The oxidation of 17.2 aldehyde with a silver oxide solution resulted in 43.2 g of metal.

The oxidation of 17.2 aldehyde with a silver oxide solution resulted in 43.2 g of metal. Determine the molecular formula of the aldehyde.

To solve the problem, it is necessary to schematically draw up an equation:
m = 17.2 g. m = 43.2 g;
1. R – COH + Ag2O = 2Ag + RCOOH – silver is released;
2. Calculations by formulas:
M (Ag) = 107.8 g / mol;
Y (Ag) = m / M = 43.2 / 107.8 = 0.4 mol.
4. Calculations by proportion:
X mol (RCOH) – 0.4 mol (Ag);
-1 mol -2 mol hence, X mol (RCOH) = 1 * 0.4 / 2 = 0.2 mol.
4. Calculate the molar mass:
M (RCOH) = m / Y = 17.2 / 0.2 = 86 g / mol.
5. Let’s establish the formula of the initial substance:
СnH2nO is the general formula of the aldehyde.
M CnH2nO = 86 g / mol;
12n + 2 + 16 = 86;
12n = 68;
n = 5.6;
C5H10O – pentanal.
Verification:
M (C5H10O) = 86 g / mol.
Answer: pentanal – С5Н10О participates in the reaction.