Chapter 21, Problem 21.23P

(a)

Interpretation Introduction

Interpretation:

Exact mass of ${}^{12}{\text{C}}_8{}^{13}{\text{CH}}_{\text{17}}{\text{N}}_5{\text{Cl}}^{+}$  with $\text{m/z}$ value of 230.1164 has to be computed.

Concept Introduction:

Molecular mass is defined as mass corresponding to mass of one molecule of substance, while nominal mass may be defined as mass of most abundant isotope of a particular ion or molecule. For instance carbon exist as two isotopes $\text{C-}12$ and $\text{C-}13$. $\text{C-}12$ is abundant as much as $98.9\text{ \%}$ while abundance of $\text{C-}13$ is only $\text{1}\text{.1 \%}$. So clearly nominal mass of carbon is $12\text{ g}$.

For instance molecular mass of benzene is calculated by exact atomic masses of $\text{C}$ and $\text{H}$ atom as follows:

  $\begin{array}{c}{\text{Molecular mass of C}}_6{\text{H}}_{\text{6}}=\left(6\right)\left(12.0107\text{ u}\right)+\left(6\right)\left(1.0078\text{ u}\right)\\ =78.111\text{ u}\end{array}$

However its nominal mass will take most abundant masses of $\text{C}$ and $\text{H}$ atoms into account and thus nominal mass is calculated as follows:

  $\begin{array}{c}{\text{Nominal mass of C}}_6{\text{H}}_{\text{6}}=\left(6\right)\left(12\text{ u}\right)+\left(6\right)\left(1\text{ u}\right)\\ =78\text{ u}\end{array}$

Explanation of Solution

Assume most abundant isotope are used for each atom in ${}^{12}{\text{C}}_9{\text{H}}_{\text{17}}{\text{N}}_5{\text{Cl}}^{+}$, mass of this ion is calculated as follows:

  $\begin{array}{c}{}^{12}{\text{C}}_9{\text{H}}_{\text{17}}{\text{N}}_5{\text{Cl}}^{+}=9\left(12\right)+17\left(1.007825\right)+5\left(14.00307\right)+\left(34.96885\right)-0.00055\\ =230.11668\end{array}$

(b)

Interpretation Introduction

Interpretation:

Exact mass of ${}^{12}{\text{C}}_8{}^{13}{\text{CH}}_{\text{17}}{\text{N}}_5{\text{Cl}}^{+}$ has to be computed and expected intensity of peak at  $\text{m/z}$ value of 231 has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since one carbon has been replaced by ${}^{13}\text{C}$, exact mass of ${}^{12}{\text{C}}_8{}^{13}{\text{CH}}_{\text{17}}{\text{N}}_5{\text{Cl}}^{+}$ ion is calculated as follows:

  $\begin{array}{c}{}^{12}{\text{C}}_8{}^{13}{\text{CH}}_{\text{17}}{\text{N}}_5{\text{Cl}}^{+}=\left[\begin{array}{c}8\left(12\right)+1\left(13.003\right)+17\left(1.007825\right)+5\left(14.003\right)+\\ \left(34.968\right)-0.00055\end{array}\right]\\ =231.12003\end{array}$

Hence mass of ${}^{12}{\text{C}}_9{\text{H}}_{\text{17}}{\text{N}}_5{}^{35}{\text{Cl}}^{+}$ is 232.11373.

Observed intensity of peaks with $\text{m/z}$ value of 231.1188 is $11.7\text{ \%}$ so corresponding relative intensity is calculated as follows:

  $\begin{array}{c}\text{Relative intensity}=\frac{11.7\text{ \%}}{100-11.7\text{ \%}}\\ =0.1325:1\end{array}$

Expected intensity for peak with  $\text{m/z}$ value of 231 relative to $\text{m/z}$ value of 230 is then calculated as follows:

  $\begin{array}{c}\frac{\text{Expected intensity}}{11.7\text{ \%}}=\frac{0.1325}{1}\\ \text{Expected intensity}=9.88\text{ \%}\\ \approx 9.9\text{ \%}\end{array}$

Thus, expected intensity is $9.9\text{ \%}$.

(c)

Interpretation Introduction

Interpretation:

Exact mass of ${}^{12}{\text{C}}_8{}^{13}{\text{CH}}_{\text{17}}{\text{N}}_5{\text{Cl}}^{+}$  with $\text{m/z}$ value of 232 has to be computed.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Exact mass of ${}^{12}{\text{C}}_9{\text{H}}_{\text{17}}{\text{N}}_5{}^{35}{\text{Cl}}^{+}$ ion is calculated as follows:

  $\begin{array}{c}{}^{12}{\text{C}}_9{\text{H}}_{\text{17}}{\text{N}}_5{}^{35}{\text{Cl}}^{+}=\left[\begin{array}{c}9\left(12\right)+17\left(1.007825\right)+5\left(14.003\right)+\\ \left(36.96590\right)-0.00055\end{array}\right]\\ =232.11373\end{array}$

Hence mass of ${}^{12}{\text{C}}_9{\text{H}}_{\text{17}}{\text{N}}_5{}^{35}{\text{Cl}}^{+}$ is 232.11373.