Chapter 4, Problem 88QP

(a)

Interpretation Introduction

Interpretation:

The percent composition of nitrogen in ${\text{NaNO}}_3$ is to be determined.

Explanation of Solution

The percent composition of any substance is calculated by dividing the mass of the substance in a sample by the total mass of the sample. This ratio is multiplied by $100\text{ \%}$ .

To determine the percent composition, let the sample size be $1\text{ mole}$ . Now, determine the mass of each element contained in $1\text{ mole}$ of the compound, ${\text{NaNO}}_3$ , as follows:

The mass of Na from the periodic table is $22.99\text{ g/mole}$ . For $1\text{ mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{22}\text{.99 g/mole}\\ =\text{22}\text{.99 g}\end{array}$

The mass of N from the periodic table is $14.01\text{ g/mole}$ . For $1\text{ mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{14}\text{.01 g/mole}\\ =\text{14}\text{.01 g}\end{array}$

The mass of O from the periodic table is $16.00\text{ g/mole}$ . For $\text{3 mole}$ , the mass is:

$\begin{array}{c}m=3\text{ mole}\times \text{16}\text{.00 g/mole}\\ =\text{48}\text{.00 g}\end{array}$

Therefore, the total mass of $1\text{ mole}$ ${\text{NaNO}}_3$ is:

$\begin{array}{c}\text{total mass}=22.99\text{ g}+14.01\text{ g}+48.00\text{ g}\\ =85.00\text{ g}\end{array}$

Now, the percent composition of N is as follows:

$\begin{array}{c}\%\text{ N}=\frac{14.01\cancel{\text{g}}}{85.00\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{16}\text{.48 \%}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The percentage of nitrogen in ${\text{NH}}_4\text{Cl}$ is to analyzed.

Explanation of Solution

To determine the percent composition, let the sample size be $1\text{ mole}$ . Now, determine the mass of each element contained in $1\text{ mole}$ of the compound, ${\text{NH}}_4\text{Cl}$ , as follows:

The mass of N from the periodic table is $14.01\text{ g/mole}$ . For $1\text{ mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{14}\text{.01 g/mole}\\ =\text{14}\text{.01 g}\end{array}$

The mass of H from the periodic table is $1.008\text{ g/mole}$ . For $4\text{ mole}$ , the mass is:

$\begin{array}{c}m=4\text{ mole}\times \text{1}\text{.008 g/mole}\\ =\text{4}\text{.032 g}\end{array}$

The mass of Cl from the periodic table is $35.45\text{ g/mole}$ . For $\text{1 mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{35}\text{.45 g/mole}\\ =\text{35}\text{.45 g}\end{array}$

Therefore, the total mass of $1\text{ mole}$ ${\text{NH}}_4\text{Cl}$ is:

$\begin{array}{c}\text{total mass}=14.01\text{ g}+4.032\text{ g}+35.45\text{ g}\\ =53.49\text{ g}\end{array}$

Now, the percent composition of N is as follows:

$\begin{array}{c}\%\text{ N}=\frac{\text{14}\text{.01 }\cancel{\text{g}}}{53.49\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{26}\text{.19 \%}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The percentage of nitrogen in ${\text{N}}_2{\text{H}}_4$ is to analyzed.

Explanation of Solution

To determine the percent composition, let the sample size be $1\text{ mole}$ . Now, determine the mass of each element contained in $1\text{ mole}$ of the compound ${\text{N}}_2{\text{H}}_4$ as follows:

The mass of N from the periodic table is $14.01\text{ g/mole}$ . For $2\text{ mole}$ , the mass is:

$\begin{array}{c}m=2\text{ mole}\times \text{14}\text{.01 g/mole}\\ =\text{28}\text{.02 g}\end{array}$

The mass of H from the periodic table is $1.008\text{ g/mole}$ . For $\text{4 mole}$ , the mass is:

$\begin{array}{c}m=4\text{ mole}\times \text{1}\text{.008 g/mole}\\ =\text{4}\text{.032 g}\end{array}$

Therefore, the total mass of $1\text{ mole}$ ${\text{N}}_2{\text{H}}_4$ is:

$\begin{array}{c}\text{total mass}=28.02\text{ g}+4.032\text{ g}\\ =32.05\text{ g}\end{array}$

Now, the percent composition of N is as follows:

$\begin{array}{c}\%\text{ N}=\frac{\text{28}\text{.02 }\cancel{\text{g}}}{32.05\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{87}\text{.43 \%}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The percentage of nitrogen in ${\text{N}}_2\text{O}$ is to analyzed.

Explanation of Solution

To determine the percent composition, let the sample size be $1\text{ mole}$ . Now, determine the mass of each element contained in $1\text{ mole}$ of the compound, ${\text{N}}_2\text{O}$ , as follows:

The mass of N from the periodic table is $14.01\text{ g/mole}$ . For $\text{2 mole}$ , the mass is:

$\begin{array}{c}m=2\text{ mole}\times \text{14}\text{.01 g/mole}\\ =\text{28}\text{.02 g}\end{array}$

The mass of O from the periodic table is $16.00\text{ g/mole}$ . For $1\text{ mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{16}\text{.00 g/mole}\\ =\text{16}\text{.00 g}\end{array}$

Therefore, the total mass of $1\text{ mole}$ ${\text{N}}_2\text{O}$ is:

$\begin{array}{c}\text{total mass}=28.02\text{ g}+16.00\text{ g}\\ =44.02\text{ g}\end{array}$

Now, the percent composition of N is as follows:

$\begin{array}{c}\%\text{ N}=\frac{28.02\cancel{\text{g}}}{44.02\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{63}\text{.65 \%}\end{array}$