Stiochoimtery Question

Question

For the reaction below determine the limiting reagent, the theoretical yield and the percent yield given an actual yield of 4.5 grams. See the Technique Book Section 3.2 on page 35 for help with setting this up. Please show all set-ups and provide answers to the correct number of significant figures

CH3CH2CH2CH2Br	+	CH3CH2O-Na+	→	CH3CH2CH=CH2	+	NaBr
1-bromobutane	sodium ethoxide	1-butene
MM 137 g/mol	MM 68.1 g/mol	MM 56 g/mol
Denisty 1.27 g/mL

The procedure specified 10.8 mL of 1-bromobutane and 10.2 grams of sodium ethoxide.

 

Answer 1

CH3CH2H2Br + CH3CH2ONa ==> CH3CH2CH=CH2 + NaBr

1Br butane..........Na ethoxide...........1-butene................sodium bromide

137 g/mol..........68.1 g/mol................56 g/mol

1.27 g/ml

Limiting reactant:

mols CH3CH2H2Br = 10.8 ml x 1.27 g/ml x 1 mol/137 g = 0.100 mols

mols CH3CH2ONa = 10.2 g x 1 mol/68.1 g = 0.150 mols

Limiting reactant is 1-bromobutane (CH3CH2H2Br)

Theoretical yield:

Assuming you are looking at the yield of 1-butene, then we have...

0.100 mols CH3CH2H2Br x 1 mol CH3CH2CH=CH2 / mol CH3CH2H2Br x 56 g CH3CH2CH=CH2/mol = 5.6 g CH3CH2CH=CH2

Percent yield:

4.5 g / 5.6 g (x100%) = 80.4% yield