Question

Halogenated compounds are particularly easy to identify by their mass spectra because chlorine and bromine occur naturally as mixtures of two abundant isotopes. •Chlorine occurs as 35Cl (75.8%) and 37Cl (24.2%); •Bromine occurs as 79Br (50.7%) and 81Br (49.3%); •Boron compounds also stand out owing to the two isotopes 10B (19.9%) and 11B (80.1%). For the compound Chlorocyclohexane, C6H11Cl: a) At what masses do the molecular ions occur? b) What are the percentages of each molecular ion?

Answer 1

sol:-

data provided the question is :-

• % abundance of 35Cl = 75.8%
• % abundance of 37Cl = 24.2%
• % abundance of 79Br = 50.7%
• % abundance of 81Br = 49.3%
• % abundancw of 10B = 19.9%
• % abundance of 11B = 80.1%

Compound = C6H11Cl

formula used :-

average atomic mass =

{(mass of isotope A * %abundance) + (mass of isotope B * % abundance)}/ 100

so,

•):- average atomic mass of Cl =

{(mass of 35Cl * % abundance) + ( mass of 37Cl * %abundance)}/100

= {(35 * 75.8) + (37 * 24.2)}/100

= (2653 + 895.4)/100

= 3548.4 / 100

= 35.484 amu

average atomic mass of Cl = 35.484 amu or g/mol

•):- average atomic mass of Br=

{(mass of 79Br * %abundance) + (mass of 81Br * %abundance)}/ 100

= {(79 * 50.7) + (81 * 49.3)}/ 100

= {(4005.3) + (3993.3)}/100

= 7998.6/100

= 79.986 amu

average atomic mass of Br = 79.986 amu or g/mol

•):- average atomic mass of B =

{(mass of 10B * %abundance) + (mass of 11B * % abundance)}/ 100

= {( 10 * 19.9) + (11 * 80.1)}/ 100

= { 199 + 881.1}/100

= 1080.1 /100

= 10.801 amu

average atomic mass of B = 10.801 amu or g/mol.

•):- atomic mass of C = 12 g/mol.

•):- atomic mass of H = 1 g/mol.

molecular mass of C6H11Cl = ( 6 * 12 g/mol + 11 * 1 g/mol + 35.484 g/mol.)

= (72 + 11 + 35.484 ) g/mol

= 118.484 g/mol.

a):- 118.484 would be the mass of compound occur.

b):- mass % of " C " = ( 72 * 100) / 118.484

= 60.77 %

mass % of " H " = (11 * 100)/118.484

= 9.28 %

mass % of " Cl " = (35.484 * 100)/118.484

= 29.95 %