Spinel is a mineral that contains 37.9\(\%\) aluminum, 17.1\(\%\) magnesium, and 45.0\(\%\) oxygen, by mass, and has a density of 3.57 $\mathrm{g} / \mathrm{cm}^{3} .\( The edge of the cubic unit cell measures 809 \)\mathrm{pm} .$ How many of each type of ion are present in the unit cell?

To find the mass of one unit cell, we first need to find the volume of the cubic unit cell, and then multiply it by the density of the mineral. The edge of the cubic unit cell measures 809 pm, which is equal to \(8.09 \times 10^{-8}\) cm. The volume of a cube is given by the formula: \(V = a^3\), where \(a\) is the edge length of the cube. The volume of the unit cell is: \(V = (8.09 \times 10^{-8} \text{ cm})^3 = 5.305 \times 10^{-22} \text{ cm}^3\) Now, we can calculate the mass of the unit cell using the formula: \(M = V \times d\), where \(M\) is the mass of the unit cell, \(V\) is its volume, and \(d\) is the density of the mineral. The mass of the unit cell is: \(M = 5.305 \times 10^{-22} \text{ cm}^3 \times 3.57\frac{\text{g}}{\text{cm}^3} = 1.893 \times 10^{-21}\) g.

To find the total number of ions present in one unit cell, we will use the fact that the ionic compound Spinel is formed by the combination of aluminum ions (\(\text{Al}^{3+}\)), magnesium ions (\(\text{Mg}^{2+}\)), and oxygen ions (\(\text{O}^{2-}\)). Thus, the ratio of charges should be equal, i.e., \(\text{Al}^{3+}\) and \(\text{Mg}^{2+}\) together have a net positive charge equal to the net negative charge of \(\text{O}^{2-}\) ions. Let \(n_{Al}\), \(n_{Mg}\), and \(n_O\) be the number of aluminum, magnesium, and oxygen ions in the unit cell, respectively. We have: \(3n_{Al}+2n_{Mg}=2n_O\) Using the mass percentage of Aluminum (\(37.9\%\)), Magnesium (\(17.1\%\)), and Oxygen (\(45.0\%\)): \(0.379M=27n_{Al}\) \(0.171M=24n_{Mg}\) \(0.450M=16n_O\) Now we will find the coefficients mentioned above by dividing each equation by the mass of one unit cell: \(n_{Al}=\frac{0.379M}{27M}=\frac{0.379}{27}\) \(n_{Mg}=\frac{0.171M}{24M}=\frac{0.171}{24}\) \(n_O=\frac{0.450M}{16M}=\frac{0.450}{16}\) Now let's calculate the sum of \(n_{Al}\), \(n_{Mg}\), and \(n_O\) for one unit cell: \(n_{total}=n_{Al}+n_{Mg}+n_O=\frac{0.379}{27}+\frac{0.171}{24}+\frac{0.450}{16}=1\) Thus, there is a total of 1 ion per unit cell.

Now that we know there is a total of 1 ion per unit cell, we can calculate the number of each type of ion using the previously calculated coefficients: \(n_{Al}=1 \times \frac{0.379}{27} = 0.014\) \(n_{Mg}=1 \times \frac{0.171}{24} = 0.007\) \(n_O=1 \times \frac{0.450}{16} = 0.028\) Note that these numbers are not integers, and we need an integer number of ions present in the unit cell. To find that, we'll multiply all the calculated values by their least common multiple (lcm): lcm(27, 24, 16) = 432 \(n'_{Al} = n_{Al} \times 432 = 6\) \(n'_{Mg} = n_{Mg} \times 432 = 3\) \(n'_O = n_O \times 432 = 12\) Therefore, there are 6 aluminum ions, 3 magnesium ions, and 12 oxygen ions present in the cubic unit cell of Spinel mineral.