Materials Science and Engineering: An Introduction

$C_{Cu}$ = 41.92 at% $C_{Zn}$ = 58.08 at%
Given: 33 g of Cu 47 g of Zn Required: composition in atom percent Solution: $A_{Cu}$ = 63.55 g/mol $A_{Zn}$ = 65.41 g/mol Using Equation 4.4, compute the number of moles: $n_{m_{Cu}} = \frac{m_{Cu}}{A_{Cu}}= \frac{33 g}{63.55 g/mol}$= 0.519 mol $n_{m_{Zn}} = \frac{m_{Zn}}{A_{Zn}}= \frac{47 g}{65.41 g/mol}$= 0.719 mol Using Equation 4.5: $C_{Cu} = \frac{n_{m_{Cu}}}{n_{m_{Cu}} + n_{m_{Zn}}} \times 100 = \frac{0.519 mol}{0.519 mol + 0.719 mol} \times 100$ = 41.92 at% $C_{Zn} = \frac{n_{m_{Zn}}}{n_{m_{Cu}} + n_{m_{Zn}}} \times 100 = \frac{0.719 mol}{0.519 mol + 0.719 mol} \times 100$ = 58.08 at%