Q.A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations AB, $AB_2$, $A_2B$ and $A_2B_3$ and show that law of multiple proportions is applicable.

The key idea is that the law of multiple proportions applies when two elements combine in different whole-number ratios by mass.

Step 1 – Fix the mass of one element.

Take a fixed mass of A, say 2 grams (the mass of one A ball).

Step 2 – Find the mass of B combining with 2 g of A in each compound.

• AB: 2 g A combines with 5 g B.
• AB₂: 2 g A combines with $2 \times 5 = 10$ g B.
• A₂B: 4 g A combines with 5 g B → for 2 g A, mass of B = $\frac{5}{4} \times 2 = 2.5$ g.
• A₂B₃: 4 g A combines with $3 \times 5 = 15$ g B → for 2 g A, mass of B = $\frac{15}{4} \times 2 = 7.5$ g.

Step 3 – Compare the masses of B. …

The law of multiple proportions says: 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.

Here, element A (red ball) weighs 2 g each. Element B (blue ball) weighs 5 g each. The compounds are AB, AB₂, A₂B, and A₂B₃. To test the law, we fix the mass of A and see how much B combines with it in each compound.

Why fix A? Because A’s mass per atom is smaller (2 g) and appears in different counts across the compounds. Fixing A lets us compare B’s mass directly.

Step-by-step reasoning

1. Find the mass of A in each compound

   • In AB: 1 atom of A → mass of A = $2$ g
   • In AB₂: 1 atom of A → mass of A = $2$ g
   • In A₂B: 2 atoms of A → mass of A = $2 \times 2 = 4$ g
   • In A₂B₃: 2 atoms of A → mass of A = $2 \times 2 = 4$ g

   The masses of A are not the same across all four. So we must fix a common mass of A — say, 2 g — and scale the B masses accordingly.

2. Scale each compound to a fixed mass of A = 2 g

   • For AB: already has 2 g A. Mass of B = $1 \times 5 = 5$ g.
   • For AB₂: already has 2 g A. Mass of B = $2 \times 5 = 10$ g.
   • For A₂B: has 4 g A. To reduce A to 2 g, take half the formula unit. Then B mass becomes $\frac{1}{2} \times 5 = 2.5$ g.
   • For A₂B₃: has 4 g A. Again take half the formula unit. Then B mass becomes $\frac{1}{2} \times (3 \times 5) = \frac{15}{2} = 7.5$ g.

3. List the masses of B that combine with 2 g of A

   Compound	Mass of A (g)	Mass of B (g) for given formula	Scaled mass of B for 2 g A
   AB	2	5	5
   AB₂	2	10	10
   A₂B	4	5	2.5

Concept: Law of Multiple Proportions

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

Method: Fixed Mass Method

Step 1: Identify the fixed element

We fix the mass of element A (red balls, 2 g each).

Let’s take 1 gram of A as the fixed mass.

Step 2: Find mass of B per gram of A in each compound

• AB → 1 atom A (2 g) + 1 atom B (5 g)

  Mass of B per 1 g of A = $\frac{5}{2} = 2.5$ g

• AB₂ → 1 atom A (2 g) + 2 atoms B (10 g)

  Mass of B per 1 g of A = $\frac{10}{2} = 5$ g

• A₂B → 2 atoms A (4 g) + 1 atom B (5 g)

  Mass of B per 1 g of A = $\frac{5}{4} = 1.25$ g

• A₂B₃ → 2 atoms A (4 g) + 3 atoms B (15 g)

  Mass of B per 1 g of A = $\frac{15}{4} = 3.75$ g

Step 3: Find the ratio of these masses

Take the smallest mass (1.25 g) as the base:

• AB : $\frac{2.5}{1.25} = 2$ …

Here's a clear breakdown of the common mistakes students make when applying the law of multiple proportions to this problem, and how to avoid each.

🧠 The Concept First

The law of multiple proportions states:

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

In this problem, the two elements are A (red balls, 2 g each) and B (blue balls, 5 g each).

The compounds are:

• AB
• AB₂
• A₂B
• A₂B₃

You must fix the mass of one element (say A) and find the masses of B that combine with it, then check if those masses are in a simple whole-number ratio.

✗ Common Mistake #1: Not fixing the mass of one element

What students do wrong:

They directly compare the masses of A and B in each compound without fixing a common reference mass.

Example of error:

They might say:

• In AB: mass ratio A:B = 2:5
• In AB₂: mass ratio A:B = 2:10
• Ratio of B masses = 5:10 = 1:2 → "So law holds."

This is incorrect because they didn't fix the mass of A to the same value across compounds — they just happened to have the same mass of A (2 g) in both, but this is not a general method and leads to confusion in other cases.

✓ How to avoid:

Always fix the mass of one element to a constant value (say 1 g or 2 g) across all compounds. Then find the mass of the other element that combines with that fixed mass.

Correct approach:

Fix mass of A = 2 g (since each A ball is 2 g). Then:

Now compare masses of B:

5 : 10 : 2.5 : 7.5 → multiply by 2 → 10 : 20 : 5 : 15 → divide by 5 → 2 : 4 : 1 : 3 → small whole numbers ✓

✗ Common Mistake #2: Forgetting to simplify the ratio to smallest whole numbers

What students do wrong:

They get a ratio like 5:10:2.5:7.5 and stop there, saying "these are not whole numbers, so law fails."

✓ How to avoid:

Always multiply by a common factor to clear decimals, then divide by the greatest common divisor to get the simplest whole-number ratio.

✗ Common Mistake #3: Confusing the law with the law of definite proportions

What students do wrong:

They check if the mass ratio of A to B is constant in all compounds. That's the law of definite proportions (same compound always has same ratio), not the law of multiple proportions.

✓ How to avoid:

Remember:

• Law of definite proportions → same compound, fixed ratio.
• Law of multiple proportions → different compounds of same two elements, masses of one element with fixed mass of other are in small whole-number ratio.

✗ Common Mistake #4: Using the wrong fixed mass (e.g., fixing B instead of A)

What students do wrong: …