Question

74 gm of a sampleon complete combustion gives 132gm co2 and 54gm of h2o. The molecular formula of the compound may be

Answer 1

Mass of C in 132g CO2 = 12/44*132 = 36g
Mass of H in 54g H2O = 2/18*54 = 6g
Missing mass is O = 74g - ( 36+6) = 32g
Divide each mass by atomic mass :
C = 36/12 = 3
H = 6/1 = 6
O = 32/16 = 2

Answer 2

Answer : The molecular formula of the compound is, $C_3H_6O_2$

Solution : Given,

Molar mass of carbon dioxide = 44 g/mole

Molar mass of water = 18 g/mole

Molar mass of hydrogen = 1 g/mole

Molar mass of carbon = 12 g/mole

Molar mass of oxygen = 16 g/mole

First we have to calculate the mass of carbon, hydrogen and oxygen.

$\text{Mass of C}=\text{Mass of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}\times \text{Molar mass of one C}$

$\text{Mass of C}=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}\times \text{Molar mass of one C}=\frac{132}{44}\times 12=36g$

$\text{Mass of H}=\text{Mass of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}\times \text{Molar mass of two H}$

$\text{Mass of H}=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}\times \text{Molar mass of two H}=\frac{54}{18}\times 2=6g$

The mass of oxygen = Total mass of sample - (Mass of carbon + Mass of hydrogen) = 74 - (36 + 6) = 32 g

Now we have to calculate the molecular formula of the compound. Divide the mass of carbon, hydrogen and oxygen by their respective molar mass.

$C=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{36}{12}=3$

$H=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{6}{1}=6$

$O=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{32}{16}=2$

Thus, the molecular formula of the compound is, $C_3H_6O_2$