Iron(II)sulfide has a primitive cubic unit cell with sulfide ions at the lattice points.The ionic radii of iron(II)ions and sulfide ions are 88 pm and 184 pm,respectively.What is the density of FeS (in g/cm³)? A) 2.56 g/cm³ B) 4.71 g/cm³ C) 5.25 g/cm³ D) 6.66 g/cm³ E) 8.97 g/cm³

Question

Quizplus · Accepted Answer

The volume of the unit cell can be calculated using the formula:

V = a^3

Where a is the edge length of the primitive cubic unit cell. From the problem statement, we know that the sulfide ions are located at the lattice points, which means that the edge length of the unit cell is equal to the diameter of a sulfide ion:

a = 2rS = 2(184 pm) = 368 pm = 3.68 Å = 3.68 × 10^-8 cm

The number of sulfide ions in one unit cell is 1, since one sulfide ion is located at each of the eight corners of the cube. The number of iron(II) ions in one unit cell is also 1, since there is only one iron(II) ion for every sulfide ion in the compound. Therefore, the total number of ions in one unit cell is 2.

The total mass of the ions in one unit cell can be calculated by adding the mass of one iron(II) ion and one sulfide ion:

m = (1 × 55.85 g/mol) + (1 × 32.06 g/mol) = 87.91 g/mol

The density of the solid can then be calculated using the formula:

ρ = m/V(Na)

Where Na is Avogadro's number. Substituting the values we have calculated, we get:

ρ = (87.91 g/mol)/(3.68 × 10^-8 cm)^3(6.022 × 10^23) = 4.71 g/cm3

Therefore, the answer is B.

Iron(II)sulfide Has a Primitive Cubic Unit Cell with Sulfide Ions