Essay about Dehydration of an Alcohol lab report

1188 Words Feb 9th, 2014 5 Pages
Experiment 9-Dehydration of 2-methylcyclohexanol
Name______________________________________________________________________
Lab Partner_________________________________________________________________
Lab Day and Time____________________________________________________________
Report appearance
(Typed, on time, in order, presentable, complete)

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Abstract

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Introduction

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Reaction Scheme and Curly arrow mechanism

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Table

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Experimental Procedure

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Calculation (percentage yield and composition of isolated materials)
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Results and
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The distillate was injected at the same time the start button was pushed. Graph results were obtained from the computer.

Calculations.
Part 1 m = d x v m = (0.9304 g/mL C7H14O)(5 mL C7H14O) = 4.652 g C7H14O moles of C7H14O = mass of C7H14O / molar mass of C7H14O = 4.652 g C7H14O / 114.19 g/mol C7H14O = 0.0407 mol C7H14O
Theoretical Yield (g) = moles of limiting reagent x molar mass of product = moles of C7H14O x molar mass of C7H12 =0.0407 mol C7H14O x 96.17 g/mol C7H12 = 3.914 g
Actual Yield of Alkene Mixture C7H12 = (liquid + vial) – vial = 9.65 g – 8.92 g = 0.73 g C7H12
Percent Yield = (Actual Yield (g) / Theorectical Yield (g)) x 100 = (0.73 g C7H12 / 3.914 g C7H12) x 100 = 18.65%
Part 2
Area of a triangle= ½ base x height
Area of Peak A= ½(1.2mm) x 0.5mm= 0.3 mm2
Area of Peak B= ½(2.3mm) x 5.4mm= 6.21 mm2
Percentage of Product A = (Area of Peak A) / (Area of Peak A + Area of Peak B) = (0.3 mm2) / (0.3 mm2+ 6.21 mm2)= 0.46 x 100= 4.6%
Percentage of Product B = (Area of Peak B) / (Area of Peak A + Area of Peak B) = (6.21 mm2) / (0.3 mm2+ 6.21 mm2)= 0.9539 x 100= 95.39%

Results and Discussion

The objective of the experiment was to carry out the dehydration of the alcohol 2-methylcyclohexanol with an acid as a catalyst. Particularly, a 4:1 mixture of sulfuric

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