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
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
#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
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.