In the Milliy sertifikat chemistry exam (45 tasks, 150 minutes) the open tasks 36-45 are almost all calculations: one condition, a) worth 1.5 points and b) worth 1.7, with a number as the answer. Most rest on three ideas: amount of substance (n = m / M), mass fraction (w = m(solute) / m(solution)) and the coefficient ratio in the reaction equation. The exam is in Uzbek; below each problem type is fully worked, with every calculation checked step by step.
Formulas and molar masses you need
Problems usually use rounded atomic masses: H 1, C 12, N 14, O 16, Na 23, Mg 24, S 32, Cl 35.5, Ca 40, Fe 56, Zn 65. If the condition gives another value, use that.
| Quantity | Formula |
|---|---|
| Amount of substance | n = m / M |
| Gas volume (STP) | V = n · 22.4 L |
| Number of particles | N = n · 6.02 · 10²³ |
| Mass fraction | w = m(solute) / m(solution) · 100% |
| Molar concentration | C = n / V(solution, L) |
| Yield | η = actual mass / theoretical mass · 100% |
| Relative density | M = D · M(reference gas), e.g. M = D(H₂) · 2 |
1. Amount of substance and particles
Condition: 4.9 g of sulfuric acid (H₂SO₄). a) Find the amount of substance. b) Find the amount of oxygen atoms in it.
Solution: M(H₂SO₄) = 2 · 1 + 32 + 4 · 16 = 98 g/mol. n = 4.9 : 98 = 0.05 mol. Each molecule has 4 O atoms: 4 · 0.05 = 0.2 mol.
Answer: a) 0.05; b) 0.2. If the number of molecules is asked: 0.05 · 6.02 · 10²³ = 3.01 · 10²².
2. Gas volume at STP
Condition: 11.2 L of carbon dioxide (CO₂) at STP. a) Find the amount of substance. b) Find the mass.
Solution: n = 11.2 : 22.4 = 0.5 mol. M(CO₂) = 12 + 2 · 16 = 44 g/mol, m = 0.5 · 44 = 22 g.
Answer: a) 0.5; b) 22. From density: if the gas's density relative to hydrogen is 22, M = 22 · 2 = 44 g/mol - that matches CO₂.
3. Mass fraction, mixing and dilution
- Making a solution: 20 g NaCl dissolved in 180 g water. Solution 20 + 180 = 200 g, w = 20 : 200 = 10%. (20 : 180 = 11.1% is the classic mistake.)
- Mixing: 200 g of a 10% and 300 g of a 20% solution. Salt: 200 · 0.1 + 300 · 0.2 = 20 + 60 = 80 g. Solution 500 g. w = 80 : 500 = 16%.
- Dilution: how much water turns 150 g of a 20% solution into 12%? The salt, 150 · 0.2 = 30 g, stays. New solution 30 : 0.12 = 250 g. Water: 250 - 150 = 100 g.
- Crystal hydrate: 24.6 g MgSO₄·7H₂O dissolved to make 200 g of solution. M = 24 + 32 + 64 + 7 · 18 = 246 g/mol, n = 0.1 mol, MgSO₄: 0.1 · 120 = 12 g. w = 12 : 200 = 6%.
4. Molar concentration
Condition: 8 g of sodium hydroxide (NaOH) dissolved to make 500 mL of solution. a) Find the amount of NaOH. b) Find the molar concentration (mol/L).
Solution: M(NaOH) = 23 + 16 + 1 = 40 g/mol, n = 8 : 40 = 0.2 mol. V = 500 mL = 0.5 L. C = 0.2 : 0.5 = 0.4 mol/L.
Answer: a) 0.2; b) 0.4. The usual mistake is leaving the volume in mL (0.2 : 500).
5. A reaction-equation problem
Condition: 13 g of zinc reacted with excess hydrochloric acid. a) Find the volume of hydrogen (L, STP). b) Find the mass (g) of 7.3% hydrochloric acid solution used.
Solution: Zn + 2HCl → ZnCl₂ + H₂. n(Zn) = 13 : 65 = 0.2 mol. n(H₂) = n(Zn) = 0.2 mol, V = 0.2 · 22.4 = 4.48 L. n(HCl) = 2 · 0.2 = 0.4 mol, m(HCl) = 0.4 · 36.5 = 14.6 g. Solution: 14.6 : 0.073 = 200 g.
Answer: a) 4.48; b) 200.
6. Excess reagent and yield
- Excess: 4 g of hydrogen and 48 g of oxygen react. 2H₂ + O₂ → 2H₂O. n(H₂) = 4 : 2 = 2 mol, n(O₂) = 48 : 32 = 1.5 mol. 2 mol H₂ need 1 mol O₂ - oxygen is in excess by 0.5 mol. Water from H₂: 2 mol = 36 g.
- Rule: always calculate from the limiting reagent (the one used up completely).
- Yield: heating 50 g CaCO₃ gave 25.2 g CaO. CaCO₃ → CaO + CO₂. n(CaCO₃) = 50 : 100 = 0.5 mol, theoretical CaO 0.5 · 56 = 28 g. η = 25.2 : 28 = 0.9 = 90%.
- Mass fraction of an element: in Fe₂O₃ iron is 2 · 56 = 112, M = 112 + 48 = 160, w(Fe) = 112 : 160 = 70%.
Writing answers and typical mistakes
- Give only the requested quantity in the requested unit: volume in L, mass in g, a percentage if a percentage is asked.
- Uzbek papers use a decimal comma: 4,48. Trylab also accepts 4.48.
- Never calculate from an unbalanced equation - the coefficient ratio decides the whole answer.
- Solution mass = solute + water; a precipitate or an escaping gas is subtracted.
- Molar concentration needs the volume in litres, mass fraction the mass in grams.
- Check every answer backwards: for example, that 200 g of a 7.3% solution really holds 14.6 g of HCl.
