THEORETICAL BACKGROUND

Stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and
metron (meaning “measure”) 1, is one of many fundamental concepts in the study of
chemistry. It focuses on the quantitative relationships between the reactants and products
of a chemical reaction under the guiding principle that matter is always conserved in a
chemical reaction. Balanced chemical equations are required to relate the amounts of
reactants and products to each other using mole-to-mole ratios and thus to perform
stoichiometric calculations. In a balanced chemical equation,
aA + bB → cC + dD,
where A and B are the reactants, C and D are the products, and a, b, c, and d are stoichiometric
coefficients, a moles of A, b moles of B, c moles of C and d moles of D are said to be
stoichiometrically equivalent. The coefficients in the balanced equation are used in various
stoichiometric calculations (Figure 4.1) 1

Figure 4.1. A summary of various computational steps involved in most reaction stoichiometry calculations.

1
Flowers, P., Theopold, T., Langley, R., and Robinson, W.R. 2019. Reaction Stoichiometry.
Chemistry 2e. Openstax. Available at https://openstax.org/books/chemistry-2e/pages/4-3-
reaction-stoichiometry [Accessed August 10, 2021]
Here is a summary of the major steps involved in solving most stoichiometry problems:
1. Write the balanced chemical equation
2. Convert the quantities of known substances into moles
3. Use the stoichiometric coefficients in the balanced equation to calculate the number
of moles of the sought quantity
4. Convert moles of the sought quantity into the desired units

Example 1. Calculating Amounts of Reactants and Products

To illustrate this, consider the Haber process, which is the production of ammonia from
hydrogen and nitrogen:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
The balanced chemical equation indicates that nitrogen reacts with hydrogen in a 1:3 ratio.
More specifically, using the mole ratio method, one mole of nitrogen gas reacts with three
moles of hydrogen gas to produce two moles of ammonia gas. In stoichiometric calculations,
one mole of N2 is stoichiometrically equivalent to three moles of H2 and two moles of NH3:
N2 ≅ 3H2 ≅ 2NH3
This relationship leads to the following conversion factors:
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁2 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁2 3 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻2
𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎
3 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻2 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝐻𝐻3 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝐻𝐻3

Suppose 4.030 g of H2 reacted completely with N2 to form NH3. a). How many grams of N2
are required to consume completely the H2 provided? b). How many grams of NH3 formed as
a result?
a) The mass of N2 required to consume completely the H2 provided can be calculated
using the molar masses and the stoichiometric mole conversion factors as follows:
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
Mass of H2 �⎯⎯⎯⎯⎯⎯⎯⎯� Moles of H2 �⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯� Moles of N2 �⎯⎯⎯⎯⎯⎯⎯⎯� Mass of N2

1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻2 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁2 28.01 𝑔𝑔 𝑁𝑁2

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑁𝑁2 (𝑔𝑔) = 4.030 𝑔𝑔 𝐻𝐻2 𝑥𝑥 𝑥𝑥 𝑥𝑥 = 18.66 𝑔𝑔
2.016 𝑔𝑔 𝐻𝐻2 3 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻2 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁2
 b) The mass of NH3 formed as a result can be calculated using the molar masses and the
stoichiometric mole conversion factors as follows:
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
Mass of H2 �⎯⎯⎯⎯⎯⎯⎯⎯� Moles of H2 �⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯� Moles of NH3 �⎯⎯⎯⎯⎯⎯⎯⎯� Mass of NH3

1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻2 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁3 42.35 𝑔𝑔 𝑁𝑁𝑁𝑁3

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑁𝑁𝑁𝑁3 (𝑔𝑔) = 4.030 𝑔𝑔 𝐻𝐻2 𝑥𝑥 𝑥𝑥 𝑥𝑥 = 56.44 𝑔𝑔
2.016 𝑔𝑔 𝐻𝐻2 3 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻2 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁3

In the above example, the objective was to find the stoichiometric amounts of reactant and
product, or more specifically, to find the exact amount of a reactant (i.e. N2) required to
consume completely the other reactant (i.e. H2) and to find the exact amount of a product
(i.e. NH3) formed as a result based on the balanced chemical equation. However, in nature or
in the laboratory, the reactants are usually not present in exact stoichiometric amounts. In
fact, in the laboratory, it is a common practice among chemists to use an excess of one of the
reactants if the other reactant present in limited amount, for example, is much more
expensive and needs to be completely used up. For example, in the Haber process of
ammonia production discussed, N2 is readily available in the atmosphere whereas H2 is
available from a more expensive process, the breakdown of natural gas to produce hydrogen
gas 22.

