Chapter 20, Problem 20.19QAP

Interpretation Introduction

(a)

Interpretation:

The type of mass spectrum used for molecular mass of compound and reason for using this mass spectrum is to be stated.

Concept introduction:

The mass spectrum of a compound is obtained by the ionization of the compound. The ionization can be done by using electron ionization or chemical ionization. The type in which the spectrum of compound obtained is simple and the process of the method is easy.

Answer to Problem 20.19QAP

The chemical ionization of the mass spectrum is used for molecular mass of a compound.

Explanation of Solution

The mass spectrum obtained by chemical ionization is used for molecular mass of compound because the molecular structure of the compound is disturbed in electron ionization. In electron ionization bombardment of electrons and photons is used to produce secondary ions. The energy consumed in electron ionization is high and the exact molecular structure is not obtained. If the molecular structure of the compound is different, then the mass spectrum obtained is also different and the molecular mass of the compound cannot be determined.

The chemical ionization process to obtain a spectrum of a compound is simple and consumes low energy. Thus, the chemical ionization of the mass spectrum is used for molecular mass of a compound.

Interpretation Introduction

(b)

Interpretation:

The type of mass spectrum used for the chemical structure of the compound and reason for using this is to be stated.

Concept introduction:

The chemical structure of the compound consists of the different orientation of atoms in a compound. The analysis of every atom is done to obtain a chemical structure of compound because a particular atom behaves differently with a different atom in the compound. The method used for the mass spectrum for the chemical structure of the compound should be such that the orientation of every atom can be analyzed separately.

Answer to Problem 20.19QAP

The electron ionization of the mass spectrum is used for the chemical structure of the compound.

Explanation of Solution

The electron ionization is used in mass spectrum for chemical structure of compound because every atom is analysed by ionizing differently and spectrum of chemical structure is very simple.

Interpretation Introduction

(c)

Interpretation:

The flight time for ion in TOF mass analyzer is to be stated.

Concept introduction:

In TOF mass analyzer positive ions are produced by bombarding the electrons or photons on molecules of a compound. These produced ions then accelerated in the drift tube due to the electric field applied. The ions in the tube are separated on the basis of their masses. The kinetic energies of all the ions in the tube are the same because all ions are under the effect of the same electric field. The velocities of ions are in an inverse of their masses.

Answer to Problem 20.19QAP

The flight time for ion in TOF mass analyzer is $\underset{\_}{0.0982\text{s}}$.

Explanation of Solution

Write the expression of flight time for ion.

$t^2=\left(\frac{m}{z}\right)\left(\frac{L^2}{2V}\right)$ ...... (I)

Here, the flight time is $t$, the mass to charge ratio of ion is $m/z$, the length of the tube is $L$, the voltage is $V$.

Substitute $58$ for $m/z$, $1\text{m}$ for $L$ and $3000\text{V}$ for $V$ in the Equation (I).

$\begin{array}{c}{t}^2=\left(58\right)\left(\frac{{\left(1\text{m}\right)}^2}{2\left(3000\text{V}\right)}\right)\\ =\sqrt{0.00966{\text{s}}^2}\\ t=0.0982\text{s}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The derivation of an equation for the differences in flight times as a function of two masses, charges and accelerating voltage is to be stated.

Concept introduction:

The ion in the tube of mass analyzer consists of kinetic energy and potential energy. The potential energy is due to charged electrons and kinetic energy due to the mass of ions. The difference in flight times is obtained by finding the flight time of each ion separately.

Answer to Problem 20.19QAP

The equation for the differences in flight times as a function of two masses, charges and accelerating voltage is $\underset{\_}{L\sqrt{\frac{\left(m_1-m_2\right)}{2zeV}}}$.

Explanation of Solution

Write the expression for potential energy of charged electrons

$P=zeV$ ...... (I)

Here, charge of electron is $e$, number of electrons is $z$, and voltage difference is $V$.

Write the expression of kinetic energy of ions

$K=\frac{1}{2}m{v}^2$ ...... (II)

Here, mass of ions is $m$, and velocity of ions is $v$

The potential energy of electrons and kinetic energy of ions is same in TOF mass analyzer. From Equation (II) and Equation (III).

$\begin{array}{c}zeV=\frac{1}{2}m{v}^2\\ v=\sqrt{\frac{2zeV}{m}}\end{array}$

Write the expression for flight time

$t_p=\frac{L}{v}$ ...... (IV)

Here, length of tube is $L$, flight time is $t_p$.

Substitute $\sqrt{\frac{2zeV}{m}}$ for $v$ in the Equation (IV)

$\begin{array}{c}t=\frac{L}{\left(\sqrt{\frac{2zeV}{m}}\right)}\\ =L\sqrt{\frac{m}{2zeV}}\end{array}$ ...... (V)

Write the expression of the flight time for ion of mass to charge ratio $m_1/z$ from the Equation (V).

$t_{p_1}=L\sqrt{\frac{m_1}{2zeV}}$

Write the expression of the flight time for ion of mass to charge ratio $m_2/z$ from the Equation (V).

$t_{p_2}=L\sqrt{\frac{m_2}{2zeV}}$

Write the expression for difference in flight time of ions

$t_p=t_{p_1}-t_{p_2}$ ...... (VI)

Substitute $L\sqrt{\frac{m_1}{2zeV}}$ for $t_{p_1}$ and $L\sqrt{\frac{m_2}{2zeV}}$ for $t_{p_2}$ in the Equation (VI).

