Question

9 volumes of a gaseous mixture contain an organic compound and oxygen sufficient for its combustion. Burning the compound yielded 4 volumes of ${ CO }_{ 2 }$, 6 volumes of ${ H }_{ 2 }O$ and 2 volumes of ${ N }_{ 2 }$. Find the molecular formula of the compound.

Answer 1

Hint:
Here the given organic compound is undergoing complete combustion to give carbon dioxide gas, steam and nitrogen gas. Since this compound is dealing with volumes instead of the number of moles we will have to take into account Gay Lussac’s law and Avogadro’s Law.

Complete step by step answer:
For solving this question we will be using two laws:

Gay Lussac’s Law of gaseous volumes: According to this law gases react with each other in volumes that bear a simple ratio to one another and to the volume of the products (if they are gaseous) if all the measurements are done at the same temperature and pressure conditions.

Avogadro’s law: According to this law if we have equal volumes of different gases at the same temperature and pressure conditions then all of these gases will contain the same numbers of atoms or molecules; or moles.
Therefore, ${ V\quad \propto \quad n\quad at\quad constant\quad T\quad and\quad P }$
Or ${ V\quad =\quad kn\quad at\quad constant\quad T\quad and\quad P }$
Where ‘V’ is the volume of the gas, ‘n’ is the number of moles of the gas and ‘k’ is the proportionality constant.

Let us understand these two laws with an example:

One mole of ${ H }_{ 2 }$ gas reacts with one mole of ${ Cl }_{ 2 }$ gas to give 2 moles of HCl gas. Therefore according to Gay Lussac’s law and Avogadro’s law, 1 volume of ${ H }_{ 2 }$ gas will react with 1 volume of ${ Cl }_{ 2 }$ gas to give 2 volumes of HCl gas.

$\begin{matrix} Chemical\quad equation: \\ According\quad to\quad Gay\quad Lussac's\quad law: \\ According\quad to\quad Avogadro's\quad law: \end{matrix}\begin{matrix} { H }_{ 2 }(g) \\ 1\quad vol. \\ n\quad molecules \end{matrix}+\begin{matrix} { Cl }_{ 2 }(g) \\ 1\quad vol. \\ n\quad molecules \end{matrix}\rightarrow \begin{matrix} 2HCl(g) \\ 2\quad vol. \\ 2n\quad molecules\quad \end{matrix}$

Now, let us solve the question:

A total of 9 volumes of the mixture is given. If a volumes of the mixture is the organic compound (${ C }_{ x }{ H }_{ y }{ N }_{ z }$) then 9-a volumes will be ${ O }_{ 2 }$ gas.
The products after the combustion process is 4 volumes of ${ CO }_{ 2 }$ gas, 6 volumes of ${ H }_{ 2 }O$ in gaseous form and 2 volumes of ${ N }_{ 2 }$ gas.
We know that 1 volume of C combines with 1 volume of ${ O }_{ 2 }$ gas to give 1 volume of ${ CO }_{ 2 }$ gas. Therefore if there are 4 volumes of ${ CO }_{ 2 }$ gas present, then 4 volumes of ${ O }_{ 2 }$ gas must have reacted with 4 volumes of C.

$\begin{matrix} C(g) \\ 1\quad vol. \end{matrix}+\begin{matrix} { O }_{ 2 }(g) \\ 1\quad vol. \end{matrix}\rightarrow \begin{matrix} { CO }_{ 2 }(g) \\ 1\quad vol. \end{matrix}$

Therefore, $\begin{matrix} 4C(g) \\ 4\quad vol. \end{matrix}+\begin{matrix} { 4O }_{ 2 }(g) \\ 4\quad vol. \end{matrix}\rightarrow \begin{matrix} { 4CO }_{ 2 }(g) \\ 4\quad vol. \end{matrix}$

Similarly, 2 volumes of hydrogen gas react with 1 volume of oxygen gas to give 2 volumes of ${ H }_{ 2 }O$ vapour. Therefore if 6 volumes of ${ H }_{ 2 }O$ vapour is present after the combustion process, then 3 volumes of oxygen gas must have reacted with 6 volumes of hydrogen gas.

$\begin{matrix} 2{ H }_{ 2 }(g) \\ 2\quad vol. \end{matrix}+\begin{matrix} { O }_{ 2 }(g) \\ 1\quad vol. \end{matrix}\rightarrow \begin{matrix} 2{ H }_{ 2 }O(g) \\ 2\quad vol. \end{matrix}$
Therefore, $\begin{matrix} 6{ H }_{ 2 }(g) \\ 6\quad vol. \end{matrix}+\begin{matrix} { 3O }_{ 2 }(g) \\ 3\quad vol. \end{matrix}\rightarrow \begin{matrix} 6{ H }_{ 2 }O(g) \\ 6\quad vol. \end{matrix}$

Therefore the total volumes of oxygen gas that has reacted is 7 volumes (4+3).
$\Rightarrow 9-a=7$
So, a=2

Therefore the combustion reaction can be written as:
$a{ C }_{ x }{ H }_{ y }{ N }_{ z }(g)+(9-a){ O }_{ 2 }(g)\rightarrow { 4CO }_{ 2 }(g)+{ 6H }_{ 2 }O(g)+{ 2N }_{ 2 }(g)$
Substituting the value of a in the above equation:
$2{ C }_{ x }{ H }_{ y }{ N }_{ z }(g)+7{ O }_{ 2 }(g)\rightarrow { 4CO }_{ 2 }(g)+{ 6H }_{ 2 }O(g)+{ 2N }_{ 2 }(g)$

Dividing the above equation by 2:
${ C }_{ x }{ H }_{ y }{ N }_{ z }(g)+\cfrac { 7 }{ 2 } { O }_{ 2 }(g)\rightarrow { 2CO }_{ 2 }(g)+{ 3H }_{ 2 }O(g)+{ N }_{ 2 }(g)$

If we compare the number of carbon atoms on the both sides then, x=2
Similarly y=6 and z=2.

The molecular formula of the compound is ${ C }_{ 2 }{ H }_{ 6 }{ N }_{ 2 }$.

Note:
Here we have not considered the combustion of nitrogen gas with oxygen. In a combustion reaction a species combines with oxygen gas to form its respective oxides. But this is not observed in the case of nitrogen atoms that simply combine with each other to give nitrogen gas which is highly stable due to the presence of strong triple bonds.