Please include step by step explanation. a) Write a molecular formula consistent with the data below INTENSITY m/e (AS PERCENT OF BASE PEAK) 14 15 18 28 29 42 43 44 100.0 (base) 73 74 75 8.0 38.6 16.3 39.7 23.4 46.6 10.7 86.1 Mt 3.2 0.2 Solution We can elucidate the molecular formula from molar mass by using Rule of 13. Rule of 13 states that molecular formula is a multiple of 13 , molecular mass of CH unit. First step is to assume that compound is made up of only carbon and hydrogen. Each CH unit has molar mass of 13. Thus divide the molar mass by 13 to get number of CH units. Then the base formula is elucidated in the following manner, M/13 = n+ r/13, where r is the remainder and base formula is Cn Hn+r Degree of unsaturation u = ( n- r + 2)/ 2 .A negative value of u indicates presence of heteroatom. A half integer value indicates presence of odd no of nitrogen atom. For each N atom once CH2 unit is deducted from base formula. For each oxygen atom, one CH4 unit is deducted. Molecular ion peak, m/z = 73 No of CH unit = 73/13 = 5 + 8/13 Base formula = C 5 H 5+8 = C 5 H 13 Degree of unsaturation u = (5-8+2)/ 2 = -1/2 Thus there is odd number of Nitrogen atoms present .Odd molecular ion mass also indicates presence of odd no of Nitrogen in the molecule. Assuming presence of 1 nitrogen atom, Base molecular formula is C 4 H 11 N Presence of M+1 peak indicates mainly carbon 13 and N-15 isotopes. Abundance of C-13 is 1.1%,. Abundance of N-15 is 0.36. Thus M+1 = (No. of C’s) 1.1% + (No. of N’s) 0.36% Given M+ = 86.1 % , M+1 peak = 3.2 % , Scaling M+ peaks to 100% , we get M+1 peaks as 3.7 % ((100/86.1)* 3.2) Thus approximate no. of carbon atoms is 3.7/1.1 = 3.378 3.7 = (No. of C’s) 1.1% + (No. of N’s) 0.36% No of carbon = 3, No of Nitrogen = 1 Abundance of O- 18 = 0.2% M+2 = (No. of O’s) 0.20% (100/86.1)*0.2 = 0.232 0.232 = (No. of O’s) 0.20% Approximate No of oxygen atoms = 1 Thus molecular formula is C 3 H 7 NO Fragments at m/z 44 corresponds to CH 2 NO- formed by loosing C 2 H 5 - fragment with m/z = 29 .

1. Please include step by step explanation.
a) Write a molecular formula consistent with the data below INTENSITY m/e (AS PERCENT
OF BASE PEAK) 14 15 18 28 29 42 43 44 100.0 (base) 73 74 75 8.0 38.6 16.3 39.7 23.4 46.6
10.7 86.1 Mt 3.2 0.2
Solution
We can elucidate the molecular formula from molar mass by using Rule of 13. Rule of 13 states
that molecular formula is a multiple of 13 , molecular mass of CH unit. First step is to assume
that compound is made up of only carbon and hydrogen. Each CH unit has molar mass of 13.
Thus divide the molar mass by 13 to get number of CH units. Then the base formula is
elucidated in the following manner,
M/13 = n+ r/13, where r is the remainder and base formula is Cn Hn+r
Degree of unsaturation u = ( n- r + 2)/ 2 .A negative value of u indicates presence of heteroatom.
A half integer value indicates presence of odd no of nitrogen atom. For each N atom once CH2
unit is deducted from base formula. For each oxygen atom, one CH4 unit is deducted.
Molecular ion peak, m/z = 73
No of CH unit = 73/13 = 5 + 8/13
Base formula = C 5 H 5+8 = C 5 H 13
Degree of unsaturation u = (5-8+2)/ 2 = -1/2
Thus there is odd number of Nitrogen atoms present .Odd molecular ion mass also indicates
presence of odd no of Nitrogen in the molecule. Assuming presence of 1 nitrogen atom,
Base molecular formula is C 4 H 11 N
Presence of M+1 peak indicates mainly carbon 13 and N-15 isotopes. Abundance of C-13 is
1.1%,. Abundance of N-15 is 0.36. Thus
M+1 = (No. of C’s) 1.1% + (No. of N’s) 0.36%

2. Given M+ = 86.1 % , M+1 peak = 3.2 % ,
Scaling M+ peaks to 100% , we get M+1 peaks as 3.7 % ((100/86.1)* 3.2)
Thus approximate no. of carbon atoms is 3.7/1.1 = 3.378
3.7 = (No. of C’s) 1.1% + (No. of N’s) 0.36%
No of carbon = 3, No of Nitrogen = 1
Abundance of O- 18 = 0.2%
M+2 = (No. of O’s) 0.20%
(100/86.1)*0.2 = 0.232
0.232 = (No. of O’s) 0.20%
Approximate No of oxygen atoms = 1
Thus molecular formula is C 3 H 7 NO
Fragments at m/z 44 corresponds to CH 2 NO- formed by loosing C 2 H 5 - fragment with m/z =
29