# 4.26: Stoichiometry: Equality Pattern and Conversions

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Upon establishing that a stoichiometric molar relationship is required to solve a problem, a corresponding stoichiometric equality must be developed. Then, using dimensional analysis, the resultant equality can be applied as a conversion factor, in order to bring about a desired unit transformation.

## Stoichiometric Equality Pattern

As stated in Section 4.1, an equality pattern contains one number and * two* units on both sides of an equal sign. One of these units, "mol," is defined on both sides of a stoichiometric equality pattern. Neither secondary unit is specified in this equality pattern. These positions, which are indicated as "blanks" in the equality pattern that is shown below, should be occupied by units that are relevant to the identities of the specific chemicals that are referenced in a given problem. In particular, the secondary unit on the left side of a stoichiometric equality should be the

*chemical formula*of the

*first chemical*that is referenced in the given problem, and the final unit position on the right side of the equality should be occupied by the

*chemical formula*of the

*second chemical*

*that is being considered. Therefore, like in a "component within" equality, the chemical formulas written on both sides of a stoichiometric equality*

*do not match*. As stated in Section 4.3, a chemical name should

*not*be used in this, or any, equality. Finally, the relative order of the two units on either side of a stoichiometric equality should not be interchanged.

A stoichiometric equality is unique from all of the previously-discussed equality patterns in that "blanks" are present in the numerical positions on *both* sides of this type of equality. Stoichiometry studies the relative molar ratios of the atoms and molecules that participate in chemical reactions, as indicated by the balancing coefficients that are associated with the reaction. Therefore, the coefficients that correspond to the first and second chemicals that are mentioned in the problem should occupy the numerical positions on the left and right side, respectively, of a stoichiometric equality.

For example, consider the following balanced chemical equation.

\(\ce{2 H_2O_2} \left( aq \right) \rightarrow \ce{2 H_2O} \left( l \right) + \ce{O_2} \left( g \right)\)

Calculate how many moles of O_{2} are produced from the decomposition of 7.53 moles of H_{2}O_{2}.

Because both of the chemicals that are referenced in the problem, O_{2} and H_{2}O_{2}, are also present in the given reaction equation, a stoichiometric equality should be developed. Since all chemical information was provided in the form of chemical formulas, the symbols "O_{2}" and "H_{2}O_{2}" are directly incorporated into the secondary unit positions on the left and right sides, respectively, of the equality that is being developed. Since the coefficient that is associated with with O_{2}, the first chemical that is referenced in the given problem, is an unwritten "1" in the corresponding chemical equation, a 1 is inserted into the numerical position on the left side of this stoichiometric equality. Finally, because the coefficient that corresponds with H_{2}O_{2}, the second chemical that is mentioned in the problem, is a "2" in the chemical equation, a 2 is inserted into the numerical position on the right side of this stoichiometric equality, as shown below.

1 mol O_{2} = 2 mol H_{2}O_{2}

## Applying Stoichiometric Equalities as Conversion Factors

Once an appropriate stoichiometric equality has been developed, the information that it contains can be re-written in the form of a conversion factor, which can then be applied to bring about a desired unit transformation. As stated previously, the quantity containing the unit that is being canceled must be written in the *denominator *of a conversion factor. This will cause the given unit, which appears in a numerator, to be divided by itself, since the same unit appears in the denominator of the conversion factor. Since any quantity that is divided by itself "cancels," orienting the conversation factor in this way results in the elimination of the undesirable unit. However, remember that both components of the equalities that are developed in this chapter contain *two *units. Therefore, in order to achieve complete unit cancelation, a conversion factor that results in the *simultaneous *elimination of *both units* must be applied.

For example, use a conversion factor based on the equality developed above to calculate how many moles of O_{2} are produced from the decomposition of 7.53 moles of H_{2}O_{2}.

As stated above, because both of the chemicals that are referenced in the problem, O_{2} and H_{2}O_{2}, are also present in the given reaction equation, a stoichiometric equality should be developed and applied to solve this problem. The equality that was generated based on the coefficients that correspond to O_{2} and H_{2}O_{2} in the given chemical equation_{ }is replicated below.

1 mol O_{2} = 2 mol H_{2}O_{2}

To create a conversion factor from this equality, the quantity on the left side of the equal sign is written in the numerator of a fraction, and the other quantity is written in the denominator. A second conversion factor can be developed by interchanging where each quantity is written, relative to the fraction bar. Both of the resultant conversion factors are shown below.

\( \dfrac{1 \text{ mol } \ce{O_2}}{2 {\text{ mol }} \ce{H_2O_2}} \) and \( \dfrac{2 {\text{ mol }} \ce{H_2O_2}}{1 \text{ mol } \ce{O_2}} \)

However, only one of these conversion factors will allow for the complete cancelation of the given unit, "moles of H_{2}O_{2}," since *both *of the units that are being canceled must be written in the denominator of the conversion factor that should be applied to solve the given problem. Since the intent of this problem is to eliminate the unit "moles of H_{2}O_{2}," the conversion factor on the left must be used. Therefore,

\( {7.53 \; \cancel{\rm{mol} \; \rm{H_2O_2}}} \times\) \( \dfrac{1 \; \rm{mol} \; \rm{O_2}}{2 \; \cancel{\rm{mol} \; \rm{H_2O_2}}}\)

The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator. Recall that, when using a calculator, each conversion factor should be entered in parentheses, * or* the "=" key should be used after

*division. In this case,*

*each*7.53 × (1 mol O_{2} ÷ 2) = 3.765 mol O_{2} ≈ 3.77 mol O_{2}

Finally, remember that the correct number of significant figures should be applied to any calculated quantity. Since the math involved in dimensional analysis is multiplication and division, the number of significant figures in each number being multiplied or divided must be counted, and the answer must be limited to the lesser count of significant figures. The numerical quantities within a stoichiometric equality are exact values, meaning that they are considered to have infinitely-many significant figures and will never limit the number of significant figures in a calculated answer. However, the given number, 7.53, is not exact, and its significant figures must be considered. As this value contains three significant figures, the final answer should be rounded to three significant digits, as shown above.