# How do we know that hydrogen gas consists of two hydrogen atoms?

Of course I know that both 1s orbitals will combine and that the electrons will occupy the bonding MO and that the two atoms will reach a lower energy state – but how do we really know that is the case?

• Well, we can measure the molar mas of hydrogen gas and it will be close to 2 g/mol which means that it indeed consist of $\ce{H2}$ molecules rather than $\ce{H}$ atoms. – Wildcat Sep 19 '14 at 12:42
• @Wildcat You mistakenly put your answer into the comment box and not the answer box further below. :P – LordStryker Sep 19 '14 at 13:30
• @LordStryker, yes, but OP mentioned orbitals and labeled equation with "quantum-chemistry" tag, so maybe his question actually is how do we now that $\ce{H}$ atomic orbitals do combine to form $\ce{H2}$ molecular orbitals? The answer would be different then. :D – Wildcat Sep 19 '14 at 14:07
• @Wildcat It would seem as though a better title to the question is merited then. – LordStryker Sep 19 '14 at 14:27

Assuming ideal gas behavior we can determine the molar mass $M$ of the hydrogen gas from its density $\rho$ (which can be measured) by the following formula $$M = {{RT\rho}\over{p}}\ \, .$$ At STP $\rho \approx 0.09 \, \mathrm{g} \, \mathrm{L}^{−1}$ so the following back-of-the-envelope calculation $$M = {{8.31 \, \mathrm{L} \, \mathrm{kPa} \, \mathrm{K}^{−1} \, \mathrm{mol}^{−1} \cdot 273.15 \, \mathrm{K} \cdot 0.09 \, \mathrm{g} \, \mathrm{L}^{-1}}\over{101.325 \, \mathrm{kPa}}}\ = 2.02 \, \mathrm{g} \, \mathrm{mol}^{−1} \, .$$

shows that $M$ is indeed close to 2 g/mol which means that hydrogen gas consist of $\ce{H2}$ molecules rather than $\ce{H}$ atoms

• I think we should try to answer this question outside the context of our current consensus reality of chemistry. There was a time when water was written HO and oxygen was thought to have atomic weight of 8. Experimentally, how did early chemists first discover that H is actually H2? – iad22agp Sep 19 '14 at 14:22
• Hmmm... But why should we answer the OP's question outside the context of our current consensus reality? The question you asked is an interesting one without a doubt, but its a different one. OP knows the electronic structure of $\ce{H}$ atom, so he surely knows its atomic mass. – Wildcat Sep 19 '14 at 14:49

I think you're asking a historical question too. This largely comes from a similar question of how we get the modern periodic table and atomic masses. The Chemical Heritage Foundation has a nice outline.

In my opinion, the big breakthrough here was due to John Dalton (the Wikipedia page is also good).

Dalton really pushed the atomic theory to molecular formulas. At the time, they knew from various analyses the various compositions - there is 8 times the mass of oxygen in water than hydrogen. So Dalton reasoned if everything is made of atoms, they must combine to form different compounds (molecules) in whole number ratios. For example, if an atom can't be divided, we can't have $H_{1.5}O$ -- that violates the idea of indivisible atoms.

So he guessed and reasoned through the different compounds and compositions known and published a table of atomic masses. As mentioned above, he was wrong about some things, assuming $\ce{HO}$ instead of $\ce{H2O}$, etc. But you can make some fairly good reasoning of the law of multiple proportions to determine stoichiometry.

Avogadro was the next step. He reasoned that equal volumes of gases should have the same number of molecules. So if you weigh equal volumes, you can get the (molar) mass. Unfortunately, his work was largely ignored. (Interesting note, Loschmidt is actually the person who worked out $N_A$, not Avogadro.)

Other people, notably Cannizzaro helped to clarify and expand Avogadro's work at the Karlsruhe Congress.

Avogadro believed that most gases were binary, i.e., $\ce{H2, O2, N2}$ not $\ce{H, O, N}$ as assumed by Dalton.

So if you carefully measure the reaction of hydrogen gas and oxygen gas to get water, you realize it's $\ce{H2O}$ based on the proportions, and then the atomic masses fall into place.

We get further proof of $\ce{H2}$ from other compounds. Sodium bicarbonate clearly has a small amount of hydrogen and you can work out $\ce{NaHCO3}$.

Put simply, this all required many people and a lot of careful measurements and deduction. But the law of multiple proportions and Avogadro's theory will eventually give you $\ce{H2}$.

• – Geoff Hutchison Sep 19 '14 at 14:59
• Nice answer, I like the historical perspective. One thing I'd add, in addition to burning hydrogen and oxygen to produce water, the reaction of hydrogen and chlorine to produce HCl was also needed (two equations, two unknowns if you will) to ascertain that the hydrogen molecule was really a diatomic. For example, see this link – ron Sep 19 '14 at 16:04
• @ron Right, thanks for the reminder. It's been ages since I taught general chemistry, so I forgot the HCl part. – Geoff Hutchison Sep 19 '14 at 16:20