Limiting Reagent
When a reaction is carried out in stoichiometrically unequivalent amounts of reactants, one
of the reactants will be completely used up first, thus limiting the amount of product formed.
The reactant that determines or limits the amount of product formed in a reaction is referred
as the limiting reagent and the other reactant that is present in amount higher than the
stoichiometric proportion necessary to react with the amount of the limiting reagent is the
excess reagent.

For any given stoichiometric problem, an important first step is determining the limiting and
excess reagents. One approach is by calculating the molar amounts of each reactant provided
and then comparing them to the stoichiometric amounts indicated in the balanced chemical
equation. The reactant present in less than the stoichiometric amount is the limiting reagent.

2
Kyriakou, V., Garagounis, I., Vourros, A., Vasileiou, E., Stoukides, M. 2020. An
Electrochemical Haber-Bosch Process. Joule. 2020; 4: 142-158. Available at
https://doi.org/10.1016/j.joule.2019.10.006 [Accessed August 10, 2021].
Alternatively, the limiting reagent can be identified by calculating the molar amounts of each
reactant provided and then using the stoichiometric molar ratio of the reactant to product
from the balanced chemical equation to compute the molar amounts of products formed. The
reactant that results in the least amount of product is the limiting reagent. i

Example 2. Determination of the Limiting Reagent

To illustrate this concept, let us examine the reaction between aluminum and iron(III) oxide:
2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2 Fe (s)
The balanced chemical equation indicates that aluminum reacts with iron(III) oxide in a 2:1
ratio. It also indicates that two moles of Al are stoichiometrically equivalent to one mole of
Fe2O3, one mole Al2O3 and two moles of Fe. This relationship leads to the following
conversion factors:
2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝑒𝑒2 𝑂𝑂3 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝑒𝑒2 𝑂𝑂3
𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝑒𝑒2 𝑂𝑂3 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑙𝑙2 𝑂𝑂3 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝐹𝐹 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑙𝑙2 𝑂𝑂3 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝐹𝐹

Suppose 130. g of Al reacted with 608. g of Fe2O3 to form Al2O3 and Fe. The limiting reagent
can be determined as follows:
1. Calculate the molar amounts of each reactant provided and then compare them to
the stoichiometric amounts indicated in the balanced chemical equation. The
reactant present in less than the stoichiometric amount is the limiting reagent.
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐴𝐴𝐴𝐴 = 130. 𝑔𝑔 𝐴𝐴𝐴𝐴 𝑥𝑥 = 4.82 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴
26.98 𝑔𝑔 𝐴𝐴𝐴𝐴

1 𝑚𝑚𝑚𝑚𝑚𝑚 Fe2 O3
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 Fe2 O3 = 608. 𝑔𝑔 Fe2 O3 𝑥𝑥 = 3.81 𝑚𝑚𝑚𝑚𝑚𝑚 Fe2 O3
159.7 𝑔𝑔 Fe2 O3

The provided Al/Fe2O3 molar ratio is:

4.82 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴 1.26 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴
=
3.81 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝑒𝑒2 𝑂𝑂3 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝑒𝑒2 𝑂𝑂3
The stoichiometric Al/Fe2O3 molar ratio is:
2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝑒𝑒2 𝑂𝑂3
Comparing the provided versus the stoichiometric Al/Fe2O3 molar ratios shows that Al is
present in a less-than-stoichiometric amount, and thus Al is the limiting reagent and Fe2O3
is the excess reagent.

2. Alternatively, calculate the molar amounts of each reactant provided and then, using
the stoichiometric molar ratio of the reactant to product from the balanced chemical
equation, compute the molar amounts of one of the products (for example Fe)
expected to form from each of the reactants. The reactant that produces the least
amount of product is the limiting reagent.

The molar amount of Fe expected to form from 130. g Al is:

1 𝑚𝑚𝑚𝑚𝑚𝑚 Al 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝐹𝐹
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝐹𝐹 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 130. 𝑔𝑔 𝐴𝐴𝐴𝐴 𝑥𝑥 𝑥𝑥 = 4.82 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝐹𝐹
26.98 𝑔𝑔 𝐴𝐴𝐴𝐴 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴

The molar amount of Fe expected to form from 608. g Fe2O3 is:

1 𝑚𝑚𝑚𝑚𝑚𝑚 Fe2 O3 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝐹𝐹
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝐹𝐹 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 608. 𝑔𝑔 Fe2 O3 𝑥𝑥 𝑥𝑥 = 7.61 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝐹𝐹
159.7 𝑔𝑔 𝐴𝐴𝐴𝐴 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝑒𝑒2 𝑂𝑂3

Since Al yields the lesser amount of product, it is the limiting reactant. Fe2O3 is the excess
reagent.

