Question

A gaseous hydrocarbon contains 80% carbon and 20% hydrogen, 1dm3 of the compound at s.t.p weighs 1.35g find the molecular formular (1 mole of any gas at s.t.p occupies 22.4dm3, C=12, H=1)​

Answer 1

C₂H₆

Step-by-step explanation:

To obtain molecular formula of the gas, you need to find first its molecular mass.

To find molecular mass you need to obtain moles of gasbecause molecular mass is the ratio between mass of the gas and the volume it occupies.

As at STP, 1 mole of a gas occupies 22.4dm³, 1dm³ of gas are:

1dm³ × (1mol / 22.4dm³) = 004464moles.

Its molecular mass is:

1.35g / 004464moles =

30.24g/mol

Now, if the gas is 80%C and 20%H, its empirical formula (Simplest ratio of atoms in a molecule) is:

80% Carbon × (1mol / 12.01g) = 6.66moles C

20% Hydrogen × (1mol / 1.01g) = 19.8moles H

Ratio of H:C is:

19.8 mol H / 6.66mol C = 3

Thus, you can know you will have 3 moles of Hydrogen per mole of Carbon, CH₃ (It weighs 15.04g/mol)

As the gas weighs 30.24 ≈ 2 CH₃

The molecular formula is:

C₂H₆