Audited 29 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 2·0 uses

Atom Economy Calculator

Percentage atom economy from a balanced equation: product mass over total reactant mass, plus by-product mass and the stoichiometric floor of the E factor.

Atom Economy Calculator

The ONE product you actually want. Everything else the equation produces counts as by-product. Hill/IUPAC notation, case sensitive; parentheses work, e.g. Ca(OH)2.
Its coefficient in the BALANCED equation. Forgetting it is the classic error: for 2 H2 + O2 -> 2 H2O the coefficient is 2, and entering 1 halves the atom economy from 100% to 50%.
Every reactant in the balanced equation: coefficient then formula, separated by commas or new lines. Leave the coefficient out and 1 is assumed. Include reagents used in stoichiometric quantities; EXCLUDE catalysts and solvents, which do not appear in the stoichiometric equation.
Atom economy
25.46
Percentage of the reactants' total mass that ends up in the desired product, assuming exact stoichiometry and complete conversion.
Product mass in the equation
44.053 g/mol-eq
Total reactant mass
173.046 g/mol-eq
By-product mass
128.993 g/mol-eq
Stoichiometric E factor
2.9281 kg/kg
Reactants counted
3
Reading
Low atom economy — most of the reactant mass is destined to become by-product before any yield loss, solvent or work-up is counted. Routes in this range are where a catalytic alternative pays off most: replacing a stoichiometric reagent with a catalyst removes its whole mass from the denominator.

Background.

Make styrene by a Wittig reaction, benzaldehyde plus the phosphorus ylide Ph₃P=CH₂, and even a perfect 100% yield sends 278.291 g of triphenylphosphine oxide to waste for every 104.152 g of styrene. No improvement in technique can change that; it is written into the equation. Atom economy is the number that exposes it. Barry Trost proposed the measure in Science in 1991: the share of the reactants' mass that ends up in the product you actually want. For the Wittig step above it is 27.23%.

The calculation needs only a balanced equation and a table of atomic weights, so it can be done before anyone goes near a flask. That is its use: comparing routes to the same molecule at the planning stage, when swapping a stoichiometric reagent for a catalyst or an addition for a substitution is still cheap. It became the second of Anastas and Warner's twelve Principles of Green Chemistry, which the US EPA phrases as designing syntheses so the final product contains the maximum proportion of the starting materials.

What atom economy leaves out matters as much as what it measures. It assumes exact stoichiometric amounts and a theoretical yield, and it ignores solvents, work-up chemicals, excess reagent and energy. A route with 100% atom economy run at low yield in a large volume of solvent is not a clean process, which is why chemists report atom economy alongside percent yield and Sheldon's E factor rather than instead of them.

What is atom economy calculator?

Atom economy, also called atom efficiency, is the mass of the desired product formed by one balanced equation divided by the total mass of the reactants that equation consumes, expressed as a percentage. It describes the route rather than any particular run of it: two chemists following the same equation get the same figure however well either of them works. Trost and later Constable, Curzons and Cunningham write the denominator as the sum of the reactants; Sheldon writes it as the sum of everything produced. For a balanced equation the two sums are equal, because mass is conserved, so the definitions agree.

How to use this calculator.

  1. Balance the equation first and decide which single product you want. Every other product counts as by-product, even if it is useful elsewhere.
  2. Type the desired product's formula in Hill-style notation (C2H4O, Ca(OH)2; symbols are case sensitive) and enter its coefficient from the balanced equation. Leaving the coefficient at 1 when the equation says 2 halves the answer.
  3. List every reactant as a coefficient followed by a formula, separated by commas or new lines, such as 1 C2H4, 0.5 O2. A reactant written without a coefficient is taken as 1.
  4. Include reagents consumed in stoichiometric amounts, including acids or bases used for hydrolysis. Leave out true catalysts and solvents, which do not appear in the stoichiometric equation.
  5. Check that the reactants-counted figure matches the number you meant to enter, then read the atom economy, the by-product mass and the stoichiometric E factor together.

The formula.

