Help with calculations of Greener Bromination of Trans-stilbene Synthesizing trans-1,2-dibromo-1,2-diphenylethane My initial mass of Trans-stilbene = 0.2033 grams

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Help with calculations of Greener Bromination of Trans-stilbene Synthesizing trans-1,2-dibromo-1,2-diphenylethane My initial mass of Trans-stilbene = 0.2033 grams At the end of the experiment, the mass of the crystal yielded 0.2211 grams Can you please help me fill in the ??? blanks? Calculations: Theoretical Yield = 0.2 grams E-Stilbene 1 mol E-Stilbene 180.25 grams E-Stilbene × 1 mol C14H12Br2 1 mol C14H12Br2 = 0.0011 moles C14H12Br2 Actual Yield = grams C14H12Br21 × 1 mol C14H12Br2 340.058 grams C14H12Br2 = 0.000159 moles C14H12Br2 Chemical Percent % Yield = Actual Yield/Theoretical Yield × 100% = 0.000159 moles/0.0011 moles × 100% = ???% Atom Economy: Atom economy = MW of products/MW of reactants × 100% = 340 grams C14H12Br2 / (180.25 grams E-Stilbene + 159.808 grams Bromine) × 100% = 100% Experimental atom economy = ????? grams C14H12Br2 / 0.2 grams E-stilbene + ?????? grams Bromine × 100% = ???% ???? ´ ???? = ????% ≈

The Correct Answer and Explanation is:

Let’s break down the calculations for your experiment step by step. You already have some values provided, so I’ll walk you through completing the missing pieces.

Theoretical Yield Calculation:

  1. Starting with Trans-stilbene (C14H12):
    • Given mass = 0.2033 grams.
    • Molar mass of Trans-stilbene = 180.25 g/mol.
    Using the formula: Moles of Trans-stilbene=Mass of Trans-stilbeneMolar mass of Trans-stilbene=0.2033 g180.25 g/mol=0.00113 mol.\text{Moles of Trans-stilbene} = \frac{\text{Mass of Trans-stilbene}}{\text{Molar mass of Trans-stilbene}} = \frac{0.2033 \, \text{g}}{180.25 \, \text{g/mol}} = 0.00113 \, \text{mol}.Moles of Trans-stilbene=Molar mass of Trans-stilbeneMass of Trans-stilbene​=180.25g/mol0.2033g​=0.00113mol.
  2. Moles of Product (C14H12Br2):
    • From the stoichiometry of the reaction, 1 mol of Trans-stilbene reacts with 1 mol of Bromine to produce 1 mol of C14H12Br2.
    • Thus, 0.00113 mol of Trans-stilbene will produce 0.00113 mol of C14H12Br2 (theoretical yield).
  3. Theoretical Yield in grams:
    • Molar mass of C14H12Br2 = 340.058 g/mol.
    Theoretical yield (grams)=Moles of product×Molar mass of product=0.00113 mol×340.058 g/mol=0.384 g.\text{Theoretical yield (grams)} = \text{Moles of product} \times \text{Molar mass of product} = 0.00113 \, \text{mol} \times 340.058 \, \text{g/mol} = 0.384 \, \text{g}.Theoretical yield (grams)=Moles of product×Molar mass of product=0.00113mol×340.058g/mol=0.384g. (You mentioned a theoretical yield of 0.2 g in your question, so double-check if this is the number you’d like to use, but this is the result based on stoichiometry).

Actual Yield Calculation:

You have the actual yield of the product, which is 0.2211 grams. To find the moles of C14H12Br2, we can use the molar mass of C14H12Br2. Moles of C14H12Br2=Mass of C14H12Br2Molar mass of C14H12Br2=0.2211 g340.058 g/mol=0.00065 mol.\text{Moles of C14H12Br2} = \frac{\text{Mass of C14H12Br2}}{\text{Molar mass of C14H12Br2}} = \frac{0.2211 \, \text{g}}{340.058 \, \text{g/mol}} = 0.00065 \, \text{mol}.Moles of C14H12Br2=Molar mass of C14H12Br2Mass of C14H12Br2​=340.058g/mol0.2211g​=0.00065mol.

Percent Yield Calculation:

Percent Yield=Actual YieldTheoretical Yield×100=0.00065 mol0.00113 mol×100=57.5%.\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 = \frac{0.00065 \, \text{mol}}{0.00113 \, \text{mol}} \times 100 = 57.5\%.Percent Yield=Theoretical YieldActual Yield​×100=0.00113mol0.00065mol​×100=57.5%.

Atom Economy Calculation:

  1. Molar mass of reactants:
    • Molar mass of Trans-stilbene (C14H12) = 180.25 g/mol.
    • Molar mass of Bromine (Br2) = 159.808 g/mol.
    • Total mass of reactants = 180.25 + 159.808 = 340.058 g/mol.
  2. Atom economy formula: Atom Economy=Molar mass of productMolar mass of reactants×100=340.058 g/mol340.058 g/mol×100=100%.\text{Atom Economy} = \frac{\text{Molar mass of product}}{\text{Molar mass of reactants}} \times 100 = \frac{340.058 \, \text{g/mol}}{340.058 \, \text{g/mol}} \times 100 = 100\%.Atom Economy=Molar mass of reactantsMolar mass of product​×100=340.058g/mol340.058g/mol​×100=100%.

Experimental Atom Economy:

For this part, let’s complete the experimental atom economy.

  • The experimental mass of C14H12Br2 = 0.2211 g (which you already have).
  • The mass of Bromine used is: Mass of Bromine=Mass of C14H12Br2−Mass of Trans-stilbene=0.2211 g−0.2033 g=0.0178 g.\text{Mass of Bromine} = \text{Mass of C14H12Br2} – \text{Mass of Trans-stilbene} = 0.2211 \, \text{g} – 0.2033 \, \text{g} = 0.0178 \, \text{g}.Mass of Bromine=Mass of C14H12Br2−Mass of Trans-stilbene=0.2211g−0.2033g=0.0178g.

Thus, the Experimental Atom Economy is: Experimental Atom Economy=0.2211 g0.2033 g+0.0178 g×100=0.22110.2211×100=100%.\text{Experimental Atom Economy} = \frac{0.2211 \, \text{g}}{0.2033 \, \text{g} + 0.0178 \, \text{g}} \times 100 = \frac{0.2211}{0.2211} \times 100 = 100\%.Experimental Atom Economy=0.2033g+0.0178g0.2211g​×100=0.22110.2211​×100=100%.

Final Summary:

  • Theoretical Yield = 0.384 grams.
  • Actual Yield = 0.2211 grams.
  • Percent Yield = 57.5%.
  • Atom Economy = 100%.
  • Experimental Atom Economy = 100%.

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