Percentage Purity in Stoichiometry Problems
Many numericals say "5.00 g of impure limestone" or "an ore containing 80% Fe2O3". The word impure is a warning. Only the pure part takes part in the reaction, so if you put the whole sample mass into the mole calculation, every answer after that is wrong. In school exams this is one of the easiest ways to lose marks in stoichiometry.
This article shows how to handle percentage purity in the right order: first find the mass of the pure substance, then use moles and the balanced equation.
The formula
Mass of pure substance = (percentage purity ÷ 100) × mass of impure sample
Here "mass of pure substance" means the mass of the compound that actually reacts. The rest of the sample (sand, clay, moisture, other salts) is called the impurity. We assume the impurity does not react with the other reagent. A good question will say this; if it does not, state the assumption in your answer.
The four-step routine
- Convert percentage purity to the mass of pure compound.
- Convert that mass to moles using the molar mass (mole = mass ÷ molar mass).
- Use the mole ratio from the balanced equation.
- Convert the answer moles to mass, volume or whatever is asked.
When the question runs the other way (find the purity), the same four steps are used in reverse and the last step becomes: divide the pure mass by the sample mass.
Worked example 1: gas obtained from impure limestone
Question. A 5.00 g sample of limestone contains 92.0% CaCO3. It is heated strongly: CaCO3 → CaO + CO2. Find the mass of CaO and the mass of CO2 formed. (Ca = 40.08, C = 12.01, O = 16.00.)
Step 1. Pure CaCO3 = 0.920 × 5.00 = 4.60 g.
Step 2. Molar mass of CaCO3 = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol. Moles = 4.60 ÷ 100.09 = 0.04596 mol.
Step 3. The equation is 1 : 1 : 1, so 0.04596 mol CaO and 0.04596 mol CO2 are formed.
Step 4. Molar mass of CaO = 56.08 g/mol, so mass of CaO = 0.04596 × 56.08 = 2.58 g. Molar mass of CO2 = 44.01 g/mol, so mass of CO2 = 0.04596 × 44.01 = 2.02 g.
Check. Pure CaCO3 mass must equal CaO + CO2: 2.58 + 2.02 = 4.60 g. ✔ (Mass is conserved, and the impurity stays behind.)
Worked example 2: finding the purity
Question. 1.50 g of impure zinc granules react with excess dilute HCl: Zn + 2HCl → ZnCl2 + H2. The hydrogen collected measures 420 mL at STP. Find the percentage purity of zinc. (Zn = 65.38; take molar volume at STP = 22.4 L/mol.)
Step 1. Moles of H2 = 0.420 L ÷ 22.4 L/mol = 0.01875 mol.
Step 2. The ratio Zn : H2 is 1 : 1, so moles of pure Zn = 0.01875 mol.
Step 3. Mass of pure Zn = 0.01875 × 65.38 = 1.226 g.
Step 4. Purity = (1.226 ÷ 1.50) × 100 = 81.7%.
Note. Some textbooks use 22.7 L/mol for STP (273.15 K and 1 bar). With 22.7 L/mol the answer would be about 80.6%. Use the value your textbook and teacher follow, and write it in your answer.
Worked example 3: how much ore is needed
Question. An ore contains 80% Fe2O3 by mass. What mass of ore is needed to obtain 10.0 kg of iron? Assume all the iron in Fe2O3 is recovered. (Fe = 55.85, O = 16.00.)
Step 1. Molar mass of Fe2O3 = 2(55.85) + 3(16.00) = 159.70 g/mol. One mole of Fe2O3 holds 2 mol Fe, which weighs 111.70 g.
Step 2. Mass of Fe2O3 that holds 10.0 kg of Fe = 10.0 × (159.70 ÷ 111.70) = 14.30 kg.
Step 3. This is only 80% of the ore, so mass of ore = 14.30 ÷ 0.80 = 17.9 kg.
Notice that here we divide by the purity, because we are going from the pure mass back to the impure sample.
Common mistakes
- Using the full sample mass. Always multiply by the purity fraction first.
- Multiplying when you should divide. Going from impure sample to pure mass: multiply by purity. Going from the required pure mass to the sample needed: divide by purity.
- Using 92 instead of 0.92. Convert the percentage to a fraction before multiplying.
- Forgetting the mole ratio. Purity does not change the equation. Balance the equation first.
- Treating the impurity as a product. The impurity stays unreacted unless the question says otherwise.
- Mixing up purity and percentage yield. Purity describes the starting sample. Yield compares the actual product with the theoretical product.
Quick reference table
| Given | Asked | First move |
|---|---|---|
| Sample mass and % purity | Product mass or volume | Pure mass = purity × sample mass |
| Sample mass and product amount | % purity | Find moles of reactant from product, then pure mass ÷ sample mass |
| Required pure product and % purity | Mass of impure raw material | Find pure reactant needed, then divide by purity |
| % purity of two reagents | Limiting reagent | Convert each to pure mass and moles, then compare with the equation |
Questions on purity usually appear inside longer mole-concept problems in Class 11 and in the ICSE and CBSE board papers. Check your own syllabus and past papers to see how they are framed.
The mass–mole step is the heart of every purity problem. Use the tool to check your moles and masses after you have worked them by hand.
Open the Mass ↔ Mole Calculator →