A monobasic carboxylic acid containing 40% carbon, 6.667% hydrogen reacts with methanol.

A monobasic carboxylic acid containing 40% carbon, 6.667% hydrogen reacts with methanol. The vapor density of this substance for oxygen is 1.875 A) find the molecular formula of this acid B) write the equation for the reaction with methanol

Let’s take the mass of hydrocarbon per 100 g.

Let’s find the mass of hydrogen in the hydrocarbon.

100 g – 100%,

m (H) – 6.667%.

m = (100 g × 6.667%): 100% = 6.667 g.

Let’s find the mass of carbon.

100 g – 100%,

m (C) – 40%.

m = (100 g × 40%): 100% = 40 g.

Let’s find the mass of oxygen.

40 g (C) + 6.667 (H) = 46.667 g.

100 g – 46.667 = 53.33 g

Let’s find the amount of matter of atoms in the compound.

n = m: M.

n (H) = 6.667: 1 = 6.667 mol.

n (C) = 40: 12 = 3.33 mol.

n (O) = 53.33: 16 = 3.33 mol.

We find the ratio of the amounts of substances.

n (C): n (H): n (O) = 3.33: 6.667: 3.33 = 1: 2: 1.

Let’s find the molar mass of the hydrocarbon.

M (O2) = 32 g / mol.

D (O2) = 1.875 × 32 = 60.

The formula of carboxylic acid CH3COOH is acetic acid.

M (CH3COOH) = 60 g / mol.

CH3COOH + CH3OH → CH3COOCH3 + H2O (methyl ester of acetic acid – methyl acetate.)

Second way.

D (O2) = 2 × 1.875 = 60.

Find the number of carbon atoms.

Let x be the number of carbon atoms. The molar mass of the substance is 60.

0.4 = 12 x: 60.

12 x = 60 × 0.4.

12 x = 24,

X = 24: 12,

X = 2.

Let’s find the number of hydrogen atoms.

0.06667 = x: 60,

X = 60 × 0.06667,

X = 4.

Let’s find the number of oxygen atoms.

0.533 = 16 x: 60,

16x = 60 × 0.533,

16 x = 32,

X = 32: 16,

X = 2.

C2H4O2.

CH3COOH – acetic (ethanic) acid.