Acetic acid (CH3COOH) reacts with isopentyl alcohol (C5H12O) to yield the ester isopentyl acetate which has the odour of bananas. If the yield from the reaction is 45%, how many grams of isopentyl acetate are formed when 3.77 g of acetic acid and 7.46 grams of isopentyl alcohol are reacted?
CH3COOH + C5H12O �¨ C7H14O2 + H2O
This is a limiting reagent problem. How do I know? Because BOTH reactants are given. Limiting reagent problems are solved, basically, by working two simple stoichiometry problems as follows:
1. Convert grams CH3COOH to moles. moles = grams/molar mass.
2. Convert grams C6H12O to moles with the same process.
3a. Using the coefficients in the balanced equation, convert moles moles CH3COOH to moles of the ester.
3b. Same for moles C6H12O.
(Note: Steps 1 and 3a are one simple stoichiometry problem; steps 2 and 3b are the other.)
(Note: Then you decide WHICH of the simple stoichiometry problems is correct.)
3c. The answer from 3a and 3b likely will be different which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
4.Using the smaller value from step 3c, convert moles to grams. g = moles x molar mass. That is the theoretical yield.
5. %yield = (actual yield/theoretical yield)* 100 = ??
You know %yield and theoretical yield, solve for actual yield.
To determine the number of grams of isopentyl acetate formed, we need to follow these steps:
Step 1: Determine the limiting reactant
To find the limiting reactant, we compare the moles of acetic acid and isopentyl alcohol used in the reaction.
First, calculate the number of moles of acetic acid:
Molar mass of acetic acid (CH3COOH) = 60.05 g/mol
Moles of acetic acid = mass of acetic acid / molar mass of acetic acid
Moles of acetic acid = 3.77 g / 60.05 g/mol = 0.0628 mol
Next, calculate the number of moles of isopentyl alcohol:
Molar mass of isopentyl alcohol (C5H12O) = 88.15 g/mol
Moles of isopentyl alcohol = mass of isopentyl alcohol / molar mass of isopentyl alcohol
Moles of isopentyl alcohol = 7.46 g / 88.15 g/mol = 0.0846 mol
Comparing the mole ratios of acetic acid and isopentyl alcohol in the balanced equation (1:1), it can be seen that acetic acid is the limiting reactant since it has fewer moles.
Step 2: Calculate the moles of isopentyl acetate formed
Since acetic acid is the limiting reactant, the number of moles of isopentyl acetate formed is equal to the number of moles of acetic acid.
Moles of isopentyl acetate formed = moles of acetic acid = 0.0628 mol
Step 3: Calculate the mass of isopentyl acetate formed
To determine the mass of isopentyl acetate formed, we need to multiply the number of moles by the molar mass of isopentyl acetate.
Molar mass of isopentyl acetate (C7H14O2) = 130.19 g/mol
Mass of isopentyl acetate = moles of isopentyl acetate formed × molar mass of isopentyl acetate
Mass of isopentyl acetate = 0.0628 mol × 130.19 g/mol = 8.1802 g
Step 4: Calculate the yield of isopentyl acetate
The yield of isopentyl acetate is given as 45%.
Yield of isopentyl acetate = (mass of isopentyl acetate formed / theoretical mass of isopentyl acetate) × 100
Theoretical mass of isopentyl acetate would be the mass calculated in step 3, so:
Yield of isopentyl acetate = (8.1802 g / 8.1802 g) × 100 = 100%
Therefore, the number of grams of isopentyl acetate formed when 3.77 g of acetic acid and 7.46 g of isopentyl alcohol are reacted is approximately 8.1802 g.
To find the grams of isopentyl acetate formed, we first need to determine the limiting reagent, which is the reactant that will be completely used up and determine the maximum amount of product that can be formed.
Step 1: Calculate the number of moles of acetic acid (CH3COOH):
Given mass of acetic acid = 3.77 g
Molar mass of acetic acid (CH3COOH) = 60.05 g/mol
Number of moles of CH3COOH = mass / molar mass
= 3.77 g / 60.05 g/mol
≈ 0.0628 mol
Step 2: Calculate the number of moles of isopentyl alcohol (C5H12O):
Given mass of isopentyl alcohol = 7.46 g
Molar mass of isopentyl alcohol (C5H12O) = 88.15 g/mol
Number of moles of C5H12O = mass / molar mass
= 7.46 g / 88.15 g/mol
≈ 0.0846 mol
Step 3: Calculate the number of moles of isopentyl acetate (C7H14O2) formed by the limiting reagent:
From the balanced chemical equation, the stoichiometric ratio between CH3COOH and C7H14O2 is 1:1.
The limiting reagent is the one that produces fewer moles of product, which is CH3COOH in this case. Therefore, the number of moles of C7H14O2 formed will be equal to the number of moles of CH3COOH.
Number of moles of C7H14O2 formed ≈ 0.0628 mol
Step 4: Calculate the mass of isopentyl acetate (C7H14O2) formed:
Molar mass of isopentyl acetate (C7H14O2) = 130.19 g/mol
Mass of C7H14O2 formed = number of moles of C7H14O2 formed × molar mass
= 0.0628 mol × 130.19 g/mol
≈ 8.20 g
Step 5: Calculate the actual yield based on the given percent yield:
Given percent yield = 45%
Actual yield = percent yield × theoretical yield
= 0.45 × 8.20 g
≈ 3.69 g
Therefore, approximately 3.69 grams of isopentyl acetate are formed when 3.77 g of acetic acid and 7.46 grams of isopentyl alcohol are reacted.