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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)
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)

Answer :

MissPhiladelphia
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

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