Chromatography Forum

how could i converte mole % to weight % for gas mixture using ideal gas law

example
gas mixture ( methane 10 mole % and argon 90 mole % )

pressure 2000 psi

volume 50 litre

if i want prepare this mixture using Gravimetric method what is the wight percent for each commponent in kg

Hi!
Sorry, but i can't understand why should we use ideal gas low?
We can calculate weight % without using volume %.
example:
Mathane's molar mass is 16 g/mol
Argon molar mass is 40 g/mol
Calculation:
weight % for methane = (10%*16 g/mol)/((10%*16g/mol)+(90%*40g/mol))
weight % for argon = (90%*40g/mol)/((10%*16g/mol)+(90%*40g/mol))
if I don't miss something in my calculations...

Ideal gas low is used for volume's calculation.

When working with gas standards, you either use units in v/v (ppbv, ppmv, %....) or units w/v (ug/m3, ug/L, mg/m3....).

You can convert each to weight by multiplying the mol% by the mol. weight for each component. Then re-normalize to 100% and you'll have the weight percent.

There's probably some way to get there with the gas law, but I do it with the mol weight.

The gas law has nothing to do with that, Axar gave the answer, why foul it up?

The gas law has nothing to do with that, Axar gave the answer, why foul it up?

Hello
Just has. In this case, and argon and methane behave as ideal gases (not liquefied). Accordingly, their volume concentration is constant. Accordingly, they can be counted by the ratio of molecular weights, if necessary - then we can give them volume and temperature to standard conditions.

DSP007, ???

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

100% - 10% - 5% = 85 %. Look up the density of argon; volume x density x % = weight needed.

Or is there some complication that I am overlooking ?

Peter

I don´t follow this last question. You have a given 85 mol% argon, but want to use argon to adjust a certain pressure? Volume and temp. stay constant? Magic.

100% - 10% - 5% = 85 %. Look up the density of argon; volume x density x % = weight needed.

Or is there some complication that I am overlooking ?

Peter

Also you can calculate it with ideal gas low with the way that DSP suggested.
I'll try to rephrase his post:
According to ideal gas low, mole% and volume% are same (at any pressure). Argon is almost ideal gas, so this way is accurate enough.

Hi Axar

I am completely at a loss to understand how what you are suggesting differs from what I presented. The ideal gas law, crudely paraphrased, says that volume fraction = mole fraction. Molecular weight is constant. Once the gas is mixed and in a container, compressing it and heating or cooling it makes no difference to the mass fraction, mole fraction or volume fraction composition of what is in there. Even if some of it turns to liquid under pressure (which might be what is worrying the OP) as soon as you open the tap it will turn back to gas, with the same composition as you started with.

Peter

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.

Hi Axar

I am completely at a loss to understand how what you are suggesting differs from what I presented.

Peter

Hi Peter!
You used density in your calculations, and DSP don't. Your way is more accurate, but needs more information.

Hi Axar

I am completely at a loss to understand how what you are suggesting differs from what I presented.

Peter

Hi Peter!
You used density in your calculations, and DSP don't. Your way is more accurate, but needs more information.

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