Question

A gaseous hydrocarbon gives upon combustion, $0.72g$ water and $3.08g$ of $C{O_2}$ . The empirical formula of the hydrocarbon is:
A. ${C_6}{H_5}$
B. ${C_7}{H_8}$
C. ${C_2}{H_4}$
D. ${C_3}{H_4}$

Answer 1

Hint: We know that the empirical formula of a chemical compound represents the positive integer ratio of the atoms present in the chemical compound. The empirical formula can be calculated using the various predefined steps. If we know the percentage or the mass of the constituents we can evaluate the empirical formula.

Complete step by step answer:
Given information:
Water, ${H_2}O = 0.72g$
$C{O_2} = 3.08g$
Also,
Number of moles of ${H_2}O$= $\dfrac{{0.72}}{{18}} = 0.04$
Number of moles of $C{O_2}$= $\dfrac{{3.08}}{{44}} = 0.07$
Calculation:
Let us assume the number of hydrocarbons to be as ${C_x}{H_y}$.
As given the reaction is a combustion reaction. We have a predefined formula of the number of hydrocarbons for combustion reactions, which is given as,
${C_x}{H_y} + (x + \dfrac{y}{4}){O_2} \to xC{O_2} + \dfrac{y}{2}{H_2}O$ (Generic empirical formula for combustion reactions)
By calculation, we have the number of moles in decimals, we will multiply it by 100 to obtain it as a whole integer as the coefficient of the product is represented in the same x and y format.
Hence,
$x = 0.07 \times 100 = 7$
$\dfrac{y}{2} = 0.04 \times 100$
On simplification we get,
$y = 8$
We will substitute the value of x and y in ${C_x}{H_y}$ .
We the empirical formula as - ${C_7}{H_8}$
Hence,the correct option is option B.

Note:
We have to remember that the method of evaluating the empirical formula using the predefined formulas for various reactions is quite easy and we can obtain the accurate formulas, though this is limited to only simple hydrocarbons. Complex ones can be a bit tough to obtain. The same method can be used to obtain the empirical formulas of unknown compounds if we have the moles or the mass contain of the reactants.