4.27: Additional Examples Last updated Save as PDF Page ID213283 \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\) Learning Objectives Use coefficients to balance a chemical equation. Classify a chemical reaction as a combination, a decomposition, a single replacement, a double replacement, or a combustion reaction. Identify the reactants and products in a chemical equation. Identify the states of matter in which chemicals participate in a chemical reaction. Calculate the molecular weight of a compound. Apply multiple conversion factors to solve problems that involve complex molar relationships. Predict the products of combination, single replacement, double replacement, and combustion reactions. Exercise \(\PageIndex{1}\) Consider the following chemical equation. ___ \(\ce{C_2H_6} \left( g \right) + \) ___ \( \ce{O_2} \left( g \right) \rightarrow\) ___ \( \ce{CO_2} \left( g \right) + \) ___ \( \ce{H_2O} \left( g \right) + \ce{E}\) Balance this equation by writing coefficients in the "blanks," as necessary. Classify this reaction as a combination, a decomposition, a single replacement, a double replacement, or a combustion reaction. State whether molecular oxygen is a reactant or a product in this equation. Calculate the molecular weight of dicarbon hexahydride, which is more commonly-known as "ethane." Represent the resultant solution as a molar equality and as a "hidden" conversion factor. Calculate how many molecules of water are generated if 76.2 grams of dicarbon hexahydride are consumed in this reaction. Answer a Coefficients are incorporated into a chemical equation in order to account for any relative differences between the chemical formulas of the reactants and products that are involved in the corresponding chemical reaction. However, energy, E, is not a chemical material and, therefore, does not contain elements or polyatomic ions. As a result, there is no need to associate a balancing coefficient with the energy that is produced in this reaction. In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of carbon, C, hydrogen, H, and oxygen, O, that are present in the given equation are summarized in the table that is shown below. Note that oxygen is present in both of the chemical products that are generated during this reaction. In order to determine the number of oxygens that are present on the product side of this equation, the subscripts on both oxygens must be added. Element or Ion Reactants Products Balanced C 2 1 H 6 2 O 2 3 None of the components of this reaction are balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. As stated above, oxygen is present in both of the chemical products that are generated during this reaction and, therefore, should not be the first element to be balanced in this equation. Because neither carbon, C, nor hydrogen, H, is present in multiple formulas on the same side of the reaction arrow or is represented in its atomic form, the balancing process could be initiated by selecting either of these elements for further analysis. In order to balance carbon, C, a coefficient should be written in the "blank" that corresponds to carbon dioxide, CO2, on the right side of the equation, as fewer carbons are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_2H_6} \left( g \right) + \) ___ \( \ce{O_2} \left( g \right) \rightarrow\) 2 \( \ce{CO_2} \left( g \right) + \) ___ \( \ce{H_2O} \left( g \right) + \ce{E}\) As carbon dioxide, CO2, contains both carbon, C, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. As stated above, because oxygen is present in both of the chemical products that are generated during this reaction, the quantities of oxygen in both of these products must be added to determine the amount of oxygen that is present on the product side of this equation. Inserting this coefficient balances carbon, C, as intended. The quantities in which hydrogen, H, and oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced C 2 \( \cancel{\rm{1} } \) 2 H 6 2 O 2 \( \cancel{\rm{3} } \) 5 In order to balance hydrogen, H, a coefficient should be written in the "blank" that corresponds to water, H2O, on the right side of the equation, as fewer hydrogens are present on this side of the reaction arrow. The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 6, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_2H_6} \left( g \right) + \) ___ \( \ce{O_2} \left( g \right) \rightarrow\) 2 \( \ce{CO_2} \left( g \right) + \) 3 \( \ce{H_2O} \left( g \right) + \ce{E}\) As water, H2O, contains both hydrogen, H, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. Again, because oxygen is present in both of the chemical products that are generated during this reaction, the quantities of oxygen in both of these products must be added to determine the amount of oxygen that is present on the product side of this equation. Inserting this coefficient balances hydrogen, H, as intended. The quantities in which oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced C 2 \( \cancel{\rm{1} } \) 2 H 6 \( \bcancel{\rm{2} } \) 6 O 2 \( \cancel{\rm{3} } \) \( \bcancel{\rm{5} } \) 7 In order to balance oxygen, O, a coefficient should be written in the "blank" that corresponds to molecular oxygen, O2, on the left side of the equation, as fewer oxygens are present on this side of the reaction arrow. The value of this coefficient, 3.5, is determined by dividing the larger quantity of this element, 7, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_2H_6} \left( g \right) + \) 3.5 \( \ce{O_2} \left( g \right) \rightarrow\) 2 \( \ce{CO_2} \left( g \right) + \) 3 \( \ce{H_2O} \left( g \right) + \ce{E}\) As oxygen, O, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amounts of carbon, C, or hydrogen, H, that are present on the reactant side of the equation. The updated quantity of this element is reflected in the table that is shown below. Inserting this coefficient balances oxygen, O, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced C 2 \( \cancel{\rm{1} } \) 2 H 6 \( \bcancel{\rm{2} } \) 6 O \( \bcancel{\rm{2} } \) 7 \( \cancel{\rm{3} } \) \( \bcancel{\rm{5} } \) 7 However, a fractional coefficient, 3.5, is written in the equation that is shown above. As a result, all of the coefficients in this equation, including the unwritten "1" that is understood to occupy the first blank, must be multiplied by 2, in order to cancel this half-fraction. The doubled coefficient values are reflected in the chemical equation that is shown below. 