# 4.26: Stoichiometry: Equality Pattern and Conversions

- Page ID
- 213202

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Use information in a given phrase or word problem to write stoichiometric equalities.
- Apply a stoichiometric conversion factor to convert between the molar quantities of two substances that participate in a chemical reaction.
- Apply multiple conversion factors to convert between a molar quantity and a particle count of two substances that participate in a chemical reaction.
- Apply multiple conversion factors to convert between a molar quantity and a mass of two substances that participate in a chemical reaction.
- Apply multiple conversion factors to convert between a mass and a particle count of two substances that participate in a chemical reaction.

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.

Consider the following balanced chemical equation.

\(\ce{3 Ag_2S} \left( s \right) + \ce{2 Al} \left( s \right) \rightarrow\ \ce{6 Ag} \left( s \right) + \ce{Al_2S_3} \left( s \right)\)

Calculate how many atoms of silver are produced if 2.4 moles of aluminum sulfide are generated in this reaction.

- Identify the indicator information in the given problem, and state which molar standard is associated with each indicator.
- Write an equality pattern that corresponds to each of the indicators identified in Part (a).
- Using dimensional analysis, apply the equalities that were developed in Part (b) as conversion factors, in order to bring about the desired unit transformation and solve the given problem.

**Answer**__Indicator Information__

The word "atoms" indicates that an Avogadro's number equality should be developed and applied to solve this problem.

Additionally, because both of the chemicals that are referenced in the problem, silver, Ag, and aluminum sulfide, Al_{2}S_{3}, are also present in the given reaction equation, a stoichiometric equality should also be developed and applied to solve this problem. In order to recognize the stoichiometric indicator, the chemical formulas of each of these chemicals must be determined from the chemical names that were provided. The subscripts in the chemical formula for aluminum sulfide, Al_{2}S_{3}, are derived by applying the Chapter 3 rules for determining ionic chemical formulas.

__Equality Patterns__

*Avogadro's Number Equality Pattern*

The word "atoms" indicates that an Avogadro's number equality should be developed. Furthermore, since "atoms" is an indicator word, this word is inserted as the second unit on the right side of this type of equality. As multiple chemicals are referenced in this problem, the chemical that is written in closest physical proximity to the indicator word "atoms," "silver," is selected for incorporation into the Avogadro's number equality. However, as chemical names should not be used in this, or any, equality, the corresponding elemental symbol, "Ag," is incorporated into both of the secondary unit positions in the equality that is being developed. The resultant Avogadro's number equality is shown below.1 mol Ag = 6.02 × 10

Because the given chemical information is an elemental name, the indicator word "atoms" appropriately corresponds to the chemical formula that is applied in this equality.^{23}Ag atoms

*Stoichiometric Equality Pattern*

Because both of the chemicals that are referenced in the problem, silver, Ag, and aluminum sulfide, Al_{2}S_{3}, are also present in the given reaction equation, a stoichiometric equality should also be developed. As chemical names should not be used in this, or any, equality, the chemical formulas for silver, Ag, and aluminum sulfide, Al_{2}S_{3}, are incorporated into the secondary unit positions on the left and right sides, respectively, of the stoichiometric equality that is being developed. Since the coefficient that is associated with silver, Ag, is a "6" in the corresponding chemical equation, a 6 is inserted into the numerical position on the lef side of this stoichiometric equality. Finally, because the coefficient that corresponds with aluminum sulfide, Al_{2}S_{3}, is an unwritten "1" in the chemical equation, a 1 is inserted into the numerical position on the right side of this stoichiometric equality, as shown below.6 mol Ag = 1 mol Al

_{2}S_{3}__Dimensional Analysis__

In order to completely eliminate the given unit, "moles of aluminum sulfide," a conversion factor based on the stoichiometric equality must be applied first. Specifically, the quantity on the right side of this equality becomes the denominator in the first conversion factor that is applied to solve the given problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor.

