Animated Solution for Chemistry - Organic Chemistry: In Carius method for estimation of halogens, $0.2\text{ g}$ of an organic compound gave $0.188\text{ g}$ of $\text{AgBr}$. The percentage of bromine in the compound is ......... (Nearest integer) [Atomic mass; $\text{Ag}=108$, $\text{Br}=80$]

The Carius method is a classic and elegant technique in organic chemistry used to determine the exact amount of halogens present in an organic compound. It might sound intimidating, but at its core, it's just a beautiful application of stoichiometry and the law of conservation of mass. Let's dive into how we can solve this problem step-by-step!

Imagine you have a mysterious organic compound, and you know it contains bromine. But how much? To find out, we use the Carius method. We take a known mass of our compound—in this case, $0.2\text{ g}$—and heat it in a sealed tube with fuming nitric acid and silver nitrate.

Why fuming nitric acid? It's a beast of an oxidizing agent! It completely destroys the organic framework, turning all the carbon into carbon dioxide and hydrogen into water. The bromine, now free, immediately reacts with the silver nitrate to form a solid, pale yellow precipitate of silver bromide ($\text{AgBr}$). We collect this precipitate, dry it, and weigh it. Our problem states we obtained $0.188\text{ g}$ of $\text{AgBr}$.

Now comes the fun part: the math. We need to find the percentage of bromine in the original compound. The logic is simple:

But wait, we don't have the mass of bromine directly; we have the mass of silver bromide. How do we extract the mass of just the bromine from it? We use their molar masses!

The atomic mass of bromine ($\text{Br}$) is $80\text{ g/mol}$, and silver ($\text{Ag}$) is $108\text{ g/mol}$. Therefore, the molar mass of $\text{AgBr}$ is $108+80=188\text{ g/mol}$. This means that in every $188\text{ g}$ of $\text{AgBr}$, there are exactly $80\text{ g}$ of bromine.

We can set up our master equation by substituting these values:

At first glance, the numbers $0.188$ and $188$ might look like a coincidence, but in JEE problems, they rarely are! This is a classic setup designed to reward students who look for smart cancellations rather than blindly multiplying decimals.

Let's convert the decimals into scientific notation to make the cancellation obvious: - $0.188$ becomes $188\times 10^{-3}$ - $0.2$ becomes $2\times 10^{-1}$

Substituting these back into our equation:

Look at that! The $188$ in the numerator and denominator cancel out perfectly. This is the moment of satisfaction in every stoichiometry problem. We are left with:

Simplifying further:

Since $10^{-2}\times 100=1$, the entire expression collapses beautifully to our final answer:

And there you have it! By understanding the chemistry behind the Carius method and applying a bit of smart algebra, we've cracked the problem. Always keep an eye out for those elegant cancellations!