Mole Concept and Stoichiometry ICSE Class 10: Laws, Formulae and Numericals

What Mole Concept and Stoichiometry is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Chemistry. Mole Concept and Stoichiometry is in the 5. Mole Concept and Stoichiometry part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Chemistry (Science Paper 2) syllabus, examination year 2027

Key concepts

Gay-Lussac's law of combining volumes

Statement: when gases react, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided all volumes are measured at the same temperature and pressure.

The ratio is simply the ratio of the coefficients in the balanced equation (for gases only). For $C{H}_4+2O_2\to C{O}_2+2H_{2}O$, the volume ratio CH₄ : O₂ : CO₂ is 1 : 2 : 1. Water is a liquid at room temperature, so its volume is ignored unless the question says steam.

When one gas is in excess, find which one runs out first, then add the leftover gas to the gaseous products for the total.

Worked example

30 cm³ of methane is burnt in 80 cm³ of oxygen: $C{H}_4+2O_2\to C{O}_2+2H_{2}O$. Find the volume of oxygen left and the total volume of gases after cooling to room temperature (all at the same temperature and pressure).

Try it first, then show the solutionSolution

1. 1 vol CH₄ needs 2 vol O₂, so 30 cm³ needs 60 cm³ O₂.
2. O₂ left = 80 - 60 = 20 cm³.
3. CO₂ formed = 30 cm³ (1 : 1). Water condenses to liquid.
4. Total gas = 20 + 30 = 50 cm³.

Answer: 20 cm³ of oxygen left; 50 cm³ of gas in total.

Avogadro's law and atomicity

Avogadro's law: equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecules.

So volume ratios are also molecule ratios. This explains why elementary gases are diatomic (atomicity 2). Take $H_2+C{l}_2\to 2HCl$: experimentally, 1 volume of hydrogen + 1 volume of chlorine gives 2 volumes of hydrogen chloride. If 1 volume holds n molecules, then n hydrogen molecules give 2n HCl molecules; every HCl molecule needs some hydrogen, so each hydrogen molecule must split into two atoms: hydrogen is H₂. The same reasoning applies to chlorine, and to nitrogen and oxygen with $N_2+O_2\to 2NO$ and $N_2+3H_2\to 2N{H}_3$. (The syllabus says this can be explained but no proof is required.)

Worked example

Two identical flasks at the same temperature and pressure contain hydrogen and carbon dioxide. Which contains more molecules, and which is heavier?

Try it first, then show the solutionSolution

1. Equal volumes at the same temperature and pressure contain equal numbers of molecules (Avogadro).
2. Each CO₂ molecule (44) is heavier than each H₂ molecule (2).

Answer: Same number of molecules; the CO₂ flask is heavier.

The mole: mass, number and molar volume

• Atomic mass refers to one atom; gram atomic mass is the mass in grams of one mole of atoms (1 gram atom). Example: 1 gram atom of carbon = 12 g.
• Molecular mass refers to one molecule; gram molecular mass is the mass of one mole of molecules (1 gram molecule). Example: CO₂ = 44 g.
• Molar volume: one mole of any gas at S.T.P. occupies 22.4 L (22,400 cm³). So the mass of 22.4 L of any gas at S.T.P. equals its gram molecular mass.

The three conversions:

$n=\frac{\text{mass}}{\text{molar mass}}=\frac{\text{volume at STP}}{22.4L}=\frac{\text{number of particles}}{6.022\times 10^{23}}$

Worked example

Find the number of molecules in 8.8 g of carbon dioxide, and the volume of 3.2 g of oxygen at S.T.P. (C = 12, O = 16)

Try it first, then show the solutionSolution

1. Molar mass of CO₂ = 12 + 32 = 44 g; $n=\frac{8.8}{44}=0.2$ mol.
2. Molecules = $0.2\times 6.022\times 10^{23}=1.2044\times 10^{23}$.
3. Molar mass of O₂ = 32 g; $n=\frac{3.2}{32}=0.1$ mol.
4. Volume = $0.1\times 22.4=2.24$ L.

Answer: 1.2044 × 10²³ molecules; 2.24 L.

Vapour density and molecular mass

Vapour density (VD) of a gas is the ratio of the mass of a certain volume of the gas to the mass of the same volume of hydrogen, at the same temperature and pressure.

Because equal volumes contain equal numbers of molecules and hydrogen is diatomic (molecular mass 2):

$\text{Molecular mass}=2\times \text{Vapour density}$

(No formal proof is needed for the 2027 syllabus.) VD has no unit. This lets you find the molecular mass of a gas from a mass comparison with hydrogen, and then the molecular formula.

