The mass spectrum of bromine, Br2, (79Br and 81Br about equally abundant) shows molecular ion peaks. At which m/z values and in what ratio?
Answer and explanation
Correct answer: D. 158, 160, 162 in the ratio 1:2:1
- Step 1: Possible molecules: 79Br79Br (158), 79Br81Br (160, two ways to arrange), 81Br81Br (162). Step 2. With equal abundance the probabilities are 1/4, 1/2, 1/4, so the ratio is 1:2:1. Answer: 158, 160, 162 in the ratio 1:2:1. Why not the others: that ignores the mixed molecule at 160; 79 and 81 are fragment ions of single atoms; 1:1:1 neglects the two ways of forming the mixed molecule.
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