Yield & atom economy
Yield & atom economy
Percentage yield
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The theoretical yield is the maximum amount of desired product predicted by the balanced equation. The actual yield is what you recover from the experiment.
- Calculate the theoretical yield from the limiting reactant.
- Divide actual yield by theoretical yield and multiply by 100.
- Compare matching quantities: mass with mass, amount with amount, or suitable gas volumes measured under matching conditions.
A reaction may be incomplete, competing reactions may form other products, or product may be lost during transfer and purification.
A result above 100% suggests an apparent product mass that is too high, for example because the product is wet or impure. It does not mean more pure product formed than the theoretical maximum.
Calcium oxide yield
10.0 g CaCO3 is heated and 4.48 g CaO is collected. M(CaCO3) = 100.1; M(CaO) = 56.1 g mol−1.
Write the balanced decomposition equation and calculate the percentage yield of calcium oxide.
Worked answer
| Quantity | CaCO3 | CaO theoretical |
|---|---|---|
| Mass / g | 10.0 | 5.60440… |
| M / g mol−1 | 100.1 | 56.1 |
| Amount / mol | n = m/M 10.0/100.1 = 0.0999001… | n = m/M n = 5.60440…/56.1 = 0.0999001… |
Percentage yield = (4.48/5.60440…) × 100
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
2NaHCO3 → Na2CO3 + CO2 + H2O. Heating 8.40 g NaHCO3 gives 4.70 g Na2CO3. M(NaHCO3) = 84.0 and M(Na2CO3) = 106.0 g mol−1.
(a) Calculate the theoretical Na2CO3 mass. (3 marks)
(b) Calculate the percentage yield. (2 marks)
(c) Give two reasons for a yield below 100%, and one reason for an apparent yield above 100%. (3 marks)
(d) Another experiment gives 4.24 g product at 80.0% yield. Calculate the theoretical mass of product in g. (1 mark)
Total: 9 marks
Show answer & marking points
(a) n(NaHCO₃) = m/M
n(NaHCO3) = 8.40/84.0 = 0.100 mol. [1] n(Na2CO3) = 0.0500 mol. [1] Theoretical mass = 0.0500 × 106.0 = 5.30 g. [1]
Percentage yield = (actual yield ÷ theoretical yield) × 100
(b) Yield = (4.70/5.30) × 100 [1] = 88.7%. [1]
(c) Any two: incomplete reaction; competing reactions; loss in transfer, separation or purification. [2] Apparent yield above 100%: wet or impure product makes its measured mass too high. [1]
Theoretical yield = actual yield/(percentage yield/100)
(d) Theoretical yield = 4.24/(80.0/100) = 5.30 g. [1]
50.0 cm3 of 0.800 mol dm−3 H2O2 decomposes: 2H2O2 → 2H2O + O2. The actual O2 volume is 435 cm3 at 102 kPa and 323 K. Use R = 8.314 J mol−1 K−1. Calculate the percentage yield. (6 marks)
Total: 6 marks
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Percentage yield = (actual yield ÷ theoretical yield) × 100
n(H₂O₂) = cV
n(H2O2) = 0.800 × 0.0500 = 0.0400 mol. [1] Theoretical n(O₂) = n(known substance) × coefficient(target)/coefficient(known substance)
n(O2) = 0.0400/2 = 0.0200 mol. [1] p = 102000 Pa and V = 435 × 10−6 m3. [1] Actual n = pV/(RT) [1] = 0.0165225… mol. [1] Yield = (0.0165225…/0.0200) × 100 = 82.6%. [1]
Atom economy and sustainable processes
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Atom economy asks how much of the reactant mass is directed into the chosen desired product by the balanced equation. Percentage yield asks how much of that product is actually recovered.
Use the formula-mass contribution of the desired product, including its coefficient, divided by the total reactant formula-mass contributions, also including coefficients. Multiply by 100.
Conservation of mass means the denominator can also be the total formula-mass contribution of all products. Do not include catalysts, solvents or arbitrary excess quantities.
A high atom economy can reduce waste, but choosing a sustainable process also needs energy use, hazards, raw-material cost and supply, renewability, rate, yield, separation and recycling to be considered. A useful by-product may also affect the choice.
Mr relative molecular or formula mass; no unit.
Reduction of copper(II) oxide
2CuO + C → 2Cu + CO2. Copper is the desired product. M(CuO) = 79.5, Ar(C) = 12.0 and Ar(Cu) = 63.5.
Calculate the percentage atom economy for producing copper.
Worked answer
Atom economy = (127.0/171.0) × 100
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Ethanol can be made by Route 1: C2H4 + H2O → C2H5OH; or Route 2: C6H12O6 → 2C2H5OH + 2CO2. Use Mr: C2H4 28, H2O 18, C2H5OH 46, C6H12O6 180.
(a) Calculate the atom economy for ethanol in each route. (4 marks)
(b) Explain whether obtaining only 80% yield changes the atom economy calculated for a route. (2 marks)
(c) Give one sustainability benefit of high atom economy and one other factor that should influence the route chosen. (2 marks)
Total: 8 marks
Show answer & marking points
(a) Route 1: 46/(28 + 18) × 100 [1] = 100%. [1] Route 2: (2 × 46)/180 × 100 [1] = 51.1%. [1]
(b) It does not change atom economy. [1] Atom economy is fixed by the balanced equation; percentage yield compares actual with theoretical recovery of the desired product. [1]
(c) High atom economy can reduce by-product waste or raw-material use. [1] Other valid factor: energy use, hazards, feedstock supply/renewability, rate, yield, purification or recycling. [1] Do not include a solvent, catalyst or arbitrary excess mass in the equation-based atom-economy calculation.