Question

Experiment for the oxidation of cyclododecanol to cyclododecanone

A) In the experiment it states, in the 50ml round bottom flask add 1.00g of cyclododecanol, 4.0 ml of acetone and 1.5 ml of acetic acid. Warm the mixtures. slowly add 12.0 ml of sodium hypochlorite.

1) From that statement. "How do you caculate the number of moles of cyclododecanol, sodium hypochlorite, and acetic acid" ?

2) Show the formula and steps to calculate the number of moles from those compounds.

B) How do you caculate the "ratio of moles" of sodium hypochlorite per mole of cyclododecanol?

C) Which was the limiting reactant?

Answer 1

1) From that statement. "How do you caculate the number of moles of cyclododecanol, sodium hypochlorite, and acetic acid" ?

Answer: moles can be calculated by dividing mass of substance in grams by molar mass of substance in gram/moles

moles = mass of substance / molar mass of substance in g/mol

2) Show the formula and steps to calculate the number of moles from those compounds.

Answer: Please see the reaction and its mechanism

Our Reaction bleach Reaction Scheme он NaOCI, H20 CH3CO2H aceto ne Cyclododecanol Cyclododecanone Mechanism (1) [CH3CO2H] (2) HOCI CI-OH2 н.tcı + NaCI H30 OH2 (5) H2O

(1) Only cyclododecanol has been given in grams, its molar mass = 184.3184 g/mol

hence moles present in 1.00g of cyclododecanol = 1.0 g / 184.3184 g/mol = 0.00542539431 moles

(2) From above reaction equation it is clear that 1 mol cyclododecanol requires 1 mol acetic acid and 1 mol sodium hypochlorite.

we know that density in g / mL = mass in g / volume in mL ==> mass in g = density in g / mL x volume in mL

(a) sodium hypochlorite density = 1.11 g/mL

mass of 12.0 mL of sodium hypochlorite = 12.0 mL x 1.11 g/mL = 13.2 g

moles sodium hypochlorite = 13.2g/74.44 g/mol = 0.17732401934 moles

(b) acetic acid density = 1.05 g/mL

mass of 1.5 mL of acetic acid = 1.5 mL x 1.05 g/mL = 1.575 g

moles acetic acid = 1.575g/60.05 g/mol = 0.02622814321 moles

B) How do you caculate the "ratio of moles" of sodium hypochlorite per mole of cyclododecanol?

Answer: From above reaction equation it is clear that 1 mol cyclododecanol requires 1 mol acetic acid and 1 mol sodium hypochlorite.

"ratio of moles" of sodium hypochlorite per mole of cyclododecanol = 1:1

C) Which was the limiting reactant?

Answer: cyclododecanol : sodium hypochlorite : acetic acid = 0.005 moles : 0.177 moles : 0.026 moles

from above ratio it is clear that cyclododecanol has lesser number of moles, hence cyclododecano is the limiting reactant.

Hope i have addressed all of your queries!

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