5.1: Chemical Recipes

Skills to Develop

• Explain the concept of stoichiometry as it pertains to chemical reactions
• Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products
• Perform stoichiometric calculations involving mass, moles, and solution molarity

A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.

The general approach to using stoichiometric relationships is similar in concept to the way people use the measurements in recipes when cooking. Suppose a recipe for making eight pancakes calls for 1 cup pancake mix, \(\dfrac{3}{4}\) cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is

\[\mathrm{1\:cup\: mix+\dfrac{3}{4}\:cup\: milk+1\: egg \rightarrow 8\: pancakes} \label{4.4.1}\]

If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is

\[\mathrm{24\: \cancel{pancakes} \times \dfrac{1\: egg}{8\: \cancel{pancakes}}=3\: eggs} \label{4.4.2}\]

Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:

\[\ce{N2}(g)+\ce{3H2}(g)\rightarrow \ce{2NH3}(g) \label{4.4.3}\]

This equation shows that ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:

\[\ce{\dfrac{2NH3 \: molecules}{3H2 \: molecules}\: or \: \dfrac{2 \: doz \: NH3\: molecules}{3\: doz\:H2 \:molecules} \: or \: \dfrac{2\: mol\: NH3\: molecules}{3\: mol\: H2\: molecules}} \label{4.4.4}\]

These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation. Recall that a balanced chemical equation describes not only numbers of atoms/molecules but also numbers of moles of reactants and products. Therefore, the stoichiometric factors in a balanced chemical equation also represent mole ratios between different species in a reaction. Numbers of moles and numbers of molecules can be directly related via Avogadro's number.

Example \(\PageIndex{1}\): Moles of Reactant Required in a Reaction

Aluminum and iodine react energetically to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. How many moles of I₂ are required to react with 0.429 mol of Al according to the following equation (see Figure \(\PageIndex{2}\))?

\[\ce{2Al + 3I2 \rightarrow 2AlI3} \label{4.4.5}\]

Solution

Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is \(\ce{\dfrac{3\: mol\: I2}{2\: mol\: Al}}\). The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this mole ratio:

\[\begin{align*} \mathrm{mol\: I_2} &=\mathrm{0.429\: \cancel{mol\: Al}\times \dfrac{3\: mol\: I_2}{2\:\cancel{mol\: Al}}} \\[5pt] &=\mathrm{0.644\: mol\: I_2} \end{align*}\]

Exercise \(\PageIndex{1}\)

How many moles of Ca(OH)₂ are required to react with 1.36 mol of H₃PO₄ to produce Ca₃(PO₄)₂ according to the equation \(\ce{3Ca(OH)2 + 2H3PO4 \rightarrow Ca3(PO4)2 + 6H2O}\) ?

Answer

2.04 mol

Example \(\PageIndex{2}\): Number of Product Molecules Generated by a Reaction

How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?

\[\ce{C3H8 + 5O2 \rightarrow 3CO2 + 4H2O} \label{4.4.6}\]

Solution

The approach here is the same as for Example \(\PageIndex{1}\), though the absolute number of molecules is requested, not the number of moles. This will simply require use of the moles-to-numbers conversion factor, Avogadro’s number.

The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:

\[\ce{\dfrac{3\: mol\: CO2}{1\: mol\: C3H8}} \label{4.4.7}\]

Using this stoichiometric factor, the provided molar amount of propane, and Avogadro’s number,

\[\mathrm{0.75\: \cancel{mol\: C_3H_8}\times \dfrac{3\: \cancel{mol\: CO_2}}{1\:\cancel{mol\:C_3H_8}}\times \dfrac{6.022\times 10^{23}\:CO_2\:molecules}{\cancel{mol\:CO_2}}=1.4\times 10^{24}\:CO_2\:molecules} \label{4.4.8}\]

Exercise \(\PageIndex{1}\)

How many NH₃ molecules are produced by the reaction of 4.0 mol of Ca(OH)₂ according to the following equation:

\[\ce{(NH4)2SO4 + Ca(OH)2 \rightarrow 2NH3 + CaSO4 + 2H2O} \label{4.4.9} \nonumber\]

Answer

4.8 × 10²⁴ NH₃ molecules

These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.

