Overall Yield Across Several Steps

Introduction

A multistep synthesis can lose material at each stage. For a simple linear route where every step yield is defined relative to the material entering that step, multiply the yield fractions to find the overall recovered fraction. The multiplication is valid only when the intermediate amounts and denominators connect consistently.

Core explanation

Suppose step one converts A to B in a 1:1 mole ratio at 80% yield, and step two converts the available B to C in a 1:1 ratio at 75% yield. Starting with 1.00 mol A, step one makes 0.800 mol usable B. Step two recovers 0.800 × 0.750 = 0.600 mol C. The overall fraction relative to the 1.00 mol ideal C maximum is 0.800 × 0.750 = 0.600, or 60.0%. Adding 80% and 75%, or averaging them, has no useful meaning for sequential survival of material.

The mole ratios need not be 1:1. Let 2A → B in step one and B → 3C in step two. One mole A could theoretically yield 0.5 mol B and then 1.5 mol C. If step yields are 80% and 75% on their respective theoretical step outputs, actual C = 1.5 × 0.800 × 0.750 = 0.900 mol. Overall yield is 0.900/1.5 = 60.0%. The stoichiometric conversion and yield fractions perform different roles: coefficients set the ideal pathway; percentages reduce the material realized.

Yield terminology can be ambiguous. “Conversion” of a reactant is the fraction consumed, not necessarily the fraction becoming desired intermediate; side reactions can consume input without making target. “Isolated yield” may count purified recovered product, incorporating workup losses. If a given 80% figure refers to reactor conversion and a 75% figure refers to purity of a mixed stream, blindly multiplying them may not give isolated target yield. Identify precisely what each percentage measures.

The simple product rule assumes a single forward stream, no recycle, no added source of the intermediate and no parallel route to final product. A recycle loop can process unused reactant again, changing the relationship between single-pass and overall yield. Multiple sources of an intermediate require adding their material flows before the next step. If separate reactions make different products that converge, build an amount balance rather than multiplying a chain of percentages.

Other reactants can limit later stages. If step one makes 0.800 mol B but step two receives only enough reagent to consume 0.500 mol B, its 75% yield should apply to the appropriate 0.500 mol theoretical step output, not the entire 0.800 mol. The unused B remains available or is lost depending on the process. A good solution writes actual moles after each stage so a hidden mismatch is visible.

Precision matters. Percentages such as 80% and 75% may be approximate experimental observations, whereas balanced ratios are exact within the model. Report final mass to a precision justified by starting amount and yield data. Clearly label whether a result is theoretical final product, formed product, transferred product or isolated product.

Step-by-step reasoning

1. Draw the sequence and balance each reaction. 2. Compute the ideal next-stage amount from available current-stage input. 3. Convert each stated step yield to a fraction and apply it at that stage. 4. Check new reactants, side streams and changes in material-flow basis. 5. Compare actual final amount with the initial ideal-pathway final amount for overall yield.

Visual explanation

Draw 1.00 mol A entering a first box marked 80% to yield 0.800 mol B. The second box is marked 75% and yields 0.600 mol C. A faint ideal line goes directly from 1.00 mol A to 1.00 mol C. The actual-to-ideal comparison labels 60% overall.

Real-world analogy

If 80% of mailed packages reach a sorting centre and 75% of those reach a final address, 0.80 × 0.75 = 0.60 of the original packages arrive. The second percentage applies to survivors of the first stage, not to the original count unless explicitly stated.

Real-world example

In a two-step preparation, a crude intermediate may be isolated, then reacted and purified into a final compound. Recording an intermediate isolation yield and a final reaction yield lets chemists estimate how much starting material is needed for a target final mass, provided the two yield definitions refer to the same transferred intermediate stream.

Why?

Why do fractions multiply for a simple chain? Each stage acts on the material that survived earlier stages. The second-stage amount is first-stage output times a second fraction. Algebraically, n(final) = n(ideal pathway) × y₁ × y₂ when both yield definitions align with that sequence.

Common misconception

“An 80% step followed by a 75% step gives 77.5% overall because that is their average.” Average performance per step does not represent final material recovered. The route retains 60% of the initial theoretical final amount in the simple chain.

Worked example

One mole starting A could ideally make 2.00 mol intermediate B, and each B could make one C. Step-one isolated yield is 90.0%, so 1.80 mol B is available. Step-two isolated yield is 80.0% relative to that B, so 1.44 mol C is recovered. The ideal final amount from the original A is 2.00 mol C. Overall yield is 1.44/2.00 × 100% = 72.0%, agreeing with 0.900 × 0.800 × 100%. If only 1.00 mol of B were actually fed to step two, the other 0.80 mol must be separately accounted for.

Quick check

1. For a simple aligned sequence with step yields 70% and 50%, what is overall yield? Answer: Multiply 0.70 × 0.50 to obtain 0.35, or 35% overall.

Exam focus

Write a mole amount after each step, not only a product of percentages. Verify whether a stated percentage is conversion, selectivity, reaction yield or isolated yield. Multiply fractions only when their denominators form a consistent chain of material.

Advanced insight

The overall yield of a long unbranched route falls rapidly even when individual steps look strong. Ten 90% steps retain only (0.90)¹⁰ ≈ 34.9% of the ideal pathway. This motivates route shortening, better separations and recycling, though recycled material requires a more detailed flow model.

Summary

For a simple linear pathway, aligned step-yield fractions multiply after the balanced mole ratios establish ideal amounts. Track actual intermediate moles at every stage. Recycle, side reactions, new limiting reagents or mismatched definitions can make naive multiplication invalid.

Practice questions

1. What is the overall yield for two aligned 80% steps? Answer: 0.80 × 0.80 = 64%. 2. Starting from 2.00 mol ideal final product, how much is recovered at 64% overall? Answer: 1.28 mol final product. 3. Why is 50% conversion not automatically 50% desired-product yield? Answer: Some consumed reactant may form side products rather than the desired intermediate. 4. Can step yields be multiplied directly if new reagent limits the second step? Answer: Not against the full intermediate stream; first calculate the second step's actual limiting input and basis.