Key Takeaways

1. Empirical Formulas

Steps to Calculate:

  1. Find mass of each element:
    • E.g., Magnesium oxide:
      • Mass of Mg = 18.24 g – 16.81 g = 1.43 g
      • Mass of O = 19.16 g – 18.24 g = 0.92 g
  2. Convert mass to moles:
    • Use n=mass (g)molar mass (g/mol)n=molar mass (g/mol)mass (g)​
    • Mg: 1.4324=0.0596 mol241.43​=0.0596mol
    • O: 0.9216=0.0575 mol160.92​=0.0575mol
  3. Simplify the ratio:
    • Divide by the smallest value:
      • Mg: 0.05960.0575≈1.04→10.05750.0596​≈1.04→1
      • O: 0.05750.0575=10.05750.0575​=1
    • Empirical formula: MgO.

Tips:

  • Always check subtraction steps (common error in part c of the example).
  • Round ratios to the nearest whole number (e.g., 1.04 ≈ 1).

2. Writing Chemical Equations

Rules:

  • Balancing charges: For ionic compounds, ensure charges cancel out.
    • E.g., Iron(III) oxide: Fe³⁺ and O²⁻ → Fe₂O₃ (2 × +3 cancels 3 × −2).
  • Balancing atoms: Adjust coefficients to balance elements.
    • E.g., Iron + Oxygen → Iron(III) oxide:
      4Fe+3O2→2Fe2O34Fe+3O2​→2Fe2​O3​

Electrolysis Calculations:

  • Use 1 mole of gas = 24 dm³ at RTP.
    • E.g., 59 cm³ of oxygen:
      n=5924,000=0.00246 moln=24,00059​=0.00246mol
    • Ratio of H₂:O₂ = 2:1 (from water: H₂O).

3. Electronic Configurations and Bonding

Key Rules:

  • Electron shells: Fill as 2, 8, 8 (e.g., chlorine: 2,8,7).
  • Ionic bonding: Transfer of electrons (e.g., Na → Na⁺ + e⁻; Cl + e⁻ → Cl⁻).
  • Covalent bonding: Sharing electrons (e.g., Cl₂: shared pair between two Cl atoms).

Examples:

  • Sodium chloride (NaCl):
    • Na loses 1 electron → Na⁺; Cl gains 1 electron → Cl⁻.
  • Methane (CH₄):
    • Carbon shares 4 electrons with 4 hydrogens.

Common Mistakes:

  • Incorrect electron placement (e.g., putting 9 electrons in Cl’s outer shell).
  • Confusing ionic (transfer) vs. covalent (sharing).

4. Data Analysis: Substance Properties

Identifying Structures:

TypeMelting PointConductivity (Solid)Conductivity (Liquid)Example
MetalHighYesYesSodium (Q)
IonicHighNoYesSodium chloride
CovalentVariableNoNoDiamond (R)

Example:

  • Diamond vs. Graphite:
    • Both have high melting points (giant covalent).
    • Graphite conducts due to delocalised electrons; diamond does not.

5. Mole Calculations

Formulas:

  • Moles from mass: n=mMn=Mm​
  • Moles from gas volume: n=V24,000n=24,000V​ (for cm³ → dm³: ÷1000).

Example:

  • 119 cm³ of hydrogen:
    n=11924,000=0.00496 moln=24,000119​=0.00496mol

Tips:

  • Watch units! Convert cm³ to dm³ by dividing by 1,000.

6. Common Pitfalls

  1. Unit errors:
    • Mixing g/kg or cm³/dm³ (e.g., 59 cm³ = 0.059 dm³).
  2. Rounding too early:
    • Keep decimals until the final ratio step.
  3. Misreading tables:
    • Double-check if conductivity refers to solid/liquid state.

50 GCSE Chemistry Questions

Empirical Formulas & Moles

  1. A student heats magnesium in a crucible. The masses are:
    • Crucible + lid = 16.81 g
    • Crucible + lid + Mg = 18.24 g
    • Crucible + lid + MgO = 19.16 g
      Calculate:
      a) Mass of magnesium used.
      b) Moles of magnesium used.
      c) Mass of oxygen gained.
      d) Moles of oxygen gained.
      e) Empirical formula of magnesium oxide.
  2. A lead oxide (PbOₓ) is heated in hydrogen, producing 12.42 g lead from 14.34 g oxide. Calculate:
    a) Mass of oxygen in the oxide.
    b) Moles of oxygen.
    c) Moles of lead.
    d) Empirical formula of lead oxide.
  3. A hydrocarbon has 14.3% hydrogen and a molecular mass of 28. Calculate:
    a) Moles of hydrogen in 100 g.
    b) Moles of carbon in 100 g.
    c) Empirical formula.
    d) Molecular formula.
  4. Anhydrous copper sulfate contains 12.7 g copper, 6.4 g sulfur, and oxygen. Calculate:
    a) Mass of oxygen.
    b) Moles of oxygen.
    c) Empirical formula.

