1 mol H2O has a mass of 18.015 grams and you can get this number by adding 2*atomic mass H + 1*atomic mass O.
In that 18.015 g there will be 6.022E23 molecules of water in one of several forms. What kind of forms? If we call 1H p for protium, 2H D for deuterium, and O16 and O18, we can get the following forms of water.
P2O16
PDO16
P2O18
PDO18
D2O16
D2O18
If I didn't miss anything that is 6 different ways of putting p and D and O16 and O18 together.
How many grams D do we have in 18.015 g H2O. That's 18.015 x 0.00015 = about 0.0027 grams D or 0.0027/2.014 = about 0.00134 mols D.
Do the same for O18 and we get
18.015 x 0.002 = 0.036 or 0.036/18 = 0.002 mols O18.
# atoms D = 0.00134*6.02E23=about 8.07E20
# atoms O18 = 0.002*6.02E23 = about 12*10^20. With a ratio of 2:1 H:O, there aren't enough H atoms to use all of the O18. Therefore, if we use all of the D atoms to form those shown above,
1/6 of those D atoms will go to form D2O18 so 1/6*8E20 = about 1.3 E20 D2O18. There will be approximately the same number of D2O16, HDO16, and HDO18 give or take a few. I don't think that uses up all of the O18 out there so there may be a few trillion of those flying around looking to form P2O18. You might want to look over this critically; as you can see I've made some assumption and some guesses.
Relative abundance of H-2 (deuterium) is 0.015% and O-18 is 0.20%. Calculate the number of D2O18 molecules in a 1mol of natural water sample?
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