To check the principle of multiple proportions, a series of pure binary compounds (P_mQ_n) were analyzed and their composition is tabulated below. The correct option(s) is (are)

Compound	Weight % of P	Weight % of Q
1	50	50
2	44.4	55.6
3	40	60

The correct options are: "If empirical formula of compound 3 is P₃Q₂ and atomic weight of element P is 20, then the atomic weight of Q is 45" and "If empirical formula of compound 2 is PQ, then the empirical formula of compound 1 is P₅Q₄."

First, let us extract the key data from the table. The weight percentages are:

Compound 1: P = 50%, Q = 50%
Compound 2: P = 44.4%, Q = 55.6%
Compound 3: P = 40%, Q = 60%

From these percentages, we can calculate the weight ratio of P to Q in each compound:

Compound 1: P : Q = 50 : 50 = 1 : 1
Compound 2: P : Q = 44.4 : 55.6 = 4 : 5 (dividing both by 11.1)
Compound 3: P : Q = 40 : 60 = 2 : 3

Now let us verify each of the two correct options step by step.

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Option B: Empirical formula of compound 3 is P₃Q₂, atomic weight of P = 20 → find atomic weight of Q.

If the empirical formula is P₃Q₂, then in one formula unit there are 3 atoms of P and 2 atoms of Q. The mass contributed by each element is:

Mass of P portion = 3 × (atomic weight of P) = 3 × 20 = 60
Mass of Q portion = 2 × (atomic weight of Q) = 2 × M_Q

The weight ratio of P to Q in compound 3 must equal 2 : 3 (from the table data above).

So we set up the equation:

\frac{3\times 20}{2\times M_Q}=\frac{2}{3}

Cross-multiplying:

3\times 60=2\times 2\times M_Q

180=4M_Q

M_Q=45

The atomic weight of Q = 45. ✓ This option is correct.

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Option C: Empirical formula of compound 2 is PQ → find empirical formula of compound 1.

If compound 2 has the empirical formula PQ, then the mole ratio of P : Q = 1 : 1, and the weight ratio of P : Q must also equal the ratio of their atomic weights:

\frac{\text{Weight\% of P}}{\text{Weight\% of Q}}=\frac{M_P}{M_Q}

From compound 2: weight ratio P : Q = 44.4 : 55.6 = 4 : 5, so:

\frac{M_P}{M_Q}=\frac{4}{5}

Let M_P = 4k and M_Q = 5k for some constant k.

Now use compound 1 where weight ratio P : Q = 50 : 50 = 1 : 1.

If compound 1 has empirical formula P_mQ_n, the weight ratio is:

\frac{m\times M_P}{n\times M_Q}=\frac{1}{1}

Substituting M_P = 4k and M_Q = 5k:

\frac{m\times 4k}{n\times 5k}=1

\frac{m}{n}=\frac{5}{4}

So the simplest whole-number ratio is m : n = 5 : 4, giving the empirical formula P₅Q₄. ✓

This option is correct.

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Why are Options A and D incorrect?

Option A says: If compound 3 is P₃Q₄, find compound 2. If compound 3 is P₃Q₄, then weight ratio P : Q = (3M_P) : (4M_Q) = 2 : 3, which gives M_P/M_Q = 8/9. For compound 2 with weight ratio 4 : 5: moles ratio = (4/M_P) : (5/M_Q) = (4 × 9) : (5 × 8) = 36 : 40 = 9 : 10. The empirical formula would be P₉Q₁₀, not P₃Q₅. So Option A is incorrect.

Option D says: Atomic weights of P = 70, Q = 35, find formula of compound 1. Weight ratio in compound 1 = P : Q = 1 : 1. Moles of P = 50/70 = 5/7, moles of Q = 50/35 = 10/7. Ratio of moles P : Q = (5/7) : (10/7) = 1 : 2. So the empirical formula is PQ₂, not P₂Q. So Option D is incorrect.

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Summary: The two correct options are B (atomic weight of Q = 45) and C (empirical formula of compound 1 is P₅Q₄), both of which are verified by correctly applying the weight-to-mole ratio logic and the principle of multiple proportions.