Dibenzalacetone, C17H14O, is a common ingredient in sunscreens that is prepared by the following chemical reaction: 2 C6H6O + C6H5CHO -> C17H14O + 2 H2O. In a lab experiment, 3.95 mL of C6H6O (molar mass 106.12 g mol-1; density = 1.094 g mL-1) and an excess of C6H5CHO were used to synthesize C17H14O (molar mass = 234.29 g mol-1). The target product was produced as a yellow solid with a mass of 3.85 g. What was the percent yield of C17H14O? Choose the closest answer. a) 56.6% b) 80.7% c) 96.6% d) 40.4% e) 20.4%
The Correct Answer and Explanation is:
To find the percent yield of C17H14O, we need to follow these steps:
Step 1: Calculate the moles of C6H6O used
The first step is to convert the volume of C6H6O into mass and then into moles. We are given:
- Volume of C6H6O = 3.95 mL
- Density of C6H6O = 1.094 g/mL
- Molar mass of C6H6O = 106.12 g/mol
First, calculate the mass of C6H6O:Mass of C6H6O=Volume×Density=3.95 mL×1.094 g/mL=4.32 g\text{Mass of C6H6O} = \text{Volume} \times \text{Density} = 3.95 \, \text{mL} \times 1.094 \, \text{g/mL} = 4.32 \, \text{g}Mass of C6H6O=Volume×Density=3.95mL×1.094g/mL=4.32g
Now, convert the mass into moles:Moles of C6H6O=MassMolar mass=4.32 g106.12 g/mol=0.0407 mol\text{Moles of C6H6O} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4.32 \, \text{g}}{106.12 \, \text{g/mol}} = 0.0407 \, \text{mol}Moles of C6H6O=Molar massMass=106.12g/mol4.32g=0.0407mol
Step 2: Determine the theoretical moles of C17H14O produced
From the balanced chemical equation:2 C6H6O+C6H5CHO→C17H14O+2 H2O2 \, \text{C6H6O} + \text{C6H5CHO} \rightarrow \text{C17H14O} + 2 \, \text{H2O}2C6H6O+C6H5CHO→C17H14O+2H2O
The molar ratio of C6H6O to C17H14O is 2:1. This means that for every 2 moles of C6H6O, 1 mole of C17H14O is produced.
So, the moles of C17H14O produced will be half the moles of C6H6O:Moles of C17H14O=0.0407 mol2=0.02035 mol\text{Moles of C17H14O} = \frac{0.0407 \, \text{mol}}{2} = 0.02035 \, \text{mol}Moles of C17H14O=20.0407mol=0.02035mol
Step 3: Calculate the theoretical mass of C17H14O
Now, convert the moles of C17H14O into mass using its molar mass:Mass of C17H14O=Moles×Molar mass=0.02035 mol×234.29 g/mol=4.77 g\text{Mass of C17H14O} = \text{Moles} \times \text{Molar mass} = 0.02035 \, \text{mol} \times 234.29 \, \text{g/mol} = 4.77 \, \text{g}Mass of C17H14O=Moles×Molar mass=0.02035mol×234.29g/mol=4.77g
Step 4: Calculate the percent yield
The percent yield is calculated by comparing the actual yield to the theoretical yield:Percent yield=(Actual yieldTheoretical yield)×100\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100Percent yield=(Theoretical yieldActual yield)×100
The actual yield is 3.85 g (given), and the theoretical yield is 4.77 g (calculated above):Percent yield=(3.85 g4.77 g)×100=80.7%\text{Percent yield} = \left( \frac{3.85 \, \text{g}}{4.77 \, \text{g}} \right) \times 100 = 80.7\%Percent yield=(4.77g3.85g)×100=80.7%
Final Answer:
The closest answer is b) 80.7%.
