Question

A sample of 4.5 mg of an unknown monohydric alcohol, R–OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is___ g/mol. [Nearest integer]

Answer 1

Correct Answer: 33

To determine the molecular weight of the unknown monohydric alcohol \( R\text{–OH} \), we analyze the reaction with methylmagnesium iodide \( \text{CH}_3\text{MgI} \). The reaction is:
\(\text{R–OH} + \text{CH}_3\text{MgI} \rightarrow \text{RH} + \text{CH}_3\text{OH}\).
The evolved gas in this scenario is methane, \( \text{CH}_4 \). Given that the gas volume collected is 3.1 mL, we convert this to moles using the ideal gas law under standard conditions: \(22.4\text{ L/mol}\) at STP. Thus:

\[ \text{moles of CH}_4 = \frac{3.1 \text{ mL}}{22400 \text{ mL/mol}} = 1.3839 \times 10^{-4} \text{ mol} \]

Since the stoichiometry of the reaction indicates a 1:1 ratio between the alcohol \( \text{R–OH} \) and methane \( \text{CH}_4 \), the moles of \(\text{R–OH}\) are also \(1.3839 \times 10^{-4} \text{ mol}\).

The molecular weight \( \text{MW} \) is calculated as:
\[ \text{MW} = \frac{\text{mass (mg)}}{\text{moles}} = \frac{4.5 \text{ mg}}{1.3839 \times 10^{-4} \text{ mol}} = 32511.88 \text{ mg/mol} = 32.51 \text{ g/mol} \]

Rounding to the nearest integer, the molecular weight of the unknown alcohol is 33 g/mol. This lies within the expected range of 33,33.