How do you calculate empirical and molecular formula?

To find the empirical formula: (1) Convert mass of each element to moles, (2) Divide all mole values by the smallest mole value to get the simplest whole-number ratio, (3) Write the formula using this ratio. To find the molecular formula: divide the molar mass by the empirical formula mass, then multiply the empirical formula by this ratio.

What is a limiting reagent?

In a chemical reaction, the limiting reagent is the reactant that is entirely consumed first and determines the maximum amount of product that can be formed. The other reactants present in excess remain partly unused after the reaction is complete.

Some Basic Concepts of Chemistry — NCERT Class 11 Chemistry

How to Use This Page

This page provides complete, step-by-step NCERT solutions for Class 11 Chemistry Chapter 1 — Some Basic Concepts of Chemistry. Every intext question and chapter-end exercise question is answered with clear working so you can follow the logic and reproduce the answers yourself. Use the section headings to jump to the specific question you need help with.

Key Concepts (Quick Recap)

Mole concept: 1 mole = 6.022 × 10²³ particles (Avogadro number, Nₐ); mass of 1 mole = molar mass in grams
Molar mass: Mass of 1 mole of a substance, numerically equal to molecular mass in unified atomic mass units (u)
Percentage composition: Mass percent of an element = (mass of element in 1 mole of compound / molar mass of compound) × 100
Empirical formula: Simplest whole-number ratio of atoms in a compound
Molecular formula: Actual number of atoms of each element; n × (empirical formula), where n = molar mass / empirical formula mass
Stoichiometry: Quantitative relationship between reactants and products in a balanced chemical equation
Limiting reagent: Reactant consumed completely first; determines the amount of product
Molarity (M): Moles of solute per litre of solution
Significant figures: Rules for determining the precision of measured quantities
Dimensional analysis: Converting quantities from one system of units to another using conversion factors
Laws of chemical combination: Law of conservation of mass, Law of definite proportions, Law of multiple proportions, Gay Lussac's law of gaseous volumes

Intext Questions

Page 1–2 (Importance of Chemistry, Classification of Matter)

Q1.1: Why is chemistry called the "central science"?

AnswerChemistry is called the central science because it bridges physics with biology and earth sciences. The principles of chemistry underlie processes in biology (metabolism, genetics), geology (mineral formation), and physics (atomic structure). Every material thing is made of chemicals, and understanding their behaviour is fundamental to all natural sciences.

Q1.2: How has chemistry contributed to human welfare?

AnswerChemistry has contributed enormously — development of life-saving drugs (antibiotics, antimalarials), fertilisers and pesticides that boosted agricultural productivity, synthetic fibres and plastics, soaps and detergents for hygiene, and materials for construction, transport, and electronics. It has improved life expectancy, food security, and quality of life.

Q1.3: Classify the following as element, compound, or mixture: (a) Sea water, (b) Distilled water, (c) Diamond, (d) Gasoline, (e) Air.

Answer(a) Sea water — Mixture (salt dissolved in water). (b) Distilled water — Compound (H₂O). (c) Diamond — Element (allotrope of carbon). (d) Gasoline — Mixture (blend of hydrocarbons). (e) Air — Mixture (N₂, O₂, CO₂, etc.).

Page 5–6 (SI Units, Mole Concept)

Q1.4: What is the SI unit of amount of substance? How is it defined?

AnswerThe SI unit of amount of substance is the mole (mol). One mole is defined as the amount of substance that contains exactly 6.02214076 × 10²³ elementary entities (atoms, molecules, ions, or other particles). This number is the Avogadro constant.

Q1.5: Calculate the molar mass of: (a) H₂O, (b) CO₂, (c) CH₄.

Answer

(a) H₂O: 2 × 1.008 + 16.00 = 18.016 g mol⁻¹
(b) CO₂: 12.01 + 2 × 16.00 = 44.01 g mol⁻¹
(c) CH₄: 12.01 + 4 × 1.008 = 16.042 g mol⁻¹

Page 12–14 (Molar Mass Calculations, Empirical Formula, Stoichiometry)

Q1.6: Calculate the mass of 0.5 mole of N₂ gas.

AnswerMolar mass of N₂ = 2 × 14.01 = 28.02 g mol⁻¹. Mass = number of moles × molar mass = 0.5 × 28.02 = 14.01 g.

Q1.7: Calculate the number of atoms in 0.5 mol of carbon.

AnswerNumber of atoms = number of moles × Avogadro number = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ atoms.

