Using Dalton's law of partial pressure to find percentages of composition 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 &quot;v&quot; (total volume) and then moles of neon and helium in terms of &quot;v&quot; from which we calculate the weight of neon and helium and calculate the weight percentage. &quot;v&quot; 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!