# Finding percent weight

I have a mixture of $$\ce{NO2(g)}$$ and $$\ce{N2O4(g)}$$ at $$63\ \mathrm{^\circ C}$$ and $$750\ \mathrm{mmHg}$$. I need to find the weight percent of the first compound. Density $$d=1.98\ \mathrm{g/L}$$.

I assumed that we have $$M=aM_1+bM_2$$ where $$M_1$$ and $$M_2$$ are the molar masses of the first and second compound. I know that $$m=nM$$ so the weight percent would be $$\frac{m_1}{m}\cdot100\ \%=\frac{aM_1}{aM_1+bM_2}\cdot100\ \%$$ My problem lies in finding the coefficients. Can someone give a hint please?

• You need to know the equilibrium constant for this temperature, and then use a similar approach as shown here: How can I calculate the percentage of dissociated dinitrogen tetroxide? Dec 3, 2019 at 8:16
• @andselisk I still haven't learnt that, so the question doesn't require that approach. thank you though :) Dec 3, 2019 at 8:40
• There is an equilibrium between both components which dictates the fraction of each, and without $K_p$ I doubt you'll be able to do much. Dec 3, 2019 at 8:45
• Personally, I don't see how this could be done with the data you posted. Could you please add a source for this problem? Dec 3, 2019 at 8:56
• My bad, I forgot to include it. It is $1.98\,g/L$. Dec 3, 2019 at 10:05

We can find the average molar mass with the density formula involving pressure and temperature $$\overline M=\frac{\rho TR}{p}$$.
We also know that, in this case, $$\overline M=\frac{n_1M_1+n_2M_2}{n_1+n_2}$$. Since we only "care" about $$M_1$$ we can write the average molar mass as $$\overline M=\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2$$ and then find $$\frac{n_1}{n_1+n_2}$$. $$\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2}^{(*)}=\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+\frac{n_2}{n_1+n_2}M_2}=\frac{n_1M_1}{n_1M_1+n_2M_2}=\frac{m_1}{m_{total}}$$ Compute $$(*)\cdot100\%$$. The development of the fraction was to show why it works.
Another possible solution is this : $$\%w_{m_1}=\frac{m_1}{\rho V}*100\%\Leftrightarrow m_1=\%w_{m_1}\rho V/100\%$$ $$\%w_{m_2}=\frac{m_2}{\rho V}*100\%=100\%-\%w_{m_1}\Leftrightarrow m_2=(100\%-\%w_{m_1})\rho V/100\%$$ $$n=\frac{m_1}{M_1}+\frac{m_2}{M_2}=\frac{\rho V}{100\%}\left(\frac{\%w_{m_1}}{M_1}+\frac{100\%-\%w_{m_1}}{M_2}\right)=\frac{PV}{RT}$$