Take C = 12, H = 1, O = 16. What is the percentage by mass of carbon in propanone, CH₃COCH₃?
Correct answer A. 62.1%
Explanation
Molar mass of C₃H₆O = 36 + 6 + 16 = 58 g/mol. Percentage of carbon = (36/58) × 100 = 62.1%. The value 27.6% is the oxygen percentage.
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