Anal Bioanal Chem DOI 10.1007/s00216-007-1707-z
ANALYTICAL CHALLENGE
Birthday chromatography challenge Juris Meija
# Crown copyright in right of Canada 2007
We would like to invite you to participate in the Analytical Challenge, a series of puzzles to entertain and challenge our readers. This special ABC feature has established itself as a truly unique quiz series, with a new scientific puzzle published every other month. Readers can access the complete collection of published problems with their solutions on the Analytical and Bioanalytical Chemistry homepage at http://www.springer.com/abc. Test your knowledge and tease your wits in diverse areas of analytical and bioanalytical chemistry by viewing this collection. In the present challenge chromatography is the topic. And please note that there is a prize to be won (a Springer book of your choice up to a value of €75). Please read on...
Meet the birthday chromatography challenge In the context of years of searching for ever-increasing column efficiency, in this challenge we invite readers to ponder upon the question: why do we really need ultra-high efficiency chromatographic columns? The specific question we ask is: what is the probability of separating all compounds from a multi-component mixture? To be more specific, how many components, relative to theoretical column capacity, can we really expect to separate? However, before readers work out the answer to this odd J. Meija (*) Institute for National Measurement Standards, National Research Council Canada, 1200 Montreal Road, Ottawa, ON K1A 0R6, Canada e-mail:
[email protected]
question, let us first analyze a similar problem that is usually referred to as the birthday paradox: what is the probability that two guests at a birthday party will share their birthdays on the same day? The probability that two randomly chosen people will share a common birthday is 1/365, neglecting leap years and rather small (± 5%) seasonal fluctuations in birth rate. The converse question allows one to estimate the probability that two people do not share a common birthday; the probability of such a circumstance is 364/365. For a greater number of people we can work out such probabilities rather simply. For example, in the case of five people (e.g., Homer, Marge, Lisa, Bart, and Maggie) we know that 364/ 365 is the probability that Homer and Marge do not share their birthdays on the same day. Now, there are 363 unique days left for Lisa, so the probability that she will have her birthday in one of those days is 363/365. For Bart, we have 362/365 and for Maggie, 361/365. To conclude, the probability that all of them have unique birthdays now is simply a product of all of the above probabilities, i.e., (364·363·362·361)/3654. Of course, such calculations are meaningful only when we in fact do not know the actual birthdays in advance. In general, the probability that no N persons share their birthday on the same day is: P¼
1 365N 1
365 ðN1Þ Y x¼3651
x¼
1 ð365 1Þ! N 1 365 ð365 N Þ!
ð1Þ
This expression can be simplified using an approximate solution obtained simply by observing that N persons form N·(N–1) ∕ 2 pairs, each pair not sharing a birthday on the same day with the probability of 364/365. Thus, Pffi
364 365
1 N ðN 1Þ 2
ð2Þ
Anal Bioanal Chem
1P ¼1
1 364! ffi 41% 19 365 345!
ð3Þ
To many, such a large probability comes as a surprise, hence the name “birthday paradox”. This example demonstrates that we tend largely to underestimate the possible coincidences in everyday life. To return to the chromatography example, such birthday coincidences correspond to chromatographically unresolved compounds, or compounds that co-elute. Now the party guests are analogous to chemical compounds, the 365 days of the year serve as analogs to the different possible retention times in the chromatographic column (theoretical plates), and the birthday of each guest corresponds to the retention time of each compound. For example, a five-component chromatogram might look like this:
very small if the number of components (N) exceeds 20 or 30. Let us look at the graph of this function (Eq. 2). 1.00 Probability of resolving all components, P
It is clear that the probability of the converse circumstance is (1−P), corresponding to the situation that the birthdays of all guests will not be all unique. Therefore, at least two of the N guests will share a common birthday. The probability of such a circumstance for a 20-guest party is:
0.75
0.50
0.25 Number of theoretical plates, m = 365 0.00 0
10
20
30 40 50 60 Number of components, N
70
From this graph one can notice a rather unexpected fact— even though our column can separate up to 365 compounds, the probability that it will separate all 50 components from a 50-compound mixture is virtually zero. The only assumptions made in arriving at this conclusion are that intrinsic peak widths are independent of retention time (obviously an approximation) and that retention times of the compounds in the mixture are randomly distributed.
The challenge
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360
Birthday (retention time, s)
We know that our imaginary chromatographic column can separate up to 365 compounds, i.e., column capacity is 365. It is clear that the intrinsic ability to separate does not necessarily mean that these compounds will actually be separated. Similarly, the simple ability to be able to tell the difference between two persons’ birthdays does not mean they will have different birthdays—they might well be born on the same day; in our chromatographic analogy, this corresponds to the overlap of the chromatographic peaks of two species. Now consider the probability of separating all components from an N-compound mixture. In birthday-partylanguage this question amounts to the probability that none of the N persons share a birthday. Mathematically it can be expressed with Eq. 1 or 2. This probability turns out to be
Considering this probability from a slightly different point of view, we ask the readers to elucidate the maximum number of single component peaks that one can expect to observe from a chromatographic column, assuming the above simplified separation model. Readers might resort to simple Monte Carlo modeling with Excel, since the exact mathematical solutions might be somewhat abstract. We invite our readers to participate in the Analytical Challenge by solving the puzzle above. Please send the correct solution to
[email protected] by February 20, 2008. Make sure you enter “Birthday chromatography challenge” in the subject line of your e-mail. The winner will be notified by e-mail and his/her name will be published on the Analytical and Bioanalytical Chemistry website at http://www.springer.com/abc and in the Journal. Readers will find the solution and a short explanation on the Analytical and Bioanalytical Chemistry website after February 20, 2008 and in the Journal (Issue 391/1). The next Analytical Challenge will be published in Issue 390/5, March 2008. If you have enjoyed solving this Challenge you are invited to try the previous puzzles on the Analytical and Bioanalytical Chemistry website.
