[retretretretretret]

Is there some kind of trick to this question?

Write the molecular formulas for compounds that show the following molecualr ions in the high-resolution mass spectra. Assume that C, H, N, and O might be present. C = 12.0000u H = 1.00783u N = 14.00307u O = 15.99491u

a) M+ = 98.0844           b) M+ = 123.0320

Is it just trial and error? Thanks.

[retretretretretret]

Asked my prof, got this systematic solution --

Example: 98.0844
 
The saturated alkane C7H16 has an exact mass of 100.1252.
C7H14 (1 degree of unsaturation) has a mass of 98.1096.
Therefore, the molecule must contain a heteroatom. The mass is even, so there is an even number of N atoms.
Try zero N's:
The saturated ether/alcohol C6H14O has a mass of 102.1045.
C6H12O (1 degree of unsaturation) has a mass of 100.0888.
C6H10O (2 degrees of unsaturation) has a mass of 98.0732.
Therefore the molecule must contain two heteroatoms.
The saturated ether/alcohol C5H12O2 has a mass of 104.0837.
C5H12O (1 degree of unsaturation) has a mass of 102.0681.
C5H10O (2 degrees of unsaturation) has a mass of 100.0524.
C5H8O (3 degrees of unsaturation) has a mass of 98.0368.
Try 2 N's:
The saturated amine C5H14N2 has a mass of 102.1157.
C5H12N2  (1 degree of unsaturation) has a mass of 100.1001.
C5H10N2  (2 degrees of unsaturation) has a mass of 98.0844. (!)

[azmanam]

Trial and error is tedious.  Use the Rule of 13s:

We're organic chemists, dealing with organic molecules... It's a pretty good bet we've got a lot of carbon and hydrogen atoms - and there are at least as many hydrogen atoms as carbon atoms.  Those are our assumptions as we start the rule of 13s

*1C and 1H has an atomic mass of 13 (12 + 1).
*Divide the [M+] peak by 13 to get an integer and a remainder
   *98/13 = 7r7

*The integer equals the number of carbon atoms
   *7 carbon atoms
*The integer plus the remainder equals the number of hydrogen atoms
   *7 + 7 = 14 hydrogen atoms
*One possible molecular formula is C₇H₁₄

*If you think there's an oxygen atom in the molecule:
   *1 Oxygen atom has atomic mass of 16... so does 1 Carbon atom and 4 Hydrogen atoms
   *If we add one oxygen atom to our molecular formula, we need to remove 1C and 4H
      *C₇H₁₄ + O - CH₄ = C₆H₁₀O
   *If you think there are 2 oxygen atoms... repeat
      *C₆H₁₀O + O - CH₄ = C₅H₆O₂

*If you think there's a nitrogen atom:
   *1 Nitrogen atom = 14 amu, so does 1C + 2H
      *C₇H₁₄ + N - CH₂ = C₆H₁₂N
   *If you think there are 2 nitrogen atoms... repeat
      *C₆H₁₂N + N - CH₂ = C₅H₁₀N₂

*Combine if you think there are mixtures of N and O
      *C₇H₁₄ + O + N - C₂H₆ = C₅H₈NO

Now, this is only guestimates and approximations.  I just used whole number masses... for exact mass to 4 decimal places you could take these possible molecular formula and see if any give the exact mass
*C₇H₁₄ = 98.1096
*C₆H₁₀O = 98.0732
*C₅H₆O₂ = 98.0368
*C₆H₁₂N even MW can't have odd  # N
*C₅H₁₀N₂ = 98.0844
*C₅H₈NO even MW can't have odd  # N

I only had to try 4 masses, not 12.  Also, if IR or ¹³C NMR or ¹H NMR tells you you definitely DO have an oxygen or definitely DO NOT have a nitrogen, you can narrow down your possible molecular formulas even more!

Can you use the Rule of 13s to narrow down a molecular formula for 123?