The cubic unit cell of rhenium trioxide (ReO₃) has Re atoms at the corners and O atoms on each of the 12 edges. The atoms touch along the edges. The radii are Re 137 pm and O 73 pm. What is the density of ReO₃? A)2.55 g/cm³ B)2.94 g/cm³ C)3.49 g/cm³ D)5.25 g/cm³ E)7.52 g/cm³

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Quizplus · Accepted Answer

The density of ReO3 can be calculated by first determining the formula unit's mass, the volume of the unit cell, and then using the density formula. Each corner atom contributes 1/8 of its mass to the unit cell, and there are 8 corners, so there is 1 Re atom per unit cell. Each edge atom (O) contributes 1/4 of its mass to the unit cell, and there are 12 edges, so there are 3 O atoms per unit cell. The mass of 1 Re atom is approximately 186.2 u (atomic mass units), and the mass of 1 O atom is approximately 16 u. Therefore, the mass of the formula unit is $186.2 u + 3(16 u) = 234.2 u$. The edge length of the unit cell can be determined by adding the radii of Re and O (since they touch along the edges), which gives $137 pm + 73 pm = 210 pm$ for half the edge length, so the full edge length is $420 pm = 420 \times 10^{-12} m = 4.20 \times 10^{-10} m$. The volume of the cubic unit cell is $(4.20 \times 10^{-10} m)^3 = 7.448 \times 10^{-29} m^3$.To convert the mass from atomic mass units to kilograms, we use the conversion factor $1 u = 1.660539 \times 10^{-27} kg$, so the mass of the formula unit in kilograms is $234.2 u \times 1.660539 \times 10^{-27} kg/u = 3.889 \times 10^{-25} kg$.Finally, the density $\rho$ is mass/volume, so $\rho = \frac{3.889 \times 10^{-25} kg}{7.448 \times 10^{-29} m^3} = 5.22 \times 10^3 kg/m^3 = 5.22 g/cm^3$, which rounds to $5.25 g/cm^3$ when considering significant figures and the approximation used in the calculation.

The Cubic Unit Cell of Rhenium Trioxide (ReO3) Has Re

The Cubic Unit Cell of Rhenium Trioxide (ReO₃) Has Re