Chapter 2, Problem 2.137P

Interpretation Introduction

Interpretation:

The masses of all possible $1+$ ions for boron trifluoride are to be calculated.

Concept introduction:

Mass spectrometry is a technique used to analyze the structure and chemical properties of different molecules. In this technique, molecules are bombarded with electrons in order to convert them into highly energetic positively charged ions, called molecular ions. The molecular ions may undergo further breakup into smaller ions, called fragment ions.

The molecular ions and the fragment ions are separated by deflection in a variable magnetic field according to their mass to charge ratio $\left(\text{m/e}\right)$.

The formula to calculate the molecular mass of a compound is,

$\text{Molecular mass}=\left[\begin{array}{l}\sum \left(\text{numer of atoms of the element}\right)\\ \left(\text{atomic mass of the element}\right)\end{array}\right]$ (1)

Answer to Problem 2.137P

The all possible molecular and fragment ions formed during the mass spectrometry of the compound ${\text{BF}}_3$ with different isotopic compositions are, ${}^{10}{\text{B}}^{19}{\text{F}}_3$, ${}^{10}{\text{B}}^{19}{\text{F}}_2$, ${}^{10}{\text{B}}^{19}\text{F}$, ${}^{10}\text{B}$, ${}^{11}{\text{B}}^{19}{\text{F}}_3$, ${}^{11}{\text{B}}^{19}{\text{F}}_2$, ${}^{11}{\text{B}}^{19}\text{F}$ and ${}^{11}\text{B}$.

The masses of the ions in $\text{amu}$ are 67, 48, 29, 10, 68, 49, 30 and 11 respectively.

Explanation of Solution

The two naturally occuring isotopes of the element boron are ${}^{10}\text{B}$ and ${}^{11}\text{B}$. The element fluorine has only one naturally occuring isotope, ${}^{19}\text{F}$.

Therefore the molecular and fragment ions for the compound ${\text{BF}}_3$ with different isotopic compositions formed during the mass spectrometry are ${}^{10}{\text{B}}^{19}{\text{F}}_3$, ${}^{10}{\text{B}}^{19}{\text{F}}_2$, ${}^{10}{\text{B}}^{19}\text{F}$, ${}^{10}\text{B}$, ${}^{11}{\text{B}}^{19}{\text{F}}_3$, ${}^{11}{\text{B}}^{19}{\text{F}}_2$, ${}^{11}{\text{B}}^{19}\text{F}$ and ${}^{11}\text{B}$.

The formula to calculate the molecular mass of ${}^{10}{\text{B}}^{19}{\text{F}}_3$ is,

$\text{Molecular mass}=\left(1\right)\left({\text{atomic mass of }}^{10}\text{B}\right)+\left(3\right)\left({\text{atomic mass of }}^{19}\text{F}\right)$ (2)

Substitute $10\text{ amu}$ for the atomic mass of ${}^{10}\text{B}$ and $19\text{ amu}$ for the atomic mass of ${}^{19}\text{F}$ in equation (2).

$\begin{array}{c}\text{Molecular mass}=\left(1\right)\left(10\text{ amu}\right)+\left(3\right)\left(19\text{ amu}\right)\\ =10\text{ amu}+57\text{ amu}\\ =67\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{10}{\text{B}}^{19}{\text{F}}_2$ is,

$\text{Molecular mass}=\left(1\right)\left({\text{atomic mass of }}^{10}\text{B}\right)+\left(2\right)\left({\text{atomic mass of }}^{19}\text{F}\right)$ (3)

Substitute $10\text{ amu}$ for the atomic mass of ${}^{10}\text{B}$ and $19\text{ amu}$ for the atomic mass of ${}^{19}\text{F}$ in equation (3).

