Chapter 3, Problem 3I.8E

(a)

Interpretation Introduction

Interpretation:

Relative number of atoms of each element in alloy with $28\text{ \% Pb}$ and $22\text{ \% Sn}$ by mass in bismuth has to be determined.

Concept Introduction:

Alloys are formed from combination of metal with other elements. It retains the properties of metal but affects their crystalline structure. Examples of alloys are stainless steel, brass, bronze, and white gold.

Answer to Problem 3I.8E

Alloy has 1.4 atoms of $\text{Sn}$ per $\text{Pb}$ atom and 1.8 $\text{Bi}$ atoms per $\text{Pb}$ atom.

Explanation of Solution

Alloy has $28\text{ \% Pb}$ and $22\text{ \% Sn}$ by mass in bismuth thus $\text{100 g}$ alloy contains $28\text{ g Pb}$, $22\text{ g Sn}$, and $50\text{ g Bi}$.

Expression to determine number of atoms present in given mass of element is as follows:

  $\text{Number of atoms}=\left(\text{Mass of element}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{\text{Molar mass}}\right)$        (1)

Substitute $28\text{ g}$ for mass of element and $207.2\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (1) to calculate number of atoms of $\text{Pb}$.

  $\text{Number of atoms}=\left(\text{28 g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{207.2\text{ g}\cdot {\text{mol}}^{-1}}\right)$        (2)

Substitute $22\text{ g}$ for mass of element and $118.71\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (1) to calculate number of atoms of $\text{Sn}$.

  $\text{Number of atoms}=\left(\text{22 g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{118.71\text{ g}\cdot {\text{mol}}^{-1}}\right)$        (3)

Substitute $50\text{ g}$ for mass of element and $208.98\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (1) to calculate number of atoms of $\text{Bi}$.

  $\text{Number of atoms}=\left(50\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{208.98\text{ g}\cdot {\text{mol}}^{-1}}\right)$        (4)

Divide equation (3) by (2) to determine number of $\text{Sn}$ atoms per $\text{Pb}$ atom as follows:

  $\begin{array}{c}\frac{\text{Number of Sn atoms}}{\text{Number of Pb atoms}}=\frac{\left(\text{22 g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{118.71\text{ g}\cdot {\text{mol}}^{-1}}\right)}{\left(\text{28 g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{207.2\text{ g}\cdot {\text{mol}}^{-1}}\right)}\\ =1.4\end{array}$

Divide equation (4) by (2) to determine number of $\text{Bi}$ atoms per $\text{Pb}$ atom as follows:

  $\begin{array}{c}\frac{\text{Number of Bi atoms}}{\text{Number of Pb atoms}}=\frac{\left(50\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{208.98\text{ g}\cdot {\text{mol}}^{-1}}\right)}{\left(\text{28 g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{207.2\text{ g}\cdot {\text{mol}}^{-1}}\right)}\\ =1.8\end{array}$

Hence, alloy has 1.4 atoms of $\text{Sn}$ per $\text{Pb}$ atom and 1.8 $\text{Bi}$ atoms per $\text{Pb}$ atom.

(b)

Interpretation Introduction

Interpretation:

Relative number of atoms of each element in alloy with $4.4\text{ \% Cu}$, $1.5\text{ \% Mg}$, and $0.6\text{ \% Mn}$ by mass in aluminum has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3I.8E

Alloy has 15 atoms of $\text{Sb}$ per $\text{Cu}$ atom and 1.2 $\text{Sb}$ atoms per $\text{Cu}$ atom.

Explanation of Solution

Alloy has $4.4\text{ \% Cu}$, $1.5\text{ \% Mg}$, and $0.6\text{ \% Mn}$ by mass in aluminum thus $\text{100 g}$ alloy contains $4.4\text{ g Cu}$, $1.5\text{ g Mg}$, $0.6\text{ g Mn}$, and $93.5\text{ g Al}$.

Expression to determine number of atoms present in given mass of element is as follows:

  $\text{Number of atoms}=\left(\text{Mass of element}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{\text{Molar mass}}\right)$        (1)

Substitute $4.4\text{ g}$ for mass of element and $63.55\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (1) to calculate number of atoms of $\text{Cu}$.

  $\text{Number of atoms}=\left(4.4\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{63.55\text{ g}\cdot {\text{mol}}^{-1}}\right)$        (5)

Substitute $1.5\text{ g}$ for mass of element and $24.31\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (1) to calculate number of atoms of $\text{Mg}$.

  $\text{Number of atoms}=\left(\text{1}\text{.5 g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{24.31\text{ g}\cdot {\text{mol}}^{-1}}\right)$        (6)

Substitute $0.6\text{ g}$ for mass of element and $54.94\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (1) to calculate number of atoms of $\text{Mn}$.

  $\text{Number of atoms}=\left(0.6\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{54.94\text{ g}\cdot {\text{mol}}^{-1}}\right)$        (7)

Substitute $93.5\text{ g}$ for mass of element and $26.98\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (1) to calculate number of atoms of $\text{Al}$.

  $\text{Number of atoms}=\left(93.5\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{26.98\text{ g}\cdot {\text{mol}}^{-1}}\right)$        (8)

Divide equation (5) by (7) to determine number of $\text{Cu}$ atoms per $\text{Mn}$ atom as follows:

  $\begin{array}{c}\frac{\text{Number of Cu atoms}}{\text{Number of Mn atoms}}=\frac{\left(4.4\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{63.55\text{ g}\cdot {\text{mol}}^{-1}}\right)}{\left(0.6\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{54.94\text{ g}\cdot {\text{mol}}^{-1}}\right)}\\ =6.3\end{array}$

Divide equation (6) by (7) to determine number of $\text{Mg}$ atoms per $\text{Mn}$ atom as follows:

  $\begin{array}{c}\frac{\text{Number of Mg atoms}}{\text{Number of Mn atoms}}=\frac{\left(\text{1}\text{.5 g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{24.31\text{ g}\cdot {\text{mol}}^{-1}}\right)}{\left(0.6\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{54.94\text{ g}\cdot {\text{mol}}^{-1}}\right)}\\ =5.6\end{array}$

Divide equation (8) by (7) to determine number of $\text{Al}$ atoms per $\text{Mn}$ atom as follows:

  $\begin{array}{c}\frac{\text{Number of Al atoms}}{\text{Number of Mn atoms}}=\frac{\left(93.5\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{26.98\text{ g}\cdot {\text{mol}}^{-1}}\right)}{\left(0.6\text{ g}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}\cdot {\text{mol}}^{-1}}{54.94\text{ g}\cdot {\text{mol}}^{-1}}\right)}\\ =317\end{array}$

Hence, alloy has 317 atoms of $\text{Al}$ per $\text{Mn}$ atom, 6.3 atoms of $\text{Cu}$ per $\text{Mn}$ atom, and 5.6 atoms of $\text{Mg}$ per $\text{Mn}$ atom.