Question 4.14: Adjusting the Quantities of Reactants in Accordance with the......
Adjusting the Quantities of Reactants in Accordance with the Percent Yield of a Reaction
When heated with sulfuric or phosphoric acid, cyclohexanol, C_{6}H_{11}OH, is converted to cyclohexene, C_{6}H_{10}. Ball-and-stick models of cyclohexanol and cyclohexene are shown here. The balanced chemical equation for the reaction is shown below.
C_{6}H_{11}OH(l) → C_{6}H_{10}(l) + H_{2}O(l) (4.7)
If the percent yield is 83%, what mass of cyclohexanol must we use to obtain 25 g of cyclohexene?
Analyze
We are given the percent yield (83%) and the actual yield of C_{6}H_{10} (25 g). We can use equation (4.6) to calculate the theoretical yield of C_{6}H_{10}. Then, we can calculate the quantity of C_{6}H_{11}OH required to produce the theoretical yield of C_{6}H_{10} by using the following series of conversions: g C_{6}H_{10} → mol C_{6}H_{10} → mol C_{6}H_{11}OH → g C_{6}H_{11}OH.
percent yield = \frac{actual yield}{theoretical yield} × 100% (4.6)
Solve
Let’s use a stepwise approach.
Rearrange equation (4.6) and calculate the theoretical yield. theoretical yield = \frac{25 g × 100\%}{83\%} = 3.0 × 10¹ g
The theoretical yield is expressed as 3.0 × 10¹ g rather than as 30 g to emphasize that there are two significant figures in the calculated result. If we expressed the result as 30 g, then the number of significant figures would be ambiguous. (See Section 1-7.) The remaining conversions are as follows.
Convert from g C_{6}H_{10} to mol C_{6}H_{10} by using the molar mass of C_{6}H_{10}. ? mol C_{6}H_{10} = 3.0 × 10¹ g C_{6}H_{10} × \frac{1 mol C_{6}H_{10}}{82.1 g C_{6}H_{10}} = 0.37 mol C_{6}H_{10}
Convert from mol C_{6}H_{10} to mol C_{6}H_{11}OH using the stoichiometric factor. ? mol C_{6}H_{11}OH = 0.37 mol C_{6}H_{10} × \frac{1 mol C_{6}H_{11}OH}{1 mol C_{6}H_{10}} = 0.37 mol C_{6}H_{11}OH
Convert from mol C_{6}H_{11}OH to g C_{6}H_{11}OH using the molar mass of C_{6}H_{11}OH. ? g C_{6}H_{11}OH = 0.37 mol C_{6}H_{11}OH × \frac{100.2 g C_{6}H_{11}OH}{1 mol C_{6}H_{11}OH} = 37 g C_{6}H_{11}OH
We could have combined the last three steps into a single line calculation. The single line calculation is shown below, starting from a theoretical yield of 3.0 × 10¹ g C_{6}H_{10}.
? g C_{6}H_{11}OH = 30 g C_{6}H_{10} × \frac{1 mol C_{6}H_{10}}{82.1 g C_{6}H_{10}} × \frac{1 mol C_{6}H_{11}OH}{1 mol C_{6}H_{10}} × \frac{100.2 g C_{6}H_{11}OH}{1 mol C_{6}H_{11}OH} = 37 g C_{6}H_{11}OH
Assess
Because we have already calculated the molar masses of C_{6}H_{10} and C_{6}H_{11}OH, it is a simple matter to work the problem in reverse by using numbers that are slightly rounded off. A 37 g sample of C_{6}H_{11}OH contains approximately 37/100 = 0.37 moles of C_{6}H_{11}OH, and we expect a maximum of 0.37 moles of C_{6}H_{10} or 0.37 × 82 ≈ 30 g C_{6}H_{10}. We obtain only 25 g C_{6}H_{10} and so the percent yield for the experiment is approximately (25/30) × 100% = 83%. By working the problem in reverse, we verify that the percent yield is 83%; thus, we are confident we have solved the problem correctly.