The vapours of organic compound was burnt in oxygen. Equal volume of both gaseous substance were taken at same pressure and temperature. After the reaction, the system was returned to the original con

Question

The vapours of organic compound was burnt in oxygen. Equal volume of both gaseous substance were taken at same pressure and temperature. After the reaction, the system was returned to the original condition and it turned out to be same volume. After combustion final mixture contain 50% CO 2 ​ ( g ) and 50% H 2 ​ O ( g ) by volume. Find the molecular weight of the organic compound.

TestHub · Accepted Answer

Let the organic compound be CₓHᵧ. Take 1 volume each of hydrocarbon and O₂.

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

The final volume equals the initial volume (2 volumes) and contains 50% CO₂ and 50% H₂O, so

$x = 1, \frac{y}{2} = 1 \Rightarrow y = 2$

Also, $x + \frac{y}{4} = 1 + \frac{1}{2} = 1.5$ , so the oxygen consumed equals the available amount after accounting for contraction, giving the compound as CH₂ (empirical formula). Hence, the molecular formula is C₂H₆.

$M o l ec u l a r ma ss = 2 (12) + 6 (1) = 30$

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