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How do you interpret peaks in a mass spectrum?

Peaks in a mass spectrum indicate the presence of various isotopes or fragments of the molecule under analysis.

In mass spectrometry, a sample is ionized to produce positive ions. These ions are subsequently accelerated and deflected in magnetic fields, with the degree of deflection being contingent upon their mass-to-charge ratio (m/z). As the ions are detected, a mass spectrum is generated. This spectrum is a graphical representation, where the relative abundance of ions—typically expressed as a percentage—is plotted on the y-axis, while the m/z ratio is displayed on the x-axis.

Each peak in the mass spectrum corresponds to an ion with a specific m/z ratio. The position of a peak on the x-axis indicates its m/z ratio, which generally corresponds to the relative atomic mass for ions bearing a single positive charge. Meanwhile, the height of each peak reflects the relative abundance of that particular ion.

The peak with the highest m/z value, referred to as the molecular ion peak or parent peak, represents the entire molecule and reveals the molecular mass of the compound. This peak is frequently accompanied by a smaller peak at m/z+1m/z + 1, attributable to the presence of the isotope carbon-13 in organic molecules.

Other peaks in the mass spectrum arise from fragment ions, which form when the molecular ion undergoes fragmentation. These peaks can yield crucial insights into the molecular structure. For instance, the observation of a peak at m/z=15m/z = 15 may indicate the presence of a methyl group (CH3\text{CH}_3) in the molecule, as the m/z ratio of a methyl cation is 1515.

In summary, interpreting the peaks in a mass spectrum entails identifying the molecular ion peak to ascertain the molecular mass, while analyzing the fragment ion peaks provides valuable information regarding the molecular structure. This analytical process necessitates a solid understanding of mass spectrometry principles as well as the characteristics of various atoms and functional groups.

Answered by: Prof. Lily Johnson
IB Chemistry Tutor
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