AE = ( ν_P × M_P ) ÷ Σ ( ν_i × M_i ) × 100

AE = (ν_P × M_P) ÷ Σ(ν_i × M_i) × 100, where ν_P and M_P are the coefficient and molar mass of the desired product and each ν_i, M_i pair belongs to one reactant. Molar masses are built from the CIAAW Abridged Standard Atomic Weights 2024 (C 12.011, H 1.008, O 15.999, Cl 35.45, Ca 40.078, S 32.06). Both sums are grams per one equation as written, so the ratio is dimensionless. Scaling every coefficient by the same factor leaves the result unchanged, which is why fractional coefficients such as ½O₂ work. The coefficient on the product is where mistakes creep in: 2 H₂ + O₂ → 2 H₂O gives 36.03 g of water from 36.03 g of reactants, exactly 100%, but entering the product coefficient as 1 gives 50%. By-product mass is the reactant total minus the product mass. The stoichiometric E factor, (100 − AE) ÷ AE, is that by-product expressed per kilogram of product, the smallest waste figure the equation allows before solvent, work-up or yield losses are added. A value above 100% cannot happen for a balanced equation, so it is reported rather than capped: it means a reactant is missing or a coefficient is wrong, and the by-product mass turns negative at the same time.

A worked example.

Example

The old chlorohydrin route to ethylene oxide runs, overall, C₂H₄ + Cl₂ + Ca(OH)₂ → C₂H₄O + CaCl₂ + H₂O. Enter C2H4O with coefficient 1 and the reactants 1 C2H4, 1 Cl2, 1 Ca(OH)2. The reactant masses are ethene 2 × 12.011 + 4 × 1.008 = 28.054 g, chlorine 2 × 35.45 = 70.9 g and calcium hydroxide 40.078 + 2 × (15.999 + 1.008) = 74.092 g, a total of 173.046 g per equation. Ethylene oxide is 2 × 12.011 + 4 × 1.008 + 15.999 = 44.053 g. Atom economy is 44.053 ÷ 173.046 × 100 = 25.46%. The remaining 173.046 − 44.053 = 128.993 g leaves as calcium chloride (110.978 g) and water (18.015 g); adding those to the ethylene oxide gives back exactly 173.046 g, which confirms the equation balances. Divided by the product mass, 128.993 ÷ 44.053 gives a stoichiometric E factor of 2.9281: nearly 3 kg of salt and water for every kilogram of ethylene oxide, before a single gram of yield loss.

product FormulaC2H4O
reactants1 C2H4, 1 Cl2, 1 Ca(OH)2
product Coefficient1

Frequently asked questions.

How does atom economy differ from percent yield?
Percent yield measures how much of the theoretically possible product you actually isolated, so it depends on conditions and skill. Atom economy measures how much of the reactant mass could ever have become product, so it depends only on the equation. A Wittig step can give an excellent yield and still score 27.23%, while an addition reaction scores 100% even if you isolate half of it.
Which reactions reach 100% atom economy?
Additions and rearrangements, where every reactant atom ends up in one product: Diels–Alder cycloadditions, hydrogenation of an alkene and Claisen rearrangements, for example. Substitutions and eliminations cannot, because they always release a leaving group. Ethylene oxide made by direct oxidation, 1 C2H4 plus 0.5 O2, scores exactly 100%, against 25.46% for the chlorohydrin route it replaced.
Does a 100% atom economy mean the process makes no waste?
No. Direct oxidation of ethylene over a silver catalyst is 100% atom-economical on paper, yet part of the ethylene burns to carbon dioxide and water instead, which is a selectivity loss the metric cannot see. Solvents, purification and energy are invisible to it too, so a perfect score describes the equation, not the plant.
What is a typical atom economy for real syntheses?
Constable, Curzons and Cunningham's survey of GSK processes found reaction classes ranging from about 64% for N-dealkylation to 100% for salt formation, with sulfonations near 89% and hydrogenations near 84%. Across 38 multi-stage pharmaceutical processes the overall figure averaged 43%, ranging from 21% to 86%, so a whole route in the 40s is ordinary.
Why is the E factor shown here far smaller than published E factors?
It counts only the by-products the balanced equation forces. Sheldon's E factor counts everything discarded per kilogram of product, including solvent losses, water and work-up chemicals, and his often-quoted sector figures run from under 0.1 for oil refining to 5–50 for fine chemicals and 25–100 or more for pharmaceuticals. The stoichiometric figure is the floor beneath those.
What happens if I forget a reactant?
The answer rises without warning unless the omission pushes it past 100%. Drop the Ca(OH)2 from the chlorohydrin example and the atom economy jumps from 25.46% to 44.52%, a plausible-looking number for an equation that no longer balances. Check the reactant count, and confirm that the products' masses add up to the reactant total.
How do I get one figure for a multi-step synthesis?
Add the step equations together and cancel every intermediate that is made in one step and consumed in a later one. The net equation, starting materials on the left and final product plus all by-products on the right, is what you enter. Its atom economy is lower than most individual steps, which is why long routes score poorly.

