Methylation of 10 g of benzene gave 9.2 g of toluene. Calculate the percentage yield of toluene ....

Q: Why do we need to calculate the theoretical yield?

The theoretical yield tells us the maximum possible amount of product we could get if the reaction was 100% efficient. We need it as a baseline to compare our actual yield against.

Q: How did you get the molar masses of 78 and 92?

Benzene is $C_6H_6$, so $(6 \times 12) + (6 \times 1) = 78 \text{ g/mol}$. Toluene is $C_6H_5CH_3$, so $(7 \times 12) + (8 \times 1) = 92 \text{ g/mol}$.

Q: Does the methyl chloride ($CH_3Cl$) affect the yield calculation?

In this problem, benzene is the limiting reagent because its mass is given and we assume $CH_3Cl$ is in excess. The yield is always based on the limiting reagent.

Q: Why is the actual yield always less than the theoretical yield?

In real experiments, side reactions occur, some product is lost during transfer or purification, and reactions may not go to completion.

Q: Can percentage yield ever be over 100%?

Theoretically, no. If you calculate a yield over 100%, it usually means your product is wet (contains solvent) or contaminated with impurities.

Question

Methylation of $10	ext{ g}$ of benzene gave $9.2	ext{ g}$ of toluene. Calculate the percentage yield of toluene ......... . (Nearest integer)

Sigma Percentile · Accepted Answer

### The Realities of the Chemistry Lab

Imagine you are standing in an organic chemistry lab, wearing your safety goggles and lab coat. You carefully weigh out exactly $10	ext{ g}$ of benzene and react it with methyl chloride.

In a perfect, mathematically ideal universe, every single molecule of benzene would transform into your desired product, toluene. This utopian scenario gives us what we call the **theoretical yield**.

However, reality is messy. Side reactions occur, some product sticks to the glassware, and purification steps always lead to minor losses. The amount you actually manage to isolate and weigh at the end is your **actual yield**. In this problem, our actual yield is given as $9.2	ext{ g}$ of toluene.

### Analyzing the Setup

The reaction we are performing is the classic **Friedel-Crafts Alkylation**.

Benzene ($	ext{C}_6	ext{H}_6$) reacts with methyl chloride ($	ext{CH}_3	ext{Cl}$) in the presence of an anhydrous aluminum chloride ($	ext{AlCl}_3$) catalyst to form toluene ($	ext{C}_6	ext{H}_5	ext{CH}_3$).

Looking at the balanced chemical equation, the stoichiometry is beautifully simple: **one mole of benzene produces exactly one mole of toluene**.

### The Master Equation

To find out how efficient our reaction was, we use the percentage yield formula.

$$ 	ext{\% Yield} = \left( \frac{	ext{Actual Yield}}{	ext{Theoretical Yield}} ight) 	imes 100 $$

We already have the actual yield ($9.2	ext{ g}$). Our mission now is to calculate the theoretical yield.

### Calculating the Theoretical Yield

First, we need to determine how many moles of benzene we started with. The molar mass of benzene is $(6 	imes 12) + (6 	imes 1) = 78	ext{ g/mol}$.

$$ n_{	ext{benzene}} = \frac{10	ext{ g}}{78	ext{ g/mol}} = \frac{10}{78}	ext{ mol} $$

Since the molar ratio is $1:1$, the theoretical moles of toluene produced will also be $\frac{10}{78}	ext{ mol}$.

Now, we convert these moles back into a mass. The molar mass of toluene is $(7 	imes 12) + (8 	imes 1) = 92	ext{ g/mol}$.

$$ 	ext{Theoretical Mass} = \left( \frac{10}{78}	ext{ mol} ight) 	imes 92	ext{ g/mol} $$

### Final Calculation

Now, we bring it all together. Let's substitute our values into the master equation.

$$ 	ext{\% Yield} = \frac{9.2}{\left( \frac{10 	imes 92}{78} ight)} 	imes 100 $$

Don't rush to multiply everything out! Let's rearrange the fraction to reveal a beautiful cancellation. The $78$ flips to the numerator.

$$ 	ext{\% Yield} = \frac{9.2 	imes 78}{92 	imes 10} 	imes 100 $$

Notice the relationship between $9.2$ and $92$?

$$ \frac{9.2}{92} = 0.1 $$

And in the denominator, we have a $10$. The $0.1$ and the $10$ perfectly cancel each other out!

$$ 	ext{\% Yield} = 78\% $$

Our final answer is **$78$**. This means $78\%$ of our starting benzene successfully converted into our isolated toluene product.

Animated Solution for Chemistry - Basic Concepts in Chemistry: Methylation of $10 g$ of benzene gave $9.2 g$ of toluene. Calculate the percentage yield of toluene ......... . (Nearest integer)

Enter Numerical Value:

Visualized Solution

Reaction Setup

Percentage Yield Formula

Moles of Benzene

Theoretical Mass of Toluene

Substituting Values

Simplifying the Expression

Final Answer

Practical Limitations

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Realities of the Chemistry Lab

Analyzing the Setup

The Master Equation

Calculating the Theoretical Yield

Final Calculation

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