# 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