Question

A hydrocarbon of 10 lit requires 40 lit O2 gas for complete combustion and gives 30 lit of CO2.
Assuming all volumes measured at same condition, molecular formula of hydrocarbon is

Answer 1

Answer:

        C₃H₄

Explanation:

Combustion Equation

$C_{x} H_{y} + (x + \frac{y}{4}) O_{2}$ ⇒ $xCO_{2} + \frac{y}{2} H_{2} O$

1 mol hydrocarbon ≡ $(x + \frac{y}{4})$ mol O₂ ≡ x mol CO₂

At STP, 1 mole of any gas occupies 22.4L

22.4L hydrocarbon ≡ $(x + \frac{y}{4})$ L O₂ ≡ x L CO₂

22.4L hydrocarbon ≡ $(x + \frac{y}{4})$ L O₂

1L hydrocarbon ≡ $(x + \frac{y}{4})$ /22.4L  O₂

10L hydrocarbon ≡ 10× $(x + \frac{y}{4})$ /22.4L O₂

10× $(x + \frac{y}{4})$ /22.4L = 40L  (from question)

$(x + \frac{y}{4})$ /22.4L = 4L

$(x + \frac{y}{4})$ = 22.4L × 4L -----{i}

22.4L hydrocarbon  ≡ x L CO₂

1L hydrocarbon ≡  x/22.4L CO₂

10L hydrocarbon ≡ 10 × x/22.4L O₂

10 × x/22.4L O₂ = 30 L   (from question)

x/22.4L O₂ = 3 L

x = 22.4L × 3L -----{ii}

from {i} and {ii}

22.4×3 + y/4 = 22.4×4

y/4 = 22.4×4 - 22.4×3

y / 4 = 22.4(4 - 3)

y / 4 = 22.4

y = 22.4 × 4L ----{iii}

{ii} / {iii}

x / y =  22.4L × 3L /  22.4 × 4L

x / y = 3 / 4

x : y = 3 : 4

Molecular formula/Empirical formula = C₃H₄