Percentage Yield and Atom Economy

Spec Mapping — OCR H432 Module 2.1.3 — Amount of substance / yield and atom economy, content statements defining and calculating percentage yield, atom economy and their importance in industrial chemistry and the design of sustainable (green) chemical processes (refer to the official OCR H432 specification document for exact wording). This is the final lesson of Module 2's quantitative-chemistry block and the conceptual gateway to industrial / green-chemistry themes that recur throughout Modules 4–6.

This lesson introduces two complementary efficiency metrics for chemical reactions. Percentage yield measures the experimental efficiency: what fraction of the limiting reagent was actually converted to isolated product, given the unavoidable losses to side reactions, equilibrium, transfer, and purification. Atom economy, by contrast, measures the inherent efficiency of the equation itself: what fraction of the mass of the reactants ends up in the desired product, regardless of how well the reaction is performed. The two metrics answer different questions and are independent — a reaction can have 100 % yield but 30 % atom economy (a wasteful equation, perfectly performed), or 100 % atom economy but 30 % yield (an elegant equation, poorly performed). The concept of atom economy was introduced by Barry Trost in 1991 in a Science paper that reshaped the design of organic syntheses, and atom economy is now Principle 2 of the Twelve Principles of Green Chemistry (Anastas and Warner, 1998). Modern pharmaceutical industries use both metrics together when choosing among competing synthetic routes — the canonical OCR case study is the Boots vs BHC routes to ibuprofen.

Key Equations: $\%\text{ yield} = \dfrac{\text{actual yield (mass or mol)}}{\text{theoretical yield (same units)}} \times 100$ $\%\text{ atom economy} = \dfrac{M_r(\text{desired product, including coefficient})}{\text{sum of } M_r \text{ over all products, weighted by coefficients}} \times 100$

Theoretical vs Actual Yield

In an idealised reaction every mole of limiting reagent would convert entirely to product. The mass calculated from the balanced equation under that assumption is the theoretical yield. In reality, chemists almost never achieve the theoretical yield because of:

Reversibility — many reactions are equilibria and do not reach completion.
Side reactions — reactants form unwanted by-products in parallel.
Impurities in starting materials.
Losses during filtration, transfer or purification.
Product trapped in apparatus or absorbed on glass surfaces.
Incomplete drying or recrystallisation.
Evaporation or sublimation of volatile products.
Decomposition of thermally unstable products.

The mass actually obtained at the end of the experiment is the actual yield, and the percentage yield is the ratio.

Percentage Yield

$\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Equivalently in moles:

$\%\text{ yield} = \frac{n(\text{product obtained})}{n(\text{product expected})} \times 100$

Note: percentage yield can never exceed 100 %. A reported yield above 100 % indicates impurity, incomplete drying, or a weighing error — the candidate should re-dry and re-weigh.

Worked Example 1 — Aspirin synthesis

$5.00\text{ g}$ of salicylic acid ( $M_r = 138.1$ ) reacts with ethanoic anhydride to give aspirin ( $M_r = 180.2$ ):

$\text{C}_7\text{H}_6\text{O}_3 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_9\text{H}_8\text{O}_4 + \text{CH}_3\text{COOH}$

Mass of aspirin obtained: $4.82\text{ g}$ . Find the % yield.

$n(\text{salicylic acid}) = 5.00/138.1 = 0.03621\text{ mol}$
1:1 ratio → $n(\text{aspirin theoretical}) = 0.03621\text{ mol}$
$m(\text{aspirin theoretical}) = 0.03621 \times 180.2 = 6.525\text{ g}$
$\%\text{ yield} = 4.82/6.525 \times 100 =$ $73.9\%$

Worked Example 2 — Non-1:1 ratio (Haber process)

$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ . $14.0\text{ g}$ N₂ with excess H₂; $12.8\text{ g}$ NH₃ obtained.

