Butane exists as three staggered minima: one anti and two equivalent gauche conformers, with the gauche conformers $3.8 kJ mo l^{- 1}$ above anti (neglect eclipsed forms; $R = 8.314 J mo l^{- 1} K^{- 1}$ ). What is the mole percent of the anti conformer at 298 K?

Answer: D

Relative population of each gauche conformer to anti is

exp (- 3800/ (8.314 \times 298)) = 0.2157

. There are two gauche conformers, so the total gauche weight is

0.4314

. The anti fraction is

1/ (1 + 0.4314) = 0.699

, i.e. 69.9 %. (82.3 % ignores the two-fold degeneracy; 30.1 % is the gauche total.)