When looking at a Chemical, what does this mean 99.9 atom % D

[nukeman]

When looking at a Chemical, what does this mean... "99.9 atom % D"

Hey,

can someone explain to me what this means.

Lets use D2o for example (Deuterium oxide)

1 store says "99.999 atom % D"

And the other store simply says "DEUTERIUM OXIDE 100% D,99.96% "

Can someone please shed some light on this for me? I am curious?

Is it its purity? So, if one says (99.99%) and the other says (99.96%), what are they talking about?

Thanks very much.

 

[Ygggdrasil]

These values are talking about D vs H atoms in the sample. 99.999 atom % D D2O has one hydrogen-1 atom for every 100,000 deuterium atoms whereas the 99.96% has one hydrogen-1 atom for every 2500 deuterium atoms.

 

[nukeman]

Hi there, thanks very much for the reply!

Can you explain how you got this?
"99.999 atom % D D2O has one hydrogen-1 atom for every 100,000 deuterium atoms whereas the 99.96% has one hydrogen-1 atom for every 2500 deuterium atoms."

so really its the concentration of Deuterium in the sample, in compared to Hydrogen?

So, if someone was to use a sample that has D of 99.999% to test an effect of deuterium on something, and then used the sample that has D of 99.96% - would there be a difference?

Is the structure of the D2o molecule different at all?

 

[Ygggdrasil]

Deuterium is twice as heavy as a proton and this has important consequences for its properties. This change in mass affects the vibrational spectra of compounds containing deuterium as well as the rate of reactions involving transfer of the hydrogen/deuterium atom thorough the kinetic isotope effect. The deuterium nucleus also has very different properties which affects its behavior in nuclear magnetic resonance spectroscopy measurements.

As to what purity your experiments require to see a difference between H and D, it depends a lot on the experiment. Experiments looking at kinetic isotope effects probably aren't going to require extremely pure D2O whereas techniques that are particularly sensitive to the presence of protons (such as NMR experiments) will require very pure D2O.

 

[nukeman]

Deuterium is twice as heavy as a proton and this has important consequences for its properties. This change in mass affects the vibrational spectra of compounds containing deuterium as well as the rate of reactions involving transfer of the hydrogen/deuterium atom thorough the kinetic isotope effect. The deuterium nucleus also has very different properties which affects its behavior in nuclear magnetic resonance spectroscopy measurements.

As to what purity your experiments require to see a difference between H and D, it depends a lot on the experiment. Experiments looking at kinetic isotope effects probably aren't going to require extremely pure D2O whereas techniques that are particularly sensitive to the presence of protons (such as NMR experiments) will require very pure D2O.

Ahh ok, that makes sense...Thanks for that!

So essentially its just the concentration of Deuterium in the sample, in compared to Hydrogen?

Can you possibly elaborate on what you said:
"99.999 atom % D D2O has one hydrogen-1 atom for every 100,000 deuterium atoms whereas the 99.96% has one hydrogen-1 atom for every 2500 deuterium atoms."

 

[Ygggdrasil]

Yes, you are correct, the value is just the concentration of deuterium in the sample compared to hydrogen. In a hypothetical sample of 99.999 atom % D D2O containing 50,000 molecules, there would be 99,999 deuterons and 1 proton.

 

[nukeman]

Yes, you are correct, the value is just the concentration of deuterium in the sample compared to hydrogen. In a hypothetical sample of 99.999 atom % D D2O containing 50,000 molecules, there would be 99,999 deuterons and 1 proton.

Great, appreciate the speedy replies, thanks!

 

[Ygggdrasil]

you say:
"This change in mass affects the vibrational spectra of compounds containing deuterium as well as the rate of reactions involving transfer of the hydrogen/deuterium atom thorough the kinetic isotope effect."

So are you saying the vibrational spectra will be different in a sample that has D=99.99% compared to a sample that has D=99.96 ?

If so, are you able to explain that in any detail?

I'm not an expert in vibrational spectroscopy, but I doubt one would see much difference between the two samples.

 

[pa5tabear]

I'm not an expert in vibrational spectroscopy, but I doubt one would see much difference between the two samples.

I'm just learning about NMR... I'm thinking the difference between the two would be a larger chemical shift in the deuterium frequency band? Does that sound right?

 

[Yanick]

Deuterium is an NMR silent nucleus (2 fermions yielding integer spin) so you would not see anything there.

Magnetic spectroscopies do not rely on weight of the nuclei, just the nuclear spin. For example carbon NMR detects the C-13 isotope, so you either enrich your sample with carbon 13 or you acquire the spectrum for a long time with the natural abundance. This is not the case with vibrational spectroscopies where the weight of the isotope can change things.

 

[Mike H]

FYI - deuterium is a quadrupolar (spin-1) nucleus which shows up at around 15% of the proton frequency (e.g., at 9.4 T, the proton Larmor frequency is ~ 400 MHz, while the deuterium frequency is ~ 60 MHz). It's a fairly popular technique in certain areas (paramagnetic systems & lipid bilayers are the two that come to my mind, at least), although most people only ever encounter it in the context of a field frequency lock. Its typical chemical shift range is comparable to proton NMR.