Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of 0.4 ∘C. A 250 mL sealed flask contains 0.75 g of butane at −22∘C.
a) How much Butane is present as a liquid (in g)
b) If the butane is warmed to 25 ∘C, how much is present as a liquid? (in g)
1 Answer

You can use the heat of vaporization and the normal boiling point temperature to calculate the vapor pressure of the gas at 22 C:
ln (P2/P1) = DeltaHvap/R (1/T1  1/T2)
ln (P2/760) = 2.244X10^4 J/mol / 8.314 J/molK (1/272.75  1/251.15 K)
ln (P2/760) = 0.8511
P2/760 = 0.4269
P2 = 324 mmHg
Now, assuming that the volume of liquid will be insignificant, use the ideal gas law to calculate moles and then mass of butane in the gas phase:
PV = n RT
P = 324.5 / 760 = 0.427 atm
0.427 atm (0.250 L) = n (0.08206 Latm/molK) (251.15K)
n = 5.18X10^3 mol butane in gas phase
Mass butane in gas phase = 5.18X10^3 mol X 58.12 g/mol = 0.30 g in gas phase
Mass butane in liquid phase = 0.75  0.30 = 0.45 g
For the second part, I believe that you would repeat all of the previous calculations to calculate the vapor pressure of butane at 25C, and then the moles and mass of gas in the vapor phase. If the mass is greater than 0.75 g, all of the butane will have vaporized.