$n(\text{N}_2) = 0.500\text{ mol}$
Ratio 1:2 → $n(\text{NH}_3 \text{ theoretical}) = 1.00\text{ mol}$
$m(\text{NH}_3 \text{ theoretical}) = 17.0\text{ g}$
$\%\text{ yield} = 12.8/17.0 \times 100 =$ $75.3\%$

Worked Example 3 — Limiting-reagent yield

$5.40\text{ g}$ Al with $16.0\text{ g}$ Br₂; $12.5\text{ g}$ AlBr₃ obtained.

$2\text{Al} + 3\text{Br}_2 \rightarrow 2\text{AlBr}_3$

$n(\text{Al}) = 5.40/27.0 = 0.200\text{ mol}$ ; $n(\text{Br}_2) = 16.0/159.8 = 0.1001\text{ mol}$
$\div$ coefficients: Al → $0.100$ ; Br₂ → $0.0334$ → Br₂ limiting
$n(\text{AlBr}_3 \text{ theoretical}) = (2/3)(0.1001) = 0.0667\text{ mol}$
$m(\text{AlBr}_3 \text{ theoretical}) = 0.0667 \times 266.7 = 17.79\text{ g}$
$\%\text{ yield} = 12.5/17.79 \times 100 =$ $70.3\%$

Worked Example 4 — Multi-step synthesis

Three-step synthesis with step yields 80 %, 70 %, 60 %:

Overall $= 0.80 \times 0.70 \times 0.60 = 0.336 =$ $33.6\%$

This illustrates why multi-step pharmaceutical syntheses, sometimes 10+ steps, can drop overall yield to single-digit percent and drive up unit cost.

Worked Example 4b — Reverse direction (given % yield, find actual mass)

A reaction has a known yield of 65 % for a 1:1 stoichiometry. Starting from $12.0\text{ g}$ of a reactant ( $M_r = 80.0$ ), what mass of product ( $M_r = 100.0$ ) would be expected?

$n(\text{reactant}) = 12.0/80.0 = 0.150\text{ mol}$
$n(\text{product theoretical}) = 0.150\text{ mol}$
$m(\text{product theoretical}) = 0.150 \times 100.0 = 15.0\text{ g}$
Actual yield = $0.65 \times 15.0 =$ $9.75\text{ g}$

Reverse-yield calculations are common in industrial planning where the target product mass is the input and the required reactant mass is the output.

Worked Example 4c — Recovering the limiting-reagent mass from yield

A reaction $\text{A} + \text{B} \rightarrow \text{C}$ runs at 80 % yield. A chemist obtains $8.00\text{ g}$ of C ( $M_r = 100$ ). What mass of A ( $M_r = 50$ ) was originally used, assuming A is the limiting reagent and the stoichiometry is 1:1:1?

$n(\text{C}) = 0.0800\text{ mol}$ (actual)
$n(\text{C theoretical}) = 0.0800 / 0.80 = 0.100\text{ mol}$
$n(\text{A}) = 0.100\text{ mol}$
$m(\text{A}) = 0.100 \times 50 =$ $5.00\text{ g}$

The inverse direction (yield-then-scale-back) is the standard pharmaceutical pricing logic: knowing the yield of each step and the cost per gram of each reactant, an industrial chemist can compute the cost per gram of product.

Atom Economy

Percentage yield measures conversion of the limiting reagent under specific experimental conditions. Atom economy measures the inherent efficiency of the equation itself — what fraction of reactant mass ends up in the desired product. Barry Trost introduced the concept in a 1991 Science paper that argued classical yield metrics conceal the inherent wastefulness of many synthetic reactions.

$\%\text{ atom economy} = \frac{M_r(\text{desired product, with coefficient})}{\sum M_r(\text{all products, with coefficients})} \times 100$

By conservation of mass, the denominator can equivalently be written as the sum of $M_r$ values across reactants weighted by coefficients — both give the same result.

A high atom economy means most of the reactant mass ends up in the desired product, not in waste by-products.

Why yield can exceed 100 % (but never should)

A reported yield above 100 % is almost always a sign of impurity in the isolated product. The most common impurities are:

Residual water from incomplete drying. Many organic products are crystallised from aqueous or alcoholic mixtures and need extensive vacuum drying to remove residual solvent. A wet crystal weighs more than a dry one.
Unreacted starting material. If the recrystallisation didn't fully separate the product from unreacted reagent, the apparent mass exceeds the theoretical.
Decomposition product. Some products are thermally unstable and partial decomposition during drying produces extra (denser) decomposition products.
Weighing error. A wet weighing boat, an unzeroed balance, or contaminated transfer apparatus all contribute.

