Wow this one has me stumped

[nlmodel]

Of all the hydrogen in the ocean, 0.0156 % of the mass is deuterium. How many deuterium nuclei could be obtained from 7.50 gal of ordinary tap water?
Can anyone give me a clue about how to go about this....the first part of the quesiton means that .0156 percent of all the hydrogen in the ocean is deuterium when measuring by mass....i think you have to account for molar masses of oxygen and 2 H but seems kinda paradoxyl but my thinking cap isnt on so need a little help and maybe an equation..thanks

 

[spelletrader]

Tap water doesn't come from the ocean. (Usually)

 

[FiberoN]

Well, don't you have to know how much hydrogen ( mass ) in the ocean you are talking about? That way, you can do a proportion. Or you can use volume of the ocean ( try to find it ) and do that too. Like - volume of ocean/vol of ocean x .0156 % = 7.5 gal / 7.5 x .0156%
Hmm... I think I have no clue wtf I'm doing, Sorry :|

 

[uart]

Well I don't know exactly the mass of a gallon so I'll give it to you per liter and you can do the conversion to 7.5 gallons.

There are approx 1000 grams of H2O per Liter, so that gives the equivalent of 0.000156 * 1000 g = 0.156 grams of Deuterium based H2O per Liter. Now each Oxygen atom has an atomic mass of approx 16 and each Deuterium atom an approx mass of 2, so about 2/20 of the above 0.156 grams (/L) is Deuterium.

So amounts to 0.0156 grams/Liter.

To find the number of Nuclei you must multiple the grams by Avogadro's number and divide by the atomic mass. That gives 0.0156 /2 * 6.02 * 10^23 or about 4.7 * 10^21 nuclei per liter.

 

[jagec]

simple, assuming tap water is the same consistancy as the ocean...

you just find the percent hydrogen in the ocean, use avagadro's number and the total volume of the ocean to find the total number of deuterium nuclei, and subtract all the nuclei that AREN'T in your container

 

[spelletrader]

The oceans mass = ~1.4*10^21 kg

 

[silverpig]

Originally posted by: uart
Well I don't know exactly the mass of a gallon so I'll give it to you per liter and you can do the conversion to 7.5 gallons.

There are approx 1000 grams of H2O per Liter, so that gives the equivalent of 0.000156 * 1000 g = 0.156 grams of Deuterium based H2O per Liter. Now each Oxygen atom has an atomic mass of approx 16 and each Deuterium atom an approx mass of 2, so about 2/20 of the above 0.156 grams (/L) is Deuterium.

So amounts to 0.0156 grams/Liter.

To find the number of Nuclei you must multiple the grams by Avogadro's number and divide by the atomic mass. That gives 0.0156 /2 * 6.02 * 10^23 or about 4.7 * 10^21 nuclei per liter.

This is the correct method. Just multiply uart's final answer by 7.5 and then again by 3.875 (litres/gal) and you've got your final answer.

 

[nlmodel]

thanks but i really need it in kg or well convert gallons to kg i guess

 

[vasdrakken]

or look deuterium up on google: Deuterium (symbol 2H) is a stable isotope of hydrogen with a natural abundance of one part in 7000 of hydrogen. The nucleus of deuterium (called a deuteron) has one proton and one neutron, whereas a normal hydrogen nucleus just has one proton. Deuterium is also called heavy hydrogen. While it is not an element in its own right, it is often given the symbol D. It occurs naturally as deuterium gas, D2 or 2H2. ummm it's waste from valcano's and nukes... of course one part in 7000 would prob give you your answer really fast...

 

[uart]

thanks but i really need it in kg or well convert gallons to kg i guess

No, your real problem is that you can't see an answer when it is staring you in the face.

 

[DKlein]

You needed to know the # of molecules right? Well that has been given, or at least the equation to get it has.

