Chapter 1, Problem 1.86P

Interpretation Introduction

Interpretation:

In modern penny, the percentage by mass of copper that is present has to be calculated.

Explanation of Solution

Density of the penny can be calculated using the equation given below.

    $d=\frac{P{d}_{Cu}+Q{d}_{Zn}}{100}$        (1)

Where,

    $P$ is the percent of copper.

    $Q$ is the percent of zinc.

    $d_{Cu}$ is the density of copper.

    $d_{Zn}$ is the density of zinc.

Volume of the penny is calculated using the equation given below.

    $V=\pi r^{2}h$        (2)

Diameter of the penny is given as $\text{19}\text{.05}\text{mm}$.  Radius can be calculated as shown below.

    $\begin{array}{c}\text{Radius}=\frac{\text{Diameter}}{\text{2}}\\ =\frac{\text{19}\text{.05}\text{mm}}{\text{2}}\\ =9.525\text{mm}\end{array}$

Height of penny is given as $\text{1}\text{.224}\text{mm}$.  Substituting these values in equation (2), the volume of penny can be calculated as shown below.

    $\begin{array}{c}V=\pi r^{2}h\\ =3.14\times {(9.525\text{mm})}^2\times 1.224\text{mm}\\ \text{=348}\text{.7}{\text{mm}}^{\text{3}}\end{array}$

Obtained volume is converted into cubic centimeter using the conversion factor as shown below.

    $\begin{array}{c}V=348.7{\text{mm}}^{\text{3}}\times \frac{0.001{\text{cm}}^{\text{3}}}{1{\text{mm}}^{\text{3}}}\\ =0.3487{\text{cm}}^{\text{3}}\end{array}$

Mass of the penny is given as $\text{2}\text{.500}\text{g}$.  Volume of penny is $0.3487{\text{cm}}^{\text{3}}$.  Density can be calculated as shown below.

    $\begin{array}{c}d=\frac{m}{V}\\ =\frac{2.500\text{g}}{0.3487{\text{cm}}^{\text{3}}}\\ =7.169\text{g}\cdot {\text{cm}}^{-3}\end{array}$

Density of copper is given as $7.13\text{g}\cdot {\text{cm}}^{-3}$ and density of zinc is given as $8.96\text{g}\cdot {\text{cm}}^{-3}$

Substituting these values in equation (1), the percentage mass of copper can be calculated as shown below.

    $\begin{array}{c}d=\frac{P{d}_{Cu}+Q{d}_{Zn}}{100}\\ =\frac{P{d}_{Cu}+(100-P)d_{Zn}}{100}\\ =\frac{P(7.13\text{g}\cdot {\text{cm}}^{-3})+(100-P)8.96\text{g}\cdot {\text{cm}}^{-3}}{100}\end{array}$

Simplifying the above equation.

    $\begin{array}{c}7.169\text{g}\cdot {\text{cm}}^{-3}=\frac{P(7.13\text{g}\cdot {\text{cm}}^{-3})+(100-P)8.96\text{g}\cdot {\text{cm}}^{-3}}{100}\\ 716.9\text{g}\cdot {\text{cm}}^{-3}=P(7.13\text{g}\cdot {\text{cm}}^{-3})+(100-P)8.96\text{g}\cdot {\text{cm}}^{-3}\\ P(7.13\text{g}\cdot {\text{cm}}^{-3})-P(8.96\text{g}\cdot {\text{cm}}^{-3})=-896\text{g}\cdot {\text{cm}}^{-3}+716.9\text{g}\cdot {\text{cm}}^{-3}\\ -P\times \text{1}\text{.83}\text{g}\cdot {\text{cm}}^{-3}=-\text{179}\text{.1}\text{g}\cdot {\text{cm}}^{-3}\\ P=\frac{\text{179}\text{.1}\text{g}\cdot {\text{cm}}^{-3}}{\text{1}\text{.83}\text{g}\cdot {\text{cm}}^{-3}}\\ =97.86\%\end{array}$

Therefore, the percentage by mass of copper in modern penny is $97.86\%$.