$\begin{array}{c}\text{Molecular mass}=\left(1\right)\left(10\text{ amu}\right)+\left(2\right)\left(19\text{ amu}\right)\\ =10\text{ amu}+38\text{ amu}\\ =48\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{10}{\text{B}}^{19}\text{F}$ is,

$\text{Molecular mass}=\left(1\right)\left({\text{atomic mass of }}^{10}\text{B}\right)+\left(1\right)\left({\text{atomic mass of }}^{19}\text{F}\right)$ (4)

Substitute $10\text{ amu}$ for the atomic mass of ${}^{10}\text{B}$ and $19\text{ amu}$ for the atomic mass of ${}^{19}\text{F}$ in equation (4).

$\begin{array}{c}\text{Molecular mass}=\left(1\right)\left(10\text{ amu}\right)+\left(1\right)\left(19\text{ amu}\right)\\ =10\text{ amu}+19\text{ amu}\\ =29\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{11}{\text{B}}^{19}{\text{F}}_3$ is,

$\text{Molecular mass}=\left(1\right)\left({\text{atomic mass of }}^{11}\text{B}\right)+\left(3\right)\left({\text{atomic mass of }}^{19}\text{F}\right)$ (5)

Substitute $11\text{ amu}$ for the atomic mass of ${}^{11}\text{B}$ and $19\text{ amu}$ for the atomic mass of ${}^{19}\text{F}$ in equation (5).

$\begin{array}{c}\text{Molecular mass}=\left(1\right)\left(11\text{ amu}\right)+\left(3\right)\left(19\text{ amu}\right)\\ =11\text{ amu}+57\text{ amu}\\ =68\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{11}{\text{B}}^{19}{\text{F}}_2$ is,

$\text{Molecular mass}=\left(1\right)\left({\text{atomic mass of }}^{11}\text{B}\right)+\left(2\right)\left({\text{atomic mass of }}^{19}\text{F}\right)$ (6)

Substitute $11\text{ amu}$ for the atomic mass of ${}^{11}\text{B}$ and $19\text{ amu}$ for the atomic mass of ${}^{19}\text{F}$ in equation (6).

$\begin{array}{c}\text{Molecular mass}=\left(1\right)\left(11\text{ amu}\right)+\left(2\right)\left(19\text{ amu}\right)\\ =11\text{ amu}+38\text{ amu}\\ =49\text{ amu}\end{array}$

The formula to calculate the molecular mass of ${}^{11}{\text{B}}^{19}\text{F}$ is,

$\text{Molecular mass}=\left(1\right)\left({\text{atomic mass of }}^{11}\text{B}\right)+\left(1\right)\left({\text{atomic mass of }}^{19}\text{F}\right)$ (7)

Substitute $11\text{ amu}$ for the atomic mass of ${}^{11}\text{B}$ and $19\text{ amu}$ for the atomic mass of ${}^{19}\text{F}$ in equation (7).

$\begin{array}{c}\text{Molecular mass}=\left(1\right)\left(11\text{ amu}\right)+\left(1\right)\left(19\text{ amu}\right)\\ =11\text{ amu}+19\text{ amu}\\ =30\text{ amu}\end{array}$

The masses of ions ${}^{10}\text{B}$ and ${}^{11}\text{B}$ are, $10\text{ amu}$ and $11\text{ amu}$ respectively.

Conclusion

The all possible molecular and fragment ions formed during the mass spectrometry of the compound ${\text{BF}}_3$ with different isotopic compositions are, ${}^{10}{\text{B}}^{19}{\text{F}}_3$, ${}^{10}{\text{B}}^{19}{\text{F}}_2$, ${}^{10}{\text{B}}^{19}\text{F}$, ${}^{10}\text{B}$, ${}^{11}{\text{B}}^{19}{\text{F}}_3$, ${}^{11}{\text{B}}^{19}{\text{F}}_2$, ${}^{11}{\text{B}}^{19}\text{F}$ and ${}^{11}\text{B}$.

The masses of the ions in $\text{amu}$ are 67, 48, 29, 10, 68, 49, 30 and 11 respectively.