Follow-up Problem 2.3: Which sample(s) best display the law of multiple proportions for bromine and fluorine compounds?

Let's solve Follow-up Problem 2.3 step by step, based on the context of the sample problem and the images provided.

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🔍 Problem Statement:

> Which sample(s) best display(s) the fact that compounds of bromine (orange) and fluorine (yellow) exhibit the law of multiple proportions? Explain.

We are given three diagrams labeled A, B, and C, each showing combinations of orange atoms (bromine, Br) and yellow atoms (fluorine, F). We need to determine which one(s) illustrate the law of multiple proportions.

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🧠 Understanding the Law of Multiple Proportions:

The law of multiple proportions states:

> When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

In simpler terms:
- If Element A combines with Element B in different ways to form different compounds,
- Then the ratios of the masses of B that combine with a fixed amount of A will be simple whole numbers.
- This is observed when different compounds are formed from the same two elements.

So, for bromine and fluorine, if they form more than one compound, and the ratios of F to Br atoms differ in simple whole number ratios, then we’re seeing the law of multiple proportions.

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🔎 Analyze Each Diagram:

Let’s examine each image:

#### 🔹 Diagram A:
- Shows single atoms of Br (orange) and F (yellow), not combined.
- No compounds formed.
- ✘ Does not show any compound formation → Not relevant.

#### 🔹 Diagram B:
- Shows two types of molecules:
- One molecule has 1 Br and 2 F atoms → BrF₂
- Another molecule has 1 Br and 3 F atoms → BrF₃
- So, two different compounds are formed: BrF₂ and BrF₃
- The ratio of F atoms per Br atom is 2:3, a simple whole number ratio.

✔ This illustrates multiple proportions: For a fixed number of Br atoms (1), the number of F atoms varies in a simple whole number ratio (2:3).

#### 🔹 Diagram C:
- All molecules are identical: each has 1 Br and 2 F atoms → all are BrF₂
- Only one compound is formed.
- Since only one compound exists, no comparison between different compounds can be made.

✘ Does not illustrate multiple proportions — only definite composition.

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✔ Answer:

> Sample B best displays the fact that compounds of bromine and fluorine exhibit the law of multiple proportions.

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📝 Explanation:

In Sample B, bromine and fluorine form two different compounds:
- One with 1 Br : 2 F (BrF₂)
- Another with 1 Br : 3 F (BrF₃)

For a fixed mass of bromine, the mass of fluorine that combines with it differs in a ratio of 2:3, which is a small whole number ratio. This is exactly what the law of multiple proportions describes.

In contrast:
- Sample A shows no compounds.
- Sample C shows only one compound, so no comparison is possible.

Thus, only Sample B demonstrates the law of multiple proportions.

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✔ Final Answer:

> Sample B best displays the law of multiple proportions because it shows two different compounds of bromine and fluorine (with Br:F ratios of 1:2 and 1:3), where the masses of fluorine combining with a fixed mass of bromine are in a simple whole number ratio.