Yield: MDCAT Chemistry notes
Yield for MDCAT: theoretical, actual and percentage yield, why actual yield is lower, reaction efficiency and worked stoichiometry problems with answers.
Three kinds of yield
- Theoretical yield: the amount of product calculated from the balanced chemical equation, assuming the limiting reactant is completely converted. It is the maximum possible.
- Actual yield: the amount of product actually obtained in the experiment.
- Percentage yield: the actual yield expressed as a percentage of the theoretical yield. It measures the efficiency of a reaction.
Key formulas
$$\%\ \text{yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100$$
$$\text{theoretical yield}=\frac{\text{actual yield}}{\%\ \text{yield}/100}\qquad \text{actual yield}=\text{theoretical yield}\times\frac{\%\ \text{yield}}{100}$$
The actual yield equals the theoretical yield only when the percentage yield is 100%.
Why the actual yield is less
- Operational (mechanical) losses: product lost during filtration, transfer, distillation or crystallisation.
- Reversibility: many reactions reach equilibrium and do not go to completion.
- Side reactions: some reactant forms unwanted by-products.
Method for yield problems
- Write the balanced equation.
- Convert the given mass of reactant to moles ($n=m/M$). If two reactants are given, find the limiting one.
- Use the mole ratio to find moles of product, then convert to grams: this is the theoretical yield.
- Apply the percentage yield formula.
Worked examples
| Problem | Working | Answer |
|---|---|---|
| Theoretical 25 g, actual 20 g | $\frac{20}{25}\times100$ | 80% |
| 80% yield gives 20 g; theoretical? | $\frac{20}{0.80}$ | 25 g |
| Theoretical 100 g, actual 70 g | $\frac{70}{100}\times100$ | 70% |
Example 1 (thermal decomposition). $CuCO_3\rightarrow CuO+CO_2$. Molar masses: $CuCO_3=123.5$, $CuO=79.5$ g/mol. 24.8 g $CuCO_3$ = 0.20 mol, giving 0.20 mol CuO = 16.0 g (theoretical). If 13.9 g is obtained: $\frac{13.9}{16.0}\times100\approx87\%$.
Example 2 (combustion). $2H_2+O_2\rightarrow2H_2O$. 4 g $H_2$ (2 mol) and 32 g $O_2$ (1 mol) are in exact ratio, so 2 mol water = 36 g can form. Obtaining 28 g gives $\frac{28}{36}\times100=77.8\%$.
Example 3 (oxidation of alcohols). Heating a primary alcohol with excess acidified $K_2Cr_2O_7$ under reflux gives the carboxylic acid; a secondary alcohol gives a ketone. 30 g ethanol ($M=46$) = 0.652 mol → 0.652 mol ethanoic acid ($M=60$) = 39.1 g; at 75% yield about 29 g. 30 g 2-propanol ($M=60$) = 0.5 mol → 0.5 mol propanone ($M=58$) = 29 g; at 75% yield 21.75 g.
Common MDCAT traps
- Percentage yield, not actual or theoretical yield alone, expresses efficiency.
- Always compute the theoretical yield from the limiting reactant.
- Use the molar mass of the product, not of the reactant, when converting moles back to grams.
- Mole ratios come from coefficients: 1 $CuCO_3$ gives 1 CuO, but 4 Al give 2 $Al_2O_3$.
- "Theoretical yield" is fixed by the equation; the amount of reactant called for is the "stoichiometric amount".
Quick revision
- Theoretical yield: from the balanced equation.
- Actual yield: obtained experimentally, usually less.
- % yield = actual/theoretical × 100.
- Losses: handling, reversibility, side reactions.
- 100% yield means actual = theoretical.