Chapter 3, Problem 3.79QP

(a)

Interpretation Introduction

Interpretation: The compound with greatest percentage of carbon by mass is to be determined. If any two of the given compounds have same empirical formula is to be stated.

Concept introduction: Empirical formula signifies the simplest or most reduced ratio of elements in a compound. Molecular formula is obtained by multiplying empirical formula with a multiple.

To determine: The percentage of carbon by mass in the given compound and its empirical formula.

Answer to Problem 3.79QP

Solution

The percentage of carbon by mass in the given compound is $\underset{\_}{93.71\%}$ and its empirical formula is ${\text{C}}_5{\text{H}}_4$.

Explanation of Solution

Explanation

The given compound is naphthalene ( ${\text{C}}_{10}{\text{H}}_8$).

The molar mass of naphthalene ( ${\text{C}}_{10}{\text{H}}_8$) is calculated by the formula,

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

There is $10$ carbon atoms and $8$ hydrogen atoms present in ${\text{C}}_{10}{\text{H}}_8$.

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

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

Substitute the values of number of carbon and hydrogen atoms and their respective masses in the above equation.

$\begin{array}{c}\text{Molar mass}=\left(10\times \text{12}\text{.01 g/mol}\right)+\left(8\times \text{1}\text{.008 g/mol}\right)\\ =128.16\text{g/mol}\end{array}$

The mass percentage of carbon ( $\text{\%}\text{C}$) is calculated by the formula,

$\text{\%}\text{C}=\frac{\text{Mass of 10}\text{C}}{\text{Molar}{\text{Mass of C}}_{10}{\text{H}}_8}\times 100$

Mass of $10$ carbon atoms $=10\times \text{12}\text{.01 g/mol}=\text{120}\text{.10 g/mol}$.

Substitute the value of mass of $10$ carbon atom and molar mass of ${\text{C}}_{10}{\text{H}}_8$ in the above equation.

$\begin{array}{c}\text{\%}\text{C}=\frac{\text{120}\text{.10 g/mol}}{128.16\text{g/mol}}\times 100\\ =93.71\%\end{array}$

Therefore, the mass percentage of carbon ( $\text{\%}\text{C}$) in ${\text{C}}_{10}{\text{H}}_8$ is $\underset{\_}{93.71\%}$.

The molecular formula of naphthalene is ${\text{C}}_{10}{\text{H}}_8$. The ratio for $\text{C:H}$ is $10:8$. The highest common multiple in the subscripts is $2$. Divide the ratio of $\text{C:H}$ by $2$. It will become,

$\frac{10}{2}:\frac{8}{2}=5:4$

The ratio of $\text{C:H}$ will be $5:4$.

Thus the empirical formula of naphthalene is ${\text{C}}_5{\text{H}}_4$.

(b)

Interpretation Introduction

To determine: The percentage of carbon by mass in the given compound and its empirical formula.

Answer to Problem 3.79QP

Solution

The percentage of carbon by mass in the given compound is $\underset{\_}{94.7\%}$ and its empirical formula is ${\text{C}}_3{\text{H}}_2$.

Explanation of Solution

Explanation

The given compound is chrysene ( ${\text{C}}_{18}{\text{H}}_{12}$).

The molar mass of chrysene ( ${\text{C}}_{18}{\text{H}}_{12}$) is calculated by the formula,

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

There is $18$ carbon atoms and $12$ hydrogen atoms present in ${\text{C}}_{18}{\text{H}}_{12}$.

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

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

Substitute the values of number of carbon and hydrogen atoms and their respective masses in the above equation.

$\begin{array}{c}\text{Molar mass}=\left(18\times \text{12}\text{.01 g/mol}\right)+\left(12\times \text{1}\text{.008 g/mol}\right)\\ =228.27\text{g/mol}\end{array}$

The mass percentage of carbon ( $\text{\%}\text{C}$) is calculated by the formula,

$\text{\%}\text{C}=\frac{\text{Mass of 18}\text{C}}{\text{Molar}{\text{Mass of C}}_{18}{\text{H}}_{12}}\times 100$

Mass of $18$ carbon atoms $=18\times \text{12}\text{.01 g/mol}=\text{216}\text{.18 g/mol}$.

Substitute the value of mass of $18$ carbon atom and molar mass of ${\text{C}}_{18}{\text{H}}_{12}$ in the above equation.

$\begin{array}{c}\text{\%}\text{C}=\frac{\text{216}\text{.18 g/mol}}{228.27\text{g/mol}}\times 100\\ =94.7\%\end{array}$

Therefore, the mass percentage of carbon ( $\text{\%}\text{C}$) in ${\text{C}}_{18}{\text{H}}_{12}$ is $\underset{\_}{94.7\%}$.

The molecular formula of chrysene is ${\text{C}}_{18}{\text{H}}_{12}$. The ratio for $\text{C:H}$ is $18:12$. The highest common multiple in the subscripts is $6$. Divide the ratio of $\text{C:H}$ by $6$. It will become,

$\frac{18}{6}:\frac{12}{6}=3:2$

The ratio of $\text{C:H}$ will be $3:2$.

Thus the empirical formula of chrysene is ${\text{C}}_3{\text{H}}_2$.

