I know that water is 2 parts hydrogen and 1 part oxygen. Hydrogen and oxygen are quite light gasses, they don't weigh very much compared to water. Water is very heavy and it made me wonder exactly how much oxygen I'd need to make a cup of water.

Please answer in terms of cubic meters of air, assuming the air is 21 % oxygen, and that my small cup contains only 100 ml. I suppose the exact question is: How many cubic meters of air contains the same amount of oxygen as found in 100 ml of water?

  • 1
    $\begingroup$ 2/3 hydrogen, 1/3 oxygen. $\endgroup$
    – Freddy
    Sep 3 '14 at 17:10
  • 1
    $\begingroup$ @Freddy - The first sentence of my question explains that I already know that. I actually wanted to know how much that is. $\endgroup$
    – Gerve
    Sep 3 '14 at 17:25

Since the molecular weight of water is $18\ \mathrm{g/mol}$, $100\ \mathrm{ml}$ (density = $1\ \mathrm{g/ml}$, so $100\ \mathrm{ml} \overset{\wedge}{=} 100\ \mathrm{g}$ of water) of water contains $$\frac{100}{18} = 5.56$$ moles of water, or 5.56 moles of atomic oxygen ($\ce{O}$), or 5.56/2 = 2.78 moles of molecular oxygen ($\ce{O2}$).

At standard temperature and pressure, 1 mole of a gas occupies 22.4 liters. Therefore, 2.78 moles of molecular oxygen will occupy 62.22 liters.

If air is 21% molecular oxygen, then we would need 62.22/0.21, or 296.3 liters of air to provide that much molecular oxygen.

Since 1 cubic meter equals 1000 liters, this volume converts to 0.3 cubic meters of air.

Therefore, 0.3 cubic meters of air contains the same amount of oxygen as is present in 100 ml of water.

  • 1
    $\begingroup$ Please note that the value $V_\mathrm m=22.4\ \mathrm{l/mol}$ for the molar volume of an ideal gas corresponds to the old definition of standard temperature and pressure (STP), which is a temperature of $T=273.15\ \mathrm K$ and a pressure of $1\ \mathrm{atm}$. Since 1982, $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ is used as the standard pressure for tabulating thermodynamic data. At this pressure, the molar volume of an ideal gas actually is $V_\mathrm m=22.710\,947(13)\ \mathrm{l/mol}$. $\endgroup$
    – user7951
    Nov 9 '15 at 12:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.