Question

What is the composition, in weight percent of an alloy that consists of 94.1 at% Ag and 5.9 at% Cu? The atomic weights for Ag and Cu
are 107.87 g/mol and 63.55 g/mol, respectively
O 9.6 wt% Ag and 90.4 wt% Cu

Answer 1

$C_{Ag} =$ 96.44%

$C_{cu}$ = 3.56%

We know that formula for Weight % of the more abundant element is given as;

C₁ = $\frac{C'_{1}A_{1} }{C'_{1}A_{1} + C'_{2}A_{2} }$

Where;

C₁ is weight % of element 1

C'₁ is at% of element 1

C'₂ is at% of element 2

A₁ is atomic weight of element 1

A₂ is atomic weight of element 2

        Applying this formula to our question, we have;

C₁ = $C_{Ag}$

$C'_{Ag}$ = 94.1

$C'_{cu}$ = 5.9

$A_{Ag}$ = 107.87 g/mol

$A_{cu}$ = 63.55 g/mol

        Plugging in these relevant values gives us;

$C_{Ag}$ = $\frac{94.1 * 107.87}{(94.1 * 107.87) + (5.9 * 63.55)}$

$C_{Ag}$ = 0.96438

In percentage gives;

$C_{Ag} =$ 96.44%

Thus; $C_{cu}$ = 100% - 96.44% = 3.56%

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