Animated Solution for Chemistry - Organic Chemistry: In an estimation of bromine by Carius method, $1.6\text{ g}$ of an organic compound gave $1.88\text{ g}$ of $\text{AgBr}$. The mass percentage of bromine in the compound is ....... . (Atomic mass, $\text{Ag}=108$, $\text{Br}=80{\text{ g mol}}^{-1}$)

A Journey into Quantitative Analysis

Welcome to a fascinating exploration of quantitative analysis! In the realm of organic chemistry, determining the exact elemental composition of an unknown compound is like solving a thrilling mystery. Today, we are going to decode the percentage of bromine in an organic compound using the elegant and highly effective Carius method.

Imagine you are handed a mysterious organic powder weighing exactly $1.6\text{ g}$. You know it contains bromine, but how much? You can't just pick the bromine atoms out with tweezers. Instead, we use chemistry to force the bromine out of hiding.

In the Carius method, we take our $1.6\text{ g}$ sample and subject it to a rigorous chemical interrogation. We place it in a sealed glass tube and heat it intensely with fuming nitric acid (${\text{HNO}}_3$) and silver nitrate (${\text{AgNO}}_3$).

This extreme environment completely destroys the carbon-hydrogen framework of the organic compound. The carbon turns into carbon dioxide, the hydrogen into water, and the halogens—in our case, bromine—are liberated. The moment the bromine atoms are free, they are captured by the silver ions from the silver nitrate, forming a solid, pale-yellow precipitate of silver bromide ($\text{AgBr}$).

Our problem states that this process yields exactly $1.88\text{ g}$ of $\text{AgBr}$ precipitate.

Here is the most crucial logical leap in this entire process: atoms cannot simply disappear.

By the fundamental principle of conservation of mass, every single atom of bromine that was originally present in our $1.6\text{ g}$ of the organic compound is now securely locked inside the $1.88\text{ g}$ of the silver bromide precipitate. There is no other place for the bromine to go!

Therefore, if we can figure out exactly how much bromine is hiding inside this precipitate, we will know exactly how much bromine was in our original sample.

To unlock the mass of bromine from the precipitate, we need to look at the molecular level. We need the molar mass of silver bromide.

A single molecule of silver bromide consists of one silver atom and one bromine atom. The problem provides us with their atomic masses: - Mass of Silver ($\text{Ag}$) = $108\text{ g/mol}$ - Mass of Bromine ($\text{Br}$) = $80\text{ g/mol}$

Adding these together gives us the total molar mass of silver bromide:

With the molar mass in hand, we can set up a simple proportion. Out of every $188\text{ g}$ of pure silver bromide, exactly $80\text{ g}$ is contributed by bromine. This is a fixed, unchangeable ratio dictated by nature.

To find the mass of bromine in our specific sample of $1.88\text{ g}$ of precipitate, we take the fraction of bromine's mass over the total molar mass, and multiply it by our sample mass:

The numbers here are beautifully designed for us. Notice the relationship between $1.88$ and $188$. If you divide $1.88$ by $188$, you get exactly $0.01$. It's a simple shift of the decimal point!

Now, we just multiply $80$ by $0.01$:

This means there is exactly $0.8\text{ g}$ of bromine in our precipitate, and consequently, in our original organic compound.

We are almost at the finish line. We know the mass of bromine is $0.8\text{ g}$, and we know the total mass of the original organic compound is $1.6\text{ g}$.

To find the mass percentage of bromine, we simply take the part over the whole and multiply by one hundred:

Substituting our values:

Look at the fraction $0.8$ over $1.6$. Since $0.8$ is exactly half of $1.6$, the fraction simplifies beautifully to $\frac{1}{2}$.

Therefore, the mass percentage of bromine in the given organic compound is exactly $50\%$.

This problem perfectly illustrates the elegance of quantitative analysis. By transforming an unknown organic compound into a known inorganic precipitate, we can precisely determine its elemental composition. Keep practicing these stoichiometric relationships, and you will master them in no time!