Answers. CH908, Problem Set 1.
1. What is the mass, in Daltons, of a 337 nm photon?
The wavelength of the electromagnetic wave became shorter when the charge was vibrating
faster, but the wave seemed to move with the same speed whether the vibration was fast or
slow. Only the wavelength changed when the oscillations became faster. All electromagnetic
radiation -- from radio waves to x-rays -- travel at the speed of light. In empty space this speed is
approximately 300,000 kilometers per second. We can predict the wavelength of an
electromagnetic wave if we know the time it takes for the charge to oscillate once, returning to its
original location. This time is called the "period", T, of the wave.
Then
c = 


Where 1/f, then c= X f
The energy of a photon is equal to Planck's constant times its frequency (this formula was
discovered by Einstein during his work on the photoelectric effect), and also energy is mass of
the particle time the square of the speed of light (usually denoted c), 299,792,458 metres per
second.
Then
mc = h f
m= h
c
m=
6.63E-34 J*s
2.99792458E8 m * 337 nm *
1m
s
10E9nm
m = 6.56E-36 Kg * 6.022E26 Da
1Kg
m = 3.951E-9 Da
2. What is the monoisotopic mass of C60?
60 * 12.00000 = 720 Da
What is the monoisotopic m/z value of C60-?
60 * 12.00000 + 0.000578 Da = 720.000548 Da
What is the monoisotopic m/z value of C60H+?
60 * 12.00000 + 1.007825 - 0.000578 Da = 721.007277 Da
3.a.Calculate the masses and isotopic distributions for:
Cs2I+= 132.9054 *2 + 126.9045 - 0.000548 = 392. 14795 Da
Cs3I2+= 132.9054 *3 + 126.9045 * 2 - 0.000548 = 652.524705 Da
Cs4I3+= 132.9054 *4 + 126.9045 *3 - 0.000548 = 912.334615 Da
Cs5I4+= 132.9054 *5 + 126.9045 * 4 - 0.000548 = 1172.144525 Da
Would these peaks have any special analytical use in mass
spectrometry?
Calibration
b. Calculate the masses and isotopic distributions for:
Na235Cl+ = 22.989770 *2 + 34.968853 - 0.000548 = 80.947845 Da
Na237Cl+ = 22.989770 *2 + 36.965903 - 0.000548 = 82.944895 Da
`m/z
80.948
Abundance
%
75
82.945
25
Na335Cl2+= 22.989770 *3 + 34.968853 * 2 - 0.000548 = 138.906468 Da
Na335Cl 37Cl+= 22.989770 *3 + 34.968853 * 1 + 34.968853 * 1 - 0.000548
Na337Cl2+ = 22.989770 *3 + 36.965903 * 2 - 0.000548 = 142.900568Da
= 140.903518 Da
Isotopic
No.
0
138.906362
Abundance
%
56.25
1
140.903462
37.50
2
142.900562
6.25
m/z
Na435Cl3+= 22.989770 *4 + 34.968853 * 3 - 0.000548 = 196.865091 Da
Na435Cl237Cl1+= 22.989770 *3 + 34.968853 *2 + 34.968853 *1 - 0.000548 = 198.862141 Da
Na435Cl137Cl2+= 22.989770 *3 + 34.968853 *1 + 34.968853 *2 - 0.000548 = 200.859191 Da
Na437Cl3+ = 22.989770 *4 + 36.965903 * 3 - 0.000548 = 202.856241 Da
Isotopic
No.
m/z
Abundance
%
0
1
2
196.864932
198.862032
200.859132
42.1875
42.1875
14.0625
3
202.856232
1.5625
Na535Cl4+= 22.989770 *5 + 34.968853 * 4 - 0.000548 = 254.823714 Da
Na535Cl337Cl1+= 22.989770 *3 + 34.968853 * 3 + 34.968853 * 1 - 0.000548 = 256.820764 Da
Na535Cl237Cl2+= 22.989770 *3 + 34.968853 * 2 + 34.968853 * 2 - 0.000548 = 258.817814 Da
Na535Cl137Cl3+= 22.989770 *3 + 34.968853 * 1 + 34.968853 * 3 - 0.000548 = 260.814864 Da
Na537Cl4+= 22.989770 *5 + 36.965903 * 4 - 0.000548 = 262.811914 Da
Isotopic
No.
