Question

10ml of a gaseous hydrocarbon on combustion gives $40ml$ of $C{{O}_{2}}$ gas and $50ml$ of vapour. The hydrocarbon is:
A. ${{C}_{4}}{{H}_{5}}$
B. ${{C}_{8}}{{H}_{10}}$
C. ${{C}_{4}}{{H}_{8}}$
D. ${{C}_{4}}{{H}_{10}}$

Answer 1

The general combustion reaction of a hydrocarbon to produce $C{{O}_{2}}$ and ${{H}_{2}}O$ is given by:
${{C}_{x}}{{H}_{y}}+(x+\dfrac{y}{4}){{O}_{2}}\to xC{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O$ . To solve this question, you will need to relate the volume of gases to their concentration in moles, the only way you can relate them is by using the fact that “all gases have volume of $22.4L$ when their concentration is $1$ mole at STP” because here the density of gases are not provided.

Complete step by step answer:
We know that, A mole of any gas at STP occupies $22.4$ litres in volume. Since, it is initially given that the initial volume of Hydrocarbon is $10ml$ . Therefore, the initial number of moles of hydrocarbon is given as: $n=\dfrac{0.01}{22.4}\approx 0.45$ millimoles.
We also observe the ratio of volume of hydrocarbon to the eventual volume of the products $C{{O}_{2}}$ and ${{H}_{2}}O$ is $10:40:50=1:4:5$
Therefore, the amount of millimoles of $C{{O}_{2}}$ and ${{H}_{2}}O$ respectively is;

& n{{~}_{C{{O}_{2}}}}=0.45\times 4=1.80mm \\\ & n{{~}_{{{H}_{2}}O}}=0.45\times 5=2.23mm \\\ \end{aligned}$$ With this, we observe that one mole of hydrocarbon produces four moles of $C{{O}_{2}}$ , Therefore, we can safely conclude that the hydrocarbon in question contains four atoms of Carbon. This helps us eliminate option b) as a possible answer straight off the bat. Let us now plug this value of x into the above formula: ${{C}_{4}}{{H}_{y}}+(4+\dfrac{y}{4}){{O}_{2}}\to 4C{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O$ Now let us try and solve for y. We observe that one mole of hydrocarbon results in the formation of $5$ moles of water. That implies, $\dfrac{y}{2}=5$ Thus, we can conclude that the value of $y$ is $10.$ Now plugging in the values of both x and y into the reaction of a general hydrocarbon with Oxygen, we observe that: ${{C}_{4}}{{H}_{10}}+(4+\dfrac{10}{4}){{O}_{2}}\to 4C{{O}_{2}}+\dfrac{10}{2}{{H}_{2}}O$ $\Rightarrow {{C}_{4}}{{H}_{10}}+6.5{{O}_{2}}\to 4C{{O}_{2}}+5{{H}_{2}}O$ **So, the correct answer is Option D.** **Note:** Remember that all hydrocarbons will give only carbon dioxide and hydrogen gas upon combustion, the amount of them will depend on the number of carbon and hydrogen atoms present in the molecule.