Chapter 8, Problem 8E

Write and balance the equation for the reaction of butane, ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$, with oxygen to form carbon dioxide and water. Use your equation to help answer the following questions. (a) Write, in words, a description of the reaction on the particulate level. (b) If you were to build physical ball-and-stick models of the reactants and products, what minimum number of balls representing atoms of each element do you need if you show both reactants and products at the same time? (c) What if the models of the reactants from Part (b) were built and then rearranged to form products? How many balls would you need? (d) Use words to interpret the equation on the molar level. (e) Use the molar-level interpretation of the equation from Part (d) and molar masses rounded to the nearest gram to show that mass is indeed conserved in this reaction.

Interpretation Introduction

(a)

Interpretation:

The description of the given reaction is to be stated on the particulate level.

Concept introduction:

In a balanced chemical equation, all the reactants and products are written with their stoichiometric coefficients and their physical states. The number of atoms of an element on both sides of a balanced chemical equation is equal.

Answer to Problem 8E

At particulate level, two molecule of butane reacts with thirteen molecules of oxygen to form eight molecules of carbon dioxide and ten molecules of water.

Explanation of Solution

The butane undergoes combustion reaction in presence of oxygen to form carbon dioxide and water. The corresponding chemical reaction is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the given chemical equation, the number of carbon atoms, oxygen atoms, and hydrogen atoms should be same on both the sides. The balanced chemical equation is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+\frac{13}{2}{\text{O}}_2\left(\text{g}\right)\to 4{\text{CO}}_2\left(\text{g}\right)+5{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the number of carbon atoms, oxygen atoms, and hydrogen atoms, the coefficient of $4$ is placed before carbon dioxide, the coefficient of $5$ is placed before water and the coefficient of $\frac{13}{2}$ is placed before oxygen gas.

The reaction is multiplied by factor of $2$ and thus, the reaction becomes,

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+13{\text{O}}_2\left(\text{g}\right)\to 8{\text{CO}}_2\left(\text{g}\right)+10{\text{H}}_2\text{O}\left(\text{g}\right)$

At particulate level, two molecules of butane reacts with thirteen molecules of oxygen to form eight molecules of carbon dioxide and ten molecules of water.

Conclusion

At particulate level, two molecule of butane reacts with thirteen molecules of oxygen to form eight molecules of carbon dioxide and ten molecules of water.

Interpretation Introduction

(b)

Interpretation:

The minimum number of balls representing atoms of each element required to show both reactants and products at the same time is to be stated.

Concept introduction:

In a balanced chemical equation, all the reactants and products are written with their stoichiometric coefficients and their physical states. The number of atoms of an element on both sides of a balanced chemical equation is equal.

Answer to Problem 8E

The minimum number of balls representing atoms of each element required to show both reactants and products at the same time are $16$ carbon atoms, $40$ hydrogen atoms and $52$ oxygen atoms.

Explanation of Solution

The butane undergoes combustion reaction in presence of oxygen to form carbon dioxide and water. The corresponding chemical reaction is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the given chemical equation, the number of carbon atoms, oxygen atoms, and hydrogen atoms should be same on both the sides. The balanced chemical equation is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+\frac{13}{2}{\text{O}}_2\left(\text{g}\right)\to 4{\text{CO}}_2\left(\text{g}\right)+5{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the number of carbon atoms, oxygen atoms, and hydrogen atoms, the coefficient of $4$ is placed before carbon dioxide, the coefficient of $5$ is placed before water and the coefficient of $\frac{13}{2}$ is placed before oxygen gas.

The reaction is multiplied by factor of $2$ and thus, the reaction becomes,

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+13{\text{O}}_2\left(\text{g}\right)\to 8{\text{CO}}_2\left(\text{g}\right)+10{\text{H}}_2\text{O}\left(\text{g}\right)$

At particulate level, two molecule of butane reacts with thirteen molecules of oxygen to form eight molecules of carbon dioxide and ten molecules of water.

In reactants, for one molecule of butane, four atoms of carbon and ten atoms of hydrogen is required and for oxygen molecule, two atoms of oxygen are required.

In products, for one molecule of carbon dioxide, one atom of carbon and two atoms of oxygen are required and for water molecule, two atoms of hydrogen and one atom of oxygen are required.

To depict the given balanced equation, reactant requires, $8$ carbon atoms, $20$ hydrogen atoms and $26$ oxygen atoms, whereas products require $8$ carbon atoms, $20$ hydrogen atoms and $26$ oxygen atoms. In total $16$ carbon atoms, $40$ hydrogen atoms and $52$ oxygen atoms are required.

