Two completely different oxides of a transition metal contain and of oxygen by mass, respectively. B — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

For the first oxide ($M_3O_4$): Mass % of Metal ($M$) = $100 - 27.6 = 72.4\%$. The atomic ratio is $M : O = 3 : 4$. $\frac{72.4 / A_M}{27.6 / 16} = \frac{3}{4} \Rightarrow \frac{72.4 \times 16}{27.6 \times A_M} = \frac{3}{4} \Rightarrow A_M = \frac{72.4 \times 16 \times 4}{27.6 \times 3} \approx 56\text{ g/mol}$ (Iron). For the second oxide: Mass % of Metal ($M$) = $100 - 30.0 = 70.0\%$. Mass % of Oxygen ($O$) = $30.0\%$. Moles of M = $\frac{70.0}{56} = 1.25\text{ moles}$. Moles of O = $\frac{30.0}{16} = 1.875\text{ moles}$. Simplest ratio $M : O = \frac{1.25}{1.25} : \frac{1.875}{1.25} = 1 : 1.5 = 2 : 3$. The empirical formula of the second oxide is $M_2O_3$. The number of oxygen atoms is 3.