Chapter 11, Problem 11.5P

(a)

Interpretation Introduction

Interpretation:

Mass $\%$ of each element in ${\text{CH}}_4$ has to be determined.

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is $\text{g/mol}$.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

Mass percent is calculated by the mass of the element divided by the mass of compound and multiplied by 100.

The formula to calculate mass percentage of substance is as follows:

  $\text{Mass \% of substance}=\left(\frac{\text{Mass of substance}}{\text{Mass of compound}}\right)\left(100\right)$

Answer to Problem 11.5P

Mass $\%$ of 1$\text{C}$ atom in ${\text{CH}}_4$ is $74.8796\text{ \%}$ and mass $\%$ of 4$\text{H}$ atoms in ${\text{CH}}_4$ is $25.1331\text{ \%}$.

Explanation of Solution

The compound is ${\text{CH}}_4$. The compound contains 1 carbon atom and4 hydrogen atoms.

Molar mass of ${\text{CH}}_4$ is $\text{16}\text{.04 g/mol}$.

There is 1$\text{C}$ atoms in ${\text{CH}}_4$.

Molar mass of $\text{C}$ is $\text{12}\text{.0107 g/mol}$.

The formula to calculate mass $\%$ of 1$\text{C}$ atom in ${\text{CH}}_4$ is as follows:

  $\text{Mass \%}=\left(\frac{\text{1}\left(\text{Mass of C atom}\right)}{{\text{Mass of CH}}_4}\right)\left(100\right)$        (1)

Substitute $\text{12}\text{.0107 g/mol}$ for mass of $\text{C}$ atoms and $\text{16}\text{.04 g/mol}$ for mass of ${\text{CH}}_4$ in equation (1).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{\text{1}\left(\text{12}\text{.0107 g/mol}\right)}{\text{16}\text{.04 g/mol}}\right)\left(100\right)\\ =74.8796\text{ \%}\end{array}$

Therefore, mass $\%$ of 1$\text{C}$ atom in ${\text{CH}}_4$ is $74.8796\text{ \%}$.

There are 4$\text{H}$ atoms in ${\text{CH}}_4$.

Molar mass of $\text{H}$ is $1.00784\text{ g/mol}$.

The formula to calculate mass $\%$ of 4$\text{H}$ atoms in ${\text{CH}}_4$ is as follows:

  $\text{Mass \%}=\left(\frac{4\left(\text{Mass of H atom}\right)}{{\text{Mass of CH}}_4}\right)\left(100\right)$        (2)

Substitute $1.00784\text{ g/mol}$ for mass of $\text{H}$ atoms and $\text{16}\text{.04 g/mol}$ for mass of ${\text{CH}}_4$ in equation (2).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{\text{4}\left(1.00784\text{ g/mol}\right)}{\text{16}\text{.04 g/mol}}\right)\left(100\right)\\ =25.1331\text{ \%}\end{array}$

Therefore, mass $\%$ of 4$\text{H}$ atoms in ${\text{CH}}_4$ is $25.1331\text{ \%}$.

(b)

Interpretation Introduction

Interpretation:

Mass $\%$ of each element in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.5P

Mass $\%$ of 2$\text{C}$ atom in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is $52.1410\text{ \%}$ mass $\%$ of 6$\text{H}$ atoms in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is $13.1257\text{ \%}$ and mass $\%$ of 1$\text{O}$ atom in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is $34.7275\text{ \%}$.

Explanation of Solution

The compound is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$. The compound contains 2 carbon atoms,6 hydrogen atoms and 1 oxygen atom.

Molar mass of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is $\text{46}\text{.07 g/mol}$.

There are 2$\text{C}$ atoms in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$.

Molar mass of $\text{C}$ is $\text{12}\text{.0107 g/mol}$.

The formula to calculate mass $\%$ of 2$\text{C}$ atoms in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is as follows:

  $\text{Mass \%}=\left(\frac{2\left(\text{Mass of C atom}\right)}{{\text{Mass of CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}}\right)\left(100\right)$        (3)

Substitute $\text{12}\text{.0107 g/mol}$ for mass of $\text{C}$ atoms and $\text{46}\text{.07 g/mol}$ for mass of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ in equation (3).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{\text{2}\left(\text{12}\text{.0107 g/mol}\right)}{\text{46}\text{.07 g/mol}}\right)\left(100\right)\\ =52.1410\text{ \%}\end{array}$

Therefore, mass $\%$ of 2$\text{C}$ atoms in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is $52.1410\text{ \%}$.

