Chapter 3, Problem 3.79QP

(a)

Interpretation Introduction

Interpretation: The percent composition of the given compounds is to be stated.

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

To determine: The percent composition of ${\text{Na}}_{\text{2}}\text{O}$.

Answer to Problem 3.79QP

Solution

Percent composition of ${\text{Na}}_{\text{2}}\text{O}$ is $\underset{\_}{74.19\%Na}$ and $\underset{\_}{25.81\%O}$.

Explanation of Solution

Explanation

The molar mass of ${\text{Na}}_{\text{2}}\text{O}$ is calculated as,

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

In ${\text{Na}}_{\text{2}}\text{O}$, number of $\text{Na}$ atoms $=2$, number of $\text{O}$ atoms $=1$.

Molar mass of $\text{Na}$ $=22.\text{98 g/mol}$ and molar mass of O $=16\text{ g/mol}$.

Substitute the number of sodium and oxygen atoms and their molar masses in the above equation.

$\begin{array}{c}\text{Molar mass}=\text{2}\times \text{22}\text{.98}+\text{1}\times \text{16}\\ =\text{61}\text{.98 g/mol}\end{array}$

The percent composition of sodium ( $\text{Na}$) is calculated by the formula,

$\text{Percent composition of Na}=\frac{\text{Mass of Na}}{{\text{Mass of Na}}_{\text{2}}\text{O}}\times 100$

Mass of $2$ atoms of $\text{Na}$ atoms $=2\times 22.9\text{9 g}=45.9\text{8 g}$.

Substitute the value of mass of $\text{Na}$ and mass of ${\text{Na}}_{\text{2}}\text{O}$ in the above equation.

$\begin{array}{c}\text{Percent composition of Na}=\frac{\text{45}\text{.98 g}}{\text{61}\text{.98 g}}\times 100\\ =\underset{\_}{74.19\%Na}\end{array}$

Therefore, the percent composition of $\text{Na}$ is $\underset{\_}{74.19\%Na}$.

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

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of Na}}_{\text{2}}\text{O}}\times 100$

Mass of $1$ atom of $\text{O}$ $=1\times 16\text{ g}=16\text{ g}$.

Substitute the value of mass of O and mass of ${\text{Na}}_{\text{2}}\text{O}$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{16 g}}{\text{61}\text{.98 g}}\times 100\\ =\underset{\_}{25.81\%O}\end{array}$

Therefore, the percent composition of $\text{O}$ is $\underset{\_}{25.81\%O}$.

Percent composition of ${\text{Na}}_{\text{2}}\text{O}$ is $\underset{\_}{74.19\%Na}$ and $\underset{\_}{25.81\%O}$.

(b)

Interpretation Introduction

To determine: The percent composition of $\text{NaOH}$.

Answer to Problem 3.79QP

Solution

Percent composition of $\text{NaOH}$ is $\underset{\_}{54.48\%Na}$ and $\underset{\_}{40.00\%O}$ and $\underset{\_}{2.52\%H}$.

Explanation of Solution

Explanation

The molar mass of $\text{NaOH}$ is calculated as,

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

In $\text{NaOH}$, number of $\text{Na}$ atoms $=1$, number of $\text{O}$ atoms $=1$ and number of $\text{H}$ atoms $=1$.

Molar mass of $\text{Na}$ $=22.\text{98 g/mol}$ and molar mass of $\text{O}$ $=16\text{ g/mol}$ and molar mass of $\text{H}$ is $\text{1}\text{.008 g/mol}$

Substitute the number of sodium, oxygen and hydrogen atoms and their molar masses in the above equation.

$\begin{array}{c}\text{Molar mass}=1\times \text{22}\text{.98}+\text{1}\times \text{16}+1\times 1.008\\ =40.00\text{ g/mol}\end{array}$

The percent composition of sodium ( $\text{Na}$) is calculated by the formula,

$\text{Percent composition of Na}=\frac{\text{Mass of Na}}{\text{Mass of NaOH}}\times 100$

Substitute the value of mass of $\text{Na}$ and mass of $\text{NaOH}$ in the above equation.

$\begin{array}{c}\text{Percent composition of Na}=\frac{\text{22}\text{.99 g}}{\text{40}\text{.00 g}}\times 100\\ =\underset{\_}{57.48\%Na}\end{array}$

Therefore, the percent composition of $\text{Na}$ is $\underset{\_}{57.48\%Na}$.

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

$\text{Percent composition of O}=\frac{\text{Mass of O}}{\text{Mass of NaOH}}\times 100$

Mass of $1$ atom of $\text{O}$ $=1\times 16\text{ g}=16\text{ g}$.