Chemical Equations & Bonding

  1. Write the formula for:
    a) Iron(III) oxide.
    b) Sodium chloride.
  2. Balance the equation for iron reacting with oxygen to form iron(III) oxide.
  3. Explain, using electron transfer, how sodium bonds with chlorine.
  4. Draw dot-and-cross diagrams for:
    a) Cl₂
    b) NaCl
    c) CH₄
    d) CO₂
  5. State the electronic configuration for:
    a) Sodium (Na)
    b) Fluorine (F)
    c) Sulfur (S)
  6. Why does oxygen gain electrons to form O²⁻?

Electrolysis & Gas Calculations

  1. In water electrolysis, 59 cm³ oxygen and 119 cm³ hydrogen are produced. Calculate:
    a) Moles of oxygen.
    b) Moles of hydrogen.
    c) Ratio of hydrogen to oxygen.
  2. If 360 cm³ of hydrogen is produced at RTP, calculate the moles.

Data Analysis: Substance Properties

  1. Use the table below to answer:SubstanceMelting Point (°C)Conducts (Solid)Conducts (Liquid)P-114NoNoQ714NoYesR3550NoNoS-39YesYesT6NoNoa) Which substance is a metal? Explain.
    b) Which has a giant covalent structure? Explain.
    c) Which is liquid at 20°C?
  2. Compare diamond and graphite:
    a) Why do both have high melting points?
    b) Why does graphite conduct electricity?

Mole Calculations & Unit Conversions

  1. Convert 250 cm³ of gas to moles at RTP.
  2. Calculate moles in 8.5 g of magnesium.

Ionic Charges & Formula Writing

  1. Lithium fluoride has ionic bonding. What are the charges on Li⁺ and F⁻?
  2. Determine the formula for calcium chloride.

Percentage Composition

  1. A compound is 40% carbon, 6.67% hydrogen, and 53.33% oxygen. Its molecular mass is 60. Find its molecular formula.

Error Analysis

  1. A student calculates the empirical formula of MgO as Mg₂O. Identify their mistake.

Detailed Answers

  1. Empirical Formula of Magnesium Oxide
    a) Mass of Mg = 18.24 g – 16.81 g = 1.43 g
    b) Moles of Mg = 1.4324=0.0596 mol241.43​=0.0596mol
    c) Mass of O = 19.16 g – 18.24 g = 0.92 g
    d) Moles of O = 0.9216=0.0575 mol160.92​=0.0575mol
    e) Ratio Mg:O = 0.0596 : 0.0575 ≈ 1:1 → MgO
  2. Lead Oxide
    a) Mass of O = 14.34 g – 12.42 g = 1.92 g
    b) Moles of O = 1.9216=0.12 mol161.92​=0.12mol
    c) Moles of Pb = 12.42207=0.06 mol20712.42​=0.06mol
    d) Ratio Pb:O = 0.06 : 0.12 = 1:2 → PbO₂
  3. Hydrocarbon
    a) Moles of H = 14.31=14.3 mol114.3​=14.3mol
    b) Moles of C = 85.712=7.14 mol1285.7​=7.14mol
    c) Ratio C:H = 7.14 : 14.3 ≈ 1:2 → CH₂
    d) Molecular formula = C₂H₄ (mass 28).
  4. Chemical Formulas
    a) Fe₂O₃ (Fe³⁺ + O²⁻)
    b) NaCl (Na⁺ + Cl⁻)
  5. Electrolysis of Water
    a) Moles of O₂ = 5924,000=0.00246 mol24,00059​=0.00246mol
    b) Moles of H₂ = 11924,000=0.00496 mol24,000119​=0.00496mol
    c) Ratio H₂:O₂ = 2:1
  6. Data Analysis
    a) S (conducts as solid and liquid → metal).
    b) R (high melting point, no conductivity → covalent).
    c) T (melting point <20°C, boiling point >20°C).
  7. Diamond vs. Graphite
    a) Both have giant covalent structures with strong bonds.
    b) Graphite has delocalised electrons between layers.