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THEORETICAL BACKGROUND
Stoichiometry, a term derived from the Greek words stoicheion (meaning โelementโ) and metron (meaning โmeasureโ)^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 โ 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:
- Write the balanced chemical equation
- Convert the quantities of known substances into moles
- Use the stoichiometric coefficients in the balanced equation to calculate the number of moles of the sought quantity
- 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:
N 2 (g) + 3 H 2 (g) โ 2 NH 3 (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 N 2 is stoichiometrically equivalent to three moles of H 2 and two moles of NH 3 :
N 2 โ
3H 2 โ
2NH 3
This relationship leads to the following conversion factors:
1 ๐๐๐๐๐๐ ๐๐ 2 3 ๐๐๐๐๐๐ ๐ป๐ป 2
Suppose 4.030 g of H 2 reacted completely with N 2 to form NH 3. a). How many grams of N (^2) are required to consume completely the H 2 provided? b). How many grams of NH 3 formed as a result?
a) The mass of N 2 required to consume completely the H 2 provided can be calculated using the molar masses and the stoichiometric mole conversion factors as follows:
Mass of H (^2)
๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐ ๏ฟฝโฏโฏโฏโฏโฏโฏโฏโฏ๏ฟฝ Moles of H (^2)
๐๐๐๐๐๐๐๐๐๐โ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ถ๐ถ๐๐๐๐๐ถ๐ถ๐ถ๐ถ๐๐๐๐๐๐๐๐๐ถ๐ถ๐๐ ๏ฟฝโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏ๏ฟฝ Moles of N 2
๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐ ๏ฟฝโฏโฏโฏโฏโฏโฏโฏโฏ๏ฟฝ Mass of N 2
2.016 ๐๐ ๐ป๐ป 2 ๐ฅ๐ฅ^
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) + Fe 2 O 3 (s) โ Al 2 O 3 (s) + 2 Fe (s)
The balanced chemical equation indicates that aluminum reacts with iron(III) oxide in a 2: ratio. It also indicates that two moles of Al are stoichiometrically equivalent to one mole of Fe 2 O 3 , one mole Al 2 O 3 and two moles of Fe. This relationship leads to the following conversion factors:
2 ๐๐๐๐๐๐ ๐ด๐ด๐๐ 1 ๐๐๐๐๐๐ ๐น๐น๐๐ 2 ๐๐ 3 ๐๐๐๐๐๐^
1 ๐๐๐๐๐๐ ๐ด๐ด๐๐ 2 ๐๐ 3 ๐๐๐๐๐๐^
Suppose 130. g of Al reacted with 608. g of Fe 2 O 3 to form Al 2 O 3 and Fe. The limiting reagent can be determined as follows:
- 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.
26.98 ๐๐ ๐ด๐ด๐๐ = 4.82^ ๐๐๐๐๐๐^ ๐ด๐ด๐๐
๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐ Fe 2 O 3 = 608. ๐๐ Fe 2 O 3 ๐ฅ๐ฅ
1 ๐๐๐๐๐๐ Fe 2 O 3 159.7 ๐๐ Fe 2 O 3 = 3.81^ ๐๐๐๐๐๐^ Fe^2 O^3
The provided Al/Fe 2 O 3 molar ratio is:
4.82 ๐๐๐๐๐๐ ๐ด๐ด๐๐ 3.81 ๐๐๐๐๐๐ ๐น๐น๐๐ 2 ๐๐ 3
The stoichiometric Al/Fe 2 O 3 molar ratio is:
2 ๐๐๐๐๐๐ ๐ด๐ด๐๐ 1 ๐๐๐๐๐๐ ๐น๐น๐๐ 2 ๐๐ 3
Comparing the provided versus the stoichiometric Al/Fe 2 O 3 molar ratios shows that Al is present in a less-than-stoichiometric amount, and thus Al is the limiting reagent and Fe 2 O 3 is the excess reagent.
- 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 ๐๐๐๐๐๐ Al 26.98 ๐๐ ๐ด๐ด๐๐ ๐ฅ๐ฅ^
The molar amount of Fe expected to form from 608. g Fe 2 O 3 is:
๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐น๐น๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ = 608. ๐๐ Fe 2 O 3 ๐ฅ๐ฅ
1 ๐๐๐๐๐๐ Fe 2 O 3 159.7 ๐๐ ๐ด๐ด๐๐ ๐ฅ๐ฅ^
Since Al yields the lesser amount of product, it is the limiting reactant. Fe 2 O 3 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:
๐๐โ๐๐๐๐๐๐๐๐๐๐๐ฆ๐ฆ๐๐๐๐๐๐ ๐ฆ๐ฆ๐ฆ๐ฆ๐๐๐๐๐๐ ๐ฅ๐ฅ^100
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:
CuSO 4 (aq) + Zn (s) โ Cu (s) + ZnSO 4 (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) โ Cu (s) + Zn 2+^ (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 (Zn 2+) 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 (CuSO 4 โข 5H 2 O) will be provided as a source of copper(II) sulfate (CuSO 4 ) 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 CuSO 4 โข 5H 2 O and CuSO 4 are equal: CuSO 4 โข 5H 2 O โ CuSO 4 + 5H 2 O
(^3) Anon, 2019. Activity Series. Available at: https://chem.libretexts.org/@go/page/
[Accessed August 10, 2021].
