Chemistry Butane Lab

Problem: To determine the molar mass of butane. Procedure: first of all we mensural the mass of the lighter. Then we filled a beaker totally with piddle and put our hand over the top. We filled a slide by half full with water and moody the beaker upside mastered and position it in the water. We took our hand finish 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 four hundred mL of the butane into the beaker. We whence took the temperature of the water. We dried off the lighter all 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:          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 printing press 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 = hoagie 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... If you want to get a full essay, baffle it on our website: OrderEssay.net

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