Sign In Sign Up
← Back to Chemistry

AS & A-Level Chemistry 02 — Moles, formulas and chemical calculations

Public

Independent Deckloop AS Chemistry study material aligned with Cambridge International 9701 (2025–2027). Deck 2 of 18: Moles, formulas and chemical calculations. Original explanations, worked applications and practice. Not affiliated with or endorsed by Cambridge International Education.

Chemistry EN A-Level
35 cards
Study this deck on Deckloop

Preview Cards

A sample of cards from this deck.

Example Explainer

A sample of the AI explainer you can generate for cards in this deck.

The Mole and the Avogadro Constant

The mole is the SI unit\mathrm{SI}\,\text{unit} for the amount of substance. One mole contains exactly 6.02×10236.02 \times 10^{23} elementary entities (which may be atoms, molecules, ions, or electrons); this specific number of particles per mole is known as the Avogadro constant (L)(L). The mass of one mole of a substance in grams is numerically equal to its relative atomic, molecular, or formula mass. Therefore, exactly 12 grams of carbon-12 contains one mole of carbon atoms. The core mathematical relationship linking amount of substance in moles (n)\text{moles}\,(n), mass in grams (m)(m), and molar mass (M)\text{molar}\,\text{mass}\,(M) is n=mMn = \frac{m}{M}. When working with chemical amounts, you must clearly distinguish between moles of molecules and moles of the individual atoms contained within those molecules.

Key points

  • One mole contains exactly 6.02×10236.02 \times 10^{23} elementary particles.
  • The Avogadro constant (L)(L) represents the number of particles per mole (6.02×1023mol−1)\text{per}\,\text{mole}\,(6.02 \times 10^{23} \mathrm{mol}^{-1}).
  • Molar mass (M)\text{Molar}\,\text{mass}\,(M) is the mass per mole of a substance, with units of gmol−1\mathrm{g} \mathrm{mol}^{-1}.
  • Amount of substance (n)=mass (m)/molar mass (M)(n) = \text{mass}\,(m) / \text{molar}\,\text{mass}\,(M).

Worked example

Question

Calculate the total number of oxygen atoms in 22.0g22.0 \mathrm{g} of carbon dioxide, CO2\mathrm{CO_{2}}. (Ar\mathrm{Ar}: C=12.0\mathrm{C} = 12.0, O=16.0\mathrm{O} = 16.0; L=6.02×1023mol−1L = 6.02 \times 10^{23} \mathrm{mol}^{-1})

Solution

1. Mr(CO2)=12.0+(2×16.0)=44.0M_{r} (\mathrm{CO_{2}}) = 12.0 + (2 \times 16.0) = 44.0. Molar mass=44.0gmol−1\text{Molar}\,\text{mass} = 44.0 \mathrm{g} \mathrm{mol}^{-1}

2. Moles of CO2=22.044.0=0.500mol\text{Moles of}\,\mathrm{CO_{2}} = \frac{22.0}{44.0} = 0.500 \mathrm{mol}

3. Number of CO2\text{Number of}\,\mathrm{CO_{2}} molecules = 0.500×6.02×1023=3.01×10230.500 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}

4. Each molecule contains 2 oxygen atoms. Number of O\text{Number of}\,\mathrm{O} atoms = 2×3.01×1023=6.02×10232 \times 3.01 \times 10^{23} = 6.02 \times 10^{23}

6.02×10236.02 \times 10^{23} atoms

Common pitfalls

  • Calculating the moles of molecules but forgetting to multiply by the number of atoms per molecule when asked for total atoms.
  • Confusing the mass of a\text{mass of}\,a single atom with the molar mass. Molar mass is the mass of 6.02×1023\text{mass of}\,6.02 \times 10^{23} atoms.

Prerequisites

  • Study Atomic structure and electron arrangement first.