Question

Calculate the percent yield from the data listed below . Thank you

oxidaHive co upling 49 2-14-18 DATA IN IAe .LoO 89 sioWlY TO10) time 9 3:33p dcep

Answer 1

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The reaction scheme for the experiment is:

$2 \ mol \ 2-naphthol + 2FeCl_3.6H_2O \rightarrow 1 \ mol \ 1,1-Bi-2-naphthol$

	Mass(gm)	Molar mass (gm/mol)	Moles (mol)
2-Naphthol	0.44657	144.17	0.0030975
FeCl3.6H2O	4.784	270.3	0.01769
1,1-Bi-2-naphthol(expt)	0.047	286.32	0.00016
1,1-Bi-2-naphthol(theory)	0.44343	286.32	0.0015487

Here, we see that 2 moles of 2-naphthol is being converted to 1 mol of 1,1-Bi-2-naphthol(BINOL)

From the table it is clear that 2-naphthol is our limiting reagent as the number of moles of it is lower than the number of moles of the other reactant i.e FeCl3 hexahydrate.

Hence, the theoretical yield of BINOL should be half the number of moles of 2-naphthol, i.e. 0.0030975/2 = 0.0015487 moles

Now, 0.0015487 moles of BINOL = 0.0015487mol*286.32gm/mol = 0.44343 gm

Percent yield is given by the following formula:

$\ percent \ yield = \frac{\ Experimental \ Yield}{\ Theoretical \ Yield} \times 100$

Here, experimental yield = 0.047 gm

hence,

percent yield = $= \frac{\ Experimental \ Yield}{\ Theoretical \ Yield} \times 100 = \frac{0.047\ gm}{0.44343\ gm}\times 100 =10.59$

Hence, the percent yield of BINOL is 10.59%.