Example \(\PageIndex{3}\): Relating Masses of Reactants and Products

What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)₂] by the following reaction?

\(\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)\)

Solution

The approach used previously in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\) is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps using molar masses are required. The calculations required are outlined in this flowchart:

\[\mathrm{16\:\cancel{g\: Mg(OH)_2} \times \dfrac{1\:\cancel{mol\: Mg(OH)_2}}{58.3\:\cancel{g\: Mg(OH)_2}}\times \dfrac{2\:\cancel{mol\: NaOH}}{1\:\cancel{mol\: Mg(OH)_2}}\times \dfrac{40.0\: g\: NaOH}{\cancel{mol\: NaOH}}=22\: g\: NaOH}\]

Exercise \(\PageIndex{3}\)

What mass of gallium oxide, Ga₂O₃, can be prepared from 29.0 g of gallium metal? The equation for the reaction is \(\ce{4Ga + 3O2 \rightarrow 2Ga2O3}\).

Answer

39.0 g

Example \(\PageIndex{4}\): Relating Masses of Reactants

What mass of oxygen gas, O₂, from the air is consumed in the combustion of 702 g of octane, C₈H₁₈, one of the principal components of gasoline?

\[\ce{2C8H18 + 25O2 \rightarrow 16CO2 + 18H2O}\nonumber\]

Solution

The approach required here is the same as for the Example \(\PageIndex{3}\), differing only in that the provided and requested masses are both for reactant species.

\(\mathrm{702\:\cancel{g\:\ce{C8H18}}\times \dfrac{1\:\cancel{mol\:\ce{C8H18}}}{114.23\:\cancel{g\:\ce{C8H18}}}\times \dfrac{25\:\cancel{mol\:\ce{O2}}}{2\:\cancel{mol\:\ce{C8H18}}}\times \dfrac{32.00\: g\:\ce{O2}}{\cancel{mol\:\ce{O2}}}=2.46\times 10^3\:g\:\ce{O2}}\)

Exercise \(\PageIndex{4}\)

What mass of CO is required to react with 25.13 g of Fe₂O₃ according to the equation \(\ce{Fe2O3 + 3CO \rightarrow 2Fe + 3CO2}\)?

Answer

13.22 g

These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure \(\PageIndex{2}\) provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.

Figure \(\PageIndex{1}\): The flowchart depicts the various computational steps involved in most reaction stoichiometry calculations.

Application: Airbags

Airbags (Figure \(\PageIndex{3}\)) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN₃. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN₃ to initiate its decomposition:

\[\ce{2NaN3}(s)\rightarrow \ce{3N2}(g)+\ce{2Na}(s)\]

This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function.

Consider the decomposition of 100. g of NaN₃, a relatively small mass. How many moles of N₂ gas will this reaction produce? Assuming that the nitrogen gas expands to approximately the same density as the surrounding air (1.2 g/L at room temperature), what volume of N₂ gas will be produced inside the airbag?

Solution

Again, we need to use the mole ratio from the balanced equation to convert between NaN₃ and N₂. Then we can use the molar mass of N₂ and the density of the gas to convert to volume in L.

Solving for moles of N₂ :

\[\mathrm{100\:\cancel{g\: NaN_3} \times \dfrac{1\:\cancel{mol\: NaN_3}}{65.0\:\cancel{g\: NaN_3}}\times \dfrac{3\:\cancel{mol\: N_2}}{2\:\cancel{mol\: NaN_3}}=2.31\: mol\: N_2}\]

Solving for volume of N₂ :

\[\mathrm{2.31\:\cancel{mol\: N_2} \times \dfrac{28.0\:\cancel{g\: N_2}}{1\:\cancel{mol\: N_2}}\times \dfrac{1\:\cancel{L\: N_2}}{1.2\:\cancel{g\: N_2}}=54\: L\: N_2}\]

100 g of NaN₃ will generate approximately 54 L of N₂, enough to inflate an airbag.

Figure \(\PageIndex{2}\): Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)

Summary

A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.

Contributors

• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).

• Anna Christianson (Bellarmine University)