Chapter 12, Problem 12.75P

Interpretation Introduction

Interpretation:

Mass percent of $\text{PbS}$ in the galena ore has to be calculated.

Concept Introduction:

Mass percent is calculated by the formula as follows:

  $\text{Mass percent}=\frac{\text{Mass of solute}}{\text{Total Mass }}\left(100\text{ \%}\right)$

Answer to Problem 12.75P

Mass percent of $\text{PbS}$ in galena ore is $88.65\text{ \%}$.

Explanation of Solution

The formula to convert mass in gram to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$        (1)

Substitute $\text{7}\text{.248 g}$ for mass and $\text{323}\text{.1937 g/mol}$ for molar mass to calculate of moles of ${\text{PbCrO}}_{\text{4}}$ in equation (1).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{7}\text{.248 g}}{\text{323}\text{.1937 g/mol}}\\ =0.022426\text{ mol}\end{array}$

Substitute $\text{6}\text{.053 g}$ for mass and $\text{239}\text{.3 g/mol}$ for molar mass to calculate moles of $\text{PbS}$ in equation (1).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{6}\text{.053 g}}{\text{239}\text{.3 g/mol}}\\ =0.0252946\text{ mol}\end{array}$

The balanced ionic reaction is given as follows:

  $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\left(aq\right)\to {\text{PbCrO}}_{\text{4}}\left(s\right)+2{\text{KNO}}_{\text{3}}\left(aq\right)$        (2)

In accordance with balanced equation (2), $\text{1 mol Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ forms ${\text{1 mol PbCrO}}_{\text{4}}$ thus moles of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ that can form $0.022426{\text{ mol PbCrO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\left(0.022426{\text{ mol PbCrO}}_{\text{4}}\right)\left(\frac{1\text{ mol Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{1{\text{ mol PbCrO}}_{\text{4}}}\right)\\ =0.022426\text{ mol Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\end{array}$

The balanced reaction that corresponds to formation of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows:

  $\text{PbS}\left(s\right)+2{\text{HNO}}_{\text{3}}\left(aq\right)\to \text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)$        (3)

In accordance with balanced equation (3), $\text{1 mol Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is formed from $\text{1 mol PbS}$ thus moles of $\text{PbS}$ required to form $0.022426\text{ mol Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}\text{Moles of PbS}=\left(0.022426\text{ mol Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)\left(\frac{1\text{ mol PbS}}{1\text{ mol Pb}{\left({\text{NO}}_{\text{3}}\right)}_2}\right)\\ =0.022426\text{ mol PbS}\end{array}$

The formula to convert moles tomass in gram is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$        (4)

Substitute $0.022426\text{ mol}$ for moles and $\text{239}\text{.3 g/mol}$ for molar mass of $\text{PbS}$ in equation (4).

  $\begin{array}{c}\text{Mass}=\left(0.022426\text{ mol}\right)\left(\text{239}\text{.3 g/mol}\right)\\ =5.3665\text{ g}\end{array}$

Mass percent is calculated by the formula as follows:

  $\text{Mass percent}=\frac{\text{Mass of solute}}{\text{Total Mass }}\left(100\text{ \%}\right)$        (5)

Substitute $5.3665\text{ g}$ for mass of solute and $\text{6}\text{.053 g}$ for total mass in equation (5) to calculate mass percent of $\text{PbS}$.

  $\begin{array}{c}\text{Mass percent}=\frac{5.3665\text{ g}}{\text{6}\text{.053 g}}\left(100\text{ \%}\right)\\ =88.65\text{ \%}\end{array}$