Calculate the % yield for the reduction of Benzil using sodium borohydride (NaBH4) Show the balanced chemical equation for the reaction and how you determined the limiting reagent. Mass of benzil is 0.404 g. Mass of sodium borohydride is 0.0801 g. Mass of hydrobenzoin product is 0.239 g. Remember that boron donates 4 H-.
The Correct Answer and Explanation is :
Let’s first start by writing the balanced chemical equation for the reduction of benzil (C₆H₅CO)₂C=O to hydrobenzoin.
The reaction can be written as:
[
\text{Benzil (C₆H₅CO)₂C=O} + 2 \text{NaBH₄} \rightarrow \text{Hydrobenzoin} (C₆H₅CHOH)₂ + 2 \text{NaB(OH)₄}
]
In this reaction, sodium borohydride (NaBH₄) donates hydride ions (H⁻), which reduce the carbonyl groups (C=O) of benzil to secondary alcohol groups (C–OH), resulting in hydrobenzoin.
Step 1: Calculate the moles of Benzil and Sodium Borohydride
- Molecular weight of Benzil (C₁₄H₁₀O₂):
[
\text{Molar mass of benzil} = (14 \times 12) + (10 \times 1) + (2 \times 16) = 180 \, \text{g/mol}
]
Moles of benzil:
[
\text{moles of benzil} = \frac{0.404 \, \text{g}}{180 \, \text{g/mol}} = 0.00224 \, \text{mol}
] - Molecular weight of Sodium Borohydride (NaBH₄):
[
\text{Molar mass of NaBH₄} = 23 + 10 + (4 \times 1) = 37 \, \text{g/mol}
]
Moles of sodium borohydride:
[
\text{moles of NaBH₄} = \frac{0.0801 \, \text{g}}{37 \, \text{g/mol}} = 0.00217 \, \text{mol}
]
Step 2: Limiting Reagent
From the balanced equation, we see that 2 moles of NaBH₄ are required for each mole of benzil. Therefore, the required moles of NaBH₄ to react with 0.00224 mol of benzil would be:
[
\text{Required moles of NaBH₄} = 0.00224 \, \text{mol} \times 2 = 0.00448 \, \text{mol}
]
Since we only have 0.00217 mol of NaBH₄, sodium borohydride is the limiting reagent.
Step 3: Theoretical Yield of Hydrobenzoin
From the balanced equation, the mole ratio between NaBH₄ and hydrobenzoin is 2:1. Thus, the theoretical moles of hydrobenzoin formed would be:
[
\text{Moles of hydrobenzoin} = \frac{0.00217 \, \text{mol}}{2} = 0.001085 \, \text{mol}
]
The molar mass of hydrobenzoin (C₆H₆(OH)₂) is:
[
\text{Molar mass of hydrobenzoin} = 2(12) + 10 + 2(16) + 2(1) = 212 \, \text{g/mol}
]
The theoretical mass of hydrobenzoin is:
[
\text{Theoretical mass of hydrobenzoin} = 0.001085 \, \text{mol} \times 212 \, \text{g/mol} = 0.230 \, \text{g}
]
Step 4: Percent Yield
The actual yield of hydrobenzoin is 0.239 g, so the percent yield is:
[
\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{0.239 \, \text{g}}{0.230 \, \text{g}} \times 100 = 103.9\%
]
This percent yield suggests an error in the measurement or that the hydrobenzoin may contain some impurities.
I will generate a visual representation of the calculations for you now.
Here is a visual representation of the moles of benzil, sodium borohydride, and hydrobenzoin, along with the theoretical and actual yields for the reduction reaction. The graph shows the comparison of each component, providing a clear view of the limiting reagent (sodium borohydride) and the percent yield.
The percent yield was calculated to be approximately 103.9%, which indicates a potential issue with measurement or impurities in the hydrobenzoin product. Let me know if you need further clarifications!
