Assumptions:
* The container holds 100 liters of soup.
* The soup may be modeled as a heterogeneous mixture in which the lightest molecule is water (18 g/mol). We bias in favor of greater likelihood of survival by assuming that the soup consists entirely of water.
* The proportion of soup retained each day from the previous day’s batch is exactly half.
* The renewal of the soup has taken place an average of 5 times per week.
* The probability of one molecule surviving from one day to the next is independent of this probability for any other molecule.
Under these conditions, 45 years = 2,340 weeks = 11,700 renewals. Therefore the proportion of soup remaining from the initial batch is 1/2^11700 ≈ 1/10^3522. The capacity of the container is 5,555 mol ≈ 3.3 * 10^27 molecules.
Since "proportion" doesn't make sense at magnitudes this small, let's switch to talking about probabilities. The likelihood of a given molecule from the original batch surviving to the present day is p = 1/10^3522 by our earlier assumptions. Using the Bernoulli trial model, the likelihood of _any_ molecule surviving is
1 – (1 – p)^(3.3 * 10^27). Switching to logarithms, we find log[(1 – p)^(3.3 * 10^27)] = (3.3 * 10^27) * log(1 – p) ≈ –3.3 * 10^(–3495). Using the approximation exp(x) ≈ 1 + x for small x, we can take 3.3 * 10^(–3495) as the probability that any molecules survived over 45 years of renewals.
This is about the same as the likelihood of you winning the Mega Millions jackpot every day for the next nine and a half years.