SME
Organic Chemistry II · Problem type
Atom economy of a reaction
Use this for “Bromoethane (108.97 g/mol) and NaOH (40.00 g/mol) give ethanol (46.07 g/mol) and NaBr. Find the atom economy.”
This is an example. Type your own values to start a new problem.
AE (%)
=
Mproduct (g/mol)
M₁ (g/mol)
+
M₂ (g/mol)
×
100
Formulas
AE
=
100
×
Mproduct
M₁
+
M₂
83.03
=
100
×
88.11
60.05 + 46.07
Assumptions
•
Atom economy is the share of the reactants’ mass that ends up in the wanted product, from the balanced equation.
•
One mole of each reactant gives one mole of product; with another coefficient, multiply that molar mass by it.
•
It counts atoms, not yield: a reaction can go in 100% yield and still waste mass as a by-product.
Step-by-step
WE KNOW
Mproduct = 88.11 g/mol, M₁ = 60.05 g/mol, M₂ = 46.07 g/mol
FIND
atom economy (AE)
Step 1 · Find atom economy (AE)
Use AE = 100 × Mproduct ÷ (M₁ + M₂)
Divide the product’s molar mass by the reactants’ total, as a percent.
AE
=
100
×
88.11
60.05 + 46.07
AE
=
100
×
88.11
106.12
AE
=
8,811
106.12
AE = 83.03% → The rest of the mass leaves as the by-product, wasted even at 100% yield.
Check
Put the numbers back in. Both sides should match.
83.03
=
100
×
88.11
60.05 + 46.07
✓
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Overall yield of a multistep route
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Conjugated dienes, Diels–Alder and UV spectroscopy
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