Question

1. An unknown hydrocarbon has a molecular ion peak at m/z 84 with a relative intensity of 31.3. The M+1 peak has a relative intensity of 2.06 and the M+2 peak has a relative intensity of 0.08. What is the possible molecular formula of the compound?

Answer 1

Firstly the relative abundances of M should be set to 100%.

Since the relative intensities of M is 31.3% and not 100%(standard), hence the relative intensities of M+1 would be (2.06/31.3)*100 = 6.5%.

And the relative intensities of M+2 would be (0.08/31.3)*100 = 0.25%.

In the M+1 spectrum, we basically determine the relative abundance of carbon.To determine the number of carbon atoms we divide the relative intensity of M+1 peak by 1.1(since the natural abundance of C-13 is 1.1%). Here, M is the molecular ion whose m/z= 84. Hence for molecular ion M+1, the m/z is one amu higher than that of M i.e m/z= 85 whose relative intensity is 6.5% (Done above).

Now, on dividing the relative intensity of M+1 peak by 1.1,

6.5/1.1= 5.98 which can be approximated to 6 carbon atoms.

Now let us solve the M+2 spectrum. In the M+2 spectrum, we basically determine the relative abundance of sulphur, chlorine and bromine. For molecular ion M+2, the m/z is one amu higher than that of M i.e m/z= 86 whose relative intensity is 0.25% (Done above). The relative intensities for sulphur, chlorine and bromine should be about 4.3%, 31.9% and 97.2%. Since the relative intensity of M+2 is 0.25% which is nowhere close to 4.3%, 31.9% and 97.2%, hence we may conclude that there are no sulphur, chlorine or bromine atoms.

To determine the number of hydrogens present, we use the relation,

Number of Hydrogens = Number of nitrogens + (2*Number of carbons) + 2.

To determine if the compound has nitrogen or not we need to check its mass. Here the m/z value of the compound is 84 which is an 'even' number, hence the number of nitrogen atoms would also be even. If the m/z value had been odd, the number of nitrogen atoms would have been odd too.

In our case we have 6 carbons. Each carbon has an atomic mass of 12 amu. So total atomic mass of 6 carbons= 6*12=72 amu. The molecular mass is 84 amu (given). Now if there had been a minimum of 2 nitrogens present (It should be even as explained above), the total atomic mass of 2 nitrogens would be 14*2=28 which would be higher than the molecular mass of the compound. Hence we may conclude that the compound contains no nitrogen. Hence our compound contains only hydrogens other than the carbon atoms.

Total atomic mass of hydrogens = Molecular mass - Total atomic mass of 6 carbons

= 84 - 72

= 12 amu.

12 amu corresponds to 12 hydrogens.

Hence the molecular formula of the compound is C₆H₁₂.