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1. The average molecular weight of lysine is 146.19 g/mol (C_{6}H_{14}N_{2}O_{2}). Calculate the monoisotopic masses of the here depicted isotopes of lysine. Note the exchanges of ^{12}C isotope against ^{13}C 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:

^{12}C = 12.00000

^{13}C = 13.00336

^{1}H = 1.00782

^{14}N = 14.00307

^{16}O = 15.99491

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.

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.

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.

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)

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

## Solution

1a, 2c, 3d

8. An analyte with the sum formula C_{5}H_{10}N_{2}O_{3 }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 C_{6}H_{9}N_{3}O_{2}. The average molecular weight is 155.15 g/mol. Calculate the monoisotopic mass using the following masses of isotopes:

^{12}C = 12.00000

^{1}H = 1.00782

^{14}N = 14.00307

^{16}O = 15.99491

## Solution

155.06941

10. An analyte has the sum formula C_{14}H_{12}O_{3}. 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