What is the weight of 1 litre of Helium at STP (or density)? And 1 litre of Hydrogen? And 1 litre of Air?

To being able to calculate how much weight a given balloon can lift  (the lifting power) based on the volume of the balloon,  it is first necessary to know what is the weight of the gas inside a balloon with a given volume. Or more technically speaking, what is the density of the gas. This post aims to answer to this last question, by considering the gas to be Helium, Hydrogen and Air, and by considering the balloon as being subject to Standard conditions for temperature and pressure (STP).

We shall now apply what we have seen in my previous post to calculate the weight of X litres of Gas.

1. Determine the volume (in litres) occupied by 1 mole of the Gas;
2. With the result in 1. determine to how much moles correspond X litres of Gas;
3. Using the molar mass of the Gas ,which is expressed in g/mol , and the result in 2, get the weight in grams.

Realizing what we have just said, lets determine the weight of 1 litre of Helium:

Step 1. Determine the volume (in litres) occupied by 1 mole of the Helium:

At STP we have:

T (Temperature)= 273.15 K
P (Absolute pressure)= 0.986 atm

Additionally we have the ideal gas constant R= 0.08205746(14) L atm K^-1 mol^-1 which we can approximate to R= 0.082 L atm K^-1 mol^-1

And n= 1 since we want to determine the volume of 1 mole of helium.

Now, by using the ideal gas law we just have to replace the variables by their correspondent values:

PV = nRT
V= (nRT) / P
V= (1 x 0.082 x 273.15) / 0.986
V= 22.398 / 0.986 = 22.7 litres

Notice that above in the ideal gas law we have user as units for n, T and P, the same units in which the R constant is expressed: mol, K, and atm respectively.

So we have just determined that 1 mole of Helium occupies 22.7 litres.

Step 2. With the result in 1. determine to how much moles correspond 1 litres of Helium:

Now we have:

1 mol of Helium ____(corresponds)_____ 22.7 litres
X mol of Helium____(corresponds)_____ 1 litres

And by applying the rule of three, which is a particular form of cross-multiplication:

X = (1 * 1) / 22.7 = 0.044 mol

So we have just determined that 1 litre of Helium contains 0.044 moles.

Step 3. Using the molar mass of Helium ,which is expressed in g/mol, and the result in 2, get the weight in grams:

And now we have reached to the last step. As seen in a previous post, the molar mass (molecular weight) of Helium is equal to its atomic weight (since Helium has only 1 atom). In this way, and knowing that the atomic weight of Helium A_r(Helium)= 4.003 g/mol, it is easy to calculate the molar mass of Helium, M(Helium):

M(Helium)= A_r(Helium)= 4.003 g/mol

In this way, we can now use once again the rule of three:

4.003 g         ____(corresponds)_____ 1 mol
X g of Helium____(corresponds)_____ 0.044 mol

X = (0.044 * 4.003) / 1 = 0.176g

Which means that the weight of 1 litre of Helium is 0.176g, which is equivalent to:
Helium density (ρ) = 0.176g/l = 0.176 Kg/m³

If we wanted to determine the weight of 1 litre of hydrogen instead of Helium, we would just have to replace the molar mass of Helium by the molar mass of Hydrogen - please be aware that hydrogen in diatomic (H2). See bellow the calculations:

Step 1:

It is the exact same calculations that were made above for Helium, meaning that:

1 mole of Hydrogen occupies 22.7 litres.

Step 2:

It is the exact same calculations that were made above for Helium, meaning that:

1 litre of Hydrogen contains 0.044 moles.

Step 3:

The molar mass (molecular weight) of Hydrogen is equal to twice its atomic weight (since  Hydrogen  has 2 atoms - diatomic). In this way, and knowing that the atomic weight of Helium A_r(Helium)= 1.008 g/mol, it is easy to calculate the molar mass of  Hydrogen , M( Hydrogen ):

M( Hydrogen )= A_r( Hydrogen ) * 2= 2.016 g/mol

In this way, we can now use once the rule of three:

2.016 g                ____(corresponds)_____ 1 mol
X g of  Hydrogen ____(corresponds)_____ 0.044 mol

X = (0.044 * 2.016) / 1 = 0.089 g

Which means that the weight of 1 litre of Hydrogen is 0.089g, which is almost half the weight of Helium, and which is equivalent to:
Hydrogen density (ρ) = 0.089g/l = 0.089 Kg/m³

If we wanted to determine the weight of 1 litre of Air, we would just have to use the molar mass of Air. See bellow the calculations:

Step 1:

It is the exact same calculations that were made before, meaning that:

1 mole of Air occupies 22.7 litres.

