To check the principle of multiple proportions, a series of pure binary compounds (PmQn) were analyz

Q: Why do we divide the weight percentage by the atomic mass?

The weight percentage gives us the mass of the element in 100g of the compound. Dividing this mass by the atomic mass gives the number of moles, which is necessary to find the empirical formula.

Q: Can we assume the atomic masses of P and Q are the same?

No, P and Q are different elements, so their atomic masses ($M_P$ and $M_Q$) will generally be different. We must treat them as variables unless specified otherwise.

Q: Why did we get an approximate ratio of 9:10 for compound 2 in option A?

Experimental data often contains slight variations. The ratio $44.4 : 49.42$ simplifies very closely to $9 : 10$ ($44.4/4.94 \approx 9$, $49.42/4.94 \approx 10$), which is standard practice in empirical formula calculations.

Q: What does the principle of multiple proportions mean here?

It states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. The different empirical formulas ($P_3Q_4$, $P_9Q_{10}$, etc.) reflect this principle.

Q: Why is option D incorrect when the math seems straightforward?

In option D, the molar ratio of P to Q is calculated as 1:2. This means for every 1 mole of P, there are 2 moles of Q, making the empirical formula $PQ_2$. The option incorrectly states it as $P_2Q$.

Question

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) \begin{array}{|c|c|c|} \hline 	ext{Compound} & 	ext{Weight \% of P} & 	ext{Weight \% of Q} \ \hline 1 & 50 & 50 \ \hline 2 & 44.4 & 55.6 \ \hline 3 & 40 & 60 \ \hline \end{array}

Options:

(A) If empirical formula of compound 3 is $P_3Q_4$, then the empirical formula of compound 2 is $P_3Q_5$.

(B) If empirical formula of compound 3 is $P_3Q_2$ and atomic weight of element P is 20, then the atomic weight of Q is 45.

(C) If empirical formula of compound 2 is PQ, then the empirical formula of the compound 1 is $P_5Q_4$.

(D) If atomic weight of P and Q are 70 and 35, respectively, then the empirical formula of compound 1 is $P_2Q$.

Sigma Percentile · Accepted Answer

### The Mystery of the Binary Compounds

Welcome to a fascinating journey into the heart of chemical composition!

Today, we are going to tackle a brilliant problem from JEE Advanced that tests our fundamental understanding of the mole concept and the principle of multiple proportions.

Imagine you are a chemist in a laboratory, and you have just synthesized three different binary compounds made of the exact same two elements, P and Q.

You run them through a mass spectrometer and obtain their weight percentages. But what are their actual chemical formulas? That is the puzzle we are going to solve.

### Decoding the Mole Concept

Before we dive into the options, let's establish our master tool.

The empirical formula of a compound represents the simplest whole-number ratio of the moles of its constituent elements.

To find the number of moles of an element in a 100g sample, we simply divide its weight percentage by its atomic mass.

Let's denote the atomic mass of element P as $M_P$ and the atomic mass of element Q as $M_Q$.

Therefore, the molar ratio of P to Q in any of these compounds will be given by the expression:

$$ 	ext{Molar Ratio} = \frac{W_P}{M_P} : \frac{W_Q}{M_Q} $$

This simple equation is the key that will unlock all four options.

### Analyzing Option A: The Domino Effect

Let's test the first hypothesis. Option A suggests that if the empirical formula of compound 3 is $P_3Q_4$, then compound 2 must be $P_3Q_5$.

Let's look at the data for compound 3. It consists of 40% P and 60% Q.

Using our master equation, the molar ratio is:

$$ \frac{40}{M_P} : \frac{60}{M_Q} = 3 : 4 $$

We can rearrange this to find a direct relationship between the atomic masses:

$$ \frac{40 M_Q}{60 M_P} = \frac{3}{4} \implies \frac{2 M_Q}{3 M_P} = \frac{3}{4} \implies 8 M_Q = 9 M_P $$

Now, let's carry this relationship over to compound 2, which has 44.4% P and 55.6% Q.

Its molar ratio is:

$$ \frac{44.4}{M_P} : \frac{55.6}{M_Q} = 44.4 M_Q : 55.6 M_P $$

Substituting $M_P = \frac{8}{9} M_Q$ into this ratio, we get:

$$ 44.4 M_Q : 55.6 \left(\frac{8}{9} M_Qight) = 44.4 : 49.42 \approx 9 : 10 $$

This means the empirical formula for compound 2 is actually $P_9Q_{10}$, not $P_3Q_5$.

**Option A is incorrect.**

### Analyzing Option B: A Direct Substitution

Now, let's evaluate Option B. It proposes that if compound 3 is $P_3Q_2$ and $M_P = 20$, then $M_Q$ must be 45.

We set up the molar ratio for compound 3 again, but this time equating it to 3:2.

$$ \frac{40}{M_P} : \frac{60}{M_Q} = 3 : 2 $$

Simplifying this gives us a new relationship:

$$ \frac{2 M_Q}{3 M_P} = \frac{3}{2} \implies 4 M_Q = 9 M_P $$

We are given that the atomic mass of P is 20. Let's substitute this value:

$$ 4 M_Q = 9(20) = 180 \implies M_Q = 45 $$

The math aligns perfectly!

**Option B is correct.**

### Analyzing Option C: The Reverse Engineering

Moving on to Option C. It states that if compound 2 is PQ, then compound 1 must be $P_5Q_4$.

If compound 2 is PQ, the molar ratio is 1:1.

$$ \frac{44.4}{M_P} : \frac{55.6}{M_Q} = 1 : 1 $$

This implies that:

$$ \frac{M_P}{M_Q} = \frac{44.4}{55.6} $$

Now, let's look at compound 1, which is a perfect 50-50 split by weight.

Its molar ratio is:

$$ \frac{50}{M_P} : \frac{50}{M_Q} = M_Q : M_P $$

Notice that this is exactly the reciprocal of the ratio we just found!

$$ M_Q : M_P = 55.6 : 44.4 \approx 5 : 4 $$

This gives us an empirical formula of $P_5Q_4$.

**Option C is correct.**

### Analyzing Option D: The Final Check

Finally, let's check Option D. It claims that if $M_P = 70$ and $M_Q = 35$, then compound 1 is $P_2Q$.

We already know compound 1 has 50% P and 50% Q. Let's plug in the given atomic masses:

$$ 	ext{Molar Ratio} = \frac{50}{70} : \frac{50}{35} $$

Simplifying the fractions:

$$ \frac{5}{7} : \frac{10}{7} = 5 : 10 = 1 : 2 $$

This means for every 1 mole of P, there are 2 moles of Q.

The empirical formula is $PQ_2$, not $P_2Q$.

**Option D is incorrect.**

### The Grand Conclusion

By systematically applying the mole concept and carefully tracking the relationships between atomic masses, we have successfully decoded the composition of these binary compounds.

The correct statements are B and C.

This problem is a beautiful reminder that in chemistry, mass is just an illusion; it is the moles that reveal the true structural reality of matter!

Select Answer:

Visualized Solution

$Analyzing the Data$

$Checking Option (A) - Part 1$

$Checking Option (A) - Part 2$

$Checking Option (B)$

$Checking Option (C) - Part 1$

$Checking Option (C) - Part 2$

$Checking Option (D)$

$Final Conclusion$

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Mystery of the Binary Compounds

Decoding the Mole Concept

Analyzing Option A

Analyzing Option B

Analyzing Option C

Analyzing Option D

The Grand Conclusion

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Data

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (A) - Part 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (A) - Part 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (B)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (C) - Part 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (C) - Part 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (D)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion