Molar Mass Determination of Ethane Produced via Wurtz Reaction

To find the molar mass of compound $(\text{Y})$, we analyze the reaction steps sequentially:

Step 1: Identification of the liberated gas ($\text{R-H}$) When a Grignard reagent ($\text{RMgI}$) reacts with water, an alkane ($\text{R-H}$) is formed: $\text{RMgI} + \text{H}_2\text{O} \rightarrow \text{R-H} + \text{Mg(OH)I}$

Given that $1\text{ g}$ of the liberated gas occupies $1.4\text{ dm}^3$ at STP, and the molar volume of an ideal gas at STP is $22.4\text{ dm}^3\text{ mol}^{-1}$: $\text{Molar mass of R-H} = \frac{22.4\text{ dm}^3\text{ mol}^{-1}}{1.4\text{ dm}^3\text{ g}^{-1}} = 16\text{ g mol}^{-1}$

The alkane with a molar mass of $16\text{ g mol}^{-1}$ is methane ($\text{CH}_4$). Thus, the alkyl group $\text{R}$ is a methyl group ($\text{CH}_3-$).

Step 2: Reaction to form compound $(\text{X})$ Methane reacts with iodine in the presence of an oxidizing agent like $\text{HIO}_3$ (which removes $\text{HI}$ to shift the reversible iodination equilibrium forward) to yield iodomethane: $\text{CH}_4 + \text{I}_2 \xrightarrow{\text{HIO}_3} \text{CH}_3\text{I} + \text{HI}$ Therefore, compound $(\text{X})$ is iodomethane ($\text{CH}_3\text{I}$).

Step 3: Reaction to form compound $(\text{Y})$ Compound $(\text{X})$ undergoes a Wurtz reaction when treated with sodium ($\text{Na}$) in dry ether: $2\text{CH}_3\text{I} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3-\text{CH}_3 + 2\text{NaI}$ Thus, compound $(\text{Y})$ is ethane ($\text{C}_2\text{H}_6$).

Step 4: Molar mass of compound $(\text{Y})$ $\text{Molar mass of C}_2\text{H}_6 = (2 \times 12) + (6 \times 1) = 30\text{ g mol}^{-1}$