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]

Solution

Chemistry ma’am · 4 steps

Each R-OH gives exactly one CH₄, so moles of gas tell you moles of alcohol directly.

1. The reaction. The Grignard reagent takes the acidic H of the OH group and releases methane. One mole of a monohydric alcohol gives one mole of CH₄.

   R-OH + CH₃MgI → CH₄ ↑ + R-OMgI

2. Moles of gas. Take the gas volume at STP, where 1 mol occupies 22400 mL. Moles of CH₄ = 3.1/22400 = 1.384 × 10⁻⁴ mol, which equals the moles of alcohol.
3. Molar mass. Convert the mass to grams first: 4.5 mg = 4.5 × 10⁻³ g. Then M = (4.5 × 10⁻³)/(1.384 × 10⁻⁴) = 32.5 g mol⁻¹ (more precisely 32.52).
4. So. To the nearest integer, the molecular weight of the alcohol is 33 g mol⁻¹.

The common slip: 32. Writing 32 because the alcohol is likely methanol ignores the calculated 32.52, which rounds to 33.

The idea to keep. Active-hydrogen counting with CH₃MgI: every acidic H (OH, NH, terminal alkyne C-H) releases one CH₄. Gas moles at STP then give you moles of compound.

Answer from the NTA final answer key. Solution by PrepWiser.