Introduction
Chemists rarely deal with just one atom or molecule at a time — reactions involve billions upon billions of particles. To make calculations manageable, chemists use the mole, a counting unit just like a "dozen" but much bigger. This lesson introduces the amount of substance (measured in moles) and the key formula you'll use throughout the stoichiometry unit: n = m / M.
The Mole and Avogadro's Constant
- The mole (mol) is the SI unit for amount of substance, symbol n.
- One mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, ions, or formula units). This number is called the Avogadro constant, N_A.
- Whether you have 1 mole of carbon atoms or 1 mole of water molecules, both contain the same number of particles — just different masses.
Molar Mass
- The molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹.
- Numerically, M equals the relative atomic mass (for elements) or relative molecular/formula mass (for compounds), found by adding up atomic masses from the periodic table.
- Example: M(H₂O) = (2 × 1.01) + 16.00 = 18.02 g mol⁻¹
Linking Mass and Moles
The key relationship connecting mass, molar mass, and amount of substance is:
n = m / M
where:
- n = amount of substance (mol)
- m = mass (g)
- M = molar mass (g mol⁻¹)
This simple equation is the foundation for almost every stoichiometry calculation you'll meet, including mass–mass and mass–gas volume problems later in this unit (Talbot, p.424–425).
Worked Example 1 – Finding moles from mass
Question: How many moles are in 9.00 g of water?
Answer:
- M(H₂O) = 18.02 g mol⁻¹
- n = m / M = 9.00 / 18.02 = 0.499 mol
Worked Example 2 – Finding mass from moles
Question: What is the mass of 0.250 mol of calcium carbonate, CaCO₃?
Answer:
- M(CaCO₃) = 40.08 + 12.01 + (3 × 16.00) = 100.09 g mol⁻¹
- m = n × M = 0.250 × 100.09 = 25.0 g
Worked Example 3 – Percentage composition using moles
Understanding moles also lets you calculate percentage composition by mass. For hydrated sodium sulfate, Na₂SO₄·10H₂O (M = 322 g mol⁻¹):
- %Na = (2 × 22.99 / 322) × 100 = 14.28%
- %S = (32.07 / 322) × 100 = 9.96%
- %O (from sulfate only) = (4 × 16.00 / 322) × 100 = 19.88%
- %H₂O = (10 × 18.02 / 322) × 100 = 55.96%
Adding these gives ≈100%, with small rounding differences expected (Talbot, p.96).
Why This Matters for Stoichiometry
Once you can convert between mass and moles, you can move on to full stoichiometry problems — comparing moles of reactants and products using the coefficients in a balanced equation. This is exactly the method used in mass–mass and mass–gas volume calculations covered in upcoming lessons (Talbot, p.424).
Key Takeaways
- The mole is a counting unit equal to 6.02 × 10²³ particles (the Avogadro constant).
- Molar mass (M) is the mass of one mole of a substance, in g mol⁻¹.
- The formula n = m / M connects mass, molar mass, and amount of substance — memorise it!
- This relationship is the starting point for all stoichiometric calculations, including finding reacting masses and gas volumes.
- Percentage composition problems are a direct application of molar mass and mole ratios.