Grade 11 Physical Sciences · Chemical Change

Calculate percentage yield or purity

Calculate percentage yield or purity, for example actual mass ÷ expected mass × 100 (8 g made out of an expected 10 g is 80%).

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Quantitative Aspects of Chemical Change: Percentage YieldKevinmathscience

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Question 7.3 · November 2025 · Paper 2 · Provincial · 2 marks

Learners crushed eggshells to determine the percentage of calcium carbonate that are found in a

1, 5

g sample. The crushed eggshells are allowed to react with EXCESS hydrochloric acid. The balanced equation for the reaction:

CaCO_{3} (s) + 2 HCl (aq) \to CaCl_{2} (aq) + CO_{2} (g) + H_{2} O (ℓ)

The calcium carbonate reacted completely in 5 minutes. The graph below shows the carbon dioxide gas that was collected at STP.

Calculate the: Diagram for this question

7.3 Percentage purity of calcium carbonate in

1, 5

g of eggshells(2)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: calculate percentage yield or purity

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3. Question 5.3.3 · May/June 2024 · Paper 2 · National (DBE) · 6 marks
The reaction between aluminium and EXCESS sulphuric acid is used to investigate factors affecting rates of reactions.

$2 A ℓ (s) + 3 H_{2} SO_{4} (aq) \to A ℓ_{2} (SO_{4})_{3} (aq) + 3 H_{2} (g)$
5.3 INVESTIGATION III

In this investigation, 5 g of the same sample of IMPURE aluminium powder reacts with an EXCESS diluted $H_{2} SO_{4}$ at $60$ °C in each of three runs. The table below summarises the conditions and the results obtained. (Assume that the impurities do not react.)

$2 A ℓ (s) + 3 H_{2} SO_{4} (aq) \to A ℓ_{2} (SO_{4})_{3} (aq) + 3 H_{2} (g)$

Answer the questions that follow.
5.3.3 Calculate the percentage purity of the aluminium. Take the molar gas volume at $60$ °C to be $27000$ cm $^{3}$ ·mol $^{- 1}$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  Substitute into the rate formulaRate $= \frac{Δ V _{H_{2}}}{Δ t}$
  $40 = \frac{Δ V _{H_{2}}}{2 , 6 ( 60 )}$
  $V (H_{2}) = 6240$ cm $^{3}$
  1 mark · Calculate reaction rate
2. ✓
  Moles of H $_{2}$ $n (H_{2}) = \frac{V}{V _{m}} = \frac{6 240}{27 000} = 0, 23$ mol
  1 mark · Calculate number of moles
3. ✓
  Mole ratio $n (A ℓ) = \frac{2}{3} n (H_{2}) = \frac{2}{3} (0, 23) = 0, 15$ mol
  1 mark · Use the mole ratio
4. ✓
  Mass of Aℓ that reacted $n (A ℓ) = \frac{m}{M}$
  $0, 15 = \frac{m}{27}$
  $m = 4, 05$ g
  1 mark · Calculate number of moles
5. ✓
  Percentage purity substitution% purity $= \frac{4 , 05}{5} (100)$
  1 mark · Calculate percentage yield or purity
6. ✓
  Final answer $= 81$ %
  1 mark · Calculate percentage yield or purity
4. Question 6.2.4 · November 2024 · Paper 2 · Provincial · 4 marks
6.2 5 g of sodium carbonate ( $Na_{2} CO_{3}$ ) reacts with $250$ cm $^{3}$ of hydrochloric acid (HC $ℓ$ ).

$Na_{2} CO_{3} (s) + 2 HC ℓ (aq) \to 2 NaC ℓ (aq) + H_{2} O (ℓ) + CO_{2} (g)$

The percentage of hydrochloric acid (HC $ℓ$ ) that reacted with sodium carbonate ( $Na_{2} CO_{3}$ ) is $76$ %.
6.2.4 The hydrochloric acid that was used in the reaction was obtained by diluting $100$ cm $^{3}$ HC $ℓ$ to $250$ cm $^{3}$ hydrochloric acid (HC $ℓ$ ) solution.

