# The mass percent of a three component gas sample is 11.0% BF_3, 74.5% CO and 14.5% PF_3. How do you calculate the partial pressure (torr) of BF_3 if the total pressure of the sample is 0.480 atm?

Jun 23, 2018

This is a good question, and embraces several gas laws....

#### Explanation:

We assume an $100 \cdot g$ mass of gas...and we interrogate the molar composition of the gas....

${n}_{\text{boron trifluoride}} = \frac{11.0 \cdot g}{67.82 \cdot g \cdot m o {l}^{-} 1} = 0.162 \cdot m o l$

${n}_{\text{carbon monoxide}} = \frac{74.5 \cdot g}{28.01 \cdot g \cdot m o {l}^{-} 1} = 2.660 \cdot m o l$

${n}_{\text{phosphorus trifluoride}} = \frac{14.5 \cdot g}{87.97 \cdot g \cdot m o {l}^{-} 1} = 0.165 \cdot m o l$

And we work out ${\chi}_{\text{the mole fraction}}$ of each component...

${\chi}_{\text{component"="moles of component"/"total moles in the mixture}}$

And so ${\chi}_{B {F}_{3}} = \frac{0.162 \cdot m o l}{\underbrace{\left(0.162 \cdot m o l + 2.660 \cdot m o l + 0.165 \cdot m o l\right)}} _ \text{2.987 mol} = 0.0542$.

${\chi}_{C O} = \frac{2.660 \cdot m o l}{0.162 \cdot m o l + 2.660 \cdot m o l + 0.165 \cdot m o l} = 0.891$.

${\chi}_{P {F}_{3}} = \frac{0.165 \cdot m o l}{0.162 \cdot m o l + 2.660 \cdot m o l + 0.165 \cdot m o l} = 0.0552$.

And note that the individual mole fractions SUM to UNITY, as required...

Now in a gaseous mixture, pressure clearly is proportional to the number of moles...and the partial pressure of EACH component is proportional to the mole fraction expressed by each component...and the constant of proportionality is the mole fraction...

${P}_{B {F}_{3}} = 0.0542 \times 0.480 \cdot a t m \times 760 \cdot m m \cdot H g \cdot a t {m}^{-} 1 = 19.8 \cdot m m \cdot H g$

${P}_{C O} = 0.891 \times 0.480 \cdot a t m \times 760 \cdot m m \cdot H g \cdot a t {m}^{-} 1 = 325.0 \cdot m m \cdot H g$

${P}_{P {F}_{3}} = 0.165 \times 0.480 \cdot a t m \times 760 \cdot m m \cdot H g \cdot a t {m}^{-} 1 = 60.0 \cdot m m \cdot H g$