Chapter 4.8, Problem 36AAP

(a)

To determine

Determine the absolute mass of silver, copper and gold metal in a 253-gram medal.

Answer to Problem 36AAP

The absolute mass of silver, copper and gold metal in a medal is $\text{239}\text{g,}\text{11}\text{.5}\text{g}\text{and}\text{2}\text{.38}\text{g}$ respectively.

Explanation of Solution

Express the relative mass of silver.

 $m_{\text{Ag}}=\frac{M_{\text{Ag}}}{N}\left(\%m_{\text{Ag}}\right)$                                                                                 (I)

Here, molecular mass of silver is $M_{\text{Ag}}$, composition of silver in medal is $\%m_{\text{Ag}}$ and Avogadro’s number is $N$.

Express the relative mass of copper.

 $m_{\text{Cu}}=\frac{M_{\text{Cu}}}{N}\left(\%m_{\text{Cu}}\right)$                                                                                  (II)

Here, molecular mass of copper is $M_{\text{Cu}}$ and composition of copper in medal is $\%m_{\text{Cu}}$.

Express the relative mass of gold.

 $m_{\text{Au}}=\frac{M_{\text{Au}}}{N}\left(\%m_{\text{Au}}\right)$                                                                                     (III)

Here, molecular mass of gold is $M_{\text{Au}}$ and composition of gold in medal is $\%m_{\text{Au}}$.

Express the absolute mass of silver in medal

 $m_{\text{Ag,medal}}=\frac{m_{\text{Ag}}}{m_{\text{Ag}}+m_{\text{Cu}}+m_{\text{Au}}}\left(m_{\text{medal}}\right)$                                                         (IV)

Here, mass of medal is $m_{\text{medal}}$.

Express the absolute mass of copper in medal

 $m_{\text{Cu,medal}}=\frac{m_{\text{Cu}}}{m_{\text{Ag}}+m_{\text{Cu}}+m_{\text{Au}}}\left(m_{\text{medal}}\right)$                                                          (V)

Express the absolute mass of gold in medal

 $m_{\text{Au,medal}}=\frac{m_{\text{Au}}}{m_{\text{Ag}}+m_{\text{Cu}}+m_{\text{Au}}}\left(m_{\text{medal}}\right)$                                                          (VI)

Conclusion:

Write the molecular mass of silver, copper and gold.

 $\begin{array}{l}{M}_{\text{Ag}}=107.99\text{g}\\ M_{\text{Cu}}=63.55\text{g}\\ M_{\text{Au}}=197\text{g}\end{array}$

Write the value of Avogadro’s number.

 $N=6.02\times {10}^{23}\text{atoms}$

Substitute $107.99\text{g}$ for $M_{\text{Ag}}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $0.92$ for $\%m_{\text{Ag}}$ in Equation (I).

 $\begin{array}{c}{m}_{\text{Ag}}=\frac{107.99\text{g}}{6.02\times {10}^{23}\text{atoms}}\left(0.92\right)\\ =1.649\times {10}^{-22}\text{g}\end{array}$

Substitute $63.55\text{g}$ for $M_{\text{Cu}}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $0.075$ for $\%m_{\text{Cu}}$ in Equation (II).

 $\begin{array}{c}{m}_{\text{Cu}}=\frac{63.55\text{g}}{6.02\times {10}^{23}\text{atoms}}\left(0.075\right)\\ =7.974\times {10}^{-24}\text{g}\end{array}$

Substitute $197\text{g}$ for $M_{\text{Au}}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $0.005$ for $\%m_{\text{Au}}$ in Equation (III).

 $\begin{array}{c}{m}_{\text{Au}}=\frac{197\text{g}}{6.02\times {10}^{23}\text{atoms}}\left(0.005\right)\\ =1.636\times {10}^{-24}\text{g}\end{array}$

Substitute $1.649\times {10}^{-22}\text{g}$ for $m_{\text{Ag}}$, $7.974\times {10}^{-24}\text{g}$ for $m_{\text{Cu}}$, $1.636\times {10}^{-24}\text{g}$ for $m_{\text{Au}}$, and $\text{253}\text{g}$ for $m_{\text{medal}}$ in Equation (IV).

 $\begin{array}{c}{m}_{\text{Ag,medal}}=\frac{1.649\times {10}^{-22}\text{g}}{\left(1.649\times {10}^{-22}\text{g}\right)+\left(7.974\times {10}^{-24}\text{g}\right)+\left(1.636\times {10}^{-24}\text{g}\right)}\left(\text{253}\text{g}\right)\\ =239\text{g}\end{array}$

Hence, the absolute mass of silver in a medal is $\text{239}\text{g}$.

