Question

What is the composition, in atom percent, of an alloy that consists of a) 91.5 wt% Ag and b) 8.5 wt% Cu?

Assume that the atomic weights for silver and copper are, respectively, 107.87 and 63.55 g/mol.
Round off your answers to 3 significant digits.

Answer 1

The alloy consists of approximately 86.4 atom percent silver (Ag) and 13.6 atom percent copper (Cu) when rounded to three significant digits, based on their given weight percentages and atomic weights.

Step-by-step explanation:

The composition of an alloy in atom percent is calculated by comparing the number of atoms of each element relative to the total number of atoms in the alloy. The atomic percent (at%) can be found by dividing the number of moles of each element by the total moles in the mixture and then multiplying by 100%. To compute this, we need to first calculate the number of moles of silver (Ag) and copper (Cu) based on their weight percentages and atomic weights.

• Number of moles of Ag = (91.5 g) / (107.87 g/mol) = 0.848 moles of Ag
• Number of moles of Cu = (8.5 g) / (63.55 g/mol) = 0.134 moles of Cu
• Total moles = 0.848 moles of Ag + 0.134 moles of Cu = 0.982 moles

The atom percent of Ag is then (0.848 moles / 0.982 moles) * 100% ≈ 86.354%.

The atom percent of Cu is (0.134 moles / 0.982 moles) * 100% ≈ 13.646%.

Rounding off to three significant digits, the alloy consists of 86.4 at% Ag and 13.6 at% Cu.