In the above reaction 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yie

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

In real-world chemistry, side reactions occur, some product is lost during purification or transfer between flasks, and some reactions simply do not go to 100% completion.

Q: How do we know the stoichiometric ratio is 1:1?

By looking at the balanced chemical equation for the nitration of benzene, one molecule of benzene reacts with one molecule of nitric acid to substitute one hydrogen atom with one nitro group, producing one molecule of nitrobenzene.

Q: What role does sulfuric acid play in this reaction?

Sulfuric acid acts as a catalyst. It protonates the nitric acid, leading to the formation of the highly reactive nitronium ion ($NO_2^+$), which is the actual electrophile that attacks the benzene ring.

Q: Can the percentage yield ever be greater than 100%?

Theoretically, no. If you calculate a yield over 100%, it usually means your product is impure (e.g., it still contains solvent or unreacted starting material), or there was a calculation error.

Q: Why do we use moles instead of just comparing masses directly?

Chemical reactions happen atom by atom, molecule by molecule, not gram by gram. Moles allow us to count the exact number of molecules reacting, ensuring we respect the law of conservation of mass correctly.

Question

In the above reaction 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is ……… % (Round off to the nearest integer).
(Given, atomic mass C : 12.0 u, H : 1.0 u, O : 16.0 u, N : 14.0 u)

Sigma Percentile · Accepted Answer

Welcome to the fascinating world of chemical stoichiometry! Imagine you are baking a cake. The recipe says you should get exactly 10 slices. But when you actually bake it, some batter sticks to the bowl, some crumbs fall off, and you end up with only 8 perfect slices. Chemistry works exactly the same way!

In this problem, we are performing the nitration of benzene. We start with a specific amount of reactant, and we want to find out how efficient our reaction was. This efficiency is what chemists call the **percentage yield**.

### The Reality of Chemical Reactions

We are given $3.9 	ext{ g}$ of benzene ($C_6H_6$). When it reacts with a mixture of nitric acid and sulfuric acid, it forms nitrobenzene ($C_6H_5NO_2$). The problem states that we actually obtained $4.92 	ext{ g}$ of nitrobenzene. This is our **actual yield**—the real-world result of our experiment.

But to know how good this result is, we need to know the **theoretical yield**. What is the absolute maximum amount of nitrobenzene we could have created if the universe were perfect and no molecules were lost?

### Calculating the Theoretical Maximum

To find the theoretical yield, we must first speak the language of chemistry: **moles**.

Let's calculate the molar mass of benzene ($C_6H_6$):
$$M_{	ext{benzene}} = (6 	imes 12.0) + (6 	imes 1.0) = 78 	ext{ g/mol}$$

Now, we find the number of moles of benzene we started with:
$$n_{	ext{benzene}} = \frac{	ext{Given Mass}}{	ext{Molar Mass}} = \frac{3.9}{78} = 0.05 	ext{ mol}$$

According to the balanced chemical equation, one mole of benzene produces exactly one mole of nitrobenzene. Therefore, if we start with $0.05 	ext{ mol}$ of benzene, we should theoretically produce $0.05 	ext{ mol}$ of nitrobenzene.

Let's convert these theoretical moles back into a tangible mass. First, we need the molar mass of nitrobenzene ($C_6H_5NO_2$):
$$M_{	ext{nitrobenzene}} = (6 	imes 12.0) + (5 	imes 1.0) + 14.0 + (2 	imes 16.0) = 123 	ext{ g/mol}$$

Now, we calculate the theoretical mass:
$$W_{	ext{theoretical}} = n 	imes M_{	ext{nitrobenzene}} = 0.05 	imes 123 = 6.15 	ext{ g}$$

This $6.15 	ext{ g}$ is our theoretical yield. It is the "10 slices of cake" from our recipe.

### The Final Verdict: Percentage Yield

We expected $6.15 	ext{ g}$, but we only got $4.92 	ext{ g}$. How efficient were we? We use the percentage yield formula:
$$\%	ext{ Yield} = \frac{W_{	ext{actual}}}{W_{	ext{theoretical}}} 	imes 100$$

Substituting our values:
$$\%	ext{ Yield} = \frac{4.92}{6.15} 	imes 100$$

When we perform this final calculation, we get exactly $80\%$.

**The percentage yield of nitrobenzene is 80%.**

This means our reaction was 80% efficient, which is actually quite a respectable yield in organic synthesis! Always remember, mastering stoichiometry isn't just about plugging numbers into formulas; it's about understanding the physical reality of atoms transforming and conserving mass.

Enter Numerical Value:

Visualized Solution

Reaction Setup

Moles of Reactant

Calculating Moles of Benzene

Moles of Benzene

Theoretical Mass of Product

Molar Mass of Nitrobenzene

Theoretical Yield

Percentage Yield Formula

Substituting Yield Values

Final Percentage Yield

Conclusion

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Reality of Chemical Reactions

Calculating the Theoretical Maximum

The Final Verdict

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