The professional response to a reported yield above 100 % is to dry the product to constant mass (repeat the weighing after additional drying cycles until the mass stops dropping) and re-weigh. If the corrected yield is still above 100 %, the product is impure and recrystallisation should be repeated. CPAC assessment criteria award a discriminator mark for explicitly recognising the source of the apparent excess yield.

Worked Example 5 — Aspirin atom economy

$\text{C}_7\text{H}_6\text{O}_3 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_9\text{H}_8\text{O}_4 + \text{CH}_3\text{COOH}$

$M_r(\text{aspirin}) = 180.2$ ; $M_r(\text{ethanoic acid}) = 60.0$
Total products $M_r = 240.2$
$\%\text{ atom economy} = 180.2/240.2 \times 100 =$ $75.0\%$

Worked Example 6 — Iron extraction

$\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$

Mass of Fe = $2 \times 55.8 = 111.6$ ; total products = $111.6 + 3 \times 44.0 = 243.6$
$\%\text{ atom economy} = 111.6/243.6 \times 100 =$ $45.8\%$

Over half the product mass is "wasted" CO₂ — an inherent inefficiency of carbon reduction and a major contributor to the global carbon footprint of steel production.

Worked Example 7 — Methanol synthesis (100 % atom economy)

$\text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH}$

Single product; atom economy = 100 %. Addition reactions with a single product always have 100 % atom economy and are ideal from the green-chemistry perspective.

Worked Example 8 — Titanium extraction (Kroll process)

$\text{TiCl}_4(\text{g}) + 2\text{Mg(l)} \rightarrow \text{Ti(s)} + 2\text{MgCl}_2(\text{s})$

$M_r(\text{Ti}) = 47.9$ ; $2M_r(\text{MgCl}_2) = 2 \times 95.3 = 190.6$ ; total = $238.5$
$\%\text{ atom economy} = 47.9/238.5 \times 100 =$ $20.1\%$

For every tonne of Ti, around 4 tonnes of MgCl₂ are produced. Electrolytic recycling of MgCl₂ back to Mg + Cl₂ is essential to the economics and environmental footprint of the Kroll process.

Worked Example 9 — Hydration vs fermentation ethanol

Industrial hydration: $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$ . Single product, atom economy = 100 %.

Fermentation: $\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$ .

$2M_r(\text{ethanol}) = 92$ ; total products $= 92 + 2(44) = 180$
Atom economy $= 92/180 \times 100 =$ $51.1\%$

Hydration has the higher atom economy; fermentation uses a renewable sugar feedstock. Both have legitimate roles depending on context.

Worked Example 10 — Ibuprofen: Boots vs BHC routes

The original Boots synthesis (1960s) used 6 stoichiometric steps with an atom economy near 40 %. The BHC route (1990s) uses 3 catalytic steps and reaches an atom economy near 77 %, with HF as a recyclable catalyst and ethanoic acid as the only major by-product. The BHC route is the canonical OCR case study of how atom economy drives industrial innovation in green chemistry.

Decision tree — route selection

flowchart TB
    A[Candidate synthetic route] --> B{High yield?}
    B -->|No| C[Investigate catalyst]
    B -->|Yes| D{High atom economy?}
    C --> A
    D -->|No| E[Investigate alternative reaction]
    D -->|Yes| F[Check cost, safety, scalability]
    E --> A
    F --> G[Industrial adoption]

Comparing Routes — Combined Efficiency

Consider routes A (yield 90 %, atom economy 40 %) and B (yield 60 %, atom economy 95 %). On the combined metric:

$\text{overall efficiency} = \text{yield} \times \text{atom economy}$

A: $0.90 \times 0.40 = 0.36 = 36\%$
B: $0.60 \times 0.95 = 0.57 = 57\%$

Route B is significantly more efficient overall. But yield and atom economy aren't the only factors — cost of reagents, availability, safety, rate, waste-treatment cost, by-product saleability, scalability and regulatory compliance all enter real industrial decisions.

