Question

3. Below are given two partial mass spectra -- specifically, just the molecular ion region. Using the table attached to the back of the exam, give the molecular formula for each compound. Mass Rel. Abundance Mass Rel. Abundance 165 (M.) 100.0 84 (M.) 100.0 166 10.5 5.3 0.3 85 167 Formula: Formula:

Answer for both molecular ion spectra

Answer 1

The first step is recognizing your M, M+1, and M+2 values. The m/z values increase by one as we go from M to M+1 because M+1 is the molecular ion whose mass is one higher than M.

• first we have to calculate number of carbons

​​​​​​​mass spectrum m/z = 165 (M; 100%), m/z = 166 (10.5%), and m/z = 167(0.3 %)

The M+1 value is 165 since we know that the definition of M+1 is a molecular ion whose mass is one amu higher than M. 166 amu is one amu higher than the m/z of M so it must be the M+1 value. Dividing the relative abundance of M+1 (8.91%) by 1.1% gives 9.5. While we can say that there are 9 or 10 carbons in the compound,

• The second step is to see if there are any contributors in the M+2 spectrum. The only isotopes that have significant peaks in this spectrum are sulfur, chlorine, and bromine. in this spectrum the abundance is very less for 167 so we can say there is no sulfur,chlorine and bromine atoms are present.

​​​​​​​now we have to calculate molecular formula

• The nitrogen rule states that when m/z for M has an even mass (even number of amu), the corresponding molecular formula has an even number of nitrogen atoms (0, 2, 4, etc.) and when m/z for M has an odd mass (odd number of amu), the corresponding molecular formula has an odd number of nitrogen atoms (1, 3, 5, etc.).
• the mass of M is 165 number of nitrogen in compound has to be odd, now we are going to determine hydrogen.
• The hydrogen rule states that for a molecule containing only hydrogen, carbon, oxygen, nitrogen, fluorine, chlorine, bromine, and iodine, the maximum number of monovalent atoms possible (max H) for a given number of carbons (C) and nitrogens (N) is given by the equation max H = 2C + N + 2. This rule is extremely important for ruling out impossible molecular formulas for a given molecule.

If we assume there are NINE carbons, then we subtract 9X12 from 165 which gives us 57 amu. so the number of hydrogens can,t be 57 so we can fit 3 nitrogens in the formula so formula should be. C9H15N3(IMMEPIP)

from this procedure we can calculate second one also.

mass spectrum m/z = 84 (M; 100%), m/z = 85 (5.3%), and m/z = 86(4.4 %)

divide 5.3/1.1 so the number of carbon is 4.8 carbon should be 8 or 9

4.4 % abundance is for sulfur so there is sulfur atom present

now number of hydrogens 4x12 = 48 minus it by 84 that is 36 so there is one sulfur atom so molecular formula should be C4H4S (THIOPHENE)

• C9H15N3
• ​​​​​​​C4H4S