Atom Economy Calculator

Atom economy —

Desired product mass (g/mol): —
Total reactant mass (g/mol): —
By-product mass (g/mol): —
Share lost to by-products: —

Where the reactant mass goes

The chart appears with your result

Atom economy is the mass of the desired product divided by the total mass of all reactants, times 100, using molar masses multiplied by their balanced-equation coefficients. For CaCO3 → CaO + CO2 it is 56.08 ÷ 100.09 × 100 = 56.03%, because the CO2 leaves as a by-product.

About this tool

Atom economy measures how much of the reactant mass a balanced equation can, in theory, turn into the product you actually want. It is one of the twelve principles of green chemistry and a standard exam calculation from GCSE upward. Enter the molar mass and coefficient of the desired product and of up to three reactants; the tool returns the atom economy, the mass that ends up in by-products, a donut chart of the split and the working line by line. It depends only on the equation, not on how well the experiment went, so it says nothing about percentage yield, solvent use or energy cost.

How to use it

Balance the equation
Write the balanced equation first; the coefficients you enter come from it.
Enter the desired product
Type its molar mass in g/mol and set its coefficient with the − / + buttons.
Enter the reactants
Fill in the molar mass and coefficient of each reactant; reactants 2 and 3 are optional.
Read the split
The result, by-product mass and the donut chart update as you type; the steps show each multiplication.

Examples

Quicklime from limestone

Product molar mass: 56.08
Product coefficient: 1
Reactant 1 molar mass: 100.09
Reactant 1 coefficient: 1

Result Atom economy: 56.03%
Desired product mass (g/mol): 56.08
Total reactant mass (g/mol): 100.09
By-product mass (g/mol): 44.01
Share lost to by-products: 43.97%

Reactant 1: 1 × 100.09 g/mol = 100.09 g/mol
Total reactant mass = 100.09 g/mol
Desired product: 1 × 56.08 g/mol = 56.08 g/mol
Atom economy = 56.08 ÷ 100.09 × 100 = 56.03%

Heating CaCO3 gives CaO and CO2; the carbon dioxide released is lost mass, so only about 56% of the limestone ends up as quicklime.

Ammonia: N2 + 3 H2 → 2 NH3

Product molar mass: 17.034
Product coefficient: 2
Reactant 1 molar mass: 28.02
Reactant 1 coefficient: 1
Reactant 2 molar mass: 2.016
Reactant 2 coefficient: 3

Result Atom economy: 100%
Desired product mass (g/mol): 34.068
Total reactant mass (g/mol): 34.068
By-product mass (g/mol): 0
Share lost to by-products: 0%

Reactant 1: 1 × 28.02 g/mol = 28.02 g/mol
Reactant 2: 3 × 2.016 g/mol = 6.048 g/mol
Total reactant mass = 34.068 g/mol
Desired product: 2 × 17.034 g/mol = 34.068 g/mol
Atom economy = 34.068 ÷ 34.068 × 100 = 100%

Every reactant atom ends up in ammonia, so the reaction scores the maximum of 100%.

Ethanol by fermentation

Product molar mass: 46.07
Product coefficient: 2
Reactant 1 molar mass: 180.16
Reactant 1 coefficient: 1

Result Atom economy: 51.14%
Desired product mass (g/mol): 92.14
Total reactant mass (g/mol): 180.16
By-product mass (g/mol): 88.02
Share lost to by-products: 48.86%

Reactant 1: 1 × 180.16 g/mol = 180.16 g/mol
Total reactant mass = 180.16 g/mol
Desired product: 2 × 46.07 g/mol = 92.14 g/mol
Atom economy = 92.14 ÷ 180.16 × 100 = 51.14%

Glucose gives 2 C2H5OH plus 2 CO2: 92.14 ÷ 180.16 means just over half the sugar mass becomes ethanol.

How it is calculated

Atom economy (%) = (n_p × M_p) ÷ Σ(n_r × M_r) × 100

n_p: coefficient of the desired product in the balanced equation
M_p: molar mass of the desired product, g/mol
n_r: coefficient of each reactant
M_r: molar mass of each reactant, g/mol

Barry Trost introduced atom economy in 1991 to compare synthetic routes by how many reactant atoms they keep. Because mass is conserved, the total reactant mass equals the total product mass, so the figure can never exceed 100%. Addition and rearrangement reactions reach 100%; substitutions and eliminations always lose some mass to by-products. If the product mass entered is larger than the reactant total, the coefficients are not balanced and the tool reports an error.

When not to use it

Use percentage yield, not atom economy, to judge how a real experiment performed.
It ignores solvents, catalysts and work-up waste; use the E-factor or process mass intensity for those.
Reactions with more than three distinct reactants need the extra masses added to one row by hand.

Common mistakes

Forgetting to multiply molar masses by the coefficients from the balanced equation.
Dividing by the total product mass but leaving out a by-product, which gives 100% every time.
Counting a catalyst as a reactant even though it is not consumed.
Using an unbalanced equation, which can push the result above 100%.

Frequently asked questions

How is atom economy different from percentage yield?

Atom economy is a theoretical figure from the balanced equation: the share of reactant mass that could end up in the product. Percentage yield compares the mass actually collected with the theoretical maximum. A reaction can have 100% atom economy and a 40% yield, or a high yield but poor atom economy.

Why can atom economy never exceed 100%?

Mass is conserved in a balanced equation, so the desired product can at most account for all of the reactant mass. A figure above 100% means the coefficients or molar masses were entered wrongly.

Which reaction types reach 100%?

Addition reactions, such as ethene plus water to ethanol or hydrogen adding across a double bond, and rearrangements, which make a single product from all the reactant atoms. The Haber process, N2 + 3 H2 → 2 NH3, is a common example.

Should I include water or oxygen as a reactant?

Yes, if it appears on the left of the balanced equation and its atoms are consumed. Leave out solvents and catalysts, which do not appear in the stoichiometry.

Why does a high atom economy matter industrially?

Mass that goes into by-products has to be separated, sold or disposed of, so a low atom economy usually means more waste and higher raw material costs per tonne of product. Finding a use for the by-product can offset this, as with CO2 captured from fermentation.

Where do I get molar masses?

Add up the standard atomic weights of each atom in the formula, for example CaCO3 = 40.08 + 12.01 + 3 × 16.00 = 100.09 g/mol. Use the same number of decimals for every species so rounding does not skew the comparison.

Cite this tool

Using this tool in an assignment, article or report? Copy a citation, or a link for your website.

APA

Tools Directory Editorial Team. (2026, October 11). Atom Economy Calculator. Tools Directory. Retrieved October 27, 2026, from https://tools.milunir.com/chemistry/chemistry-reference/atom-economy-calculator/

MLA

Tools Directory Editorial Team. “Atom Economy Calculator.” Tools Directory, 11 Oct. 2026, https://tools.milunir.com/chemistry/chemistry-reference/atom-economy-calculator/. Accessed 27 Oct. 2026.

Chicago

Tools Directory Editorial Team. “Atom Economy Calculator.” Tools Directory. Last modified October 11, 2026. Accessed October 27, 2026. https://tools.milunir.com/chemistry/chemistry-reference/atom-economy-calculator/.

Link

<a href="https://tools.milunir.com/chemistry/chemistry-reference/atom-economy-calculator/">Atom Economy Calculator – Tools Directory</a>