Question

Which of the following pair of compounds demonstrates the law of multiple proportions?

• A. CH$_4$, CCl$_4$
• B. BF$_3$, NH$_3$
• C. CO, CO$_2$
• D. NO$_2$, CO$_2$

Hint:

The Law of Multiple Proportions applies to pairs of compounds formed from the *same two elements*. You then check if the ratios of the masses of one element, combined with a fixed mass of the other, are simple whole numbers.

Answer 1

The Correct Option is C

Concept:
The Law of Multiple Proportions states:

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

So, the compounds must contain the same two elements only. 
Step 1: Check Option A: $CH_4$ and $CCl_4$
\[ CH_4 \rightarrow C, H \] \[ CCl_4 \rightarrow C, Cl \] Different second elements are present. \[ \Rightarrow \text{Does not satisfy law} \] 
Step 2: Check Option B: $BF_3$ and $NH_3$
\[ BF_3 \rightarrow B, F \] \[ NH_3 \rightarrow N, H \] Different elements present. \[ \Rightarrow \text{Does not satisfy law} \] 
Step 3: Check Option C: $CO$ and $CO_2$
Both compounds contain same two elements: \[ C \text{ and } O \] Masses combining with fixed $12\,g$ carbon: \[ CO: 12\,g\,C + 16\,g\,O \] \[ CO_2: 12\,g\,C + 32\,g\,O \] Ratio of oxygen masses: \[ 16:32 = 1:2 \] This is a simple whole number ratio. \[ \Rightarrow \text{Satisfies law} \] 
Step 4: Check Option D: $NO_2$ and $CO_2$
\[ NO_2 \rightarrow N, O \] \[ CO_2 \rightarrow C, O \] Different first elements are present. \[ \Rightarrow \text{Does not satisfy law} \] 
Step 5: Final Answer
The pair which illustrates the Law of Multiple Proportions is: \[ \boxed{CO \text{ and } CO_2} \] 
Quick Tip:
For this law, always check that both compounds are formed from the same two elements only.