Question #21663

The metallic iron have a cubic structure. The edge length of the unit cell is 287 pm. The density of the solid is 7.87 g/cm3. Calculate the number of Fe atoms per unit cell

Expert's answer

Volume of the unit cell = (287 pm)^3 = 2.364 * 10^7 pm^3

(2.364 * 10^7 pm^3)(10^-10 cm/pm)^3 = 2.364 × 10^-23 cm^3

mass of one mole of unit cells = (7.87 g/cm^3)(2.364 × 10^-23 cm^3)(6.022 ×10^23 /mol) = 112.04 g/mol of cells

Since the mass of a mole of iron is 55.845 g/mol, 112.04 g is 2 mol of iron.

There are 2 iron atoms per cell.

(2.364 * 10^7 pm^3)(10^-10 cm/pm)^3 = 2.364 × 10^-23 cm^3

mass of one mole of unit cells = (7.87 g/cm^3)(2.364 × 10^-23 cm^3)(6.022 ×10^23 /mol) = 112.04 g/mol of cells

Since the mass of a mole of iron is 55.845 g/mol, 112.04 g is 2 mol of iron.

There are 2 iron atoms per cell.

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