Question

In a mass spectrometer, positive ions are produced when a gaseous mixture is ionized byelectron bombardment produced by an electric discharge. When the electric-discharge voltage is low, singly positive ions are produced and the following peaks are observed in the mass spectrum:

Mass(u)	Relative Intensity
32	0.3743
34	0.0015
40	1.0000

When the electric discharge is increased, still only singly charged ions are produced, but now the peaks observed in the mass spectrum are

Mass(u)	Relative Intensity
16	0.7500
18	0.0015
40	1.0000

What does the gas mixture consist of, and what is the percent composition by isotope of the mixture?

Short answer

The gas mixture consistof ¹⁶O,¹⁸O, and ⁴⁰Ar and The percent compositionby isotope of the mixture is ¹⁶O = 42.82%, ${}^{18}O=8.6\times {10}^{-2}\%$, and⁴⁰Ar = 57.094%

Step-by-step solution

Definition

The mass percent of an atom in a compound is equal to the ratio of the atom's total mass to the complex'stotal mass multiplied by 100. As a result, it may be stated as follows:

Mass percent of an atom$=\left(\frac{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{atom}}{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{substance}}\right)100$

Explanation for using ions

To tackle this problem, we need to know how discharge voltage affects ions. Because there is enough energy to break the bonds when the discharge voltage is high, the ions are in the form of individual atoms. There isn't enough energy when the discharge voltage is low. As a result, the ions exist in the form of molecules.

We may state that the ions present are ¹⁶O⁺, ¹⁸O⁺ and ⁴⁰Ar⁺ when we look at the data for increased electric discharge. These ions can explain the peaks observed when the discharge voltage was low.

The peak for mass 32 must be for ¹⁶O⁺, ¹⁶O⁺ the peak for mass 34 must be for ¹⁶O⁺, ¹⁸O⁺ , and the peak for mass 40 must be for ⁴⁰O⁺ .

Determination of percent composition

Finally, the relative intensity data may be used to calculate the percent content of each isotope in the mixture, as shown:

$\begin{array}{rcl}\%{}^{16}O& =& \frac{0.7500}{0.7500+0.0015+1.0000}\times 100\%\\ & =& 42.82\%{}^{16}O\\ \%{}^{18}O& =& \frac{0.0015}{0.7500+0.0015+1.0000}\times 100\%\\ & =& 8.6\times {10}^{-2}\%{}^{18}O\\ \%{}^{40}Ar& =& \frac{1.0000}{0.7500+0.0015+1.0000}\times 100\%\\ & =& 57.094\%{}^{40}Ar\end{array}$

Therefore the percentage composition from the mixture of each isotopeis,

$$\begin{gathered} {}^{16}O=42.82\% \\ {}^{18}O=8.6\times {10}^{-2}\% \\ {}^{40}Ar=57.094\% \end{gathered}$$