Stoichiometry problems involving impure reactants require one additional step compared with ordinary chemical calculations: determining how much of the sample is actually the substance that reacts. A sample may contain the desired chemical compound along with moisture, sand, unreacted materials, or other impurities. Only the reactive substance contributes to the reaction in the amount predicted by the balanced chemical equation.
The key is to calculate the mass of the pure reactant before converting it to moles and using stoichiometric ratios. Once the amount of pure substance is known, the rest of the calculation follows the same principles used in ordinary stoichiometry problems.
What impure reactants mean in stoichiometry
A pure substance contains only one chemical substance, apart from trace contaminants that are negligible for the problem at hand. An impure reactant contains the desired chemical compound mixed with other materials.
For example, a sample of limestone may contain calcium carbonate (CaCO₃), along with sand and other minerals. If calcium carbonate reacts with hydrochloric acid, the calcium carbonate participates in the reaction, while the sand generally does not.
This distinction matters because the total mass of a sample is not necessarily the mass of the reactant available for the chemical reaction. If a 100-gram sample contains 80% calcium carbonate by mass, only 80 grams of the sample consists of calcium carbonate. The remaining 20 grams represents impurities.
Using the entire 100 grams as though it were pure calcium carbonate would overestimate the amount of reactant and produce an incorrect theoretical yield.
The solution is to account for the sample’s purity before applying the balanced chemical equation.
Understand percentage purity
Percentage purity expresses the mass of the desired substance as a percentage of the total mass of the sample.
The general equation is:
Percentage purity = (Mass of pure substance ÷ Total mass of sample) × 100%
Rearranging this equation gives the formula used most often in stoichiometry problems:
Mass of pure reactant = Total mass of impure sample × (Percentage purity ÷ 100)
The percentage must be divided by 100 before being used as a multiplier.
For example, suppose a 50-gram sample of zinc is 80% pure. The mass of pure zinc is:
Mass of pure zinc = 50 g × (80 ÷ 100)
Mass of pure zinc = 40 g
The sample contains 40 grams of zinc and 10 grams of impurities.
The 40 grams, rather than the full 50 grams, is the quantity used in subsequent stoichiometric calculations, assuming the impurities do not contribute to the reaction being studied.
If a problem gives the impurity percentage instead of the purity percentage, first subtract the impurity percentage from 100%.
For example, a sample that is 15% impure is 85% pure:
Percentage purity = 100% − 15% = 85%
This conversion prevents a common mistake: using the impurity percentage as though it represented the amount of reactant available.
Follow the standard stoichiometry process
Solving a stoichiometry problem with an impure reactant involves a sequence of calculations. Each step provides the information needed for the next.
First, determine the mass of the impure sample and its percentage purity. Use these values to calculate the mass of the pure reactant.
Second, convert the mass of the pure reactant into moles using its molar mass. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).
Third, use the coefficients in the balanced chemical equation to determine the mole ratio between the reactant and the substance being calculated.
Finally, convert the resulting number of moles into the required quantity, such as grams of product, volume of gas under specified conditions, or moles of another reactant.
The overall process can be summarized as:
Impure sample mass → Pure reactant mass → Reactant moles → Product or reactant moles → Required quantity
The purity correction comes first because stoichiometric ratios apply to the actual chemical substances participating in the reaction, not to the total mass of a mixture.
Worked example: Calculate the mass of a product
Consider a 50-gram sample of calcium carbonate that is 80% pure. The sample reacts with excess hydrochloric acid according to the balanced chemical equation:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Calculate the theoretical mass of carbon dioxide produced.
Theoretical yield is the maximum amount of product predicted by stoichiometric calculations, assuming the reaction proceeds according to the equation and the limiting reactant is completely consumed.
Step 1: Calculate the mass of pure calcium carbonate.
The sample has a total mass of 50 grams and a purity of 80%.
Mass of pure CaCO₃ = 50 g × (80 ÷ 100)
Mass of pure CaCO₃ = 40 g
Therefore, the reaction has 40 grams of calcium carbonate available.
Step 2: Convert the pure reactant mass into moles.
The molar mass of calcium carbonate is approximately 100.09 g/mol.
Moles of CaCO₃ = Mass ÷ Molar mass
Moles of CaCO₃ = 40 g ÷ 100.09 g/mol
Moles of CaCO₃ ≈ 0.400 mol
Step 3: Apply the mole ratio from the balanced equation.
