5.10: Stoichiometry by the Mole Ratio Method

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The calculation of quantities of materials involved in chemical reactions is called stoichiometry. To illustrate stoichiometric calculations consider a typical chemical reaction, in this case the heat-producing combustion of ethane, a hydrocarbon fuel with a chemical formula of C₂H₆:

\[\ce{2C2H6 + 7O2 \rightarrow 4CO2 + 6H2O}\]

Rather than viewing this reaction in terms of individual molecules, it is possible to scale up to moles. Recall from Section 4.8 that the mole is a fundamental unit for quantity of material and that each mole contains Avogadro’s number (6.022×10²³) of formula units (molecules of covalently bound compounds). This equation simply says that 2 moles of C₂H₆ react with 7 moles of O₂ to yield 4 moles of CO₂ and 6 moles of H₂O. Now we can examine the equation in more detail to do some quantitative calculations. Before doing that, however, review the following two terms

Formula mass: The sum of the atomic masses of all the atoms in a formula unit of a compound. Although the average masses of atoms and molecules may be expressed in atomic mass units (amu or u), formula mass is generally viewed as being relative and without units.

Molar mass: Where X is the formula mass, the molar mass is X grams of an element or compound, that is, the mass in grams of 1 mole of the element or compound

Given the atomic masses H 1.0, C 12.0, and O 16.0 the molar mass of C₂H₆ is 2×12.0 + 6×1.0 = 30.0 g/mol, that of O₂ is 2×16.0 = 32.0 g/mol, that of CO₂ is 12.0 + 2×16.0 = 44.0 g/mol, and that of H₂O is 2×1.0 + 16.0 = 18.0 g/mol. Now consider the chemical equation

\[\ce{2C2H6 + 7O2 \rightarrow 4CO2 + 6H2O}\]

in terms of the minimum whole number of moles reacting and produced and the masses in grams of these quantities. The equation states that 2 moles of C₂H₆ with a mass of 2×30.0 g = 60.0 g of C₂H₆ react with 7 moles of O₂ with a mass of 7×32.0 g = 224 g of O₂ to produce 4 moles of CO₂ with a mass of 4×44.0 g = 176 g of CO₂ and 6 moles of H₂O with a mass of 6×18.0 g = 108 g of H₂O. The total mass of reactants is

\[\textrm{60.0 g of } C_{2}H_{6} + \textrm{ 224 g of } O_{2} = \textrm{284.0g of reactants}\]

and the total mass of products is

\[\textrm{176 g of } CO_{2} + \textrm{108g of } H_{2}O = \textrm{284 g of products}\]

Note that, as in all chemical reactions, the total mass of products equals the total mass of reactants. Stoichiometry, the calculation of quantities of materials involved in chemical reactions, is based upon the law of conservation of mass which states that the total mass of reactants in a chemical reaction equals the total mass of products, because matter is neither created nor destroyed in chemical reactions. The basic premise of the mole ratio method of stoichiometric calculations is that the relative numbers of moles of reactants and products remain the same regardless of the total quantity of reaction. To illustrate stoichiometric calculations, consider again the following reaction:

\[\ce{2C2H6 + 7O2 \rightarrow 4CO2 + 6H2O}\]

As noted above, this equation states that 2 moles C₂H₆ react with 7 moles of O₂ to produce 4 moles of CO₂ and 6 moles of H₂O. The same ratios hold true regardless of how much material reacts. So for 10 times as much material, 20 moles C₂H₆ react with 70 moles of O₂ to produce 40 moles of CO₂ and 60 moles of H₂O

Suppose that it is given that 18.0 g of C₂H₆ react. What is the mass of O₂ that will react with this amount of C₂H₆? What mass of CO₂ is produced? What mass of H₂O is produced? This problem can be solved by the mole ratio method. Mole ratios are, as the name implies, simply the ratios of various moles of reactants and products to each other as shown by a chemical equation. Mole ratios are obtained by simply examining the chemical equation in question; the three that will be used in solving the problem posed are the following:

\(\frac{7 \text{mol O}_{2}}{\textrm{2 mol C}_{2} \textrm{H}_{6}} \: \: \: \frac{\textrm{4 mol CO}_{2}}{\textrm{2 mol C}_{2} \textrm{H}_{6}} \: \: \: \frac{\textrm{6 mol H}_{2} \textrm{O}}{\textrm{2 mol C}_{2} \textrm{H}_{6}}\)

To solve for the mass of O₂ reacting the following steps are involved:

A. Mass of C₂H₆ reacting B. Convert to moles of C₂H₆ C. Convert to moles of O₂ D. Convert to mass of O₂

In order to perform the calculation, it will be necessary to have the molar mass of C₂H₆, stated earlier as 30.0 g/mol, the molar mass of O₂(18.0 g/mol) and the mole ratio relating moles of O₂ reactant to moles of C₂H₆, 7 mol O₂/2 mol C₂H₆. The calculation becomes the following

