⚖️ Relative Atomic Mass (Ar) & Relative Formula Mass (Mr)
Relative Atomic Mass (Ar)
- Average mass of an atom compared to 1/12 of a carbon-12 atom.
- Found on the periodic table.
- Example: H → 1, O → 16, C → 12
Relative Formula Mass (Mr)
- Sum of the relative atomic masses of all atoms in a compound.
- Example: H₂O → (2 × 1) + (1 × 16) = 18
- Example: CO₂ → (1 × 12) + (2 × 16) = 44
💡 Tip: Mr is used for compounds, Ar is used for single elements.
🧪 Moles – The Chemist’s Dozen
1 mole = 6.02 × 10²³ particles (Avogadro’s constant).
- Can be atoms, molecules, ions, or electrons.
Mole Calculations:
- Moles = mass ÷ Mr
- Moles = concentration × volume (for solutions)
📦 Example 1: How many moles are in 36 g of water?
- Mr H₂O = 18 → Moles = 36 ÷ 18 = 2 mol
📦 Example 2: 0.5 moles of NaCl → Mass = ?
- Mr NaCl = 58.5 → Mass = 0.5 × 58.5 = 29.25 g
🔢 Chemical Equations & Stoichiometry
Balanced chemical equations tell you the ratio of moles of each substance.
Example:
2H₂ + O₂ → 2H₂O
- 2 moles H₂ react with 1 mole O₂ to produce 2 moles H₂O
- Ratios are important for calculating masses and volumes.
💡 Tip: Always balance equations first before any calculations.
⚗️ Conservation of Mass
- Total mass of reactants = total mass of products (atoms are not lost or created).
- Law is true except for nuclear reactions, where tiny amounts of mass convert to energy.
Example:
Mg + 2HCl → MgCl₂ + H₂
- Mass of Mg + HCl = Mass of MgCl₂ + H₂
📏 Limiting Reactants
- When one reactant runs out, the reaction stops.
- Other reactants are in excess.
- Calculations often involve:
- Moles of each reactant
- Mole ratio from the balanced equation
- Determine which is limiting
Example:
- 4 moles H₂ + 2 moles O₂ → 4 moles H₂O
- If 3 moles H₂ + 2 moles O₂ → Only 2 moles H₂O formed
- H₂ is the limiting reactant
💧 Concentration of Solutions
Concentration = Amount of solute ÷ Volume of solution
- Concentration (mol/dm³) = moles ÷ volume (dm³)
- Remember to convert cm³ → dm³ (1000 cm³ = 1 dm³)
Example:
- 0.5 moles of NaOH in 250 cm³ solution
- Volume in dm³ = 250 ÷ 1000 = 0.25 dm³
- Concentration = 0.5 ÷ 0.25 = 2 mol/dm³
🔥 Percentage Yield
Percentage yield = (actual yield ÷ theoretical yield) × 100
- Theoretical yield: Maximum amount of product possible (from calculations).
- Actual yield: Amount you actually get in the lab.
💡 Reasons yield is <100%:
- Side reactions
- Loss during transfer
- Impurities in reactants
Example:
- Theoretical yield = 10 g, actual yield = 8 g → % yield = 8 ÷ 10 × 100 = 80%
💨 Concentration of Gases (using volume)
- At room temperature and pressure: 1 mole of any gas = 24 dm³
- Moles of gas = Volume ÷ 24 dm³
Example:
- 48 dm³ of oxygen → Moles = 48 ÷ 24 = 2 mol
🧱 Atom Economy
Atom economy = (Mr of desired product ÷ Mr of all reactants) × 100
- Measures how much of the reactants end up in the product
- High atom economy → greener, less waste
Example:
- N₂ + 3H₂ → 2NH₃
- Mr NH₃ = 17, Mr reactants = 28 + 6 = 34
- Atom economy = (34 ÷ 34) × 100 = 100% (all atoms used in product)
💡 Tip: Used in industry to select sustainable reactions.