The balanced equation shows that one mole of calcium carbonate produces one mole of carbon dioxide.
Therefore:
Moles of CO₂ = 0.400 mol
The mole ratio is 1:1, so the number of moles of carbon dioxide produced equals the number of moles of calcium carbonate consumed.
Step 4: Convert the carbon dioxide moles into grams.
The molar mass of carbon dioxide is approximately 44.01 g/mol.
Mass of CO₂ = Moles × Molar mass
Mass of CO₂ = 0.400 mol × 44.01 g/mol
Mass of CO₂ ≈ 17.6 g
Answer: The theoretical yield is approximately 17.6 grams of carbon dioxide.
The calculation uses 40 grams of calcium carbonate rather than the original 50-gram sample. Treating the entire sample as pure calcium carbonate would predict a larger quantity of carbon dioxide than the available reactant could produce.
Worked example: Calculate product yield when the reactant has a different mole ratio
Not every reaction has a 1:1 mole ratio. The balanced equation must always determine the relationship between the reactant and product.
Suppose a 25-gram sample of potassium chlorate (KClO₃) is 80% pure. When heated, potassium chlorate decomposes according to the equation:
2KClO₃ → 2KCl + 3O₂
Calculate the theoretical mass of oxygen produced, assuming complete decomposition of the pure potassium chlorate.
Step 1: Calculate the mass of pure potassium chlorate.
Mass of pure KClO₃ = 25 g × (80 ÷ 100)
Mass of pure KClO₃ = 20 g
Step 2: Convert the pure reactant mass into moles.
The molar mass of potassium chlorate is approximately 122.55 g/mol.
Moles of KClO₃ = 20 g ÷ 122.55 g/mol
Moles of KClO₃ ≈ 0.163 mol
Step 3: Use the balanced equation to calculate oxygen moles.
The equation shows that two moles of potassium chlorate produce three moles of oxygen.
Moles of O₂ = 0.163 mol × (3 mol O₂ ÷ 2 mol KClO₃)
Moles of O₂ ≈ 0.245 mol
Step 4: Convert oxygen moles into mass.
The molar mass of oxygen gas (O₂) is approximately 32.00 g/mol.
Mass of O₂ = 0.245 mol × 32.00 g/mol
Mass of O₂ ≈ 7.84 g
Answer: The theoretical yield is approximately 7.84 grams of oxygen.
This example illustrates why the purity calculation and the mole-ratio calculation must remain separate. Purity determines how much potassium chlorate is present, while the balanced equation determines how much oxygen that potassium chlorate can produce.
Potassium chlorate is a strong oxidizer, and heating it can present serious hazards. This example is intended for theoretical calculations, not as a recommendation to conduct the reaction outside an appropriately equipped laboratory.
How to identify the limiting reactant when samples are impure
Some problems provide two or more impure reactants rather than one impure reactant reacting with an excess of another substance. In these cases, the limiting reactant must be identified before the theoretical yield can be calculated.
The limiting reactant is the substance that is consumed first according to the balanced chemical equation. It determines the maximum amount of product that can form.
The presence of impurities changes how much of each reactant is available. Therefore, the mass of each impure sample must be corrected for purity before comparing the reactants.
Consider the reaction:
N₂ + 3H₂ → 2NH₃
Suppose a problem provides a 28-gram sample of nitrogen that is 80% pure and a 6-gram sample of hydrogen that is 90% pure.
First, calculate the mass of pure nitrogen:
Mass of pure N₂ = 28 g × 0.80 = 22.4 g
Then calculate the mass of pure hydrogen:
Mass of pure H₂ = 6 g × 0.90 = 5.4 g
Convert both masses into moles. Using molar masses of approximately 28.02 g/mol for N₂ and 2.016 g/mol for H₂:
Moles of N₂ = 22.4 g ÷ 28.02 g/mol ≈ 0.800 mol
Moles of H₂ = 5.4 g ÷ 2.016 g/mol ≈ 2.68 mol
The balanced equation requires three moles of hydrogen for every mole of nitrogen.
To react completely with 0.800 mol of nitrogen, the required hydrogen amount is:
Required H₂ = 0.800 mol × 3 = 2.40 mol
Because approximately 2.68 mol of hydrogen is available, there is enough hydrogen to react with all the nitrogen.
