Under normal conditions, 5.6 liters of alkane has a mass of 4.0 g. Derive its molecular formula.

Let’s implement the solution:

In accordance with the condition of the problem, we will write the data:
ShNu – alkane;

m (CxHy) = 4 g;

V (CxHy) = 5.6 L;

Determine the hydrocarbon formula.

Proportion:
1 mol of gas at normal level – 22.4 liters;

X mol (CxHy) -5.6 L from here, X mol (CxHy) = 1 * 5.6 / 22.4 = 0.25 mol.

Calculations by the formula:
Y = m / M; M (CxHy) = m / Y = 4 / 0.25 = 16 g / mol.

Let’s set the formula:
СnH2n + 2 is the general formula of saturated hydrocarbons (alkanes).

14n + 2 = 16;

14n = 14;

n = 1;

CH4 – methane;

M (CH4) = 12 + 4 = 16 g / mol.

Answer: CH4 – methane (alkanes).