A reaction was performed in which 0.47 g of 2-naphthol was reacted with a slight excess of 1-bromobutane to make 0.29 g of 2-butoxynaphthalene. Calculate the theoretical yield and percent yield for this reaction.
2-naphthol + 1-bromobutane --> 2-butoxynaphthalene
Convert 0.47g 2-naphthol to mols. mol = g/molar mass.
Using the coefficients in the balanced equation, convert mols 2-naphthol to mols 2-butoxynaphthalene.
Convert mol 2-butoxynaphthalene to grams. g = mols x molar mass. This is the theoretical yield.
%yield = (actual yield/theor yield)*100 = ?
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To calculate the theoretical yield and percent yield for the reaction, we need to compare the actual yield (0.29 g) with the theoretical yield.
1. Calculate the molar mass of 2-naphthol:
2-naphthol (C10H8O) = 144.18 g/mol
2. Calculate the number of moles of 2-naphthol:
moles = mass / molar mass = 0.47 g / 144.18 g/mol = 0.00326 mol
3. Balance the chemical equation:
2-naphthol + 1-bromobutane → 2-butoxynaphthalene
4. Calculate the molar mass of 2-butoxynaphthalene:
2-butoxynaphthalene (C12H12O) = 172.23 g/mol
5. Use the balanced equation to determine the stoichiometry between 2-naphthol and 2-butoxynaphthalene. From the balanced equation, we see that 1 mol of 2-naphthol produces 1 mol of 2-butoxynaphthalene.
6. Calculate the theoretical yield of 2-butoxynaphthalene:
theoretical yield = moles of 2-naphthol × molar mass of 2-butoxynaphthalene
= 0.00326 mol × 172.23 g/mol = 0.560 g
7. Calculate the percent yield:
percent yield = (actual yield / theoretical yield) × 100
= (0.29 g / 0.560 g) × 100
= 51.79%
Therefore, the theoretical yield is 0.560 g and the percent yield is 51.79%.
To calculate the theoretical yield for this reaction, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, we need to convert the given masses of reactants into moles. We can do this by dividing the mass of each reactant by its molar mass. The molar mass of 2-naphthol can be found in the periodic table or a chemical reference book.
Molar mass of 2-naphthol (C10H8O) = 10(12.01 g/mol) + 8(1.01 g/mol) + 16.00 g/mol = 144.18 g/mol
Moles of 2-naphthol = 0.47 g / 144.18 g/mol = 0.00326 mol
Next, we need to determine the moles of 1-bromobutane. Similarly, we can find the molar mass of 1-bromobutane (C4H9Br) and use it to convert the mass of 1-bromobutane into moles.
Molar mass of 1-bromobutane = 4(12.01 g/mol) + 9(1.01 g/mol) + 79.90 g/mol = 137.03 g/mol
Moles of 1-bromobutane = (mass of 1-bromobutane) / (molar mass of 1-bromobutane) = (assumed mass - mass of product) / (molar mass of 1-bromobutane)
= (0.47 g - 0.29 g) / 137.03 g/mol = 0.00131 mol
Next, we can write the balanced chemical equation for the reaction:
2-naphthol + 1-bromobutane --> 2-butoxynaphthalene
From the balanced equation, we can see that the stoichiometric ratio between 2-naphthol and 2-butoxynaphthalene is 1:1. This means that every one mole of 2-naphthol reacts to produce one mole of 2-butoxynaphthalene.
Therefore, the theoretical yield of 2-butoxynaphthalene can be calculated using the moles of 2-naphthol reacted:
Theoretical yield = moles of 1-bromobutane x (molar mass of 2-butoxynaphthalene) / (molar mass of 1-bromobutane)
= 0.00131 mol x (156.19 g/mol) / (137.03 g/mol)
= 0.00149 mol
Now, we can convert the theoretical yield from moles to grams by multiplying it by the molar mass of 2-butoxynaphthalene:
Theoretical yield = moles of 2-butoxynaphthalene x (molar mass of 2-butoxynaphthalene)
= 0.00149 mol x 156.19 g/mol
= 0.233 g
The theoretical yield for this reaction is 0.233 g.
To calculate the percent yield, we need to divide the actual yield (0.29 g) by the theoretical yield (0.233 g) and multiply by 100:
Percent yield = (actual yield / theoretical yield) x 100
= (0.29 g / 0.233 g) x 100
= 124.46%
Therefore, the percent yield for this reaction is approximately 124.46%.