Question

Spinel is a mineral that contains 37.9% aluminum,17.1% magnesium, and45.0% oxygen, by mass, and has a density of 3.57 g/cm³. The edge of the cubic unit cell measures 809 pm. How many of each type of atom are present in the unit cell?

Short answer

Aluminum = 16

Magnesium = 8

Oxygen = 32

Step-by-step solution

Determining the empirical formula

Assume 100 g of spinel

$$\begin{gathered} 37.9 g Al \times \frac{1 mol Al}{26.98 g Al} = 1.40 mol Al \\ 17.1 g Mg \times \frac{1 mol Mg}{24.31 g Mg} = 0.703 mol Mg \\ 45 g O \times \frac{1 mol O}{16 g O} = 2.81 mol O \\ The mole ratio are 2:1:4 \\ Empirical formula = A{l}_{2}Mg{O}_4 \end{gathered}$$

Determining formula unit with the help of density

Let each unit cell contain x of Al₂MgO₄ formula units.

It has

$$\begin{gathered} mass\text{=}24.31+2\left(26.98\right)+4\left(16.00\right) \\ =142.27g/mol. \end{gathered}$$

$$\begin{gathered} Density = \frac{x \times \frac{1 mol}{6.022 \times {10}^{23} formula units } \times \frac{142.27 g}{1 mol}}{{(8.09 \times {10}^{-8} cm)}^3} \\ = 3.57 gc{m}^{-3} \\ on solving , x=8 \end{gathered}$$

Determining the number of atoms

Each unit cell has 8 formula units of Al₂MgO₄ or 16 Al, 8 Mg, and 32 O atoms.