Chapter 4, Problem 87QP

What is the percent composition of each of the following compounds?

$\begin{array}{l}\left(\text{a}\right){\text{ SO}}_{\text{2}}\\ \left(\text{b}\right){\text{ CuCl}}_{\text{2}}\\ \left(\text{c}\right){\text{ Na}}_{\text{3}}{\text{PO}}_{\text{4}}\\ \left(\text{d}\right)\text{ Mg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\end{array}$

Interpretation Introduction

Interpretation:

The percent composition of the given compound 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 as follows:

The mass of S from the periodic table is $32.06\text{ g/mole}$ . For $1\text{ mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{32}\text{.06 g/mole}\\ =\text{32}\text{.06 g}\end{array}$

The mass of O from the periodic table is $16.00\text{ g/mole}$ . For $\text{2 mole}$ , the mass is:

$\begin{array}{c}m=2\text{ mole}\times \text{16}\text{.00 g/mole}\\ =\text{32}\text{.00 g}\end{array}$

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

$\begin{array}{c}\text{total mass}=32.06\text{ g}+32.00\text{ g}\\ =64.06\text{ g}\end{array}$

Now, the percent composition of S is as follows:

$\begin{array}{c}\%\text{ S}=\frac{32.06\cancel{\text{g}}}{64.06\cancel{\text{g}}}\times 100\text{ \%}\\ \text{=50}\text{.05 \%}\end{array}$

Now, the percent composition of O is as follows:

$\begin{array}{c}\%\text{ O}=\frac{32.00\cancel{\text{g}}}{64.06\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{49}\text{.95 \%}\end{array}$

Interpretation Introduction

Interpretation:

The percent composition 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, ${\text{CuCl}}_2$ , as follows:

The mass of Cu from the periodic table is $63.55\text{ g/mole}$ . For $1\text{ mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{63}\text{.55 g/mole}\\ =\text{63}\text{.55 g}\end{array}$

The mass of Cl from the periodic table is $35.45\text{ g/mole}$ . For $\text{2 mole}$ , the mass is:

$\begin{array}{c}m=2\text{ mole}\times \text{35}\text{.45 g/mole}\\ =\text{70}\text{.90 g}\end{array}$

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

$\begin{array}{c}\text{total mass}=63.55\text{ g}+70.90\text{ g}\\ =134.45\text{ g}\end{array}$

Now, the percent composition of Cu is as follows:

$\begin{array}{c}\%\text{ Cu}=\frac{\text{63}\text{.55 }\cancel{\text{g}}}{134.45\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{47}\text{.27 \%}\end{array}$

Now, the percent composition of Cl is as follows:

$\begin{array}{c}\%\text{ Cl}=\frac{70.90\cancel{\text{g}}}{134.45\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{52}\text{.73 \%}\end{array}$

Interpretation Introduction

Interpretation:

The percent composition 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, ${\text{Na}}_3{\text{PO}}_4$ , as follows:

The mass of Na from the periodic table is $22.99\text{ g/mole}$ . For $3\text{ mole}$ , the mass is:

$\begin{array}{c}m=3\text{ mole}\times \text{22}\text{.99 g/mole}\\ =\text{68}\text{.97 g}\end{array}$

The mass of P from the periodic table is $30.97\text{ g/mole}$ . For $\text{1 mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{30}\text{.97 g/mole}\\ =\text{30}\text{.97 g}\end{array}$

The mass of O from the periodic table is $16.00\text{ g/mole}$ . For $4\text{ mole}$ , the mass is:

$\begin{array}{c}m=4\text{ mole}\times \text{16}\text{.00 g/mole}\\ =\text{64}\text{.00 g}\end{array}$

Therefore, the total mass of $1\text{ mole}$ ${\text{Na}}_3{\text{PO}}_4$ is:

$\begin{array}{c}\text{total mass}=68.97\text{ g}+30.97\text{ g}+64.00\text{ g}\\ =163.94\text{ g}\end{array}$

Now, the percent composition of Na is as follows:

$\begin{array}{c}\%\text{ Na}=\frac{\text{68}\text{.97 }\cancel{\text{g}}}{163.94\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{42}\text{.07 \%}\end{array}$

Now, the percent composition of P is as follows:

$\begin{array}{c}\%\text{ P}=\frac{30.97\cancel{\text{g}}}{163.94\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{18}\text{.89 \%}\end{array}$

Now, the percent composition of O is as follows:

$\begin{array}{c}\%\text{ O}=\frac{\text{64}\text{.00 }\cancel{\text{g}}}{163.94\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{39}\text{.04 \%}\end{array}$

Interpretation Introduction

Interpretation:

The percent composition 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, $\text{Mg}{\left({\text{NO}}_3\right)}_2$ , as follows:

The mass of Mg from the periodic table is $24.31\text{ g/mole}$ . For $1\text{ mole}$ , the mass is:

$\begin{array}{c}m=1\text{ mole}\times \text{24}\text{.31 g/mole}\\ =\text{24}\text{.31 g}\end{array}$

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 $6\text{ mole}$ , the mass is:

$\begin{array}{c}m=6\text{ mole}\times \text{16}\text{.00 g/mole}\\ =\text{96}\text{.00 g}\end{array}$

Therefore, the total mass of $1\text{ mole}$ $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is:

$\begin{array}{c}\text{total mass}=24.31\text{ g}+28.02\text{ g}+96.00\text{ g}\\ =148.33\text{ g}\end{array}$

Now, the percent composition of Mg is as follows:

$\begin{array}{c}\%\text{ Mg}=\frac{\text{24}\text{.31 }\cancel{\text{g}}}{148.33\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{16}\text{.39 \%}\end{array}$

Now, the percent composition of N is as follows:

$\begin{array}{c}\%\text{ N}=\frac{28.02\cancel{\text{g}}}{148.33\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{18}\text{.89 \%}\end{array}$

Now, the percent composition of O is as follows:

$\begin{array}{c}\%\text{ O}=\frac{\text{96}\text{.00 }\cancel{\text{g}}}{148.33\cancel{\text{g}}}\times 100\text{ \%}\\ =\text{64}\text{.72 \%}\end{array}$