Chemistry Butane Lab

Problem: To determine the molar mass of butane. Procedure: origin we mensural the mass of the lighter. Then we filled a beaker solely with piddle and put our hand over the top. We filled a slide by half full with water and sour the beaker upside mastered and place it in the water. We took our hand shoot of the beaker making accredited that no air got into the beaker. Then we placed the lighter downstairs the beaker opening in the water and emptied cd mL of the butane into the beaker. We whence took the temperature of the water. We dried off the lighter exclusively and measured it afterward the loss of the butane. Data:                  Mass of visible radiation 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 insistency 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 compress 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 = grinder Volume of Butane at STP 4) n = m / M n = .63 / 58 n = .011 molecules = .011 x (6.02x1023)                   molecules = 6.54                  5) molar leger = .4 / .011                   molar volume = 36.36 L                                                      Questions 1)         M = n / m M = .011 / .63 M = .175                  3) .63 g / .4... If you want to get a full essay, baffle it on our website: OrderEssay.net

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