Question

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.

Short answer

The formula of the titanium oxide compound is \(\mathrm{Ti_2O_3}\), with one \(\mathrm{Ti}^{2+}\) and one \(\mathrm{Ti}^{3+}\) ion.

Step-by-step solution

Calculate the mass fraction of titanium

Since the titanium oxide compound is \(28.31 \%\) oxygen by mass, we can determine the mass fraction of titanium by subtracting the mass fraction of oxygen from \(100 \%\): Mass fraction of titanium = \(100 \% - 28.31 \% = 71.69 \%\)

Determine the mole ratio of titanium to oxygen

To find the mole ratio of titanium to oxygen, we will use their molar masses (atomic masses in grams): Molar mass of Ti = \(47.87 \, \text{g/mol}\) Molar mass of O = \(16.00 \, \text{g/mol}\) Let's use \(m_\text{Ti}\) and \(m_\text{O}\) to represent the mass of titanium and oxygen in \(100 \, \text{g}\) of titanium oxide compound. Now we can calculate the moles of titanium and oxygen in the compound: Moles of Ti = \(\frac{m_\text{Ti}}{M_\text{Ti}} = \frac{71.69}{47.87}\) Moles of O = \(\frac{m_\text{O}}{M_\text{O}} = \frac{28.31}{16.00}\) Divide the moles of Ti by the moles of O to find the mole ratio of titanium to oxygen: Mole ratio of Ti to O = \(\frac{\frac{71.69}{47.87}}{\frac{28.31}{16.00}}\)

Simplify the mole ratio

Calculate and simplify the mole ratio of Ti to O found in step 2 to get the whole-number ratio: Mole ratio of Ti to O = \(\frac{71.69}{47.87} \times \frac{16.00}{28.31} \approx 2:3\) Hence, in the titanium oxide compound, there are 2 moles of Ti for every 3 moles of O.

Determine the relative numbers of \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ti}^{3+}\) ions

Let the number of \(\mathrm{Ti}^{2+}\) ions be \(x\) and the number of \(\mathrm{Ti}^{3+}\) ions be \(y\). The total charge contributed by \(x\) \(\mathrm{Ti}^{2+}\) ions and \(y\) \(\mathrm{Ti}^{3+}\) ions should balance the charge of 3 O\(^{2-}\) ions. Accounting for the charges: \(2x + 3y = 6\) Now, based on the mole ratio obtained in Step 3, we have 2 moles of Ti in total. So: \(x + y = 2\)

Solve the linear equations and find the values of \(x\) and \(y\)

To find the values of \(x\) and \(y\), we can use the method of substitution or elimination. Here, we will use the elimination method. Multiply the second equation by 2 and subtract the first equation: \(2(x + y) - (2x + 3y) = -y = 2 \times 2 - 6 \Rightarrow y = 1\) Now, substitute the value of \(y\) back into the second equation: \(x + 1 = 2 \Rightarrow x = 1\)

Determine the formula of the titanium oxide compound

Since we found \(x = 1\) and \(y = 1\), there is one \(\mathrm{Ti}^{2+}\) ion and one \(\mathrm{Ti}^{3+}\) ion for every 3 O\(^{2-}\) ions. The formula of the titanium oxide compound is: \(\mathrm{Ti}^{2+}\mathrm{Ti}^{3+}(\mathrm{O}^{2-})_3 \Rightarrow \mathrm{Ti_2O_3}\) The formula of the titanium oxide compound is \(\mathrm{Ti_2O_3}\), with one \(\mathrm{Ti}^{2+}\) and one \(\mathrm{Ti}^{3+}\) ion.