Worked example

A cylinder holds 12 g of hydrogen. An identical cylinder at the same temperature and pressure holds 384 g of gas X. Find the molecular mass and vapour density of X.

Try it first, then show the solutionSolution

1. Moles of H₂ = $\frac{12}{2}=6$.
2. Same volume, temperature and pressure: X also contains 6 mol (Avogadro).
3. Molecular mass of X = $\frac{384}{6}=64$.
4. VD = $\frac{64}{2}=32$.

Answer: Molecular mass 64; vapour density 32.

Percentage composition, empirical and molecular formula

Percentage of an element = $\frac{\text{mass of element in 1 mol}}{\text{molar mass}}\times 100$.

The empirical formula shows the simplest whole-number ratio of atoms; the molecular formula shows the actual number of atoms in a molecule. Molecular formula = n × empirical formula, where $n=\frac{\text{molecular mass}}{\text{empirical formula mass}}$.

Method (use a table): element | % | % ÷ atomic mass (relative moles) | ÷ smallest | whole-number ratio. If a ratio comes out like 1.5, multiply all by 2.

Worked example

A compound contains C 40%, H 6.67% and O 53.33%. Its molecular mass is 180. Find its empirical and molecular formulae.

Try it first, then show the solutionSolution

1. C: 40 ÷ 12 = 3.33; H: 6.67 ÷ 1 = 6.67; O: 53.33 ÷ 16 = 3.33.
2. Divide by the smallest (3.33): C 1, H 2, O 1. Empirical formula CH₂O.
3. Empirical formula mass = 12 + 2 + 16 = 30.
4. n = 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆.

Answer: Empirical formula CH₂O; molecular formula C₆H₁₂O₆.

Calculations from chemical equations

A balanced equation tells you the mole ratio, and from it the mass ratio and (for gases) the volume ratio.

Method: (1) Write the balanced equation. (2) Convert the given quantity to moles. (3) Use the mole ratio. (4) Convert to mass (× molar mass), volume at S.T.P. (× 22.4 L) or number (× 6.022 × 10²³).

For $CaC{O}_3\to CaO+C{O}_2$: 1 mol (100 g) of CaCO₃ gives 1 mol (56 g) of CaO and 1 mol (22.4 L at S.T.P.) of CO₂.

Worked example

Find the mass of potassium chlorate needed to produce 6.72 L of oxygen at S.T.P.: $2KCl{O}_3\to 2KCl+3O_2$. (K = 39, Cl = 35.5, O = 16)

Try it first, then show the solutionSolution

1. Moles of O₂ = $\frac{6.72}{22.4}=0.3$.
2. Mole ratio KClO₃ : O₂ = 2 : 3, so KClO₃ = $0.3\times \frac{2}{3}=0.2$ mol.
3. Molar mass of KClO₃ = 39 + 35.5 + 48 = 122.5 g.
4. Mass = $0.2\times 122.5=24.5$ g.

Answer: 24.5 g

Chemistry in real life: Mole Concept and Stoichiometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Calculations from Chemical Equations
• Mole, Avogadro Number and Molar Volume
• Empirical and Molecular Formula

Mole Concept and Stoichiometry formula sheet

Moles from mass

$n=\frac{m}{M}$

M = gram molecular (or atomic) mass.

Moles from gas volume at S.T.P.

$n=\frac{V}{22.4L}$

22.4 L = 22,400 cm³.

Moles from number of particles

$n=\frac{N}{6.022\times 10^{23}}$

Use the value given in the question if it differs.

Vapour density

$M_r=2\times VD$

No unit for VD.

Percentage of an element

$\%X=\frac{\text{mass of X in 1 mol}}{M}\times 100$

Add all percentages to check they make 100.

Molecular formula

$n=\frac{\text{molecular mass}}{\text{empirical formula mass}}$

Molecular formula = n × empirical formula.

Mole Concept and Stoichiometry: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Mole Concept and Stoichiometry

The question patterns to expect, and the method that works for each.

Tricks to remember Mole Concept and Stoichiometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

• The mole bridge

  Every road goes through MOLES: grams ÷ M, litres ÷ 22.4, particles ÷ 6.022 × 10²³. Cross the bridge, use the ratio, then cross back.

  Use it when: Every numerical.

• VD doubles to M

  Vapour Density is half the molecular mass because hydrogen's molecule weighs 2.

  Use it when: VD questions.

• Percent → divide → divide → double if needed

  % ÷ atomic mass, ÷ smallest, then multiply all by 2 if you see .5.

  Use it when: Empirical formula tables.

