The substance contains C-26.1%, H-4.35%, O-69.55%. Define a formula, give a name.

Let’s write down the solution:
1. This organic substance belongs to oxygen-containing, we give its general formula: CxHyOz;
2. M (CxOyOz) = 26.1 + 4.35 + 69.55 = 100 g / mol;
3. Let’s calculate the number of moles of carbon, hydrogen, oxygen:
Y (C) = 26.1 / 12 = 2.17 mol;
Y (H) = 4.35 / 1 = 4.35 mol;
Y (O) = 69.55 / 16 = 4.34 mol;
4. Ratio X: Y: Z = C: H: O = 2.17: 4.35: 4.34 = 1: 2: 2. Considering the ratio;
5. General formula of organic matter: CnH2n + 1COOH;
6. Determine the value – n:
7.12n + 2 + 1 + 12 + 16 * 2 + 1 = 100;
12n = 52;
n = 4;
Molecular formula: С4Н9СООН – pentanoic acid or valeric acid.
Answer: С4Н9СООН – a monobasic carboxylic acid called valeric acid.