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 turn out that its volume has not changed. The product of combustion contain 50% CO 2 ​ ( g ) and 50% H 2 ​ O ( g ) by volume and no other gas. Find the molecular weight of organic compound (in gram/mol) in question.

TestHub · Accepted Answer

Let the organic compound be CₓHᵧO𝓏. C x ​ H y ​ O z ​ + ( x + 4 y ​ − 2 z ​ ) O 2 ​ ⟶ x CO 2 ​ ( g ) + 2 y ​ H 2 ​ O ( g ) V ( x + 4 y ​ − 2 z ​ ) V x V V = ( x + 4 y ​ − 2 z ​ ) V ⇒ x + 4 y ​ − 2 z ​ = 1 x = 2 y ​ ⇒ y = 2 x 1 + ( x + 4 y ​ − 2 z ​ ) = x + 2 y ​ From (i), (ii), and (iii), x = 1 , y = 2 , z = 1 . This implies that the compound is CH₂O, with a molar mass of 30 g / m o l .

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