Quiz: Mass spectrometry

Multiple choice: Check your answers by clicking on them.

1. Choose the correct statements.
Electrospray ionization is a hard ionization technique, which is likely to generate analyte fragments.
ESI can be used for thermally labile analytes.
Adjusting the pH might enhance the generation of ions.
It is very unlikely for adducts to occur during ESI.
2. Which statements regarding full scan mode and selected ion mode (SIM) are true? Select one or more:
During SIM it is possible to observe lower concentrations of an analyte compared to full scan mode.
SIM allows the detection with higher sensitivity compared to full scan.
You need to know the molecular weight of the analyte to use SIM.
3. Decide whether a statment about resolution (R) is true or not.
The higher the mass spectrometric resolution, the worse the distinction between two only slightly differing m/z.
The higher the mass spectrometric resolution, the better the distinction between two only slightly differing m/z.
Assuming a resolution of 1000, you can distinguish between m/z 10.00 and m/z 10.01.
Assuming a resolution of 1000, you can distinguish between m/z 1000.00 and m/z 1000.01.
4. Select those options, which represent mass spectrometric ion sources. Select one or more:
GCI
EI
ESI
SPE
APCI
5. A sample containing a compound of interest with the monoisotopic weight 146.10552 is analyzed with mass spectrometric detection. The analyte is observed to carry two positive charges in the form of two protons (= monoisotopic weight of a proton 1.0078). Which one of the provided options is the mass spectrometrically detected m/z of the doubly charged analyte? Select one or more:
146.10552
74.06056
148.12112
73.05276
147.11332

Fill in the gaps. Toggle down to check your results.

1. The average molecular weight of lysine is 146.19 g/mol (C6H14N2O2). Calculate the monoisotopic masses of the here depicted isotopes of lysine. Note the exchanges of 12C isotope against 13C isotope, which are marked in the image below. Match the numbers in the boxes with the possible solutions below.

Masses of the most abundant isotopes of C, H, N and O:

12C = 12.00000

13C = 13.00336

1H = 1.00782

14N = 14.00307

16O = 15.99491

monoisotopic masses lysin isoptopes

a) 148.100552 Da, b) 146.10552 Da, c) 145.09770 Da, d) 147.10552 Da

Solution

1b, 2d, 3a

 

2. The process of mass spectrometric analysis consists of different steps. Match the provided options with the possible solutions below.

ms process

a) detector, b) vacuum pump, c) recording, d) mass analyzer, e) ion source, f) sample inlet

Solution

1f, 2e, 3b, 4d, 5a, 6c

 

3. What does TIC?

Solution

Total Ion Chromatogram

 

4. Write out EIC.

Solution

Ectracted Ion Chromatogram

 

5. Match the numbers in hte boxes with a „molecular ion“-label as well as the „charge state“ of the lysines. The green circle depicts a positive and the red one a negative charge.charge state_molecular ions_ lysine

a) singly charged (2x), b) doubly charged c) M, d) [M+2H]2+, e) [M+2H-1H]+, f) [M+H]+, g) uncharged

Solution

1gc, 2af, 3bd, 4ae

 

6. Each ionization technique is suitable for specific applications with regard to the analyte’s molecular weight and polarity. Match the provided ionization techniques with the respective field.

ionization techniques

a) APCI, b) ESI, c) EI/ CI, d) APPI

Solution

1c, 2d, 3a, 4b

 

7. Label the depicted graphs, which are obtained using an LC-MS experimental setup. (Software used for data evaluation: MassHunter Qualitative Analysis B. 03.01, Agilent Technologies)

TIC_EIC_mass spectrum

a) TIC, b) RIC, c) EIC, d) mass spectrum, d) ionogram

Solution

1a, 2c, 3d

 

8. An analyte with the sum formula C5H10N2O3 has an monoisotopic mass of 146.06913. Calculate:

a) [M+H]+

b) [M+2H]2+

c) [M-H]–

d) [M+Na]+

e) [M-2H+Na]–

Use:

H = 1.00782, Na = 22.98980

Solution

a) 147.07695, b) 74.042385, c) 145.06131, d) 169.05893, e) 167.04329

 

9. An analyte has the sum formula C6H9N3O2. The average molecular weight is 155.15 g/mol. Calculate the monoisotopic mass using the following masses of isotopes:

12C = 12.00000

1H = 1.00782

14N = 14.00307

16O = 15.99491

Solution

155.06941

 

10. An analyte has the sum formula C14H12O3. The calculated monoisotopic weight is 228.07863. It is mass spectrometrically detected to carry a single positive charge in the form of a proton, i.e. [M+H]+. The calculated accurate m/z would therefore be 229.08645. The actual mass spectrometrically detected m/z is however 229.08603.

Calculate the mass deviation between calculated and detected m/z in ppm (parts per million) using the following equation:

Mass deviation = [(calculated−detected) calculated]×1000000
Give the calculated mass deviation with two decimal places (i.e. 0.00).

Solution

1.83

 

11. The m/z´s of analytes in your sample only differ by 0.1 m/z, i.e 1140 and 1140.1. What mass spectrometric resolution is required at the very least to resolve the signals.
Use the following equation for the calculation: Resolution R = m/(m2-m1)

Solution

11400

 

12. a) Which ionization technique uses a beam of high energy electrons to generate radical cations, i.e. M•+ (M= the analyte)?
M + e- → M•+ + 2 e-

b) Which one generates radical anions, i.e. M•-?
M + e- → M•-

Solution

a) electron ionization, b) electron capture ionization