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If I look at any tables or calculators, I find that negative ion formate adduct has a monoisotopic mass of +44.9982 units. If I try to calculate myself, I get slightly different result:

H + C + O + O = 
1.007825 + 12.0107 + 15.9994 + 15.9994 = 
45.017325

The difference is 0.019, what is already not indifferent, as usually I calculate with a tolerance of +- 0.02 units, which means this can cause me miss many hits. I could not find out what may cause this difference.

Later had similar experiences with the acetate, as in tables +59.013851 can be found, but if I calculate:

3 x H + 2 x C + 2 X O = 
3 x 1.007825 + 2 x 12.0107 + 2 x 15.9994 =
59.043748

Which means even more, 0.0298 difference.

My question is what causes the difference, and what is the correct wat to calculate these masses?

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In your calculation, you are using the average molecular mass of the structure, where atomic masses are based on the natural abundance of all isotopes of the element.

H:  1.00794 u
C: 12.01070 u
O: 15.99940 u
O: 15.99940 u
─────────────
   45.01744 u

However, for the exact molecular mass (or monoisotopic mass) of the structure, the atomic masses of each atom are based on the most common isotope for the element.

H-1:   1.00783 u
C-12: 12.00000 u
O-16: 15.99491 u
O-16: 15.99491 u
────────────────
      44.99765 u

Furthermore, you are missing the mass of the extra electron of the formate ion. The electron mass is $m_\mathrm e=0.000548579909070(16)\ \mathrm u$. Thus,

H-1:   1.00783 u
C-12: 12.00000 u
O-16: 15.99491 u
O-16: 15.99491 u
e−:    0.00055 u
────────────────
      44.99820 u
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  • $\begingroup$ Thank you! I was believing I have already the monoisotopic masses in my table... so I need to look for another table. And special thanks for pointing out the extra electrons should be added. $\endgroup$ – deeenes Nov 6 '15 at 15:40
  • $\begingroup$ And if anybody else needs the monisotopic masses, here is a nice table: ciaaw.org/atomic-masses.htm $\endgroup$ – deeenes Nov 6 '15 at 16:03

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