2 \(\ce{C_2H_6} \left( g \right) + \) 7 \( \ce{O_2} \left( g \right) \rightarrow\) 4 \( \ce{CO_2} \left( g \right) + \) 6 \( \ce{H_2O} \left( g \right) + \ce{E}\) By multiplying all of the coefficients in this equation by 2, the quantities in which carbon, C, hydrogen, H, and oxygen, O, are present in the equation have changed, as shown in the table below, but their relative ratios have not. Therefore, all of the components of this equation are still balanced. Element or Ion Reactants Products Balanced C \( \cancel{\rm{2} } \) 4 \( \cancel{\rm{1} } \) \( \cancel{\rm{2} } \) 4 H \( \cancel{\rm{6} } \) 12 \( \bcancel{\rm{2} } \) \( \cancel{\rm{6} } \) 12 O \( \bcancel{\rm{2} } \) \( \cancel{\rm{7} } \) 14 \( \cancel{\rm{3} } \) \( \bcancel{\rm{5} } \) \( \cancel{\rm{7} } \) 14 Finally, these coefficients cannot be divided, as they do not all share a common divisor that would result in the calculation of four whole number coefficients. Therefore, the final equation that is presented above is chemically-correct, as written. Answer b This reaction is classified as a combustion because energy, "E," is generated as a product. Furthermore, a combustion reaction requires that molecular oxygen, O2, participate in the reaction as a reactant and that carbon dioxide, CO2, and water, H2O, be generated as products. All of these chemicals are present in the given equation. Answer c The chemical formula for molecular oxygen, O2, is derived by applying the Chapter 3 rules for determining diatomic covalent chemical formulas. Molecular oxygen, O2, is a reactant in this reaction, as its chemical formula is written on the left side of the reaction arrow. Answer d In order to calculate the molecular weight of a compound, the mass contribution of each element that is found within that compound must first be determined. The chemical formula for dicarbon hexahydride, C2H6, is determined based on the numerical prefixes in the given chemical name, in alignment with the Chapter 3 rules for determining covalent chemical formulas. Because dicarbon hexahydride, C2H6, contains two elements, the mass contributions of both carbon, C, and hydrogen, H, must be calculated. 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; the subscript for the element that is being considered 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 each resultant calculated value is reported to the hundredths place. The mass contribution calculations for dicarbon hexahydride, C2H6, are shown below. \( {\text {2}}\) \({\cancel{\rm{mol} \; \rm{C}}}\) × \( \dfrac{12.01 \; \rm{g} \; \rm{C}}{1\; \cancel{\rm{mol} \; \rm{C}}}\) = \( {\text {24.02}}\) \({\rm{g} \; \rm{C}}\) \( {\text {6}}\) \({\cancel{\rm{mol} \; \rm{H}}}\) × \( \dfrac{1.01 \; \rm{g} \; \rm{H}}{1\; \cancel{\rm{mol} \; \rm{H}}}\) = \( {\text {6.06}}\) \({\rm{g} \; \rm{H}}\) The numerical value of the molecular weight of dicarbon hexahydride, C2H6, 30.08, is determined by adding the mass contributions that are shown above and should be reported to the hundredths place. 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 the entire compound is utilized as the secondary unit on both sides of the resultant equality, as shown below. 1 mol C2H6 = 30.08 g C2H6 Finally, this equality can be rewritten as a "hidden" conversion factor by dividing the quantity on the right side of the equal sign by the information on the left side, removing the "1," and condensing the primary unit to "g/mol." The chemical formula of the entire compound is still utilized as the secondary unit in this representation of molecular weight, as shown below. 30.08 g/mol C2H6 Answer e Indicator Information & Equality Patterns The word "molecules" indicates that an Avogadro's number equality should be developed and applied to solve this problem. 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 formula for the substance that is written in closest physical proximity to the indicator word "molecules," "water," H2O, is incorporated into both of the secondary unit positions in this equality, as shown below. Because the given chemical information is the name of a covalent molecule, the indicator word "molecules" appropriately corresponds to the chemical formula that is applied in this equality. 1 mol H2O = 6.02 × 1023 H2O molecules Additionally, 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 formula for the substance that is written in closest physical proximity to the indicator word "grams," "dicarbon hexahydride," C2H6, is incorporated into the mass-based equality. Since dicarbon hexahydride, C2H6, is a compound, a molecular weight equality for this chemical, which was developed in Part (d) and is replicated below, should be applied to solve this problem. 1 mol C2H6 = 30.08 g C2H6 Finally, because both of the chemicals that are referenced in the problem, water, H2O, and dicarbon hexahydride, C2H6, are also present in the given reaction equation, a stoichiometric equality should also be developed and applied to solve this problem. The chemical formulas for water, H2O, and dicarbon hexahydride, C2H6, are incorporated into the secondary unit positions on the left and right sides, respectively, of this equality. Since the coefficient that is associated with water, H2O, is a "6" in the chemical equation that was balanced in Part (a), a 6 is inserted into the numerical position on the left side of this equality. Finally, because the coefficient that corresponds to dicarbon hexahydride, C2H6, is a "2" in the balanced chemical equation, a 2 is inserted into the numerical position on the right side of this equality, as shown below. 6 mol H2O = 2 mol C2H6 Dimensional Analysis In order to completely eliminate the given unit, "grams of dicarbon hexahydride," 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 this conversion factor, and the remaining portion of the equality is written in the numerator. However, the unit that results upon the cancelation of "grams of dicarbon hexahydride" is "mol C2H6," which is not the desired final unit. Therefore, a conversion factor based on the stoichiometric equality must be applied. In order to completely cancel the intermediate unit "mol C2H6," 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, and the remaining portion of the equality is written in the numerator. The unit that results upon the cancelation of the intermediate unit "mol C2H6" is "mol H2O," which still is not the desired final unit. Therefore, the Avogadro's number equality must be applied as a third conversion factor. In order to completely cancel the intermediate unit "mol H2O," 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. 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. The remaining portion of the equality is written in the numerator in the resultant conversion factor, as shown below. \( {\text {76.2}} {\cancel{\rm{g} \; \rm{C_2H_6}}} \times\) \( \dfrac{1 \; \bcancel{\rm{mol} \; \rm{C_2H_6}}}{30.08 \; \cancel{\rm{g} \; \rm{C_2H_6}}}\) × \( \dfrac{6 \; \cancel{\rm{mol} \; \rm{H_2O}}}{2 \; \bcancel{\rm{mol} \; \rm{C_2H_6}}}\) × \( \dfrac{6.02 \times 10^{23} \; \rm{molecules} \; \rm{H_2O}}{1 \; \cancel{\rm{mol} \; \rm{H_2O}}}\) = \( {\text {4.57503989...}} \times {\text{10}^{24}}\) \({\rm{molecules} \; \rm{H_2O}}\) ≈ \( {\text {4.58}} \times {\text{10}^{24}}\) \({\rm{molecules} \; \rm{H_2O}}\) 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, or the "=" key should be used after each 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. Exercise \(\PageIndex{2}\) Consider the following chemical equation. ___ \(\ce{Si_3N_4} \left( s \right) \rightarrow\) ___ \( \ce{Si} \left( s \right) + \) ___ \( \ce{N_2} \left( g \right)\) Balance this equation by writing coefficients in the "blanks," as necessary. Classify this reaction as a combination, a decomposition, a single replacement, a double replacement, or a combustion reaction. Identify the state of matter in which molecular nitrogen participates in this reaction. Calculate the molecular weight of trisilicon tetranitride. Represent the resultant solution as a molar equality and as a "hidden" conversion factor. Calculate how many grams of trisilicon tetranitride must be consumed if 7.6 x 1023 atoms of silicon are generated in this reaction. Answer a In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of silicon, Si, and nitrogen, N, that are present in the given equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced Si 3 1 N 4 2 Neither of the components of this reaction are balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. Neither