Ever wondered why chemists talk about a "dozen" of atoms but use a completely different number?
💡 In Simple Words: A mole is just a handy way to count huge numbers of tiny particles, like atoms or molecules. One mole always means 6.022×10²³ of them, and stoichiometry tells us how many of each you need to make a chemical reaction happen.
What is a mole?
Think of a mole as a chemist’s version of a dozen. Instead of 12 items, a mole is 6.022×10²³ items – that’s Avogadro's number, named after the scientist who first estimated it. The number is so big that you can’t picture it, but it’s useful because it links the microscopic world (atoms) to the macroscopic world (grams you can weigh).
Key terms
- Avogadro's number – 6.022×10²³, the exact count of particles in one mole.
- Molar mass – the mass (in grams) of one mole of a substance; you get it from the periodic table.
- Balanced chemical equation – a recipe that shows the correct number of each reactant and product, obeying the law of conservation of mass.
Molar mass – the bridge between grams and moles
Take water (H₂O). Hydrogen’s atomic mass is about 1 g mol⁻¹ and oxygen’s is about 16 g mol⁻¹. Add them up: 2×1 + 16 = 18 g mol⁻¹. That means 18 g of water contains exactly one mole (6.022×10²³ molecules) of H₂O.
Stoichiometry – the recipe of reactions
Stoichiometry is just a fancy word for “how much of each thing you need or get in a reaction”. The coefficients in a balanced equation are the mole ratios. For example, in the combustion of methane:
CH₄ + 2 O₂ → CO₂ + 2 H₂O
One mole of methane reacts with two moles of oxygen to give one mole of carbon dioxide and two moles of water.
Step‑by‑step to solve a stoichiometry problem
Here’s a quick recipe you can follow every time:
- Write a balanced chemical equation.
- Convert the given mass (or volume) of a reactant into moles using molar mass.
- Use the mole ratio from the equation to find moles of the desired product.
- Convert those moles back into grams (or liters) if the question asks for it.
Worked example
Problem: How many grams of CO₂ are produced when 50 g of CH₄ combusts completely?
Solution:
- Balanced equation: CH₄ + 2 O₂ → CO₂ + 2 H₂O
- Molar mass of CH₄ = 12 + 4×1 = 16 g mol⁻¹. Moles of CH₄ = 50 g ÷ 16 g mol⁻¹ = 3.125 mol.
- From the equation, 1 mol CH₄ gives 1 mol CO₂. So moles of CO₂ = 3.125 mol.
- Molar mass of CO₂ = 12 + 2×16 = 44 g mol⁻¹. Mass of CO₂ = 3.125 mol × 44 g mol⁻¹ = 137.5 g.
Answer: 137.5 g of carbon dioxide.
Quick comparison table
| Concept | What you do | Typical unit |
|---|---|---|
| Mole | Count particles | mol |
| Molar mass | Convert between grams and moles | g mol⁻¹ |
| Stoichiometric coefficient | Read mole ratios from the equation | unitless |
| Avogadro's number | Number of particles in one mole | 6.022×10²³ |
📝 Likely Exam Questions
- Define the mole and state Avogadro's number.
Answer: A mole is the amount of substance containing exactly 6.022×10²³ elementary entities. - Calculate the number of moles in 25 g of NaCl (Molar mass = 58.5 g mol⁻¹).
Answer: 25 g ÷ 58.5 g mol⁻¹ = 0.428 mol. - Given the reaction 2 Al + 3 Cl₂ → 2 AlCl₃, how many grams of AlCl₃ are formed from 10 g of Al?
Answer: Moles Al = 10 g ÷ 27 g mol⁻¹ = 0.370 mol. Mole ratio Al:AlCl₃ = 2:2, so moles AlCl₃ = 0.370 mol. Mass AlCl₃ = 0.370 mol × 133.5 g mol⁻¹ ≈ 49.4 g. - Explain why a balanced chemical equation is essential for stoichiometric calculations.
Answer: It ensures the mole ratios reflect the true conservation of atoms, allowing accurate conversion between reactants and products. - Convert 3.5×10²⁴ molecules of O₂ to grams. (Molar mass of O₂ = 32 g mol⁻¹)
Answer: Moles = 3.5×10²⁴ ÷ 6.022×10²³ = 5.81 mol. Mass = 5.81 mol × 32 g mol⁻¹ ≈ 186 g.