Question

In a magnetic sector (single-focusing) mass spectrometer, it might be reasonable under some circumstances to monitor one $m/z$value, to then monitor a second $m/z$, and to repeat this pattern in a cyclic manner.

Rapidly switching between two accelerating voltages while keeping all other conditions constant is called

peak matching.

(a) Derive a general expression that relates the ratio of the accelerating voltages to the ratio of the

corresponding $m/z$values.

(b) Use this equation to calculate $m/z$of an unknown peak if $m/z$of the ion used as a standard, ${C{F}_3}^{+}$is $69.00$and the ratio of $V_{unknown}/V_{s\tan dard}$is $0.965035$.

(c) Based on your answer in part (b), and the assumption that the unknown is an organic compound that has a mass of $143$, draw some conclusions about your answer in part (b), and about the compound.

Short answer

The solution for the following questions are :

(a) The general expression that relates the ratio of accelerating voltages to the ratio of corresponding $m/z$value is $\frac{V_u}{V_{s_{}}}$.

(b) $m/z$ratio of unknown peak is $71.50$.

(c) The calculations suggest that the ion was doubly charged and unknown compound contain odd number of nitrogen atoms.

Step-by-step solution

Mass spectrometer

A device that measures the flight path of charged particles or ions across a system of magnetic and electric fields to calculate their exact mass.

:The general expression relating the ratio of accelerating voltage to the corresponding  mzvalue can be derived by : 

The equation revealing mass spectra can be acquired by varying one of the three variables $\left(B,Vorr\right)$and holding the remaining two as constant by :

$\frac{m}{z}=\frac{B^2{r}^{2}e}{2V}$

This can also be written as,

${\left(\frac{m}{z}\right)}_s=\frac{K}{V_s}$and${\left(\frac{m}{z}\right)}_u=\frac{K}{V_u}$

where,

$B$and $r$are constant,

Subscript $s$as standard value,

Subscript $u$as unknown value.

On dividing the equations, we get the following relationship -

$$\begin{gathered} \frac{{\left({\displaystyle \frac{m}{z}}\right)}_s}{{\left({\displaystyle \frac{m}{z}}\right)}_u}=\frac{{\displaystyle \frac{K}{V_s}}}{{\displaystyle \frac{K}{V_u}}} \\ =\frac{V_u}{V_s} \end{gathered}$$

Hence, this is the required general expression relating the ratio of accelerating voltages to the corresponding $\frac{m}{z}$values.

Calculating mz of an unknown peak, :

According to the information given-

$$\begin{gathered} {\left(\frac{m}{z}\right)}_s=69.00 \\ \frac{V_{unknown}}{V_{s\tan dard}}=0.965035 \end{gathered}$$

By substituting the values, we get-

$$\begin{gathered} \frac{{\left({\displaystyle \frac{m}{z}}\right)}_s}{{\left({\displaystyle \frac{m}{z}}\right)}_u}=\frac{V_u}{V_s} \\ \frac{69.00}{{\left({\displaystyle \frac{m}{z}}\right)}_u}=0.965035 \\ {\left(\frac{m}{z}\right)}_u=\frac{69.00}{0.965035} \\ =71.499 \\ \approx 71.50 \end{gathered}$$

Hence, the $\frac{m}{z}$ratio of unknown peak is 71.50.

: Finding the conclusions :

The following conclusions are-

The ion in part (b) was doubly charged as per the calculated$\frac{m}{z}$value, where the given molecular mass is$143$.

And the unknown compound contains odd number of nitrogen atoms.