Step 2:

It is the exact same calculations that were made before, meaning that:

1 litre of Air contains 0.044 moles.

Step 3:

The determination of the molar mass (molecular weight) of Air is complicated  because the chemical composition of air varies. The most notable variation is related with the presence of water vapour (H₂O), which is the gas phase of water. Following what is stated in Wikipedia, the water vapour represents 0.40% of the atmosphere and at the Earth's surface may represent 1% to 4%. Here, you can read a very interesting article on the air compositions and properties, including information regarding the water vapour and air density. But since the water vapour in the Air varies, let's simplify things and do the calculation for dry Air.
And as stated in the IUPAC Gold Book, "the composition of the major components in dry air is relatively constant (percent by volume given): nitrogen, 78.084; oxygen, 20.946; argon, 0.934; carbon dioxide, 0.033; neon, 0.0018; helium, 0.000524; methane, 0.00016; krypton, 0.000114; hydrogen 0.00005; nitrous oxide, 0.00003; xenon, 0.0000087". See bellow the table that presents and overview of these values and after the calculations to determine the molar mass of dry air.

Gas	Chemical symbol	% by volume	Atomic Weight	Molecular Weight
Nitrogen	N2	78.084	N: 14.007	28.014
Oxygen	O2	20.946	O: 15.999	31.998
Argon	Ar	0.934	Ar: 39.95	39.95
Carbon dioxide	CO2	0.033	C: 12.011 O: 15.999	63.996
Neon	Ne	0.0018	Ne: 20.18	20.18
Helium	He	0.000524	He: 4.003	4.003
Methane	CH4	0.00016	C: 12.011 H: 1.008	16.043
Krypton	Kr	0.000114	Kr: 83.80	83.80
Hydrogen	H2	0.00005	H: 1.008	2.016
Nitrus Oxide	N2O	0.00003	N: 14.007 O: 15.999	44.013
Xenon	Xe	0.0000087	Xe: 131.3	131.3

The atomic weights of the table above were retrieved from table 4 of the IUPAC Recommended Atomic Weights of 2009, where the atomic weights are abridged to four significant digits, and which apply to elements of natural terrestrial origin (as the gases in the air do). The values that are provided in an interval, like hydrogen, were retrieved from table 6 instead.

So now to determine the molecular weight of Air it is just a question of performing a weighted multiplication as shown bellow:

M(Air)= 0.78084 * 28.014 + 0.20946 * 31.998 + 0.00934 * 39.95 + 0.00033 * 63.996 + 0.000018 * 20.18 + 0.00000524 * 4.003 + 0.0000016 * 16.043 + 0.00000114 * 83.80 + 0.0000005 * 2.016 + 0.0000003 * 44.013 + 0.000000087 * 131.3 = 28,9715355715 = 28.972 g/mol

In this way, we can now use once the rule of three:

28.972 g             ____(corresponds)_____ 1 mol
X g of  Air           ____(corresponds)_____ 0.044 mol

X = (0.044 * 28.9715) / 1 = 1.275 g

Which means that the weight of 1 litre of Air is 1.275g, which is equivalent to:
Air density (ρ) = 1.275g/l = 1.275 Kg/m³


Quicker way to reach a value almost as accurate as the above of 1.275g:


If for the determination of the molecular weight of Air we used only the three elements most existing in Air, Nitrogen, Oxygen, and Argon, the final result would be very similar: 

M(Air)= 0.78084 * 28.014 + 0.20946 * 31.998 + 0.00934 * 39.95 = 28.950 g/mol

In this way, we can now use once again the rule of three:

28.950 g              ____(corresponds)_____ 1 mol
X g of  Air           ____(corresponds)_____ 0.044 mol

X = (0.044 * 28.950) / 1 = 1.274 g

A difference of just 0.001g

Of course, and before concluding, there is an even easier way to know the densities of helium and hydrogen - by consulting the many existing tables and charts, like this one in Wikipedia. For air, less sources exist.

To provide with a quick overview, the table bellow shows some of the values we have just determined for Helium, Hydrogen and Air.

Gas / Mixture	Molecular Weight / Molar Mass (g/mol)	Weight of 1 litre (g) / Density(ρ)
Helium	4.003	0.176
Hydrogen	2.016	0,089
Air	28,972	1,275

Please notice that for all the above calculations we have the following approximation:

- We are considering standard conditions for temperature and pressure (STP).