An excess of an ammoniacal solution of silver oxide was added to 2.2 g of a certain limiting aldehyde.

An excess of an ammoniacal solution of silver oxide was added to 2.2 g of a certain limiting aldehyde. This formed a precipitate weighing 10.8 g. Determine the formula for the starting aldehyde and name it.

Let’s find the amount of silver substance Ag.

n (Ag) = m: M.

M (Ag) = 108 g / mol.

n (Ag) = 10.8 g: 108 g / mol = 0.1 mol.

The radical of the aldehyde is denoted by X.

We get the formula X – COH.

Let’s compose the reaction equation for the “silver mirror”:

X – COH + Ag2O NH3 X – COOH + 2 Ag. (silver precipitates).

The quantitative ratio of aldehyde and silver is 1: 2, so the amount of the aldehyde substance will be 2 times less than the amount of the silver substance.

n (X – COH) = 0.1 mol: 2 = 0.05 mol.

The mass of a substance is found by the formula:

m = n × M, where M is the molar mass, n is the amount of substance.

M = m: n,

M (X – COH) = x + 12 + 16 + 1 = x + 29.

X + 29 = 2.2: 0.05,

X + 29 = 44,

X = 44 – 29,

X = 15.

Let us find the radical for the aldehyde, knowing that it consists of hydrogen and carbon atoms.

M (C) = 12g / mol,

M (H) = 1 g / mol.

The radical is CH3.

The formula of the CH3 – СОН aldehyde is acetaldehyde.