(c)

Interpretation Introduction

To determine: The percentage of carbon by mass in the given compound and its empirical formula.

Answer to Problem 3.79QP

Solution

The percentage of carbon by mass in the given compound is $\underset{\_}{94.93\%}$ and its empirical formula is ${\text{C}}_{11}{\text{H}}_7$.

Explanation of Solution

Explanation

The given compound is pentacene ( ${\text{C}}_{22}{\text{H}}_{14}$).

The molar mass of pentacene ( ${\text{C}}_{22}{\text{H}}_{14}$) is calculated by the formula,

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

There is $22$ carbon atoms and $14$ hydrogen atoms present in ${\text{C}}_{22}{\text{H}}_{14}$.

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

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

Substitute the values of number of carbon and hydrogen atoms and their respective masses in the above equation.

$\begin{array}{c}\text{Molar mass}=\left(22\times \text{12}\text{.01 g/mol}\right)+\left(14\times \text{1}\text{.008 g/mol}\right)\\ =278.33\text{g/mol}\end{array}$

The mass percentage of carbon ( $\text{\%}\text{C}$) is calculated by the formula,

$\text{\%}\text{C}=\frac{\text{Mass of 22}\text{C}}{\text{Molar}{\text{Mass of C}}_{22}{\text{H}}_{14}}\times 100$

Mass of $22$ carbon atoms $=22\times \text{12}\text{.01 g/mol}=\text{264}\text{.22 g/mol}$.

Substitute the value of mass of $22$ carbon atom and molar mass of ${\text{C}}_{22}{\text{H}}_{14}$ in the above equation.

$\begin{array}{c}\text{\%}\text{C}=\frac{\text{264}\text{.22 g/mol}}{278.33\text{g/mol}}\times 100\\ =94.93\%\end{array}$

Therefore, the mass percentage of carbon ( $\text{\%}\text{C}$) in ${\text{C}}_{22}{\text{H}}_{14}$ is $\underset{\_}{94.93\%}$.

The molecular formula of pentacene is ${\text{C}}_{22}{\text{H}}_{14}$. The ratio for $\text{C:H}$ is $22:14$. The highest common multiple in the subscripts is $2$. Divide the ratio of $\text{C:H}$ by $2$. It will become,

$\frac{22}{2}:\frac{14}{2}=11:7$

The ratio of $\text{C:H}$ will be $11:7$.

Thus the empirical formula of pentacene is ${\text{C}}_{11}{\text{H}}_7$.

(d)

Interpretation Introduction

To determine: The percentage of carbon by mass in the given compound and its empirical formula.

Answer to Problem 3.79QP

Solution

The percentage of carbon by mass in the given compound is $\underset{\_}{95.01\%}$ and its empirical formula is ${\text{C}}_8{\text{H}}_5$.

Explanation of Solution

Explanation

The given compound is pyrene ( ${\text{C}}_{16}{\text{H}}_{10}$).

The molar mass of pyrene ( ${\text{C}}_{16}{\text{H}}_{10}$) is calculated by the formula,

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

There is $16$ carbon atoms and $10$ hydrogen atoms present in ${\text{C}}_{16}{\text{H}}_{10}$.

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

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

Substitute the values of number of carbon and hydrogen atoms and their respective masses in the above equation.

$\begin{array}{c}\text{Molar mass}=\left(16\times \text{12}\text{.01 g/mol}\right)+\left(10\times \text{1}\text{.008 g/mol}\right)\\ =202.24\text{g/mol}\end{array}$

The mass percentage of carbon ( $\text{\%}\text{C}$) is calculated by the formula,

$\text{\%}\text{C}=\frac{\text{Mass of 16}\text{C}}{\text{Molar}{\text{Mass of C}}_{16}{\text{H}}_{10}}\times 100$

Mass of $16$ carbon atoms $=16\times \text{12}\text{.01 g/mol}=\text{192}\text{.16 g/mol}$.

Substitute the value of mass of $16$ carbon atoms and molar mass of ${\text{C}}_{16}{\text{H}}_{10}$ in the above equation.

$\begin{array}{c}\text{\%}\text{C}=\frac{\text{192}\text{.16 g/mol}}{202.24\text{g/mol}}\times 100\\ =95.01\%\end{array}$

Therefore, the mass percentage of carbon ( $\text{\%}\text{C}$) in ${\text{C}}_{16}{\text{H}}_{10}$ is $\underset{\_}{95.01\%}$.

The molecular formula of pyrene is ${\text{C}}_{16}{\text{H}}_{10}$. The ratio for $\text{C:H}$ is $16:10$. The highest common multiple in the subscripts is $2$. Divide the ratio of $\text{C:H}$ by $2$. It will become,

$\frac{16}{2}:\frac{10}{2}=8:5$

The ratio of $\text{C:H}$ will be $8:5$.

Thus the empirical formula of pyrene is ${\text{C}}_8{\text{H}}_5$.

Therefore, the compound with greatest percentage of carbon by mass is pyrene ( ${\text{C}}_{16}{\text{H}}_{10}$). None of the compounds have same empirical formula.

Conclusion

The compound with greatest percentage of carbon by mass is pyrene ( ${\text{C}}_{16}{\text{H}}_{10}$). None of the compounds have same empirical formula