When analyzing the substance, it was found that it contains 40% carbon, 6.66% hydrogen, 53.34% oxygen.

When analyzing the substance, it was found that it contains 40% carbon, 6.66% hydrogen, 53.34% oxygen. Find the molar formula and name this substance if it is known that the density of its vapor in air is 2.07

1. We assume that the mass of the analyte is 100 g, then, respectively, the masses of the elements:

m (C) = w (C) * 100 = 0.4 * 100 = 40 g;

m (H) = w (H) * 100 = 0.0666 * 100 = 6.66 g;

m (O) = w (O) * 100 = 0.5334 * 100 = 53.34 g;

2. Calculate the number of moles per element:

n (C) = m (C): Ar (C) = 40: 12 = 3.33 mol;

n (H) = m (H): Ar (H) = 6.66: 1 = 6.66 mol;

n (O) = m (O): Ar (O) = 53.34: 16 = 3.33 mol;

3. Let us establish the ratio of elements in the simplest formula of a substance:

n (C): n (H): n (O) = 3.33: 6.66: 3.33;

n (C): n (H): n (O) = 1: 2: 1;

4. Let’s calculate the molar mass of the empirical formula:

M (CH2O) = 12 + 2 + 16 = 30 g / mol;

5. Knowing the value of the relative density of the substance in the air, we find the molar mass of the actual formula of the substance:

M (CxHyOx) = Dair (CxHyOx) * M (air) = 2.07 * 29 = 60 g / mol;

6. The molar mass of the actual formula is twice as large as the simplest, that is, to write the molecular formula of a substance, it is necessary to double the number of elements in the empirical formula:

M (CxHyOx): M (CH2O) = 60: 30 = 2;

CxHyOx = C2H4O2.

Answer: acetic acid CH3COOH.