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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Tessy on August 09, 2005, 09:30:09 PM

Title: [Q'a] Experiment of Ideal gases
Post by: Tessy on August 09, 2005, 09:30:09 PM

Hello viewers! Since I'm doing my first lab experiment I have several variables that I know but some others that I wasn`t capable of answering myself, so if someone can give me a clue on how to do my calculations I will aprecciate that.

The problem that I have is the following:

I have to mix He and Ne in an erlenmeyer of 356cm³ and it weights 0.1480g and this experiment is carried at 20ºC and 748 torr, how can I calculate the mass and the molar fraction of He?

Title: Re:[Q'a] Experiment of Ideal gases
Post by: Donaldson Tan on August 09, 2005, 09:58:17 PM

assuming ideal gas behavior,

n = PV/RT

the total number of moles (n) of He-Ne can be calculated using the above equation.

u know the mass of n moles of the He-Ne mixture. let this mass be m, ie. m = 0.1480g

let mole fraction of He in He-Ne mixture be X

M_He: Molar Mass of Helium
M_Ne: Molar Mass of Neon

X.n.M_He + (1-X).n.M_Ne = m
X.M_He + (1-X).M_Ne = m/n
X.M_He + M_Ne - X.M_Ne = m/n
X.(M_He - M_Ne)  = m/n - M_Ne

solve for X

Title: Re:[Q'a] Experiment of Ideal gases
Post by: Tessy on August 09, 2005, 11:17:22 PM

Well I wanna thank you for this...I was a little confused in how to use the X in the equation, but now it's done! ;D

Title: Re:[Q'a] Experiment of Ideal gases
Post by: Donaldson Tan on August 10, 2005, 12:00:34 PM

the gist of the method is that the molar mass of the mixture is the summation of the products of the molar mass of each component and its corresponding molar fraction, ie.

m/n = X.M_He + (1-X).M_Ne