However, the unit that results upon the cancelation of "moles of aluminum sulfide" is "mol Ag," which is not the desired final unit. Therefore, the Avogadro's number equality must be applied as a second conversion factor. In order to completely cancel the intermediate unit "mol Ag," the quantity on the left side of the Avogadro's number equality becomes the denominator in the second conversion factor that is applied to solve the given problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor. While reversing the order of the two units on the right side of an Avogadro's number equality is not absolutely necessary, doing so more clearly illustrates that the answer will ultimately be expressed in the desired unit, as shown below.\( {2.4 \; \cancel{\rm{mol} \; \rm{Al_2S_3}}} \times\) \( \dfrac{6 \; \bcancel{\rm{mol} \; \rm{Ag}}}{1 \; \cancel{\rm{mol} \; \rm{Al_2S_3}}}\) × \( \dfrac{6.02 \times 10^{23} \; \rm{atoms} \; \rm{Ag}}{1 \; \bcancel{\rm{mol} \; \rm{Ag}}}\) = \( {\text {8.6688}} \times {\text{10}^{24}}\) \({\rm{atoms} \; \rm{Ag}}\)

≈ \( {\text {8.67}} \times {\text{10}^{24}}\) \({\rm{atoms} \; \rm{Ag}}\)

The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator. When using a calculator, each conversion factor should be entered in parentheses,the "=" key should be used after*or*division, and any quantity that is expressed in scientific notation should be offset by an additional set of parentheses. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above.*each*

Consider the following balanced chemical equation.

\(\ce{2 C_1_2H_2_6} \left( l \right) + \ce{37 O_2} \left( g \right) \rightarrow \ce{24 CO_2} \left( g \right) + \ce{26 H_2O} \left( g \right) + \ce{E}\)

Calculate how many moles of water are produced if 632 grams of molecular oxygen are consumed in this reaction.

- Identify the indicator information in the given problem, and state which molar standard is associated with each indicator.
- Write an equality pattern that corresponds to each of the indicators identified in Part (a).
- Using dimensional analysis, apply the equalities that were developed in Part (b) as conversion factors, in order to bring about the desired unit transformation and solve the given problem.

**Answer**__Indicator Information__

Because both of the chemicals that are referenced in the problem, water, H_{2}O, and molecular oxygen, O_{2}, are also present in the given reaction equation, a stoichiometric equality should be developed and applied to solve this problem. In order to recognize the stoichiometric indicator, the chemical formulas of each of these chemicals must be determined from the chemical names that were provided. The subscript in the chemical formula for molecular oxygen, O_{2}, is derived by applying the Chapter 3 rules for determining diatomic covalent chemical formulas.

Additionally, the mass unit "grams" indicates that a mass-based equality should also be developed and applied to solve this problem. As multiple chemicals are referenced in this problem, the chemical that is written in closest physical proximity to the indicator word "grams," "molecular oxygen," is selected for incorporation into the mass-based equality. In order to determine whether an atomic weight equality or a molecular weight equality should be applied to solve this problem, the chemical formula for molecular oxygen must be determined from the chemical name that was provided. As stated above, the subscript in the chemical formula for molecular oxygen, O_{2}, is derived by applying the Chapter 3 rules for determining diatomic covalent chemical formulas. Therefore, since molecular oxygen, O_{2}, is a compound, a molecular weight equality should be developed and applied to solve this problem.

__Equality Patterns__

*Stoichiometric Equality Pattern*

Because both of the chemicals that are referenced in the problem, water, H_{2}O, and molecular oxygen, O_{2}, are also present in the given reaction equation, a stoichiometric equality should be developed. As chemical names should not be used in this, or any, equality, the chemical formulas for water, H_{2}O, and molecular oxygen, O_{2}, are incorporated into the secondary unit positions on the left and right sides, respectively, of the stoichiometric equality that is being developed. Since the coefficient that is associated with water, H_{2}O, is a "26" in the corresponding chemical equation, a 26 is inserted into the numerical position on the left side of this stoichiometric equality. Finally, because the coefficient that corresponds with molecular oxygen, O_{2}, is a "37" in the chemical equation, a 37 is inserted into the numerical position on the right side of this stoichiometric equality, as shown below.26 mol H

_{2}O = 37 mol O_{2}*Molecular Weight Equality Pattern*

The mass unit "grams," coupled with the knowledge that molecular oxygen, O_{2}, is a compound, indicates that a mass-based molecular weight equality should also be developed. However, molecular weight is unique among the molar standards, in that its equality pattern cannot be immediately developed from readily-available information. Instead, the mass contribution of each element that is found within the compound that is being considered must first be calculated. The compound's molecular weight, which is determined by adding each of the resultant mass contributions, can then be rewritten as an equality.

Because molecular oxygen, O_{2}, contains one element, the mass contribution pattern is applied once for oxygen, O. In a mass contribution calculation,- the chemical formula of the element that is being considered is written as the secondary unit for all numerical quantities, as chemical names should
be used in this, or any, molar equality or calculation;*not* - the subscript for the element that is being considered, as found in the compound's chemical formula, is inserted into the "component within" portion of the mass contribution pattern;
- the atomic mass average of the element that is being considered is recorded to the hundredths place in the numerator of the atomic weight conversion factor; and
- the resultant calculated value is reported to the hundredths place.

