Loav · VCE Chemistry · Unit 4 / Area of Study 1
U4A1-05 · Current 2026 curriculum · Original teaching guide · Study guide
1. Percentage yield asks what was actually obtained
The theoretical yield is the maximum amount of desired product predicted by the balanced equation from the limiting reactant, assuming complete conversion through the stated reaction. The actual yield is the amount of that product obtained. Compare the same quantity on the same basis: mass with mass, or moles with moles.
Percentage yield = actual yield ÷ theoretical yield × 100%
Start by converting each supplied reactant quantity to moles. Divide each mole amount by its stoichiometric coefficient; the smallest value identifies the limiting reactant and the available reaction extent. Multiply that extent by the product coefficient, then convert to mass if needed. Do not assume the smaller reactant mass is limiting: molar masses and coefficients matter.
Equilibrium limitations, competing reactions and product losses during isolation can reduce yield. A reported value above 100% usually indicates that the recovered material is not pure and dry, or that measurement/calculation assumptions are wrong. It does not demonstrate creation of matter. If an impure product mass is given, use its purity to calculate the actual mass of the desired substance before reporting a chemically meaningful yield.
Worked example 1: limiting reactant and recovered product
Illustrative data: 6.00 g ethanoic acid and 4.60 g ethanol react to form ethyl ethanoate and water. A dry, pure ester mass of 7.04 g is recovered. For this exercise use molar masses 60.0, 46.0, 88.0 and 18.0 g mol⁻¹ respectively.
Equation: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. Acid: 6.00/60.0 = 0.100 mol. Alcohol: 4.60/46.0 = 0.100 mol. Their 1:1 ratio gives a theoretical 0.100 mol ester, or 8.80 g. Percentage yield = 7.04/8.80 × 100 = 80.0%. This is an equilibrium reaction, but the theoretical yield calculation still uses the complete-conversion stoichiometric maximum unless the question defines another basis.
2. Atom economy asks where the reactant atoms go
Atom economy is a theoretical property of the balanced reaction with a specified desired product. It measures what fraction of the stoichiometric reactant mass is incorporated into that product. It is not calculated from the experimentally recovered mass and does not depend on the observed percentage yield.
Atom economy = (coefficient × molar mass of desired product) ÷ Σ(coefficient × molar mass of each reactant) × 100%
Include stoichiometric coefficients. Include all net reactants, not merely the limiting one. A catalyst regenerated in the overall reaction is not a net reactant, and a solvent is not included unless it is chemically consumed as a reactant. Excess reactant charged to an actual vessel affects resource efficiency but does not change the conventional balanced-equation atom economy.
For the esterification above, atom economy for the ester is 88.0/(60.0 + 46.0) × 100 = 83.0%. The missing 17.0% is incorporated into water. Even if the isolated yield improved to 95%, the equation’s atom economy would remain 83.0%. A reaction with only one product can have 100% atom economy yet still give a poor isolated yield or require hazardous conditions.
3. Multi-step pathways: keep the two calculations separate
If each step has a yield based on the material actually entering it, multiply the fractional yields to find the overall yield, provided the pathway’s stoichiometry is accounted for. Two successive steps at 80.0% and 75.0% produce an overall 0.800 × 0.750 × 100 = 60.0% yield. Adding the percentages or averaging them is wrong. More steps create more opportunities for reaction and separation losses.
For overall atom economy, first balance the steps and add them, cancelling intermediates made in one step and consumed in another. Use the resulting net reaction and its desired final product. Do not multiply individual atom economies; their mass bases can differ. When a reagent is regenerated, cancel it only if the actual overall pathway genuinely regenerates it in the required quantity.
Worked example 2: net pathway atom economy
Consider a model two-step route from ethene to ethanol through bromoethane:
C₂H₄ + HBr → C₂H₅Br
C₂H₅Br + NaOH → C₂H₅OH + NaBr
Net: C₂H₄ + HBr + NaOH → C₂H₅OH + NaBr
Use exercise molar masses C₂H₄ = 28.0, HBr = 80.9, NaOH = 40.0, ethanol = 46.0 and NaBr = 102.9 g mol⁻¹. The bromoethane cancels. Atom economy for ethanol = 46.0/(28.0 + 80.9 + 40.0) × 100 = 30.9%. Direct hydration, C₂H₄ + H₂O → C₂H₅OH, has 46.0/(28.0 + 18.0) × 100 = 100%. This comparison concerns theoretical atom use; deciding which practical process is preferable also requires energy, selectivity, safety and separation evidence.
4. Evaluate greener chemistry with specific evidence
High atom economy can reduce raw-material consumption per unit product and decrease coproduct generation. This may lower purchasing, separation and disposal costs for industry and reduce resource use and waste burdens for society. But a coproduct may have useful applications, and a high atom-economy route can still use toxic reagents or energy-intensive conditions.
For the VCAA focus on renewable feedstocks, ask whether a raw material can be replenished on a relevant timescale and how it is sourced. Plant-derived material is not automatically impact-free: land use, agricultural inputs, transport and competition with other uses matter. Waste biomass may reduce demand for a dedicated crop, but collection and processing still have consequences.
