Express in mole percent the composition of the alloy containing 36% by weight of silver and 64% by weight of copper.

Given:
ω (Ag) = 36%
ω (Cu) = 64%

To find:
x (Ag) -?
x (Cu) -?

Decision:
1) Let m alloy = 100 g;
2) m (Ag) = ω (Ag) * m alloy / 100% = 36% * 100/100% = 36 g;
3) n (Ag) = m / M = 36/108 = 0.33 mol;
3) m (Cu) = ω (Cu) * m alloy / 100% = 64% * 100/100% = 64 g;
4) n (Cu) = m / M = 64/64 = 1 mol;
5) Σn = n (Ag) + n (Cu) = 0.33 + 1 = 1.33 mol;
6) x (Ag) = n (Ag) / Σn = 0.33 / 1.33 = 0.25;
7) x (Cu) = n (Cu) / Σn = 1 / 1.33 = 0.75.

Answer: The mole fraction of Ag is 0.25; Cu – 0.75.