Question

A reaction was performed in which 1.7 g of camphor was reduced by an excess of sodium borohydride to make 1.3 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction.

Answer 1

Concepts and reason

The given problem is based on the concept of moles to calculate the theoretical yield of the reaction. The percent yield of the reaction is calculated using the theoretical and actual yield of the reaction.

Fundamentals

The percent yield of the reaction is calculated using the theoretical and actual values of the yields.

Theoretical yield is defined as the yield expected after the completion of the reaction. It is calculated using the number of moles of reactant forming the products.

The formula for calculating the theoretical yield is as follows:

${\rm{Theoretical yield}} = {\rm{Number}}\,{\rm{of}}\,{\rm{moles}} \times {\rm{Molar mass}}$

Actual yield is defined as the exact yield obtained after the complete reaction has taken place.

There is generally a difference between theoretical and actual yield. The difference in the two yields is observed because some inconsistencies might creep into the experiment being done. There are a few types of errors which results in the difference between the two yields.

The formula for calculating the percent yield is as follows:

${\rm{Percent yield}} = \frac{{{\rm{Actual}}\,{\rm{yield}}}}{{{\rm{Theoretical}}\,{\rm{yield}}}} \times 100$

Calculate the number of moles of camphor as follows:

$\begin{array}{c}\\{\rm{Number}}\,{\rm{of}}\,{\rm{moles}} = \frac{{{\rm{Mass}}}}{{{\rm{Molar}}\,{\rm{mass}}}}\\\\ = \frac{{1.7\,{\rm{g}}}}{{152.2\,{\rm{g}}/{\rm{mol}}}}\\\\ = 0.0112\,{\rm{mol}}\\\end{array}$

Calculate the theoretical yield of isoborneol as follows:

$\begin{array}{c}\\{\rm{Theoretical yield}} = {\rm{Number}}\,{\rm{of}}\,{\rm{moles}} \times {\rm{Molar mass}}\,{\rm{of}}\,{\rm{isoborneol}}\\\\ = 0.0112\,{\rm{mol}} \times 154.{\rm{3}}\,{\rm{g}}/{\rm{mol}}\\\\ = 1.73\,{\rm{g}}\\\end{array}$

Calculate the percent yield of the reaction as follows:

$\begin{array}{c}\\{\rm{Percent yield}} = \frac{{{\rm{Actual}}\,{\rm{yield}}}}{{{\rm{Theoretical}}\,{\rm{yield}}}} \times 100\\\\ = \frac{{1.3\,{\rm{g}}}}{{1.73\,{\rm{g}}}} \times 100\\\\ = 75.1\% \\\end{array}$

Ans:

The percent yield of the reaction is $75.1\%$ .