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CHEM 121 Studioby Learn4Less · UBC CHEM 121

CHEM 121 glossary

Atom economy

The percentage of the total mass of reactants that ends up in the desired product: M(desired product)∑M(reactants)×100%\dfrac{M(\text{desired product})}{\sum M(\text{reactants})} \times 100\%, with each molar mass multiplied by its coefficient. It is fixed by the balanced equation, so even a 100% yield cannot raise it.

Example

Fermenting glucose to ethanol, CX6HX12OX6→2 CX2HX5OH+2 COX2\ce{C6H12O6 -> 2C2H5OH + 2CO2}: 2×46.07180.16×100%=51.1%\dfrac{2 \times 46.07}{180.16} \times 100\% = 51.1\%. The rest of the mass leaves as COX2\ce{CO2}.

See alsoByproductPercent yieldGreen chemistryAddition reaction