Chapter 3, Problem 3.10VP

(a)

Interpretation Introduction

Interpretation: The questions based upon the given representations are to be answered.

Concept introduction: Molar mass of a compound is sum of the atomic mass of each elements present in the compound.

To determine: The compounds that have the same empirical formula.

Answer to Problem 3.10VP

Solution

The compounds that have the same empirical formula are shown in Figure 1.

Explanation of Solution

Explanation

Molecular formula of compound shown in representation [A] is ${\text{C}}_{\text{6}}{\text{H}}_6$. The ratio of $\text{C:H}$ is $1:1$. Thus the empirical formula of ${\text{C}}_{\text{6}}{\text{H}}_6$ is $\text{CH}$. Molecular formula of compound shown in representation [H] is ${\text{C}}_2{\text{H}}_2$. The ratio of $\text{C:H}$ is $1:1$. Thus the empirical formula of ${\text{C}}_2{\text{H}}_2$ is $\text{CH}$. The compounds that have the same empirical formula are ${\text{C}}_{\text{6}}{\text{H}}_6$ and ${\text{C}}_2{\text{H}}_2$.

Figure 1

(b)

Interpretation Introduction

To determine: The compound that has a molecular mass of $\text{180}\text{amu}$.

Answer to Problem 3.10VP

Solution

The compound that has a molecular mass of $\text{180}\text{amu}$ is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$.

Explanation of Solution

Explanation

Molecular mass of ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$ is calculated by the formula,

$\begin{array}{c}\text{Molecular mass of} {\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}=\left(\text{Number}\text{of}\text{C}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{C}\right)+\\ \left(\text{Number}\text{of}\text{H}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{H}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}$

Number of carbon atoms is $9$.

Number of hydrogen atoms is $8$.

Number of oxygen atoms is $4$.

Molar mass of carbon is $\text{12}\text{.01}\text{g/mol}$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Moalr mass of hydrogen is $\text{1}\text{.008}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of carbon, hydrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}=\left(\text{9}\times \text{12}\text{.01}\text{g/mol}\right)+\left(\text{8}\times \text{1}\text{.008}\text{g/mol}\right)+\left(4\times 16.00\text{g/mol}\right)\\ =180\text{amu}\end{array}$

The compound that has a molecular mass of $\text{180}\text{amu}$ is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$ shown in the figure given below.

Figure 2

(c)

Interpretation Introduction

To determine: The compound that has a molar mass of $\text{180}\text{g}$.

Answer to Problem 3.10VP

Solution

The compound that has a molar mass of $\text{180}\text{g}$ is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$.

Explanation of Solution

Explanation

The molar mass of a compound in grams per mol or g/mol is equal to molecular mass expressed in amu. The difference lies in units only. Therefore, the compound that has a molar mass of $\text{180}\text{g}$ is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$ which is shown in Figure 2.

(d)

Interpretation Introduction

To determine: The compound that has the largest percent oxygen by mass.

Answer to Problem 3.10VP

Solution

The compound that has the largest percent oxygen by mass is ${\text{H}}_2{\text{O}}_2$.

Explanation of Solution

Explanation

Compounds in which oxygen is present are: ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$, ${\text{H}}_2{\text{O}}_2$ and ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$.

The percent composition of oxygen (O) in ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}}\times 100$

There are four oxygen atoms present in ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$. Mass of $4$ atoms of O

$=4\times 16\text{ amu}=64\text{ amu}$

Substitute the value of mass of O atoms and mass of ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{64 amu}}{\text{180}\text{amu}}\times 100\\ =35.55\text{\% O}\end{array}$

Molecular mass of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ is calculated by the formula,

$\begin{array}{c}{\text{Molecular mass ofC}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\left(\text{Number}\text{of}\text{C}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{C}\right)+\\ \left(\text{Number}\text{of}\text{H}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{H}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}$

Number of carbon atoms is $12$.

Number of hydrogen atoms is $22$.

Number of oxygen atoms is $11$.

Molar mass of carbon is $\text{12}\text{.01}\text{g/mol}$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Molar mass of hydrogen is $\text{1}\text{.008}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of carbon, hydrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\left(12\times \text{12}\text{.01}\text{g/mol}\right)+\left(22\times \text{1}\text{.008}\text{g/mol}\right)+\left(11\times 16.00\text{g/mol}\right)\\ =342.3\text{amu}\end{array}$

The percent composition of oxygen (O) in ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{\text{Mass of}{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}\times 100$

There are $11$ oxygen atoms present in ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$. Mass of $11$ atoms of O

$=11\times 16\text{ amu}=176\text{ amu}$

Substitute the value of mass of O atoms and mass of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{176 amu}}{342.3\text{amu}}\times 100\\ =51.4\text{\% O}\end{array}$

Molecular mass of ${\text{H}}_2{\text{O}}_2$ is calculated by the formula,

$\begin{array}{c}\text{Molecular mass of}{\text{H}}_2{\text{O}}_2=\left(\text{Number}\text{of}\text{H}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{H}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}$

Number of hydrogen atoms is $2$.