_{2}, is shown below.\( {\text {2}}\) \({\cancel{\rm{mol} \; \rm{O}}}\) × \( \dfrac{16.00 \; \rm{g} \; \rm{O}}{1\; \cancel{\rm{mol} \; \rm{O}}}\) = \( {\text {32.00}}\) \({\rm{g} \; \rm{O}}\)

As only a single mass contribution calculation was necessary for this compound, the resultant numerical solution, 32.00, corresponds to the molecular weight of molecular oxygen, O_{2}.

In order to develop a molecular weight equality, the calculated molecular weight of the compound is equated to 1 mol of the compound, and the chemical formula of theis utilized as the secondary unit on both sides of the resultant equality. The molecular weight equality for molecular oxygen, O*entire compound*_{2}, is shown below.1 mol O

_{2}= 32.00 g O_{2}__Dimensional Analysis__

In order to completely eliminate the given unit, "grams of molecular oxygen," a conversion factor based on the molecular weight equality must be applied first. Specifically, the quantity on the right side of this equality becomes the denominator in the first conversion factor that is applied to solve the given problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor.

However, the unit that results upon the cancelation of "grams of molecular oxygen" is "mol O_{2}," which is not the desired final unit. Therefore, a conversion factor based on the stoichiometric equality must be applied as a second conversion factor. In order to completely cancel the intermediate unit "mol O_{2}," the quantity on the right side of the stoichiometric equality becomes the denominator in the second conversion factor that is applied to solve the given problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor, as shown below.\( {632 \; \cancel{\rm{g} \; \rm{O_2}}} \times\) \( \dfrac{1 \; \bcancel{\rm{mol} \; \rm{O_2}}}{32.00 \; \cancel{\rm{g} \; \rm{O_2}}}\) × \( \dfrac{26 \; \rm{mol} \; \rm{H_2O}}{37 \; \bcancel{\rm{mol} \; \rm{O_2}}}\) = \( {\text {13.87387387...}}\) \({\rm{mol} \; \rm{H_2O}}\) ≈ \( {\text {13.9}}\) \({\rm{mol} \; \rm{H_2O}}\)

The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator. When using a calculator, each conversion factor should be entered in parentheses,the "=" key should be used after*or*division. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above.*each*- the chemical formula of the element that is being considered is written as the secondary unit for all numerical quantities, as chemical names should

Consider the following balanced chemical equation.

\(\ce{2 Co} \left( s \right) + \ce{3 Br_2} \left( g \right) \rightarrow \ce{2 CoBr_3} \left( s \right)\)

Calculate how many grams of cobalt must be present to completely consume 1.38 x 10^{24} molecules of molecular bromine in this reaction.

- Identify the indicator information in the given problem, and state which molar standard is associated with each indicator.
- Write an equality pattern that corresponds to each of the indicators identified in Part (a).
- Using dimensional analysis, apply the equalities that were developed in Part (b) as conversion factors, in order to bring about the desired unit transformation and solve the given problem.

**Answer**__Indicator Information__

The mass unit "grams" indicates that a mass-based equality should be developed and applied to solve this problem. As multiple chemicals are referenced in this problem, the chemical that is written in closest physical proximity to the indicator word "grams," "cobalt," is selected for incorporation into the mass-based equality. Therefore, since cobalt, Co, is an element, an atomic weight equality should be developed and applied to solve this problem.

Additionally, the word "molecules" indicates that an Avogadro's number equality should also be developed and applied to solve this problem.

Finally, because both of the chemicals that are referenced in the problem, cobalt, Co, and molecular bromine, Br_{2}, are also present in the given reaction equation, a stoichiometric equality should also be developed and applied to solve this problem. In order to recognize the stoichiometric indicator, the chemical formulas of each of these chemicals must be determined from the chemical names that were provided. The subscript in the chemical formula for molecular bromine, Br_{2}, is derived by applying the Chapter 3 rules for determining diatomic covalent chemical formulas.

