Answer
e) $C_3H_6O_2$ is the correct answer.
Work Step by Step
1. Assume a 100 g sample and calculate the amount of moles.
C:
$$ 48.6 \space g \times \frac{1 \space mole}{ 12.01 \space g} = 4.05 \space moles$$
H:
$$ 8.2 \space g \times \frac{1 \space mole}{ 1.008 \space g} = 8.1 \space moles$$
O:
$$ 43.2 \space g \times \frac{1 \space mole}{ 16.00 \space g} = 2.70 \space moles$$
2. The formula is $C_{4.05}H_{8.1}O_{2.70}$.
3. Dividing all numbers by 2.70: $C_{1.5}H_{3}O_1$.
In order to eliminate the decimal part, multiply the numbers by 2:
$C_3H_6O_2$
