Question

For \[500mL\] of hydrocarbon gas , burnt in excess of oxygen , yields $2500mL$ $C{O_2}$ and 3 liters of water vapour, all volumes being measured at the same temperature and pressure, what is the formula of the hydrocarbon?
A.${C_4}{H_8}$
B.${C_5}{H_{12}}$
C.${C_5}{H_{10}}$${C_5}{H_{10}}$
D.${C_2}{H_6}$

Answer 1

Hint: Hydrocarbons are compounds that contain elements of carbon and hydrogen. Hydrocarbons are organic compounds. Although they are composed of hydrogen and carbon, they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. These compounds can be saturated like alkanes or unsaturated like alkenes, alkynes. They can be represented in the expanded formula, condensed formula, or skeletal structure.

Complete step by step answer:
The formula can be determined by
Given,
The volume of hydrocarbons =$500ml$
The volume of $C{O_2}$=$2500ml$
The volume of water vapour = $3L$=$3000ml$
The general formula of hydrocarbons = ${C_x}{H_y}$
The reaction is as follows:
$ \Rightarrow $${C_x}{H_y}$+ $x + \dfrac{y}{4}$$x$${O_2}$$ \to $$xC{O_2}$+ $\dfrac{y}{2}$
By looking at the reaction, one molecule of the hydrocarbon produces one molecule of carbon dioxide and $\dfrac{y}{2}$of water.
$ \Rightarrow 500ml+ x + \dfrac{y}{4}{O_2} \to (2500ml) + \dfrac{y}{2}x500=3000ml$
Solving for $x$,
$ \Rightarrow $$x = \dfrac{{2500}}{{500}} = 5$
Solving for$y$ ,
$ \Rightarrow $$\dfrac{y}{2}x$$500$= $3000$
$ \Rightarrow $$y = 12$
Thus, option B is the right answer.
\[500mL\] of hydrocarbon gas , burnt in excess of oxygen , yields $CO{_2}$ and 3 liters of water vapour, all volumes being measured at the same temperature and pressure, the formula of the hydrocarbon is ${C_5}{H_{12}}$.

Note:
Alkanes are $s{p^3}$ hybridized. Whereas, Alkenes and alkynes are \[s{p^2}\] and \[sp\] hybridized respectively. Examples of alkanes are methane, ethane, propane. They have a bond angle close to ${120^0}$, hence, they are zigzag in shape. These hydrocarbons exhibit isomeric forms. Carbon atoms are free to rotate around a single bond but not around a double bond; a double bond is rigid. This makes it possible to have two isomers of 2-butene, one with both methyl groups on the same side of the double bond and one with the methyl groups on opposite sides. When structures of butene are drawn with ${120^0}$ bond angles around the \[s{p^2}\] hybridized carbon atoms participating in the double bond, the isomers are apparent.