An alkanol containing 60% carbon by mass would have a molecular formula [H=1.0. C-12.0, O= 16.0]

Analyze the question: We need the alkanol (alcohol) whose percentage of carbon by mass is 60%.

Concept: Percentage of carbon = (mass of C in the formula ÷ molar mass) × 100

Solution:

• A. CH₃OH: M = 12 + 4 + 16 = 32. %C = 12/32 × 100 = 37.5%
• B. C₂H₅OH: M = 24 + 6 + 16 = 46. %C = 24/46 × 100 = 52.2%
• C. C₃H₇OH: M = 36 + 8 + 16 = 60. %C = 36/60 × 100 = 60%
• D. C₄H₉OH: M = 48 + 10 + 16 = 74. %C = 48/74 × 100 = 64.9%

Only C₃H₇OH (propanol) gives 60%.

Check using the general formula: For CₙH₂ₙ₊₁OH, M = 14n + 18. Setting 12n/(14n + 18) = 0.6 gives 12n = 8.4n + 10.8, so 3.6n = 10.8 and n = 3. This confirms propanol.

Answer: C. C₃H₇OH

Exam tip: For "percentage by mass" questions with options, just calculate %C for each option. It takes under a minute. Also memorise the alkanol formula CₙH₂ₙ₊₁OH (M = 14n + 18) to speed up the checks. Note that C₃H₇OH has a molar mass of exactly 60, which makes the percentage easy to compute: 36/60 = 0.6.