D2.1 Use mole, stoichiometry, limiting-reagent, and yield terminology

Ontario Grade 11 Chemistry

Quantities in Chemical Reactions

A Grade 11 guide to the language of reaction amounts

A reaction may stop producing visible product even though some starting material remains. The particle model helps explain why: one reactant may have run out, so the reaction cannot continue. Chemists use the mole and stoichiometry to describe the amounts involved. They use limiting-reagent and yield terminology to describe which reactant controls product formation and how much product is obtained.

What you will learn

Define the mole and explain how it helps describe amounts of particles.
Explain stoichiometry using a balanced chemical equation.
Identify the limiting reagent and describe how it affects the amount of product.
Distinguish theoretical yield, actual yield, and percent yield.

1. From particles to the mole

Before studying reaction amounts, recall that a chemical formula shows which atoms make up a substance and how many of each atom are present. A balanced chemical equation has the same number of each type of atom on both sides. Its coefficients show the relative amounts of particles that react or form.

Atoms and molecules are too small to count one by one in a classroom sample. A mole is a counting unit for particles, just as a dozen is a counting unit for objects. One mole contains approximately

6.02 \times 10^{23}

particles. The particles may be atoms, molecules, or ions, depending on the substance.

A mole also connects particle amounts to measurable mass. The molar mass is the mass of one mole of a substance, written in grams per mole. For example, the molar mass of magnesium is about

24.3\,\mathrm{g/mol}

. This lets you convert between a measured mass and an amount in moles.

n = \frac{m}{M}

A mole is a counting unit for particles.
Molar mass connects mass in grams to amount in moles.
Balanced equations conserve the number of each type of atom.

2. Stoichiometry: amounts in a balanced equation

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. A reactant is a starting substance in a reaction. A product is a substance formed by the reaction.

The coefficients in a balanced equation give mole ratios. A mole ratio compares the amounts of two substances in moles. For example, in the reaction of hydrogen with oxygen to form water, two moles of hydrogen react with one mole of oxygen to form two moles of water. These ratios come from the balanced equation; they are not mass ratios.

Use a balanced equation before interpreting amounts. If an equation is not balanced, its coefficients do not give the correct mole ratios. The equation also does not say that equal masses of reactants are needed. Different substances have different molar masses.

2\,\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{H_2O(l)}

Stoichiometry describes amount relationships in reactions.
Coefficients in a balanced equation give mole ratios.
Use moles—not grams—when applying a coefficient ratio.

3. Limiting reagent and product amount

When reactants are mixed, they may not be present in exactly the amounts required by the balanced equation. The limiting reagent is the reactant that is used up first. Once it is gone, the reaction cannot form more product by that reaction. The excess reagent is any reactant left over.

A particle-level picture can help. Imagine that each product unit requires a fixed group of reactant particles, as shown by the equation’s coefficients. If one kind of particle runs out while other particles remain, no more complete groups can form. That reactant is limiting.

To identify the limiting reagent, compare each reactant’s available amount in moles with the amount required by the balanced equation. You can ask how much product each reactant could make on its own. The reactant that could make the smaller amount of product is limiting. This comparison must use the equation’s mole ratios.

The limiting reagent is consumed first and controls the maximum product amount.
An excess reagent remains after the limiting reagent is used up.
Compare amounts using the balanced equation’s mole ratios.

4. Yield: predicted and obtained product

Yield describes the amount of product. The theoretical yield is the greatest amount of product predicted from the limiting reagent, assuming the reaction follows the balanced equation and the calculation. It is a calculated amount, not a measurement.

The actual yield is the amount of product obtained in a real process and measured. Actual yield may be less than theoretical yield. Percent yield compares actual yield with theoretical yield as a percentage. Both yields must use the same units, such as grams or moles.

In calculations, keep units with the numbers and round the final result to a sensible number of significant figures. Significant figures are the digits that reflect the precision of a measurement. Do not round intermediate values more than necessary, because early rounding can change the final answer.

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

Theoretical yield is calculated from the limiting reagent.
Actual yield is the product amount measured.
Percent yield compares actual yield with theoretical yield.

Worked example

Find the limiting reagent and percent yield

Magnesium reacts with oxygen to form magnesium oxide. A mixture contains

4.86\,\mathrm{g}

of magnesium and

4.80\,\mathrm{g}

of oxygen gas. The reaction produces an actual yield of

7.25\,\mathrm{g}

of magnesium oxide. Determine the limiting reagent, theoretical yield, and percent yield. Use molar masses of

24.3\,\mathrm{g/mol}

for magnesium,

32.0\,\mathrm{g/mol}

for oxygen gas, and

40.3\,\mathrm{g/mol}

for magnesium oxide.

