# Question #52fdb

May 10, 2015

Just randomly take one trillion water molecules. (These molecules will weigh as few as 30 picograms)

Water molecules with Oxygen-18 atoms will be present in the 0.20% of one trillion, that is 2 billions water molecules, with 4 billions hydrogen atoms.

The probability that there is a deuterium atom D is 0,015% that is 0.015/100 = 0.00015. Four billions multiplied by 0.00015 gives only six hundred thousands of water molecules with one D atom on average (a few molecules may have two, and some others none).

Now, in those six hundred thousands molecules, having on average one D atom, the probability that there is a second D is independent by the presence or not of a "twin D" in the molecule and it is - once again - 0.00015. Therefore, by multiplying 600,000 molecules by this probability we get 90 molecules of ${D}_{2}^{\text{18}} O$ in a trillion.

Finally, let's scale this number to a mole, that is one Avogadro's number of molecules, roughly $6 \cdot {10}^{\text{23}}$, that is equal to $600 \cdot {10}^{9} \cdot {10}^{\text{12}}$, i.e. six hundred billions of trillions.

Given that there were 90 ${D}_{2}^{\text{18}} O$ molecules in a trillion, we just have to multiply these 90 molecules by six hundred billions to get the final result of 54 trillions of "ultra-heavy" water molecules, with molar mass 22 a.m.u. (instead of 18 a.m.u.)