Alkyne weighing 0.54 g completely reacted with chlorine of 0.448 dm3. Determine the molecular formula of the alkyne.

Let’s write down the solution:
1. According to the condition of the problem, let’s outline the process:
m = 0.54 g. V = 0.448 dm3
CxHy + Cl2;
2. Let’s make the calculations:
V (Cl2) = 0.448 dm3 = 0.448 l;
Proportion: 1 mol of gas at normal level – 22.4 liters;
X mol (Cl2) – 0.448 L. hence, X mol (Cl2) = 1 * 0.448 / 22.4 = 0.02 mol.
3. Determine the hydrocarbon formula:
Y (CxHy) = m / M;
M (CxHy) = m / Y = 0.54 / 0.02 = 27 g / mol;
General formula of alkynes: СnH2n – 2;
12n + 2 – 2 = 27;
12n = 27;
n = 2;
C2H2 – acetylene;
Checking: M (C2H2) = 12 * 2 + 2 = 26 g / mol.
Answer: molecular formula C2H2 (ethine).