Percentage Yield and Atom Economy

Learning Objectives (OCR Spec 2.1.3)

• Define and calculate percentage yield
• Define and calculate atom economy
• Explain the importance of both quantities in the chemical industry and in sustainable (green) chemistry
• Compare synthetic routes on the basis of atom economy and yield
• Recognise trade-offs between yield, atom economy, cost and environmental impact
• Appreciate the twelve principles of green chemistry

Theoretical vs Actual Yield

In a perfect chemistry world, every mole of limiting reagent would convert entirely to product. The mass calculated from the stoichiometric equation assumes this perfect outcome — the theoretical yield.

In reality, chemists almost never achieve the theoretical yield. Reasons include:

• The reaction is reversible and does not reach completion (equilibrium limitations)
• Side reactions consume reactants to form undesired products
• Impurities in starting materials
• Losses during filtration, transfer or purification
• Product trapped in apparatus or absorbed on surfaces
• Incomplete purification or drying
• Sublimation or evaporation of volatile products
• Decomposition of thermally unstable products

The mass actually obtained at the end of the experiment is the actual yield.

Percentage Yield

$\text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$Percentage yield=theoretical yieldactual yield​×100

Yields can also be calculated in moles:

$\text{Percentage yield} = \frac{\text{moles of product obtained}}{\text{moles of product expected}} \times 100$Percentage yield=moles of product expectedmoles of product obtained​×100

Note: percentage yield can never exceed 100%. If it does, the product has not been fully dried, or contains impurities, or there has been a weighing error. A yield over 100% is a sign the student should dry the product more thoroughly before weighing.

Worked Example 1: Percentage Yield from Masses

In the preparation of aspirin, 5.00 g of salicylic acid (Mr = 138.1) is converted into aspirin (Mr = 180.2). The experiment produces 4.82 g of pure aspirin. Calculate the percentage yield.

Equation: C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH

Step 1: Theoretical yield

• n(salicylic acid) = 5.00 / 138.1 = 0.03621 mol
• Ratio 1:1 → n(aspirin, theoretical) = 0.03621 mol
• m(aspirin, theoretical) = 0.03621 × 180.2 = 6.525 g

Step 2: Percentage yield

$\%\text{ yield} = \frac{4.82}{6.525} \times 100 = 73.9\%$% yield=6.5254.82​×100=73.9%

Worked Example 2: Yield with a Non-1:1 Ratio

The Haber process reacts hydrogen and nitrogen to form ammonia:

N₂(g) + 3H₂(g) → 2NH₃(g)

If 14.0 g of N₂ is reacted with excess H₂ and 12.8 g of NH₃ is obtained, calculate the percentage yield.

Theoretical yield:

• n(N₂) = 14.0 / 28.0 = 0.500 mol
• Ratio 1 N₂ : 2 NH₃ → n(NH₃, theoretical) = 1.000 mol
• m(NH₃, theoretical) = 1.000 × 17.0 = 17.0 g

Percentage yield:

$\%\text{ yield} = \frac{12.8}{17.0} \times 100 = 75.3\%$% yield=17.012.8​×100=75.3%

Worked Example 3: Yield in a Limiting-Reagent Problem

5.40 g of aluminium reacts with 16.0 g of bromine:

2Al + 3Br₂ → 2AlBr₃

The actual yield of AlBr₃ is 12.5 g. Calculate the percentage yield.

Find limiting reagent:

• n(Al) = 5.40 / 27.0 = 0.200 mol
• n(Br₂) = 16.0 / 159.8 = 0.1001 mol
• Divide by coefficients: Al → 0.200/2 = 0.100; Br₂ → 0.1001/3 = 0.03337
• Br₂ is limiting

Theoretical yield of AlBr₃:

• Ratio 3 Br₂ : 2 AlBr₃ → n(AlBr₃) = (2/3) × 0.1001 = 0.0667 mol
• m(AlBr₃, theoretical) = 0.0667 × 266.7 = 17.79 g

Percentage yield:

$\%\text{ yield} = \frac{12.5}{17.79} \times 100 = 70.3\%$% yield=17.7912.5​×100=70.3%

Worked Example 4: Multi-Step Synthesis

A three-step organic synthesis has yields of 80%, 70% and 60% at each step. Calculate the overall percentage yield.