Nitrogen is the limiting reactant.
The theoretical amount of ammonia is calculated using the nitrogen-to-ammonia mole ratio of 1:2:
Moles of NH₃ = 0.800 mol × 2 = 1.60 mol
The molar mass of ammonia is approximately 17.03 g/mol.
Mass of NH₃ = 1.60 mol × 17.03 g/mol ≈ 27.2 g
The theoretical yield is approximately 27.2 grams of ammonia.
The important principle is that limiting-reactant calculations must use the available amounts of pure substances, not the total masses of their impure samples.
Distinguish theoretical yield from actual yield
Stoichiometric calculations predict the theoretical yield, but an actual chemical process may produce less product. This difference can arise from incomplete reactions, competing reactions, product losses during separation, or other practical limitations.
Percentage yield compares the actual amount of product obtained with the theoretical amount:
Percentage yield = (Actual yield ÷ Theoretical yield) × 100%
For example, if the theoretical yield is 17.6 grams of carbon dioxide but an experiment produces only 15.0 grams, the percentage yield is:
Percentage yield = (15.0 g ÷ 17.6 g) × 100%
Percentage yield ≈ 85.2%
This calculation measures how the actual product amount compares with the theoretical prediction. It does not measure the purity of the original reactant.
Purity and percentage yield describe different aspects of a chemical process. Purity tells you how much of the starting sample consists of the desired substance. Percentage yield tells you how much product was obtained relative to the theoretical maximum.
A sample can be highly pure while a reaction gives a low percentage yield. Conversely, an impure starting material can still give a high percentage yield when the theoretical yield is correctly calculated from the amount of pure reactant.
If the final product is itself impure, its measured mass should not automatically be treated as the mass of pure product when calculating percentage yield. If product purity is known, the mass of pure product must first be determined.
Common mistakes when solving impurity problems
Several errors can lead to incorrect answers even when the balanced equation and molar masses are correct.
Using the total sample mass instead of the pure reactant mass. This is the most fundamental mistake. Always calculate the pure mass using the percentage purity before converting the reactant into moles.
Confusing purity with impurity percentage. A sample that is 20% impure is 80% pure. Using 20% as the purity would greatly underestimate the amount of reactant.
Applying percentage purity twice. Once the pure mass has been calculated, use that value throughout the stoichiometric calculation. Do not multiply by the purity percentage again unless the problem introduces another separate purity correction.
Ignoring the balanced chemical equation. The coefficients determine the mole ratios. Even when the mass of pure reactant is correct, an incorrect mole ratio produces an incorrect result.
Comparing impure sample masses to identify the limiting reactant. The reactant with the smaller sample mass is not necessarily limiting. Different substances have different molar masses, and their samples may have different purities. Calculate the available moles of each pure reactant and compare them with the balanced equation.
Confusing theoretical yield with actual yield. The theoretical yield comes from stoichiometry and the amount of limiting reactant. The actual yield is the quantity obtained in practice. Percentage yield connects these two values but does not replace either one.
Rounding too early. Keep several digits during intermediate calculations and round the final answer according to the precision of the supplied data. This reduces avoidable rounding errors.
A reliable method for every stoichiometry problem with impure reactants
The same general approach works for most textbook problems involving impure reactants, whether the desired result is a product mass, a gas volume, or the amount of another reactant needed.
Begin by writing and balancing the chemical equation. Identify the given sample masses and percentage purities, converting any impurity percentages into purity percentages where necessary.
Calculate the mass of each pure reactant by multiplying its total sample mass by its purity expressed as a decimal. Convert each pure reactant mass into moles using its molar mass.
If only one reactant is specified and the other is present in excess, use the balanced equation to calculate the required product or reactant quantity. If multiple reactants are supplied in limited amounts, use their available moles and the balanced equation to identify the limiting reactant before calculating the theoretical yield.
Finally, convert the calculated moles into the units requested by the problem. If an actual yield or product purity is provided, apply the appropriate additional calculation without confusing it with the original reactant purity.
The central principle is straightforward: stoichiometry depends on the amount of each chemical substance available to react, not on the total mass of a mixture containing that substance. Correcting for purity at the beginning ensures that every subsequent calculation reflects the actual quantities of reactants participating in the chemical reaction.
