A sample of $$4.5 \text{ mg}$$ 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 $$3.1 \text{ mL}$$. The molecular weight of the unknown alcohol is ______ g/mol.

The balanced chemical equation for the Zerewitinoff determination is:

$$\mathrm{R-OH + CH_3MgI \longrightarrow CH_4\uparrow + R-OMgI}$$

From the stoichiometry of the reaction:

$$\mathrm{1\ mol\ Alcohol \longrightarrow 1\ mol\ CH_4}$$

Volume of methane evolved:

$$\mathrm{3.1\ mL}$$

Molar volume of gas:

$$\mathrm{22700\ mL\ mol^{-1}}$$

Moles of methane evolved:

$$\mathrm{Moles\ of\ CH_4 = \frac{3.1}{22700}}$$

Since the molar ratio is $$\mathrm{1:1}$$:

$$\mathrm{Moles\ of\ R-OH = \frac{3.1}{22700}\ mol}$$

Mass of alcohol given:

$$\mathrm{4.5\ mg = 4.5\times10^{-3}\ g}$$

Molecular weight of alcohol:

$$\mathrm{Molecular\ Weight = \frac{Mass}{Moles}}$$

$$\mathrm{= \frac{4.5\times10^{-3}}{\frac{3.1}{22700}}}$$

$$\mathrm{= \frac{4.5\times10^{-3}\times22700}{3.1}}$$

$$\mathrm{= \frac{102.15}{3.1}}$$

$$\mathrm{\approx 32.95\ g\ mol^{-1}}$$

Rounding to the nearest whole number:

$$\boxed{\mathrm{33\ g\ mol^{-1}}}$$