Find the molecular formula for an alkyne with a carbon mass fraction of 90%.

Find the molecular formula for an alkyne with a carbon mass fraction of 90%. The relative hydrogen vapor density of this substance is 20.

Let’s implement the solution, formulate the condition of the problem:
Given: СnH2n – 2 – the general formula of alkynes;
W (C) = 90%;
D (H2) = 20;
M (H2) = 2 g / mol;
Determine: the molecular formula of the hydrocarbon CxNy:
1. Find the molecular weight by the formula:
M (CxHy) = M (H2) * D (H2) = 2 * 20 = 40 g / mol;
2. Determine the number of moles and mass of hydrogen, carbon:
m (H) = 72 * 0.1667 = 12 g;

m (C) = 72-12 = 60 g;

Y (C) = m / M = 60/12 = 5 mol;
Y (H) = m / M = 12/1 = 12 mol;
3. The ratio X: Y = C: H = 5: 12 = 1: 2. The simplest formula is CH2.
4. Let’s make calculations according to the general formula:
CnH2n – 2 = 40 g / mol;
12n + 2 – 2 = 40;
12n = 40;
n = 3.3
5. Molecular formula: C3H4 (propyne).
Answer: alkadiene hydrocarbon formula: C3H4