Question

0.22 g of a monohydric alcohol (A) librates 0.04 g of CH₄ at STP on reaction with CH₃MgBr. The molecular weight (in g/mole) of alcohol which satisfy these conditions is _____________________________________________________________________________________.

Answer 1

88

Answer 2

The reaction between a monohydric alcohol (R-OH) and methylmagnesium bromide (CH₃MgBr) is an acid-base reaction where the acidic hydrogen from the alcohol reacts with the methyl group of the Grignard reagent to produce methane gas.

The balanced chemical equation for the reaction is: $\text{R-OH} + \text{CH}_3\text{MgBr} \rightarrow \text{R-OMgBr} + \text{CH}_4$

From the stoichiometry of the reaction, 1 mole of monohydric alcohol reacts to produce 1 mole of methane (CH₄).

Given: Mass of CH₄ liberated = 0.04 g Molar mass of CH₄ = 12.01 (C) + 4 * 1.008 (H) ≈ 16.04 g/mol (often approximated as 16 g/mol for calculation purposes). Let's use 16 g/mol.

1. Calculate the moles of CH₄ liberated: $\text{Moles of CH}_4 = \frac{\text{Mass of CH}_4}{\text{Molar mass of CH}_4} = \frac{0.04 \text{ g}}{16 \text{ g/mol}} = 0.0025 \text{ mol}$

2. Determine the moles of alcohol (A): Since the reaction involves a 1:1 molar ratio between the alcohol and CH₄, the moles of alcohol are equal to the moles of CH₄ liberated. $\text{Moles of alcohol (A)} = 0.0025 \text{ mol}$

3. Calculate the molecular weight of the alcohol (A): Given mass of alcohol (A) = 0.22 g $\text{Molecular weight of alcohol (A)} = \frac{\text{Mass of alcohol (A)}}{\text{Moles of alcohol (A)}} = \frac{0.22 \text{ g}}{0.0025 \text{ mol}} = 88 \text{ g/mol}$

The molecular weight of the alcohol is 88 g/mol.