Equilibrium JEE 2022: Empirical Formulae PYQ Solution

What was asked in the equilibrium JEE 2022 multi-correct question on empirical formulae?

Three pure binary compounds of elements P and Q have weight percentages: compound 1 exactly 50 % P and 50 % Q, compound 2 44.4 % P and 55.6 % Q, compound 3 40 % P and 60 % Q. The task is to verify which of the four statements about their empirical formulae are consistent with the given data and the law of multiple proportions.

Option A claimed that an empirical formula of P3Q4 for compound 3 forces compound 2 to be P3Q5.

Option B claimed that an empirical formula of P3Q2 for compound 3 combined with atomic weight 20 for P fixes the atomic weight of Q at 45.

Option C claimed that an empirical formula of PQ for compound 2 forces compound 1 to be P5Q4.

Option D claimed that atomic weights of 70 for P and 35 for Q make the empirical formula of compound 1 equal to P2Q.

How does the official solution walk through every step of this equilibrium JEE 2022 question?

The official solution lets the atomic weights of P and Q be $M_{P}$ and $M_{Q}$ respectively. For any compound the empirical mole ratio is obtained from $\frac{w_{P}}{M_{P}} : \frac{w_{Q}}{M_{Q}}$ .

Mass ratios are obtained directly from the percentages. Compound 1 gives P:Q = 50:50 = 1:1. Compound 2 gives P:Q ≈ 44.4:55.6 = 4:5. Compound 3 gives P:Q = 40:60 = 2:3.

Option A assumes compound 3 has empirical formula P3Q4. Its mass ratio must satisfy $3 M_{P} : 4 M_{Q} = 2 : 3$ . This simplifies to $\frac{3 M_{P}}{4 M_{Q}} = \frac{2}{3}$ , so $9 M_{P} = 8 M_{Q}$ and $\frac{M_{P}}{M_{Q}} = \frac{8}{9}$ .

For compound 2 the mass ratio 4:5 produces mole ratio $\frac{4}{M_{P}} : \frac{5}{M_{Q}}$ , which rearranges to $4 M_{Q} : 5 M_{P}$ . Substitute $\frac{M_{P}}{M_{Q}} = \frac{8}{9}$ to obtain $4 M_{Q} : 5 \times \frac{8}{9} M_{Q}$ or $4 : \frac{40}{9}$ . Multiply through by 9 to reach 36:40 or 9:10. The empirical formula is therefore P9Q10, not P3Q5. Option A is rejected.

Option B assumes compound 3 has empirical formula P3Q2 and $M_{P} = 20$ . Its mass ratio satisfies $3 \times 20 : 2 M_{Q} = 2 : 3$ or $60 : 2 M_{Q} = 2 : 3$ . Cross-multiplication yields $180 = 4 M_{Q}$ so $M_{Q} = 45$ exactly. Option B is correct.

Option C assumes compound 2 has empirical formula PQ. Its mass ratio 4:5 then requires $M_{P} : M_{Q} = 4 : 5$ . For compound 1 the mass ratio 1:1 produces mole ratio $\frac{1}{M_{P}} : \frac{1}{M_{Q}}$ , which equals $M_{Q} : M_{P}$ . Substituting the 4:5 ratio gives 5:4, so the empirical formula is P5Q4. Option C is correct.

Option D assumes $M_{P} = 70$ and $M_{Q} = 35$ . For compound 1 the mole ratio is $\frac{50}{70} : \frac{50}{35} = \frac{5}{7} : \frac{10}{7} = 1 : 2$ . This corresponds to PQ2, not P2Q. Option D is rejected.

The correct options are B and C.

What single reasoning error leads to wrongly accepting option A?

The method mistake that leads students to accept option A is using the mass ratio of compound 2 directly as mole ratio after finding $M_{P} / M_{Q}$ from compound 3 without performing the full $(4 / M_{P}) : (5 / M_{Q})$ conversion and subsequent simplification with the 8/9 ratio.

This shortcut yields an apparent 3:5 instead of the correct 9:10, leading to the false acceptance of P3Q5.

