Figure out how many deuterium atoms in a 3 microgram ice core sample

I'm in an introductory Chemistry class and ran into a Pearson problem, that no-one in our class including our professor has been able to truly figure out. 


I know the solution is $4.69 \times 10^{13}$. 


Attached are images. 


[![enter image description here](https://i.stack.imgur.com/x6NFx.png)](https://i.stack.imgur.com/x6NFx.png)


[![enter image description here](https://i.stack.imgur.com/gb1cF.png)](https://i.stack.imgur.com/gb1cF.png)


So far, I have tried two methods. First, I tried to calculate the answer by mass. 


1) Calculate mass of deuterium


$3\times10^{-6} \times .0156 \times 10^{-2} = 4.68 \times 10^{-10}$


2) Calculate moles of deuterium


$\frac{4.68 \times 10^{-10}}{2} = 2.34 \times 10^{-10}$


3) Calculate atoms of deuterium


$2.34 \times 10^{-10} \times 6.02214 \times 10^{23} = 1.41 \times 10^{14}$


This answer is 3 times more than the actual solution. 


Now, for the real B.S. that is this problem. 


Let's try solving this problem by moles/atoms (as in percentage is percent deuterium by atoms)


1) Calculate moles of deuterium 


$\frac{3\times10^{-6}}{18} \times 0.0156 \times 10^{-2} = 2.6 \times 10^{-11}$


2) Calculate atoms of deuterium


$2.6 \times 10^{-11} \times 6.02214 \times 10^{23} = 1.57 \times 10^{13}$


This answer is 3 times less than the real answer. 


Help. What is 3?




How many deuterium atoms (symbol D for deuterium) are present in $\pu {3.00 x 10^{-6} g}$ of water?


**Answer:**


$$\pu{3.00 x 10^{-6} g} \;\ \ce{H2O} \times \frac{\pu{1 mol} \;\ce{H2O}}{\pu{18.015 g} \;\ce{H2O}}\times \frac{\pu{2 mol} \;\text{H}}{\pu{1 mol} \;\ce{H2O}}\times 0.000156 \times \frac{N\_\mathrm A \;\text{D atoms}}{\pu{1 mol} \;\text{D atoms}} = \pu{3.12 x 10^{13} D atoms}$$


where $N\_\mathrm A = \pu{6.0221415 x 10^{23}}$ is Avogadro's number and the deuterium fractional isotopic abundance (from the wiki article on "Deuterium") is 1 deuterium atom per 6420 atomic number 1 species, i.e., 1/6420 = 0.000156 = 0.0156%. The units can be mol D per mol H.


**Alternate solution:** The probability that a water molecule is HOD, rather than HOH, is 2/6420, i.e., 1/3210 = 0.000312 = 0.0312%. This is simply due to ordinary (unsubstituted) water having two hydrogens per molecule, i.e., it is HOH, so it has two independent chances to have H replaced by D. So, *on average*, there is one HOD molecule out of every 3210 water molecules. Likewise, every 3210 moles of water contains, *on average*, 1 mole of HOD molecules.


The probability that a water molecule is DOD, rather than HOH, is the square of 1/6420, i.e., $ \pu{2.426 x 10^{-8}}$. This is negligible compared with 1/3210, so it is ignored in what follows. Therefore:


$$\pu{3.00 x 10^{-6} g} \;\ \ce{H2O} \times \frac{\pu{1 mol} \;\ce{H2O}}{\pu{18.015 g} \;\ce{H2O}}\times \frac{\pu{1 mol} \;\ce{HOD}}{\pu{3210 mol} \;\text{HOH}}\times \frac{N\_\mathrm A \;\text{D atoms}}{\pu{1 mol} \;\text{HOD}} = \pu{3.12 x 10^{13} D atoms}$$


 2

I go back to your first calculations. $3.00$ microgram water does not contain $3.00$ microgram Hydrogen, as you have calculated. No ! It contains $3.00/18$ micromole of water, and also $2·3.00/18 = 0.333 $ micromole $H$. This is also $0.333$ microgram $H$. I think you can go ahead with this information. Try to calculate the mass of Deuterium in 0.333 microgram $H$. 


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