Question

The percentage of carbon in ethanoic acid $\left(C{H}_{3}COOH\right)$ is

• A.
  $16\%$
• B.
  $40\%$
• C.
  $24\%$
• D.
  $60\%$

Answer 1

The correct option is B

$40\%$
Molecular mass of ethanoic acid = 60 u

Number of atoms of carbon in one molecule of ethanoic acid = 2

Atomic mass of carbon = 12 u

Mass of carbon in one molecule of ethanoic acid $=2\times 12=24u$

Percentage of carbon in $C{H}_{3}COOH=\frac{\text{Mass of carbon in}C{H}_{3}COOH}{\text{Molecular mass of}C{H}_{3}COOH}\times 100=\frac{24}{60}\times 100=40\%$

= Hence, the correct answer is option (b).