1. Calculate the percent yield of 1- bromobutane obtained when 15.4 g of butan-1-ol was used, and 22.5 g of 1-bromobutane was obtained after purification (NB. The limiting reagent in the production of 1-bromobutane is 1-butanol).
how many moles in 15.4g?
how many moles expected? how many actually produced?
% yield = produced/expected
14
To calculate the percent yield, you need to use the following formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
In this case, the actual yield is given as 22.5 g of 1-bromobutane and the limiting reagent is 1-butanol. To calculate the theoretical yield, you need to determine the stoichiometry of the reaction between 1-butanol and 1-bromobutane.
The balanced equation for the reaction is:
1-butanol + HBr -> 1-bromobutane + H2O
From the equation, you can see that the molar ratio between 1-butanol and 1-bromobutane is 1:1.
Step 1: Convert the mass of 1-butanol to moles.
Molar Mass of 1-butanol = 74.12 g/mol
Moles of 1-butanol = Mass / Molar Mass
Moles of 1-butanol = 15.4 g / 74.12 g/mol
Step 2: Convert the moles of 1-butanol to moles of 1-bromobutane.
Moles of 1-bromobutane = Moles of 1-butanol
Step 3: Convert the moles of 1-bromobutane to grams.
Molar Mass of 1-bromobutane = 137.02 g/mol
Mass of 1-bromobutane = Moles of 1-bromobutane * Molar Mass of 1-bromobutane
Step 4: Calculate the percent yield.
Percent Yield = (Actual Yield / Theoretical Yield) x 100
Percent Yield = (22.5 g / Mass of 1-bromobutane) x 100
Now, substitute the values into the equation and solve for the percent yield.
To calculate the percent yield of 1-bromobutane, we need to compare the actual yield (22.5 g) to the theoretical yield. The theoretical yield is the amount of 1-bromobutane that should be produced based on the stoichiometry of the reaction and assuming complete conversion of the limiting reagent.
1. First, determine the balanced chemical equation for the reaction. From the information given, it can be deduced that the balanced equation for the conversion of 1-butanol to 1-bromobutane is:
C4H9OH → C4H9Br
2. Next, calculate the molar mass of 1-butanol and 1-bromobutane.
Molar mass of 1-butanol (C4H9OH) = 74.12 g/mol
Molar mass of 1-bromobutane (C4H9Br) = 137.03 g/mol
3. Convert the given masses of butan-1-ol and 1-bromobutane into moles.
Moles of 1-butanol = mass / molar mass = 15.4 g / 74.12 g/mol
Moles of 1-bromobutane = mass / molar mass = 22.5 g / 137.03 g/mol
4. The stoichiometry of the balanced equation tells us that the molar ratio between 1-butanol and 1-bromobutane is 1:1. Therefore, the moles of 1-butanol used should be the same as the moles of 1-bromobutane produced.
5. Since the limiting reagent is 1-butanol, the moles of 1-bromobutane produced should be equal to the moles of 1-butanol used.
6. Now, we can calculate the theoretical yield of 1-bromobutane.
Theoretical yield = moles of 1-bromobutane produced × molar mass of 1-bromobutane
7. Calculate the theoretical yield using the formula,:
Theoretical yield = moles of 1-bromobutane × molar mass of 1-bromobutane
8. Finally, calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.
Percent yield = (Actual yield / Theoretical yield) × 100
By following these steps and plugging in the given values, you can calculate the percent yield of 1-bromobutane obtained in this reaction.