Percent Yield Formula

Percent yield is the ratio of the actual yield obtained in an experiment to the theoretical yield predicted by stoichiometry, expressed as a percentage.

The Formula

\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

When to use: How much of the possible product you actually got — 100% is perfect, real reactions are always less.

Quick Example

Theoretical: 10g. Actual: 8g. Percent yield

= \frac{8}{10} \times 100 = 80\%

Notation

\%

denotes percent yield.

m_{\text{actual}}

is the mass of product obtained experimentally.

m_{\text{theoretical}}

is the maximum mass predicted by stoichiometry.

What This Formula Means

The ratio of the actual yield obtained in an experiment to the theoretical yield predicted by stoichiometry, expressed as a percentage.

How much of the possible product you actually got — 100% is perfect, real reactions are always less.

Formal View

Percent yield quantifies reaction efficiency:

\%\text{yield} = \frac{m_{\text{actual}}}{m_{\text{theoretical}}} \times 100\%

, where

m_{\text{theoretical}}

is computed from the stoichiometry of the balanced equation using the limiting reactant. Values range from 0% (no product) to 100% (perfect conversion).

Worked Examples

Example 1

easy

A reaction has a theoretical yield of

50.0

g but only

42.0

g of product is obtained. Calculate the percent yield.

Answer

84.0\%

First step

Start with the percent-yield formula:

\%\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

Full solution

2
Substitute the given masses: $\%\text{yield} = \frac{42.0}{50.0} \times 100\%$ .
3
Compute the ratio and convert to a percent: $\%\text{yield} = 84.0\%$ .

Percent yield measures the efficiency of a reaction. It is always

\leq 100\%

because side reactions, incomplete reactions, and losses during purification reduce the actual yield.

Example 2

hard

In the reaction

2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3

10.0

g of Al reacts with excess

\text{Cl}_2

to produce

40.0

g of

\text{AlCl}_3

. Calculate the percent yield.

Example 3

medium

\text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2

, 24 g Mg (M=24) reacts with excess

\text{Cl}_2

to give 76 g

\text{MgCl}_2

(M=95). Find the percent yield.

Common Mistakes

Dividing theoretical yield by actual yield instead of actual by theoretical — the formula is $\frac{\text{actual}}{\text{theoretical}} \times 100\%$ - Fix this by naming the substances or sample, checking "Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts?", and attaching units, formulas, states, or evidence to the final statement. - Fix this by naming the substances or sample, checking "Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts?", and attaching units, formulas, states, or evidence to the final statement.
Using the mass of a reactant as the theoretical yield — the theoretical yield must be calculated from stoichiometry for the product - Fix this by naming the substances or sample, checking "Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts?", and attaching units, formulas, states, or evidence to the final statement. - Fix this by naming the substances or sample, checking "Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts?", and attaching units, formulas, states, or evidence to the final statement.
Forgetting to identify the limiting reactant first — theoretical yield must be based on the limiting reactant, not just any reactant - Fix this by naming the substances or sample, checking "Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts?", and attaching units, formulas, states, or evidence to the final statement. - Fix this by naming the substances or sample, checking "Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts?", and attaching units, formulas, states, or evidence to the final statement.
Using percent yield from a keyword alone - Signal words like mole, grams, particles only point to a possible model; the substances and evidence must match too. - Fix this by naming the substances or sample, checking "Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts?", and attaching units, formulas, states, or evidence to the final statement.

Why This Formula Matters

Percent Yield is the bridge between invisible particles and measurable lab amounts. It lets students weigh, count, compare, and predict chemical amounts with units instead of guessing from coefficients alone.

Frequently Asked Questions

What is the Percent Yield formula?

The ratio of the actual yield obtained in an experiment to the theoretical yield predicted by stoichiometry, expressed as a percentage.

How do you use the Percent Yield formula?

How much of the possible product you actually got — 100% is perfect, real reactions are always less.

What do the symbols mean in the Percent Yield formula?

$\%$ denotes percent yield. $m_{\text{actual}}$ is the mass of product obtained experimentally. $m_{\text{theoretical}}$ is the maximum mass predicted by stoichiometry.

Why is the Percent Yield formula important in Chemistry?

What do students get wrong about Percent Yield?

Students often know a formula related to percent yield but skip the recognition step: Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts? That leads to a correct-looking substitution attached to the wrong chemical model.

What should I learn before the Percent Yield formula?

Before studying the Percent Yield formula, you should understand: theoretical yield.

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Related Concepts

Theoretical Yield