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Halogenated compounds are particularly easy to identify by
Chapter 12, Problem 12(choose chapter or problem)
Halogenated compounds are particularly easy to identify by their mass spectra because both chlorine and bromine occur naturally as mixtures of two abundant isotopes. Recall that chlorine occurs as \(\mathrm{^{35}Cl}\) (75.8%) and \(\mathrm{^{37}Cl}\) (24.2%); and bromine occurs as \(\mathrm{^{79}Br}\) (50.7%) and \(\mathrm{^{81}Br}\) (49.3%). At what masses do the molecular ions occur for the following formulas? What are the relative percentages of each molecular ion?
(a) Bromomethane, \(\mathrm{CH_3Br}\)
(b) 1-Chlorohexane, \(\mathrm{C_6H_{13}Cl}\)
Questions & Answers
QUESTION:
Halogenated compounds are particularly easy to identify by their mass spectra because both chlorine and bromine occur naturally as mixtures of two abundant isotopes. Recall that chlorine occurs as \(\mathrm{^{35}Cl}\) (75.8%) and \(\mathrm{^{37}Cl}\) (24.2%); and bromine occurs as \(\mathrm{^{79}Br}\) (50.7%) and \(\mathrm{^{81}Br}\) (49.3%). At what masses do the molecular ions occur for the following formulas? What are the relative percentages of each molecular ion?
(a) Bromomethane, \(\mathrm{CH_3Br}\)
(b) 1-Chlorohexane, \(\mathrm{C_6H_{13}Cl}\)
ANSWER:Step 1 of 2
(a) Bromomethane, \(\mathrm{CH}_{3} \mathrm{Br}\)
1st we have to figure out the molecular weight of each isotope of bromine. In this question two
isotopes of bromines are given i.e \({ }^{79} \mathrm{Br}(50.7 \%)\) and \({ }^{81} \mathrm{Br}(49.3 \%)\).
The relative percentages can be calculated as,
For \(\left.{ }^{79} \mathrm{Br} \text { in } \mathrm{CH}_{3} \mathrm{Br}=(1 \times 12.01)+(3 \times 1.008)+79=94.03 \text { (at } 50.7 \%\right)\)
For \(\left.{ }^{81} \ma