The correct procedure demands substituting the atomic-mass ratio into the mole-ratio expression before reducing to smallest integers.

What two related practice questions reuse the identical ratio technique?

Related question 1: Given two oxides of an element X with 50 % and 60 % X by mass, find possible empirical formulae assuming the lower oxide is X2O and verify consistency with multiple proportions.

Let atomic mass of X be A and of oxygen be 16. For the lower oxide (50 % X) taken as X2O, mass ratio X:O = 50:50 = 1:1 gives $2 A : 16 = 1 : 1$ so A = 8. For the second oxide (60 % X), mass ratio X:O = 60:40 = 3:2. Mole ratio becomes $\frac{3}{8} : \frac{2}{16} = 3 : 1$ , yielding empirical formula X3O. Fixing 16 g oxygen, X combines in 16 g and 24 g ratio (2:3), a simple multiple. The data are consistent.

Related question 2: Three compounds of A and B show mass ratios 1 : 1, 1.5 : 1 and 2 : 1; if the first is AB, deduce the formulae of the other two and the atomic mass ratio.

For the first compound, mass ratio A:B = 1:1 and formula AB imply mole ratio 1:1, so atomic mass ratio $M_{A} : M_{B} = 1 : 1$ . For the second, mass ratio 1.5:1 = 3:2 produces mole ratio $\frac{3}{M_{A}} : \frac{2}{M_{B}} = 3 : 2$ (since masses equal), giving A3B2. For the third, mass ratio 2:1 produces mole ratio 2:1, giving A2B. The atomic mass ratio is 1:1 throughout.

Use the past-paper archive to locate every JEE Advanced stoichiometry question from 2007 onward.

How can the official solution habits raise speed on similar JEE Advanced questions?

First reduce all percentage data to simplest integer mass ratios (50:50 → 1:1, 44.4:55.6 → 4:5, 40:60 → 2:3).

Treat each option as an independent assumption and propagate the derived $M_{P} : M_{Q}$ through every compound before comparing.

Never stop at the first matching ratio. Always complete the cross-multiplication and integer scaling to confirm the final empirical formula.

You can expect the official route to take 2.5 minutes once these habits are internalised, based on the official solution, versus 5+ minutes with inconsistent substitutions that force rechecking.

How does this PYQ from equilibrium JEE 2022 fit into the rest of your JEE Advanced preparation?

This question appears in Paper 2 of JEE Advanced 2022 and tests stoichiometry tools that occasionally surface in equilibrium problems involving percentage composition of reactants.

When you meet a similar composition question in a mock test, photograph any question you are stuck on and get a step-by-step solution (20 free).

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Frequently asked questions

What are the correct options in the equilibrium JEE 2022 empirical formulae question?

Options B and C are correct. For B, P3Q2 for compound 3 with atomic weight 20 for P fixes atomic weight of Q at exactly 45 to match the 2:3 mass ratio. For C, PQ for compound 2 sets MP:MQ as 4:5 which forces compound 1 to P5Q4 from its 1:1 mass ratio.

Why is option A wrong in equilibrium JEE 2022?

Assuming P3Q4 for compound 3 gives MP/MQ = 8/9. Substituting into compound 2's 4:5 mass ratio yields the mole ratio 36:40 or 9:10 after clearing fractions, so the empirical formula is P9Q10 and not P3Q5. Option A is therefore incorrect.

What is the common mistake in the equilibrium JEE 2022 question?

Students treat the mass ratio of compound 2 directly as its mole ratio after deriving MP/MQ from compound 3. This shortcut falsely yields 3:5 instead of performing the full (4/MP):(5/MQ) substitution and scaling, which actually simplifies to 9:10.

How do you convert mass percentage to empirical formula in JEE Advanced questions?

First reduce percentages to simplest mass ratios: 50:50 becomes 1:1, 44.4:55.6 becomes 4:5, and 40:60 becomes 2:3. Assume an empirical formula for one compound to find the atomic mass ratio MP:MQ, then propagate this ratio to the other compounds and compute their mole ratios to check consistency.

Equilibrium JEE 2022 Empirical Formulae Multi-Correct Solution