Q1.8: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

Answer

Element	%	Atomic mass	Moles	Ratio
C	40	12	40/12 = 3.33	1
H	6.7	1	6.7/1 = 6.7	2
O	53.3	16	53.3/16 = 3.33	1

Dividing by the smallest (3.33): C = 1, H ≈ 2, O = 1. Empirical formula = CH₂O.

Q1.9: What mass of CO₂ is produced when 6 g of carbon is burnt in excess oxygen?

AnswerC + O₂ → CO₂. 1 mol C (12 g) produces 1 mol CO₂ (44 g). For 6 g C: (6/12) × 44 = 22 g CO₂.

Chapter-End Exercise

Q1. Calculate the molecular mass of the following: (i) H₂O, (ii) CO₂, (iii) CH₄.

Answer

(i) H₂O = 2 × 1.008 + 16.00 = 18.016 u
(ii) CO₂ = 12.01 + 2 × 16.00 = 44.01 u
(iii) CH₄ = 12.01 + 4 × 1.008 = 16.042 u

Q2. Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).

AnswerMolar mass = 2 × 23 + 32 + 4 × 16 = 142 g mol⁻¹

Na: (46/142) × 100 = 32.39%
S: (32/142) × 100 = 22.54%
O: (64/142) × 100 = 45.07%

Q3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.

Answer

Element	%	At. mass	Moles	Ratio
Fe	69.9	55.85	1.25	2
O	30.1	16	1.88	3

Empirical formula = Fe₂O₃.

Q4. Calculate the amount of carbon dioxide that could be produced when 1 mole of carbon is burnt in: (i) 1 mole of O₂, (ii) 16 g of O₂.

AnswerC + O₂ → CO₂. 1 mol C reacts with 1 mol O₂ to give 1 mol CO₂ (44 g).

(i) 1 mol O₂ is sufficient. CO₂ produced = 44 g.
(ii) 16 g O₂ = 16/32 = 0.5 mol. Here O₂ is the limiting reagent. CO₂ produced = 0.5 × 44 = 22 g.

Q5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution.

AnswerMolar mass of CH₃COONa = 12 + 3 × 1 + 12 + 16 + 16 + 23 = 82 g mol⁻¹. Moles = M × V(L) = 0.375 × 0.500 = 0.1875 mol. Mass = 0.1875 × 82 = 15.375 g.

Q6. Calculate the concentration of nitric acid in moles per litre if the density of 69% (by mass) nitric acid is 1.41 g mL⁻¹.

AnswerMass of 1 L solution = 1000 × 1.41 = 1410 g. Mass of HNO₃ = 0.69 × 1410 = 972.9 g. Moles of HNO₃ = 972.9 / 63 = 15.445 mol. Concentration = 15.44 M.

Q7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

AnswerMolar mass of CuSO₄ = 63.5 + 32 + 64 = 159.5 g mol⁻¹. Mass of Cu from 100 g = (63.5/159.5) × 100 = 39.81 g.

Q8. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9% and 30.1% respectively. (Given: molar mass = 159.8 g mol⁻¹)

AnswerEmpirical formula (from Q3) = Fe₂O₃. Empirical formula mass = 2 × 55.85 + 3 × 16 = 159.7. n = 159.8 / 159.7 ≈ 1. Molecular formula = Fe₂O₃.

Q9. Calculate the number of atoms in each of the following: (i) 52 moles of Ar, (ii) 52 u of He, (iii) 52 g of He.

Answer

(i) 52 × 6.022 × 10²³ = 3.13 × 10²⁵ atoms
(ii) 52 u of He. Atomic mass of He = 4 u. Number of atoms = 52/4 = 13 atoms
(iii) Moles of He = 52/4 = 13 mol. Atoms = 13 × 6.022 × 10²³ = 7.83 × 10²⁴ atoms

Q10. If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer2H₂ + O₂ → 2H₂O. 10 vol H₂ need 5 vol O₂ (exact stoichiometry). Product = 10 vol H₂O (vapour). Answer: 10 volumes.

Q11. Convert the following into basic units: (i) 28.7 pm, (ii) 15.15 pm, (iii) 25365 mg.

Answer

(i) 28.7 pm = 28.7 × 10⁻¹² m
(ii) 15.15 pm = 15.15 × 10⁻¹² m
(iii) 25365 mg = 25365 × 10⁻⁶ kg = 0.025365 kg

Q12. Which has the maximum number of atoms? (i) 24 g of C, (ii) 56 g of Fe, (iii) 27 g of Al.

