Question

A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula. (2016 ICSE board question)

Answer 1

Molecular mass = Vapour Density x 2 = 29 x 2 = 58

One mole of given hydrocarbon has 58 grams of mass. In that 82.76% is carbon. So mass of of carbon is 58 X 82.76 /100 = 48 g

Now find the number of moles present is 48g of carbon. Number of moles = Mass/GramMolecularMass = 48/12 = 4

Therefore, there are 4 moles of Carbon atoms in 1 mole of given Hydrocarbon molecules.

Same way , calculate number of moles of Hydrogen. { Hydrocarbon mass is 58 and carbon mass is 48. So hydrogen mass is 10. Since Gram Molecular Mass of Hydrogen is one, number of moles is 10/1 = 10 }

So there are 4 moles of Carbon atoms and 10 moles of Hydrogen atoms in one mole of hydrocarbon molecule.

Therefore molecular formula is given hydrocarbon is C4H10.

Please tell if I am wrong. Edit suggestions are greatly appreciated.

Answer 2

A haseous hydrocarbon contains 82.76 of carbon. given that vapour density is 30 find its molecular formula