# A gas sample containing only SO2, PF3, and CO has the following mass percentage: 29.1 % SO2, 61.8 % PF3. How would you calculate the partial pressure in torr, of CO if the total pressure of sample is 684 torr?

Dec 1, 2015

${P}_{C O} = \text{150. torr}$

#### Explanation:

The idea here is that the partial pressure exerted by a gas that's part of a gaseous mixture is proportional to its mole fraction.

This means that in order to find the partial pressure of carbon monoxide, you need to find

• the number of moles of carbon dioxide found in the mixture
• the total number of moles present in the mixture

So, you know that the percent composition of the mixture is 29.1% ${\text{SO}}_{2}$ and "61.8% ${\text{PF}}_{3}$. Since the mixture only contains three gases, you can say that the percent composition of carbon monoxide will be equal to

100% - (29.1% + 61.8%) = 9.1%

Your strategy now will be to pick a sample of this mixture and determine the mass of each gas it contains. To make the calculations easier, let's say that you have a $\text{100.0-g}$ sample of this mixture.

Using the percentages given, you can say that this sample will contain

• ${\text{29.1 g SO}}_{2}$
• ${\text{61.8 g PF}}_{3}$
• $\text{9.1 g CO}$

Use the respective molar masses of the three gases to find their number of moles

${\text{For SO"_2:" " 29.1 color(red)(cancel(color(black)("g"))) * "1 mole SO"_2/(64.06color(red)(cancel(color(black)("g")))) = "0.4543 moles SO}}_{2}$

${\text{For PF"_3: " " 61.8 color(red)(cancel(color(black)("g"))) * "1 mole PF"_3/(87.97color(red)(cancel(color(black)("g")))) = "0.7025 moles PF}}_{3}$

$\text{For CO": " " 9.1color(red)(cancel(color(black)("g"))) * "1 mole CO"/(28.01color(red)(cancel(color(black)("g")))) = "0.3249 moles CO}$

The total number of moles present in the mixture will be

${n}_{\text{total" = 0.4543 + 0.7025 + 0.3249 = "1.4817 moles}}$

Now, the mole fraction of carbon monoxide will be equal to the number of moles of carbon monoxide divided by the total number of moles present in the mixture

${\chi}_{C O} = {n}_{C O} / {n}_{\text{total}}$

${\chi}_{C O} = \left(0.3249 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(1.4817 color(red)(cancel(color(black)("moles}}}}\right) = 0.2193$

The partial pressure of carbon monoxide will thus be

${P}_{C O} = {\chi}_{C O} \times {P}_{\text{total}}$

P_(CO) = 0.2193 * "684 torr" = color(green)("150. torr")

The answer is rounded to three sig figs.