Question

At normal temperatures and pressures a certain metal element M forms a crystal with a bcc unit cell and lattice constant a = 0.533 nm. The density of Mis measured to be 0.8560 g/cm³?

Using only this information, Identify the element M and write its chemical symbol.

Answer 1

Using the given density and lattice constant measurements for a metal with a bcc unit cell, calculations reveal that the metal element M is chromium (Cr), with a molar mass closest to the calculated 77.97 g/mol.

Step-by-step explanation:

To identify the metal element M, which forms a crystal with a bcc unit cell and lattice constant a = 0.533 nm (0.533 x 10-7 cm), and a density of 0.8560 g/cm³, we can use the following steps and the given properties:

• Calculate the volume of the unit cell: V = a3 = (0.533 x 10-7 cm)3 = 1.51177 x 10-22 cm³.
• Determine the number of atoms per unit cell for BCC: 2 atoms.
• Calculate mass per unit cell: mass = density x volume = 0.8560 g/cm³ x 1.51177 x 10-22 cm³ = 1.29511 x 10-22 g.
• Calculate the molar mass (atomic weight) of element M: molar mass = mass per unit cell x Avogadro's number = 1.29511 x 10-22 g x 6.022 x 1023 = 77.97 g/mol.

By comparing the calculated molar mass to the known atomic weights of elements with BCC structure, we find that M likely corresponds to an element with an atomic weight close to 77.97 g/mol. Isomorphous metals with a BCC structure at room temperature include K, Ba, Cr, Mo, W, and Fe. The metal that has a molar mass nearest to 77.97 g/mol out of these elements is chromium (Cr).

Therefore, the element M with a bcc unit cell, lattice constant a = 0.533 nm, and a density of 0.8560 g/cm³ is chromium, and its chemical symbol is Cr.