called the limiting reagent. The excess reactant has leftover unconsumed material at the end. Every yield calculation must be done with reference to the limiting reagent — never the excess one.

Identifying the limiting reagent takes two arithmetic steps. Convert each reactant to moles using molar mass. Then, for each reactant, calculate how many moles of product it would form if fully consumed. The reactant that gives the smaller yield is limiting. NCERT's worked example uses the Haber synthesis of ammonia.

Reactions in solution introduce one more concentration measure beyond mass-per-cent and mole fraction: molarity (mol of solute per litre of solution) and molality (mol of solute per kg of solvent). Molarity is the more commonly used in NEET stoichiometry problems; molality, being independent of temperature, appears in colligative-property calculations.

NEET PYQ Snapshot

Real NEET previous-year questions — solve before moving on.

NEET 2023

The right option for the mass of CO₂ produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40). [CaCO₃ → CaO + CO₂]

  1. 1.32 g
  2. 1.12 g
  3. 1.76 g
  4. 2.64 g
Answer: (3) 1.76 g

Why: Mass of pure CaCO₃ = 20 × 20/100 = 4 g. From the balanced equation, 100 g CaCO₃ gives 44 g CO₂. So 4 g CaCO₃ gives (4 × 44) / 100 = 1.76 g CO₂. Apply purity first, then the mole ratio.

NEET 2022

What mass of 95% pure CaCO₃ will be required to neutralise 50 mL of 0.5 M HCl solution? CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + CO₂(g) + 2 H₂O(l). [Calculate up to second decimal.]

  1. 1.32 g
  2. 3.65 g
  3. 9.50 g
  4. 1.25 g
Answer: (1) 1.32 g

Why: Moles of HCl = 0.050 × 0.5 = 0.025 mol. Moles of CaCO₃ required = 0.025 / 2 = 0.0125 mol. Pure mass = 0.0125 × 100 = 1.25 g. Sample is 95% pure, so impure mass = 1.25 × 100/95 = 1.32 g.

NEET 2021

An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C is 12, H is 1]

  1. CH₄
  2. CH
  3. CH₂
  4. CH₃
Answer: (4) CH₃

Why: Moles of C = 78/12 = 6.5. Moles of H = 22/1 = 22. Ratio H : C = 22 / 6.5 ≈ 3.38 ≈ 3. So the simplest ratio of C : H is 1 : 3, giving empirical formula CH₃.

NEET 2020

Which one of the following has the maximum number of atoms?

  1. 1 g of Mg(s) [Atomic mass of Mg = 24]
  2. 1 g of O₂(g) [Atomic mass of O = 16]
  3. 1 g of Li(s) [Atomic mass of Li = 7]
  4. 1 g of Ag(s) [Atomic mass of Ag = 108]
Answer: (3) 1 g of Li

Why: Number of atoms = (mass / molar mass) × NA × atomicity. For Li (atomicity 1): (1/7) NA ≈ 0.143 NA. For Mg: (1/24) NA. For O₂ (atomicity 2): (1/32) × 2 × NA = (1/16) NA. For Ag: (1/108) NA. Lithium wins because of its small molar mass.

NEET 2018

In which case is the number of molecules of water maximum?

  1. 18 mL of water
  2. 0.18 g of water
  3. 0.00224 L of water vapours at 1 atm and 273 K
  4. 10⁻³ mol of water
Answer: (1) 18 mL of water

Why: 18 mL water = 18 g (density 1 g/mL) = 1 mol = NA molecules. 0.18 g = 0.01 mol. 0.00224 L vapour at STP = 0.00224 / 22.4 = 10⁻⁴ mol. 10⁻³ mol = 10⁻³ NA. The 18 mL option wins by orders of magnitude.

Expert FAQs

Questions NEET asks from this chapter, answered straight.

What is the value of Avogadro's number?
Avogadro's number is 6.02214076 × 10²³ mol⁻¹, usually written as 6.022 × 10²³. It is the number of elementary entities (atoms, molecules, ions, electrons) contained in exactly one mole of any substance and is the fixed numerical value of the Avogadro constant in the 2019 SI redefinition.
What is the difference between empirical formula and molecular formula?
The empirical formula represents the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in one molecule. For glucose, the empirical formula is CH₂O and the molecular formula is C₆H₁₂O₆. Molecular formula = n × empirical formula, where n = molar mass ÷ empirical formula mass.
What is a limiting reagent?
The limiting reagent is the reactant that is completely consumed first in a chemical reaction, thereby limiting the amount of product formed. The other reactants — present in excess — remain partially unreacted. Stoichiometric calculations of yield are always done with reference to the limiting reagent.
How many atoms are present in one mole of any substance?
One mole of any monatomic substance contains 6.022 × 10²³ atoms. For diatomic molecules such as O₂ or H₂, one mole contains 6.022 × 10²³ molecules but 2 × 6.022 × 10²³ atoms. The mole counts elementary entities — whatever entity is specified.
What is the Law of Multiple Proportions?
Proposed by Dalton in 1803, this law states that when two elements combine to form more than one compound, the masses of one element which combine with a fixed mass of the other are in the ratio of small whole numbers. Hydrogen + oxygen forms water (H:O = 2:16) and hydrogen peroxide (H:O = 2:32) — the oxygen masses 16 and 32 are in a 1:2 ratio.
What is molar mass and how is it different from molecular mass?
Molecular mass is the sum of atomic masses in one molecule, expressed in unified atomic mass units (u). Molar mass is the mass of one mole of a substance, expressed in grams per mole (g mol⁻¹). The two are numerically equal — water has molecular mass 18.02 u and molar mass 18.02 g mol⁻¹ — but their units are different.
Why is the mole concept needed in chemistry?
Atoms and molecules are so small and so numerous that handling them individually is impossible. The mole provides a unit of convenient magnitude — 6.022 × 10²³ entities per mole — that lets chemists weigh out a definite number of particles. It bridges the macroscopic scale of laboratory measurements (grams, litres) with the microscopic world of atoms.
What is the postulate of Dalton's atomic theory that explains the Law of Definite Proportions?
Dalton's postulate that compounds are formed when atoms of different elements combine in a fixed ratio directly explains the Law of Definite Proportions. If a compound always contains the same atoms in the same ratio, then the mass ratio of its elements must also be fixed — regardless of how or where the compound was prepared.

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Drill into the subtopics that NEET asks most often.