What is the mass fraction of trichloroethene in the liquid phase using vapour pressure and mole fraction in gas phase?

Question

Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene ($\ce{C2HCl3}$) and trichloromethane ($\ce{CHCl3}$) is established at $\pu{25 °C}$. The mole fraction of $\ce{CHCl3}$ in the vapour phase is $0.73$. What is the mass fraction of $\ce{C2HCl3}$ in the liquid phase? Round your answer to two significant figures.

The vapour pressures of trichloroethene and trichloromethane at $\pu{25 °C}$ are:

$$\begin{align}P_\text{vap}(\ce{C2HCl3}) &= \pu{73.0 mmHg}\\[0.5em] P_\text{vap}(\ce{CHCl3}) &= \pu{199.1 mmHg}\end{align}$$

So, what I did was I found mole fraction of $\ce{C2HCl3}$ and then used the two mole fractions along with the vapour pressures to find the total pressure of the solution.

$$P_\text{vap}= \frac{0.73}{199.1}+\frac{0.27}{73} = \pu{165.053mmHg}$$

Then, from Raoult's Law I know that the mole fraction in liquid phase is equal to mole fraction in vapour phase, multiplied by vapour pressure, divided by total pressure. From that, I found the mole fraction of both things in liquid phase. I use the mole fraction to find mass of both, and then did mass of $\ce{C2HCl3}$ divided by the total mass that I calculated. I got an answer of

$\frac{15.68}{120}=0.13$ but it says that it's wrong. I'm not sure where I messed up?

Answer 1

There is actually small error in the calculation part of total pressure and in the Raoult’s law formulae.

According to Raoult’s law: $$ P_A= P^0 _ A \chi_A $$

i.e. partial pressure of a component in a solution($P_A$) is the product of the vapour pressure of the pure solvent A at that temperature($ P^0_A$) and it's mole fraction in liquid phase($\chi_A$).

Also, we have $$ P_A= P_{total}\chi^‘_A$$

where: $P_A$=partial pressure of component A

$P_{total}$=total vapour pressure of the solution

$\chi^{’}_A$=mole fraction of component A in vapour phase

Considering these equations along with Dalton’s law of partial pressure (total vapour pressure of a solution is the sum of vapour pressures of its individual components)$$P_{total}=P_A+P_B$$

we get: $$P^0_A\chi_A=P_{total}\chi^{‘}_A$$

Substituting the respective values:

$$ (199.1){\chi_{\ce{CHCl3}}}=P_{total}(0.73)$$ and $$ (73){\chi_{\ce{C2HCl3}}}=P_{total}(0.27)$$

Since $${\chi_{\ce{CHCl3}}}+{\chi_{\ce{C2HCl3}}}=1$$

we have two equation with two unknowns which upon solving we get ${\chi_{\ce{CHCl3}}}=0.4978$ and ${\chi_{\ce{C2HCl3}}}=0.5022$ and you shall continue in the same way to get the mass fraction.

Hope it helps.

Answer 2

As far as the question is concerned, you do not need the vapor pressure. In order to calculate the mass fraction of $\ce{C2HCl3}$, we first have to calculate the molar mass of $\ce{C2HCl3}$ and $\ce{CHCl3}$. $$M(\ce{C2HCl3}) = \pu{131.38 g mol-1}$$ $$M(\ce{CHCl3}) = \pu{119.37 g mol-1}$$ With these two calculated we can calculate the absolute mass relative to $\pu{1 mol}$, due to the fact of ${\pu{0.73 mol}}/{\pu{1 mol}}$ being relative to a total of $\pu{1 mol}$ of molecules. Now if we calculate the absolute masses, we get: $$M(\ce{C2HCl3}) = \pu{131.38 g mol-1}\times\pu{0.27\frac{mol}{mol}} = \pu{35.47 g mol-1}$$ $$M(\ce{CHCl3}) = \pu{119.37 g mol-1}\times\pu{0.73\frac{mol}{mol}} = \pu{87.14 g mol-1}$$ And the total mass in the system results in: $$m_\text{total}=\pu{35.47 g mol-1}+\pu{87.14 g mol-1}=\pu{122.61 g mol-1}$$ And with the total relative mass given, we can calculate the mass fraction of $\ce{C2HCl3}$, as follows: $$w(\ce{C2HCl3})=\frac{\pu{35.47 g mol-1}}{\pu{122.61 g mol-1}}=0.29$$ $$w(\ce{CHCl3})=\frac{\pu{87.14 g mol-1}}{\pu{122.61 g mol-1}}=0.71$$