Question

A gold atom is a sphere with a diameter of 272 pm and a mass of 3.27 x 10-13 ng. Calculate the density of one gold atom (in g/cm3). Note the Vsphere = 4/3 π r3.

Answer 1

Density =mass/volume

Mass is 3.27 x 10^-13ng.

Since diameter is 272 pm, the radius is half of that: 137pm.
Now solve the volume: Vsphere=4/3π(137^3)≈1091317pm³
Then solve density:
3.27x10^-13ng/1091317pm³ ≈3.00x10^-19 ng/pm³
Convert units: 3.00x10^-19ng x(1 gram/1x10^9 grams)x(1 pm/1x10^-10cm)³=300g/cm³

Answer 2

Answer is: density is 3.88 g/cm³.

r(Au) = 272 pm = 272·10⁻¹² m.

r(Au) = 2.72·10⁻¹⁰ m · 10²cm/m.

r(Au) = 2.72·10⁻⁸ cm; diameter of gold atom.

V(Au) = 4/3 · (2.72·10⁻⁸ cm)³ · 3.14.

V(Au) = 8.43·10⁻²³ cm³; volume of gold atom.

m(Au) = 3.27·10⁻¹³ ng · 10⁻⁹ g/ng.

m(Au) = 3.27·10⁻²² g; mass of gold atom.

d(Au) = m(Au) ÷ V(Au).

d(Au) = 3.27·10⁻²² g ÷ 8.43·10⁻²³ cm³.

d(Au) = 3.88 g/cm³; density of one gold atom.