Conclusion

The minimum number of balls representing atoms of each element required to show both reactants and products at the same time are $16$ carbon atoms, $40$ hydrogen atoms and $52$ oxygen atoms.

Interpretation Introduction

(b)

Interpretation:

The minimum number of balls representing atoms of each element required to show reactants which are then rearranged to products are to be stated.

Concept introduction:

In a balanced chemical equation, all the reactants and products are written with their stoichiometric coefficients and their physical states. The number of atoms of an element on both sides of a balanced chemical equation is equal.

Answer to Problem 8E

The minimum number of balls representing atoms of each element required to show reactants which are then rearranged to products are $8$ carbon atoms, $20$ hydrogen atoms and $26$ oxygen atoms.

Explanation of Solution

The butane undergoes combustion reaction in presence of oxygen to form carbon dioxide and water. The corresponding chemical reaction is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the given chemical equation, the number of carbon atoms, oxygen atoms, and hydrogen atoms should be same on both the sides. The balanced chemical equation is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+\frac{13}{2}{\text{O}}_2\left(\text{g}\right)\to 4{\text{CO}}_2\left(\text{g}\right)+5{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the number of carbon atoms oxygen atoms, and hydrogen atoms, the coefficient of $4$ is placed before carbon dioxide, the coefficient of $5$ is placed before water and the coefficient of $\frac{13}{2}$ is placed before oxygen gas.

The reaction is multiplied by factor of $2$ and thus, the reaction becomes,

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+13{\text{O}}_2\left(\text{g}\right)\to 8{\text{CO}}_2\left(\text{g}\right)+10{\text{H}}_2\text{O}\left(\text{g}\right)$

At particulate level, two molecule of butane reacts with thirteen molecules of oxygen to form eight molecules of carbon dioxide and ten molecules of water.

In reactants, for one molecule of butane, four atoms of carbon and ten atoms of hydrogen is required and for oxygen molecule two atoms of oxygen are required.

In products, for one molecule of carbon dioxide, one atom of carbon and two atoms of oxygen are required and for water molecule two atoms of hydrogen and one atom of oxygen are required.

To depict the given balanced equation reactant requires, $8$ carbon atoms, $20$ hydrogen atoms, $26$ oxygen atoms. The numbers of atoms are balanced on both the sides of the reaction. Thus, in total $8$ carbon atoms, $20$ hydrogen atoms and $26$ oxygen atoms are required.

Conclusion

The minimum number of balls representing atoms of each element required to show reactants which are then rearranged to products are $8$ carbon atoms, $20$ hydrogen atoms and $26$ oxygen atoms.

Interpretation Introduction

(d)

Interpretation:

The description of the given reaction is to be stated on the molar level.

Concept introduction:

In a balanced chemical equation, all the reactants and products are written with their stoichiometric coefficients and their physical states. The number of atoms of an element on both sides of a balanced chemical equation is equal.

Answer to Problem 8E

At molar level, one mole of butane reacts with $\frac{13}{2}$ moles of oxygen to form four moles of carbon dioxide and five moles of water.

Explanation of Solution

The butane undergoes combustion reaction in presence of oxygen to form carbon dioxide and water. The corresponding chemical reaction is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the given chemical equation, the number of carbon atoms, oxygen atoms, and hydrogen atoms should be same on both the sides. The balanced chemical equation is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+\frac{13}{2}{\text{O}}_2\left(\text{g}\right)\to 4{\text{CO}}_2\left(\text{g}\right)+5{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the number of carbon atoms oxygen atoms, and hydrogen atoms, the coefficient of $4$ is placed before carbon dioxide, the coefficient of $5$ is placed before water and the coefficient of $\frac{13}{2}$ is placed before oxygen gas.

At molar level, one mole of butane reacts with $\frac{13}{2}$ moles of oxygen to form four moles of carbon dioxide and five moles of water.

Conclusion

At molar level, one mole of butane reacts with $\frac{13}{2}$ moles of oxygen to form four moles of carbon dioxide and five moles of water.

Interpretation Introduction

(e)

Interpretation:

The validation of the fact that the mass is conserved for the given reaction is to be stated.

Concept introduction:

In a balanced chemical equation, all the reactants and products are written with their stoichiometric coefficients and their physical states. The number of atoms of an element on both sides of a balanced chemical equation is equal.

Answer to Problem 8E

The mass of reactants and products is same and thus, the mass of the given reaction is conserved.

