Animated Solution for Chemistry - Organic Chemistry: When 0.15 g of an organic compound was analysed using Carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is ....... (Nearest integer) [Atomic mass : Silver = 108, bromine = 80]

Imagine you have a mysterious organic compound, and you know it contains some bromine. But how much? You can't just pick the bromine atoms out with tweezers! This is where the Carius method comes to the rescue.

It is a brilliant piece of chemical detective work. We take a known mass of our organic compound—in this case, $0.15\text{ g}$—and subject it to some extreme conditions. We heat it in a sealed hard glass tube with fuming nitric acid (${\text{HNO}}_3$) and silver nitrate (${\text{AgNO}}_3$).

The fuming nitric acid acts as a ruthless oxidizer, destroying the organic carbon-hydrogen framework and turning it into carbon dioxide and water. Meanwhile, the liberated bromine atoms have nowhere to hide. They immediately react with the silver ions from the silver nitrate to form a solid, pale-yellow precipitate of silver bromide ($\text{AgBr}$).

By weighing this precipitate, we can trace back exactly how much bromine was in the original sample. It is a perfect application of the Law of Conservation of Mass!

Our experiment yielded $0.2397\text{ g}$ of $\text{AgBr}$ precipitate. But this isn't pure bromine; it's a combination of silver and bromine. To find the mass of just the bromine, we need to look at the molar masses.

The atomic mass of silver ($\text{Ag}$) is $108\text{ g/mol}$, and the atomic mass of bromine ($\text{Br}$) is $80\text{ g/mol}$. Adding these together gives us the molar mass of silver bromide:

This tells us a crucial fact: out of every $188\text{ g}$ of $\text{AgBr}$, exactly $80\text{ g}$ is pure bromine. We can use this ratio to find the mass of bromine in our specific precipitate:

Now that we have an expression for the mass of bromine, we can find its percentage in the original organic compound. The formula for percentage composition is straightforward:

Let's substitute our values into this master equation. The mass of the organic compound was given as $0.15\text{ g}$.

To make the calculation cleaner, let's bring the $188$ down to the denominator:

Now, it's just a matter of careful arithmetic. Multiplying the numerator gives $19.176$, and multiplying the denominator gives $28.2$.

Dividing these values yields $0.68$. Finally, multiplying by $100$ gives us our answer:

The organic compound contains exactly $68\%$ bromine by mass. A beautiful result from a classic analytical technique!