__Equality Patterns__

*Atomic Weight Equality Pattern*

The mass unit "grams," coupled with the knowledge that cobalt, Co, is an element, indicates that a mass-based atomic weight equality should be developed. The chemical name that is referenced in the given statement, "cobalt," should not be used in this, or any, equality. Instead, the corresponding elemental symbol, "Co," is incorporated into both of the secondary unit positions in the equality that is being developed. Finally, the atomic mass average of cobalt, Co, is 58.933, based on the numerical value that is written underneath its elemental symbol on the periodic table. However, this value is recorded to the thousandths place, which exceeds the decimal place standard for reporting atomic weights. Therefore, when this value is inserted into the numerical position on the right side of an atomic weight equality, it must be rounded to the hundredths place, as shown below.1 mol Co = 58.93 g Co

*Avogadro's Number Equality Pattern*

The word "molecules" indicates that an Avogadro's number equality should also be developed. Furthermore, since "molecules" is an indicator word, this word is inserted as the second unit on the right side of this type of equality. As multiple chemicals are referenced in this problem, the chemical that is written in closest physical proximity to the indicator word "molecules," "molecular bromine," is selected for incorporation into the Avogadro's number equality. However, as chemical names should not be used in this, or any, equality, the corresponding chemical formula, "Br_{2}," is incorporated into both of the secondary unit positions in the equality that is being developed. The resultant Avogadro's number equality is shown below.1 mol Br

Because the given chemical information is the name of a diatomic covalent molecule, the indicator word "molecules" appropriately corresponds to the chemical formula that is applied in this equality._{2}= 6.02 × 10^{23}Br_{2}molecules

*Stoichiometric Equality Pattern*

Because both of the chemicals that are referenced in the problem, cobalt, Co, and molecular bromine, Br_{2}, are also present in the given reaction equation, a stoichiometric equality should also be developed. As chemical names should not be used in this, or any, equality, the chemical formulas for cobalt, Co, and molecular bromine, Br_{2}, are incorporated into the secondary unit positions on the left and right sides, respectively, of the stoichiometric equality that is being developed. Since the coefficient that is associated with cobalt, Co, is a "2" in the corresponding chemical equation, a 2 is inserted into the numerical position on the left side of this stoichiometric equality. Finally, because the coefficient that corresponds with molecular bromine, Br_{2}, is a "3" in the chemical equation, a 3 is inserted into the numerical position on the right side of this stoichiometric equality, as shown below.2 mol Co = 3 mol Br

_{2}__Dimensional Analysis__

In order to completely eliminate the given unit, "molecules of molecular bromine," the Avogadro's number equality must be applied first. Specifically, the quantity on the right side of the Avogadro's number equality becomes the denominator in the first conversion factor that is applied to solve the given problem. While reversing the order of the two units on the right side of an Avogadro's number equality is not absolutely necessary, doing so more clearly illustrates the desired unit cancelation for this particular problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor.

However, the unit that results upon the cancelation of "molecules of molecular bromine" is "mol Br_{2}," which is not the desired final unit. Therefore, a conversion factor based on the stoichiometric equality must be applied as a second conversion factor. In order to completely cancel the intermediate unit "mol Br_{2}," the quantity on the right side of this equality becomes the denominator in the second conversion factor that is applied to solve the given problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor.

The unit that results upon the cancelation of the intermediate unit "mol Br_{2}" is "mol Co," which still is not the desired final unit. Therefore, a conversion factor based on the atomic weight equality must be applied as a third conversion factor. In order to completely cancel the intermediate unit "mol Co," the quantity on the left side of this equality becomes the denominator in the third conversion factor that is applied to solve the given problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor, as shown below.\( {\text {1.38}} \times {\text{10}^{24}}\) \({\cancel{\rm{molecules} \; \rm{Br_2}}} \times\) \( \dfrac{1 \; \bcancel{\rm{mol} \; \rm{Br_2}}}{6.02 \times 10^{23} \; \cancel{\rm{molecules} \; \rm{Br_2}}}\) × \( \dfrac{2 \; \cancel{\rm{mol} \; \rm{Co}}}{3 \; \bcancel{\rm{mol} \; \rm{Br_2}}}\) × \( \dfrac{58.93 \; \rm{g} \; \rm{Co}}{1 \; \cancel{\rm{mol} \; \rm{Co}}}\) = \( {\text {90.059136...}}\) \({\rm{g} \; \rm{Co}}\)

≈ \( {\text {90.1}}\) \({\rm{g} \; \rm{Co}}\)

The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator. When using a calculator, each conversion factor should be entered in parentheses,the "=" key should be used after*or*division, and any quantity that is expressed in scientific notation should be offset by an additional set of parentheses. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above.*each*