Question

5.6 g of an organic compound on burning with excess of oxygen gave 17.6 g of CO2 and 7.2 g of H2O . THE ORGANIC COMPOUND ???

Answer 1

Check this attachment :-)

ANSWER : C4H8

Answer 2

Answer : The organic compound will be, $C_4H_8$

Explanation : Given,

Mass of $CO_2$ = 17.6 g

Molar mass of $CO_2$ = 44 g/mole

Mass of $H_2O$ = 7.2 g

Molar mass of $H_2O$ = 18 g/mole

The general balanced chemical reaction of combustion will be,

$C_xH_y+\frac{1}{2}(x+\frac{y}{2})O_2\rightarrow xCO_2+\frac{y}{2}H_2O$

First we have to calculate the moles of $CO_2$ and $H_2O$

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}=\frac{17.6g}{44g/mole}=0.4mole$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{7.2g}{18g/mole}=0.4mole$

From the balanced reaction, we conclude that

The value of x = 0.4

The value of y = 0.8   $(\frac{y}{2}=0.4\\\\y=0.4\times 2=0.8$

The ratio of x : y = 4 : 8

The organic compound = $C_xH_y=C_4H_8$

Therefore, the organic compound will be, $C_4H_8$