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Chemistry Term 1  STPM  Chapter 1 Atoms, Molecules and Stoichiometry
               Question 12

                                                                m
                The mass spectrum of dichloromethane consists of peaks at — values of 84, 86
                                                                e
                and 88. Only  H,  C,  Cl and  Cl isotopes are present in the molecule.
                           1
                                35
                             12
                                       37
                (a)  Write the formula of ions responsible for the above peaks.
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                (b)  Calculate the percent abundance of the above peaks if the natural abundance
                   ratio of  Cl and  Cl is 3 : 1.
                          35
                                37
                (c)  Sketch and label a mass spectrum for 1,1-dichloroethane showing the above
                   peaks.
              Answer:
         Term
        1           Exam Tips
                    Exam Tips
                    Write all the possible combination of chlorine isotopes in the dichloromethane
                                                                    mass
                    molecule, then add up their respective atomic mass to get the —–—– ratio,
                     m                                             charge
                    —–.
                    e
                     m    12  1  35  + m  12  1  35  37  + m  12  1  37  +
                     e
                                                       e
                                     e
                 (a)  —  84:  C H 2 Cl 2 , — 86:  C H 2 Cl Cl , — 88:  C H 2 Cl 2
                     m     3   3
                 (b)  —  84: —  × —  = 0.5625
                     e
                           4
                               4
                     m
                                  1
                              3
                     —  86: 2 × —  × —  = 0.375
                     e
                              4
                                  4
                     m
                               1
                           1
                     —  88: —  × —  = 0.0625
                     e
                           4
                               4
                     m
                          m
                                m
                     —  84: —  86: —  88  =  56.25% : 37.50% : 6.25%
                     e
                           e
                                e
                 (c)   Percent abundance
                      56.25
                      37.50
                      6.25
                               84     86    88  m/e

                                              10




         01 STPM Q&A Chem T1 layout.indd   10                                  4/19/19   9:28 AM
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