AnswerAtoms = (mass / atomic mass) × Nₐ

C: (24/12) × Nₐ = 2Nₐ
Fe: (56/56) × Nₐ = 1Nₐ
Al: (27/27) × Nₐ = 1Nₐ

24 g of C has the maximum number of atoms.

Q13. How many significant figures are present in: (i) 0.0025, (ii) 208, (iii) 5005, (iv) 126.0?

Answer

(i) 0.0025 → 2 significant figures (leading zeros not significant)
(ii) 208 → 3 significant figures
(iii) 5005 → 4 significant figures
(iv) 126.0 → 4 significant figures (trailing zero after decimal is significant)

Q14. Round up the following to three significant figures: (i) 34.216, (ii) 10.4107, (iii) 0.04597.

Answer

(i) 34.216 → 34.2
(ii) 10.4107 → 10.4
(iii) 0.04597 → 0.0460

Q15. Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to make 250 mL of the solution.

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

Q16. The density of 3 M solution of NaCl is 1.25 g mL⁻¹. Calculate the molality of the solution.

Answer3 M means 3 mol NaCl per litre. Mass of NaCl = 3 × 58.5 = 175.5 g. Mass of solution = 1000 × 1.25 = 1250 g. Mass of water = 1250 − 175.5 = 1074.5 g = 1.0745 kg. Molality = 3/1.0745 = 2.79 m.

Q17. Calculate the mass of NaCl required to prepare 250 mL of 0.5 M solution.

AnswerMoles = 0.5 × 0.250 = 0.125 mol. Mass = 0.125 × 58.5 = 7.3125 g.

Q18. In a reaction: A + B₂ → AB₂, identify the limiting reagent if 2 mol of A and 3 mol of B are present.

Answer1 mol A reacts with 1 mol B₂. With 2 mol A, we need 2 mol B₂. Since 3 mol B₂ is available and only 2 is needed, A is entirely consumed first. A is the limiting reagent.

Q19. Calculate the mass of 0.5 mol of: (i) H₂O, (ii) NH₃, (iii) H₂SO₄.

Answer

(i) 0.5 × 18 = 9 g
(ii) 0.5 × 17 = 8.5 g
(iii) 0.5 × 98 = 49 g

Q20. If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions.

AnswerMoles of Na⁺ = 92/23 = 4 mol. Mass of water = 1 kg. Molality = 4/1 = 4 m.

Q21. 18 g of water is electrolysed to give H₂ and O₂. Calculate the mass of O₂ produced.

Answer2H₂O → 2H₂ + O₂. 36 g H₂O gives 32 g O₂. Mass of O₂ = (32/36) × 18 = 8 g.

Q22. A sample of NaNO₃ weighing 0.38 g is placed in a 50 mL volumetric flask and diluted to the mark. What is the molarity?

AnswerMolar mass NaNO₃ = 85 g mol⁻¹. Moles = 0.38/85 = 0.00447 mol. Volume = 0.050 L. Molarity = 0.00447/0.050 = 0.089 M.

Q23. Calculate the number of atoms in: (i) 0.5 mol of H atoms, (ii) 0.5 mol of H₂ molecules.

Answer

(i) 0.5 × 6.022 × 10²³ = 3.011 × 10²³ atoms
(ii) 0.5 mol H₂ = 0.5 × 6.022 × 10²³ molecules = 3.011 × 10²³ molecules. Each H₂ has 2 H atoms. Total atoms = 6.022 × 10²³ atoms

Q24. What is the concentration of sugar (C₁₂H₂₂O₁₁) if a 20 g sugar solution contains 2 g of sugar?

AnswerMass percent = (2/20) × 100 = 10%. If the solution volume is needed, we would need the density. As mass percent: 10% (w/w).

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Frequently Asked Questions

A mole is the amount of substance that contains as many particles (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12. This number, called Avogadro's number, is 6.022 × 10²³. One mole of any substance always contains 6.022 × 10²³ particles.

Some Basic Concepts of Chemistry — NCERT Class 11 Chemistry

How to Use This Page

Key Concepts (Quick Recap)

Intext Questions

Page 1–2 (Importance of Chemistry, Classification of Matter)

Page 5–6 (SI Units, Mole Concept)

Page 12–14 (Molar Mass Calculations, Empirical Formula, Stoichiometry)

Chapter-End Exercise

Internal Links

Frequently Asked Questions

Other question types — same chapter

CBSE Class 11 Chemistry Chapter 1 Notes: Some Basic Concepts of Chemistry

Chemical Bonding and Molecular Structure — NCERT Class 11 Chemistry

Classification of Elements and Periodicity — NCERT Class 11 Chemistry