conversion of mole % to weight %

[Peter Apps]

You can do all the calculations you want and they are simple. I just don't understand the logic of calculating weight % for gases. The question is confusing.

Not half as confusing as some of the answers

The reason for using weight is that accurate gas mixtures are often made up by filling given weights of gas into cylinders (using fancy comparator balances with an empty cylinder on the other side). This is more accurate than measuring both the internal volume of the cylinder, and the high pressure to which they are filled, and gets around the problems that pressurizing the gas to get it into the cylinder heats it up and increases the pressure, and that at cylinder pressure nothing is ideal any more. If I recall correctly the top end filling machines can put measured milligrams of gas into cylinders that weight tens of kilograms.

Peter

[Axar]

So substituting we go from: "100% - 10% - 5% = 85 %. Look up the density of argon; volume x density x % = weight needed."

To:
100% - 10% - 5% = 85 %. Look up the molecular weight of argon; volume / 22.4l x molecular weight x % = weight needed.

Kind of simialr, no ?

Peter

With the second method of calculations we do not need density. Molecular weight is easier to find becouse I don't know chemists without periodic table at hand accessibility.

Well, at nowadays almost every chemist has access to internet were he can find almost everything. So, from this point of view they are similar indeed.

[Peter Apps]

And if you know the MW from your periodic table, divide it by 22.4 L and you have the density, so all the calculations are exactly equivalent.

Peter

[tom jupille]

The OP explicitly asked how to go from mole % to weight %. The correct answer is to multiply the mole %s by the corresponding molecular weights and renormalize to 100%. Volume, density, and pressure don't enter into it.

[Peter Apps]

That's true Tom, but the goalposts have moved from weight % to weight (my highlight):

thanks to all replys

let me change the question , what about if the customer need the mixture in argon balance(i.e. solvent) instead of 90 % at 130 bar, so , how could i calculate the required weight of argon to reach that pressure with 10 mole % methane and 5mole% ehane in 50 L. volume

Regard
Amr

Peter