Chapter 14, Problem 14.8P

Interpretation Introduction

Interpretation:

How the given compounds (${\text{C}}_{\text{6}}{\text{H}}_{\text{10}}\text{O}$ and ${\text{C}}_7{\text{H}}_{\text{14}}$)can be distinguished by the $\text{m/z}$ ratio of their molecular ions in high-resolution mass spectrometry has to be shown.

Concept Introduction:

High resolution mass spectrometry: This method allows $\text{m/z}$ to be measured with up to four decimal points.

Masses are generally not whole number integers, for example 1 proton = $\text{1}\text{.0073}\text{amu}$ and 1 neutron = $\text{1}\text{.0086}\text{amu}$

One ${}^{12}\text{C}$ atom =exactly $\text{12}\text{.0000}\text{amu}$ because the amu scale is based on the mass of ${}^{12}\text{C}$.

All atoms other than ${}^{12}\text{C}$ will have a mass in amu that can be measured to 4 decimal places by a high resolution mass spectral instrument.

Nominal mass: The nominal mass for an element is the mass number of it is most abundant naturally occurring stable isotope and for an ion or given molecule the nominal mass is the sum of the nominal masses of the constituent atoms.

Example: The exact mass of the most abundant isotopes of each element

For example hydrogen  $\text{H}\text{=}1.007825$

Average atomic mass value: ${}^1{\text{H and }}^2\text{D}$ with relative abundance of 100% and 0.0115% $\text{H}\text{=}1.007825$