Thought experiment

Given two glasses of water, how would one detect which glass contains heavy water, and which contains potable water without using complicated laboratory equipment?

Something like the way a piece of litmus paper (yes, it is laboratory equipment and easily available to the layman at that!) may be used to test whether a solution is an acid, or alkali.

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    $\begingroup$ heavy water has higher boiling and melting points and is slightly toxic for eukaryotes. It is also heavier (1.1 g/ml) then normal water. $\endgroup$ – permeakra Oct 21 '12 at 5:43
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    $\begingroup$ This is just a hunch, but I imagine if you filled a small waterbomb from each glass, and placed it in the opposite glass, you would see one sinking (in the normal water) and one floating (in the heavy water) $\endgroup$ – wim Oct 21 '12 at 11:05
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    $\begingroup$ Not sure how big your glass is, but would a 7 dollar hydrometer be "complicated laboratory equipment"? A small volumetric flask (10 ml) and a balance would be less than $50. $\endgroup$ – MaxW Oct 27 '15 at 3:58
  • $\begingroup$ Why would you oppose heavy water and potable water, to begin with? $\endgroup$ – Ivan Neretin Jun 3 '18 at 18:17

There's a very simple test for $\ce{D_{2}O}$ that springs to mind - ice cubes made with heavy water sink in light water. I assume you're talking about differentiating a glass of essentially pure heavy water from, say, tap water, in which case this test should work rather well and requires equipment no more sophisticated than a freezer, ice cube tray and some glasses.

  • $\begingroup$ Cool (+: But could there be something still simpler - akin to the boiled egg floats in salt-water? As an extreme example, a person out trekking may not have a freezer ^ power-source handy. $\endgroup$ – Everyone Oct 21 '12 at 4:52
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    $\begingroup$ @Everyone - You could make a custom float that floats on heavy water but sinks in light water. Quantitative floats similar to this are used, for instance, to estimate alcohol concentrations in distillates from hobby stills. $\endgroup$ – Richard Terrett Oct 21 '12 at 5:43
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    $\begingroup$ There is a much simpler test. Just weigh each one, or even lift up each one and feel which is heavier. $\endgroup$ – Canageek Oct 21 '12 at 23:15

Mass. Heavy water is noticeably heavier then light water. Simply weigh an equal amount of each water and see which weights more.

Light water weights 1 g/mL (0.99700 g/mL at 25 °C)

Heavy water weights 1.107 g/mL at 25 °C

So if you have 100 mL you will see an 11 g difference, assuming pure samples.

(The Science and Technology museum in Ottawa actually has samples you can lift to feel this).


If you don't have access to a freezer, you could allow sun light to pass through your glass vessels and then through a prism. The heavy water will absorb less orange red light and the difference might be observable.

enter image description here

One of these might help (not complicated).

P.S. I did not want to post this as an answer because I'm not sure if it's realistic but I can not see were to add a comment.

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    $\begingroup$ You can't add comments unless you have earned enough reputation. Besides that, I think this qualifies as an answer, although you will need a considerable layer thickness to see the differences. $\endgroup$ – Antimon Oct 21 '12 at 20:53
  • $\begingroup$ @Antimon Thanks for the clarification Sb. I was uncertain if the naked eye would be able to tell the difference between H2O and D2O. Do you think this might work better if one substituted the sun with an artificial light source that emitted at a wavelength that was absorbed, Like a 660nm LED? $\endgroup$ – Cargo Oct 21 '12 at 21:19
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    $\begingroup$ The problem lies in the Beer-Lambert law and the low extinction coefficient of water, not in the light source. As you can see in the spectra you posted, there is no significant absorption or characteristic bands in the visible region. $\endgroup$ – Antimon Oct 21 '12 at 21:50
  • $\begingroup$ Pity one could not make use of that bump at around 950 nm, like with a long darkened tube with a InGaAs photodiode at one end and a GaAs IR LED at the other. $\endgroup$ – Cargo Oct 22 '12 at 0:11
  • $\begingroup$ Why couldn't you? $\endgroup$ – Antimon Oct 22 '12 at 8:40

(Completely untested since, sadly, I have no access to heavy water.) Slowly pour the contents of one glass into the other. The index of refraction of heavy water is 1.328, that of light water is 1.333. Not much, but maybe just enough to see the inhomogeneities. If the decanted glass's contents sink, it had the heavy water. Otherwise, light water.

But if you have even a cheap kitchen scale, that would probably be the better test, since the densities differ by about 10%, as opposed to the index of refraction, which differs by far less.

Or, just add a drop of food colouring to one of the glasses. Or a drop of milk. Just to help out with the contrast.

Do the test with just a small sample from each glass if you want to avoid contaminating the expensive heavy water.


If you have two glasses—knowing that one contains deuterated water, the other contains potable water; one might determine the contents by freezing both samples in a freezer at $\pu{0 ^\circ C}$. Since the melting point of the potable water is $\pu{0 ^\circ C}$ and the melting point of the deuterated water, $\ce{D2O}$, is $\pu{3.8 ^\circ C}$, the $\ce{D2O}$ will take longer to melt assuming they are side-by-side.

I would add, having done this, practically speaking: this means you can hold the glass of frozen $\ce{D2O}$ in one hand and $\ce{H2O}$ in the other and while the regular water melts, the frozen $\ce{D2O}$ takes noticeably longer to melt. And on a side note: imagine some of the toxicity in Eukaryotes is due to the subtle differences in thermal properties between these two water species.

Be aware: $\ce{D2O}$ must be in an enclosed container, otherwise it will exchange heavy hydrogen in the water for atmospheric light hydrogen in atmospheric moisture. If left uncovered, the experiment may work a couple times at first; however, it will become less effective (as the deuterium exchanges and evaporates).


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