Calculate the concentration of the concentrated hydrochloric acid (HC $ℓ$ ).(4)
Show the memo (4 ticks)Memo
1. ✓
  Initial moles of HC $ℓ$ $n (HC ℓ)_{initial} = \frac{0 , 094 \times 100}{76} = 0, 124$ mol
  1 mark · Calculate percentage yield or purity
2. ✓
  Formula $c = \frac{n}{V}$
  1 mark · Calculate concentration of a solution
3. ✓
  Substitution $c = \frac{0 , 124}{0 , 1}$
  1 mark · Calculate concentration of a solution
4. ✓
  Answer with unit $c = 1, 24$ mol·dm $^{- 3}$
  1 mark · Calculate concentration of a solution
5. Question WE8 · Siyavula Grade 11 Physical Sciences, 8.2 Stoichiometric calculations, Worked example 8 · 4 marks
WE8 Sulfuric acid ( $H_{2} SO_{4}$ ) reacts with ammonia ( $NH_{3}$ ) to produce the fertiliser ammonium sulfate ( $(NH_{4})_{2} SO_{4}$ ) according to the following equation:
$H_{2} SO_{4} (aq) + 2 NH_{3} (g) \to (NH_{4})_{2} SO_{4} (aq)$
A factory worker carries out the above reaction (using $2, 0$ kg of sulfuric acid and $1, 0$ kg of ammonia) and gets $2, 5$ kg of ammonium sulfate. What is the percentage yield of the reaction?(4)
Show the memo (4 ticks)Memo
1. ✓
  Limiting reagentSulfuric acid is the limiting reagent ( $20, 38$ mol $H_{2} SO_{4}$ can give only $20, 38$ mol product, while $58, 72$ mol $NH_{3}$ could give $29, 36$ mol). The number of moles of ammonium sulfate that can be produced is $20, 38$ mol.
  1 mark · Identify the limiting reagent
2. ✓
  Theoretical yield $m = n M = (20, 38) (132, 18) = 2693, 8$ g $= 2, 694$ kg
  1 mark · Calculate number of moles
3. ✓
  Percentage yield substitution $% yield = \frac{actual yield}{theoretical yield} \times 100$
  $= \frac{2 , 5}{2 , 694} \times 100$
  1 mark · Calculate percentage yield or purity
4. ✓
  Answer $% yield = 92, 8 %$
  1 mark · Calculate percentage yield or purity
6. Question WE11 · Siyavula Grade 11 Physical Sciences, 8.2 Stoichiometric calculations, Worked example 11 · 4 marks
WE11 Limestone is mostly calcium carbonate ( $CaCO_{3}$ ). Jake wants to know how much calcium carbonate is in a sample of limestone. He finds that the sample weighs $3, 5$ g. He then adds concentrated hydrochloric acid ( $HCl$ ) to the sample. The equation for this reaction is:
$CaCO_{3} (s) + 2 HCl (aq) \to CO_{2} (g) + CaCl_{2} (aq) + H_{2} O (l)$
If the mass of calcium chloride produced is $3, 6$ g, what is the percent purity of the limestone sample?(4)
Show the memo (4 ticks)Memo
1. ✓
  Moles of calcium chloride $n = \frac{m}{M} = \frac{3 , 6}{111} = 0, 03243...$ mol
  1 mark · Calculate number of moles
2. ✓
  Moles of calcium carbonateThe mole ratio of calcium chloride to calcium carbonate is $1 : 1$ , so $n (CaCO_{3}) = 0, 032$ mol
  1 mark · Use the mole ratio
3. ✓
  Mass of calcium carbonate $m = n M = (0, 03243...) (100) = 3, 243...$ g
  1 mark · Calculate number of moles
4. ✓
  Percent purity $% purity = \frac{3 , 243...}{3 , 5} \times 100$
  $\approx 92, 66 %$
  1 mark · Calculate percentage yield or purity

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