Question

0.90g of an organic compound containing only carbon,hydrogen and nitrogen on combustion gave 1.1 g CO2 and 0.3 g water. What is %C,%H and %N in the organic compound??

Answer 1

$carbon = \frac{1100}{9} \%$
$hydrogen = \frac{100}{3} \%$
$nitrogen = \frac{521}{7} \%$

Answer 2

Answer: Percentage composition of Carbon is 33.33 %, of Hydrogen is 3.66 % and that of Nitrogen is 63 %.

Explanation:

We are given:

Mass of $H_2O$ = 0.3 g

Mass of $CO_2$ = 1.1 g

Molar mass of $H_2O$ = 18 g/mol

Molar mass of $CO_2$ = 44 g/mol

• For $H_2O$

In 18 g of water, amount of hydrogen present is 2 grams

So, in 0.3 g of water, amount of hydrogen present will be = $\frac{2}{18}\times 0.3=0.033g$ of hydrogen

• For $CO_2$

In 44 g of carbon dioxide, amount of carbon present is 12 grams

So, in 1.1 g of carbon dioxide, amount of carbon present will be = $\frac{12}{44}\times 1.1=0.3g$ of carbon.

Mass of nitrogen in compound = (0.9 - 0.3 - 0.033) = 0.567 g

To calculate the percentage composition of an element in compound, the formula used is:

$\%\text{ composition}=\frac{\text{Mass of element}}{\text{Mass of compound}}\times 100$

$\%\text{ composition of Hydrogen}=\frac{0.033}{0.9}\times 100=3.66\%\\\\\%\text{ composition of Carbon}=\frac{0.3}{0.9}\times 100=33.33\%\\\\\%\text{ composition of Nitrogen}=\frac{0.567}{0.9}\times 100=63\%$

Hence, the percentage composition of Carbon is 33.33 %, of Hydrogen is 3.66 % and that of Nitrogen is 63 %.