What is the molecular mass of an alkanoic acid, if 0.5 mole of the acid weighs 44g?

Analyze the question: We're told 0.5 mole of an alkanoic acid has a mass of 44 g. We need its molar (molecular) mass.

Concept: Number of moles = Mass ÷ Molar mass So, Molar mass = Mass ÷ Moles

Solution:

Molar mass = 44 g ÷ 0.5 mol = 88 g/mol

Check with the alkanoic acid formula: Alkanoic acids have the general formula CₙH₂ₙ₊₁COOH, or CₘH₂ₘO₂.

• Using CₘH₂ₘO₂: 12m + 2m + 32 = 14m + 32
• Setting 14m + 32 = 88 gives 14m = 56, so m = 4
• The acid is C₄H₈O₂, which is butanoic acid (C₃H₇COOH). ✔

Check the other options:

• A (72): 14m + 32 = 72 gives m = 2.86, which is not a whole number, so it is not an alkanoic acid.
• C (60): m = 2, which is ethanoic acid (CH₃COOH). But 0.5 mole of it would weigh 30 g, not 44 g.
• D (46): m = 1, which is methanoic acid (HCOOH). But 0.5 mole would weigh 23 g, not 44 g.

Answer: The molecular mass is 88, which is option B.

Exam tip: When a question gives you a mass for a fraction of a mole, just divide the mass by the number of moles (or double it if it is 0.5 mole). Then check with the general formula CₘH₂ₘO₂ (molar mass = 14m + 32) to confirm that the result is a genuine alkanoic acid.