Consider the sodium borohydride reduction of camphor to isoborneol. A reaction was performed in which 1.400 1.400 g of camphor was reduced by an excess of sodium borohydride to make 1.040 1.040 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction.

The percent yield of isoborneol is 73.24 %

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of camphor = 1.400 g

Molar mass of camphor = 152.23 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of camphor and sodium borohydride follows:

By Stoichiometry of the reaction:

1 mole of camphor produces 1 mole of isoborneol

So, of camphor will produce = of isoborneol

Now, calculating the mass of isoborneol from equation 1, we get:

Molar mass of isoborneol = 154.25 g/mol

Moles of isoborneol = moles

Putting values in equation 1, we get:

To calculate the percentage yield of isoborneol, we use the equation:

Experimental yield of isoborneol = 1.040 g

Theoretical yield of isoborneol = 1.42 g

Putting values in above equation, we get:

Hence, the percent yield of isoborneol is 73.24 %

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)