A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2 . The empirical formula of the hydrocarbon is

Question

A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2 . The empirical formula of the hydrocarbon is (A) C6H5 (B) C3H4 (C) C2H4 (D) C7H8

Tardigrade Admin · Accepted Answer

CxHy+(x + (y/4))O2 arrow xCO2+(y/2)H2O Number of moles of CO2=(3.08/44)=0.07 So x=0.07 Number of moles of H2O=(0.72/18)=0.04 So y=0.08 x:y=7:8 Therefore, empirical formula of hydrocarbon is C7H8 .

Q. A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of $C O_{2}$ . The empirical formula of the hydrocarbon is

$C_{6} H_{5}$

$C_{3} H_{4}$

$C_{2} H_{4}$

$C_{7} H_{8}$

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

Number of moles of $C O_{2} = \frac{3.08}{44} = 0.07$

So $x = 0.07$

Number of moles of $H_{2} O = \frac{0.72}{18} = 0.04$

So $y = 0.08$

$x : y = 7 : 8$

Therefore, empirical formula of hydrocarbon is $C_{7} H_{8}$ .

Q. A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2​ . The empirical formula of the hydrocarbon is

C6​H5​

C3​H4​

C2​H4​

C7​H8​