Molecular Weight of #2 Diesel??? 3

By Dalton's Law of partial pressure, and assuming you are around sea level, the concentration of vapor is 0.04*100/14.7=0.27 mol %. This is consistent with the flash point data which suggests that the vapour will not be in the flammable range until the tank temperature gets into the 125-150 deg F range.

HAZOP at

Thanks for the info. Any hints on how to convert mol % to % volume or % mass??

Mol%=vol%. For a ballpark estimate of mass %, one can assume the hydrocarbon vapor composed of a mix of C[sub]8[/sub] and C[sub]10[/sub] or thereabouts (MW~130).
A concentration of 0.27% mol would convert into:


I took 29 as the molecular mass of air.

For these low concentrations, a simplified formula, such as


would have been sufficiently accurate.

The US EPA lists "Distillate fuel oil #2" as having a liquid molecular weight of 188 and a vapor molecular weight of 130. This is from their "Tanks" program to estimate evaporation losses from above-ground storage tanks, freely available from

It gives me satisfaction to see that my estimate wasn't far off.

Thanks for the helpful info. I also posted a similar question to the "Chemical Engineering - Other" forum where it got some good comments, including a link posted by 25362 to a good Chevron tutorial on diesel fuel

OK, let me get this straight.
The equations seem straightforward enough, but as a guy who learned to do engineering with a slide rule, I always feel a bit better if the answer seems to make sense.

For: #2 Diesel fuel Air @ 60F
Partial Pressure: .0074 14.7 psi
MW 130 29
Vapor Density: 8x .0102 lb/ gal
Liquid Density: 7.1 lb/gal -
Tank Volume: 1000 gal

Molar Concentration = .0074 psi / 14.7 psi = .0503%

Mass Concentration ~= .0503%*130 / 100*29 = .00226%

For 1000 gal tank:
Air Mass = 1000 gal * .0102 lb/gal = 10.2 lbm

Diesel vapor mass = 10.2 lbm * (.00226% / 100) = .000231 lbm = 0.10 g

If this vapor were liquefied:
Diesel liquid mass = .000231 lb / 7.1 lb/gal = 3.25e-5 gal = 0.123 cc

Therefore, the most diesel fuel that I can evaporate into a 1000 gal tank of 60F air is 0.1 cc.
???
This doesn’t seem to make a lot of sense to a poor ME.
Does it make any sense to any of my learned colleagues??

You are absolutely right and I failed in simple arithmetics. In my previous posting I should have deleted the % symbol when converting from mol% to mass fraction (not %).

Thus your results, as estimated, should be multiplied by 100. Besides, the density of the lighter diesel fraction, when "liquified" would be ~ 6.3 lb/gal.

Thanks for the comment.
Duh, I should have caught that - Thanks for graciously refraining from pointing that out. As they say, I get by with a little help from my friends.

So, the mass of diesel evaporated in the tank is more correctly 0.0231 lbm, and to further beat this dead horse:

Diesel Vapor:
Density: ~= 8 x 0.0102 = .0816 lbm/gal
Volume = 0.0231 lbm / .0816 lb/gal = 0.238 gal
Concentration = 0.238 gal / 999.8 gal = 0.0238%

You should explain where is the -apparently incorrect- 8x density factor stemming from.

The ideal gas law PV=nRT stipulates that at given P,T conditions n, number of moles, is proportional to V. Using metric units when n=1 mol, the volume is 22.4 m[sup]3[/sup].

For air with a molecular mass of 29 gr, the density at STP would be 29/22.4=1.295 g/m[sup]3[/sup], and for the hydrocarbon vapor 130/22.4=5.804 g/m[sup]3[/sup].

Thus, if the mol mass of the hydrocarbon vapor is 130 vs air's 29, the ratio of densities would have to be as the ratio of mol masses, considering that all mols have the same volume at standard P,T conditions.

Thus the ratio is not 8, but 130/29=4.48.

The density would be 4.48*0.0102 lb/gal=0.0457 lb/gal
The volume: 0.0231/0.0457=0.505 gal
The concentration: 0.505*100/1000=0.05%, which equals the molar concentration you estimated yourself in your May 7 posting. As once said mol%=vol%. QED

Thanks, once again.
I got the 8x specific gravity of diesel vapor from an MSDS. Your explanation makes much more sense.

The air and vapor densities are, of course, in kg/m[sup]3[/sup], not as erroneously indicated.
A fact that doesn't change the conclusion, but most certainly makes the star granted me as totally undeserved.

Hey, if you're a star, your're a star.

What I conclude from this little discussion is that
for something approximating an ideal gas,
the % volume is equal to the ratio of partial pressures,
whether I know the molecular weight or not.

You are referring to what is generally recognized as Dalton's law (applicable at moderate pressures). This law states that in a mixture of gases the "partial" pressures of the components are those they exert when they each occupy the container alone. The sum of these "partial" pressures gives the total pressure.

P[sub]A[/sub] + P[sub]B[/sub] + ... = P[sub]total[/sub] = (RT/V) (n[sub]A[/sub] + n[sub]B[/sub] + ...)

where

P[sub]A[/sub] = n[sub]A[/sub](RT/V); P[sub]B[/sub] = n[sub]B[/sub](RT/V) and so forth.

Rephrasing your statement: the ratio of partial pressures is equal to the ratio of molar contents. Or one can say that "partial" pressures and moles of a particular component are proportional; the proportionality factor being RT/V.

One can also say that in a mixture of (unreacting) gases, the ratio of the "partial" pressure of a particular component to the system's total pressure equals the mol fraction of that particular component:


It's always a pleasure to be helpful.