$\begin{array}{c}{t}_p=L\sqrt{\frac{m_1}{2zeV}}-L\sqrt{\frac{m_2}{2zeV}}\\ =L\sqrt{\frac{\left(m_1-m_2\right)}{2zeV}}\end{array}$

Interpretation Introduction

(e)

Interpretation:

The difference between flight time for ions of mass to charge ratios $m_1/z$ and $m_2/z$ is to be stated.

Concept introduction:

The difference in flight time for ions depends on mass of ions, charge of electrons and accelerating voltage in the tube. It is given by the formula $t^2=\left(\frac{m}{z}\right)\left(\frac{L^2}{2V}\right)$.

Answer to Problem 20.19QAP

The difference between flight time for ions of mass to charge ratios $m_1/z$ and $m_2/z$ is $\underset{\_}{0.0008\text{s}}$.

Explanation of Solution

Write the expression of flight time of ions

$t_1{}^2=\left(\frac{m_1}{z}\right)\left(\frac{L^2}{2V}\right)$ ....... (VII)

Here, the mass of ions are $m_1$ ,, number of electrons is $z$, and voltage difference is $V$, length of tube is $L$, flight time is $t_p$.

Substitute $59$ for $m_1/z$, $58$ for $m_2/z$

$1\text{m}$ for $L$ and $3000\text{V}$ for $V$, $1.6\times {10}^{-19}\text{C}$ for $e$ in the Equation (VII).

$\begin{array}{c}{t}_1{}^2=\left(58\right)\left(\frac{{\left(1\text{m}\right)}^2}{2\left(3000\text{V}\right)}\right)\\ =\sqrt{0.00966{\text{s}}^2}\\ t_1=0.0982\text{s}\end{array}$

Write the expression of flight time of ions

$t_2{}^2=\left(\frac{m_2}{z}\right)\left(\frac{L^2}{2V}\right)$ ...... (VIII)

Substitute $59$ for $m_2/z$

$1\text{m}$ for $L$ and $3000\text{V}$ for $V$, $1.6\times {10}^{-19}\text{C}$ for $e$ in the Equation (VIII).

$\begin{array}{c}{t}_2{}^2=\left(59\right)\left(\frac{{\left(1\text{m}\right)}^2}{2\left(3000\text{V}\right)}\right)\\ =\sqrt{0.00949{\text{s}}^2}\\ t_2=0.0974\text{s}\end{array}$

Write the expression for differences in flight time

$t=t_1-t_2$ ...... (IX)

Substitute $0.0982\text{s}$ for $t_1$ and $0.0974\text{s}$ for $t_2$ in the Equation (IX).

$\begin{array}{c}t=t_1-t_2\\ =0.0982\text{s}-0.0974\text{s}\\ =0.0008\text{s}\end{array}$

Interpretation Introduction

(f)

Interpretation:

The type of ionization source in tendom mass spectrometer and reason for using this is to be stated.

Concept introduction:

The mass spectrum is the plot between the relative abundance and mass to charge ratio. The mass spectrum is different for the same substance. The relative abundance represents the y-axis and the mass to charge ratio represents the x-axis. The heights of the peaks represent the relative intensity. The peak with the highest value in the plot is called the base peak. All the calculations are made with respect to this base peak. The structural information can be obtained by fragmentation of molecules in the sample.

Answer to Problem 20.19QAP

The electronic ionization is used to get the structural information of the compound because the single unpaired electron is obtained by electronic ionization.

Explanation of Solution

The fragmentation pattern is obtained by passing the vaporized organic sample from the ionization chamber. This vaporized organic sample is then bombarded by the stream of electrons and positive ions are obtained. This positive ion is called molecular ion and consists of a single unpaired electron. This single unpaired electron is obtained by the electronic ionization process. The peak in the mass spectra with value $58$ for $m/z$ is the base peak. The molecule corresponding to the base peak is of $100\%$ relative abundance. This molecule absorbs a large amount of ionization. The fragmentation of molecules occurs due to electronic ionization because the vaporized organic sample is fragmented in the ionization chamber.

Interpretation Introduction

(g)

Interpretation:

The types of mass spectra obtained from the MS/MS experiment is to be stated.

Concept introduction:

The mass spectrometry is a technique used to find the structure and chemical properties of different molecules. In the process of mass spectrometry, different molecules are converted into ions. The ions are separated by the mass to charge ratio and detected by relative abundance in the sample. The mass spectrum of ion is obtained by plotting the mass to charge ratio and relative abundance of the ion in a sample.

Answer to Problem 20.19QAP

The type of mass spectra obtained from MS/MS experiment by:

(1) Holding the first analyzer constant and scanning the second analyzer is product ion spectra.

(2) Scanning both analyzers with a small $m/z$ offset between them is neutral loss spectra.

(3) Scanning the first analyzer and holding the second analyzer constant is precursor ion spectra.

(4) Scanning the second mass analyzer for every mass-selected by the first mass analyzer is three-dimensional spectra.

Explanation of Solution

(1) The ion spectra obtained in the MS/MS experiment is very different from each other for any compound. The product ion spectra of the compound are obtained in the MS/MS experiment by holding the first analyzer constant and scanning the second analyzer.

(2).When the difference between mass to charge ratio to different ions is very small, by scanning both analyzers, then ions are detected by analyzing the neutral loss spectrum.

(3) The compounds with the same products in ions are identified by determining the concentration of ions in each compound. If ions are detected by scanning the first mass analyzer and keeping the second mass analyzer constant, then precursor ion spectra are obtained in the MS/MS experiment.

(4) If ions are detected by selecting ions in the first mass analyzer and scanned by the second mass analyzer, then the three-dimensional spectrum is obtained in the MS/MS experiment.