# M (COH) = 8.6 g, m (Ag) = 21.6 g. We need to find the formula for the aldehyde

According to the condition of the problem, we schematically write the process:

Aldehyde formula: R – COH. General formula: CnH2n + 1COH.

RCOH + Ag2O = 2Ag + R – COOH – oxidation of aldehyde, silver is released;

M (Ag) = 107.8 g / mol;

Determine the amount of moles of silver:

Y (Ag) = m / M = 21.6 / 107.8 = 0.2 mol;

Let’s make the proportion:

X mol (RCOH) – 0.2 mol (Ag);

-1 mol – 2 mol hence, X mol (RCOH) = 1 * 0.2 / 2 = 0.1 mol;

Let’s calculate the molar mass of the aldehyde:

Y = m / M; M (RCOH) = m / Y; M = 8.6 / 0.1 = 86 g / mol;

Let’s define the molecular formula of the aldehyde:

CnH2n + 1COH = 86 g / mol;

12n + 2 + 1 + 12 + 16 + 1 = 86;

12n = 54;

n = 4.5 = 5;

Molecular formula: C5H10O (pentanal);

M (C5H10O) = 12 * 5 + 1 * 10 + 16 = 86 g / mol.

Answer: the formula for the aldehyde C5H10O is pentanal.