Question

The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent composition of the solution? Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature.

Answer 1

• The mass percent of Pentane in solution is 16.49%

• The mass percent of Hexane in solution is 83.51%

Explanation:

• Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.

• Convert these values to mol% using their molecular weights:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

• From Raoult's law we know:  

Pp = xp $\times$ Pp - vap = yp $\times$ Pt                                       (1)

Ph = xh $\times$ Ph - vap = yh $\times$ Pt                                       (2)

• Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) $\times$ Ph - vap = (1 - yp) $\times$ Pt                                   (3)

• Substituting (1) into (3) we get:

(1-xp) $\times$ Ph - vap = (1 - yp) $\times$ xp $\times$ Pp - vap / yp            (4)

• Rearrange for xp:

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap)      (5)

• Subbing in the values we find:

Pentane mol% in solution: xp = 19.08%  

Hexane mol% in solution: xh = 80.92%

• Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol $\times$ 72.15 g/mol

     = 13.766 g

mh = 0.8092 mol $\times$ 86.18 g/mol

     = 69.737 g

• Mass% of Pentane solution = 13.766/(13.766+69.737)

                                                       = 16.49%

• Mass% of Hexane solution  = 83.51%