Sample Purity and Mixture Stoichiometry

Percent purity is the percentage of a sample’s mass that consists of the desired pure substance, and mixture stoichiometry uses reaction data to find the composition of a mixture or the purity of a sample. In brief: if a 5.00 g sample of limestone gives 1.000 L of CO₂ (0.04087 mol) at 25 °C and 1 atm with excess acid, the mass of pure CaCO₃ is 0.04087 × 100.09 = 4.09 g, and the purity is 4.09 ÷ 5.00 × 100% = 81.8%.

Real materials are rarely pure. Ores contain waste rock, tablets contain fillers, and reagents contain traces of other substances. When a sample is analyzed through a reaction, only the pure component takes part, and the stoichiometry gives its amount. This article defines purity, explains how to use stoichiometry for impure samples, shows how to solve for the composition of a mixture of two components, includes tables and worked examples with gases, precipitates and mass loss, and gives some history of analysis.

What is percent purity?

percent purity = (mass of pure substance ÷ total mass of the sample) × 100%

A sample that is “92% pure” has 92 g of the substance in 100 g of sample. The remaining 8% is impurities, which are assumed not to take part in the reaction under consideration (or to react differently). The same idea is used in ore grade, in the potency of a medicine and in the concentration of commercial acids.

How do you do stoichiometry with an impure reactant?

• From the mass of an impure sample to the product: first multiply the mass by the purity (as a fraction) to find the mass of the pure reactant, then do the normal stoichiometry.
• From the product to the purity: calculate the amount of pure reactant that would produce the measured amount of product, then divide by the mass of the sample.

How do you find the composition of a mixture?

For a mixture of two components, there are two unknown quantities, so two pieces of information are needed. One is the total mass of the mixture. The other comes from an experiment, such as the volume of gas produced, the mass of a precipitate or the loss of mass on heating. Let x be the mass of one component, so that (total − x) is the mass of the other. Write the measured quantity in terms of x using the stoichiometry of each component, and solve the equation.

Methods for analyzing mixtures

Worked examples

1. Purity from gas volume. A 5.00 g sample of limestone reacts with excess HCl, CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, and produces 1.000 L of CO₂ at 25 °C and 1 atm. n(CO₂) = 1.000 ÷ 24.465 = 0.04087 mol = n(CaCO₃). m(CaCO₃) = 0.04087 × 100.09 = 4.09 g. Purity = 4.09 ÷ 5.00 × 100% = 81.8%.

2. Product from an impure ore. How much iron can be obtained from 500 kg of an ore that is 80.0% Fe₂O₃? Pure Fe₂O₃: 500 × 0.800 = 400 kg. n = 400,000 ÷ 159.69 = 2,505 mol. From Fe₂O₃ + 3CO → 2Fe + 3CO₂, n(Fe) = 5,010 mol, and m = 5,010 × 55.845 = 279,800 g = 280 kg.

3. Mixture: mass loss on heating. A 5.00 g mixture of NaHCO₃ and NaCl is heated until the mass is constant. Only the hydrogencarbonate decomposes: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂. The loss in mass is 0.62 g. For 2 mol (168.01 g) of NaHCO₃, the mass loss is 18.015 + 44.009 = 62.02 g, so the fraction lost is 62.02 ÷ 168.01 = 0.3691. The mass of NaHCO₃ is 0.62 ÷ 0.3691 = 1.68 g. The composition is 1.68 ÷ 5.00 × 100% = 33.6% NaHCO₃ and 66.4% NaCl.
4. Mixture of metals. A 1.000 g alloy of magnesium and aluminum reacts completely with acid and gives 0.04981 mol of H₂. Mg + 2HCl → MgCl₂ + H₂ (1 mol H₂ per mol Mg) and 2Al + 6HCl → 2AlCl₃ + 3H₂ (1.5 mol H₂ per mol Al). Let x be the mass of Mg. Then x ÷ 24.305 + 1.5 × (1.000 − x) ÷ 26.982 = 0.04981. This gives 0.041144x + 0.055593 − 0.055593x = 0.04981, so −0.014449x = −0.005783 and x = 0.4002 g. The alloy is 40.0% Mg and 60.0% Al.
5. Purity of a precipitate-forming salt. A 1.000 g sample of impure sodium chloride gives 2.250 g of AgCl with excess AgNO₃. n(AgCl) = 2.250 ÷ 143.32 = 0.01570 mol = n(NaCl). m(NaCl) = 0.01570 × 58.44 = 0.9175 g. The purity is 91.8%.
6. Purity from a titration. A 0.500 g sample of impure oxalic acid dihydrate is dissolved in water and titrated with 0.1000 mol/L NaOH, H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. The end point is reached after 76.2 mL. n(NaOH) = 0.0762 × 0.1000 = 7.62 × 10⁻³ mol, so n(acid) = 3.81 × 10⁻³ mol. With M(H₂C₂O₄·2H₂O) = 126.07 g/mol, the mass of pure acid is 3.81 × 10⁻³ × 126.07 = 0.480 g. The purity is 0.480 ÷ 0.500 × 100% = 96.1%.

Where is purity analysis used?

• Mining: the grade of an ore determines whether it is profitable to mine.
• Pharmaceuticals: the purity of drugs must meet strict standards, such as 98.0 to 101.0% of the labeled amount.
• Agriculture: the lime content of agricultural limestone, which is valued by its calcium carbonate equivalent.
• Food and water testing: determination of vitamin C, salt or alkalinity.
• Forensic science: testing the purity of seized drugs.
• Industry: checking the quality of raw materials and products.

How did quantitative analysis develop?

Gravimetric analysis, which measures masses of precipitates, was perfected by Jöns Jacob Berzelius in the early 19th century. Volumetric analysis with burettes was developed by Descroizilles and Gay-Lussac. Carl Remigius Fresenius published the first textbook of quantitative analysis in 1846, and set up an analytical laboratory that trained many chemists. Today, instrumental methods supplement the classical ones, but stoichiometry remains the basis of many analyses.

Which mistakes are common?

Mistake 1: Using the mass of the impure sample in the stoichiometry

Only the pure component reacts. Multiply by the purity first.

Mistake 2: Assuming that impurities react

Check that the impurity is inert under the conditions used.

Mistake 3: Forgetting that the percentages must add up to 100%

For a two-component mixture, the second percentage is 100% minus the first.

Mistake 4: Using mass fractions as mole fractions

Purity is based on mass. Convert to moles before using the equation.