Question

A compound has the following percentage competition by mass: carbon 14.4% and chlorine 84.5%.

a. The relative molecular mass of this compound is 168, so what is its molecular formula?

b. By what type of reaction could this compound be obtained from ethyne?

Answer 1

Step 1: Write the given terms:

Relative molecular mass = 168

Element	Percentage weight	Atomic mass	Relative no. of moles	Simple ratio of atoms	Whole number ratio
C	14.4	12	14.4/12 = 1.2	1.2/1.2=1	1
H	1.2	1	1.2/1 = 1.2	1.2/1.2=1	1
Cl	84.5	35.5	84.5/35.5= 2.38	2.38/1.2= 1.98	2

Hence, the empirical formula is ${\mathrm{CHCl}}_2$.

Step 2: Find the molecular formula

Empirical formula mass = 12 x 1 + 1 x 1 + 35.5 x 2 = 84

Relative molecular mass = 168
n = $\frac{Molecularweight}{Empiricalformulaweight}$

n = $\frac{168}{84}$ = 2

Hence, the molecular formula is ${\left({\mathrm{CHCl}}_2\right)}_{\mathrm{n}}$

= ${\left({\mathrm{CHCl}}_2\right)}_2$= ${\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{Cl}}_4$

Step 3:

Hence, the molecular formula is ${\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{Cl}}_4$

b.

An addition reaction is the type of reaction from which this compound is formed from ethyne.

${\mathrm{C}}_2{\mathrm{H}}_2+{\mathrm{Cl}}_2\to {\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{Cl}}_2\stackrel{{\mathrm{Cl}}_2}{\to }{\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{Cl}}_4$