Question

What mass of propane will occupy the same volume as 0.057 mol of butane at 1.00 atm and 298 K?

My answer: 0.057m x 0.0821Latm/Kmol x 298K = 1.4 L

Suppose a student performed a similar experiment using pentane and hexane as his gases.

Pentane Hexane

Molar Mass (g/mol) 72.15 86.18

Mass (g) 0.541 0.646

Given the data in the figure above, which gas will occupy a larger volume at STP? (This is the question I am stuck on.)

Answer 1

volume occupied by butane = nRT/P

             = 0.057*0.0821*298/1

             = 1.4 L

no of mol of propane must be taken (n) = PV/RT

           = 1*1.4/(0.0821*298)

           = 0.057 mol

mass of propane = n*mwt = 0.057*44  

                  = 2.508 g

                      Pentane      Hexane

Molar Mass (g/mol)      72.15      86.18

Mass (g)                0.541      0.646

volume of Pentane at STP = nRT/P

   n = w/mwt

   = (w/mwt)RT/P   (STP : P= 1 atm , T = 273k)

   = (0.541/72.15)*0.0821*273/1

   = 0.168 L

volume of hexane at STP = nRT/P

   n = w/mwt

   = (w/mwt)RT/P   (STP : P= 1 atm , T = 273k)

   = (0.646/86.18)*0.0821*273/1

   = 0.168 L

both the gases occupies same volume at STP