of the unbalanced components in this reaction is present in multiple formulas on the same side of the reaction arrow. However, as silicon, Si, is represented in its atomic form, the balancing process should not be initiated using this element. Therefore, nitrogen, N, should be the first element that is balanced in this equation. In order to balance nitrogen, N, a coefficient should be written in the "blank" that corresponds to molecular nitrogen, N2, on the right side of the equation, as fewer nitrogens are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 4, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{Si_3N_4} \left( s \right) \rightarrow\) ___ \( \ce{Si} \left( s \right) + \) 2 \( \ce{N_2} \left( g \right)\) As nitrogen, N, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amount of silicon, Si, that is present on the product side of the equation. The updated quantity of this element is reflected in the table that is shown below. Inserting this coefficient balances nitrogen, N, as intended. The quantities in which silicon, Si, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced Si 3 1 N 4 \( \cancel{\rm{2} } \) 4 In order to balance silicon, Si, a coefficient should be written in the "blank" that corresponds to this element on the right side of the equation, as fewer silicons are present on this side of the reaction arrow. The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 3, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{Si_3N_4} \left( s \right) \rightarrow\) 3 \( \ce{Si} \left( s \right) + \) 2 \( \ce{N_2} \left( g \right)\) As silicon, Si, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amount of nitrogen, N, that is present on the product side of the equation. The updated quantity of this element is reflected in the table that is shown below. Inserting this coefficient balances silicon, Si, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced Si 3 \( \cancel{\rm{1} } \) 3 N 4 \( \bcancel{\rm{2} } \) 4 No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1" that is understood to occupy the first "blank" in this equation, these coefficients cannot be divided. Therefore, the final equation that is presented above is chemically-correct, as written. Answer b This reaction is classified as a decomposition because only a single molecule, trisilicon tetranitride, Si3N4, is present on the left side of the reaction arrow. Answer c The chemical formula for molecular nitrogen, N2, is derived by applying the Chapter 3 rules for determining diatomic covalent chemical formulas. Molecular nitrogen, N2, participates in this reaction as a gas, as indicated by the "\( \left( g \right)\)" that is associated with its formula in the given chemical equation. Answer d In order to calculate the molecular weight of a compound, the mass contribution of each element that is found within that compound must first be determined. The chemical formula for trisilicon tetranitride, Si3N4, is determined based on the numerical prefixes in the given chemical name, in alignment with the Chapter 3 rules for determining covalent chemical formulas. Because trisilicon tetranitride, Si3N4, contains two elements, the mass contributions of both silicon, Si, and nitrogen, N, must be calculated. 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; the subscript for the element that is being considered 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 each resultant calculated value is reported to the hundredths place. The mass contribution calculations for trisilicon tetranitride, Si3N4, are shown below. \( {\text {3}}\) \({\cancel{\rm{mol} \; \rm{Si}}}\) × \( \dfrac{28.09 \; \rm{g} \; \rm{Si}}{1\; \cancel{\rm{mol} \; \rm{Si}}}\) = \( {\text {84.27}}\) \({\rm{g} \; \rm{Si}}\) \( {\text {4}}\) \({\cancel{\rm{mol} \; \rm{N}}}\) × \( \dfrac{14.01 \; \rm{g} \; \rm{N}}{1\; \cancel{\rm{mol} \; \rm{N}}}\) = \( {\text {56.04}}\) \({\rm{g} \; \rm{N}}\) The numerical value of the molecular weight of trisilicon tetranitride, Si3N4, 140.31, is determined by adding the mass contributions that are shown above and should be reported to the hundredths place. 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 the entire compound is utilized as the secondary unit on both sides of the resultant equality, as shown below. 1 mol Si3N4 = 140.31 g Si3N4 Finally, this equality can be rewritten as a "hidden" conversion factor by dividing the quantity on the right side of the equal sign by the information on the left side, removing the "1," and condensing the primary unit to "g/mol." The chemical formula of the entire compound is still utilized as the secondary unit in this representation of molecular weight, as shown below. 140.31 g/mol Si3N4 Answer e Indicator Information & Equality Patterns 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 formula for the substance that is written in closest physical proximity to the indicator word "grams," "trisilicon tetranitride," Si3N4, is incorporated into the mass-based equality. Since trisilicon tetranitride, Si3N4, is a compound, a molecular weight equality for this chemical, which was developed in Part (d) and is replicated below, should be applied to solve this problem. 1 mol Si3N4 = 140.31 g Si3N4 Additionally, the word "atoms" indicates that an Avogadro's number equality should be developed and applied to solve this problem. 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 formula for the substance that is written in closest physical proximity to the indicator word "atoms," "silicon," Si, is incorporated into both of the secondary unit positions in this equality, as shown below. Because the given chemical information is the name of an element, the indicator word "atoms" appropriately corresponds to the chemical formula that is applied in this equality. 1 mol Si = 6.02 × 1023 Si atoms Finally, because both of the chemicals that are referenced in the problem, trisilicon tetranitride, Si3N4, and silicon, Si, are also present in the given reaction equation, a stoichiometric equality should also be developed and applied to solve this problem. The chemical formulas for trisilicon tetranitride, Si3N4, and silicon, Si, are incorporated into the secondary unit positions on the left and right sides, respectively, of this equality. Since the coefficient that is associated with trisilicon tetranitride, Si3N4, is an unwritten "1" in the chemical equation that was balanced in Part (a), a 1 is inserted into the numerical position on the left side of this equality. Finally, because the coefficient that corresponds to silicon, Si, is a "3" in the balanced chemical equation, a 3 is inserted into the numerical position on the right side of this equality, as shown below. 1 mol Si3N4 = 3 mol Si Dimensional Analysis In order to completely eliminate the given unit, "atoms of silicon," a conversion factor based on the Avogadro's number equality must be applied first. Specifically, the quantity on the right side of this equality becomes the denominator in this 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 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 "atoms of silicon" is "mol Si," which is not the desired final unit. Therefore, a conversion factor based on the stoichiometric equality must be applied. In order to completely cancel the intermediate unit "mol Si," 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, and the remaining portion of the equality is written in the numerator. The unit that results upon the cancelation of the intermediate unit "mol Si" is "mol Si3N4," which still is not the desired final unit. Therefore, the molecular weight equality must be applied as a third conversion factor. In order to completely cancel the intermediate unit "mol Si3N4," 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, as shown below. \( {\text {7.6}} \times {\text{10}^{23}}\) \({\cancel{\rm{atoms} \; \rm{Si}}} \times\) \( \dfrac{1 \; \bcancel{\rm{mol} \; \rm{Si}}}{6.02 \times 10^{23} \; \cancel{\rm{atoms} \; \rm{Si}}}\) × \( \dfrac{1 \; \cancel{\rm{mol} \; \rm{Si_3N_4}}}{3 \; \bcancel{\rm{mol} \; \rm{Si}}}\) × \( \dfrac{140.31 \; \rm{g} \; \rm{Si_3N_4}}{1 \; \cancel{\rm{mol} \; \rm{Si_3N_4}}}\) = \( {\text {59.04518...