References& sources.

  1. [1]Trost, B. M. (1991). 'The Atom Economy — A Search for Synthetic Efficiency'. Science 254(5037), 1471–1477, doi:10.1126/science.1962206. PRIMARY SOURCE: the paper that introduced the metric, defining it as the fraction of reactant atoms appearing in the product and arguing that methods combining building blocks with any additional reagent needed only catalytically represent the highest atom economy. Retrieved 2026-07-29. Publisher page paywalled; the record, abstract and DOI are open, and the definition is independently restated in the two open sources below.
  2. [2]Constable, D. J. C., Curzons, A. D. & Cunningham, V. L. (2002). 'Metrics to "green" chemistry — which are the best?' Green Chemistry 4(6), 521–527, doi:10.1039/B206169B. Royal Society of Chemistry, peer-reviewed. States the definition against the REACTANTS ('atom economy is a calculation of how much of the reactants remain in the final product'), states the assumptions ('kept deliberately simple by making certain key assumptions, ignoring reaction yield and molar excesses of reactants... does not account for solvents and reagents'), gives the inclusion rule for stoichiometric reagents versus catalysts, and supplies both the reference figures used in this page's assessment bands — Table 1 (atom economy by reaction class, 64–100%) and Table 2 (38 multi-stage processes, overall AE averaging 43%, range 21–86%). Its worked example (benzyl alcohol FW 108.1 + tosyl chloride FW 190.65 → sulfonate ester FW 262.29 + HCl) is reproduced by this calculator. Retrieved 2026-07-29. Publisher page paywalled; an open copy was consulted.
  3. [3]Sheldon, R. A. (2018). 'Metrics of Green Chemistry and Sustainability: Past, Present, and Future'. ACS Sustainable Chemistry & Engineering 6(1), 32–48, doi:10.1021/acssuschemeng.7b03505. Peer-reviewed; the author's TU Delft repository copy is open access. RECORDED SOURCE VARIATION: Sheldon phrases the denominator as the sum of the molecular weights of all substances PRODUCED in the stoichiometric equation, where Trost and Constable phrase it as the reactants. For a balanced equation the two are identical by conservation of mass, and this page demonstrates that identity in its worked example. Sheldon is also the source for the statement that 'AE is a theoretical number which assumes the use of exact stoichiometric quantities of starting materials and a theoretical chemical yield and disregards substances, such as solvents and auxiliary chemicals which do not appear in the stoichiometric equation', and for the E-factor definition and its solvent-dominated composition. Retrieved 2026-07-29.
  4. [4]US Environmental Protection Agency, 'Basics of Green Chemistry' — the 12 Principles of Green Chemistry of Anastas and Warner, of which Principle 2 is 'Maximize Atom Economy: design syntheses so that the final product contains the maximum proportion of the starting materials. Waste few or no atoms.' Page last updated 29 January 2026; retrieved 2026-07-29. Open access, US government publication.
  5. [5]Meija, J. et al. (2021). 'Atomic weights of the elements 2021 (IUPAC Technical Report)'. Pure and Applied Chemistry 93(5), 573–600, doi:10.1515/pac-2019-0603, as maintained in the CIAAW table 'Abridged Standard Atomic Weights 2024' (named revision: 2024, incorporating the Gd, Lu and Zr revisions on the Atomic Weights 2021 base). The named, versioned source of every molar mass this calculator computes — C 12.011, H 1.0080, O 15.999, Cl 35.45, Ca 40.078, S 32.06. Retrieved 2026-07-29. Open access.

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Method
AE = ( ν_P × M_P ) ÷ Σ ( ν_i × M_i ) × 100
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