SOLIDS: Determining the Mass of a Golden Sphere

Q: Gold has an atomic mass of 197 amu. If a golden sphere has a diameter of 0.10 m, how many atoms does it contain?

A: In order to determine what volume ( V ) is occupied by the sphere, we must obtain a value for the sphere’s radius ( r ). Fortunately, the radius of a circle is ½ the diameter ( D ) in question:

r = ½ D

r = 0.05 m

Additionally,

V = ⁴/₃𝜋r³

Substitution into this equation gives us the volume value we desire:

V = ⁴/₃𝜋( 0.05 m )³ 

V = 0.524 x 10^-3 m³

We must now determine the mass contained by the volume of gold at hand. Afterward, we can use relationships we’ve established between the grams ( g ) of a substance to the mol, and we subsequently use the relationship between moles and Avogadro’s number to solve the question. We begin by multiplying the density of gold by the volume at hand:

m = ⍴V

m = ( 19.3 x 10³ kg / m³ )( 0.524 x 10^-3 m³ )

m = 10 kg

There are 197 amu in an atom of gold, so a mol of gold contains 197 grams:
( 10 kg )( 1,000 g / 1 kg )( 1 mol / 197 g )( 6.02 x 10²³ atoms / 1 mol ) = 3.1 x 10²⁵ atoms.