44g of a sample on complete combustion gives 88 gm CO2 and 36 gm of H2O. The molecular formula of the compound may be:

Question

44g of a sample on complete combustion gives 88 gm CO2 and 36 gm of H2O. The molecular formula of the compound may be: (A) C4H6 (B) C2H6O (C) C2H4O (D) C3H6O

Tardigrade Admin · Accepted Answer

CO2 = (88/44) = 2 mole of CO2 = 2 mole of C H2O = (36/18) = 2 mole of H2O = 4 mole of H Mass of C + Mass of H + Mass of O = 44 ⇒ 24 + 4 + x = 44; x = 16 ∴ mole of O = 1 and molecular formula is C2H4O.

Q. $44 g$ of a sample on complete combustion gives $88$ $g m$ $C O_{2}$ and $36$ $g m$ of $H_{2} O$ . The molecular formula of the compound may be:

$C_{4} H_{6}$

$C_{2} H_{6} O$

$C_{2} H_{4} O$

$C_{3} H_{6} O$

$C O_{2} = \frac{88}{44} = 2$ mole of $C O_{2} = 2$ mole of $C$
$H_{2} O = \frac{36}{18} = 2$ mole of $H_{2} O = 4$ mole of $H$
Mass of $C$ + Mass of $H$ + Mass of $O = 44$
$\Rightarrow 24 + 4 + x = 44; x = 16$
$∴$ mole of $O = 1$ and molecular formula is $C_{2} H_{4} O$ .

Q. 44g of a sample on complete combustion gives 88 gm CO2​ and 36 gm of H2​O. The molecular formula of the compound may be:

C4​H6​

C2​H6​O

C2​H4​O

C3​H6​O