Question

Methane $\left({\mathrm{CH}}_4\right)$ can be successively chlorinated by substituting its hydrogen atoms with chlorine atoms

1. How many different formulas can result? Write them.
2. Calculate the molecular weight corresponding to each chemical formula.

Answer 1

1. Substitution reaction

• Methane $\left({\mathrm{CH}}_4\right)$ can be successively chlorinated by substituting its hydrogen atoms with chlorine atoms. this reaction is termed a substitution reaction.
• The reaction occurs in the presence of U V light.
• The reaction proceeds as:

$$\begin{gathered} \underset{\mathrm{Methne}}{{\mathrm{CH}}_4}+\underset{\mathrm{Chlorine}}{{\mathrm{Cl}}_2}\stackrel{\mathrm{h\nu }}{\to }\underset{\mathrm{Chloromethane}}{{\mathrm{CH}}_3\mathrm{Cl}}+\underset{\mathrm{Hydrogen}\mathrm{chloride}}{\mathrm{HCl}} \\ \underset{\mathrm{Chloromethne}}{{\mathrm{CH}}_3\mathrm{Cl}}+\underset{\mathrm{Chlorine}}{{\mathrm{Cl}}_2}\stackrel{\mathrm{h\nu }}{\to }\underset{\mathrm{Dichloromethane}}{{\mathrm{CH}}_2{\mathrm{Cl}}_2}+\underset{\mathrm{Hydrogen}\mathrm{chloride}}{\mathrm{HCl}} \\ \underset{\mathrm{Dichloromethane}}{{\mathrm{CH}}_2{\mathrm{Cl}}_2}+\underset{\mathrm{Chlorine}}{{\mathrm{Cl}}_2}\stackrel{\mathrm{h\nu }}{\to }\underset{\mathrm{Trichloromethane}}{{\mathrm{CHCl}}_3}+\underset{\mathrm{Hydrogen}\mathrm{chloride}}{\mathrm{HCl}} \\ \underset{\mathrm{Trichloromethne}}{{\mathrm{CHCl}}_3}+\underset{\mathrm{Chlorine}}{{\mathrm{Cl}}_2}\stackrel{\mathrm{h\nu }}{\to }\underset{\mathrm{Tetrachloromethane}}{{\mathrm{CCl}}_4}+\underset{\mathrm{Hydrogen}\mathrm{chloride}}{\mathrm{HCl}} \end{gathered}$$

2. The molecular weight corresponding to each chemical formula is

• ${\mathrm{CH}}_4$ = Mass of C + 4 x Mass of H = 12 + 4 = 16 g
• ${\mathrm{CH}}_3\mathrm{Cl}$ = Mass of C + 3 x Mass of H + Mass of Cl = 12 + 3 + 35.5 = 50.5 g
• ${\mathrm{CH}}_2{\mathrm{Cl}}_2$ = Mass of C + 2 x Mass of H + 2 x Mass of Cl = 12 + 2 + 2 x 35.5 = 85 g
• ${\mathrm{CHCl}}_3$ = Mass of C + Mass of H + 3 x Mass of Cl = 12 + 1 + 3 x 35.5 = 119.5 g
• ${\mathrm{CCl}}_4$ = Mass of C + 4 x Mass of Cl = 12 + 4 x 35.5 = 154 g