Answer:
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The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.
<h3>
What is relative abundance?</h3>
- The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
- When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
- Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.
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Answer:
<em>For both cases the answer is C</em>
Explanation:
We can see that the orbitals are not filled in the order of increasing energy and the Pauli exclusion principle is violated because it does not follow the correct order of the electron configuration; In the first exercise after the 2s2 orbital, the 2p2 orbital follows.
For the second exercise, you must start in order with level 1 and correctly filling each of the sublevels corresponding to each level until reaching level 7 and thus completing the desired number of electrons.
The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment
Answer:
Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z
Explanation:
The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.
The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.
Compound 2 (2,5-dimethylhexane) structure shows that the cleavage of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.