There are 6$\text{H}$ atoms in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$.

Molar mass of $\text{H}$ is $1.00784\text{ g/mol}$.

The formula to calculate mass $\%$ of 6$\text{H}$ atoms in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is as follows:

  $\text{Mass \%}=\left(\frac{\text{6}\left(\text{Mass of H atom}\right)}{{\text{Mass of CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}}\right)\left(100\right)$        (4)

Substitute $1.00784\text{ g/mol}$ for mass of $\text{H}$ atoms and $\text{46}\text{.07 g/mol}$ for mass of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ in equation (4).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{6\left(1.00784\text{ g/mol}\right)}{\text{46}\text{.07 g/mol}}\right)\left(100\right)\\ =13.1257\text{ \%}\end{array}$

Therefore, mass $\%$ of 6$\text{H}$ atoms in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is $13.1257\text{ \%}$.

There is 1$\text{O}$ atom in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$.

Molar mass of $\text{O}$ is $15.999\text{ g/mol}$.

The formula to calculate mass $\%$ of 1$\text{O}$ atom in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is as follows:

  $\text{Mass \%}=\left(\frac{1\times \text{Mass of O atom}}{{\text{Mass of CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}}\right)\left(100\right)$        (5)

Substitute $15.999\text{ g/mol}$ for mass of $\text{O}$ atoms and $\text{46}\text{.07 g/mol}$ for mass of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ in equation (5).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{1\times 15.999\text{ g/mol}}{\text{46}\text{.07 g/mol}}\right)\left(100\right)\\ =34.7275\text{ \%}\end{array}$

Therefore, mass $\%$ of 1$\text{O}$ atom in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is $34.7275\text{ \%}$.

(c)

Interpretation Introduction

Interpretation:

Mass $\%$ of each element in ${\text{CH}}_{\text{3}}\text{COOH}$ has to be determined.

Concept Introduction:

Refer to part (a)

Answer to Problem 11.5P

Mass $\%$ of 2$\text{C}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ is $40.0009\text{ \%}$,mass $\%$ of 4$\text{H}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ is $6.7131\text{ \%}$ and mass $\%$ of 2$\text{O}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ is $52.2838\text{ \%}$.

Explanation of Solution

The compound is ${\text{CH}}_{\text{3}}\text{COOH}$. The compound contains 2 carbon atom, 4 hydrogen atoms and 2 oxygen atoms.

Molar mass of ${\text{CH}}_{\text{3}}\text{COOH}$ is $\text{60}\text{.052 g/mol}$.

There are 2$\text{C}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$.

Molar mass of $\text{C}$ is $\text{12}\text{.0107 g/mol}$.

The formula to calculate mass $\%$ of 2$\text{C}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ is as follows:

  $\text{Mass \%}=\left(\frac{2\times \text{Mass of C atom}}{{\text{Mass of CH}}_{\text{3}}\text{COOH}}\right)\left(100\right)$        (6)

Substitute $\text{12}\text{.0107 g/mol}$ for mass of $\text{C}$ atoms and $\text{60}\text{.052 g/mol}$ for mass of ${\text{CH}}_{\text{3}}\text{COOH}$ in equation (6).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{\text{2}\times \text{12}\text{.0107 g/mol}}{\text{60}\text{.052 g/mol}}\right)\left(100\right)\\ =40.0009\text{ \%}\end{array}$

Therefore, mass $\%$ of 2$\text{C}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ is $40.0009\text{ \%}$.

There are 4$\text{H}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$.

Molar mass of $\text{H}$ is $1.00784\text{ g/mol}$.

The formula to calculate mass $\%$ of 4$\text{H}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ is as follows:

  $\text{Mass \%}=\left(\frac{\text{4}\times \text{Mass of H atom}}{{\text{Mass of CH}}_{\text{3}}\text{COOH}}\right)\left(100\right)$        (7)

Substitute $1.00784\text{ g/mol}$ for mass of $\text{H}$ atoms and $\text{60}\text{.052 g/mol}$ for mass of ${\text{CH}}_{\text{3}}\text{COOH}$ in equation (7).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{4\times 1.00784\text{ g/mol}}{\text{60}\text{.052 g/mol}}\right)\left(100\right)\\ =6.7131\text{ \%}\end{array}$

Therefore, mass $\%$ of 4$\text{H}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ is $6.7131\text{ \%}$.

