Purity in Reaction Calculations

Introduction

An ore, commercial reagent, or damp powder may contain material besides the compound of interest. Reaction coefficients apply to moles of the actual reactive substance, not total sample mass. Purity converts the weighed sample to that substance's mass before molar mass and equation ratios are applied.

Core explanation

If a 10.0 g sample is 80.0% by mass CaCO₃, it contains 8.00 g CaCO₃ and 2.00 g other material. Its reactive amount is 8.00 g divided by M(CaCO₃), not 10.0 g divided by that molar mass. Whether the other material affects the reaction must be specified; introductory purity problems usually treat impurities as inert for the stated reaction.

Purity fraction p = mass desired compound / mass sample, so desired compound mass = p × sample mass. Conversely sample mass required = desired pure-compound mass/p. Use p as a decimal, such as 0.800 rather than 80.0, when multiplying. Percent purity is p × 100%. This is a composition ratio of a starting material.

Yield is a separate ratio involving product recovery. A sample can be 80% pure and a reaction using its active portion can still have 90% yield. To predict actual product from sample mass, first multiply by 0.80 to find reactive mass, then calculate theoretical product, then multiply by 0.90 if an actual-yield estimate is requested. Combining the two percentages before understanding their different bases can conceal errors, especially when multiple reactants or product purification are involved.

The purity definition should be read carefully. Mass percent of an active compound, mass percent of an element, and percent recovery from a purification procedure are different quantities. A 75% pure NaCl sample contains 75 g NaCl per 100 g sample; it does not contain 75 g sodium element. The formula gives sodium's fraction within the NaCl portion if that is later needed.

If an impurity also reacts to form the measured product, the simple inert-impurity assumption fails. One must model its reaction separately or use selective analysis. Similarly, if a sample contains water of hydration as part of the stated compound formula, that water is not automatically an impurity; the full hydrate formula determines its molar mass.

Step-by-step reasoning

1. Identify exactly what the purity percentage measures. 2. Multiply sample mass by purity fraction to find active-compound mass. 3. Convert active mass to moles and apply the balanced equation. 4. Apply product yield only if separately given, using the correct product basis.

Visual explanation

Draw a sample-mass bar split into an active segment and an impurity segment. Only the active segment enters the mole-and-reaction calculation arrow.

Real-world analogy

A bag labeled 80% rice by mass contains husks or other material in the remaining 20%. A recipe calling for rice grains must start from the rice portion, not the whole bag weight.

Real-world example

Limestone may contain non-carbonate minerals. Predicting CO₂ from heating it requires the mass of CaCO₃ specifically, rather than treating every gram of rock as calcium carbonate.

Why?

Why adjust mass before calculating moles? The molar mass belongs to a particular compound, so dividing a mixed sample's total mass by it invents formula units for the impurities.

Common misconception

“Purity and percentage yield are interchangeable.” Purity describes the starting sample's composition, while yield compares collected product with a reaction-based theoretical product amount.

Worked example

Heat 25.0 g of limestone that is 80.0% CaCO₃. Active CaCO₃ mass is 20.0 g. For CaCO₃ → CaO + CO₂, n(CaCO₃) = 20.0/100.09 ≈ 0.200 mol. Theoretical CO₂ mass is 0.200(44.01) ≈ 8.79 g. Using the whole 25.0 g would overpredict CO₂.

Quick check

1. What mass of pure reagent is in 50.0 g of a 60.0%-pure sample? Answer: 0.600 × 50.0 = 30.0 g of the stated reagent.

Exam focus

Name the numerator and denominator of every percentage. Calculate pure reactive mass before mole ratios and keep a separate yield factor for product recovery.

Advanced insight

An assay may report purity on a dry basis or as-received basis. Moisture changes the denominator; technical reports specify the basis so that masses and percentages refer to the same sample condition.

Summary

Purity is the fraction of a sample that is the named substance. Apply it to weighed mass before converting to moles, and distinguish that starting-material adjustment from product percentage yield.

Practice questions

1. How much sample is needed for 12.0 g pure compound if purity is 75.0%? Answer: 12.0/0.750 = 16.0 g sample. 2. In 20 g of 90%-pure CaCO₃, what mass enters the stoichiometric calculation? Answer: 18 g CaCO₃; the remaining 2 g is outside that compound under the stated purity model. 3. Does a 90% pure reactant necessarily give 90% reaction yield? Answer: No. Purity and yield have different definitions and may vary independently.