Question

DEPARTMENT OF CHEMISTRY in State University CHE 231, ORGANIC CHEMISTRY LAB I This assignment is due at the beginning of your la period in the next week pts:__/15 pts Name: Consider the following reaction between cyclohexene and diatomic bromine. H CH2Cl2 + Brz + Brz 40 mL 35 mL Cyclohexene (C.H.2): molecular weight = 82.14 g/mol; density -0.811 g/mL Bromine (Br): molecular weight - 159.81 g/mol; density 3.12 g/mL (1 point each) 1. How many grams of cyclohexene are involved in this reaction? 2. How many grams of bromine are involved in this reaction? How many moles of cyclohexene are involved? How many moles of bromine are involved? 5. Which compound is the limiting reagent? What is the maximum number of moles of product that can form? What is the maximum number of grams of products that can form? 8. What would be the percent yield if 57 grams of product are isolated?

Answer 1

Q1. volume of cyclohexene = 40 mL

mass of cyclohexene = (volume of cyclohexene) * (density of cyclohexene)

mass of cyclohexene = (40 mL) * (0.811 g/mL)

mass of cyclohexene = 32.44 g

Q2. volume of bromine = 35 mL

mass of bromine = (volume of bromine) * (density of bromine)

mass of bromine = (35 mL) * (3.12 g/mL)

mass of bromine = 109.2 g

Q3. moles of cyclohexene = (mass of cyclohexene) / (molar mass of cyclohexene)

moles of cyclohexene = (32.44 g) / (82.14 g/mol)

moles of cyclohexene = 0.395 mol

Q4. moles of bromine = (mass of bromine) / (molar mass of bromine)

moles of bromine = (109.2 g) / (159.81 g/mol)

moles of bromine = 0.683 mol

Q5. Since moles of cyclohexene involved are less, therefore, cyclohexene is the limiting reagent.

Q6. Product is 1,2-dibromocyclohexane

maximum number of moles of product formed = moles of cyclohexene involved

maximum number of moles of product formed = 0.395 mol

Q7. maximum mass of product = (maximum moles of product) * (molar mass 1,2-dibromocyclohexane)

maximum mass of product = (0.395 mol) * (241.95 g/mol)

maximum mass of product = 95.55 g

Q8. percent yield = (amount of product isolated / maximum mass of product) * 100

percent yield = (57 g / 95.55 g) * 100

percent yield = (0.597) * 100

percent yield = 59.7 %