A 100 mL solution of CH₃ CH₂ MgBr on treatment with methanol produces 2.24 mL of a gas at STP. The weight of gas produced is _____________ mg. [nearest integer]

Solution

Chemistry ma’am · 4 steps

The Grignard's ethyl group grabs the H from methanol, so the gas is ethane, not methane.

1. Identify the gas. Methanol has an acidic O–H hydrogen. The ethyl group of the Grignard reagent acts as a strong base, takes that hydrogen, and leaves as ethane gas.

   CH₃CH₂MgBr + CH₃OH → CH₃CH₃ ↑ + CH₃OMgBr

2. Moles of gas. At STP one mole of gas occupies 22400 mL. So n = 2.24/22400 = 1 × 10⁻⁴ mol of ethane.
3. Convert to mass. The molar mass of C₂H₆ is 2 × 12 + 6 × 1 = 30 g mol⁻¹. Mass = 1 × 10⁻⁴ × 30 = 3 × 10⁻³ g.
4. So. 3 × 10⁻³ g is 3 mg, so the answer is 3.

The common slip: 2. Taking the gas as methane, CH₄, from the methanol gives 1.6 mg, which rounds to 2. The alkane comes from the Grignard's ethyl group, so it is ethane.

The idea to keep. A Grignard reagent RMgX with any O–H or N–H compound always gives the alkane R–H from its own alkyl group. Then it is just volume to moles to mass.

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