Using Dalton's law of partial pressure to find percentages of composition


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> 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.
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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% : $17.43\%$




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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