Quantitative calculations
Domain 7 - arvutused - rewards a fixed routine: convert units, find moles, use the equation ratio, convert back. Marks are lost on units and on unbalanced equations, not on arithmetic.
Core formulas
- n = m/M (mol), with M summed from the periodic table: M(H₂O) = 2×1 + 16 = 18 g/mol.
- Concentration c = n/V, with V in dm³ (1 dm³ = 1000 cm³ = 1 L).
- Gas amount from volume: n = V/22.4 dm³/mol at 0 °C and 101.3 kPa. Some tables use 24.0 dm³/mol at 25 °C - always use the value printed in the task.
- Mass fraction w = m(solute)/m(solution) × 100 %; particle number N = n × 6.02×10²³.
- Dilution: c₁V₁ = c₂V₂, because the amount of solute stays the same while the volume grows.
Stoichiometry routine
- Write and balance the equation.
- Convert the given data to moles.
- Use the coefficient ratio to get moles of the wanted substance.
- Convert to mass, volume or concentration, and state the unit.
Worked example
4.6 g of sodium reacts completely with water. What volume of hydrogen (at 0 °C, 101.3 kPa) forms? 2Na + 2H₂O → 2NaOH + H₂. n(Na) = 4.6 / 23 = 0.20 mol. Ratio Na : H₂ = 2 : 1, so n(H₂) = 0.10 mol. V = 0.10 × 22.4 = 2.24 dm³ (2240 cm³). Note the ratio step: taking n(H₂) = n(Na) would give 4.48 dm³ - a classic error caused by skipping the coefficients.
Limiting reactant and yield
- The reactant that runs out first limits the product; compare available moles divided by the coefficients.
- Percentage yield = actual mass / theoretical mass × 100 %.
Traps that cost marks
- Using cm³ in c = n/V without converting to dm³.
- Taking the ratio from the formulas instead of the balanced equation.
- Using 22.4 dm³/mol at room temperature - check the stated conditions.
- Forgetting that the molar mass of Cl₂ is 71 g/mol, not 35.5 g/mol.