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

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{16 g}}{\text{40}\text{.00 g}}\times 100\\ =\underset{\_}{40.00\%O}\end{array}$

Therefore, the percent composition of $\text{O}$ is $\underset{\_}{40.00\%O}$.

The percent composition of hydrogen ( $\text{H}$) is calculated by the formula,

$\text{Percent composition of H}=\frac{\text{Mass of H}}{\text{Mass of NaOH}}\times 100$

Mass of $1$ atom of $\text{H}$ $=1\times 1.008\text{ g}=1.008\text{ g}$.

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

$\begin{array}{c}\text{Percent composition of H}=\frac{\text{1}\text{.008 g}}{\text{40}\text{.00 g}}\times 100\\ =\underset{\_}{2.52\%H}\end{array}$

Therefore, the percent composition of $\text{H}$ is $\underset{\_}{2.52\%H}$.

Percent composition of $\text{NaOH}$ is $\underset{\_}{54.48\%Na}$ and $\underset{\_}{40.00\%O}$ and $\underset{\_}{2.52\%H}$.

(c)

Interpretation Introduction

To determine: The percent composition of ${\text{NaHCO}}_3$.

Answer to Problem 3.79QP

Solution

Percent composition of ${\text{NaHCO}}_3$ is $\underset{\_}{27.37\%Na}$, $\underset{\_}{1.20\%H}$, $\underset{\_}{14.30\%C}$ and $\underset{\_}{57.14\%O}$

Explanation of Solution

Explanation

The molar mass of ${\text{NaHCO}}_3$ is calculated as,

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

In ${\text{NaHCO}}_3$, number of $\text{Na}$ atoms $=1$, number of $\text{O}$ atoms $=3$, number of C atoms $=1$ and number of $\text{H}$ atoms $=1$.

Molar mass of $\text{Na}$ $=22.\text{98 g/mol}$, molar mass of C $=12\text{ g/mol}$, molar mass of $\text{O}$ $=16\text{ g/mol}$ and molar mass of $\text{H}$ is $\text{1}\text{.008 g/mol}$

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

$\begin{array}{c}\text{Molar mass}=1\times \text{22}\text{.98}+3\times \text{16}+1\times 1.008+1\times 12\\ =84.01\text{ g/mol}\end{array}$

The percent composition of sodium ( $\text{Na}$) is calculated by the formula,

$\text{Percent composition of Na}=\frac{\text{Mass of Na}}{{\text{Mass of NaHCO}}_3}\times 100$

Substitute the value of mass of $\text{Na}$ and mass of ${\text{NaHCO}}_3$ in the above equation.

$\begin{array}{c}\text{Percent composition of Na}=\frac{\text{22}\text{.99 g}}{\text{84}\text{.01 g}}\times 100\\ =\underset{\_}{24.37\%Na}\end{array}$

Therefore, the percent composition of $\text{Na}$ is $\underset{\_}{24.37\%Na}$.

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

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of NaHCO}}_3}\times 100$

Mass of $3$ atom of $\text{O}$ $=3\times 16\text{ g}=48\text{ g}$.

Substitute the value of mass of $\text{O}$ and mass of ${\text{NaHCO}}_3$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{48 g}}{\text{84}\text{.01 g}}\times 100\\ =\underset{\_}{57.14\%O}\end{array}$

Therefore, the percent composition of $\text{O}$ is $\underset{\_}{57.14\%O}$.

The percent composition of hydrogen ( $\text{H}$) is calculated by the formula,

$\text{Percent composition of H}=\frac{\text{Mass of H}}{{\text{Mass of NaHCO}}_3}\times 100$

Mass of $1$ atom of $\text{H}$ $=1\times 1.008\text{ g}=1.008\text{ g}$.

Substitute the value of mass of $\text{H}$ and mass of ${\text{NaHCO}}_3$ in the above equation.

$\begin{array}{c}\text{Percent composition of H}=\frac{\text{1}\text{.008 g}}{\text{84}\text{.01 g}}\times 100\\ =\underset{\_}{1.20\%H}\end{array}$

Therefore, the percent composition of $\text{H}$ is $\underset{\_}{1.20\%H}$.

The percent composition of carbon (C) is calculated by the formula,

$\text{Percent composition of C}=\frac{\text{Mass of C}}{{\text{Mass of NaHCO}}_3}\times 100$

Mass of $1$ atom of C $=1\times 12\text{ g}=12\text{ g}$.

Substitute the value of mass of C and mass of ${\text{NaHCO}}_3$ in the above equation.