0
1
254.823502
256.820602
Abundance
%
31.6406
21.0938
2
258.817702
0.3906
3
4
260.814802
262.811902
4.6875
42.1875
m/z
c. Calculate the masses and abundances of the first three isotopic peaks
for C60, assuming a 13C abundance of 1.1 %.
Isotopic No.
0
1
2
m/z
720
721.0034
722.0067
Abundance %
51.4965
34.3657
11.2757
d. Calculate the masses and abundances of the most abundant 10
isotopic peaks of cholesterol, C27H46O.
Isotopic No.
m/z
Abundance %
0
1
2
3
4
5
6
7
8
9
386.355
387.358
388.360
387.361
389.362
388.363
388.359
389.361
387.359
389.365
74.0302
22.2315
3.3665
0.51
0.30
0.15
0.15
0.05
0.03
0.02
Heavy isotope
composition
monoisotopic
13C
1
13C
2
2H
1
13C
3
13C 2H
1
1
18O
1
13C 18O
1
1
17O
1
13C 2H
2
1
4. Estimate the mass resolving power for the following examples:
Mass Resolving Power (FWHM)
=
=
m
m FW HM
868.54548
= 1447.6
868.3  868.9
Mass Resolving Power (FWHM)
=
=
m
m FW HM
868.50183
= 72375.2
868.501  868.49
5. What elemental compositions are possible for a positive ion molecule
of mass 386.3542 +/- 5 ppm? Assume Carbon, Hydrogen, Oxygen,
Nitrogen, and Sulfur.
C20H44N5O2
C22H48N3S
C22H46N2O3
C24H50OS
C25H44N3
C27H46O
386.34949800 – 0.00470200
386.35689300 + 00269300
386.3508400 - 0.00336000
386.35823500 + 0.00403500
386.35352200 - 0.00067800
386.35486400 - 0.00066400
What about a neutral fragment loss of 43.0184 +/- 100 ppm?
HN3
C2H3O
43.017047 - 0.001353
43.018389 - 0.000011
Explain how you determine the possible compositions.
First you need to calculate the upper limits:
Carbon =
386.3542
= 32.2
12
Nitrogen =
386.3542
= 28
14
Hydrogen =
386.3542
= 386.3542
1
Sulfur =
Oxygen =
386.3542
= 24
16
386.3542
= 12
32
With all tose limits you shall put them on a worksheet until you get the value +/- 5 ppm.
6. Calculate the R+DB of each elemental composition listed in question
5.
r+db = X 
Y Z
 1,
2 2
with X = C, Si
Y = H, Cl, F
Z = N, P
Structure
C20H44N5O2
C22H48N3S
C22H46N2O3
C24H50OS
C25H44N3
C27H46O
HN3
C2H3O
X
Y
20
22
22
24
25
27
0
2
Z
44
48
46
50
44
46
1
3
5
3
2
0
3
0
3
0
dbe
1.5
0.5
1
0
5.5
5
2
1.5
7. How many hydrogen, oxygen, carbon, nitrogen, chlorine, bromine,
phosphorus, and sulfur atoms are possible in the following example
isotopic patterns?
This spectrum could indicate a molecule of mass 38 which could form
an abundant fragment ion y the loss of two hydrogen atoms (m/z 36,
[M-2]+). This hints in the text should lead to the observation in Table
2.1 of the 3/1 abundance ratio of the natural chlorine isotopes,
masses 35 and 37. Tus, the spectrum is a mixture of H35Cl and H37Cl,
in proportions correspondinf to the sotopic abundantes of 35Cl and
37
Cl.
Hydrogen Cloride
Methyl bromide
Sulfur dioxide
C9H8O
The abundante of the m/z 78 peak shows immediately that no “A+1”
elements, Table2.2 tell us that the posibilitéis are C3 and C2N3. The
m/z 76 molecular ion can not contain an add number of nitrogen
atoms so that c3 is correct. A posible combination of “A” elements is
H9P. (C3H2F2 could not fragment to yield m/z 73)
C3H9P