Determine the molecular formula of an organic substance that contains 62.1% carbon and 10.3% hydrogen

Determine the molecular formula of an organic substance that contains 62.1% carbon and 10.3% hydrogen, its vapor density for helium is 14.5. Organic matter gives a “silver mirror” reaction.

Let’s execute the solution:
1. According to the condition of the problem, organic matter participates in the reaction of a silver mirror, which means it belongs to aldehydes. Let’s write down its formula for making calculations schematically: CxHyOz.
2. Let’s calculate the molar mass, if the density for helium is known:
M (CxHyOz) = D (He) * M (He) = 14.5 * 4 = 58 g / mol.
3. Determine the amount of moles of carbon, hydrogen, oxygen:
Y (C) = 62.1 / 12 = 5.17 mol;
Y (H) = 10.3 / 1 = 10.3 mol;
m (C) = Y * M = 5.17 * 12 = 62.1 g;
m (H) = Y * M = 10.3 * 2 = 20.6 g;
m (O) = 100 – (62.1 + 20.6) = 17.3 g;
Y (H) = m / M = 17.3 / 2 = 8.65 mol.
4. The ratio: X: Y: Z = C: H: O = 5.17: 17.3: 8.65 = 1: 3: 1.
Simplest formula: CH3O
5. We derive the molecular formula: C3H6O (propanal)
M (C3H6O) = 3 * 12 + 1 * 6 + 16 = 58 g / mol.
Answer: C3H6O is propionic aldehyde.




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