Dibenzalacetone, C17H14O, is a common ingredient in sunscreens that can be prepared by the following chemical reaction.2 C7H6O + C3H6O ? C17H14O + 2 H2OIn a lab experiment, 4.25 g of C7H6O and 1.10 g of C3H6O were used to synthesize C17H14O. The target product was produced as a yellow solid with a 78.4% yield.What mass of C17H14O was isolated? Choose the closest answer. Relevant molecular masses:C7H6O: 106.12 g mol?1 C3H6O: 58.08 g mol?1 C17H14O: 234.29 g mol?1
The Correct Answer and Explanation is:
To solve this problem, we need to calculate the mass of C17H14O that was isolated from the reaction, considering the yield percentage. Here’s the step-by-step process:
Step 1: Write the reaction
The balanced chemical reaction is:2 C7H6O+C3H6O⟶C17H14O+2 H2O2 \, C_7H_6O + C_3H_6O \longrightarrow C_{17}H_{14}O + 2 \, H_2O2C7H6O+C3H6O⟶C17H14O+2H2O
Step 2: Calculate the moles of reactants
We are given the masses of C7H6O (4.25 g) and C3H6O (1.10 g). To find the moles of each reactant, we use the formula:moles=massmolar mass\text{moles} = \frac{\text{mass}}{\text{molar mass}}moles=molar massmass
- Molar mass of C7H6O = 106.12 g/mol
- Molar mass of C3H6O = 58.08 g/mol
For C7H6O:moles of C7H6O=4.25 g106.12 g/mol=0.0400 mol\text{moles of C7H6O} = \frac{4.25 \, \text{g}}{106.12 \, \text{g/mol}} = 0.0400 \, \text{mol}moles of C7H6O=106.12g/mol4.25g=0.0400mol
For C3H6O:moles of C3H6O=1.10 g58.08 g/mol=0.0189 mol\text{moles of C3H6O} = \frac{1.10 \, \text{g}}{58.08 \, \text{g/mol}} = 0.0189 \, \text{mol}moles of C3H6O=58.08g/mol1.10g=0.0189mol
Step 3: Determine the limiting reagent
From the balanced equation, we see that 2 moles of C7H6O react with 1 mole of C3H6O. Therefore, the molar ratio of C7H6O to C3H6O is 2:1.
- For 0.0400 mol of C7H6O, we would need:
0.0400 mol2=0.0200 mol of C3H6O\frac{0.0400 \, \text{mol}}{2} = 0.0200 \, \text{mol of C3H6O}20.0400mol=0.0200mol of C3H6O
- But we only have 0.0189 mol of C3H6O, so C3H6O is the limiting reagent.
Step 4: Calculate the theoretical yield of C17H14O
From the balanced equation, 1 mole of C3H6O produces 1 mole of C17H14O. Therefore, the moles of C17H14O produced will be equal to the moles of C3H6O used, which is 0.0189 mol.
To find the theoretical mass of C17H14O:mass of C17H14O=moles×molar mass of C17H14O\text{mass of C17H14O} = \text{moles} \times \text{molar mass of C17H14O}mass of C17H14O=moles×molar mass of C17H14Omass of C17H14O=0.0189 mol×234.29 g/mol=4.43 g\text{mass of C17H14O} = 0.0189 \, \text{mol} \times 234.29 \, \text{g/mol} = 4.43 \, \text{g}mass of C17H14O=0.0189mol×234.29g/mol=4.43g
Step 5: Account for the percent yield
The reaction yields 78.4% of the theoretical product. So, the actual mass of C17H14O isolated is:actual mass=4.43 g×78.4100=3.47 g\text{actual mass} = 4.43 \, \text{g} \times \frac{78.4}{100} = 3.47 \, \text{g}actual mass=4.43g×10078.4=3.47g
Final Answer:
The mass of C17H14O isolated is 3.47 g.
