Question
A gaseous hydrocarbon gives upon combustion $$0.72g$$ of water and $$3.08g$$ of $$C{O_{2.}}$$ The empirical formula of the hydrocarbon is :
A.
$${C_2}{H_4}$$
B.
$${C_3}{H_4}$$
C.
$${C_6}{H_5}$$
D.
$${C_7}{H_8}$$
Answer :
$${C_7}{H_8}$$
Solution :
$$\eqalign{
& \because \,\,18g,\,{H_2}O\,{\text{contains}} = 2gm\,H \cr
& \therefore \,\,0.72\,gm\,{H_2}O\,{\text{contains}} = \frac{2}{{18}} \times 0.72gm = 0.08\,gm\,H \cr
& \because \,\,44\,gm\,C{O_2}\,{\text{contains}} = 12\,gm\,C \cr
& \therefore 3.08\,gm\,C{O_2}\,{\text{contains}} = \frac{{12}}{{44}} \times 3.08 = 0.84\,gm\,C \cr
& \therefore C:H = \frac{{0.84}}{{12}}:\frac{{0.08}}{1} = 0.07:0.08 = 7:8 \cr
& \therefore \,\,{\text{Empirical formula}} = {C_7}{H_8} \cr} $$