A mixture of helium and neon gases is collected over water at $28.0 \rm~^\circ C$ and $745\rm~ mmHg$. If the partial pressure of helium is $368\rm~ mmHg$, what is the partial pressure of neon? (Vapor pressure of water at $28\rm~^\circ C =28.3~mmHg$). Calculate the weight percentage of He and Ne in the original mixture.

The question is based on Dalton's law of partial pressure. I got the first part right.In the second part, since we have to deal only with the original mixture, total pressure should be $$745-28.7=\pu{716.7 mmHg}$$ So we calculate the total moles in terms of "v" (total volume) and then moles of neon and helium in terms of "v" from which we calculate the weight of neon and helium and calculate the weight percentage. "v" will eventually cancel out and we get weight percentage of neon as $34.54\%$ and that of helium as $65.45\%$ but that is not the right answer.

In the solution to this question, they took the total pressure while dealing with the second part as $\pu{748 mmHg}$ only but why would we do that. We are only talking about the original mixture. Where am I wrong?

Answer in my textbook : Neon weight% = $82.58 \%$, helium weight% : $17.43\%$

  • $\begingroup$ Sorry, I wrote hydrogen in place of neon. The question does not mention any other detail besides these, and I don't think we need any more details. Rest of the variables can be assumed as they will eventually cancel out. $\endgroup$
    – user23923
    Feb 26, 2016 at 12:23
  • $\begingroup$ OK. What do you get for the mole fractions of helium and neon in the original mixture (excluding the water)? $\endgroup$ Feb 26, 2016 at 12:58
  • $\begingroup$ Weight % = 100*(weight of substance/ weight of SOLUTION),so whats your doubt? $\endgroup$ Feb 26, 2016 at 12:59
  • $\begingroup$ @SujithSizon The question is not about simply applying the weight % formula. All I want to know is why should we take total pressure as 748 mm Hg when we are just concerned about the original mixture. The vapour pressure of water was just given to solve the first part, it should not have any relation with original mixture. $\endgroup$
    – user23923
    Feb 26, 2016 at 16:29
  • $\begingroup$ @ChesterMiller I already mentioned it; weight percentage of Neon=34.54% and Helium=65.45% $\endgroup$
    – user23923
    Feb 26, 2016 at 16:30

1 Answer 1


First find the mole fraction of each gas in the original mixture using the formula, P(gas) = X * P(tot). For He:

$$X = \frac{368}{716} = 0.514$$

For Ne:

$$X = 1 - X(\ce{He}) = 0.486$$

Say you have 1 mol of mixture then you have 0.514 mol He and 0.486 mol Ne. Converting these to weights using periodic table:

$$\rm 0.514~mol~He = 2.06~g~He \\\ 0.486~mol~Ne = 9.81~g~Ne$$

The total mass of your mixture is the sum of these two masses, 11.9 g. Divide the individual masses by the total mass to get mass percent.

$$\rm\% w. He = 2.06/11.9 = 17.3 \%$$

Subtracting this from one for Ne (or you could apply the formula again):

$$\rm\% w. Ne = 1 - 17.3\% = 82.7\%$$

Note: I didn't carefully carry sig figs but you should!


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