Answer
$C_{Ag}$ = 44.80%
$C_{Au}$ = 46.14%
$C_{Cu}$ = 9.06%
Work Step by Step
Given:
44.5 lbm Ag
83.7 lbm Au
5.3 lbm Cu
Required:
atom percent
Solution:
$A_{Ag}$ = 107.87 g/mol
$A_{Au}$ = 196.97 g/mol
$A_{Cu}$ = 63.55 g/mol
For easier computation, convert the given weights in grams:
$m_{Ag} = (44.5 lbm)(453.6 g/lbm) $= 20,185.2 g
$m_{Au} = (83.7 lbm)(453.6 g/lbm) $= 37,966.32 g
$m_{Cu} = (5.3 lbm)(453.6 g/lbm) $= 2,404.08 g
Using Equation 4.4:
$n_{m_{Ag}} = \frac{m_{Ag}}{A_{Ag}} = \frac{20,185.2 g}{107.87 g/mol} = 187.13 mol$
$n_{m_{Au}} = \frac{m_{Au}}{A_{Au}} = \frac{37,966.32 g}{196.97 g/mol} = 192.75 mol$
$n_{m_{Cu}} = \frac{m_{Cu}}{A_{Cu}} = \frac{2,404.08 g}{63.55 g/mol} = 37.83 mol$
Using a modified form of Equation 4.5:
$C_{Ag}=\frac{n_{m_{Ag}}} {{{n_{m_{Ag}}}+n_{m_{Au}}}+n_{m_{Cu}}} \times 100 = \frac{187.13}{187.13+192.75+37.83}$ = 44.80%
$C_{Au}=\frac{n_{m_{Au}}} {{{n_{m_{Ag}}}+n_{m_{Au}}}+n_{m_{Cu}}} \times 100 = \frac{192.75}{187.13+192.75+37.83}$ = 46.14%
$C_{Cu}=\frac{n_{m_{Cu}}} {{{n_{m_{Ag}}}+n_{m_{Au}}}+n_{m_{Cu}}} \times 100 = \frac{37.83}{187.13+192.75+37.83}$ = 9.06%
