IB Chemistry Option A: Materials
Search
⌃K
Comment on page

A8.8 Density of Pure metal

The density of a pure metal can be determined by using the measurements of the unit cell for that metal.
You need various pieces of information:
1. The mass of the atom (divide the relative atomic mass by the Avogadro constant).
2. The type of unit cell – and from this the number of atoms per unit cell.
3. The cell dimensions (see diagrams below for different options).

Worked example
🖊

Vanadium has a body centred structure. Its unit cell has dimensions of 303pm
(3.3×10−10m)\textsf{(3.3}\times\textsf{10}{^{-10}}\textsf{m)}
. Calculate the density of vanadium metal.
1. Ar of vanadium = 50.94 So 1 atom has a mass of
50.946.02×1023 = 8.46×10-23g\dfrac{\textsf{50.94}}{\textsf{6.02}\times\textsf{10}{^\textsf{23}}}\textsf{ = 8.46}\times\textsf{10}{^\textsf{-23}} \textsf{g}
2. BCC structure has 2 atoms per unit cell So total mass of 1 unit cell
8.46×10-23× 2 = 1.692×10-22g\textsf{8.46}\times\textsf{10}{^\textsf{-23}} \times\textsf{ 2}\textsf{ = 1.692}\times\textsf{10}{^\textsf{-22}} \textsf{g}
3. Volume of the unit cell =
(3.3×10-10)3= 3.594×10-29m-3\textsf{(3.3}\times\textsf{10}{^\textsf{-10}}\textsf{)}^\textsf{3} \textsf{= 3.594}\times\textsf{10}^\textsf{-29}\textsf m^\textsf{-3}
 Density = mass/volume
= 1.692×10-22 3.594×10-29 = 4.704× 106 g m−3=\dfrac{\textsf{ 1.692}\times \textsf{10} {^\textsf{-22}}} {\textsf{ 3.594}\times \textsf{10}{^\textsf{-29}}} \textsf{ = 4.704}\times \textsf{ 10}{^\textsf{6}} \textsf{ g m} {^{-3}}
Since there are
106 cm3\textsf{10}{^\textsf{6}} \ \textsf{cm}{^\textsf{3}}
in
1m3\textsf{1m}{^{3}}
, convert the answer to
gcm-3\textsf{gcm}{^\textsf{-3}}
, divide by 1 =
4.6 g cm-3\textsf {4.6 g cm}{^\textsf{-3}}
Previous
A8.7 X-Ray Diffraction
Next
A8.9 Using the Ionic Radius to Calculate Density of Metals
Last modified 2yr ago