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496 CHAPTER 14 
 
This means that the (M+1)+ peak is 4.6% as tall as the 
(M)+ peak. Recall that each carbon atom in the 
compound contributes 1.1% to the height of the (M+1)+ 
peak, so we must divide by 1.1% to determine the 
number of carbon atoms in the compound: 
 
 
 
The compound cannot have 4.15 carbon atoms. It must 
be a whole number, so we round to the nearest whole 
number, which is 4. That is, the compound contains four 
carbon atoms, which account for 4  12 = 48 amu 
(atomic mass units), but the molecular weight is 54 amu. 
So we must account for the remaining 54 – 48 = 6 amu. 
This indicates six hydrogen atoms, giving the following 
proposed molecular formula: C4H6. 
 
(d) In order to determine the number of carbon atoms in 
the compound, we compare the relative heights of the 
(M+1)+ peak and the (M)+ peak, like this: 
 
 
 
 
This means that the (M+1)+ peak is 7.9% as tall as the 
(M)+ peak. Recall that each carbon atom in the 
compound contributes 1.1% to the height of the (M+1)+ 
peak, so we must divide by 1.1% to determine the 
number of carbon atoms in the compound: 
 
 
 
The compound cannot have 7.2 carbon atoms. It must be 
a whole number, so we round to the nearest whole 
number, which is 7. That is, the compound contains 
seven carbon atoms, which account for 7  12 = 84 amu 
(atomic mass units), but the molecular weight is 96 amu. 
So we must account for the remaining 96 – 84 = 12 amu. 
This indicates twelve hydrogen atoms, giving the 
following proposed molecular formula: C7H12. 
 
 
14.21. The molecular formula of the first structure is 
C10H18O, while the molecular formula of the second 
structure is C9H14O2. Each compound is expected to 
exhibit a parent ion at m/z = 154, so the location of the 
parent signal cannot be used to differentiate these 
compounds. Instead, we can use the relative abundance 
of the (M+1)+• peak. The following calculations show 
that the observed data are more consistent with a 
compound containing 10 carbon atoms, rather than 9. 
 
 
 
The first structure is expected to give an (M+1)+• peak 
with a relative abundance of 2.0%. Therefore, we 
conclude that the first structure is cis-rose oxide: 
 
 
 
When comparing the relative abundance of the (M+1)+• 
peaks in this particular case, the expected difference is 
quite small (1.8% vs. 2.0%), so other techniques will be 
more useful for differentiating these compounds, 
including high-resolution mass spectrometry (Section 
14.13). 
As a side note, these compounds can be easily 
distinguished by their strong odors: cis-rose oxide 
smells like roses, and the other compound 
(bicyclononalactone) smells like coconuts! 
 
 
 
14.22. 
(a) This fragment is M – 79, which is formed by loss of 
a Br. So the fragment does not contain Br. Also, the 
fragment has a smaller mass than a single bromine atom 
(77