Reaction Yield
The theoretical yield of a reaction is the maximum amount of product that may form per
the stoichiometry of the balanced chemical equation if all the limiting reagent completely
reacts. In practice, the amount of product actually obtained, called the actual yield, is usually
less than the theoretical yield. The reasons for this include incomplete reactions, reversible
reactions, side reactions or recovery loss. The percent yield, which is the proportion of the
actual to the theoretical yield, is used to determine the efficiency of the reaction and the
techniques used to carry out the reaction and is determined as follows:

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = 𝑥𝑥 100
𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
Example 3. Determination of Theoretical Yield and Percent Yield
To demonstrate this, let us examine the synthesis of aspirin, C9H8O4, from the reaction of
salicylic acid, C7H6O3 and acetic anhydride, C4H6O3:
2 C7H6O3 (s) + C4H6O3 (l) → 2 C9H8O4 (s) + H2O (l)
The balanced chemical equation indicates that salicylic acid, C7H6O3, reacts with acetic
anhydride, C4H6O3 in a 2:1 ratio. Suppose 7.20 g of C7H6O3 reacted with excess amount of
C4H6O3 to form 3.13 g of C9H8O4. (a). What is the theoretical yield? (b). What is the percent
yield

a) Salicylic acid, C7H6O3, is the limiting reagent since acetic anhydride, C4H6O3, is
available in excess. Therefore, using C7H6O3 as the limiting reagent, the theoretical
yield can be calculated as illustrated in Example 1 (i.e. calculating the molar amount
of the limiting reagent and then using the stoichiometric conversion factors) and as
shown below:

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
Mass of C7H6O3 �⎯⎯⎯� Moles of C7H6O3 �⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯� Moles of C9H8O4 �⎯⎯⎯⎯� Mass of C9H8O4

1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶7 𝐻𝐻6 𝑂𝑂3 2 𝑚𝑚𝑚𝑚𝑚𝑚 C9 H8 O4 180.2 𝑔𝑔 C9 H8 O4

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 C9 H8 O4 (𝑔𝑔) = 7.20 𝑔𝑔 𝐶𝐶7 𝐻𝐻6 𝑂𝑂3 𝑥𝑥 𝑥𝑥
138.1 𝑔𝑔 𝐶𝐶7 𝐻𝐻6 𝑂𝑂3 2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶7 𝐻𝐻6 𝑂𝑂3 1 𝑚𝑚𝑚𝑚𝑚𝑚 C9 H8 O4

= 9.39 𝑔𝑔

b) Using this theoretical yield and the actual yield provided, the percent yield is
calculated as:
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = 𝑥𝑥 100
𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

3.13 𝑔𝑔 C9 H8 O4
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = 𝑥𝑥 100 = 33.3%
9.39 𝑔𝑔 C9 H8 O4

Experimental Background
This experiment is designed to quantitatively evaluate the techniques used to carry out a
single replacement reaction between zinc and copper(II) sulfate. It entails (1). determining
the limiting and excess reagents, and (2). determining the percent yield of the reaction. A
single replacement (displacement) reaction is a redox reaction in which an ion in solution is
replaced (or displaced) via the oxidation of another more active metal. In this experiment,
copper(II) ion in solution is replaced via the oxidation of zinc:
CuSO4 (aq) + Zn (s) → Cu (s) + ZnSO4 (aq)
According to the activity series 3, zinc is more active than copper and it can replace copper in
copper(II) sulfate by losing two electrons and reducing copper (II) ions as shown in the
following net ionic equation:
Cu2+ (aq) + Zn (s) → Cu (s) + Zn2+ (aq)
This reaction may be observed by placing solid zinc in an aqueous solution containing
copper(II) sulfate. As copper (II) ions (Cu2+), in the copper(II) sulfate aqueous solution are
reduced to elemental copper (Cu) and the resulting zinc ions (Zn2+) dissolve in the solution,
the characteristic blue color of copper(II) sulfate aqueous solution turns clear. The elemental
solid copper that forms precipitates out.
In this experiment, copper(II) sulfate pentahydrate (CuSO4 • 5H2O) will be provided as a
source of copper(II) sulfate (CuSO4) and thus copper(II) ions (Cu2+) and metal zinc powder,
as a source of metal zinc. The quantities of both reactants are assigned and are used to
determine the limiting and excess reagents. Using the amount of metal copper recovered, the
percent yield of the reaction can then be determined. Note: In stoichiometric calculations,
remember that the molar amounts of CuSO4 • 5H2O and CuSO4 are equal: CuSO4 • 5H2O →
CuSO4 + 5H2O

3
Anon, 2019. Activity Series. Available at: https://chem.libretexts.org/@go/page/188845
[Accessed August 10, 2021].