Number of oxygen atoms is $2$.

Molar mass of oxygen is $\text{16}\text{.00}\text{g/mol}$.

Molar mass of hydrogen is $\text{1}\text{.008}\text{g/mol}$.

Substitute the value of number of atoms and molar masses of hydrogen and oxygen in the above equation.

$\begin{array}{c}\text{Molecular mass of} {\text{H}}_{\text{2}}{\text{O}}_{\text{2}}=\left(2\times \text{1}\text{.008}\text{g/mol}\right)+\left(2\times 16.00\text{g/mol}\right)\\ =34.01\text{amu}\end{array}$

The percent composition of oxygen (O) in ${\text{H}}_2{\text{O}}_2$ is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{\text{Mass of}{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}\times 100$

There are $2$ oxygen atoms present in ${\text{H}}_2{\text{O}}_2$. Mass of $2$ atoms of O

$=2\times 16\text{ amu}=32\text{ amu}$

Substitute the value of mass of O atoms and mass of ${\text{H}}_2{\text{O}}_2$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{32 amu}}{34.01\text{amu}}\times 100\\ =94.08\text{\% O}\end{array}$

Therefore, the compound that has the largest percent oxygen by mass is ${\text{H}}_2{\text{O}}_2$. It is shown in the figure given below.

Figure 3

(e)

Interpretation Introduction

To determine: whether the gold bar or silver bar contains more atoms.

Answer to Problem 3.10VP

Solution

The $\text{31}\text{kg}$ silver bar will contain more atoms than $\text{12}\text{.4}\text{kg}$ gold bar.

Explanation of Solution

Explanation

The mass of one gold bar is $\text{12}\text{.4}\text{kg}$.

The mass of one silver bar is $\text{31}\text{kg}$.

The conversion of kg to g is done as,

$\text{1}\text{kg}=1000\text{g}$

Therefore, the conversion of $\text{12}\text{.4}\text{kg}$ to g is done as,

$12.4\text{kg}=12400\text{g}$

The conversion of $\text{31}\text{kg}$ to g is done as,

$\text{31}\text{kg}=31000\text{g}$

Number of atoms in $\text{196}\text{.66}\text{g}$ of gold is equal to $\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{atoms}$.

Therefore, number of atoms in $12400\text{g}$ of gold is equal to $\begin{array}{c}\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{atoms}}{\text{196}\text{.66}\text{g}}\times 12400\text{g}\\ =3.79\times {\text{10}}^{\text{25}}\text{atoms}\end{array}$

Number of atoms in $\text{107}\text{.868}\text{g}$ of gold is equal to $\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{atoms}$.

Therefore, number of atoms in $31000\text{g}$ of gold is equal to $\begin{array}{c}\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{atoms}}{\text{107}\text{.868}\text{g}}\times 31000\text{g}\\ =1.73\times {\text{10}}^{\text{26}}\text{atoms}\end{array}$

Hence, $\text{31}\text{kg}$ silver bar will contain more atoms than $\text{12}\text{.4}\text{kg}$ gold bar.

(f)

Interpretation Introduction

To determine: Which compound will produce more moles of carbon dioxide-benzene [A] or table sugar [C].

Answer to Problem 3.10VP

Solution

Table sugar will produce more moles of carbon dioxide.

Explanation of Solution

Explanation

The molecular formula of benzene is ${\text{C}}_{\text{6}}{\text{H}}_6$. Six moles of carbon dioxide is produced per mole of ${\text{C}}_{\text{6}}{\text{H}}_6$ combusted. Moelcualr formula of table sugar is ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$. Twelve moles of carbon dioxide is produced per mole of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$. Therefore, table sugar will produce more moles of carbon dioxide.

Conclusion

1. a) Table sugar will produce more moles of carbon dioxide.
2. b) The compound that has a molecular mass of $\text{180}\text{amu}$ is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$.
3. c) The compound that has a molar mass of $\text{180}\text{g}$ is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$.
4. d) The compound that has the largest percent oxygen by mass is ${\text{H}}_2{\text{O}}_2$.
5. e) The $\text{31}\text{kg}$ silver bar will contain more atoms than $\text{12}\text{.4}\text{kg}$ gold bar.
6. f) Table sugar will produce more moles of carbon dioxide