How to Calculate the Percent Weight of Components in a Gas Mixture?

[archaic]

Homework Statement

A mixture of $\mathrm{NO_{2(g)}}$ and $\mathrm{N_2O_{4(g)}}$ at $63\,\mathrm{C^°}$ and $750\,\mathrm{mmHg}$ pressure has a density of $1.98\,\mathrm{g/L}$. What is the weight% of $\mathrm{NO_2}$ in the mixture?

Relevant Equations

$d=\frac{MP}{RT}$

The total molar mass is $M=\frac{dRT}{P}=\frac{1.98*0.08206*(63+273.15)}{750/760}=55.3454786\,g/mol$.
We have $M=aM_1+bM_2$ where $M_1$ and $M_2$ are the molar masses of $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$ respectively.
If I consider $n$ moles, I'd have $m=nM=naM_1+nbM_2=m_1+m_2$ and so the weight percentage would be $\frac{m_1}{m}*100\%=\frac{aM_1}{aM_1+bM_2}*100\%$, but how can I find $a$ and $b$?

 

[Borek]

Would assuming you have an (arbitrary) volume V of mixture help?

 

[archaic]

Would assuming you have an (arbitrary) volume V of mixture help?

The I'd have $1.98V\,g$ of matter but I don't see how that can help.

 

[mjc123]

If M = aM1 + bM2, what is a+b? (Remember molar mass is an intensive property.)

 

[archaic]

If M = aM1 + bM2, what is a+b? (Remember molar mass is an intensive property.)

Unfortunately I couldn't solve this on my own. $a$ would be $n_1/(n_1+n_2)$ and $b=1-a$.
I am comforting myself by saying that I didn't know about the average molar mass formula and then didn't put much thought in seeing that $\frac{aM_1}{aM_1+(1-a)M_2}=\frac{n_1M_1}{n_1M_1+n_2M_2}$
I now know, however.

 

[Borek]

The I'd have $1.98V\,g$ of matter but I don't see how that can help.

If there is 1L of gas, how many moles is that?

Basically the information given allows you to write system of two linear equations - one to express total mass in terms of sum of masses, the other to express total number number of moles as sum of numbers of moles.

Using average molar mass is equivalent to one of possible "middle" stages of the math involved you can reach when solving this system of equations.

 

[archaic]

If there is 1L of gas, how many moles is that?

Basically the information given allows you to write system of two linear equations - one to express total mass in terms of sum of masses, the other to express total number number of moles as sum of numbers of moles.

Using average molar mass is equivalent to one of possible "middle" stages of the math involved you can reach when solving this system of equations.

Yes, I also read this from the solution, I understand better now, although I am disappointed that I couldn't find the solution.
$$\%w_{m_1}=\frac{m_1}{\rho V}*100\%\Leftrightarrow m_1=\%w_{m_1}\rho V/100\%\\
\%w_{m_2}=\frac{m_2}{\rho V}*100\%=100\%-\%w_{m_1}\Leftrightarrow m_2=(100\%-\%w_{m_1})\rho V/100\%\\
n=\frac{m_1}{M_1}+\frac{m_2}{M_2}=\frac{\rho V}{100\%}\left(\frac{\%w_{m_1}}{M_1}+\frac{100\%-\%w_{m_1}}{M_2}\right)=\frac{PV}{RT}$$
We can then solve for the percent weight.