Ask A Scientist Chemistry Archive

name         Van R.
status       other
age          40s

Question -   My son's chemistry text has a problem I am not sure about. We have an alloy of 
gold and silver. We are given the mass and volume of the alloy as well as the density of 
gold and of silver. The problem is to calculate the amount of gold by mass. It is assumed 
that the volume does not change. Can we use the densities of gold and silver to calculate 
percent gold in the alloy?
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The given mass of each component: M(Au) and M(Ag) and their respective atomic weights A(Au) 
and A(Ag) of 196.96654 and 107.8682 you can calculate the number of moles of each:

m(Au) = M(Au) / A(Au) and m(Ag) = M(Ag) / A(Ag).

From this you can calculate the mole fraction of each:

 X(Au) = M(Au) /(M(Au) + M(Ag)) and
M(Ag) = M(Ag) / (M(Au) + M(Ag)) respectively. Note that X(Au) + X(Ag) = 1.

The molar volumes, V(Au) and V(Ag) can be calculated from the density of the components, 
D(Au) and D(Ag), respectively:
V(Au) = M(Au) / D(Au) and V(Ag) = M(Ag) / D(Ag).
The constant volume assumption implies there is no change in volume upon mixing. This, by 
the way is not always the case. An example, is water + ethanol.

The volume of the alloy of gold and silver is then the sum of the fraction contributed by 
each:
V(alloy) = V(Au) * X(Au)  + V(Ag) * X(Ag) = V(Au) * X (Au)  + X(Ag) * [ (1 - X(Au) ]

You know every quantity in this formula except X(Au). Rearrange and solve for X(Au). 
Then:
X(Ag) = 1 - X(Au). You can then calculate the masses: M(Au) = A(Au) * X(Au)
and
M(Ag) = A(Ag) * X(Ag).

The weight % then is: W%(Au)  = 100 * M(Au) / (M(Au) + M(Ag)) and W%(Ag) = 100 - W%(Au).

Note that the atomic weight of gold is known with more precision than for silver. This is 
because gold has only one naturally occurring isotope, and silver has two. The relative 
amounts of the isotopes of silver vary slightly, depending upon where the silver comes 
from (and how it is purified).

Vince Calder
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