}}\) \({\rm{g} \; \rm{Si_3N_4}}\) ≈ \( {\text {59}}\) \({\rm{g} \; \rm{Si_3N_4}}\) 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, or the "=" key should be used after each 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. Exercise \(\PageIndex{3}\) Consider the following chemical equation. ___ \(\ce{Sn(Cr_2O_7)_2} \left( aq \right) + \) ___ \( \ce{Na} \left( s \right) \rightarrow\) ___ \( \ce{Sn} \left( s \right) + \) ___ \( \ce{Na_2Cr_2O_7} \left( aq \right)\) Balance this equation by writing coefficients in the "blanks," as necessary. (Hint: Cr2O7–2, which is known as the "dichromate ion," is a polyatomic anion.) Classify this reaction as a combination, a decomposition, a single replacement, a double replacement, or a combustion reaction. Identify which chemicals participate in this reaction in the aqueous state of matter. Calculate the molecular weight of tin (IV) dichromate, Sn(Cr2O7)2. Represent the resultant solution as a molar equality and as a "hidden" conversion factor. Calculate how many grams of sodium must be present to completely consume 1.59 kilograms of tin (IV) dichromate, Sn(Cr2O7)2. Answer a In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. As indicated in the hint that is provided, the dichromate ion, Cr2O7–2, is a polyatomic anion. Because this polyatomic ion is present in both a reactant and a product in the given equation, the dichromate ion, Cr2O7–2, should be balanced as a single entity. Recall that parentheses are utilized in ionic chemical formulas to offset a polyatomic ion as a unit, and the subscript that is written outside of the parentheses indicates the quantity in which that ion is present within the ionic compound. If parentheses are not explicitly-written around the formula of a polyatomic ion, an unwritten subscript of "1" is understood to correspond to that polyatomic unit. Finally, tin, Sn, and sodium, Na, are also present in the given reaction equation. The quantities in which these chemicals are present in the given equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced Cr2O7–2 2 1 Sn 1 1 Na 1 2 Since both sides of the equation contain equal amounts of tin, Sn, that element is balanced. Neither the dichromate ion, Cr2O7–2, nor sodium, Na, is balanced, because each of these components is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. Neither of the unbalanced components in this reaction is present in multiple formulas on the same side of the reaction arrow. However, as sodium, Na, is represented in its atomic form, the balancing process should not be initiated using this element. Therefore, the dichromate ion, Cr2O7–2, should be the first chemical that is balanced in this equation. In order to balance the dichromate ion, Cr2O7–2, a coefficient should be written in the "blank" that corresponds to sodium dichromate, Na2Cr2O7, on the right side of the equation, as fewer the dichromate ions are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this ion, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{Sn(Cr_2O_7)_2} \left( aq \right) + \) ___ \( \ce{Na} \left( s \right) \rightarrow\) ___ \( \ce{Sn} \left( s \right) + \) 2 \( \ce{Na_2Cr_2O_7} \left( aq \right)\) As sodium dichromate, Na2Cr2O7, contains both sodium, Na, and the dichromate ion, Cr2O7–2, incorporating this coefficient alters the amounts of both of these components on the product side of the equation. The updated quantities of these components are reflected in the table that is shown below. Inserting this coefficient balances the dichromate ion, Cr2O7–2, as intended. The quantities in which sodium, Na, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced Cr2O7–2 2 \( \cancel{\rm{1} } \) 2 Sn 1 1 Na 1 \( \cancel{\rm{2} } \) 4 In order to balance sodium, Na, a coefficient should be written in the "blank" that corresponds to this element on the left side of the equation, as fewer sodiums are present on this side of the reaction arrow. The value of this coefficient, 4, is determined by dividing the larger quantity of this element, 4, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{Sn(Cr_2O_7)_2} \left( aq \right) + \) 4 \( \ce{Na} \left( s \right) \rightarrow\) ___ \( \ce{Sn} \left( s \right) + \) 2 \( \ce{Na_2Cr_2O_7} \left( aq \right)\) As sodium, Na, is present in its atomic form, incorporating this coefficient only alters the amount of this element on the reactant side of the equation, as reflected in the table that is shown below. Inserting this coefficient balances sodium, Na, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced Cr2O7–2 2 \( \cancel{\rm{1} } \) 2 Sn 1 1 Na \( \bcancel{\rm{1} } \) 4 \( \cancel{\rm{2} } \) 4 No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1"s that are understood to occupy the first and third "blanks" in this equation, these coefficients cannot be divided. Therefore, the final equation that is presented above is chemically-correct, as written. Answer b This reaction is classified as a single replacement because an elemental reactant, sodium, Na, and the metallic portion of a compound, tin, Sn, exchange positions. Answer c Tin (IV) dichromate, Sn(Cr2O7)2, and sodium dichromate, Na2Cr2O7, participate in this reaction in the aqueous state of matter, as indicated by the "\( \left( aq \right)\)"s that are associated with their formulas in the given chemical equation. Answer d In order to calculate the molecular weight of a compound, the mass contribution of each element that is found within that compound must first be determined. Because tin (IV) dichromate, Sn(Cr2O7)2, contains three elements, the mass contributions of tin, Sn, chromium, Cr, and oxygen, O, must all be calculated. 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; the amount in which the element that is being considered is present in the compound 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 each resultant calculated value is reported to the hundredths place. Additionally, recall that parentheses are utilized in ionic chemical formulas to offset a polyatomic ion as a unit, and the subscript that is written outside of the parentheses indicates the quantity in which that ion is present within the compound. Any subscripts that are written inside of the parentheses correspond to an element that is found within that polyatomic ion. Therefore, in order to determine the amounts in which both chromium, Cr, and oxygen, O, are present in tin (IV) dichromate, Sn(Cr2O7)2, the subscript that is associated with each elemental symbol must be multiplied by the subscript that is written outside of the parentheses. The mass contribution calculations for tin (IV) dichromate, Sn(Cr2O7)2, are shown below. \( {\text {1}}\) \({\cancel{\rm{mol} \; \rm{Sn}}}\) × \( \dfrac{118.71 \; \rm{g} \; \rm{Sn}}{1\; \cancel{\rm{mol} \; \rm{Sn}}}\) = \( {\text {118.71}}\) \({\rm{g} \; \rm{Sn}}\) \( {\text {4}}\) \({\cancel{\rm{mol} \; \rm{Cr}}}\) × \( \dfrac{51.97 \; \rm{g} \; \rm{Cr}}{1\; \cancel{\rm{mol} \; \rm{Cr}}}\) = \( {\text {207.88}}\) \({\rm{g} \; \rm{Cr}}\) \( {\text {14}}\) \({\cancel{\rm{mol} \; \rm{O}}}\) × \( \dfrac{16.00 \; \rm{g} \; \rm{O}}{1\; \cancel{\rm{mol} \; \rm{O}}}\) = \( {\text {224.00}}\) \({\rm{g} \; \rm{O}}\) The numerical value of the molecular weight of tin (IV) dichromate, Sn(Cr2O7)2, 550.59, is determined by adding the mass contributions that are shown above and should be reported to the hundredths place. 