Answer
$C_{Cu} $ = 41.92 at%
$C_{Zn} $ = 58.08 at%
Work Step by Step
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%