.000156 x atomic weight deuterium + 1-.000156 x atomic weight hydrogen = atomic weight of hydrogen* in this example.
Mass of 7.5Gal H2O (do this yourself) x ratio hydrogen to water by mass [H*/(H* + 2 x O)] = mass H* in sample:
(.000156D+.999844H)/[(.000156D+.999844H)+2xO]x[(7.5Gal converted to g) = mass H*
H* mass in sample x .000156 = mass Deuterium in sample:
(.000156D+.999844H)/[(.000156D+.999844H)+2xO]x[(7.5Gal converted to g)x.000156 = Dg
mass Deuterium divided by atomic weight Deuterium = moles Deuterium
moles Deuterium x avog's # = molecules Deuterium:
(.000156D+.999844H)/[(.000156D+.999844H)+2xO]x(7.5Gal converted to g)x.000156/Dxavog's# = molecules D
or
[.000156x(7.5gal->g)x(.000156D+.999644H)]/{[(.000156D+.999844H)+2xO]xDxAvogadro's #} = molecules D
Maybe you could simplify this, but I'm done for now. All you have to do is find out what 7.5Gal is in grams! I hope I didn't make any mistakes... I nearly failed chemisty in HS so check it for yourself. Plus I got 4 hours of sleep.

 

[DKlein]

According to your explanation of the problem you need to know the number of molecules of Deuterium in the sample of 7.5Gal of water, whose hydrogen contains .0156% Deuterium by mass, assuming that this is the same as is present in the ocean. It MAY be a trick though, as tap water probably has regular hydrogen (1.008g I think), not the specially weighted ocean water. In that case take out all the .000156 and .999844s like so:
[(7.5gal->g)x(H)x(ratio of deuterium to hydrogen in regular hydrogen)]/{[(H)+2xO]xDxAvogadro's #} = molecules D

I hate trick questions...:|
Check to see if that .0156% thing works out first for the standard atomic weight, if they're nice it does. Otherwise you have to find that ratio in regular water.

 

[gsellis]

Water is water, but sea water has other stuff. Otherwise you are assuming that water with Deuterium has a different evaporation rate, which I doubt it does. Tap water and sea water should be the same for Deuterium.

 

[nlmodel]

hmm thanks for the replies but i tried all your suggestions and basically got the same answer each time but nope they were all wrong..see it is a wonder how anyone can pass these classes

 

[uart]

Originally posted by: nlmodel
hmm thanks for the replies but i tried all your suggestions and basically got the same answer each time but nope they were all wrong..see it is a wonder how anyone can pass these classes

Ok so what answer do you currently hold as correct and from what source did you obtain that answer. Please post the answer you currently have.


Also when you say "wrong", do you mean off by a few percent or are they orders of magnetude (multiples of ten) different. An approx solution is understood to be ok in this problem as neither temperature nor pressure were specified, so you have to use "typical" values anyway. I would accept answers to two or three significant figures.

If you take my answer of 7.5 * 3.785 * 4.7 * 10^21, it comes out to about 1.3 * 10^23 Nuclei.


Also note that there are multiple conficting definitions of "Gallon". A US Gallon is 3.785 Liters whereas a British Gallon is 4.55 Liters. If you are using British Gallons then the results would be greater, about 1.6 * 10^23 Nuclei. That's the reason I was reluctant to calculate it per gallon and instead left the gallon to liter conversion to you.

BTW, Scientists don't use units like gallons, pounds or inches, they use Standard International units like Liter, Kg and meter.

 

[uart]

Ok I did make a slightly unjustified approximation with the mass ratio. I took the 0.0156% ratio as the ratio of whole molecules instead of the mass ratio of hydrogen to deuterium as specifically asked. I figured this wouldn't make too much difference and it actually it makes about 11% error (as that's about how much the mass of the total water molecule changes with the switch from hydrogen to deuterium).

Re-Doing it properly it get the following :

- Approx 1000/18 * 6 * 10^23 = 3.34 * 10^25 water molecules per liter.

