# Homework Help: Finding Mass Through Percent by Volume

1. Sep 18, 2014

### JeweliaHeart

A gas analysis on a gaseous mixture gives 60% methane and 40% ethylene by volume. You need to store 12.3 kg of the mixture in a cylinder of volume 0.0514 m^3 at a maximum temperature of 45°C. Determine the pressure (kPa) inside the cylinder by:

a. assuming that the mixture obeys ideal gas law

b. using the compressibility factor method

I approached part a of this problem by taking both 60 and 40 percent of 12.3 kg to find the mass of the methane and ethylene, respectively in the ideal gas mixture. A friend has informed me that this is not the proper approach because 60% and 40% are percentages by volume, not mass fractions, and therefore he recommended that I should instead take 60% and 40% of the molar masses of methane and ethylene instead.

I don't understand what percent volume has to do with molar mass. Why would I multiply the percentage by volume times the respective molar masses if I want to find the number of moles methane and ethylene in the compound?

2. Sep 19, 2014

### Staff: Mentor

I am not sure I follow your friend advice (I am not saying it is wrong).

Assuming ideal gas behavior volume ratio and molar ratio are identical. That means you have

$$\frac{n_{methane}}{n_{ethane}}=\frac{6}{4}$$

$$M_{methane}\times{n_{methane}}+M_{ethane}\times{n_{ethane}}=12.3~kg$$

Can you go from there?

3. Sep 19, 2014

### JeweliaHeart

Yes. That makes sense above. I am just trying to figure out how what you are saying translates over to multiplying the molar masses of methane and ethylene by the volume percentages, which is what my friend was doing.

I tried solving part a using what you did above and what my friend did and got nearly identical answers using both methods. However, his method is still not as intuitive to me.

4. Sep 19, 2014

### JeweliaHeart

My friend advises I do it his way:

0.6(16.04 kg/kmol of Methane) +0.4(28.04 kg/kmol of Ethylene)

to get the average molecular weight of the combined gas mixture.

After that, he uses 12.3 kg to find the number of moles of gas in the mixture and uses pv=nrt to find pressure.

Last edited: Sep 19, 2014
5. Sep 19, 2014

### Staff: Mentor

Using weighted average molar mass should yield the same answer (it is like calculating number of moles of air using molar mass of 29).

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