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Consider the sodium borohydride reduction of camphor to isoborneol. A reaction was performed in which 1.600 g of camphor was reduced by an excess of sodium borohydride to make 1.062 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction. Theoretical yield: g Percent yield:

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Answer:

62.5%

Explanation:

Number of moles of camphor reacted= 1.6g/152.23 g/mol = 0.011 moles

Since the reaction is 1:1,  then 0.011 moles of isoborenol is also produced.

Theoretical yield of isoborenol produced = 154.25 g/mol * 0.011 moles = 1.698 g

% yield = actual yield/theoretical yield * 100

% yield =  1.062 g/1.698 g * 100

% yield =  62.5%

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