Chapter 3, Problem 133AP

A sample containing ${\text{NaCl, Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{, and NaNO}}_{\text{3}}$ gives the following elemental analysis: 32.08 percent Na, 36.01 percent O, 19.51 percent Cl. Calculate the mass percent of each compound in the sample.

Interpretation Introduction

Interpretation:

The mass percent of each compound in the sample is to be calculated.

Concept introduction:

The percent composition of an element by its mass in a compound can be calculated as:

$\text{percent by mass of an element}=\frac{n\times \text{ atomic mass of an element}}{\text{molecular mass of the compound}}\times 100\%.$

Here, n represents the element’s number of atoms in the given molecule or compound.

The number of moles is defined as the ratio of mass to the molar mass.

$n=\frac{m}{M}$.

Here, $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.

Answer to Problem 133AP

Solution: The percent composition by mass of $\text{NaCl}$, ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$, and ${\text{NaNO}}_{\text{3}}$ is $32.165\%$, $20.09\%,$ and $47.75\%,$ respectively.

Explanation of Solution

Given information: The percent composition by mass of $\text{Na}$ is $32.08\%$, $\text{O}$ is $36.01\%$, and $\text{Cl}$ is $19.51\%$.

Let the mass of sample be $100\text{ g}\text{.}$

The mass of sodium is $32.08\text{ g}\text{.}$

The mass of oxygen is $36.0\text{ g}\text{.}$

The mass of chlorine is $\text{19}\text{.51 g}\text{.}$

The molar mass of sodium is $22.99\text{ g}/\text{mol}\text{.}$

The molar mass of oxygen is $16\text{ g}/\text{mol}\text{.}$

The molar mass of chlorine is $35.45\text{ g}/\text{mol}\text{.}$

Calculate the number of moles of sodium as follows:

$n_{\text{Na}}=\frac{m_{\text{Na}}}{M_{\text{Na}}}$.

Substitute, $22.99\text{ g}/\text{mol}$ for $M_{\text{Na}}$ and $32.08\text{ g}$ for $m_{\text{Na}}$ in the above equation.

$\begin{array}{c}{n}_{\text{Na}}=\frac{32.08\cancel{\text{g}}}{22.99\cancel{\text{g}}/\text{mol}}\\ =1.395\text{ mol}\text{.}\end{array}$

Calculate the number of moles of oxygen as follows:

$n_{\text{O}}=\frac{m_{\text{O}}}{M_{\text{O}}}$.

Substitute, $\text{16}\text{.00 g}/\text{mol}$ for $M_{\text{O}}$ and $36.0\text{ g}$ for $m_{\text{O}}$ in the above equation.

$\begin{array}{c}{n}_{\text{O}}=\frac{36.0\cancel{\text{g}}}{\text{16 }\cancel{\text{g}}/\text{mol}}\\ =2.251\text{ mol}\text{.}\end{array}$

Calculate the number of moles of chlorine as follows:

$n_{\text{Cl}}=\frac{m_{\text{Cl}}}{M_{\text{Cl}}}$.

Substitute, $\text{35}\text{.45 g}/\text{mol}$ for $M_{\text{Cl}}$ and $\text{19}\text{.51 g}$ for $m_{\text{Cl}}$ in the above equation.

$\begin{array}{c}{n}_{\text{Cl}}=\frac{19.51\cancel{\text{ g}}}{35.45\cancel{\text{g}}/\text{mol}}\\ =0.5504\text{ mol}\text{.}\end{array}$

The element $\text{Cl}$ is present in only one out of the three compoundspresent in the sample $\text{NaCl}$. So, the element $\text{Na}$ also has the same number of moles in $\text{NaCl}$, that is, $0.5504\text{ mol}$.

