Chemistry Butane Lab

Problem: To determine the molar mass of butane. Procedure: number one we mensural the mass of the lighter. Then we filled a beaker altogether with piddle and put our hand over the top. We filled a slide by half full with water and glum the beaker upside mastered and fit(p) it in the water. We took our hand kill 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 entirely and measured it afterward the loss of the butane. Data:                  Mass of igniter before:          17.63 g                           Mass of Lighter after:                   17.00 g                           Mass of Butane released:    Â Â Â Â Â Â  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 kP! ais the pressure level 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 ledger = .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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