Question

Please Show work:

Please draw out the Relative abundace Table and explain all calculations. Also, draw structure and provide molecular formula.

Exercise 2: An unknown substance shows a molecular ion peak at m/2 = 170 with a relative intensity of 100. The M + 1 peak has an intensity of 13.2, and the M + 2 peak has an intensity of 1.00. What is the molecular formula of the unknown? Total, MW (3 = 170=1t real (even ) Nithrale relatire intensity 100% Isotope = MH Peak = Intensity - 13,2

Answer 1

ains An unknown substance shows a moleculat son peak at mg=170 with a relative intensity of loo. The nte peak has an intensity of 13-2 and the M+2 Deak has an intensity of 1.00 What is the molecular formula of the unknown ? I The molecular ion peak or parent peak at m = 170 with a relative intensity of too 2 Mt is base peak with 100% intensity Mtt peak represents the presence of fe in a the sample. If you going to measure the peak height of the Mt.1 peak as a percentage of the peak height of the m+ peak , that gives the numbee of carbon atoms in the compound.

mth peak has an intensity 13.2 M+D Deak = 13:2 tit = 12 (mt) Peak = +2 carbons. M+2 peak has an intensity of 1.00, that means too% of m+ (base peak? So Mt2 Peak Presence indicate the - 2 m units greated than the parent peak 2 in theie masses spectra, due to Dresence of small amount of too (which is the most abundant isotope of oxygen too is another isotope Thus mt2 peal with too% of Mt indicates no sulphul chlorine, Bromine

This Mt - (+2) (12) = 170 - 12(12) = 26 Hence G heg is possible, but it cannot have two Nitrogen and Mt2 peak presence might indicate oxyoen Hence, M+- (+2) (12) - 16 170 - (144) - 16 = So, G H o is the possible structure Hence molecular formula for Unknown is CHO The structure of Can Heo is OH 4-Phenyl-phenol (C12H100)

Relative abundanas formula is - CM,) (a) + (M2) (1-x) = Me Where me is the atomic mass of element M, is mass of the isotope for which we know the abundance, Me is mass of unknown isotope x is the relative abundance of known isotope for carbon, it has different naturally occucing isotopes which are 12C, IBe +46 Hence, suppose the has a weight of 12.010 amu Where 13c has weight of 13 army Hence the atomic weight of carbon is 12.000 Now, from formula put all this values. (12.010) (x + (13) (1-x) = 12.000 12.000 x + ( 13 - 13 x = 12.010 12:0102 - 1330 = 12.010-13 8= 0.989 Hence the pool 12c isotope abundance is 98.9% and +3 c isotope abundance is lit% Also Abundance of hydrogen isotope Eelatives are eH= gg.gg / 24 = 0,015 % 34 = 10-18 % I = 99.76%, Ito = 0.04 % 180=0.2%.