Determine the mass of butane gas that was released from the lighter and collected in the gas collection tube. use the given below to derive the answer.
given:
mass of lighter before (g)
32.105 g
volume of butane gas collected (ml)
90. ml
mass of lighter after (g)
31.844 g
room temperature (k)
297 k
air pressure (atm)
1.2 atm
so (90)(1.2) = (297)

There are several information's already given in the question. Based on those information's, the answer can be easily deduced. 
Gas constant = 0.0821
Now
PV=nRT
(1.2) * (0.09) = n * (0.0821) * (297)
n = 0.00443
Mass of Butane = 32.105 - 31. 844
                          = 0.261
Then
Molar mass of Butane gas = 0.261/.00443
                                 = 58.92

H tricky one let me think and get back to u