Question

banana oil (c₇h₁₄o₂) is created by adding acetic acid (c₂h₄o₂) and isopentanol (c₅h₁₂o) together, see the reaction below. what is the percent yield if you combine 55.85 g of acetic acid with 84.35 g of isopentanol to get 93.29 g of banana oil? solve to 4 significant figures.
c₂h4o₂ + c₅h₁₂o → c₇h₁₄o₂ + h₂o
acetic acid isopentanol banana oil
table of molar masses
mₐₙₑₜᵢₑ ₐₙₑ = 60.049 g/mol
mᵢₛₒₚₑₙₜₐₙₒₗ = 88.143 g/mol
mᵦₐₙₐₙₐ ₒₗₗ = 130.177 g/mol
mₕ₂ₒ = 18.015 g/mol
answer

Explanation:

Step1: Calculate moles of reactants

• Moles of acetic acid: $n_{acetic\ acid}=\frac{m}{M}=\frac{55.85\ g}{60.049\ g/mol}\approx0.9301\ mol$
• Moles of isopentanol: $n_{isopentanol}=\frac{m}{M}=\frac{84.35\ g}{88.143\ g/mol}\approx0.9570\ mol$

Step2: Determine limiting reactant

From the balanced equation, the mole ratio of acetic acid to isopentanol is \(1:1\). Since \(n_{acetic\ acid}

Step3: Calculate theoretical yield of banana oil

Using the mole ratio of acetic acid to banana oil (\(1:1\)), moles of banana oil \(n_{banana\ oil}=n_{acetic\ acid} = 0.9301\ mol\)
Theoretical mass of banana oil: \(m_{theoretical}=n\times M=0.9301\ mol\times130.177\ g/mol\approx121.1\ g\)

Step4: Calculate percent yield

Percent yield \(=\frac{m_{actual}}{m_{theoretical}}\times100\%=\frac{93.29\ g}{121.1\ g}\times100\%\approx77.04\%\)

Answer:

\(77.04\%\)