Animated Solution for Chemistry - Organic Chemistry: In Carius method of estimation of halogen 0.172 g of an organic compound showed presence of 0.08 g of bromine. Which of these is the correct structure of the compound?

Imagine you are a detective in a chemistry lab. You are handed a mysterious organic compound weighing exactly $0.172\text{ g}$. Your mission? Identify its exact molecular structure.

Through the Carius method, a classic analytical technique, you extract all the bromine from this compound and find that it weighs $0.08\text{ g}$. This is our crucial clue. Every chemical compound has a unique "fingerprint" based on the mass percentage of its constituent elements. By calculating the percentage of bromine in our mystery sample, we can match it against the theoretical percentages of our suspects (the given options).

To find the mass percentage of bromine, we use a very straightforward formula:

Let's substitute our experimental values into this equation:

When we crunch the numbers, we get $46.51\%$. This is our target! The correct molecular structure must have exactly $46.51\%$ bromine by mass.

Now, let's test our options one by one. We calculate the theoretical percentage of bromine for each using their molecular formulas.

Suspect A: Methyl Bromide (${\text{CH}}_3\text{Br}$) The molecular mass is $12+3+80=95$. The percentage of bromine is $\frac{80}{95}\times 100=84.21\%$. This is way too high. Option A is innocent.

Suspect B: Ethyl Bromide (${\text{C}}_2{\text{H}}_5\text{Br}$) The molecular mass is $24+5+80=109$. The percentage of bromine is $\frac{80}{109}\times 100=73.39\%$. Still not a match.

Suspect C: p-Bromoaniline (${\text{C}}_6{\text{H}}_6\text{NBr}$) Let's calculate its molecular mass carefully. Six carbons ($72$), six hydrogens ($6$), one nitrogen ($14$), and one bromine ($80$). The total mass is $172$. The percentage of bromine is $\frac{80}{172}\times 100=46.51\%$. Bingo! We have a perfect match.

Suspect D: Dibromoaniline (${\text{C}}_6{\text{H}}_5{\text{NBr}}_2$) Just to be absolutely thorough, let's check the last option. With two bromine atoms, the molecular mass jumps to $251$, and the bromine mass is $160$. The percentage is $\frac{160}{251}\times 100=63.74\%$. Definitely not our compound.

The experimental percentage of bromine perfectly aligns with the theoretical percentage of p-bromoaniline. Therefore, we can confidently conclude that option (c) is the correct structure of our organic compound. Always trust the math, it never lies!