Question

In an experiment to in which Oxone was used to oxidize Borneol to Camphor, Calculate the theoretical yield of camphor. 0.4381g of Borneol 1.2333g of Oxone 0.0523g and 0.0254g of NaCl

Answer 1

Ans. Balanced reaction: Borneol + 0.6 Oxone ----NaCl (catalyst)----> Camphor

Stoichiometry: 1 mol Borneol reacts with 0.6 mol Oxone to form 1 mol Camphor.

Theoretical molar ration of reactants: Borneol : Oxone = 1 : 0.6

# Moles of Borneol = Mass /Molar mass = 0.4381 g / (154.25 g/ mol) = 0.00284 mol

Moles of Oxone = Mass /Molar mass = 0.1.2333 g / (614.6 g/ mol) = 0.00200 mol

Experimental molar ratio of reactants = moles of Borneol : moles of Oxone

                                                            = 0.00284 mol : 0.00200 mol

                                                            = 1 : 0.70                  

Comparing the theoretical and experimental molar ratio of reactants, the moles of Oxone are greater than theoretical value 0.6 whereas that of Broneol is kept constant at 1.0 mol

Therefore, Oxone is the reagent in excess. And, Broneol is the limiting reactant.

# The formation of product follows the stoichiometry of balanced reaction.

According to the stoichiometry of balanced reaction, 1 mol Broneol produces 1 mol Camphor.

So,

            Moles of Camphor formed = Moles of Broneol consumed

Hence, Moles of Camphor formed = 0.00284 mol

# Now, theoretical yield of reaction = Moles of Camphor formed x Molar mass

                                                            = 0.00284 mol x (152.23 g/ mol)

                                                            = 0.4323 g