Question

Four grams of hydrocarbon (cxhy)on complete combustion gave 12 grams of co2 . what is its empirical formula of hydrocarbon?

Answer 1

Reaction of combustion of hydrocarbon is given by
$\bold{C_xH_y+(x + \frac{y}{4})O_2\implies xCO_2 + \frac{y}{2}H_2O}$
Here you can see 1 mole of hydrocarbon gives x mole of CO₂
∴ (12x + y) gm of hydrocarbon gives 44x of CO₂
∴ 1 gm of hydrocarbon gives 44x/(12x + y) gm of CO₂
∴ 4gm of hydrocarbon gives 44x/(12x + y) × 4 gm of CO₂

But A/C to question,
4gm of hydrocarbon gives 12 gm of CO₂
so, 44x × 4/(12x + y) = 12
⇒ 44x/(12x + y) = 3
⇒44x = 36x + 3y
⇒8x = 3y
⇒x/y = 3/8

Hence, emperical formula of hydrocarbon is C₃H₈

Answer 2

Answer:

Step-by-step explanation:

I think will be C3H8so X is 3 and y is 8