Question

Q- 0.202 g of a carbon compound, on combustion, gave 0.361 g of carbon dioxide and 0.47 g of water. Calculate the percentage composition of carbon.

Answer 1

Answer :  The % composition of Carbon is 48.743%.

Solution : Given,

Mass of carbon compound = 0.202 g

Mass of carbon dioxide = 0.361 g

Mass of water = 0.47 g

Molar mass of carbon = 12 g/mole

Molar mass of Carbon dioxide = 44 g/mole

First we have to calculate the moles of $CO_2$.

Moles of $CO_2$ = $\frac{\text{ Given mass of }CO_2}{\text{ Molar mass of }CO_2}= \frac{0.361g}{44g/mole}=0.008205moles$

The moles of $CO_2$ = 0.008205 moles

In $CO_2$, there are 1 carbon atom and 2 atom molecules are present.

So, the moles of $C$ is equal to the moles of $CO_2$.

The moles of $C$ = 0.008205 moles

Now we have to calculate the mass of Carbon.

Mass of Carbon = Moles of Carbon × Molar mass of Carbon

Mass of Carbon = 0.008205 moles × 12 g/mole = 0.09846 g

Now we have to calculate the % composition of Carbon.

% composition of Carbon = $\frac{\text{ Mass of carbon}}{\text{ Total mass of carbon compound}}\times 100=\frac{0.09846g}{0.202g}\times 100=48.743\%$

Therefore, the % composition of Carbon is 48.743%.