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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)

  1. 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)

Verified Answer :

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