For catalysis, explain the mechanism of benefit at the appropriate level: a catalyst provides a lower-activation-energy pathway, increasing reaction rate and potentially enabling less energy-intensive conditions or greater selectivity. It does not shift the equilibrium position at fixed temperature. Evaluate catalyst recovery, lifetime and toxicity rather than assuming every catalyst is harmless.
For designing safer chemicals, focus on preserving the intended function while reducing hazards to people and ecosystems. This includes the properties of the designed product, not merely adding protective equipment for workers. Compare actual hazard and performance evidence. A balanced conclusion can recommend a route for its reduced waste while acknowledging missing toxicity or energy data. These principles are consistent with the US EPA overview of green chemistry.
Practice
- A reaction predicts 12.5 g product but yields 9.50 g pure product. Calculate yield.
- Step yields are 90.0%, 80.0% and 50.0%. Find overall yield.
- Why can hydrogenation of an alkene have 100% atom economy and only 70% isolated yield?
- A recovered 10.0 g sample is 80.0% desired product. The theoretical product yield is 12.0 g. Calculate corrected yield.
- Explain why “renewable feedstock, therefore completely sustainable” is an unsupported conclusion.
Answers with reasoning
- 9.50/12.5 × 100 = 76.0%. Use actual divided by theoretical, not the reverse.
- 0.900 × 0.800 × 0.500 × 100 = 36.0%. Each step passes only a fraction of its input to the next.
- In the balanced alkene + H₂ → alkane equation, all atoms enter the one desired product. In practice, incomplete conversion, side reactions or recovery losses can reduce the obtained amount.
- Pure product mass = 10.0 × 0.800 = 8.00 g. Yield = 8.00/12.0 × 100 = 66.7%. Using 10.0 g would count impurities as product.
- Renewability addresses feedstock replacement. It does not establish low land use, low energy demand, low toxicity or low waste. These need separate evidence before an overall judgement.
Go deeper: choose the correct denominator
Three different mass questions are easy to confuse. Percentage yield compares recovered desired product with its stoichiometric maximum. Atom economy compares desired-product mass with all stoichiometric net-reactant mass. The mass fraction of product in a recovered sample compares pure product with the whole sample. Write the words beside the numbers before inserting a percentage into any formula.
For example, an isolated oily layer may contain residual alcohol, water or unreacted triglyceride. Its total mass is not automatically biodiesel mass. Separation loss reduces isolated yield, but contamination can inflate the apparent yield. Neither changes the atom economy of the proposed balanced reaction. Similarly, adding excess alcohol may improve equilibrium conversion while increasing material handling and recovery requirements; this can improve yield without improving the equation’s atom economy.
Worked example 3 · Apply: coefficient-aware biodiesel calculations
An original theoretical model uses a triglyceride with three identical C₁₄ fatty-acid chains. Its formula is C₄₅H₈₆O₆. Transesterification with methanol gives glycerol and three molecules of C₁₅H₃₀O₂. Use rounded exercise molar masses 722, 32.0, 92.0 and 242 g mol⁻¹ respectively.
C₄₅H₈₆O₆ + 3CH₃OH ⇌ C₃H₈O₃ + 3C₁₅H₃₀O₂
Atom economy for the methyl esters = (3 × 242)/(722 + 3 × 32.0) × 100 = 88.8%. Both numerator and denominator need the coefficient three. The mass balance is 818 g of reactants giving 92 g glycerol and 726 g esters. Ignoring glycerol because it can be sold would not change this desired-product-specific atom economy.
Suppose 14.44 g triglyceride reacts with sufficient methanol and 12.342 g pure, dry methyl esters is recovered. Triglyceride amount = 14.44/722 = 0.02000 mol. Theoretical ester amount = 0.06000 mol, so theoretical ester mass = 0.06000 × 242 = 14.52 g. Yield = 12.342/14.52 × 100 = 85.0%, reported to three significant figures because the illustrative molar masses are rounded. The ester mass can exceed the starting oil mass because methanol contributes atoms; this is not a violation of mass conservation.
Application practice: recommend with evidence
- For the same model reaction, 7.22 g triglyceride reacts with 0.480 g methanol. Using the given rounded molar masses, identify the limiting reactant and find the theoretical methyl-ester mass.
- Two hypothetical routes make the same product. Route A has 95% atom economy, uses a persistent hazardous reagent and consumes 12 MJ per kg product. Route B has 85% atom economy, uses a less hazardous reagent and consumes 5 MJ per kg product. What can you conclude, and what additional evidence would you request?
Application answers and checks
- Triglyceride: 7.22/722 = 0.0100 mol. Methanol: 0.480/32.0 = 0.0150 mol. Divide by coefficients: 0.0100/1 versus 0.0150/3 = 0.00500 mol reaction extent. Methanol limits conversion. Theoretical ester amount = 3 × 0.00500 = 0.0150 mol; mass = 0.0150 × 242 = 3.63 g. Half of the initial triglyceride would remain if only this reaction occurred and conversion reached the limiting-reactant maximum.
- A incorporates a greater fraction of stoichiometric reactant mass into the desired product. B has advantages on the supplied energy and hazard evidence. Those facts do not establish one universal sustainability score. Request actual yields, solvent use and recovery, feedstock sourcing, catalyst lifetime, waste treatment and comparable toxicity/exposure information. Recommend conditionally according to the stated objective and identify missing evidence; do not treat either a renewable label or one percentage as a complete evaluation.
