Yield, Conversion and Selectivity

Introduction

A reaction can consume most starting material but make little desired product. Another can form the desired product efficiently but lose it during isolation. Conversion, selectivity and recovered yield describe these different stages; one percentage cannot explain them all without additional measurements.

Core explanation

Reactant conversion asks what fraction of a named starting reactant was used: X = (initial moles A − final moles A)/initial moles A. If 1.00 mol A enters a closed batch and 0.20 mol A remains, 0.80 mol A was consumed and conversion is 80%. This does not say which product used A. A might react through the desired path or through side reactions.

To make the product accounting explicit, consider two hypothetical one-to-one parallel paths, A → P (desired) and A → Q (undesired). Suppose 1.00 mol A initially yields 0.60 mol P and 0.20 mol Q, with 0.20 mol A left. Atom balance is understood schematically; a real chemical example would need fully specified balanced formulas. The consumed A is 0.80 mol, so conversion is 80%. The fraction of converted A taking the desired path is 0.60/0.80 = 75% fractional selectivity for P on this one-to-one molar basis. The undesired fraction is 25%. Both fractions sum to 100% because these are the only two reaction paths stipulated.

Terminology for selectivity varies across chemical fields. Some sources use a relative selectivity, desired product divided by undesired product, which in this example is 0.60/0.20 = 3.0. That ratio is not 75%. State the definition before calculating. For reactions with unequal stoichiometric coefficients, compare equivalent reaction extents or an explicit atom basis rather than raw product moles. A selectivity number without its denominator and basis can be misleading.

The desired product formed is 0.60 mol from a maximum of 1.00 mol P if all feed A followed A → P. On the simple feed basis, formed-product yield is 60%. If only 0.54 mol pure P is isolated, recovered-product yield is 0.54/1.00 = 54%. Product recovery is 0.54/0.60 = 90%. Under these stated one-to-one assumptions, 0.80 conversion × 0.75 fractional selectivity × 0.90 recovery = 0.54 recovered yield fraction. This factorization is an accounting identity for the chosen definitions and complete product inventory, not a universal formula to apply without checking stoichiometry.

School percentage yield commonly compares actual recovered pure target product with theoretical maximum from the limiting reagent. That is the 54% recovered value here. It does not by itself reveal whether unreacted A, side product Q or isolation loss caused the shortfall. Measuring all three quantities allows the causes to be separated. The example illustrates why high conversion is not sufficient: at 100% conversion but poor selectivity, much feed could become Q instead of P.

In a real process, product may undergo further reactions, catalysts may affect pathways, and recycles may change the basis of “feed.” Specify a system boundary and whether quantities refer to a single pass, whole batch or overall process. Percentages can then be compared consistently. For this introductory page, the simple closed one-to-one batch avoids those complications while teaching the distinctions.

Step-by-step reasoning

1. Define the initial and final amounts of the named reactant A; compute fraction consumed. 2. Specify all desired and undesired pathways and their stoichiometric basis. 3. Calculate desired-path share of consumed A for fractional selectivity. 4. Compare pure desired product recovered with the theoretical maximum from feed. 5. If formed product is separately measured, calculate recovery and explain any gap.

Visual explanation

Draw a flow from “1.00 mol A feed” splitting into “0.20 mol A unreacted,” “0.60 mol P formed” and “0.20 mol Q formed.” Split the P branch again into “0.54 mol P recovered” and “0.06 mol P lost in work-up.” Brackets label 80% conversion, 75% desired fractional selectivity and 90% P recovery.

Real-world analogy

A workshop processes eighty of one hundred raw pieces. Of those eighty, sixty become the requested model and twenty become another model. Only fifty-four requested pieces reach the customer after packing loss. “Processed,” “made the right model” and “delivered” describe three different rates, just as conversion, selectivity and recovery do.

Real-world example

In an organic synthesis, reactant may disappear while forming both desired and undesired products. Measuring only reactant disappearance can overstate progress toward the target. A chromatographic product analysis and an isolated-product mass answer different questions: chemical selectivity and recovered yield.

Why?

Why can yield equal the product of conversion, selectivity and recovery in the simplified example? Every starting A unit either remains or enters one of two one-to-one paths. Of the converted units, a fraction becomes P, and of P formed, a fraction is recovered. Multiplying those nested fractions gives recovered P per initial A.

Common misconception

“Eighty percent conversion means eighty percent yield.” Converted A may make an undesired product, and desired P may be lost after formation. The example has 80% conversion but only 54% recovered yield of P.

Worked example

Start with 2.00 mol A in a stipulated pair of one-to-one paths A → P and A → Q. At the end, 0.50 mol A remains, 1.20 mol P formed and 0.30 mol Q formed; 1.08 mol P is isolated. A consumed is 2.00 − 0.50 = 1.50 mol, so conversion is 1.50/2.00 = 75.0%. Fractional selectivity for P among converted A is 1.20/1.50 = 80.0%; relative P:Q selectivity is 1.20/0.30 = 4.0 if that convention is requested. P recovery is 1.08/1.20 = 90.0%. The theoretical P maximum from the 2.00 mol feed is 2.00 mol, so recovered yield is 1.08/2.00 = 54.0%. Check: 0.750 × 0.800 × 0.900 = 0.540.

Quick check

1. Can a batch with 80% reactant conversion have only 54% recovered desired-product yield? Answer: Yes. Side-product formation and product loss can both reduce recovered target below the consumed-reactant fraction.

Exam focus

State the denominator for every percentage. Say whether selectivity is a desired fraction of converted feed or a desired-to-undesired ratio, and use stoichiometric equivalents for unequal coefficients. Do not infer mechanisms of loss from one overall yield number.

Advanced insight

IUPAC usage recognizes more than one selectivity convention, including product ratios and fractional selectivity. Process reports may also use carbon, mass or mole bases. Comparing published values therefore requires reading definitions and system boundaries, not merely matching a word printed beside a percentage.

Summary

Conversion measures reactant use, selectivity measures preference among product pathways on a stated basis, recovery measures isolated fraction of formed product, and percentage yield compares actual desired product with its theoretical maximum. A high value of one does not guarantee high values of the others.

Practice questions

1. If 1.00 mol A starts and 0.20 mol remains, what is conversion? Answer: 80% of A was consumed. 2. If 0.60 mol P and 0.20 mol Q form from 0.80 mol consumed A, what is P fractional selectivity? Answer: 75% on the stipulated one-to-one basis. 3. What is relative P:Q product ratio for those amounts? Answer: 3.0 moles P per mole Q. 4. If 0.54 mol P is recovered from 0.60 mol formed, what is recovery? Answer: 90% of formed P is recovered. 5. What is recovered yield from 1.00 mol initial A if 0.54 mol P is isolated and maximum is 1.00 mol? Answer: 54% of the feed-based theoretical P amount.