Why Both Matter

Economic

Expensive starting materials wasted as by-products reduce profit.
Low yield means more starting material is required per kg of product.
Waste disposal is regulated and costs money.
Higher atom economy means smaller reactor volumes and less solvent.

Environmental

Low atom economy generates by-products that may be hazardous and require disposal.
High atom economy reduces raw-material consumption per unit product.
Reduces separation/purification energy.
Lower carbon footprint.
Conserves finite resources.

Green Chemistry Principles

Atom economy is Principle 2 of the Twelve Principles of Green Chemistry (Anastas and Warner, 1998). Other principles include waste prevention, safer chemicals and solvents, energy efficiency, renewable feedstocks, and the use of catalysts in preference to stoichiometric reagents. The principles collectively shift chemistry from "clean up afterwards" to "don't make the mess in the first place".

Worked Example 11 — Combining yield and atom economy

The overall material efficiency of a single-step synthesis:

$\text{overall efficiency} = \text{yield} \times \text{atom economy}$

For Route A (90 % × 40 %): 36 %. For Route B (60 % × 95 %): 57 %. Route B wins on combined efficiency despite lower yield.

Worked Example 12 — Benzene nitration

$\text{C}_6\text{H}_6 + \text{HNO}_3 \rightarrow \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O}$ — $10.0\text{ g}$ benzene with excess HNO₃ gives $12.3\text{ g}$ pure nitrobenzene.

(a) Percentage yield:

$n(\text{benzene}) = 10.0/78.0 = 0.1282\text{ mol}$
$n(\text{nitrobenzene theoretical}) = 0.1282\text{ mol}$
$m(\text{nitrobenzene theoretical}) = 0.1282 \times 123.0 = 15.77\text{ g}$
$\%\text{ yield} = 12.3/15.77 \times 100 =$ $78.0\%$

(b) Atom economy:

$M_r(\text{nitrobenzene}) = 123.0$ ; $M_r(\text{H}_2\text{O}) = 18.0$
Total = $141.0$
$\%\text{ atom economy} = 123.0/141.0 \times 100 =$ $87.2\%$

High atom economy (water is the only by-product), moderate yield (side reactions and purification losses).

Worked Example 13 — Substitution vs addition

Consider two alternative routes to chloroethane (C₂H₅Cl):

Route 1 (substitution): $\text{C}_2\text{H}_6 + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{HCl}$

$M_r$ (products) $= 64.5 + 36.5 = 101.0$
Atom economy $= 64.5/101.0 \times 100 =$ $63.9\%$

Route 2 (addition): $\text{C}_2\text{H}_4 + \text{HCl} \rightarrow \text{C}_2\text{H}_5\text{Cl}$

Single product, atom economy $=$ $100\%$

Percentage Yield and Atom Economy

Percentage Yield and Atom Economy

Theoretical vs Actual Yield

Percentage Yield

Worked Example 1 — Aspirin synthesis

Worked Example 2 — Non-1:1 ratio (Haber process)

Worked Example 3 — Limiting-reagent yield

Worked Example 4 — Multi-step synthesis

Worked Example 4b — Reverse direction (given % yield, find actual mass)

Worked Example 4c — Recovering the limiting-reagent mass from yield

Atom Economy

Why yield can exceed 100 % (but never should)

Worked Example 5 — Aspirin atom economy

Worked Example 6 — Iron extraction

Worked Example 7 — Methanol synthesis (100 % atom economy)

Worked Example 8 — Titanium extraction (Kroll process)

Worked Example 9 — Hydration vs fermentation ethanol

Worked Example 10 — Ibuprofen: Boots vs BHC routes

Decision tree — route selection

Comparing Routes — Combined Efficiency

Why Both Matter

Economic

Environmental

Green Chemistry Principles

Worked Example 11 — Combining yield and atom economy

Worked Example 12 — Benzene nitration

Worked Example 13 — Substitution vs addition

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