Animated Solution for Chemistry - Organic Chemistry: In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is (at. mass Ag = 108, Br = 80)

A Journey of Mass Conservation

Welcome to a fascinating application of quantitative analysis in organic chemistry! The Carius method is a classic and highly reliable technique used to determine the exact percentage of halogens (like chlorine, bromine, or iodine) present in an organic compound.

Imagine you have a mysterious organic compound, and you need to know exactly how much bromine is hiding inside its molecular structure. You can't just weigh the bromine directly because it's chemically bonded to carbon and hydrogen. This is where the genius of the Carius method comes into play.

Trapping the Halogen

In our problem, we start with exactly $250\mathrm{mg}$ of an organic compound. The first step is to completely destroy the organic framework (the carbon and hydrogen) without losing the precious bromine. We do this by heating the compound in a sealed tube with fuming nitric acid (${\mathrm{HNO}}_3$) and silver nitrate (${\mathrm{AgNO}}_3$).

The fuming nitric acid acts as a powerful oxidizing agent, converting all the carbon into carbon dioxide gas and all the hydrogen into water. Meanwhile, the bromine is liberated and immediately reacts with the silver nitrate to form a solid, insoluble precipitate of silver bromide ($\mathrm{AgBr}$).

According to the law of conservation of mass, every single atom of bromine from the original compound is now trapped inside this silver bromide precipitate. We are told that this precipitate weighs $141\mathrm{mg}$.

Before we can find the percentage of bromine in the original compound, we need to figure out what fraction of that $141\mathrm{mg}$ precipitate is actually pure bromine. To do this, we look at the molar mass of silver bromide.

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

This tells us a crucial fact: out of every $188\mathrm{g}$ of silver bromide, exactly $80\mathrm{g}$ is pure bromine.

Using a simple unitary method, we can determine the mass of bromine in our specific precipitate. If $188\mathrm{g}$ of $\mathrm{AgBr}$ contains $80\mathrm{g}$ of $\mathrm{Br}$, then $141\mathrm{mg}$ of $\mathrm{AgBr}$ will contain:

Now, to find the percentage of bromine in the original organic compound, we divide this mass of bromine by the total mass of the organic compound ($250\mathrm{mg}$) and multiply by $100$:

Notice a beautiful mathematical convenience here: because we are taking a ratio of two masses (the mass of bromine in milligrams divided by the mass of the compound in milligrams), the units cancel out perfectly. There is absolutely no need to convert milligrams to grams!

Let's simplify the expression:

Multiplying the numerators and denominators gives:

Dividing these numbers yields exactly $24\%$.

And there we have it! The organic compound contains exactly $24\%$ bromine by mass. The Carius method not only gives us a precise numerical answer but also beautifully demonstrates how chemical reactions can be used to isolate and measure specific elements within complex molecules.