Determine the molecular formula of a hydrocarbon given its percentage composition and the density of its gas at STP.

Assume 100 g of the compound.
Mass of Carbon (C) = 80 g
Mass of Hydrogen (H) = 20 g

Moles of Carbon = Mass / Molar mass = 80 g / 12 g/mol = 6.67 mol
Moles of Hydrogen = Mass / Molar mass = 20 g / 1 g/mol = 20 mol

Divide by the smallest number of moles to find the simplest ratio:
For C: 6.67 / 6.67 = 1
For H: 20 / 6.67 ≈ 3

The empirical formula is CH₃.

Given that 1 dm³ of the gas at S.T.P. weighs 1.35 g, and 1 mole of gas at S.T.P. occupies 22.4 dm³.

Molar mass = (Mass of 1 dm³) × (Molar volume at S.T.P.)
Molar mass = 1.35 g/dm³ × 22.4 dm³/mol
Molar mass = 30.24 g/mol

First, calculate the empirical formula mass for CH₃:
Empirical formula mass = (1 × Atomic mass of C) + (3 × Atomic mass of H)
Empirical formula mass = (1 × 12 g/mol) + (3 × 1 g/mol) = 12 + 3 = 15 g/mol

Now, find the factor 'n' by which the empirical formula must be multiplied:
n = Molar mass / Empirical formula mass
n = 30.24 g/mol / 15 g/mol
n ≈ 2.016 ≈ 2 (round to the nearest whole number)

Finally, multiply the empirical formula by 'n' to get the molecular formula:
Molecular formula = (CH₃)₂ = C₂H₆

The molecular formula of the hydrocarbon is C₂H₆.