Determine the molecular formula of the alcohol containing 21.6% oxygen by weight.

Since the condition does not say otherwise, we will assume that the unknown alcohol is limiting and monohydric. Then its general formula is: CnH2n + 1OH, where CnH2n + 1 is an unknown alkyl radical.

Let’s find the molecular weight of the alcohol:

M (CnH2n + 1OH) = M (O) * w (CnH2n + 1OH) / w (O) = 16 * 100 / 21.6 = 74 (amu).

Let’s find the molecular weight of the alkyl radical:

M (CnH2n + 1) = M (CnH2n + 1OH) – M (OH) = 74 – 17 = 57 (amu).

Let’s compose and solve the equation:

12 * n + 1 * (2n + 1) = 57,

n = 4, whence the molecular formula of alcohol: C4H9OH.

Answer: C4H9OH.