CHM112 Home Page  Reference Materials  Homework Assignments  Lectures  Exams & Quizzes  Grades  Study Aids  URI Home Page 
The equilibrium constant for the isomerization of butane at 25 ^{o}C is K_{c} = 7.94.
If 5.00 g butane is introduced into a 12.5L flask at 25 ^{o}C, what mass of isobutane will be present when equilibrium is reached?
Convert the mass of butane (abbreviated B) to molarity, then set up a table of concentrations to solve for the equilibrium concentration of isobutane (abbreviated iB). Then convert this molarity back to a mass of isobutane.
B: C_{4}H_{10} molar mass = 4(12.01) + 10(1.01) = 58.14 g/mol






Initial 



Change 



Equilibrium 


[iB]_{e} = 6.11×10^{–3} M
iB: C_{4}H_{10} so the molar mass is the same as butane (which must be true if they are isomers) = 58.14 g/mol.
So the mass of isobutane at equilibrium is:
mass = Molarity×Volume×molar mass = 6.11×10^{–3} mol/L × 12.5 L × 58.14 g/mol = 4.44 g