Answer
$C_{Ag}$ = 33.34 wt%
$C_{Au}$ = 62.69 wt%
$C_{Cu}$ = 3.97 wt%
Work Step by Step
Given:
44.80 at% Ag
46.14 at% Au
9.06 at% Cu
Required:
Weight percent
Solution:
$A_{Ag}$ = 107.87 g/mol
$A_{Au}$ = 196.97 g/mol
$A_{Cu}$ = 63.55 g/mol
Using Equation 4.7a:
$C_{Ag}= \frac{C_{Ag}A_{Ag}}{C_{Ag}A_{Ag} +C_{Au}A_{Au} +C_{Cu}A_{Cu} } \times 100 = \frac{(44.8)(107.87 g/mol)}{(44.8)(107.87 g/mol)+(46.14)(196.97)+(9.06)(63.55 g/mol)} \times 100$ = 33.34 wt%
$C_{Au}= \frac{C_{Au}A_{Au}}{C_{Ag}A_{Ag} +C_{Au}A_{Au} +C_{Cu}A_{Cu} } \times 100 = \frac{(46.14)(196.97 g/mol)}{(44.8)(107.87 g/mol)+(46.14)(196.97)+(9.06)(63.55 g/mol)} \times 100$ = 62.69 wt%
$C_{Cu}= \frac{C_{Cu}A_{Cu}}{C_{Ag}A_{Ag} +C_{Au}A_{Au} +C_{Cu}A_{Cu} } \times 100 = \frac{(9.06)(63.55 g/mol)}{(44.8)(107.87 g/mol)+(46.14)(196.97)+(9.06)(63.55 g/mol)} \times 100$ = 3.97 wt%