- Double this gives approx 6.69 * 10^25 total hydrogen nuclei.

- By mass 0.0156% are deuterium, so by number we have half this percentage = 0.000156/2 * 6.69*10^25 = 5.22 * 10^21 deuterium nuclei per liter.

- Finally times by 3.785 Liters per US Gallon and times 7.5 Gallons gives approx 1.5*10^23 Deuterium nuclei present.

BTW nlmodel. If you had the correct answer all along but just didn't know how it was calculated then you should have noticed my figure was only off by about 10% and posted this info. Then I could easily have figured out what accounted for that discrepancy.

 

[nlmodel]

uart i really appreciate you helping me..i see where you are coming from..the answer can only be .001 percent ...ive used your answers and they get me no where..i dont know why they would be using gallons either..seems like a trick question to me

 

[uart]

uart i really appreciate you helping me..i see where you are coming from..the answer can only be .001 percent ...ive used your answers and they get me no where..i dont know why they would be using gallons either..seems like a trick question to me

Ok, but do you currently have a numerical answer to this question against which you are comparing other answers ? If so then what is that value ?

 

[jai6638]

Originally posted by: uart
Now each Oxygen atom has an atomic mass of approx 16 and each Deuterium atom an approx mass of 2, so about 2/20 of the above 0.156 grams (/L) is Deuterium.

i was wondering... when u say 2 / 20, have u just rounded of the at. mass of oxygen from 16 to 20????

 

[uart]

Originally posted by: jai6638

Originally posted by: uart
Now each Oxygen atom has an atomic mass of approx 16 and each Deuterium atom an approx mass of 2, so about 2/20 of the above 0.156 grams (/L) is Deuterium.

i was wondering... when u say 2 / 20, have u just rounded of the at. mass of oxygen from 16 to 20????

No jai, that 20 is the molecular weight of the entire water molecule (one oxygen plus two deuterium). Anyway look at the new derivation, it's simpler and more accurate. Actually I should have just used 18 as the atomic weight (regular water) in that quoted line. That is actually where the 9/10 error occurred in that first derivation.

 

[uart]

Anyway, it looks like the whole thing is a trick question. Different isotopes often have both physical and chemical properties that are slightly different from each other. In particular I wouldn't mind betting that D2O (heavy water) would have a slightly higher boiling point and correspondingly lower vapor pressure as compared to ordinary water, due to the differences in molecular mass.

This would make the concentrations of heavy water in the atmospheric water vapor somewhat less than the concentration found in sea water. This effect might be slightly offset by heavy water being slightly more prone to condense into rain drops as compared to normal water, however the net effect is probably a lower concentration of deuterium in rain water than compared with sea water. Just how much less I'm not sure, but I suspect this "trick" is part of the problem.

 

[DrPizza]

Originally posted by: uart
Anyway, it looks like the whole thing is a trick question. Different isotopes often have both physical and chemical properties that are slightly different from each other. In particular I wouldn't mind betting that D2O (heavy water) would have a slightly higher boiling point and correspondingly lower vapor pressure as compared to ordinary water, due to the differences in molecular mass.

This would make the concentrations of heavy water in the atmospheric water vapor somewhat less than the concentration found in sea water. This effect might be slightly offset by heavy water being slightly more prone to condense into rain drops as compared to normal water, however the net effect is probably a lower concentration of deuterium in rain water than compared with sea water. Just how much less I'm not sure, but I suspect this "trick" is part of the problem.

Interesting point... thus when determining the amount of deuterium in "tap water", one should expect a lower percent than in the ocean.\

Incidentally, your bet would win... boiling point of Deuterium Oxide at .1013325MPa is 101.42 degrees C, according to my Handbook of Chemistry and Physics. Oddly, I've used that big thick book more times for posts on Anandtech, than I have for anything else this year. (2 to 0)

 

[DKlein]

Just realized I had it as HO2... remember to fix that when you calculate, that might be it