Substitute $1.649\times {10}^{-22}\text{g}$ for $m_{\text{Ag}}$, $7.974\times {10}^{-24}\text{g}$ for $m_{\text{Cu}}$, $1.636\times {10}^{-24}\text{g}$ for $m_{\text{Au}}$, and $\text{253}\text{g}$ for $m_{\text{medal}}$ in Equation (V).

 $\begin{array}{c}{m}_{\text{Cu,medal}}=\frac{7.974\times {10}^{-24}\text{g}}{\left(1.649\times {10}^{-22}\text{g}\right)+\left(7.974\times {10}^{-24}\text{g}\right)+\left(1.636\times {10}^{-24}\text{g}\right)}\left(\text{253}\text{g}\right)\\ =11.5\text{g}\end{array}$

Hence, the absolute mass of copper in a medal is $\text{11}\text{.5}\text{g}$.

Substitute $1.649\times {10}^{-22}\text{g}$ for $m_{\text{Ag}}$, $7.974\times {10}^{-24}\text{g}$ for $m_{\text{Cu}}$, $1.636\times {10}^{-24}\text{g}$ for $m_{\text{Au}}$, and $\text{253}\text{g}$ for $m_{\text{medal}}$ in Equation (VI).

 $\begin{array}{c}{m}_{\text{Cu,medal}}=\frac{1.636\times {10}^{-24}\text{g}}{\left(1.649\times {10}^{-22}\text{g}\right)+\left(7.974\times {10}^{-24}\text{g}\right)+\left(1.636\times {10}^{-24}\text{g}\right)}\left(\text{253}\text{g}\right)\\ =2.38\text{g}\end{array}$

Hence, the absolute mass of gold in a medal is $\text{2}\text{.38}\text{g}$.

(b)

To determine

Determine the absolute mass of silver and copper metal in a 253-gram medal.

Answer to Problem 36AAP

The absolute mass of silver and copper in a medal is $241\text{g}\text{and}\text{11}\text{.6}\text{g}$ respectively.

Explanation of Solution

Express the relative mass of silver.

 $m_{\text{Ag}}=\frac{M_{\text{Ag}}}{N}\left(\%m_{\text{Ag}}\right)$                                                                                 (VII)

Express the relative mass of copper.

 $m_{\text{Cu}}=\frac{M_{\text{Cu}}}{N}\left(\%m_{\text{Cu}}\right)$                                                                                 (VIII)

Express the absolute mass of silver in medal

 $m_{\text{Ag,medal}}=\frac{m_{\text{Ag}}}{m_{\text{Ag}}+m_{\text{Cu}}}\left(m_{\text{medal}}\right)$                                                                 (IX)

Express the absolute mass of copper in medal

 $m_{\text{Cu,medal}}=\frac{m_{\text{Cu}}}{m_{\text{Ag}}+m_{\text{Cu}}}\left(m_{\text{medal}}\right)$                                                                (X)

Conclusion:

Substitute $107.99\text{g}$ for $M_{\text{Ag}}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $0.925$ for $\%m_{\text{Ag}}$ in Equation (VII).

 $\begin{array}{c}{m}_{\text{Ag}}=\frac{107.99\text{g}}{6.02\times {10}^{23}\text{atoms}}\left(0.925\right)\\ =1.658\times {10}^{-22}\text{g}\end{array}$

Substitute $63.55\text{g}$ for $M_{\text{Cu}}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $0.075$ for $\%m_{\text{Cu}}$ in Equation (VIII).

 $\begin{array}{c}{m}_{\text{Cu}}=\frac{63.55\text{g}}{6.02\times {10}^{23}\text{atoms}}\left(0.075\right)\\ =7.974\times {10}^{-24}\text{g}\end{array}$

Substitute $1.658\times {10}^{-22}\text{g}$ for $m_{\text{Ag}}$, $7.974\times {10}^{-24}\text{g}$ for $m_{\text{Cu}}$ and $\text{253}\text{g}$ for $m_{\text{medal}}$ in Equation (IX).

 $\begin{array}{c}{m}_{\text{Ag,medal}}=\frac{1.658\times {10}^{-22}\text{g}}{\left(1.658\times {10}^{-22}\text{g}\right)+\left(7.974\times {10}^{-24}\text{g}\right)}\left(\text{253}\text{g}\right)\\ =241\text{g}\end{array}$

Hence, the absolute mass of silver in a medal is $\text{241}\text{g}$.

