Question

300 ml of a gaseous hydrocarbon when burnt in excess of o2 gave 2.4 lit. of co2 and 2.7 lit. of water vapour under same condition. the molecular formula of hydrocarbon is :-

Answer 1

CₓHₐ + (x + a/4)O₂ ------> xCO₂ + a/2H₂O
Here it is clear that , 1 vol of CₓHₐ reacts with (x + a/4)vol of O₂ is formed x vol of CO₂ and a/2 vol of H₂O .
∴ 300mL of CₓHₐ reacts with 300(x + a/4)mL of O₂ will form 300x mL of CO₂ and 300a/2 = 150a mL
Given, 300xmL = 2.4L ⇒300xmL = 2400mL[∵ 1 L = 1000mL ]
⇒ x = 8

Similarly, given, 150a mL= 2.7L ⇒150a mL = 2700 mL
⇒a = 2700/150 = 18

Hence, molecular formula of hydrocarbon is C₈H₁₈

Answer 2

We first write an imaginary equation of the reaction.

CH + O₂ ------> CO₂(g) + H₂O (g)

The CH (hydrogencarbon) = 300ml

CO₂ = 2.4 litres = 2.4 × 1000 = 2400ml

H₂O = 2.7 litres = 2.7 × 1000 = 2700ml

To get the moles produced and the reacting moles, we divide by the smallest volume :

2400 / 300 = 8moles of CO₂ (g)

2700 / 300 = 9 moles of H₂O (g)

Inserting this in the equation we get :

CH + O₂ -----> 8CO₂(g) + 9H₂O (g)

Balancing the equation we get :

C₈H₁₈ + 12.5 O₂ (g) —> 8 CO₂ (g) + 9 H₂O (g)

Therefore the molecular formula of the compound is :

C₈H₁₈