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?

1 Answer
Dec 1, 2015

Answer:

#P_(CO) = "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%# #"SO"_2# and #"61.8%# #"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 #"100.0-g"# sample of this mixture.

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

  • #"29.1 g SO"_2#
  • #"61.8 g PF"_3#
  • #"9.1 g CO"#

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

#"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#

#"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#

#"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_"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_(CO) = n_(CO)/n_"total"#

#chi_(CO) = ( 0.3249 color(red)(cancel(color(black)("moles"))))/(1.4817 color(red)(cancel(color(black)("moles")))) = 0.2193#

The partial pressure of carbon monoxide will thus be

#P_(CO) = chi_(CO) xx P_"total"#

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

The answer is rounded to three sig figs.