Some ionic compounds contain a mixture of different charged cations. For example, some titanium oxides contain a mixture of \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ti}^{3+}\) ions. Consider a certain oxide of titanium that is \(28.31 \%\) oxygen by mass and contains a mixture of \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ti}^{3+}\) ions. Determine the formula of the compound and the relative numbers of \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ti}^{3+}\) ions.

Given that the given compound is \(28.31\%\) oxygen by mass, we can calculate the mass percentage of titanium. Since it's a binary compound (titanium and oxygen), the mass percentage of titanium will be: \% Titanium = 100 - \% Oxygen. Therefore: % Titanium = 100 - 28.31 % Titanium = 71.69 The compound is \(71.69\%\) titanium by mass.

To obtain the mole ratio of the elements in the compound, we need to convert the mass percentages to moles. We'll use the molar masses of titanium and oxygen (PTable: 47.87 g/mol for titanium and 16.00 g/mol for oxygen) to make the conversions. Given that the percentage represents mass over 100 g, we get: moles of titanium = (71.69 g Ti / 47.87 g/mol Ti) moles of oxygen = (28.31 g O / 16.00 g/mol O) Calculate these values to find the moles of titanium and oxygen.

To determine the empirical formula of the compound, we need to find the simplest whole number ratio of titanium and oxygen in the compound. Divide the moles of each element by the smaller value to obtain the ratio: \(Ratio_{Ti} = \frac {moles \: of \: Ti} {Smaller \: total}\) \(Ratio_{O} = \frac {moles \: of \: O} {Smaller \: total}\) Calculate these ratios, and if necessary, multiply them by an integer value to obtain the whole number ratios.

Combine the whole number ratios from the previous step to generate the empirical formula of the compound, written as: Empirical Formula = Ti_xO_y Where x and y are the whole number ratios calculated in Step 3.

To find the relative numbers of two different cations in the compound, assume that there is a certain portion (p) of \(\mathrm{Ti}^{2+}\) ions and (1-p) portion of \(\mathrm{Ti}^{3+}\) ions. With this assumption, we can write the charge balance equation for the compound: (\(Ti^2+ ions \: charge \: contribution) + (\: Ti^{3+} ions \: charge \: contribution) = (\: charge \: contributed \: by \: O^2- ions\) Substitute the assumptions and empirical formula ratios we found earlier: \(p(Ti^{2+} ions \: \times \: 2) + (1-p)(Ti^{3+} ions \: \times \: 3) = (O^{2-} ions \: \times \: -2)\) Solve this equation for p, which will indicate the relative ratio of Ti²⁺ ions. Then, use (1-p) to find the relative ratio of Ti³⁺ ions. Thus, you'll obtain the formula of the compound and the relative numbers of \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ti}^{3+}\) ions.