Question

A gaseous hydrocarbon gives up on combustion 0.72 gram of water and 3.0 gram of carbon dioxide the empirical formula and molecular of the hydrocarbon is

Answer 1

no of mole of H2O is 0.72/18 = 0.04

no of moles of H = 0.08 ( 2 x 0.04)

no of moles of CO2 = 3/44 = 0.07

no of moles of C = 0.08

C: H mole ratio 7:8

The empirical formula is C7H8 i.e C6H5-CH3 = Toluene.

Answer 2

Hey there!

We know,

Hydrocarbon contains C and H. Let he formula be $C_xH_y$

 And

18 g of H₂O consist of 2 moles of H atoms.

Then, 0.72 g of H₂O consist of:

2/18 x 0.72 = 0.08

So, 0.72 g of H₂O consist of 0.08 moles of H atoms.

We know,

44 g of CO₂ consist of 1 mole of C atom.

Then, 3.08 g of CO₂ consist of:

1/44 x 3.08 = 0.07 

So, 3.08 g of CO₂ consist of 0.07 mole of C atom.

Then, ratio of x and y :

= 0.07/0.08 

= 7 : 8

So, empirical formula is C₇H₈.

Hope It Helps You!