Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carius method. Reason

NCERT Exemplar · Q59

Q. Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carius method. Reason (R): Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid.

(i) Both A and R are correct and R is the correct explanation of A.

(ii) Both A and R are correct but R is not the correct explanation of A.

(iii) Both A and R are not correct.

(iv) A is not correct but R is correct.

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Concept understanding — Empirical Formula Calculation

Empirical Formula Calculation: From Intuition to Precision

Imagine you have a jar of marbles — some red, some blue. You don't know how many marbles are in the jar, but you know that for every 2 red marbles, there are 3 blue ones. That ratio — 2:3 — is the simplest description of the mixture. It doesn't tell you the total count, but it captures the essential relationship between the two types.

That's exactly what an empirical formula does for a chemical compound. It tells you the simplest whole-number ratio of atoms of each element present in the compound.

The Core Idea

When a new compound is discovered, chemists first find out what elements are in it and in what proportions by mass. But mass alone doesn't tell you the atomic ratio — because different atoms have different masses. A gram of hydrogen contains far more atoms than a gram of carbon.

The empirical formula is the bridge from "how much mass of each element" to "how many atoms of each element, in the simplest ratio."

Note

The empirical formula is not the same as the molecular formula. For hydrogen peroxide, the empirical formula is $HO$ (ratio 1:1), but the molecular formula is $H_2O_2$ . The empirical formula is the reduced fraction; the molecular formula is the actual molecule.

The Step-by-Step Process

Let's work through a concrete example. Suppose a compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

Step 1: Assume 100 g of the compound. This converts percentages directly into grams. So we have:

Carbon: 40.0 g
Hydrogen: 6.7 g
Oxygen: 53.3 g

Step 2: Convert each mass to moles. Use the atomic masses from the periodic table:

Moles of C = $\frac{40.0 \text{ g}}{12.0 \text{ g/mol}} = 3.33 \text{ mol}$
Moles of H = $\frac{6.7 \text{ g}}{1.0 \text{ g/mol}} = 6.7 \text{ mol}$
Moles of O = $\frac{53.3 \text{ g}}{16.0 \text{ g/mol}} = 3.33 \text{ mol}$

Step 3: Divide each mole value by the smallest mole value. This normalises the ratio:

C: $\frac{3.33}{3.33} = 1$
H: $\frac{6.7}{3.33} \approx 2$
O: $\frac{3.33}{3.33} = 1$

Step 4: If needed, multiply to get whole numbers. Here we already have 1:2:1, so the empirical formula is $CH_2O$ .

Watch out

Never round 0.5 to 1 or 0.33 to 0.3. If you get 1.5, multiply everything by 2. If you get 1.33, multiply by 3. The ratio must be exact whole numbers.

Why This Works

The key insight is that moles directly count atoms. One mole of any element contains the same number of atoms ( $6.022 \times 10^{23}$ ). So when you find the mole ratio, you're finding the atom ratio. Dividing by the smallest number just reduces that ratio to its simplest form.

\text{Empirical formula} = \text{simplest whole-number ratio of moles of each element}

Common Pitfalls …

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Assertion (A) is correct because the Carius method is a standard quantitative technique for sulphur estimation. Reason (R) is incorrect because the precipitate formed, barium sulphate, is a white solid, not a light yellow solid. None of the provided options accurately reflect this.

The question asks us to evaluate an Assertion (A) and a Reason (R) related to the Carius method for sulphur estimation and choose the correct option describing their relationship.

Concept and Intuition: Empirical Formula Calculation (Carius Method for Sulphur)

The Carius method is a classical quantitative analytical technique used to determine the percentage of certain elements (like halogens, sulphur, or phosphorus) present in an organic compound. The core idea is to convert the element of interest into a stable, weighable inorganic compound.

For sulphur, the organic compound is heated strongly with fuming nitric acid in a sealed tube (Carius tube). This process oxidises all the sulphur present in the organic compound into sulphuric acid ( $H_2SO_4$ ). After cooling, barium chloride ( $BaCl_2$ ) solution is added to the contents of the tube. Barium chloride reacts with the sulphuric acid to form a precipitate of barium sulphate ( $BaSO_4$ ), which is highly insoluble in water. This precipitate is then filtered, washed, dried, and weighed. From the known mass of $BaSO_4$ and its molar mass, the mass of sulphur can be calculated, and subsequently, its percentage in the original organic compound.

Step-by-step Analysis

Analyze Assertion (A): "Sulphur present in an organic compound can be estimated quantitatively by Carius method."
- As explained above, the Carius method is indeed a well-established and reliable quantitative method for estimating sulphur in organic compounds. The sulphur is converted to sulphuric acid and then precipitated as barium sulphate, which is weighed to determine the amount of sulphur.
- Therefore, Assertion (A) is correct.
Analyze Reason (R): "Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid."
- Let's break this reason into two parts:
  - Part 1: "Sulphur is separated easily from other atoms in the molecule." In the Carius method, sulphur is oxidised to sulphate ions, which are then precipitated as barium sulphate. This process effectively separates sulphur from the other components of the organic molecule for quantitative analysis. So, this part of the statement is generally true in the context of the method.
  - Part 2: "and gets precipitated as light yellow solid." This part of the statement is incorrect. The precipitate formed in the Carius method for sulphur estimation is barium sulphate ( $BaSO_4$ ). Barium sulphate is a white crystalline solid, not a light yellow solid. Elemental sulphur is yellow, but it is not precipitated in its elemental form during this estimation. …

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Q.Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carius method. Reason (R): Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid.

Concept and Intuition: Empirical Formula Calculation (Carius Method for Sulphur)

Step-by-step Analysis

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Q.Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carius method.
Reason (R): Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid.

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