Question

For $44g$ of a sample on complete combustion gives $88g$ of $C{O_2}$ and $36g$ of ${H_2}O$.
The molecular formula of compound may be:
A ${C_4}{H_6}$
B ${C_2}{H_6}O$
C ${C_2}{H_4}O$
D ${C_3}{H_6}O$

Answer 1

To answer to the given above question, we need to apply the Gay-Lussac’s law. It states that in a given equation, all of the reactants react in the fixed proportion to produce a fixed proportion of the product.

Complete step by step answer:
The concept of the above question is according to the stoichiometry in chemistry, few important concepts of stoichiometry are the molecular formula, relative molecular formula, moles, Avogadro’s number, constant proportion etc. one should be through with following concepts while studying chemistry
So first we need to know what is provided in the given question. We are given the mass of the reactant that is $44g$, which in turn produces $88g$ of carbon dioxide and $36g$ of water. It is a combustion reaction where hydrocarbons react with oxygen to form carbon dioxide and water molecules.
First, we will find that how many volumes of carbon dioxide being used;
$Volum{e_{C{O_2}}} = total{\text{ - }}mass/molecular - mass$
Therefore,
$\Rightarrow$ $\;Volum{e_{C{O_2}}} = 88/\left( {12 + 16*2} \right)$
So,
$\Rightarrow$ $Volum{e_{C{O_2}}} = 88/\left( {12 + 32} \right)$
On solving,
$\Rightarrow$ $Volum{e_{C{O_2}}} = 88/44$
We get ,
$\Rightarrow$ $Volum{e_{C{O_2}}} = 2$
The volumes of water being used:
$Volum{e_{{H_2}O}} = total - mass/molecular - mass$
$\Rightarrow$ $Volum{e_{{H_2}O}} = 36/\left( {2 + 16} \right)$
$\Rightarrow$ $Volum{e_{{H_2}O}} = 36/18$
$\Rightarrow$ $Volum{e_{{H_2}O}} = 2$
Now from the above data, we tried to write a balanced equation, where $A$ stand for the hydrocarbon:
$A + {O_2} \to 2C{O_2} + 2{H_2}O$
By observing the above equation, we can say that the compound air contains two carbon and four hydrogen atoms. Since the total mass of the substance is $44g$, hence,
$mass ‘C’ + mass ‘H’ + mass ‘O’ = 44$
$\Rightarrow$ $2*12 + 4*1 + massO = 44$
$\Rightarrow$ $mass’O’ = 44 - 24 - 4$
$\Rightarrow$ $mass’O’ = 44 - 28$
\Rightarrow $$$mass’O’ = 16$$ Since, Number of oxygen =\dfrac{{mass}}{{molecular,mass}}$Number of oxygen =$\dfrac{{16}}{{16}}$
Number of oxygen = one
The formula of the compound is ${C_2}{H_4}O$
Therefore , the correct option is C.

Note:
In such types of questions, always remember that you need a balanced chemical reaction first. Also, in the combustion of the organic compound, you always get carbon dioxide, a water molecule as a product. Also, a balanced equation helps you to pretend many other phenomena of chemistry.