An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5

Q: Why is the $5.0\text{ g}$ of chlorine given in the question?

It is a classic distractor! That amount was used to synthesize Compound A, but it doesn't tell us how much actually reacted. We only care about the $0.5\text{ g}$ sample taken for analysis.

Q: How does the Carius method work?

The organic compound is heated with fuming nitric acid and silver nitrate. The carbon and hydrogen oxidize to gases, while the halogen reacts with silver to form a solid precipitate like $\text{AgCl}$.

Q: Why do we divide by $143.37$?

$143.37\text{ g/mol}$ is the molar mass of $\text{AgCl}$. Dividing the atomic mass of $\text{Cl}$ ($35.5$) by this gives the mass fraction of chlorine in the precipitate.

Q: Is there a way to estimate the answer without a calculator?

Yes! $35.5 / 143.37$ is roughly $1/4$. $0.3849$ is roughly $0.4$. So, $(1/4) \times 0.4 = 0.1\text{ g}$ of $\text{Cl}$. Then, $(0.1 / 0.5) \times 100 = 20\%$. The exact answer is $19\%$, so estimation gets you very close!

Q: What if the compound contained bromine instead?

The process is identical, but you would get an $\text{AgBr}$ precipitate. You would use the atomic mass of bromine and the molar mass of $\text{AgBr}$ in your calculations.

Question

An organic compound is subjected to chlorination to get compound A using $5.0	ext{ g}$ of chlorine. When $0.5	ext{ g}$ of compound A is reacted with $	ext{AgNO}_3$ [Carius method], the percentage of chlorine in compound A is ............ when it forms $0.3849	ext{ g}$ of $	ext{AgCl}$. (Round off to the nearest integer)
(Atomic masses of $	ext{Ag}$ and $	ext{Cl}$ are $107.87$ and $35.5$ respectively)

Sigma Percentile · Accepted Answer

### The Trap of Extra Information

Imagine you are in a chemistry lab, synthesizing a new organic compound. You use a whole $5.0	ext{ g}$ of chlorine gas to get the reaction going.

But here is the catch: when it is time to analyze the compound to see how much chlorine actually made it inside, you only take a tiny $0.5	ext{ g}$ sample.

That $5.0	ext{ g}$ mentioned in the problem? It is a classic distractor! It is just the amount used during synthesis, not the amount present in our analytical sample. We must focus entirely on the $0.5	ext{ g}$ of Compound A.

### The Magic of the Carius Method

So, how do we find out how much chlorine is hiding in that $0.5	ext{ g}$ sample? Enter the **Carius Method**.

By heating the organic compound with fuming nitric acid and silver nitrate, we completely destroy the organic structure. The carbon and hydrogen burn off as gases, but the chlorine is trapped.

It reacts with the silver to form a solid, white precipitate of silver chloride ($	ext{AgCl}$). The problem tells us we collected exactly $0.3849	ext{ g}$ of this precipitate. Because of the law of conservation of mass, we know that **every single atom of chlorine** in that precipitate came directly from our $0.5	ext{ g}$ sample.

### Decoding the Molar Masses

To find the mass of chlorine in the precipitate, we need to look at the molar masses.

The atomic mass of silver ($	ext{Ag}$) is $107.87	ext{ g/mol}$, and chlorine ($	ext{Cl}$) is $35.5	ext{ g/mol}$.

Adding these together, the molar mass of $	ext{AgCl}$ is $143.37	ext{ g/mol}$. This means that in every $143.37	ext{ g}$ of silver chloride, there is exactly $35.5	ext{ g}$ of chlorine.

We can use this ratio to find the exact mass of chlorine in our specific $0.3849	ext{ g}$ precipitate:

$	ext{Mass of Cl} = \frac{35.5}{143.37} 	imes 0.3849	ext{ g}$

### The Final Percentage

Now we have the mass of the chlorine. But the question asks for the **percentage** of chlorine in the original organic compound.

To find this, we divide the mass of the chlorine by the total mass of the organic compound sample ($0.5	ext{ g}$), and multiply by $100$:

$\% 	ext{ of Cl} = \frac{\frac{35.5}{143.37} 	imes 0.3849}{0.5} 	imes 100$

Let's carefully calculate the values. Multiplying the numerators gives us approximately $13.66395$, and the denominator is $71.685$.

Dividing these gives $0.1906$. Multiplying by $100$ yields **$19.06\%$**.

The question specifically asks us to round off to the nearest integer. Rounding $19.06$ gives us our final, elegant answer: **$19$**.

Enter Numerical Value:

Visualized Solution

The Carius Method Principle

Molar Masses of $AgCl$

Mass of $Cl$ in Precipitate

Percentage of $Cl$

Final Calculation

The Way Forward

The Sigma Insight: Nomenclature and Characterisation

The Trap of Extra Information

The Magic of the Carius Method

Decoding the Molar Masses

The Final Percentage

Similar Questions

In Carius method for estimation of halogens, $0.2 g$ of an organic compound gave $0.188 g$ of $AgBr$ . The percentage of bromine in the compound is ......... (Nearest integer) [Atomic mass; $Ag = 108$ , $Br = 80$ ]

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

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]

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)

In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of

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?

In the sulphur estimation, $0.471 g$ of an organic compound gave $1.44 g$ of barium sulphate. The percentage of sulphur in the compound is ......... %. (Nearest integer) (Atomic mass of $Ba = 137 u$ )

29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is