Isotope ratio in mass spectrometry. The example in the book on p. 580 (6 th
edition of the textbook) shows how to solve the ratio for (M+1). + to M. + to peaks.
Isotope ratio in mass spectrometry The example in the book on p. 580 (6th edition of the textbook) shows how to solve the ratio for (M+1)+ to M+ to peaks. It calculates the values for C6H4N2O4 and C12H24. The two molecules have identical mass and the purpose of the exercise is to show that a mass spectrometry measurement can distinguish between the two based on the (M+1)+ to M+ ratio. The conclusion obtained from the calculation is correct, the values obtained from the calculation are almost right and allow to make the correct conclusion, but they are not done quite right. In essence, according to the calculation, the number of the isotope with value higher by one (for carbon, hydrogen, nitrogen and oxygen) increase linearly with the number of those elements in the molecule. Thus, in the case of carbon, with abundance 1.08 % the percentage of 13C will be 1.08 % for one carbon and six times as much (6.48 %) for six carbons. Following the calculation it would be 108 % for 100 carbons in a molecule. Whereas the actual number would be quite high, it would not be that high. The reasoning is that as the number of carbons is increasing, so is increasing the number of carbons which will be the 13C isotopes; the trouble is, with more carbons, some molecules will have multiple 13C, and those will not contribute to the (M+1)+ to M+ ratio, rather to (M+2)+ to M+ or higher ratios. An example in class was given about rolling several dice and calculating the probability that on one (and only one) of the die the number will be six. The probability that this will be true for one die is 1/6. According to the approach of the calculation used in the book, for six dice if would be 6x(1/6). If that is not alarming, for seven, it would be 7x(1/6), which is more than one, i.e., more than certainty. Clearly, there is something wrong the way we are doing it for the dice and there is the same principle wrong the way it is done in the text for the carbon isotopes. The proper method for calculating the dice issue goes like this: For one it is simple ratio of 1/6. There is one favorable result (the six) and six total results, so the probability is 1/6. For two dice (A and B) there are two favorable results (six on A or six on B). Keep in mind that 6 on both A and B in not favorable. So how many favorable results are there. Not two! When A has a six on it, it can be accompanied by five (1, 2, 3, 4, 5, but not six) results on the other die. Likewise, when B has a six on it, there are five favorable results on die A. Thus, the number of favorable results is 2x5. The number of all possible results is 6x6. For three dice (A, B, C), there are three cases when only one “six” shows, either on A, or B, or C. Each such case is accompanied by five possibilities on B and C, if A is the one
with the six. The five possibilities each equate to 5x5, i.e, 25. The number of all possible results is 6x6x6. If one continues the derive the values of probability for more and more dice, the formula, for n dice is n 1 (6 1)n 1 probability 6n You can check that this works also for n=1 (five to the power of zero is one), so the probability is 1/6. It is interesting to calculate the probabilities for several dice: # of dice 1 2 3 4 5 6 7
probability 0.1666 0.2777 0.34722 0.385802 0.401877 0.401877 0.3907
Notice that the probability, as one would expect, increases as the number of rolled dice goes up, but only to 5and 6 dice; afterwards, it decreases. The reason why the probability does not keep going up is, that we are figuring out chances that “some” six will roll. We are specifically asking for condition in which only one will roll. And with more and more dice the probability of that is of course decreasing. The same calculation and reasoning can be applied to a molecule with certain number of atoms (dice). The only difference is that the favorable states are not neat whole numbers, rather, just fractions. Consider the probability that a molecule C12H24 will have one carbon 13C and the other carbons and all the hydrogens will be the regular isotopes. The isotope ratio (as given in the book, is 1.08 %, or 0.0108 for 13C. The count of total carbons is 100 % or 1. For hydrogen it is 0.00015. To modify the equation for dice we will start with 12 carbons, where one of the carbons can be as carbon-13, and multiply by the possibilities the other eleven carbons can have, but we have to also multiply by the number of possibilities all the 24 hydrogens can have. Thus, we will get for the favorable (one carbon-13) case the following: carbon-13:
12 (0.0108) (1 0.0108)11 (1 0.00015)24 = 0.1146 or 11.46 %
because the total number of all cases is on this scale 1, we divide by one (to whatever appropriate power, which is still one) so the above number, 11.46 % is the percent of
(M+1)+ to M+ due to carbon-13. The book gives 12.96 %, based on the simplified calculation. Similar calculation would be dome for contribution by hydrogen: hydrogen-2:
24 (0.00015) (1 0.00015)23 (1 0.0108)12 = 0.003149 or 0.315 %
The book gives 0.36 %. The reason why the book values are larger is, that because the calculation there does not assume that multiple higher isotopes will exist in the molecule. The error will become more pronounced with larger number of atom in the molecule. Petr Vanýsek, DeKalb, 27 March 2012.
