JEE Main Chemistry · Chapter 1

Some Basic Concepts of Chemistry

Mole concept, empirical and molecular formula, stoichiometry, limiting reagent and concentration terms.

NCERT Class 11JEE Main: 1 Q per shift 9 min read

1. Chapter at a Glance

This chapter teaches the arithmetic of chemistry. Atoms and molecules are far too small to count one by one, so chemists count them in bundles called moles. Once you are fluent with the mole, you can convert between mass, number of particles and gas volume, balance quantities in reactions (stoichiometry) and express concentrations. Nearly every numerical problem in physical chemistry uses these ideas, and JEE Main usually asks one direct question.

What JEE asks most
  • Mole calculations: mass ↔ moles ↔ number of particles ↔ volume of gas.
  • Limiting reagent and the amount of product formed.
  • Empirical and molecular formula from percentage composition.
  • Concentration terms: molarity, molality, mole fraction, and conversion between them.

2. Matter and Laws of Chemical Combination

Matter is classified into pure substances (elements and compounds) and mixtures (homogeneous or heterogeneous). A compound has a fixed composition; a mixture does not.

LawStatementGiven by
Conservation of massMass is neither created nor destroyed in a chemical reaction.Lavoisier
Definite proportionsA compound always contains the same elements in the same fixed ratio by mass.Proust
Multiple proportionsWhen two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a simple whole-number ratio.Dalton
Gaseous volumesGases react in simple whole-number ratios by volume (same temperature and pressure).Gay-Lussac
Avogadro's lawEqual volumes of gases at the same temperature and pressure contain equal numbers of molecules.Avogadro

Example of multiple proportions: in CO and CO₂, the masses of oxygen combining with 12 g of carbon are 16 g and 32 g — a ratio of 1 : 2.

3. Atomic Mass and the Mole Concept

  • Atomic mass unit: 1 u = 1/12 of the mass of one carbon-12 atom ≈ 1.66 × 10−24 g.
  • Average atomic mass accounts for isotopes: Σ(fractional abundance × isotopic mass). Chlorine (75% ³⁵Cl, 25% ³⁷Cl) has average atomic mass 35.5 u.
  • One mole contains exactly 6.022 × 10²³ entities (the Avogadro constant NA).
  • Molar mass is the mass of one mole in grams; it is numerically equal to the atomic or molecular mass in u.
  • One mole of an ideal gas occupies 22.4 L at 273 K and 1 atm (22.7 L at 273.15 K and 1 bar, the newer STP).
MOLES Mass (g) No. of particles Gas volume (L) ÷ molar mass× molar mass × 6.022 × 10²³÷ 6.022 × 10²³ × 22.4 / ÷ 22.4
Convert everything to moles first; the mole is the bridge between mass, particles and gas volume.

4. Empirical and Molecular Formula

The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number. Molecular formula = n × empirical formula, where n = molar mass ÷ empirical formula mass.

Method: (1) take the percentages as grams in 100 g; (2) divide each by its atomic mass to get moles; (3) divide by the smallest value; (4) multiply by a small integer if needed to get whole numbers.

Worked example

C 40%, H 6.67%, O 53.3%: moles C = 40/12 = 3.33, H = 6.67/1 = 6.67, O = 53.3/16 = 3.33. Ratio 1 : 2 : 1, so the empirical formula is CH₂O. If the molar mass is 180 g mol−1, n = 180/30 = 6 and the molecular formula is C₆H₁₂O₆ (glucose).

5. Stoichiometry and Limiting Reagent

A balanced equation gives the ratio of moles (not grams) in which substances react. For 2H₂ + O₂ → 2H₂O, 2 mol H₂ react with 1 mol O₂ to give 2 mol H₂O.

  • The limiting reagent is the reactant that is used up first; it decides how much product forms. The other reactant is in excess.
  • To find it, divide the moles of each reactant by its coefficient in the balanced equation — the smallest value marks the limiting reagent.
  • Percentage yield = (actual yield ÷ theoretical yield) × 100.
  • Percentage purity = (mass of pure substance ÷ mass of impure sample) × 100.

6. Concentration Terms

TermDefinitionDepends on temperature?
Mass percent (w/w)(mass of solute/mass of solution) × 100No
Mole fraction (x)moles of component/total molesNo
Molarity (M)moles of solute per litre of solutionYes (volume changes)
Molality (m)moles of solute per kg of solventNo
ppm(mass of solute/mass of solution) × 10⁶No
  • Dilution: M1V1 = M2V2. Mixing: M = (M1V1 + M2V2)/(V1 + V2).
  • Relation between molarity and molality (density d in g mL−1, solute molar mass MB): m = 1000M/(1000d − M·MB).

Significant figures

All non-zero digits are significant; zeros between them are significant; leading zeros are not; trailing zeros after a decimal point are. So 0.00250 has 3 significant figures. In multiplication/division, the answer keeps the least number of significant figures; in addition/subtraction, the least number of decimal places.