There are 2$\text{O}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$.

Molar mass of $\text{O}$ is $15.999\text{ g/mol}$.

The formula to calculate mass $\%$ of 2$\text{O}$ atom in ${\text{CH}}_{\text{3}}\text{COOH}$ is as follows:

  $\text{Mass \%}=\left(\frac{2\times \text{Mass of O atom}}{{\text{Mass of CH}}_{\text{3}}\text{COOH}}\right)\left(100\right)$        (8)

Substitute $15.999\text{ g/mol}$ for mass of $\text{O}$ atoms and $\text{60}\text{.052 g/mol}$ for mass of ${\text{CH}}_{\text{3}}\text{COOH}$ in equation (8).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{2\times 15.999\text{ g/mol}}{\text{60}\text{.052 g/mol}}\right)\left(100\right)\\ =53.2838\text{ \%}\end{array}$

Therefore, mass $\%$ of 2$\text{O}$ atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ is $52.2838\text{ \%}$.

(d)

Interpretation Introduction

Interpretation:

Mass $\%$ of each element in ${\text{C}}_6{\text{H}}_6$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.5P

Mass $\%$ of 6$\text{C}$ atoms in ${\text{C}}_6{\text{H}}_6$ is $92.2598\text{ \%}$ and mass $\%$ of 6$\text{H}$ atoms in ${\text{C}}_6{\text{H}}_6$ is $7.7416\%$.

Explanation of Solution

The compound is ${\text{C}}_6{\text{H}}_6$. The compound contains 6 carbon atoms and6 hydrogen atoms.

Molar mass of ${\text{C}}_6{\text{H}}_6$ is $\text{78}\text{.11 g/mol}$.

There are 6$\text{C}$ atoms in ${\text{C}}_6{\text{H}}_6$.

Molar mass of $\text{C}$ is $\text{12}\text{.0107 g/mol}$.

The formula to calculate mass $\%$ of 6$\text{C}$ atoms in ${\text{C}}_6{\text{H}}_6$ is as follows:

  $\text{Mass \%}=\left(\frac{6\times \text{Mass of C atom}}{{\text{Mass of C}}_6{\text{H}}_6}\right)\left(100\right)$        (9)

Substitute $\text{12}\text{.0107 g/mol}$ for mass of $\text{C}$ atoms and $\text{78}\text{.11 g/mol}$ for mass of ${\text{C}}_6{\text{H}}_6$ in equation (9).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{\text{6}\times \text{12}\text{.0107 g/mol}}{\text{78}\text{.11 g/mol}}\right)\left(100\right)\\ =92.2598\text{ \%}\end{array}$

Therefore, mass $\%$ of 6$\text{C}$ atoms in ${\text{C}}_6{\text{H}}_6$ is $92.2598\text{ \%}$.

There are 6 $\text{H}$ atoms in ${\text{C}}_6{\text{H}}_6$.

Molar mass of $\text{H}$ is $1.00784\text{ g/mol}$.

The formula to calculate mass $\%$ of 6$\text{H}$ atoms in ${\text{C}}_6{\text{H}}_6$ is as follows:

  $\text{Mass \%}=\left(\frac{\text{6}\times \text{Mass of H atom}}{{\text{Mass of C}}_6{\text{H}}_6}\right)\left(100\right)$        (10)

Substitute $1.00784\text{ g/mol}$ for mass of $\text{H}$ atoms and $\text{78}\text{.11 g/mol}$ for mass of ${\text{C}}_6{\text{H}}_6$ in equation (10).

  $\begin{array}{c}\text{Mass \%}=\left(\frac{\text{6}\times 1.00784\text{ g/mol}}{\text{78}\text{.11 g/mol}}\right)\left(100\right)\\ =7.7416\text{ \%}\end{array}$

Therefore, mass $\%$ of 6$\text{H}$ atoms in ${\text{C}}_6{\text{H}}_6$ is $7.7416\%$.