\(\textrm{Mass of O}_{2} = \textrm{18.0 g C}_{2} \textrm{H}_{6} \times \frac{\textrm{1 mol C}_{2} \textrm{H}_{6}}{\textrm{30.0 g C}_{2} \textrm{H}_{6}} \times \frac{\textrm{7 mol O}_{2}}{\textrm{2 mol C}_{2} \textrm{H}_{6}} \times \frac{\textrm{32.0 g O}_{2}}{\textrm{1 mol O}_{2}} = \textrm{67.2 g O}_{2}\)

Note that in this calculation units cancel above and below the line, starting with units of g C₂H₆. Now that the mass of O₂ reacting has been calculated, it is possible using the appropriate mole ratios and molar masses to calculate the masses of CO₂ and of H₂O produced as follows:

\(\textrm{Mass of CO}_{2} = \textrm{18.0 g C}_{2} \textrm{H}_{6} \times \frac{\textrm{1 mol C}_{2}\textrm{H}_{6}}{\textrm{30.0 g C}_{2} \textrm{H}_{6}} \times \frac{\textrm{6 mol H}_{2} \textrm{O}}{\textrm{2 mol C}_{2} \textrm{H}_{6}} \times \frac{\textrm{18.0 g H}_{2} \textrm{O}}{\textrm{1 mol H}_{2} \textrm{O}} = \textrm{32.4 g H}_{2} \textrm{O}\)

\(\textrm{Mass of H}_{2} \textrm{O} = \textrm{18.0 g C}_{2} H_{6} \times \frac{\textrm{1 mol C}_{2} \textrm{H}_{6}}{\textrm{30.0gC}_{2} \textrm{H}_{6}} \times \frac{\textrm{6 mol H}_{2}\textrm{O}}{\textrm{2 mol C}_{2} \textrm{H}_{6}} \times \frac{\textrm{18.0 g H}_{2} \textrm{O}}{\textrm{1 mol H}_{2} \textrm{O}} = \textrm{32.4 g H}_{2} \textrm{O}\)

Are the masses calculated above correct? A good check is to compare the total mass of reactants, 18.0 g C₂H₆+ 67.2 g O₂= 85.2 g of reactants, with the total mass of products, 52.8 g CO₂+ 32.4 g H₂O = 85.2 g of products. The fact that the total mass of reactants is equal to the total mass of products gives confidence that the calculations are correct.

As one more example consider the reaction of 15.0 g of Al with Cl₂ to give AlCl₃:

\[\ce{2Al + 3Cl2 \rightarrow 2AlCl3}\]

What mass of Cl₂ reacts and what is the mass of AlCl₃ produced? The atomic mass of Al is 27.0 and that of Cl is 35.5. Therefore, the molar mass of Cl₂ is 71.0 g/mol and the molar mass of AlCl₃ is 133.5 g/mole. The mass of Cl₂ reacting is

\(\textrm{Mass of Cl}_{2} = \textrm{15.0 g Al} \times \frac{\textrm{1 mol Al}}{\textrm{27.0 g Al}} \times \frac{\textrm{3 mol Cl}_{2}}{\textrm{2 mol Al}} \times \frac{\textrm{71.0 g Cl}_{2}}{\textrm{1 mol Cl}_{2}} = \textrm{59.2 g Cl}_{2}\)

\(\textrm{Mass of AlCl}_{3} = \textrm{15.0 g Al} \times \frac{\textrm{1 mol Al}}{\textrm{27.0 g Al}} \times \frac{\textrm{2 mol AlCl}_{3}}{\textrm{2 mol Al}} \times \frac{\textrm{133.5 g AlCl}_{3}}{\textrm{1 mol AlCl}_{3}} = \textrm{74.2 g AlCl}_{3}\)

As a check, 15.0 g Al + 59.2 g Cl₂ reactant gives a total of 74.2 g of reactants equal to the mass of the AlCl₃ product.

Exercise

Calculate the mass of CH₄ that reacts and the masses of the products when 25.0 g of Fe₂O₃ undergo the reaction below. The atomic masses involved are H 1.0, C12.0, O 16.0, Fe 55.8

\(\ce{4Fe2O3 + 3CH4 \rightarrow 8Fe + 3CO2 + 6H2O}\)

Answer: Answer: 1.88 g CH4, 17.5 g Fe, 5.2 g CO2, 4.2 g H2O

Exercise

Calculate the mass of O₂ that reacts and the masses of the products when 100 g of benzoic acid, C₇H₆O₂undergo the reaction below. The atomic masses involved are H 1.0, C 12.0, and O 16.0.

\(\ce{2C7H6O2 + 15O2 \rightarrow 14CO2 + 6H2O}\)

Answer: 197 g O₂, 252 g CO₂, 44.3 g H₂O