Mole Concept and Stoichiometry Solved Examples (Class 11 Chemistry)
The mole concept is fundamental to chemistry, relating the number of particles to measurable quantities like mass and volume. These examples demonstrate mo
TL;DR: The mole concept is fundamental to chemistry, relating the number of particles to measurable quantities like mass and volume. These examples demonstra…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
The mole concept is fundamental to chemistry, relating the number of particles to measurable quantities like mass and volume. These examples demonstrate mo
Mole Concept and Stoichiometry — Solved Numerical Examples (Step by Step)
Example 1: Calculate the molar mass of calcium carbonate (CaCO3). Atomic masses: Ca = 40, C = 12, O = 16
Solution: CaCO3 contains: 1 Ca atom, 1 C atom, 3 O atoms
Molar mass = (1 × 40) + (1 × 12) + (3 × 16)
Molar mass = 40 + 12 + 48 = 100 g/mol
Example 2: How many moles are present in 25 g of NaCl? (Na = 23, Cl = 35.5)
Solution: Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Number of moles = mass / molar mass
Number of moles = 25 / 58.5 = 0.427 mol
Example 3: How many atoms are present in 4 g of oxygen gas (O2)? (Atomic mass of O = 16, Avogadro's number = 6.022 × 10²³)
Solution: Molar mass of O2 = 2 × 16 = 32 g/mol
Number of moles = mass / molar mass = 4 / 32 = 0.125 mol
Number of molecules = moles × Avogadro's number
= 0.125 × 6.022 × 10²³ = 7.53 × 10²² molecules
Number of atoms = Number of molecules × 2 atoms per molecule
= 7.53 × 10²² × 2 = 1.506 × 10²³ atoms
Example 4: In the reaction 2H2 + O2 → 2H2O, how many grams of H2O are produced from 8 g of H2? (H = 1, O = 16)
Solution: Molar mass of H2 = 2 × 1 = 2 g/mol
Molar mass of H2O = 2 + 16 = 18 g/mol
Number of moles of H2 = 8 / 2 = 4 mol
From the balanced equation: 2 mol H2 produces 2 mol H2O
So 4 mol H2 produces 4 mol H2O
Mass of H2O = 4 × 18 = 72 g
Example 5: Calculate the percentage composition of nitrogen in ammonium nitrate (NH4NO3). (H = 1, N = 14, O = 16)
Solution: Molar mass of NH4NO3 = 14 + 4(1) + 14 + 3(16)
= 14 + 4 + 14 + 48 = 80 g/mol
Total mass of nitrogen = 14 + 14 = 28 g (2 N atoms)
Percentage of nitrogen = (28 / 80) × 100
= 35%
Example 6: What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen? (C = 12, H = 1, O = 16)
Solution: Assume 100 g of compound:
C: 40 g ÷ 12 = 3.33 mol
H: 6.7 g ÷ 1 = 6.7 mol
O: 53.3 g ÷ 16 = 3.33 mol
Divide by smallest number (3.33):
C: 3.33 ÷ 3.33 = 1
H: 6.7 ÷ 3.33 = 2
O: 3.33 ÷ 3.33 = 1
Empirical formula = CH2O
Tips
- Always identify the molar mass correctly by summing the atomic masses of all atoms in the formula.
- Use the mole ratio from the balanced chemical equation to convert between reactants and products in stoichiometry problems.
- For percentage composition, divide the total mass of the element by the molar mass of the compound and multiply by 100.
- In empirical formula problems, convert mass percentages to moles, then find the simplest whole number ratio.
Frequently Asked Questions
What is the difference between molar mass and molecular mass?
Molecular mass (or molecular weight) is the mass of a single molecule expressed in atomic mass units (amu). Molar mass is the mass of one mole of that substance expressed in grams per mole (g/mol). Numerically they are the same, but the units and scale differ.
Why is Avogadro's number exactly 6.022 × 10²³?
Avogadro's number is defined such that exactly 12 g of carbon-12 contains one mole of atoms. Since the atomic mass unit is defined relative to carbon-12, this number naturally emerges from that definition. It bridges the atomic scale and the macroscopic scale.
More Chemistry Solved Examples
- Atomic Structure (Numericals)
- Electrochemistry (Numericals)
- Chemical Equations and Balancing
- Thermodynamics
- Solutions - Previous Year Questions
- Mole Concept and Stoichiometry
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