Question

At \[300K\] and 1 atmospheric pressure, \[10{\text{ }}mL\] of a hydrocarbon requires \[55{\text{ }}mL\]of ${O_2}$ for complete combustion and \[40{\text{ }}mL\] of $C{O_2}$ is formed. Then, the formula of the hydrocarbon is:
A. ${C_4}{H_8}$
B. ${C_4}{H_7}Cl$
C. ${C_4}{H_{10}}$
D. ${C_4}{H_6}$

Answer 1

Hint: We all know that hydrocarbon is an organic compound which consists of carbon and hydrogen atoms in particular. Both of these atoms are bonded with covalent bonds. So, while solving the above question we need to consider different types of chemical reactions and try to balance the x and y components of a hydrocarbon.

Complete step by step answer:
As we all have studied there are different types of chemical reactions. So, let’s consider the general reaction to be
$A + B \to C + D$
Now, according to the question let’s place the values of A, B, C, and D in the reaction.
\[Hydrocarbon{\text{ }} + {\text{ }}Oxygen\; \to Carbon{\text{ }}dioxide{\text{ }} + {\text{ }}Water\]
Now, let’s write their chemical formulas
${C_x}{H_y} + {O_2} \to C{O_2} + {H_2}O$
So, to balance the x of a hydrocarbon we have to balance the whole chemical equation.
The equation used:
${C_x}{H_y}(g) + (x + y/4){O_2} \to xC{O_2} + 4/2{H_2}O$
In the question, they given that the hydrocarbon =\[10{\text{ }}mL\], Oxygen =\[55{\text{ }}mL\], Carbon dioxide = \[40{\text{ }}mL\]
According to the question let’s consider the reactants side to calculate the products mol value.
Firstly, we will take Hydrocarbon:
\[1{\text{ }}mL\] of the hydrocarbon = \[x{\text{ }}mL\] of $C{O_2}$ produced which implies that
\[1{\text{ }}mL\] of the hydrocarbon = \[10\left( x \right){\text{ }}mL\] of $C{O_2}$produced
Thus, \[10{\text{ }}x{\text{ }} = {\text{ }}40{\text{ }}mL\]
So, \[X = {\text{ }}4{\text{ }}mL\]
Therefore, we found the value of x.
Now, let’s consider for Oxygen:
$(x + y/4){O_2}$ ml of oxygen is required = produces x mL of $C{O_2}$
\[55ml = \dfrac{x}{{(x + y/4)}}\; \times 55{\text{ }}mL{\text{ }}ofC{O_2}\]
We can simplify this step by cross multiplication.
$ \Rightarrow \dfrac{x}{{(x + y/4)}} \times 55 = 40$
\[ \Rightarrow \;\;55x{\text{ }} = {\text{ }}40x{\text{ }} + {\text{ }}10y\]
\[ \Rightarrow \;55x{\text{ }}-{\text{ }}40x{\text{ }} = {\text{ }}10y\]
\[ \Rightarrow 15x{\text{ }} = {\text{ }}10{\text{ }}y\]
Here, we can substitute the value of x.
As\[x{\text{ }} = {\text{ }}4{\text{ }}mL\], so the equation becomes
\[ \Rightarrow 15\; \times 4{\text{ }} = {\text{ }}10y\]
\[ \Rightarrow 60{\text{ }} = {\text{ }}10y\]
\[ \Rightarrow y = {\text{ }}6{\text{ }}mL\]
But according to the question we had to determine the chemical formula of a hydrocarbon. So, we will substitute the respective values of x and y in the formula to complete it.
Formula of hydrocarbon = ${C_x}{H_y}$
Hence, the formula of hydrocarbon becomes ${C_4}{H_6}$
$\therefore $Option D is the correct answer.

Note:
We must know that combustion analysis is a common analysis method for hydrocarbon. We can use this to analyze samples of unknown chemical formulas. It requires only milligrams of a sample. So we have to weigh the sample and then fully combust at a high temperature in the presence of excess oxygen, which produces carbon dioxide and water.