2.1.5—Atom economy
- Syllabus
- First assessment 2025
- Objective
- 2.1.5
- Level
- SL
atomeconomy=(Mrofdesiredproduct/totalMrofreactants)×100%
Atom economy measures how much of the reactant mass is represented by the desired product. Low atom economy means more reactant mass becomes by-products and waste.
Apply stoichiometric coefficients to every formula mass before forming the ratio. Atom economy is fixed by the chosen equation and desired product, whereas percentage yield measures experimental recovery; a reaction can have high atom economy but poor yield, or the reverse.
Worked atom-economy example: 4CHX3OH+2CO+OX22(CHX3O)X2CO+2HX2O, with dimethyl carbonate as the desired product. Atom economy=4(32.05)+2(28.01)+32.002(90.09)×100=83.33%. Coefficients multiply every molar mass. The remaining 16.67% of reactant mass becomes the water by-product for this equation; actual percentage yield is a separate experimental measure.
Representative question
Calculate the atom economy for the synthesis of ethyl ethanoate by the reaction below. Use sections 1 and 7 of the data booklet.
88.12 g mol−1×1002×30.08gmol−1+70.90gmol−1+3/2×32.00gmol−1/(88.12/179.06)
49.21\%
Marking guidance:
Award [2] for correct final answer.
Retrieve the route: balance the equation, convert through the coefficient mole ratio, identify the limiting reactant, calculate theoretical and percentage yield, then assess atom economy.
Check formula subscripts, state symbols, ratio direction, units, the smallest normalized reactant amount, the theoretical-yield denominator, and the total reactant mass used for atom economy.