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 the entire compound is utilized as the secondary unit on both sides of the resultant equality, as shown below. 1 mol Sn(Cr2O7)2 = 550.59 g Sn(Cr2O7)2 Finally, this equality can be rewritten as a "hidden" conversion factor by dividing the quantity on the right side of the equal sign by the information on the left side, removing the "1," and condensing the primary unit to "g/mol." The chemical formula of the entire compound is still utilized as the secondary unit in this representation of molecular weight, as shown below. 550.59 g/mol Sn(Cr2O7)2 Answer e Indicator Information & Equality Patterns 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 formula for the substance that is written in closest physical proximity to the indicator word "grams," "sodium," Na, is incorporated into the mass-based equality. Since sodium, Na, is an element, an atomic weight equality for this chemical should be developed by equating the atomic weight of the element, which should be reported to the hundredths place, to 1 mol of the element, and utilizing an elemental symbol as the secondary unit on both sides of the resultant equality, as shown below. 1 mol Na = 22.99 g Na Additionally, the mass unit "kilograms" indicates that a second mass-based equality should be developed and applied to solve this problem. As multiple chemicals are referenced in this problem, the chemical formula for the substance that is written in closest physical proximity to the indicator word "kilograms," "tin (IV) dichromate," Sn(Cr2O7)2, is incorporated into the second mass-based equality. Since tin (IV) dichromate, Sn(Cr2O7)2, is a compound, a molecular weight equality for this chemical, which was developed in Part (d) and is replicated below, should be applied to solve this problem. 1 mol Sn(Cr2O7)2 = 550.59 g Sn(Cr2O7)2 Finally, because both of the chemicals that are referenced in the problem, sodium, Na, and tin (IV) dichromate, Sn(Cr2O7)2, are also present in the given reaction equation, a stoichiometric equality should also be developed and applied to solve this problem. The chemical formulas for sodium, Na, and tin (IV) dichromate, Sn(Cr2O7)2, are incorporated into the secondary unit positions on the left and right sides, respectively, of this equality. Since the coefficient that is associated with sodium, Na, is a "4" in the chemical equation that was balanced in Part (a), a 4 is inserted into the numerical position on the left side of this equality. Finally, because the coefficient that corresponds to tin (IV) dichromate, Sn(Cr2O7)2, is an unwritten "1" in the balanced chemical equation, a 1 is inserted into the numerical position on the right side of this equality, as shown below. 4 mol Na = 1 mol Sn(Cr2O7)2 Dimensional Analysis None of the equalities that were developed above can be used to completely eliminate the given unit, "kilograms of tin (IV) dichromate, Sn(Cr2O7)2," because the mass unit that is referenced in the problem, "kilograms," and the unit that is included in both mass-based equalities, "grams," do not match. Therefore, a prefix modifier equality that relates these two units must first be developed and applied. As discussed in Section 1.8, a unit that contains a prefix modifier, such as "kilograms," can be related back to its base unit, "grams," by replacing the prefix modifier, "kilo" (k), with its meaning, "103". The base unit, gram (g), is written with both the prefix modifier and its meaning. Finally, since their base units (grams) match, and the prefix modifier and its meaning are equivalent, these two quantities can be equated to one another, as shown below. kg = 103 g While not absolutely necessary, a prefix modifier equality can be simplified by rewriting the numerical value in decimal format, as follows. kg = 1,000 g Before applying one of these relationships to cancel the given unit, the chemical formula for tin (IV) dichromate, Sn(Cr2O7)2, must be added as a secondary unit on each side of the equal sign, as all equalities in this chapter must contain two units. After incorporating this elemental symbol, either equality can be applied to completely cancel the given unit, "kilograms of tin (IV) dichromate, Sn(Cr2O7)2." Regardless of which equality is selected, the quantity on the left side of the equal sign becomes the denominator in the first conversion factor that is applied to solve the given problem, and the remaining portion of the equality is written in the numerator. However, the unit that results upon the cancelation of "kilograms of tin (IV) dichromate, Sn(Cr2O7)2" is "g Sn(Cr2O7)2," which is not the desired final unit. Therefore, a conversion factor based on the molecular weight of tin (IV) dichromate, Sn(Cr2O7)2, must be applied. In order to completely cancel the intermediate unit "g Sn(Cr2O7)2," the quantity on the right side of this equality becomes the denominator in this conversion factor, and the remaining portion of the equality is written in the numerator. The unit that results upon the cancelation of "g Sn(Cr2O7)2" is "mol Sn(Cr2O7)2," which is still not the desired final unit. Therefore, a conversion factor based on the stoichiometric equality must be applied. In order to completely cancel the intermediate unit "mol Sn(Cr2O7)2," the quantity on the right side of this equality becomes the denominator in the third conversion factor that is applied to solve the given problem, and the remaining portion of the equality is written in the numerator. The unit that results upon the cancelation of the intermediate unit "mol Sn(Cr2O7)2" is "mol Na," which still is not the desired final unit. Therefore, the atomic weight equality must be applied as a fourth conversion factor. In order to completely cancel the intermediate unit "mol Na," the quantity on the left side of this equality becomes the denominator in the fourth conversion factor that is applied to solve the given problem, and the remaining portion of the equality is written in the numerator, as shown below. \( {\text {1.59}} {\cancel{\rm{kg} \; \rm{Sn(Cr_2O_7)_2}}} \times\) \( \dfrac{1,000\; \bcancel{\rm{g} \; \rm{Sn(Cr_2O_7)_2}}}{ \; \cancel{\rm{kg} \; \rm{Sn(Cr_2O_7)_2}}}\) × \( \dfrac{1 \; \cancel{\rm{mol} \; \rm{Sn(Cr_2O_7)_2}}}{550.59 \; \bcancel{\rm{g} \; \rm{Sn(Cr_2O_7)_2}}}\) × \( \dfrac{4 \; \bcancel{\rm{mol} \; \rm{Na}}}{1 \; \cancel{\rm{mol} \; \rm{Sn(Cr_2O_7)_2}}}\) × \( \dfrac{22.99 \; \rm{g} \; \rm{Na}}{1 \; \bcancel{\rm{mol} \; \rm{Na}}}\) = \( {\text {265.5631...}}\) \({\rm{g} \; \rm{Na}}\) ≈ \( {\text {266}}\) \({\rm{g} \; \rm{Na}}\) 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, or the "=" key should be used after each division. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above. Exercise \(\PageIndex{4}\) Complete the following double replacement reaction by determining the chemical formula(s) of the product(s) that are predicted to form. Balance the resultant equation by writing coefficients in the "blanks," as necessary. (States of matter are not required.) ___ \(\ce{AuCl_3 }\) + ___ \( \ce{H_2O} \rightarrow \) Answer In a double replacement reaction, either the cationic or the anionic portions of two compounds exchange their relative positions. Executing a double replacement reaction through a cation exchange has the same net result as repositioning the corresponding anions. Based on the Chapter 3 rules for determining chemical formulas, the symbol of the cationic component of an ionic compound is always written before the symbol of the corresponding anion. Therefore, gold, Au, and hydrogen, H, are the cations in the first and second reactants, respectively, in the partial chemical equation that is shown above. The first product that would result upon exchanging these reaction components would contain hydrogen, H, and chlorine, Cl, and the second product would be comprised of gold, Au, and oxygen, O. However, the subscripts that are present within a chemical formula are solely dependent on the elemental, ionic, or covalent nature of the corresponding substance. Therefore, the chemical formulas of the compounds that are expected to form in the reaction that is shown above cannot be obtained simply by rewriting the chemical information, as given, after exchanging the relative positions of the appropriate components. Because hydrogen, H, and chlorine, Cl, are both non-metals, the formula of the molecule that contains these elements can only be established by applying the Chapter 3 rules for determining covalent chemical formulas. Therefore, a corresponding Lewis structure must first be drawn by writing the electron dot structures for each element and then correctly pairing the unpaired electrons on each symbol. The chemical formula of the molecule that results, which contains one hydrogen and one chlorine, is HCl. Because gold, Au, is a metal, and oxygen, O, is a non-metal, the formula of the compound that contains these elements can only be established by applying the Chapter 3 rules for determining ionic chemical formulas. Gold, Au, is a transition metal and, therefore, can ionize to form several unique cations. In the reaction that is shown above, the gold (III) ion, Au+3, must be present in the first reactant, in order to form AuCl3 upon bonding with the chloride ion, Cl–1. As the charge of an ion does not change during a chemical reaction, the gold (III) ion, Au+3, bonds with the oxide ion, O–2, to produce gold (III) oxide, Au2O3. Writing these chemical formulas on the product side of the reaction arrow results in the unbalanced chemical equation that is shown below. ___ \(\ce{AuCl_3 }\) + ___ \( \ce{H_2O} \rightarrow \) ___ \(\ce{HCl }\) + ___ \( \ce{Au_2O_3}\) In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of gold, Au, chlorine, Cl, hydrogen, H, and oxygen, O, that are present in the given equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced Au 1 2 Cl 3 1 H 2 1 O 1 3 None of the components of this reaction are balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. None of the unbalanced components in this reaction are present in multiple formulas on the same side of the reaction arrow or are represented in an atomic form. Therefore, the balancing process could be initiated by selecting gold, Au, chlorine, Cl, hydrogen, H, or oxygen, O, for further analysis. In order to balance gold, Au, a coefficient should be written in the "blank" that corresponds to gold (III) chloride, AuCl3, on the left side of the equation, as fewer golds are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 2 \(\ce{AuCl_3 }\) + ___ \( \ce{H_2O} \rightarrow \) ___ \(\ce{HCl }\) + ___ \( \ce{Au_2O_3}\) As gold (III) chloride, AuCl3, contains both gold, Au, and chlorine, Cl, incorporating this coefficient alters the amounts of both of these elements on the reactant side of the equation. The updated quantities of these elements are reflected in the table that is shown below. Inserting this coefficient balances gold, Au, as intended. The quantities in which chlorine, Cl, hydrogen, H, and oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced Au \( \cancel{\rm{1} } \) 2 2 Cl \( \cancel{\rm{3} } \) 6 1 H 2 1 O 1 3 In order to balance chlorine, Cl, a coefficient should be written in the "blank" that corresponds to hydrogen chloride, HCl, on the right side of the equation, as fewer chlorines are present on this side of the reaction arrow. The value of this coefficient, 6, is determined by dividing the larger quantity of this element, 6, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 2 \(\ce{AuCl_3 }\) + ___ \( \ce{H_2O} \rightarrow \) 6 \(\ce{HCl }\) + ___ \( \ce{Au_2O_3}\) As hydrogen chloride, HCl, contains both hydrogen, H, and chlorine, Cl, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. Inserting this coefficient balances chlorine, as intended. The quantities in which hydrogen, H, and oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced Au \( \cancel{\rm{1} } \) 2 2 Cl \( \cancel{\rm{3} } \) 6 \( \bcancel{\rm{1} } \) 6 H 2 \( \bcancel{\rm{1} } \) 6 O 1 3 In order to balance hydrogen, H, a coefficient should be written in the "blank" that corresponds to water, H2O, on the left side of the equation, as fewer hydrogens are present on this side of the reaction arrow. The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 6, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. 2 \(\ce{AuCl_3 }\) + 3 \( \ce{H_2O} \rightarrow \) 6 \(\ce{HCl }\) + ___ \( \ce{Au_2O_3}\) As water, H2O, contains both hydrogen, H, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. Inserting this coefficient balances hydrogen, as intended, and also balances oxygen. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced Au \( \cancel{\rm{1} } \) 2 2 Cl \( \cancel{\rm{3} } \) 6 \( \bcancel{\rm{1} } \) 6 H \( \cancel{\rm{2} } \) 6 \( \bcancel{\rm{1} } \) 6 O \( \cancel{\rm{1} } \) 3 3 No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1" that is understood to occupy the fourth "blank" in this equation, these coefficients cannot be divided. Therefore, the final equation that is presented above is chemically-correct, as written. Exercise \(\PageIndex{5}\) Complete the following combination reaction by determining the chemical formula(s) of the product(s) that are predicted to form. Balance the resultant equation by writing coefficients in the "blanks," as necessary. (States of matter are not required.) ___ \(\ce{K}\) + ___ \( \ce{O_2} \rightarrow \) Answer In a combination reaction, the components of two or more reactants bond with one another to form a single product. Therefore, only a single chemical formula, which must contain both potassium, K, and oxygen, O, can be written on the right side of the reaction arrow. However, the subscripts that are present within a chemical formula are solely dependent on the elemental, ionic, or covalent nature of the corresponding substance. Therefore, the chemical formula of the compound that is expected to form in the reaction that is shown above cannot be obtained simply by removing the plus sign from the reactant side of the equation and rewriting the remaining chemical information as a single formula. Because potassium, K, is a metal, and oxygen, O, is a non-metal, the formula of the compound that contains these elements can only be established by applying the Chapter 3 rules for determining ionic chemical formulas. In this reaction, the potassium ion, K+1, bonds with the oxide