Mole Concept and Stoichiometry: ICSE Class 10 Chemistry Study Notes

The mole is the bridge between the invisible world of atoms and the measurable world of grams and litres. This study note explains the mole concept and stoichiometry from first principles, with the exact reasoning needed for ICSE numericals. Master these ideas once, and every mole-based problem becomes a simple conversion.

Why Do We Need the Mole?

Atoms and molecules are far too small to count directly, yet chemical reactions depend on exact particle ratios. The mole is the chemist's counting unit: one mole contains 6.023 × 10^23 particles, a number called the Avogadro constant. It is like a 'dozen' for chemistry, except the number is enormous because particles are tiny.

Using moles lets us weigh out a sample and know exactly how many particles it contains. One mole of carbon atoms weighs 12 g, and one mole of oxygen atoms weighs 16 g, but both contain the same number of atoms. The mass differs because each atom has a different mass, not because the count differs.

• Mole: the amount of substance that contains 6.023 × 10^23 particles.
• Avogadro constant: 6.023 × 10^23 per mole.
• Particles: atoms, molecules, ions, or formula units, depending on the substance.

Molar Mass, Gram Atomic Mass and Gram Molecular Mass

The mass of one mole of a substance in grams is its molar mass. For an element, the gram atomic mass is the atomic mass expressed in grams; for example, the atomic mass of oxygen is 16 u, so one mole of oxygen atoms weighs 16 g. For a molecule, the gram molecular mass is the sum of the atomic masses of all atoms in one molecule.

Take water, H2O: two hydrogen atoms contribute 2 × 1 g and one oxygen atom contributes 16 g, so the molar mass is 18 g/mol. This means one mole of water molecules weighs 18 g and contains 6.023 × 10^23 molecules. The molar mass is numerically equal to the relative molecular mass, but the molar mass carries the unit g/mol.

• Gram atomic mass: mass of one mole of atoms of an element.
• Gram molecular mass: mass of one mole of molecules of a compound.
• Molar mass: mass in grams of one mole of a substance; unit g/mol.

The Mole Conversion Roadmap

Most mole problems use three simple conversions. If you know mass, use n = m/M, where m is mass in grams and M is molar mass in g/mol. If you know the number of particles, use n = N/6.023 × 10^23. If the substance is a gas at STP, use n = V/22.4, where V is volume in litres.

Worked example: calculate the moles, number of molecules, and STP volume of 88 g of carbon dioxide. The molar mass of CO2 is 12 + 2(16) = 44 g/mol, so n = 88/44 = 2 mol. The number of molecules is 2 × 6.023 × 10^23 = 1.2046 × 10^24. The volume at STP is 2 × 22.4 = 44.8 L. One mole amount is thus expressed in three different but equivalent ways.

• Mass to moles: n = m/M.
• Particles to moles: n = N/6.023 × 10^23.
• Gas volume at STP to moles: n = V/22.4 L.

Stoichiometry: Balanced Equations as Mole Ratios

A balanced chemical equation tells you the mole ratio of reactants and products, not the mass ratio. In 2H2 + O2 → 2H2O, the coefficients mean 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. To solve a stoichiometry problem, first convert the given quantity to moles, then use the ratio from the equation, and finally convert to the required unit.

Worked example: how many grams of oxygen are needed to burn 4 g of hydrogen? The molar mass of H2 is 2 g/mol, so 4 g gives n = 4/2 = 2 mol H2. From the equation, H2 : O2 = 2 : 1, so oxygen needed is 1 mol. Since the molar mass of O2 is 32 g/mol, the mass required is 32 g. If more than one reactant amount is given, find the limiting reagent: the reactant that produces the least product controls the reaction.

• Mole ratio: coefficients in a balanced equation.
• Mass-mass: convert mass to moles, use ratio, convert back to mass.
• Mass-volume: convert mass to moles, use ratio, then multiply by 22.4 L at STP for gases.

Percentage Composition and Empirical and Molecular Formula

Percentage composition tells you the mass percentage of each element in a compound. For a compound with formula AxBy, the percentage of A is (mass of A in one mole / molar mass of compound) × 100. In water, hydrogen contributes 2 g out of 18 g, so H = 11.11% and O = 88.89%.

The empirical formula is the simplest whole-number ratio of atoms in a compound. To find it from percentage data, assume a 100 g sample, convert each percentage to grams, divide by the atomic mass to get moles, and then divide all mole values by the smallest value. The molecular formula is a whole-number multiple of the empirical formula: n = molar mass / empirical formula mass.

Worked example: a compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen, and its molar mass is 60 g/mol. In 100 g, moles are C = 40/12 = 3.33, H = 6.67/1 = 6.67, O = 53.33/16 = 3.33. Dividing by 3.33 gives C : H : O = 1 : 2 : 1, so the empirical formula is CH2O. Its empirical mass is 12 + 2 + 16 = 30, so n = 60/30 = 2, and the molecular formula is C2H4O2.

• Percentage composition: (mass of element / molar mass) × 100.
• Empirical formula: simplest whole-number ratio of atoms.
• Molecular formula: empirical formula × n, where n = molar mass / empirical formula mass.

Gas Volumes, Avogadro's Law and Vapour Density

At STP, defined as 0°C and 1 atmosphere pressure, one mole of any gas occupies 22.4 litres. This is called molar volume. Therefore, the volume of a gas at STP is simply moles × 22.4 L. Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules; this is why gas coefficients in a balanced equation can also be read as volume ratios, as Gay Lussac's law states.

Vapour density is another useful relation for gases. It is the ratio of the density of a gas to the density of hydrogen gas measured under the same conditions. Because hydrogen molecules are H2 with molecular mass 2, the molecular mass of a gas is twice its vapour density: molecular mass = 2 × vapour density. For example, a gas with vapour density 14 has molecular mass 28, which matches nitrogen, N2.

• Molar volume at STP: 22.4 L per mole.
• Avogadro's law: equal volumes of gases at same T and P contain equal numbers of molecules.
• Vapour density: molecular mass = 2 × vapour density.

Key takeaways

• A mole is 6.023 × 10^23 particles; it connects the microscopic particle count to measurable mass and volume.
• Molar mass in g/mol equals gram atomic or gram molecular mass; use n = m/M for mass-to-mole conversions.
• At STP, one mole of any gas occupies 22.4 L; use n = V/22.4 for gas volume conversions.
• Balanced equation coefficients are mole ratios; convert given data to moles first, then apply the ratio.
• Percentage composition leads to the empirical formula; the molecular formula is a whole-number multiple of the empirical formula.
• For gases, molecular mass = 2 × vapour density, and density at STP = molar mass / 22.4 g/L.

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