$\begin{array}{c}\text{Percent composition of C}=\frac{\text{12}\text{.01 g}}{\text{84}\text{.01 g}}\times 100\\ =\underset{\_}{14.30\%C}\end{array}$

Therefore, the percent composition of C is $\underset{\_}{14.30\%C}$.

Percent composition of ${\text{NaHCO}}_3$ is $\underset{\_}{27.37\%Na}$, $\underset{\_}{1.20\%H}$, $\underset{\_}{14.30\%C}$ and $\underset{\_}{57.14\%O}$

(d)

Interpretation Introduction

To determine: The percent composition of ${\text{Na}}_2{\text{CO}}_3$.

Answer to Problem 3.79QP

Solution

Percent composition of ${\text{Na}}_2{\text{CO}}_3$ is $\underset{\_}{43.38\%Na}$, $\underset{\_}{11.33\%C}$ and $\underset{\_}{45.28\%O}$.

Explanation of Solution

Explanation

The molar mass of ${\text{Na}}_2{\text{CO}}_3$ is calculated as,

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

In ${\text{Na}}_2{\text{CO}}_3$, number of $\text{Na}$ atoms $=2$, number of O atoms $=3$, number of C atoms $=1$.

Molar mass of Na $=22.\text{98 g/mol}$, molar mass of C $=12\text{ g/mol}$, molar mass of $\text{O}$ $=16\text{ g/mol}$.

Substitute the number of sodium, carbon, oxygen atoms and their molar masses in the above equation.

$\begin{array}{c}\text{Molar mass}=2\times \text{22}\text{.98}+3\times \text{16}+1\times 12\\ =106\text{ g/mol}\end{array}$

The percent composition of sodium ( $\text{Na}$) is calculated by the formula,

$\text{Percent composition of Na}=\frac{\text{Mass of Na}}{{\text{Mass of Na}}_2{\text{CO}}_3}\times 100$

Mass of $\text{2 Na}$ atoms $2\times 22.98\text{g}=45.98\text{ g}$.

Substitute the value of mass of $\text{Na}$ and mass of ${\text{Na}}_2{\text{CO}}_3$ in the above equation.

$\begin{array}{c}\text{Percent composition of Na}=\frac{\text{45}\text{.98 g}}{\text{106 g}}\times 100\\ =\underset{\_}{43.38\%Na}\end{array}$

Therefore, the percent composition of $\text{Na}$ is $\underset{\_}{43.38\%Na}$.

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

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of Na}}_2{\text{CO}}_3}\times 100$

Mass of $3$ atom of $\text{O}$ $=3\times 16\text{ g}=48\text{ g}$.

Substitute the value of mass of $\text{O}$ and mass of ${\text{Na}}_2{\text{CO}}_3$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{48 g}}{\text{106 g}}\times 100\\ =\underset{\_}{45.28\%O}\end{array}$

Therefore, the percent composition of $\text{O}$ is $\underset{\_}{45.28\%O}$.

The percent composition of carbon ( $\text{C}$) is calculated by the formula,

$\text{Percent composition of C}=\frac{\text{Mass of C}}{{\text{Mass of Na}}_2{\text{CO}}_3}\times 100$

Mass of $1$ atom of C $=1\times 12\text{ g}=12\text{ g}$.

Substitute the value of mass of $\text{C}$ and mass of ${\text{Na}}_2{\text{CO}}_3$ in the above equation.

$\begin{array}{c}\text{Percent composition of C}=\frac{\text{12}\text{.01 g}}{\text{106 g}}\times 100\\ =\underset{\_}{11.33\%C}\end{array}$

Therefore, the percent composition of $\text{C}$ is $\underset{\_}{11.33\%C}$.

Conclusion

1. a. Percent composition of ${\text{Na}}_{\text{2}}\text{O}$ is $\underset{\_}{74.19\%Na}$ and $\underset{\_}{25.81\%O}$.
2. b. Percent composition of $\text{NaOH}$ is $\underset{\_}{54.48\%Na}$ and $\underset{\_}{40.00\%O}$ and $\underset{\_}{2.52\%H}$.
3. c. Percent composition of ${\text{NaHCO}}_3$ is $\underset{\_}{27.37\%Na}$, $\underset{\_}{1.20\%H}$, $\underset{\_}{14.30\%C}$ and $\underset{\_}{57.14\%O}$.
4. d. Percent composition of ${\text{Na}}_2{\text{CO}}_3$ is $\underset{\_}{43.38\%Na}$, $\underset{\_}{11.33\%C}$ and $\underset{\_}{45.28\%O}$