Question 12.32: Expressing Gaseous Mixture Composition in Mole Percent, Pres......
Expressing Gaseous Mixture Composition in Mole Percent, Pressure Percent, and Volume Percent
A 17.92 L flask, at STP, contains 0.200 mole of O _2 , 0.300 mole of N _2 , and 0.300 mole of Ar. For this gaseous mixture, calculate the
a. mole percent of O _2 present.
b. pressure percent of O _2 present.
c. volume percent of O _2 present.
a. The total number of moles of gas present is 0.800 mole (0.200 mole + 0.300 mole + 0.300 mole). The mole percent O _2 is
moles % O _2 = \frac{n_{\textrm{O}_2}}{n_{\textrm{total}}} \times 100 = \frac{0.200 \cancel{\textrm{mole}}}{0.800 \cancel{\textrm{mole}}} \times 100 = 25 % (calculator answer)
= 25.0% (correct answer)
b. Since conditions are specified as STR the total pressure in the flask is 1.00 atm. The partial pressure of the O _2 is
P_{\textrm{O}_2} = X_{\textrm{O}_2} \times P_{\textrm{total}}= \frac{0.200 \cancel{\textrm{mole}}}{0.800 \cancel{\textrm{mole}}} \times 1.00 atm
= 0.25 atm (calculator answer)
= 0.250 atm (correct answer)
The pressure percent O _2 is
pressure % O _2 = \frac{P_{\textrm{O}_2}}{P_{\textrm{total}}} \times 100 = \frac{0.250 \cancel{\textrm{atm}}}{1.00 \cancel{\textrm{atm}}} \times 100
= 25% (calculator answer)
= 25.0% (correct answer)
c. The total volume of the flask is given as 17.92 L. The volume of the oxygen present, if it were alone at STP conditions in a different container, is
0.200 \cancel{\textrm{mole O}_2} \times \frac{22.41 \textrm{ L O}_2}{1 \cancel{\textrm{mole O}_2}} = 4.482 L O _2 (calculator answer)
= 4.48 L O _2 (correct answer)
The volume percent O _2 is
volume % O _2 = \frac{V_{\textrm{O}_2}}{V_{\textrm{total}}} \times 100 = \frac{4.48 \cancel{\textrm{L}}}{17.92 \cancel{\textrm{L}}} \times 100
= 25% (calculator answer)
= 25.0% (correct answer)
Note, from the answers to parts (a), (b), and (c), that for O _2 the
mole percent = pressure percent = volume percent