Answer to Question #133326 in Inorganic Chemistry for Nisaa

Question

A small bar of an unknown metal was left on the bench. The density of the metal is 11.5 g/cm3. An X-ray diffraction experiment was conducted to measures the edge of the face-centered cubic unit cell as 4.06 x 10-10 m. Find the atomic weight of this metal and tentatively identify this metal.

Answer 1

$Density=$ $Mass \over Volume$ $=11.5g/cm^3$

$Volume(i^3)=$ $(4.06\times 10^{-8}cm^3)^3$

$=6.692\times 10^{-23}cm^3$

$11.5g/cm^3=$ $Mass of unit cell \over 6.692\times 10^{-23}cm^3$

$\therefore$ Mass of unit cell$=7.696\times 10^{-22}g$

Since there are $4$ units per unit cell, the mass of one atom is mass of unit cell/ 4. Converting the mass of one atom to the mass of 1 mole therefore is;

$7.696\times 10^{-22}g\over 4atoms$ $=$ $1.924\times 10^{-22}g\over atom$ $\times 6.022\times 10^{23} atoms$

$=115.86g/mol$

The compound is thus $FeCO_3$