Question

0.90g of an organic compound containing only carbon, hydrogen and nitrogen on combustion gave $1.1g$ $C{O_2}$ and $0.3g$ water. What is $\% C,{\text{ }}\% H$ and $\% N$ in the organic compound?

Answer 1

When the combustion of a particular organic compound takes place, it yields carbon, nitrogen and hydrogen. The percentage composition of a component in a compound is the percent of the total mass of the compound that is due to that component.

Complete answer:
Let the organic compound provided be ${C_x}{H_y}{N_z}$ .
When the combustion takes place, the reaction that takes place here is as follows:
${C_x}{H_y}{N_z} + {O_2} \to xC{O_2} + \dfrac{y}{2}{H_2}O +$ some nitrogen compound
Here, we can see that one mole of organic compound gives $x$ moles of carbon dioxide and $\dfrac{y}{2}$ moles of water.
Given weight of carbon dioxide = ${w_{C{O_2}}} = 1.1g$
Molecular weight of carbon dioxide = ${M_{C{O_2}}} = 44g$
Number of moles of carbon dioxide = ${n_{C{O_2}}} = \dfrac{{{w_{C{O_2}}}}}{{{M_{C{O_2}}}}} = \dfrac{{1.1}}{{44}}$
Similarly, given weight of water = $0.3g$
Molecular weight of water = $18g$
Number of moles of water = ${n_{{H_2}O}} = \dfrac{{0.3}}{{18}}$
Number of moles of organic compound = $n' = \dfrac{{{w_{compound}}}}{{{M_{compound}}}} = \dfrac{{0.90}}{{12x + y + 14z}}$
From the chemical equation of the reaction, we have:
$x \times$ moles of organic compound = moles of $C{O_2}$
$\Rightarrow \dfrac{{0.9x}}{{12x + y + 14z}} = \dfrac{{1.1}}{{44}}$
$\Rightarrow 36x = 12x + y + 14z$
Thus, on solving, we have the final equation as:
$24x = y + 14z$ ….(i)
Similarly, $\dfrac{y}{2}$ moles of organic compound = moles of ${H_2}O$
$\Rightarrow \dfrac{{0.9\left( {\dfrac{y}{2}} \right)}}{{12x + y + 14z}} = \dfrac{{0.3}}{{18}}$
$\Rightarrow 27y = 12x + y + 14z$
Thus, on solving, we have the final equation as:
$26y = 12x + 14z$ ….(ii)
Thus, from equation (i) and (ii), we have:
$\Rightarrow 24x - y = 26y - 12x$
Thus, on solving, we have:
$4x = 3y$ ….(iii)
Substituting the value of equation (iii) in equation (i), we have:
$\Rightarrow 24x = y + 14z$
$\Rightarrow 24 \times \dfrac{3}{4}y = y + 14z$
$\Rightarrow 17y = 14z \Rightarrow y = \dfrac{{14}}{{17}}z$
From equation (iii), we have:
$\Rightarrow 4x = 3y$
$4x = 3\left( {\dfrac{{14}}{{17}}z} \right) \Rightarrow 68x = 42z$
Also, $4x = 3y = \dfrac{{42}}{{17}}z \Rightarrow 68x = 51y = 42z$
Thus, the organic compound is = ${C_{68}}{H_{51}}{N_{42}}$
$\% C$ in the organic compound = $\dfrac{{12 \times 68}}{{(12 \times 68) + (51 \times 1) + (42 \times 14)}} \times 100 = 56\%$
$\% H$ in the organic compound = $\dfrac{{51 \times 1}}{{(12 \times 68) + (51 \times 1) + (42 \times 14)}} \times 100 = 3.6\%$
$\% N$ in the organic compound = $\dfrac{{42 \times 14}}{{(12 \times 68) + (51 \times 1) + (42 \times 14)}} \times 100 = 40.4\%$

Note:
The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100. Here, the quantity is measured in terms of grams of the elements present. The percentage composition of any compound expresses its composition in terms of all the elements present. Thus, it helps in chemical analysis of the given compound.