• Gay-Lussac counts only gases

  Coefficients of gases = volumes; cross out water (l) and solids before adding up.

  Use it when: Total-volume questions.

Mole Concept and Stoichiometry: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 2 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Mole Concept and Stoichiometry

The slips examiners see most often, and what to write instead.

• ✗Mistake: Final answer with no working

  CISCE examiners' comments, 2024 and 2025: candidates wrote final answers such as 500 g or 98 g without the calculation, and lost the method marks.

  Wrong working

  Mass of CaCO₃ = 500 g

  Right working

  Moles of CaCO₃ = 5 (from the equation); mass = 5 × 100 = 500 g

  ✓Do this: Write each conversion as a separate line with units.

  Usually costs: Up to all the marks of the part.

• ✗Mistake: Dividing atomic mass by percentage

  CISCE examiners' comments, 2025: in an empirical formula question, candidates divided atomic weights by percentages, did not divide by the smallest number, and did not multiply the empirical formula by n.

  Wrong working

  C: 12 ÷ 47.4

  Right working

  C: 47.4 ÷ 12 = 3.95

  ✓Do this: Always % ÷ atomic mass. Then ÷ smallest. Then × n.

  Usually costs: 2-3 marks.

• ✗Mistake: Using 22.4 L when there are several moles

  CISCE examiners' comments, 2024: candidates wrote 22.4 L for carbon dioxide instead of 5 × 22.4 L.

  Wrong working

  Volume of CO₂ = 22.4 L

  Right working

  Volume of CO₂ = 5 × 22.4 = 112 L

  ✓Do this: 22.4 L is one mole; multiply by the number of moles.

  Usually costs: 1 mark.

• ✗Mistake: Not finding the total volume left

  CISCE examiners' comments, 2025: in a methane-oxygen question, candidates calculated the gases formed but not the total volume remaining (there, oxygen ran out and the unused methane had to be added).

  Wrong working

  Volume of gas after reaction = volume of CO₂ only.

  Right working

  Total = CO₂ formed + whichever gas is left over in excess, methane or oxygen (water is liquid).

  ✓Do this: List every gas present after the reaction before adding.

  Usually costs: 1 mark.

• ✗Mistake: Incomplete definitions

  CISCE examiners' comments, 2024 and 2025: candidates omitted 'at S.T.P.' and 'gas' from molar volume, and 'under the same temperature and pressure' from Avogadro's law.

  Wrong working

  Avogadro's law: equal volumes of gases contain equal numbers of molecules.

  Right working

  Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.

  ✓Do this: Every gas law needs its conditions.

  Usually costs: 1 mark.

• ✗Mistake: Confusing VD with molecular mass

  CISCE examiners' comments, 2025: in a cylinder question comparing hydrogen with gas G, candidates confused the vapour density and molecular mass relationship.

  Wrong working

  A flask holds 2 g of hydrogen or 32 g of G: M of G = 32 ÷ 2 = 16

  Right working

  Mass ratio with hydrogen = VD = 16; M of G = 2 × 16 = 32 (2 g H₂ is 1 mol, so 32 g of G is also 1 mol).

  ✓Do this: Mass ratio with hydrogen = VD; M = 2 × VD.

  Usually costs: 1-2 marks.

Practice questions on Mole Concept and Stoichiometry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The number of moles in 4.9 g of sulphuric acid (H = 1, S = 32, O = 16) is:
   (a) 0.5
   (b) 0.05
   (c) 0.1
   (d) 0.01

   Hint

   M = 98.

   Answer

   (b) 0.05

2. Q2 · 1 mark · MCQBoard level

   The vapour density of sulphur dioxide (S = 32, O = 16) is:
   (a) 64
   (b) 16
   (c) 32
   (d) 48

   Hint

   Half of 64.

   Answer

   (c) 32

3. Q3 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): 1 g of hydrogen and 16 g of oxygen occupy the same volume at S.T.P.
   Reason (R): Both contain 0.5 mol of molecules.
   (a) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (b) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
   (c) (A) is true and (R) is false.
   (d) (A) is false and (R) is true.

   Hint

   1 ÷ 2 and 16 ÷ 32.

   Answer

   (a): each is 0.5 mol, so each occupies 11.2 L.

4. Q4 · 2 marks · Short answerBoard level

   Calculate the number of atoms in 0.5 mol of ammonia. (Avogadro's number 6.022 × 10²³)

   Hint

   4 atoms per molecule.

   Answer

   0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules; × 4 = 1.2044 × 10²⁴ atoms.

