Purity Corrections before Stoichiometry

Introduction

A bottle labelled as a chemical may contain water, another mineral, or an intentionally added carrier. If a problem says the sample is 80% pure by mass, only 80% of its mass belongs to the named reactant. Stoichiometric coefficients apply to reacting moles, not the gross mass weighed. A purity correction must therefore come before conversion to moles and before a limiting-reagent comparison.

Core explanation

For a sample with mass m sample and mass-purity fraction p, active mass is m active = p m sample. Here 80% means p = 0.80, not 80. Moles of active reactant are n active = p m sample/M active, where M active is the molar mass of the named species. The remaining sample mass, (1 − p)m sample, is impurity. If the statement calls it inert under the reaction conditions, it does not consume another reagent and does not create the named product. It still contributes to the mass of a recovered solid mixture unless separated.

For example, heating impure calcium carbonate according to CaCO₃(s) → CaO(s) + CO₂(g), a 10.0 g sample at 80.0% CaCO₃ contains 8.00 g of CaCO₃. At about 100.1 g mol⁻¹ this is 0.0799 mol, so the ideal maximum CO₂ is 0.0799 mol. Using the entire 10.0 g as CaCO₃ would overpredict gas by 25%, because 10.0/8.00 = 1.25. If the 2.00 g of impurity is a stable, nonvolatile solid, it remains in the solid residue along with CaO. If it reacts or volatilises, the residue calculation needs its chemistry; “impure” by itself does not establish what happens to it.

Purity can be inferred in reverse. If an impure sample is the limiting source of a product and the reaction is quantitatively complete, measured product moles divided by stoichiometric product per mole of pure reactant gives the active-reactant moles. Multiply by molar mass to get active mass, then divide by original sample mass. This inference depends on complete conversion and no other source of product. Low gas recovery could otherwise imitate low sample purity.

Purity and yield are separate factors. Purity describes starting material composition; percent yield compares actual product with theoretical product from the active reactants under the chosen reaction. If the theoretical product is calculated from an uncorrected gross mass, the resulting percent yield is misleading. In a sequential problem, apply purity to input, identify the limiter, calculate theoretical product, and only then apply reaction yield. If multiple reagents each have stated purity, correct each independently. A volume fraction or mole fraction is not automatically a mass fraction, and a solution labelled 80% w/w differs from one labelled 80% v/v.

Step-by-step reasoning

1. Identify the purity basis: mass percentage, volume percentage, mole fraction, or concentration. 2. Multiply the gross sample amount by the fraction of the stated active substance on that same basis. 3. Convert active mass to moles using the active substance's molar mass. 4. Compare coefficient-adjusted moles with those of other required reactants; find the limiting extent. 5. Calculate theoretical products, then apply a separate yield factor if one is supplied.

Visual explanation

Draw a 10.0 g bar split into 8.00 g active CaCO₃ and 2.00 g inert material. Only the active segment feeds an arrow labelled ÷100.1 g mol⁻¹ into the mole-ratio calculation. The inert segment bypasses that arrow; it may appear later in a residue inventory but not as CaCO₃ consumed.

Real-world analogy

A bag labelled “rice mix” weighs 1.00 kg but contains 0.80 kg rice and 0.20 kg spices. A recipe that needs a known mass of rice cannot treat the whole bag as rice. Likewise, a balanced chemical equation counts units of the reacting compound, while the scale reading may include other material.

Real-world example

Ore processing reports the grade of a mineral rather than assuming every tonne of rock is valuable mineral. A copper ore at a stated mass fraction needs a composition calculation before the expected amount of copper is estimated. Moisture can also alter an as-received purity: drying a sample changes its gross mass and therefore its reported percentage basis.

Why?

Why is purity applied before a mole ratio? A coefficient describes a count of reacting formula units. Inert matter supplies none of those units. Multiplying a coefficient by gross sample moles invents reactant that is not present and propagates the error into every product and leftover calculation.

Common misconception

“An 80% pure reactant gives an 80% yield.” Purity is a property of the feed. A reaction of the pure portion can still have 100% yield relative to that portion, or less than 100% because of incomplete conversion and losses. These percentages answer different questions.

Worked example

A 25.0 g limestone sample is 84.0% CaCO₃ by mass. It is heated under conditions where CaCO₃ decomposes completely, and the remaining matter is inert and nonvolatile. Active CaCO₃ mass is 25.0 × 0.840 = 21.0 g. With M(CaCO₃) = 100.1 g mol⁻¹, n = 0.210 mol. From CaCO₃ → CaO + CO₂, the theoretical CO₂ amount is 0.210 mol, or 9.24 g using M(CO₂) = 44.0 g mol⁻¹. CaO formed is 0.210 mol, about 11.8 g. The ideal dry residue is 11.8 g CaO plus 4.00 g inert matter = 15.8 g. Initial 25.0 g equals residue plus 9.24 g gas within rounding.

Quick check

1. A 12.0 g sample is 75.0% pure by mass. What active mass enters the mole calculation? Answer: 12.0 × 0.750 = 9.00 g of the named reactant; the other 3.00 g is impurity.

Exam focus

Label the purity basis and write active mass = gross mass × purity fraction before converting to moles. State what happens to impurity only if the problem supplies enough chemistry. Keep percent yield as a separate, later operation.

Advanced insight

An assay from product mass is an inverse mass-balance problem. If product collection is incomplete, purity and collection efficiency are confounded: one measurement can determine their product but not each factor separately. An independent recovery test or another analytical measurement is needed to separate them.

Summary

Correct each impure feed to its active amount before applying reaction coefficients. Track inert matter separately if final residue or total mass is requested. Purity measures input composition, whereas yield measures recovered product relative to a theoretical amount calculated from the corrected inputs.

Practice questions

1. What CaCO₃ mass is present in 40.0 g of rock at 62.5% CaCO₃ by mass? Answer: 40.0 × 0.625 = 25.0 g CaCO₃. 2. For that rock, what is the ideal amount of CO₂ on complete decomposition? Answer: 25.0/100.1 = 0.250 mol CaCO₃, so 0.250 mol CO₂ forms by the 1:1 equation. 3. If only 0.200 mol CO₂ is collected, does that alone prove the rock was 50.0% pure? Answer: No. Gas may have been lost or decomposition incomplete. Without a recovery assumption, collected gas alone cannot uniquely determine purity.