Balance the equation
The balanced equation gives the mole ratio needed for the comparison. Two moles of magnesium react with one mole of oxygen gas.
$2\,\mathrm{Mg(s)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{MgO(s)}$
Convert each reactant mass to moles
Use amount equals mass divided by molar mass. Keep the units in the conversion so grams cancel and the result is in moles.
$n(\mathrm{Mg}) = \frac{4.86\,\mathrm{g}}{24.3\,\mathrm{g/mol}} = 0.200\,\mathrm{mol}; n(\mathrm{O_2}) = \frac{4.80\,\mathrm{g}}{32.0\,\mathrm{g/mol}} = 0.150\,\mathrm{mol}$
Identify the limiting reagent
The equation requires two moles of magnesium for each mole of oxygen. The available $0.150\,\mathrm{mol}$ of oxygen would require $0.300\,\mathrm{mol}$ of magnesium. Only $0.200\,\mathrm{mol}$ of magnesium is available, so magnesium runs out first.
$0.150\,\mathrm{mol\ O_2} \times \frac{2\,\mathrm{mol\ Mg}}{1\,\mathrm{mol\ O_2}} = 0.300\,\mathrm{mol\ Mg}$
Calculate theoretical yield
Use the limiting reagent and the equation’s mole ratio to find the greatest amount of magnesium oxide. Convert that amount to grams using the product’s molar mass.
$0.200\,\mathrm{mol\ Mg} \times \frac{2\,\mathrm{mol\ MgO}}{2\,\mathrm{mol\ Mg}} \times \frac{40.3\,\mathrm{g\ MgO}}{1\,\mathrm{mol\ MgO}} = 8.06\,\mathrm{g\ MgO}$
Calculate percent yield
Compare the measured actual yield with the theoretical yield. The result is a percentage, so include the percent sign.
$\frac{7.25\,\mathrm{g}}{8.06\,\mathrm{g}} \times 100\% = 89.95\% \approx 90.0\%$

Answer: Magnesium is the limiting reagent. The theoretical yield is

8.06\,\mathrm{g}

of magnesium oxide, and the percent yield is 90.0%.

Check: The mole ratio shows that the available oxygen could react with more magnesium than is present. Magnesium therefore runs out first. The actual yield is below the theoretical yield, as expected for these stated values.

Common mistakes and how to avoid them

Treating equation coefficients as mass ratios.

Correction: Coefficients give ratios of particles or moles. Convert mass to moles before using a coefficient ratio.

Calling the reactant with the smaller mass the limiting reagent.

Correction: Compare available mole amounts with the balanced equation’s required ratio. Mass alone does not identify the limiting reagent.

Using actual yield to calculate the theoretical yield.

Correction: Calculate theoretical yield from the limiting reagent. Use actual yield only when comparing the measured product amount with that prediction.

Lesson summary

A mole counts particles, and molar mass connects moles to measured mass.
Stoichiometry uses balanced-equation coefficients to relate amounts of reactants and products.
The limiting reagent is used up first and determines theoretical yield.
Actual yield is measured; percent yield compares actual yield with theoretical yield.

Check your understanding

Question 1

What does a coefficient in a balanced chemical equation tell you in a stoichiometry calculation?

The mass in grams of each substance
The mole ratio between substances
The actual yield of the reaction
Which substance has the greater mass

Show answer and explanation

The mole ratio between substances

Balanced-equation coefficients give the relative numbers of particles, and therefore the mole ratios, for the reaction.

Question 2

A reaction mixture still contains some excess reagent after the reaction stops. What does this indicate?

The excess reagent was the limiting reagent.
The limiting reagent was used up first.
The theoretical yield equals the actual yield.
The balanced equation no longer applies.

Show answer and explanation

The limiting reagent was used up first.

The limiting reagent is consumed first. Another reactant may remain in excess.

Question 3

A reaction has a theoretical yield of

10.0\,\mathrm{g}

and an actual yield of

8.00\,\mathrm{g}

. What is its percent yield?

20.0%
80.0%
125%
2.00%

Show answer and explanation

80.0%

Divide actual yield by theoretical yield and multiply by 100%:

\frac{8.00\,\mathrm{g}}{10.0\,\mathrm{g}} \times 100\% = 80.0\%

.

Key terms

Mole: A counting unit equal to approximately $6.02 \times 10^{23}$ particles.
Molar mass: The mass of one mole of a substance, commonly measured in grams per mole.
Stoichiometry: The study of quantitative relationships between reactants and products in a chemical reaction.
Limiting reagent: The reactant that is used up first and limits the amount of product that can form.
Excess reagent: A reactant that remains after the limiting reagent has been used up.
Theoretical yield: The greatest calculated amount of product predicted from the limiting reagent.
Actual yield: The amount of product obtained and measured.
Percent yield: The actual yield expressed as a percentage of the theoretical yield.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation D2.1. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.1 · Use mole, stoichiometry, limiting-reagent, and yield terminology

Ontario Grade 11 Chemistry

A Grade 11 guide to the language of reaction amounts

What you will learn

1. From particles to the mole

2. Stoichiometry: amounts in a balanced equation

3. Limiting reagent and product amount

4. Yield: predicted and obtained product

Find the limiting reagent and percent yield

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SCH3U

About this lesson and its review