5. Q5 · 2 marks · Short answerBoard level

   Find the volume occupied at S.T.P. by 3.2 g of oxygen and the mass of 5.6 L of ammonia. (O = 16, N = 14, H = 1)

   Hint

   Through moles.

   Answer

   0.1 mol × 22.4 = 2.24 L; 0.25 mol × 17 = 4.25 g.

6. Q6 · 3 marks · Short answerHarder board level

   An oxide of iron contains 70% iron. Find its empirical formula. (Fe = 56, O = 16)

   Hint

   O by difference; ratio 1 : 1.5.

   Answer

   Fe 70 ÷ 56 = 1.25; O 30 ÷ 16 = 1.875; ratio 1 : 1.5 = 2 : 3; Fe₂O₃.

7. Q7 · 2 marks · Short answerHarder board level

   50 cm³ of carbon monoxide burns in 50 cm³ of oxygen: $2CO+O_2\to 2C{O}_2$. Find the composition and total volume of the gas mixture afterwards (same temperature and pressure).

   Hint

   Oxygen in excess.

   Answer

   50 cm³ CO₂ + 25 cm³ O₂ = 75 cm³.

8. Q8 · 2 marks · Short answerBoard level

   Calculate the percentage of nitrogen in ammonium sulphate, (NH₄)₂SO₄. (N = 14, H = 1, S = 32, O = 16)

   Hint

   M = 132.

   Answer

   28 ÷ 132 × 100 = 21.21%

9. Q9 · 3 marks · Short answerHOTS / competency

   An identical flask holds 0.4 g of hydrogen or 12.8 g of gas Y at the same temperature and pressure. Find the vapour density and molecular mass of Y. If Y contains sulphur and oxygen only and one sulphur atom per molecule, suggest its formula. (S = 32, O = 16)

   Hint

   VD = mass ratio with hydrogen.

   Answer

   VD = 12.8 ÷ 0.4 = 32; M = 64; 64 - 32 = 32 = two O: SO₂.

10. Q10 · 3 marks · Short answerHarder board level

    $2KCl{O}_3\to 2KCl+3O_2$. What mass of potassium chloride remains when enough KClO₃ is heated to give 6.72 L of oxygen at S.T.P.? (K = 39, Cl = 35.5)

    Hint

    KCl : O₂ = 2 : 3.

    Answer

    O₂ = 0.3 mol; KCl = 0.2 mol × 74.5 = 14.9 g.

11. Q11 · 3 marks · Short answerHOTS / competency

    A gaseous hydrocarbon X needs 3 volumes of oxygen to burn 1 volume of X completely and gives 2 volumes of CO₂ (same temperature and pressure). Use Gay-Lussac's law to find the formula of X and write the balanced equation.

    Hint

    2 C atoms; O left after CO₂ goes into water.

    Answer

    2 vol CO₂ means 2 C atoms. 3 O₂ = 6 O atoms; 4 used in 2CO₂, 2 left form 2H₂O, so 4 H atoms: C₂H₄. $C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$

12. Q12 · 2 marks · Short answerHarder board level

    Write the gram molecular mass of each and state which contains the greater number of molecules: 22 g of CO₂ or 16 g of O₂. (C = 12, O = 16)

    Hint

    Compare moles.

    Answer

    CO₂ 44 g, O₂ 32 g; 22 g CO₂ = 0.5 mol; 16 g O₂ = 0.5 mol. Equal numbers of molecules.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State Avogadro's law.Show answer

   Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.

2. 2.State Gay-Lussac's law.Show answer

   Gases react in volumes that bear a simple whole-number ratio to each other and to the volumes of gaseous products, at the same temperature and pressure.

3. 3.Value of Avogadro's number?Show answer

   6.022 × 10²³.

4. 4.Molar volume at S.T.P.?Show answer

   22.4 L (22,400 cm³).

5. 5.Relation between molecular mass and VD?Show answer

   Molecular mass = 2 × vapour density.

6. 6.Define empirical formula.Show answer

   The formula showing the simplest whole-number ratio of atoms of each element in a compound.

7. 7.Gram molecular mass of H₂SO₄?Show answer

   98 g.

8. 8.Atomicity of nitrogen?Show answer

   2 (diatomic).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state Avogadro's and Gay-Lussac's laws with all conditions?
• ☐Can you convert grams to moles to litres to molecules?
• ☐Can you find which gas is in excess and the total volume afterwards?
• ☐Can you find an empirical formula when a ratio comes out as 1.5?
• ☐Can you use VD to get the molecular formula?
• ☐Do you show units and each step in every numerical?

Mole Concept and Stoichiometry: frequently asked questions