Overall yield = 0.80 × 0.70 × 0.60 = 0.336 = 33.6%

This illustrates why multi-step syntheses suffer dramatic losses: even modest yields at each step multiply together. Pharmaceutical syntheses of 10+ steps may give overall yields of just a few per cent, making the drugs very expensive to produce.

Atom Economy

Percentage yield measures how efficiently the limiting reagent is converted to product in a specific experiment. It says nothing about whether the equation itself is efficient — some reactions produce large amounts of by-product waste even at 100% yield. The concept of atom economy was introduced by Barry Trost in 1991 to quantify this. Trost's original paper argued that traditional yield metrics hid the inherent wastefulness of many classical synthetic reactions.

$\text{Atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all products}} \times 100$Atom economy=sum of Mr​ of all productsMr​ of desired product​×100

Equivalently, for a balanced equation where we consider the total mass of products equals total mass of reactants (conservation of mass):

$\text{Atom economy} = \frac{M_r \text{ of desired product}}{\text{total } M_r \text{ of all reactants (including coefficients)}} \times 100$Atom economy=total Mr​ of all reactants (including coefficients)Mr​ of desired product​×100

A high atom economy means most of the atoms from the reactants end up in the desired product, rather than in waste by-products.

Worked Example 5: Atom Economy of Aspirin Synthesis

Aspirin synthesis:

C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH

• Mr(aspirin, C₉H₈O₄) = 180.2
• Mr(ethanoic acid, CH₃COOH) = 60.0
• Total Mr of products = 180.2 + 60.0 = 240.2

$\%\text{ atom economy} = \frac{180.2}{240.2} \times 100 = 75.0\%$% atom economy=240.2180.2​×100=75.0%

Worked Example 6: Atom Economy of Iron Extraction

Industrial iron production in a blast furnace:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

• Mass of Fe (desired product) = 2 × 55.8 = 111.6
• Total Mr of all products = 2(55.8) + 3(44.0) = 111.6 + 132.0 = 243.6

$\%\text{ atom economy} = \frac{111.6}{243.6} \times 100 = 45.8\%$% atom economy=243.6111.6​×100=45.8%

Over half the mass in the products is "wasted" CO₂ — an inherent inefficiency of carbon reduction. This is one reason why iron production is such a large source of CO₂ emissions globally.

Worked Example 7: Atom Economy of Methanol Synthesis

CO + 2H₂ → CH₃OH

• Mr(CH₃OH) = 32.0
• Total Mr of products = 32.0 (only one product)

$\%\text{ atom economy} = \frac{32.0}{32.0} \times 100 = 100\%$% atom economy=32.032.0​×100=100%

Addition reactions with a single product always have 100% atom economy. This is ideal from a green chemistry perspective.

Worked Example 8: Titanium Extraction

Titanium is extracted by the Kroll process:

TiCl₄(g) + 2Mg(l) → Ti(s) + 2MgCl₂(s)

• Mr(Ti) = 47.9
• Mr(MgCl₂) = 95.3
• Total Mr of products = 47.9 + 2(95.3) = 238.5

$\%\text{ atom economy} = \frac{47.9}{238.5} \times 100 = 20.1\%$% atom economy=238.547.9​×100=20.1%

The extraction is atom-inefficient — for every tonne of titanium, 4 tonnes of waste MgCl₂ are produced. Recycling the MgCl₂ back to Mg and Cl₂ (by electrolysis) is essential for the economics and environmental impact of the process.

Worked Example 9: Alkene Hydration (Ethanol Production)

Ethanol is now mainly produced by hydration of ethene:

CH₂=CH₂ + H₂O → CH₃CH₂OH

• Mr(ethanol) = 46.0
• Only one product
• Atom economy = 100%

Compare with the alternative fermentation route:

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

• Mr(ethanol) = 46.0; 2 × 46.0 = 92.0
• Total products Mr = 92.0 + 2(44.0) = 180.0
• Atom economy = (92.0/180.0) × 100 = 51.1%