Explanation of Solution

The butane undergoes combustion reaction in presence of oxygen to form carbon dioxide and water. The corresponding chemical reaction is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the given chemical equation, the number of carbon atoms, oxygen atoms, and hydrogen atoms should be same on both the sides. The balanced chemical equation is,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\left(l\right)+\frac{13}{2}{\text{O}}_2\left(\text{g}\right)\to 4{\text{CO}}_2\left(\text{g}\right)+5{\text{H}}_2\text{O}\left(\text{g}\right)$

To balance the number of carbon atoms oxygen atoms, and hydrogen atoms, the coefficient of $4$ is placed before carbon dioxide, the coefficient of $5$ is placed before water and the coefficient of $\frac{13}{2}$ is placed before oxygen gas.

At molar level, one mole of butane reacts with $\frac{13}{2}$ moles of oxygen to form four moles of carbon dioxide and five moles of water.

The molar mass of carbon is $12.01\text{g}/\text{mol}$.

The molar mass of hydrogen is $1.008\text{g}/\text{mol}$.

The molar mass of oxygen is $16.00\text{g}/\text{mol}$.

The mass of reactants is calculated by the sum of mass of each reactant.

The molar mass of butane is calculated as follows,

$\text{Molar}\text{mass}\text{of}\text{butane}=\left(4\times \text{Molar}\text{mass}\text{of}\text{carbon}+10\times \text{Molar}\text{mass}\text{of}\text{hydrogen}\right)$

Substitute the molar mass of carbon and hydrogen in the above formula.

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{butane}=\left(4\times \text{12}\text{g}/\text{mol}+10\times \text{1}\text{g}/\text{mol}\right)\\ \text{Molar}\text{mass}\text{of}\text{butane}=\left(48\text{g}/\text{mol}+10\text{g}/\text{mol}\right)\\ \text{Molar}\text{mass}\text{of}\text{butane}=58\text{g}/\text{mol}\end{array}$

The molar mass of oxygen is calculated as follows,

$\text{Molar}\text{mass}\text{of}\text{oxygen}=2\times \text{Molar}\text{mass}\text{of}\text{oxygen}$

Substitute the molar mass of oxygen in the above formula.

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{oxygen}=2\times \text{16}\text{g}/\text{mol}\\ \text{Molar}\text{mass}\text{of}\text{oxygen}=32\text{g}/\text{mol}\end{array}$

The total mass of oxygen is,

$\begin{array}{l}\text{Mass}\text{of}\text{oxygen}=6.5\text{mol}\times 32\text{g}/\text{mol}\\ \text{Mass}\text{of}\text{oxygen}=208\text{g}\end{array}$

Total mass of reactants is $\text{208}\text{g}+\text{58}\text{g}=266\text{g}$.

The molar mass of carbon dioxide is calculated as follows,

$\text{Molar}\text{mass}\text{of}\text{carbon}\text{dioxide}=\left(1\times \text{Molar}\text{mass}\text{of}\text{carbon}+2\times \text{Molar}\text{mass}\text{of}\text{oxygen}\right)$

Substitute the molar mass of carbon and oxygen in the above formula.

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{carbon}\text{dioxide}=\left(1\times \text{12}\text{g}/\text{mol}+2\times \text{16}\text{g}/\text{mol}\right)\\ \text{Molar}\text{mass}\text{of}\text{carbon}\text{dioxide}=\left(12\text{g}/\text{mol}+32\text{g}/\text{mol}\right)\\ \text{Molar}\text{mass}\text{of}\text{carbon}\text{dioxide}=44\text{g}/\text{mol}\end{array}$

The molar mass of water is calculated as follows,

$\text{Molar}\text{mass}\text{of}\text{water}=1\times \text{Molar}\text{mass}\text{of}\text{oxygen}+2\times \text{Molar}\text{mass}\text{of}\text{hydrogen}$

Substitute the molar mass of oxygen and hydrogen in the above formula.

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{water}=1\times \text{16}\text{g}/\text{mol}+2\times \text{1}\text{g}/\text{mol}\\ \text{Molar}\text{mass}\text{of}\text{water}=18\text{g}/\text{mol}\end{array}$

The total mass of carbon dioxide is,

$\begin{array}{l}\text{Mass}\text{of}\text{carbon}\text{dioxide}=4\text{mol}\times 44\text{g}/\text{mol}\\ \text{Mass}\text{of}\text{carbon}\text{dioxide}=176\text{g}\end{array}$

The total mass of water is,

$\begin{array}{l}\text{Mass}\text{of}\text{water}=5\text{mol}\times 18\text{g}/\text{mol}\\ \text{Mass}\text{of}\text{water}=90\text{g}\end{array}$

Total mass of products is $176\text{g}+\text{90}\text{g}=266\text{g}$.

The mass of reactants and products is same, thus, the mass is conserved.

Conclusion

The mass of reactants and products is same and thus, the mass of the given reaction is conserved.