Reactant Purity and Effective Mass

Introduction

A balance measures the entire sample placed on it. An ore, commercial reagent or damp solid may contain other substances besides the chemical named in an equation. If purity is stated, find the mass of the reactive compound first; only that mass belongs in the molar-mass conversion.

Core explanation

Mass purity is often written as a percentage: purity % = (mass of named pure substance / total sample mass) × 100. As a decimal fraction f, the effective mass is m(pure reactant) = f × m(sample). A 25.0 g limestone sample that is 80.0% CaCO₃ by mass contains 0.800 × 25.0 = 20.0 g CaCO₃. The remaining 5.00 g is other material under the simple composition model. It must not be counted as CaCO₃ when using M(CaCO₃) ≈ 100.09 g mol⁻¹.

For CaCO₃ → CaO + CO₂, the 20.0 g pure CaCO₃ portion is about 0.200 mol. Complete decomposition by this equation could yield about 0.200 mol CO₂ and 0.200 mol CaO. Using the full 25.0 g as if it were pure CaCO₃ would predict about 0.250 mol of each, a 25% overstatement relative to the correct 0.200 mol prediction. The overstatement factor is 1/0.800 = 1.25; purity correction should occur before moles and coefficients are applied.

Impurity may be assumed inert for a school problem, but real samples can contain other reactive compounds. A mineral mixture may have several carbonates that all release CO₂ on heating. If only the CaCO₃ percentage is given and the rest is explicitly inert, the simple calculation is justified. If other components also react, their identities and amounts are needed to predict total gas. “Impure” does not automatically mean every non-target component is chemically irrelevant; read the problem's stated assumptions.

Purity can be expressed on different bases. “80.0% by mass CaCO₃” means the fraction of total sample mass due to CaCO₃. It is not 80.0% of moles, and it is not a percentage yield. A solution labeled 80% by mass has a different interpretation from 0.800 mol L⁻¹; the former requires a solution mass to find solute mass, while molarity uses solution volume to find solute moles. Units and denominator must be stated.

If both reactants are impure, correct each sample separately before comparing limiting capacities. Suppose 10.0 g sample A is 90.0% A and 10.0 g sample B is 60.0% B; pure masses are 9.00 g A and 6.00 g B. Those still cannot be compared directly across substances. Divide each by its own molar mass, then by its coefficient. The sequence is sample mass → pure mass → pure moles → reaction capacity.

The calculation can be reversed when composition analysis is the target. If a reaction yield allows one to infer 12.0 g reactive compound in a 15.0 g sample, purity is 12.0/15.0 × 100 = 80.0%, provided the analysis uniquely attributes the measured product to that compound and accounts for recovery. A low actual product yield could otherwise be mistaken for low reactant purity. Distinguish composition from reaction performance.

Step-by-step reasoning

1. Identify what the sample mass includes and what substance's purity is stated. 2. Convert the purity percentage to a fraction by dividing by 100. 3. Multiply total sample mass by that fraction to obtain effective reactive mass. 4. Divide effective mass by the pure compound's molar mass to obtain its moles. 5. Apply balanced coefficients, limiting-reagent logic and yield information only afterward.

Visual explanation

Draw a 25.0 g rectangle partitioned into an 80.0% CaCO₃ region labeled 20.0 g and a 20.0% other-material region labeled 5.00 g. Only the 20.0 g arrow enters the “÷ 100.09 g mol⁻¹” box. This makes the excluded mass visible before any reaction arithmetic.

Real-world analogy

A sack labeled “80% grain by mass” weighs 25 kg, but a recipe that uses grain has only 20 kg available from it. The rest of the sack still has mass; it simply is not the ingredient named in the recipe. A reaction equation likewise counts only the specified chemical amount after composition is accounted for.

Real-world example

A 12.0 g metal powder is 75.0% Mg by mass and the rest is specified inert. It contains 9.00 g Mg, or about 0.370 mol Mg using 24.31 g mol⁻¹. In Mg + 2HCl → MgCl₂ + H₂ with excess acid, the ideal hydrogen amount is about 0.370 mol. Using all 12.0 g as magnesium would overpredict gas.

Why?

Why apply purity before molar mass? The molar mass of CaCO₃ converts CaCO₃ grams into CaCO₃ moles . It cannot convert the mass of unrelated material into CaCO₃ particles. The purity factor isolates the grams to which the chemical formula actually applies.

Common misconception

“An 80% pure sample has 80% reaction yield.” Purity describes starting composition; yield compares actual target product with its theoretical maximum. A pure reactant can give low yield, and an impure reactant can be converted efficiently after its pure amount is correctly identified.

Worked example

What theoretical CO₂ mass follows from 25.0 g limestone that is 80.0% CaCO₃ by mass, assuming other material is inert and decomposition completes? Effective CaCO₃ mass is 25.0 × 0.800 = 20.0 g. Amount is 20.0 g / 100.09 g mol⁻¹ = 0.19982 mol CaCO₃. Balanced CaCO₃ → CaO + CO₂ gives 0.19982 mol CO₂. With M(CO₂) = 44.01 g mol⁻¹, theoretical mass is 8.79 g to three significant figures. A direct use of 25.0 g would give about 11.0 g, which improperly treats impurity as carbonate. The 5.00 g other material remains outside this stated reaction balance unless its chemistry is supplied.

Quick check

1. How much CaCO₃ is present in a 25.0 g sample that is 80.0% CaCO₃ by mass? Answer: The sample contains 20.0 g CaCO₃, and only that portion enters the CaCO₃ mole conversion.

Exam focus

Write the purity fraction beside the total sample mass and name the pure substance. Apply the factor before n = m/M and before identifying a limiting reagent. Check whether the remaining material is declared inert or could itself contribute to the measured product.

Advanced insight

Assay-based purity may depend on what species the analytical method counts. An acid titration could respond to multiple basic components, so translating its result into “percent CaCO₃” requires assumptions about other substances. Stoichiometry is exact for a chosen model, but sample characterization determines whether that model matches reality.

Summary

Reactant purity converts a whole sample mass into effective mass of the named reacting compound. Multiply by the mass fraction first, then use molar mass and balanced ratios. Purity is a composition measure, distinct from reaction yield, and the behavior of other sample components must be stated or assessed.

Practice questions

1. What pure mass is in 10.0 g of a reagent that is 90.0% by mass? Answer: 9.00 g of the named reagent. 2. What fraction corresponds to 75.0% purity? Answer: 0.750 of total sample mass. 3. How much Mg is in 12.0 g powder at 75.0% Mg by mass? Answer: 9.00 g Mg. 4. Does the impurity mass disappear from a conservation account? Answer: No. It remains material, though it may be inert for the specified reaction. 5. What step follows the effective pure mass calculation? Answer: Divide by the named compound's molar mass to obtain its mole amount.