Let's begin with the assumption that our homeopathic medium is water. What we need to begin is the molecular mass of water; this is easy to calculate, if you have a periodic table handy. Water (H

_{2}0) contains two hydrogen atoms (one proton each) and one oxygen atom (eight protons and eight neutrons, usually), making for atomic masses of 1 u and 16 u, respectively. This makes the molecular mass of water 1 + 1 + 16 = 18 u.

Now, if you remember high school chemistry (I won't blame you if you don't), 1 unified atomic mass unit (I just found out that chemical and physical amu were deprecated!) is equal to 1 gram/mole. This means that water, at an atomic mass of 18 u, has a mass of 18 grams per mole of particles. And here's where we come to the importance of Avagadro's constant: one mole is approximately 6.02 x 10

^{23}(this number was arrived at by calculating the number of atoms in 12 grams of carbon-12, in case you're interested).

If we want to take a hard look at our dilution, we need the number of molecules in our dilution medium before we begin. So if we have 18 grams per mole (which, for water, is 18 millilitres per mole—thank you, metric system!) and one mole is 6.02 x 10

^{23}... Well, let's do the math!

18 g/mol = 0.0556 mol/g

0.0556 mol/g = 0.0556 x 6.02 x 10^{23}/g

0.0556 x 6.02 x 10^{23}/g = 3.34 x 10^{22}/g

Or you could just have WolframAlpha calculate it for you. But that wouldn't be nearly as fun!

How about alcohol? Ethanol being more popular (and less deadly) than methanol, we'll use that:

46 g/mol = 0.0217 mol/g

0.0217 mol/g = 0.0217 x 6.02 x 10^{23}/g

0.0217 x 6.02 x 10^{23}/g = 1.03 x 10^{22}/g

So what does this tell us? Well, a standard homeopathic dilution is 30C (or 60X), which means one part "active" ingredient in 100

^{30}(or 10

^{60}) parts water or alcohol. But here's the kicker: we just calculated the maximum number of "parts" into which it is possible to divide water! Unless you go splitting the atom, you can't divide 1 cc of water into more than 3.34 x 10

^{22}parts. Let's say we're dealing with a full litre of the stuff (approximately two pints, for you Americans): we're still only dealing with 3.34 x 10

^{25}molecules—we're still off by a whopping thirty-five orders of magnitude!

So what's the maximum dilution at which you're more likely than not to have at least one molecule of the original substance? You use logarithms! Assuming one litre of water (1 kg):

log_{10}3.34 x 10^{25}= 25.5

Alternatively, assuming 1.27 litres of ethanol (1 kg):

log_{10}1.03 x 10^{25}= 25.0

That would be a dilution of 25X (approximately 13C). At dilutions above 13C, you are unlikely to have a single molecule of the original substance left!

The solution? (Pun unintentional, but appreciated!) Water has memory! For an erudite discussion of this subject matter, may I suggest Storm, by Tim Minchin?

beautifull

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