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If theres a mixture of components we can calculate the mole fraction
mole fraction can be calculated as follows
mole fraction of component =
number of moles of ethanol - 3.00 mol
total number of moles in mixture - 3.00 + 5.00 = 8.00 mol
mole fraction of ethanol =
mole fraction of ethanol is 0.375
mole fraction can be calculated as follows
mole fraction of component =
number of moles of ethanol - 3.00 mol
total number of moles in mixture - 3.00 + 5.00 = 8.00 mol
mole fraction of ethanol =
mole fraction of ethanol is 0.375
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.
