Chapter 24, Problem 1Q

Interpretation Introduction

Interpretation:

The compounds with their correct molecular masses should be matched. Also the way odd and even molecular mass of compounds help to identify the compounds should be stated.

Concept introduction:

Mass spectrometry is an important spectroscopic method to determine the molecular weights and molecular formulas of the compounds. It also can help in the identification of compound or determination of its structure.

A mass spectrograph is plotted between the intensity of ions on y-axis and $\text{m/z}$ on x axis. Mass spectrometry works as follows:

• A sample is converted into a gaseous ion by an ionization technique in the mass spectrometer.
• Then on the basis of mass-to-charge $\left(\text{m/z}\right)$ ratios, gaseous ions are arranged.
• These ions generate electric current at detector and a mass spectrum is formed.

Explanation of Solution

To assign the molecular weights to compounds the rule of thirteen is utilized in mass spectrometry. Thirteen here is obtained by the addition of the mass of carbon and hydrogen.

Steps to determine the molecular weight through the rule of thirteen is as follows:

• $\text{m/z}$value is first divided by 13 to obtain the number of carbon present in the compound, then integer obtained in the quotient is the number of carbons.
• This integer is added to the reminder to obtain the number of hydrogen present in the compound.
• If some heteroatom is present like oxygen or nitrogen then some number of carbon and hydrogens are substracted accordingly to get the appropriate m/z value.
• An important rule for the presence of nitrogen is used which states that a compound whose molecular ion contains an even m/z value will contain either no or even nitrogen atoms and if a compound has odd m/z value then the molecular ion contains odd nitrogen atoms.

The molecular weight of azobenzene is 182, pyridine is 79, and ethanol is 46.

Since azobenzene has an even $\text{m/z}$ value so it will have even number of nitrogen atom that is 2 and pyridine has an odd $\text{m/z}$ value so it will have odd number of nitrogen that is 1. So it is easier to determine the number of nitrogen atom present in the compound by their molecular weights.