ion, O–2, to produce potassium oxide, K2O. Writing this chemical formula on the product side of the reaction arrow results in the unbalanced chemical equation that is shown below. ___ \(\ce{K}\) + ___ \( \ce{O_2} \rightarrow \) ___ \(\ce{K_2O}\) In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of potassium, K, and oxygen, O, that are present in the given equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced K 1 2 O 2 1 Neither of the components of this reaction are balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. Neither of the unbalanced components in this reaction is present in multiple formulas on the same side of the reaction arrow. However, as potassium, K, is represented in its atomic form, the balancing process should not be initiated using this element. Therefore, oxygen, O, should be the first element that is balanced in this equation. In order to balance oxygen, O, a coefficient should be written in the "blank" that corresponds to potassium oxide, K2O, on the right side of the equation, as fewer oxygens are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{K}\) + ___ \( \ce{O_2} \rightarrow \) 2 \(\ce{K_2O}\) As potassium oxide, K2O, contains both potassium, K, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. Inserting this coefficient balances oxygen, O, as intended. The quantities in which potassium, K, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced K 1 \( \cancel{\rm{2} } \) 4 O 2 \( \cancel{\rm{1} } \) 2 In order to balance potassium, K, a coefficient should be written in the "blank" that corresponds to this element on the left side of the equation, as fewer potassiums are present on this side of the reaction arrow. The value of this coefficient, 4, is determined by dividing the larger quantity of this element, 4, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 4 \(\ce{K}\) + ___ \( \ce{O_2} \rightarrow \) 2 \(\ce{K_2O}\) As potassium, K, is present in its atomic form, incorporating this coefficient only alters the amount of this element on the reactant side of the equation, as reflected in the table that is shown below. Inserting this coefficient balances potassium, K, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced K \( \bcancel{\rm{1} } \) 4 \( \cancel{\rm{2} } \) 4 O 2 \( \cancel{\rm{1} } \) 2 No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1" that is understood to occupy the second "blank" in this equation, these coefficients cannot be divided. Therefore, the final equation that is presented above is chemically-correct, as written. Exercise \(\PageIndex{6}\) Complete the following single replacement reaction by determining the chemical formula(s) of the product(s) that are predicted to form. Balance the resultant equation by writing coefficients in the "blanks," as necessary. (States of matter are not required.) ___ \(\ce{Li}\) + ___ \( \ce{Cu_3PO_4} \rightarrow \) Answer In a single replacement reaction, an elemental reactant exchanges positions with a portion of a compound that has an identical metallic classification. Therefore, since the elemental reactant in the given equation is a metal, lithium, Li, must exchange with the metallic portion of copper (I) phosphate, Cu3PO4, which is an ionic compound. Based on the Chapter 3 rules for determining chemical formulas, the symbol of the cationic component of an ionic compound, which is usually generated when a metal loses valence electrons, is always written before the symbol of the corresponding anion. Therefore, copper, Cu, is the metallic portion of this compound that exchanges positions with lithium, Li, in this reaction. The resultant products of this reaction are elemental copper, Cu, and a compound that contains both lithium, Li, and the unaltered reaction component, the phosphate ion, PO4–3. However, the subscripts that are present within a chemical formula are solely dependent on the elemental, ionic, or covalent nature of the corresponding substance. Therefore, the chemical formulas of the compounds that are expected to form in the reaction that is shown above cannot be obtained simply by rewriting the chemical information, as given, after exchanging the relative positions of the appropriate components. Because lithium, Li, is a metal, and the phosphate ion, PO4–3, is a polyatomic anion, the formula of the compound that contains these elements can only be established by applying the Chapter 3 rules for determining ionic chemical formulas. In this reaction, the lithium ion, Li+1, bonds with the phosphate ion, PO4–3, to produce lithium phosphate, Li3PO4. Writing these chemical formulas on the product side of the reaction arrow results in the unbalanced chemical equation that is shown below. ___ \(\ce{Li}\) + ___ \( \ce{Cu_3PO_4} \rightarrow \) ___ \(\ce{Cu}\) + ___ \( \ce{Li_3PO_4}\) In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. The phosphate ion, PO4–3, is a polyatomic anion. Because this polyatomic ion is present in both a reactant and a product in the given equation, the phosphate ion, PO4–3, should be balanced as a single entity. Recall that parentheses are utilized in ionic chemical formulas to offset a polyatomic ion as a unit, and the subscript that is written outside of the parentheses indicates the quantity in which that ion is present within the ionic compound. If parentheses are not explicitly-written around the formula of a polyatomic ion, an unwritten subscript of "1" is understood to correspond to that polyatomic unit. Finally, lithium, Li, and copper, Cu, are also present in the reaction equation that is shown above. The quantities in which these chemicals are present in the reaction equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced PO4–3 1 1 Li 1 3 Cu 3 1 Since both sides of the equation contain equal amounts of the phosphate ion, PO4–3, that ion is balanced. Neither lithium, Li, nor copper, Cu, is balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. Neither of the unbalanced components in this reaction is present in multiple formulas on the same side of the reaction arrow. Typically, the balancing process should not begin by determining a coefficient for a chemical that is present in its atomic form. However, because both of the unbalanced components in this reaction, lithium, Li, and copper, Cu, are represented in their atomic forms, the balancing process could be initiated by selecting either of these elements for further analysis. In order to balance lithium, Li, a coefficient should be written in the "blank" that corresponds to this element on the left side of the equation, as fewer lithiums are present on this side of the reaction arrow. The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 3, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 3 \(\ce{Li}\) + ___ \( \ce{Cu_3PO_4} \rightarrow \) ___ \(\ce{Cu}\) + ___ \( \ce{Li_3PO_4}\) As lithium, Li, is present in its atomic form, incorporating this coefficient only alters the amount of this element on the reactant side of the equation, as reflected in the table that is shown below. Inserting this coefficient balances lithium, Li, as intended. The quantities in which copper, Cu, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced PO4–3 1 1 Li \( \cancel{\rm{1} } \) 3 3 Cu 3 1 In order to balance copper, Cu, a coefficient should be written in the "blank" that corresponds to this element on the right side of the equation, as fewer coppers are present on this side of the reaction arrow. The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 3, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 3 \(\ce{Li}\) + ___ \( \ce{Cu_3PO_4} \rightarrow \) 3 \(\ce{Cu}\) + ___ \( \ce{Li_3PO_4}\) As copper, Cu, is present in its atomic form, incorporating this coefficient only alters the amount of this element on the product side of the equation, as reflected in the table that is shown below. Inserting this coefficient balances copper, Cu, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced PO4–3 1 1 Li \( \cancel{\rm{1} } \) 3 3 Cu 3 \( \bcancel{\rm{1} } \) 3 No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1"s that are understood to occupy the second and fourth "blanks" in this equation, these coefficients cannot be divided. Therefore, the final equation that is presented above is chemically-correct, as written. Exercise \(\PageIndex{7}\) Write a complete reaction for the combustion of ethanol, C2H5OH. Balance the resultant equation by writing coefficients, as necessary. (States of matter are not required.) Answer In a combustion reaction, a chemical is burned in the presence of molecular oxygen to form carbon dioxide, water, and energy. Therefore, molecular oxygen, O2, must participate in the reaction as a reactant, and carbon dioxide, CO2, water, H2O, and energy, E, must be generated as products. As a result, the only variable component of a combustion reaction is the chemical that is being burned. Writing these chemical formulas on the appropriate sides of the reaction arrow results in the unbalanced chemical equation that is shown below. ___ \(\ce{C_2H_5OH}\) + ___ \( \ce{O_2} \rightarrow \) ___ \(\ce{CO_2 }\) + ___ \( \ce{H_2O}\) + E Coefficients are incorporated into a chemical equation in order to account for any relative differences between the chemical formulas of the reactants and products that are involved in the corresponding chemical reaction. However, energy, E, is not a chemical material and, therefore, does not contain elements or polyatomic ions. As a result, there is no need to associate a balancing coefficient with the energy that is produced in this reaction. In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of carbon, C, hydrogen, H, and oxygen, O, that are present in the given equation are summarized in the table that is shown below. Note that hydrogen, H, appears in two distinctive places within the chemical formula for ethanol, C2H5OH. In order to determine the number of hydrogens that are present within this molecule, the subscripts on both hydrogens must be added. Furthermore, oxygen, O, is present in both of the reactants shown in this equation and is contained in both of the chemical products that are generated during this reaction. In order to determine the number of oxygens that are present on both sides of this equation, the subscripts on the relevant oxygens must be added. Element or Ion Reactants Products Balanced C 2 1 H 6 2 O 3 3 Since both sides of the equation contain equal amounts of oxygen, O, that element is balanced. Neither carbon, C, nor hydrogen, H, are balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. Neither of the unbalanced components in this reaction is present in multiple formulas on the same side of the reaction arrow or is represented in its atomic form. Therefore, the balancing process could be initiated by selecting either carbon, C, or hydrogen, H, for further analysis. In order to balance carbon, C, a coefficient should be written in the "blank" that corresponds to carbon dioxide, CO2, on the right side of the equation, as fewer carbons are present on this side of the reaction arrow. The value of this coefficient, 2, is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_2H_5OH}\) + ___ \( \ce{O_2} \rightarrow \) 2 \(\ce{CO_2 }\) + ___ \( \ce{H_2O}\) + E As carbon dioxide, CO2, contains both carbon, C, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. As stated above, because oxygen is present in both of the chemical products that are generated during this reaction, the quantities of oxygen in both of these products must be added to determine the amount of oxygen that is present on the product side of this equation. Inserting this coefficient balances carbon, C, as intended. The quantities in which hydrogen, H, are present are still unequal. Additionally, the quantities in which oxygen, O, are present, which had been identical, are now unequal. Therefore, this element must now be balanced, as well, and one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced C 2 \( \cancel{\rm{1} } \) 2 H 6 2 O 3 \( \cancel{\rm{3} } \) 5 In order to balance hydrogen, H, a coefficient should be written in the "blank" that corresponds to water, H2O, on the right side of the equation, as fewer hydrogens are present on this side of the reaction arrow. The value of this coefficient, 3, is determined by dividing the larger quantity of this element, 6, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_2H_5OH}\) + ___ \( \ce{O_2} \rightarrow \) 2 \(\ce{CO_2 }\) + 3 \( \ce{H_2O}\) + E As water, H2O, contains both hydrogen, H, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. Again, because oxygen is present in both of the chemical products that are generated during this reaction, the quantities of oxygen in both of these products must be added to determine the amount of oxygen that is present on the product side of this equation. Inserting this coefficient balances hydrogen, H, as intended. The quantities in which oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced C 2 \( \cancel{\rm{1} } \) 2 H 6 \( \bcancel{\rm{2} } \) 6 O 3 \( \cancel{\rm{3} } \) \( \bcancel{\rm{5} } \) 7 In order to balance oxygen, O, a coefficient should be written in a "blank" on the left side of the equation, as fewer oxygens are present on this side of the reaction arrow. However, oxygen, O, is present in both of the reactants in this equation. As the first reactant, ethanol, C2H5OH, also contains carbon, C, and hydrogen, H, which have already been balanced, a coefficient should not be placed in the "blank" that is associated with this molecule. Instead, a coefficient should be written in the "blank" that corresponds to molecular oxygen, O2. The value of this coefficient, 3, is determined by first subtracting 1 from the larger quantity of this element, 7, in order to account for the oxygen, O, that is present in ethanol, C2H5OH, and then dividing the resultant quantity, 6, by the smaller count of this element, 2. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_2H_5OH}\) + 3 \( \ce{O_2} \rightarrow \) 2 \(\ce{CO_2 }\) + 3 \( \ce{H_2O}\) + E As oxygen, O, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amounts of carbon, C, or hydrogen, H, that are present on the reactant side of the equation. The updated quantity of this element is reflected in the table that is shown below. Because oxygen is present in both of the reactants, the quantities of oxygen in both of these molecules must be added to determine the amount of oxygen that is present on the reactant side of this equation. Inserting this coefficient balances oxygen, O, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced C 2 \( \cancel{\rm{1} } \) 2 H 6 \( \bcancel{\rm{2} } \) 6 O \( \bcancel{\rm{3} } \) 7 \( \cancel{\rm{3} } \) \( \bcancel{\rm{5} } \) 7 No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1" that is understood to occupy the first "blank" in this equation, these coefficients cannot be divided. Therefore, the final equation that is presented above is chemically-correct, as written.
