To determine the molecular formula of the hydrocarbon gas, we need to analyze the information provided and use the concept of stoichiometry.

Given:
- Initial volume of hydrocarbon gas = 7.5 mL
- Contraction in volume = 15 mL
- Vapour density of the hydrocarbon = 14

Let's break down the problem into steps:

Step 1: Determine the initial volume of the hydrocarbon gas in moles.
To do this, we need to convert the volume of gas into moles using the ideal gas law equation:
PV = nRT

Since the volume and pressure are not given, we can assume constant pressure and temperature, so the equation simplifies to:
V = nRT/P

Given that the vapour density (VD) is the ratio of the molar mass of the hydrocarbon to the molar mass of hydrogen gas (VD = Molar mass of hydrocarbon/Molar mass of hydrogen), we can calculate the molar mass of the hydrocarbon:
Molar mass of hydrocarbon = VD * Molar mass of hydrogen

Substituting the values, we get:
Molar mass of hydrocarbon = 14 * 2 = 28 g/mol

Using the molar mass, we can calculate the initial moles of the hydrocarbon gas:
n = mass/molar mass
n = 7.5 mL * 28 g/mol = 210 g/mol

Step 2: Determine the final volume of the hydrocarbon gas in moles.
Since the contraction in volume is 15 mL, the final volume of the hydrocarbon gas can be calculated by subtracting the contraction from the initial volume:
Final volume = Initial volume - Contraction
Final volume = 7.5 mL - 15 mL = -7.5 mL

Step 3: Determine the final moles of the hydrocarbon gas.
We can calculate the final moles of the hydrocarbon gas using the ideal gas law equation:
n = PV/RT

Substituting the values, we get:
n = (-7.5 mL * 14 g/L) / (22.4 L/mol * 273 K)
n = -0.348 mol

Step 4: Determine the molecular formula.
To find the molecular formula, we need to compare the ratio of the initial and final moles of the hydrocarbon gas.

The initial moles of the hydrocarbon gas (n = 210 g/mol) and the final moles of the hydrocarbon gas (n = -0.348 mol) have a ratio of approximately -603.

Since the ratio is negative, it suggests that the initial hydrocarbon gas had a higher number of moles than the final hydrocarbon gas. This indicates that the hydrocarbon gas must have undergone combustion, resulting in the release of carbon dioxide and water.

The molecular formula of the hydrocarbon gas can be determined by considering the stoichiometry of the combustion reaction. Since the ratio of moles is approximately -603, it suggests that the hydrocarbon gas can be represented by the formula C603H.

Therefore, the molecular formula of the hydrocarbon gas is C603H.

methane CH4