Substitute $1.658\times {10}^{-22}\text{g}$ for $m_{\text{Ag}}$, $7.974\times {10}^{-24}\text{g}$ for $m_{\text{Cu}}$ and $\text{253}\text{g}$ for $m_{\text{medal}}$ in Equation (X).

 $\begin{array}{c}{m}_{\text{Cu,medal}}=\frac{7.974\times {10}^{-24}\text{g}}{\left(1.658\times {10}^{-22}\text{g}\right)+\left(7.974\times {10}^{-24}\text{g}\right)}\left(\text{253}\text{g}\right)\\ =11.6\text{g}\end{array}$

Hence, the absolute mass of copper in a medal is $\text{11}\text{.6}\text{g}$.

(c)

To determine

Determine the absolute mass of tin and copper metal in a 253-gram medal.

Answer to Problem 36AAP

The absolute mass of tin and copper in a medal is $27.5\text{g}\text{and}\text{225}\text{g}$ respectively.

Explanation of Solution

Express the relative mass of tin.

 $m_{\text{Sn}}=\frac{M_{\text{Sn}}}{N}\left(\%m_{\text{Sn}}\right)$                                                                                  (XI)

Here, molecular mass of tin is $M_{\text{Sn}}$ and composition of tin in medal is $\%m_{\text{Sn}}$.

Express the relative mass of copper.

 $m_{\text{Cu}}=\frac{M_{\text{Cu}}}{N}\left(\%m_{\text{Cu}}\right)$                                                                                  (XII)

Express the absolute mass of tin in medal

 $m_{\text{Sn,medal}}=\frac{m_{\text{Sn}}}{m_{\text{Sn}}+m_{\text{Cu}}}\left(m_{\text{medal}}\right)$                                                                  (XIII)

Express the absolute mass of copper in medal

 $m_{\text{Cu,medal}}=\frac{m_{\text{Cu}}}{m_{\text{Sn}}+m_{\text{Cu}}}\left(m_{\text{medal}}\right)$                                                                (XIV)

Conclusion:

Write the molecular mass of tin.

 $M_{\text{Sn}}=118.7\text{g}$

Substitute $118.7\text{g}$ for $M_{\text{Sn}}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $0.1$ for $\%m_{\text{Sn}}$ in Equation (XI).

 $\begin{array}{c}{m}_{\text{Sn}}=\frac{118.7\text{g}}{6.02\times {10}^{23}\text{atoms}}\left(0.1\right)\\ =1.972\times {10}^{-23}\text{g}\end{array}$

Substitute $63.55\text{g}$ for $M_{\text{Cu}}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $0.9$ for $\%m_{\text{Cu}}$ in Equation (XII).

 $\begin{array}{c}{m}_{\text{Cu}}=\frac{63.55\text{g}}{6.02\times {10}^{23}\text{atoms}}\left(0.9\right)\\ =1.613\times {10}^{-22}\text{g}\end{array}$

Substitute $1.972\times {10}^{-23}\text{g}$ for $m_{\text{Sn}}$, $1.613\times {10}^{-22}\text{g}$ for $m_{\text{Cu}}$ and $\text{253}\text{g}$ for $m_{\text{medal}}$ in Equation (XIII).

 $\begin{array}{c}{m}_{\text{Sn,medal}}=\frac{1.972\times {10}^{-23}\text{g}}{\left(1.972\times {10}^{-23}\text{g}\right)+\left(1.613\times {10}^{-22}\text{g}\right)}\left(\text{253}\text{g}\right)\\ =27.5\text{g}\end{array}$

Substitute $1.972\times {10}^{-23}\text{g}$ for $m_{\text{Sn}}$, $1.613\times {10}^{-22}\text{g}$ for $m_{\text{Cu}}$ and $\text{253}\text{g}$ for $m_{\text{medal}}$ in Equation (XIV).

 $\begin{array}{c}{m}_{\text{Cu,medal}}=\frac{1.613\times {10}^{-22}\text{g}}{\left(1.972\times {10}^{-23}\text{g}\right)+\left(1.613\times {10}^{-22}\text{g}\right)}\left(\text{253}\text{g}\right)\\ =225\text{g}\end{array}$

Hence, the absolute mass of tin and copper in a medal is $27.5\text{g}\text{and}\text{225}\text{g}$ respectively