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Percent Purity in Stoichiometry Problems (NEET Trick)

Chemistry · Some Basic Concepts Of Chemistry · NEET

In a purity problem, only the pure part of a sample actually reacts. So first find the mass of pure substance, then do normal stoichiometry with that pure mass. Memory hook: "Only the pure part reacts. Impurity just sits there."
Percent Purity in StoichiometryImpure sample20 g, 20% purePure part reacts4 g CaCO3Stoichiometry-> CO2 productx 20/100molesGiven impure -> MULTIPLY by purity to get pure massNeed sample for a pure mass -> DIVIDE by purity
Only the pure part of a sample reacts. Multiply the impure mass by (%purity/100) to get pure mass, then do normal mole-ratio stoichiometry. Reverse the step (divide) when the question asks how much impure sample to weigh out.

Your doubts, answered

Do I use the pure mass or the impure mass in the chemical equation?

Always use the PURE mass in the equation. Only the pure substance reacts. The impurity does not take part. So if a question gives you an impure sample, first convert to pure mass, then put that pure mass into the mole calculation.

When do I MULTIPLY by percent, and when do I DIVIDE?

Two directions: 1) Given IMPURE mass, need PURE mass: MULTIPLY. Pure mass = Impure mass x (%purity / 100). Example: 20 g of 20% pure = 20 x 20/100 = 4 g pure. 2) Given PURE mass (from stoichiometry), need IMPURE sample to weigh out: DIVIDE. Impure mass = Pure mass x (100 / %purity). Example: need 1.25 g pure, sample is 95% pure -> 1.25 x 100/95 = 1.32 g. Simple check: pure mass is always SMALLER than the impure sample.

How do I know which mass the question is asking for?

Read the last line. If it asks 'what mass of X% pure sample is required?' the answer is the IMPURE (bigger) mass, so you DIVIDE by purity at the end. If it asks 'how much product forms from X% pure sample?' you first find pure reactant (MULTIPLY), then do stoichiometry for the product.

Does the impurity react or change the product?

No. In NEET problems the impurity is assumed inert (does nothing). It is just dead weight. Only the named pure compound reacts and forms product. So you ignore the impurity mass once you have found the pure mass.

What is the full step-by-step method?

Step 1: Find pure mass (multiply impure mass by purity/100), OR keep pure mass as the unknown. Step 2: Convert pure mass to moles (mass / molar mass). Step 3: Use the balanced equation mole ratio to find moles of the other species. Step 4: Convert back to mass or volume. Step 5: If the question wants the impure sample, divide the pure mass by purity/100 at the end.

Why does percent purity matter for NEET?

NEET loves this because it adds one extra step that trips rushing students. Students do perfect stoichiometry, get the PURE answer (a common option), and forget the purity step. The exam sets that pure value as a trap option. Knowing when to multiply or divide by purity is often the difference between the right and wrong option.

⚠️ The NEET trap
✗ For '95% pure CaCO3 needed to neutralise 50 mL of 0.5 M HCl', stopping at the pure mass 1.25 g (option A).
✓ 1.25 g is only the PURE CaCO3. The sample is 95% pure, so the sample you weigh out = 1.25 x 100/95 = 1.32 g (option B).
🧠 The pure answer is always given as a trap option. If the question says '% pure sample required', you must DIVIDE by purity for one more step.

Real NEET questions

NEET 2022

What mass of 95% pure CaCO3 is required to neutralise 50 mL of 0.5 M HCl according to: CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + 2H2O(l)?

A · 1.25 g
B · 1.32 g ✓
C · 3.65 g
D · 9.50 g
Solution: Moles of HCl = 0.5 x 50/1000 = 0.025 mol. From the equation, 2 mol HCl react with 1 mol CaCO3, so moles CaCO3 = 0.025 / 2 = 0.0125 mol. Pure mass = 0.0125 x 100 = 1.25 g. The sample is only 95% pure, so mass of sample needed = 1.25 x 100/95 = 1.32 g. Option A (1.25 g) is the trap: it ignores purity.
NEET 2023 Phase 1

The mass of CO2 produced by heating 20 g of 20% pure limestone is (at. mass Ca = 40): CaCO3 -> CaO + CO2

A · 1.76 g ✓
B · 2.64 g
C · 1.32 g
D · 1.12 g
Solution: Here you are GIVEN an impure sample, so multiply. Pure CaCO3 = 20 x 20/100 = 4 g. M(CaCO3) = 100 g/mol, so moles = 4/100 = 0.04 mol. From CaCO3 -> CaO + CO2, moles CO2 = 0.04 mol. Mass CO2 = 0.04 x 44 = 1.76 g (option A by direct calculation). Note: the official NEET key marked (B) 2.64 g; work through both so you can match whichever key your source uses, but the method above is the standard purity approach.

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Frequently asked

Is percent purity the same as percent yield?

No. Percent purity is about the REACTANT sample (how much of the sample is the real compound). Percent yield is about the PRODUCT (how much you actually got compared to the theoretical maximum). Do not mix them.

Does impurity ever affect molar mass?

No. You always use the molar mass of the pure compound only. The impurity mass is separate dead weight and never enters the mole calculation.

What if the question gives purity as a fraction like 0.8 instead of 80%?

Same idea. Pure mass = impure mass x fraction. 0.8 already means 80/100, so just multiply directly (impure x 0.8).

Can a sample be more than 100% pure?

No. Purity ranges from 0% to 100%. If your working gives a purity above 100%, you made an arithmetic slip, usually multiplying when you should divide.

Related concepts:
Finding Atomic Weights from Two Compounds of Two ElementsThermal Decomposition of Limestone (CaCO3 to CaO + CO2)Mole-to-Mole Relationships in Chemical ReactionsStoichiometry of Reactions in Solution Using MolarityHow to Identify the Limiting Reagent from Given Amounts
Practise Chemistry for NEET:
Chemistry MCQs (free, chapter-wise)Some Basic Concepts Of Chemistry PYQs