2.1.4—Percentage yield
- Syllabus
- First assessment 2025
- Objective
- 2.1.4
- Level
- SL
percentageyield=(experimentalyield/theoreticalyield)×100%
Find theoretical yield from the limiting reactant and stoichiometric ratio before comparing it with the measured experimental yield. Do not divide the product mass by a starting mass directly.
If stoichiometry predicts 10.0 g but 8.20 g is isolated, percentage yield is 82.0%. A value above 100% signals wet or impure product, measurement error or an incorrect theoretical yield; it is not evidence that the reaction created extra conserved matter.
Representative question
1.72 g of methyl methanoate is produced from 2.83 g of methanoic acid and excess of the other reagent. Determine the percentage yield.
ALTERNATIVE 1
expected yield <=2.83×46.0360.06>=3.69 «g»
percentage yield « =100×3.691.72»=46.6<%≫
ALTERNATIVE 2
«amount of methanoic acid used =46.032.83=»0.0615 «mol»
«expected amount of methyl methanoate =0.0615 mol »
«actual amount of methyl methanoate =60.061.72=0.0286 mol »
percentage yield μ=0.06150.0286×100»=46.5%
Marking guidance:
Award [2] for correct final answer.
Award [0] for 60.8\% (simple ratio of starting and final masses).
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.