Step-by-step Solution

1. Data Analysis and Law of Multiple Proportions

First, let's analyze the composition of the three compounds and verify the law of multiple proportions. The law states that when two elements (P and Q) form multiple compounds, the masses of one element (Q) that combine with a fixed mass of the other element (P) are in a ratio of small whole numbers.

Let's calculate the mass of Q that combines with 1 gram of P for each compound.

• Compound 1: Weight % of P = 50, Weight % of Q = 50. Mass of Q per 1 g of P = 50 g / 50 g = 1.00

• Compound 2: Weight % of P = 44.4, Weight % of Q = 55.6. Mass of Q per 1 g of P = 55.6 g / 44.4 g ≈ 1.25

• Compound 3: Weight % of P = 40, Weight % of Q = 60. Mass of Q per 1 g of P = 60 g / 40 g = 1.50

The ratio of the masses of Q that combine with a fixed mass of P is: 1.00 : 1.25 : 1.50

To simplify this ratio to whole numbers, we can express the decimals as fractions: 1 : 5/4 : 3/2

Multiplying the entire ratio by 4 (the least common multiple of the denominators) gives a simple whole number ratio: 4 : 5 : 6

This ratio of masses $(m_Q / m_P)$ is directly proportional to the ratio of atoms $(n_Q / n_P)$ in the empirical formulas. Therefore, we have: $(n_Q/n_P)_1 : (n_Q/n_P)_2 : (n_Q/n_P)_3 = 4 : 5 : 6$

Now, let's evaluate each option using this relationship.

2. Evaluation of Option A

• Statement: If empirical formula of compound 3 is P₃Q₄, then the empirical formula of compound 2 is P₃Q₅.
• For compound 3 (P₃Q₄), the ratio of moles is $(n_Q/n_P)_3 = 4/3$.
• We know $(n_Q/n_P)_2 / (n_Q/n_P)_3 = 5/6$.
• Therefore, $(n_Q/n_P)_2 = (5/6) * (n_Q/n_P)_3 = (5/6) * (4/3) = 20/18 = 10/9$.
• So, the empirical formula for compound 2 should be $P₁Q_{10/9}$ or P₉Q₁₀.
• The option suggests the formula P₃Q₅, which has a mole ratio $n_Q/n_P = 5/3$.
• Since 10/9 ≠ 5/3, option A is incorrect.

3. Evaluation of Option B

• Statement: If empirical formula of compound 3 is P₃Q₂ and atomic weight of element P is 20, then the atomic weight of Q is 45.
• For compound 3, the ratio of weights is mass(P) / mass(Q) = 40 / 60 = 2/3.
• The empirical formula P₃Q₂ implies the ratio of moles is $n_P / n_Q = 3/2$.
• Using the relation n = mass / M (where M is atomic weight): $n_P / n_Q = (mass(P)/M_P) / (mass(Q)/M_Q) = 3/2$ $(mass(P)/mass(Q)) * (M_Q/M_P) = 3/2$
• Substituting the known values: $(2/3) * (M_Q / 20) = 3/2$ $M_Q / 30 = 3/2$ $M_Q = 30 * (3/2) = 45$
• The calculated atomic weight of Q is 45. Thus, option B is correct.

4. Evaluation of Option C

• Statement: If empirical formula of compound 2 is PQ, then the empirical formula of the compound 1 is P₅Q₄.
• For compound 2 (PQ), the ratio of moles is $(n_Q/n_P)_2 = 1/1 = 1$.
• We know $(n_Q/n_P)_1 / (n_Q/n_P)_2 = 4/5$.
• Therefore, $(n_Q/n_P)_1 = (4/5) * (n_Q/n_P)_2 = (4/5) * 1 = 4/5$.
• This means the empirical formula for compound 1 has a mole ratio $n_Q:n_P$ of 4:5. The formula is P₅Q₄.
• Thus, option C is correct.

5. Evaluation of Option D

• Statement: If atomic weight of P and Q are 70 and 35, respectively, then the empirical formula of compound 1 is P₂Q.
• For compound 1, weight % of P = 50 and weight % of Q = 50.
• Let's find the ratio of moles ($n_P : n_Q$) assuming a 100g sample: $n_P = mass(P) / M_P = 50 / 70 = 5/7$ $n_Q = mass(Q) / M_Q = 50 / 35 = 10/7$
• The mole ratio $n_P : n_Q$ is (5/7) : (10/7).
• Simplifying the ratio by multiplying by 7 gives 5 : 10, which reduces to 1 : 2.
• The empirical formula is P₁Q₂ or PQ₂.
• The statement says the formula is P₂Q. Thus, option D is incorrect.

Conclusion:

The correct options are B and C.