Question

The compound \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) (red lead) contains a mixture of lead(II) and lead(IV) oxidation states. What is the mole ratio of lead(II) to lead(IV) in \(\mathrm{Pb}_{3} \mathrm{O}_{4} ?\)

Short answer

The mole ratio of lead(II) to lead(IV) in the compound \(Pb_3 O_4\) (red lead) is 2:1.

Step-by-step solution

Assign oxidation states of oxygen and lead in the compound

In order to determine the oxidation states of the lead atoms in Pb₃O₄, we first need to know the oxidation state of oxygen. Oxygen usually has an oxidation state of -2, and we can assume that to be the case here. Each oxygen atom contributes -2, making the total oxidation charge from oxygen to be (-2) * 4 = -8. Now let's assume there are x moles of Pb(II) and y moles of Pb(IV) in the compound. Then the sum of the oxidation states of the lead atoms can be given as: 2x + 4y.

Write an equation for the total oxidation charge balance

Since the compound is neutral (electrically balanced), the sum of all the oxidation states must be equal to zero. Therefore, we can write the equation: 2x + 4y = 8 (total charge from the oxygen atoms)

Simplify the equation

To simplify the equation, divide both sides by 2: x + 2y = 4

Use the Pb atoms count in Pb₃O₄ to find the mole ratio

Since there are three lead atoms, x and y must add up to 3: x + y = 3

Solve the equations for the mole ratio of Pb(II) to Pb(IV)

Now we have a system of two linear equations: 1. x + 2y = 4 2. x + y = 3 Solving equation (2) for x: x = 3 - y Now, substitute this expression for x in equation (1): (3 - y) + 2y = 4 Simplify the equation, and solve for y: 3 + y = 4 y = 1 Substitute the value of y back into the expression for x: x = 3 - 1 x = 2 Our result shows that there are 2 moles of Pb(II) and 1 mole of Pb(IV) in the Pb₃O₄ compound. The mole ratio of lead(II) to lead(IV) is therefore 2:1.