Question

In the experiment of the reduction of camphor to isoborneol, how is the percent yield calculated?

0.1038 g of camphor is initially used.

0.0447 g of NaBH4 is the reducing agent.

0.0524 g of product is obtained.

Per the manual, stoichiometry is 4 moles of camphor to one mole of NaBH4

MW of camphor is 152.23

MW of isoborneol is 154.24

Please help explain. Thank you

Answer 1

4 Ketone(camphor) + 1 NaBH4 --------> 4 Alcohol ( isoborneol)

mass 0.1038 g 0.0447g 0.0524 g

M.Wt 152.23 g/mol 37.83 g/mol 154.24 g/mol

Moles (mass/M.wt) 0.00068186 0.0011816 0.00033973

percent Yield = (exp amount / theoretical amount) x 100

since reactant and product both are in 1 : 1 ratio

we can expect all these 0.00068186 moles of reactant should give 0.00068186 moles of product

i.e, (0.00068186 / 0.00068186) x 100 = 100 %

but we got only 0.00033973 amount of product

So % Yield = (0.00033973 / 0.00068186) x 100 = 49.82 %