7. Formula Master Table

QuantityFormula
Moles from massn = m/M
Number of particlesN = n × 6.022 × 10²³
Moles of gas at STP (1 atm)n = V(L)/22.4
MolarityM = nsolute/Vsolution(L)
Molalitym = nsolute/mass of solvent (kg)
Mole fractionxA = nA/(nA + nB); xA + xB = 1
DilutionM1V1 = M2V2
Molecular formulan × empirical formula, n = M/(EF mass)
Average atomic massΣ (abundance × mass)

8. Mind Map

Some Basic Concepts of Chemistry
Laws
  • Conservation of mass
  • Definite & multiple proportions
  • Gay-Lussac, Avogadro
Mole
  • 6.022 × 10²³
  • Molar mass
  • 22.4 L at STP
Formulae
  • Empirical
  • Molecular = n × EF
Reactions
  • Mole ratio
  • Limiting reagent
  • Yield, purity
Concentration
  • M, m, x, ppm
  • Dilution

9. Tips and Tricks

Shortcuts
  • Always convert grams to moles before using a balanced equation — never compare masses directly.
  • For the limiting reagent, compare (moles ÷ coefficient) for each reactant.
  • In empirical-formula problems, ratios like 1 : 1.5 mean "multiply by 2"; 1 : 1.33 means "multiply by 3".
  • Molality does not change with temperature, which is why it is used for colligative properties.
Mistakes students make
  • Using the volume of solvent instead of the volume of solution for molarity.
  • Using grams of solution instead of kg of solvent for molality.
  • Applying 22.4 L to liquids or solids — it is only for gases (at 273 K, 1 atm).

10. Important One-liners

  • 1 mole of any substance contains 6.022 × 10²³ particles of that substance.
  • 1 mol of H₂O has a mass of 18 g and contains 3 × 6.022 × 10²³ atoms.
  • The empirical and molecular formulas of water and CO₂ are the same.
  • Molarity decreases slightly with rise in temperature because the solution expands.
  • The sum of the mole fractions of all components of a solution is 1.
  • One atomic mass unit is 1/12 of the mass of a C-12 atom.

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main. (Atomic masses: H = 1, C = 12, O = 16, Na = 23, Ca = 40.)

Q1Easy

The number of molecules in 4.4 g of CO₂ is:

  1. 6.022 × 10²²
  2. 6.022 × 10²³
  3. 1.2 × 10²³
  4. 3.0 × 10²²
Show solution

Molar mass of CO₂ = 44 g mol−1, so 4.4 g = 0.1 mol. Molecules = 0.1 × 6.022 × 10²³ = 6.022 × 10²².

Answer: (A)

Q2Numerical

4 g of NaOH is dissolved in water to make 250 mL of solution. Find its molarity.

Show solution

Moles of NaOH = 4/40 = 0.1 mol. Molarity = 0.1/0.250 L = 0.4 M.

Answer: 0.4

Q3Medium

A compound contains 40% C, 6.67% H and 53.3% O by mass. Its empirical formula is:

  1. CHO
  2. CH₂O
  3. C₂H₄O
  4. CH₃O
Show solution

Moles in 100 g: C = 40/12 = 3.33, H = 6.67/1 = 6.67, O = 53.3/16 = 3.33. Dividing by 3.33 gives 1 : 2 : 1.

Answer: (B) CH₂O

Q4Numerical

4 g of H₂ and 16 g of O₂ are allowed to react to form water. Find the mass (in g) of water formed.

Show solution

2H₂ + O₂ → 2H₂O. Moles: H₂ = 2, O₂ = 0.5. Moles ÷ coefficient: H₂ = 1, O₂ = 0.5, so O₂ is limiting.

0.5 mol O₂ gives 1 mol H₂O = 18 g. (1 mol H₂ remains unused.)

Answer: 18

Q5Easy

18 g of glucose (C₆H₁₂O₆) is dissolved in 500 g of water. The molality of the solution is:

  1. 0.1 m
  2. 0.2 m
  3. 0.5 m
  4. 2 m
Show solution

Molar mass of glucose = 180 g mol−1; moles = 0.1. Molality = 0.1/0.5 kg = 0.2 m.

Answer: (B)

Q6Medium

The mole fraction of ethanol in a solution containing 46 g of ethanol (C₂H₅OH) and 36 g of water is:

  1. 0.25
  2. 0.33
  3. 0.50
  4. 0.67
Show solution

Moles of ethanol = 46/46 = 1; moles of water = 36/18 = 2. x(ethanol) = 1/(1 + 2) = 1/3 ≈ 0.33.

Answer: (B)

Q7Numerical

What volume of oxygen (in litres, at 273 K and 1 atm) is needed for the complete combustion of 1 mol of methane?

Show solution

CH₄ + 2O₂ → CO₂ + 2H₂O. 1 mol CH₄ needs 2 mol O₂ = 2 × 22.4 = 44.8 L.

Answer: 44.8

Q8Hard

12.5 g of an impure sample of CaCO₃ on strong heating gives 4.4 g of CO₂. Assuming the impurities do not decompose, the percentage purity of the sample is:

  1. 70%
  2. 80%
  3. 88%
  4. 90%
Show solution

CaCO₃ → CaO + CO₂. 100 g CaCO₃ gives 44 g CO₂, so 4.4 g CO₂ comes from 10 g of pure CaCO₃.

Purity = (10/12.5) × 100 = 80%.

Answer: (B)

12. Quick Revision Summary

Remember before the exam
  • n = m/M = N/NA = V/22.4 (gas, 273 K, 1 atm).
  • Balanced equations give mole ratios; find the limiting reagent by moles ÷ coefficient.
  • Empirical formula: % → moles → simplest ratio; molecular = n × EF.
  • Molarity (per litre of solution) changes with temperature; molality (per kg of solvent) does not.
  • M1V1 = M2V2 for dilution; mole fractions add to 1.
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