The remaining value of $\text{Na}$ in the two other compounds of the sample, after subtracting the value in $\text{NaCl}$ is:

$\begin{array}{c}\text{number of remaining moles of Na}=1.395\text{ mol}-0.5504\text{ mol}\\ =0.8446\text{ mol}\text{.}\end{array}$

In ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$, there are two atoms of $\text{Na}$ and four atoms of oxygen, and in ${\text{NaNO}}_{\text{3}}$, there is one atom of $\text{Na}$ and three atoms of oxygen. So, to calculate the moles of these two compounds, the following equations are formed for ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{NaNO}}_{\text{3}}$, by assuming the moles of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ to be A and moles of ${\text{NaNO}}_{\text{3}}$ to be B, as follows:

$\begin{array}{c}2A+B=\text{mol of Na;}\\ 4A+3B=\text{mol of Na}\text{.}\end{array}$

Equation (1) is as follows:

$2A+B=0.8446$…… (1)

Equation (2) is as follows:

$4A+3B=2.251$…… (2)

Multiply equation (1) by two and solve as follows:

$2\times \left(2A+B=0.8446\right)=4A+2B=1.6892$. …… (3)

Subtract equation (2) from equation (3) as follows:

$\begin{array}{l}4A+3B=2.251\\ 4A+2B=1.6892\\ ---\\ \overline{B=0.5618}\end{array}$

The value of A can be calculated from value of B by using equation (1) as follows:

$\begin{array}{l}2A+B=0.8446\\ 2A+0.5618=0.8446\\ 2A=0.8446-0.5618\\ A=\mathrm{0.1414.}\end{array}$

So, the number of moles of $\text{NaCl}$ is $0.5504\text{ mol}$.

The number of moles of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $0.1414\text{ mol}$.

So, the number of moles of ${\text{NaNO}}_{\text{3}}$ is $0.5618\text{ mol}$.

The molar mass of $\text{NaCl}$ is $\text{58}\text{.44 g}/\text{mol}$.

The molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{142}\text{.1 g}/\text{mol}$.

The molar mass of ${\text{NaNO}}_{\text{3}}$ is $85\text{ g}/\text{mol}$.

The mass of each compound in the sample can be calculated as:

$\text{percent by mass of an compound}=\frac{n\times \text{ molecular mass of the compound}}{\text{mass of sample}}\times 100\%$ ……(4)

Calculate the percent by mass of $\text{NaCl}$ as follows:

Substitute, $0.5504\text{ mol}$ for n, $58.44\text{ g}$ for molecular mass of the compound, and $100\text{ g}$ mass of sample in the equation(4).

$\begin{array}{c}\text{percent by mass of an NaCl}=\frac{0.5504\text{ mol}\times 58.44\cancel{\text{g}}}{100\cancel{\text{g}}}\times 100\%\\ =32.165\%.\end{array}$

Calculate the percent by mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ as follows:

Substitute, $0.1414\text{ mol}$ for n, $\text{142}\text{.1 g}$ for molecular mass of the compound and $100\text{ g}$ mass of sample in the equation (4).

$\begin{array}{c}{\text{percent by mass of an Na}}_{\text{2}}{\text{SO}}_{\text{4}}=\frac{0.1414\text{ mol}\times 142.1\cancel{\text{g}}}{100\cancel{\text{g}}}\times 100\%\\ =20.09\%.\end{array}$

Calculate the percent by mass of ${\text{NaNO}}_{\text{3}}$ as follows:

Substitute, $0.5618\text{ mol}$ for n, $\text{85 g}$ for molecular mass of the compound and $100\text{ g}$ mass of sample in the equation (4).

$\begin{array}{c}{\text{percent by mass of an NaNO}}_{\text{3}}=\frac{0.5618\text{ mol}\times 85\cancel{\text{g}}}{100\cancel{\text{g}}}\times 100\%\\ =47.75\%.\end{array}$

Conclusion

Hence, the percent composition by mass of $\text{NaCl}$ is $32.165\%$, ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $20.09\%$, and ${\text{NaNO}}_{\text{3}}$ is $47.75\%$.