Question

A compound 'X' consists of 4.8% carbon and 95.2% bromine by mass.
(i) Determine the empirical formula of this compound working correct to one decimal place.
(C = 12, Br = 80)
(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?
(iii) Name the type of chemical reaction by which X can be prepared from ethane.

Answer 1

(i)

Element	Atomic mass	Percentage	Relative number of atoms	Simplest ratio	Whole number ratio
C	12	4.8	$\frac{4.8}{12}=0.4$	$\frac{0.4}{0.4}=1$	1
Br	80	95.2	$\frac{95.2}{80}=1.19$	$\frac{1.19}{0.4}=2.97$	3

Empirical formula of the compound is CBr₃.

(ii) Molecular mass of the compound = 2 $\times$ Vapour density of the compound
= 2 $\times$ 252 = 504 g
Empirical formula mass of the compound = 12 + 3(80) = 252 g
Molecular formula = (Empirical formula)_n
$n=\frac{\mathrm{Molecular}\mathrm{formula}\mathrm{mass}}{\mathrm{Empirical}\mathrm{formula}\mathrm{mass}}=\frac{504}{252}=2$
Molecular formula = (CBr₃)₂ = C₂Br₆

(iii) Substitution reaction is the type of reaction in which X (C₂Br₆) is obtained from ethane.
${\mathrm{C}}_2{\mathrm{H}}_6\underset{-\mathrm{HBr}}{\overset{{\mathrm{Br}}_2/\mathrm{h}\nu }{\to }}{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}\underset{-\mathrm{HBr}}{\overset{{\mathrm{Br}}_2/\mathrm{h}\nu }{\to }}{\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{Br}}_2\underset{-\mathrm{HBr}}{\overset{{\mathrm{Br}}_2/\mathrm{h}\nu }{\to }}{\mathrm{C}}_2{\mathrm{H}}_3{\mathrm{Br}}_3\underset{-\mathrm{HBr}}{\overset{{\mathrm{Br}}_2/\mathrm{h}\nu }{\to }}{\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{Br}}_4\underset{-\mathrm{HBr}}{\overset{{\mathrm{Br}}_2/\mathrm{h}\nu }{\to }}{\mathrm{C}}_2{\mathrm{HBr}}_5\underset{-\mathrm{HBr}}{\overset{{\mathrm{Br}}_2/\mathrm{h}\nu }{\to }}{\mathrm{C}}_2{\mathrm{Br}}_6$