Problem: To determine the molar mass of butane.
Procedure: First we measured the mass of the lighter. Then we filled a beaker completely with water and put our hand over the top. We filled a sink half full with water and turned the beaker upside down and placed it in the water. We took our hand off of the beaker making sure that no air got into the beaker. Then we placed the lighter under the beaker opening in the water and emptied 400 mL of the butane into the beaker. We then took the temperature of the water. We dried off the lighter completely and measured it after the loss of the butane.
Data: Mass of Lighter before: 17.63 g Mass of Lighter after: 17.00 g Mass of Butane released: 0.63 g Temperature of water: 20.00 oC Volume of Butane 400 mL Air Pressure on Nov. 2 101.4 kPa Conclusions: Calculations 1) 17.63
"" 17.00 = 0.63 grams of butane used 2) 101.3 "" 2.33 = 99.07 kPais the pressure of the butane 3) P1 x V1 / T1 = P2 x V2 / T2 99.07 x 36.36 / 293 = 101.3 x V2 / 273 12.29 x 273 / 101.3 = V2 33.12 = Molar Volume of Butane at STP 4) n = m / M n = .63 / 58 n = .011 molecules = .011 x (6.02x1023) molecules = 6.54 5) molar volume = .4 / .011 molar volume = 36.36 L Questions 1) M = n / m M = .011 / .63 M = .175 3) .63 g / .4 L = 1.58 g/L