Indium oxide contains \(4.784 \mathrm{~g}\) of indium for every \(1.000 \mathrm{~g}\) of oxygen. In 1869 , when Mendeleev first presented his version of the periodic table, he proposed the formula \(\mathrm{In}_{2} \mathrm{O}_{3}\) for indium oxide. Before that time it was thought that the formula was InO. What values for the atomic mass of indium are obtained using these two formulas? Assume that oxygen has an atomic mass of \(16.00\).

Using the given mass ratio 4.784 g In : 1.000 g O, we first determine the mass of each indium oxide sample for both proposed formulas. - For InO: 1 In atom and 1 O atom - For In₂O₃: 2 In atoms and 3 O atoms Let's calculate the mass of both compounds: - InO mass: Since there is 1 In atom and 1 O atom in InO, the mass of In will be 4.784 grams, and the mass of O will be 1.000 grams. - In₂O₃ mass: Since there are 2 In atoms and 3 O atoms in In₂O₃, the mass of In will be 2 * 4.784 = 9.568 grams, and the mass of O will be 3 * 1.000 = 3.000 grams.

Now, let's calculate the moles of indium and oxygen in both samples. To calculate moles, we use the formula: Moles = mass / molar mass - Moles of O in InO: Moles = 1.000 g / 16.00 g/mol = 0.0625 mol - Moles of O in In₂O₃: Moles = 3.000 g / 16.00 g/mol = 0.1875 mol

To calculate the atomic mass of indium using the given mass ratio and the moles of In and O in both samples, we will use the formula: Atomic mass of In = mass of In / moles of In Since the ratio of In:O is the same in both samples, we can directly calculate moles of In: - Moles of In in InO: Moles of In = Moles of O = 0.0625 mol - Moles of In in In₂O₃: Moles of In = 0.5 * Moles of O = 0.5 * 0.1875 = 0.09375 mol Now, we can calculate the atomic mass of In using both formulas: - Atomic mass of In (InO): Atomic mass = 4.784 g / 0.0625 mol ≈ 76.54 g/mol - Atomic mass of In (In₂O₃): Atomic mass = 9.568 g / 0.09375 mol ≈ 102.05 g/mol So, the values for the atomic mass of indium using the formulas InO and In₂O₃ are approximately 76.54 g/mol and 102.05 g/mol, respectively.