Yield with an Impure Reactant

Introduction

Purity and yield affect opposite ends of a reaction calculation. Purity tells how much of the starting sample is the named reactant. Percentage yield compares the pure product actually obtained with the theoretical product from that corrected reactant amount. Keeping the two percentages separate prevents a misleading result.

Core explanation

Suppose a 10.0 g iron powder is 80.0% Fe by mass and the remainder is inert. Effective iron mass is 10.0 × 0.800 = 8.00 g Fe, not 10.0 g. Under 4Fe + 3O₂ → 2Fe₂O₃ with oxygen excess, n(Fe) = 8.00/55.85 = 0.143 mol Fe, so theoretical Fe₂O₃ amount is half that, about 0.0716 mol. With M(Fe₂O₃) = 159.70 g mol⁻¹, theoretical pure oxide mass is about 11.4 g. If 9.15 g dry pure oxide is obtained, percentage yield is about 80.0% after using unrounded intermediates. The coincidence between 80.0% starting purity and about 80.0% product yield is not a chemical identity; different samples could have any combination of these values.

The reliable chain is sample grams × purity fraction → pure reactant grams ÷ reactant molar mass → reactant moles × balanced coefficient ratio → theoretical product moles × product molar mass → theoretical product grams. Then divide actual pure product grams by theoretical product grams. The purity factor belongs near the start, not at the very end as a substitute for a yield correction. If the actual product is itself impure, a separate product-purity measurement may be needed to find its pure target mass before the final ratio.

If another reactant has a quantified supply, correct its amount too and check the limiting reagent after both conversions. An impure sample may appear abundant by gross mass but supply fewer reactive moles than expected. Conversely, a second reactant might limit even after a purity correction. Do not assume that the impure reagent is automatically limiting; compare normalized capacities from all supplied reactants.

A theoretical product mass can exceed the pure reactant mass without violating conservation. Iron oxide includes oxygen drawn from the excess O₂ source. In the example, 8.00 g Fe can ideally give about 11.4 g oxide because oxygen contributes roughly 3.44 g to the product. The 2.00 g inert part of the original 10.0 g powder also remains somewhere in the material inventory; it is not included in the oxide unless evidence shows incorporation or contamination.

Actual product yield below theoretical could reflect incomplete conversion of the pure Fe, another iron product, oxide loss during transfer or an analytical difference in product form. The percentage does not identify the cause. If the recovered material is a different oxide, a hydrated compound or a mixture, the stated Fe₂O₃ theoretical mass is not directly comparable with the total recovered mass. Formula and purity checks precede any valid yield interpretation.

Reverse problems are possible. A measured actual product and known percent yield can infer the theoretical target mass; the equation can then infer the pure starting reagent needed; dividing by its purity fraction gives total impure sample required. This reverse chain is useful for planning but depends on the assumed yield being appropriate to the same process and recovery conditions.

Step-by-step reasoning

1. Convert each starting sample mass to its pure reactive mass using the stated purity fraction. 2. Convert pure masses to moles and identify the limiting reagent if more than one is quantified. 3. Use balanced coefficients and product molar mass to obtain theoretical pure-product mass. 4. Confirm the measured actual mass refers to the same pure product and chemical form. 5. Compute actual/theoretical × 100 and interpret the two percentages separately.

Visual explanation

Draw a pipeline with two percentage signs at different positions: “10.0 g powder” → “× 0.800 purity” → “8.00 g Fe” → “reaction ratios” → “11.4 g theoretical Fe₂O₃” → “compare 9.15 g actual” → “about 80.0% yield.” Label the first percent as a sample-composition factor and the last as an outcome comparison.

Real-world analogy

A delivery may weigh 100 kg but contain only 80 kg usable raw material. A workshop's plan predicts finished goods from those 80 kg. The fraction of finished goods actually recovered relative to that plan is a separate performance measure. Multiplying two percentages without defining their denominators hides the distinction.

Real-world example

A 50.0 g limestone sample is 60.0% CaCO₃ and other material is inert. Pure CaCO₃ mass is 30.0 g, which could theoretically release about 13.2 g CO₂ under CaCO₃ → CaO + CO₂. If 10.6 g dry CO₂ equivalent is measured, the product yield is about 80%, based on the 30.0 g reactive portion, not all 50.0 g rock.

Why?

Why correct purity before calculating the theoretical yield? Stoichiometry counts molecules of the reacting compound. The impurity mass does not supply those molecules in the stated model, so using whole-sample mass would overestimate the theoretical denominator and understate the calculated yield.

Common misconception

“Multiply the starting purity and the product percentage yield together to get the final yield.” Their denominators differ: purity references total starting sample, while yield references the theoretical pure product from the reactive portion. A combined factor may be useful for a particular mass-flow prediction, but it is not itself the standard percentage yield.

Worked example

An 18.0 g sample is 75.0% Mg by mass, with inert remainder, and reacts with excess oxygen by 2Mg + O₂ → 2MgO. Pure Mg mass is 18.0 × 0.750 = 13.5 g. With M(Mg) = 24.31 g mol⁻¹, n(Mg) = 0.555 mol. The Mg:MgO ratio is 1:1, giving theoretical oxide amount 0.555 mol and theoretical mass 0.555 × 40.31 ≈ 22.4 g. If 19.0 g dry pure MgO is recovered, percentage yield is about 84.9% using unrounded values. The 4.50 g inert starting material is not counted as Mg, and the oxide's mass includes oxygen from the air.

Quick check

1. What mass enters the Mg mole conversion for an 18.0 g powder that is 75.0% Mg? Answer: Use 13.5 g Mg, obtained by multiplying total sample mass by the 0.750 purity fraction.

Exam focus

Mark starting purity and product yield as distinct percentages. Correct initial sample mass before using M, then check the limiting reagent. Compare pure target product with theoretical pure target product, not a wet or mixed sample with an anhydrous prediction.

Advanced insight

If yield and purity are known independently, a process target can be worked backward: desired recovered product ÷ yield fraction gives theoretical product requirement, stoichiometry gives pure reactant need, and division by reactant purity gives gross purchase mass. This planning use depends on keeping each factor tied to its own material stream.

Summary

An impure starting sample supplies only its purity-corrected mass of reactive compound. That pure amount sets theoretical yield after molar-mass, limiting-reagent and coefficient steps. Actual pure product divided by this corrected theoretical mass gives percentage yield; purity and yield remain separate properties.

Practice questions

1. How much Mg is in 18.0 g powder at 75.0% Mg by mass? Answer: 13.5 g Mg. 2. What theoretical MgO mass follows from that Mg with oxygen excess? Answer: About 22.4 g MgO using the 1:1 mole ratio. 3. What yield follows if 19.0 g pure MgO is collected? Answer: About 84.9% using the unrounded theoretical mass. 4. Does 75.0% starting purity require a 75.0% product yield? Answer: No. They describe independent composition and recovery comparisons. 5. Why can theoretical MgO mass exceed the pure Mg mass used? Answer: Oxygen from a second reactant contributes mass to the oxide.