We know that terrestrial Cl is 75% 35Cl and 25% 37Cl.
A molecule containing one Cl would show the same distribution, i.e., 75% H35Cl and 25% H37Cl. This would appear in mass spectrometry as two molecular ion peaks with an intensity ratio of 3:1.
An interesting question is therefore to ask what pattern would we see if there are two chlorine atoms in a molecule?
The best way to think about this is in terms of probability and choosing an isotope for each Cl in the molecule. The probability of any particular outcome, e.g., a Cl2 molecule with one 35Cl and one 37Cl, depends on two things:
- The abundances of each isotope. The probabilities are independent so we just multiply them (using either decimals or fractions).
- The number of different ways of getting the same outcome. If there are multiple ways of getting the same outcome, we multiply our result by this number.
The probability that a Cl2 molecule has one 35Cl and one 37Cl is worked out using 0.75 x 0.25 x 2= 0.375.
0.75 = The probability of choosing a 35Cl for the 1st Cl.
0.25 = The probability of choosing a 37Cl for the 2nd Cl.
2 = the number of ways of getting a Cl2 molecule with one 35Cl and one 37Cl; it could either be 35Cl37Cl or 37Cl35Cl
Here are two further examples where all possible outcomes are considered:
| Example: CH2Cl2 contains 2 Cl atoms. Describe the appearance of the molecular ion peaks in the mass spectra. Use the following abundances of Cl in decimal form: 35Cl = 0.75 37Cl = 0.25 Answer We must work systematically through each possible outcome to determine the appearance of the molecular ion peaks. If there are 2 Cl atoms, the molecule can contain either 0, 1 or 2 37Cl atoms. This means there are three outcomes so there would be three molecular ion peaks.
Outcome 1: CH235Cl35Cl Mass = 84 Probability = 0.75 x 0.75 = 0.5625
Outcome 2: CH237Cl35Cl Mass = 86 Probability = 2 x 0.25 x 0.75 = 0.375 There are two different ways having one 37Cl in the molecule. It can either be in the 1st position (i.e., CH237Cl35Cl) or 2nd position (i.e., CH235Cl37Cl). This means that to calculate the probability, and abundance, we multiply by 2.
Outcome 3: CH237Cl37Cl Mass = 86 Probability = 0.25 x 0.25 = 0.0625
This means that there would be 3 molecular ion peaks in the mass spectra of CH2Cl2 with an intensity ratio of 9:6:1. |
A challenging extension of this would be to consider what happens if there are 3 Cl atoms in the molecule:
| Example: CHCl3 contains 3 Cl atoms. Describe the appearance of the molecular ions peaks in the mass spectra. Answer Again, we must work systematically through each possible outcome to determine the apperance of the molecular ion peaks. If there are 3 Cl atoms, the molecule can contain either 0, 1, 2 or 3 37Cl atoms. This means there are four outcomes so there would be four molecular ion peaks. Outcome 1: CH35Cl35Cl35Cl Mass = 118 Probability = 0.75 x 0.75 x 0.75 = 0.422
Outcome 2: CH37Cl35Cl35Cl Mass = 120 Probability = 3 x 0.25 x 0.75 x 0.75 = 0.422 There are three different ways having one 37Cl. It can either be in the 1st, 2nd or 3rd position. This means that to calculate its abundance, we multiply by 3.
Outcome 3: CH37Cl37Cl35Cl Mass = 122 Probability = 3 x 0.25 x 0.25 x 0.75 = 0.141 There are three different ways having one 35Cl. It can either be in the 1st, 2nd or 3rd position. This means that to calculate its abundance, we multiply by 3.
Outcome 4: CH37Cl37Cl37Cl Mass = 124 Probability = 0.25 x 0.25 x 0.25 = 0.0156
This means that there would be 4 molecular ion peaks in the mass spectra of CHCl3 with an intensity ratio of 27:27:9:1. An interesting result from this calculation is finding that having zero or one 37Cl atom in the molecule is equally likely! |
If you are also studying A Level Maths, you might know where the 1, 2,1 and 1, 3, 3, 1 come from! If not, look up Pascal’s triangle and binomial expansions…