ZIMSEC A Level · Chemistry · Form 5
Key relationships
- n = m/M; n = c × V (V in dm³); gas volume = n × 24.0 dm³ at r.t.p.; ideal gas equation pV = nRT (R = 8.31 J K⁻¹ mol⁻¹, p in Pa, V in m³, T in K).
- Avogadro constant L = 6.02 × 10²³ mol⁻¹.
Empirical and molecular formulae
Worked example: A compound contains 40.0% C, 6.7% H and 53.3% O by mass; Mr = 60. Find its molecular formula.
C: 40.0/12 = 3.33; H: 6.7/1 = 6.7; O: 53.3/16 = 3.33 → ratio 1 : 2 : 1 → empirical formula CH₂O (Mr 30)
Molecular formula = 2 × CH₂O = C₂H₄O₂
Titrations
Worked example: 25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.150 mol dm⁻³ H₂SO₄. Find the concentration of NaOH.
n(H₂SO₄) = 0.150 × 0.0200 = 3.00 × 10⁻³ mol; ratio 2NaOH : 1H₂SO₄ → n(NaOH) = 6.00 × 10⁻³ mol
c = 6.00 × 10⁻³ ÷ 0.0250 = 0.240 mol dm⁻³
Yield and atom economy
- Percentage yield = actual ÷ theoretical × 100.
- Atom economy = Mr of desired product ÷ total Mr of reactants × 100.
Limiting reagents
The reactant that runs out first determines the amount of product.
Exam tip: Convert cm³ to dm³ (÷ 1000) and use the mole ratio from the balanced equation every time.
Check yourself
- How many moles are in 250 cm³ of 0.400 mol dm⁻³ HCl?
- Find the volume of 0.0500 mol of gas at r.t.p.
- Calculate the number of molecules in 3.6 g of water.
Show answers
- 0.100 mol
- 1.20 dm³
- 0.200 mol × 6.02 × 10²³ = 1.20 × 10²³
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