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 $\mathrm{Pb}_{3} \mathrm{O}_{4}$ (red lead) is 2:1.

Step-by-step solution

Determine the oxidation number of Pb in the compound

First, we need to determine the oxidation number of each lead atom in the compound. It is given that red lead contains a mixture of lead(II) and lead(IV) oxidation states. Let x represent the number of moles of lead(II) with oxidation state +2 and y represent the number of moles of lead(IV) with oxidation state +4. There are a total of 3 moles of lead in Pb3O4, so we have: x + y = 3.

Determine the total charge in the compound

Next, we need to determine the total charge in the compound. The total charge of the compound is the sum of the charges of all the elements in the compound. Oxygen has an oxidation state of -2. Since there are four oxygen atoms in the compound, the total charge due to oxygen is -8. The total charge due to lead(II) is +2x and due to lead(IV) is +4y. The total charge in the compound will be 0 since it is a neutral compound. Combining all these charges, we get the equation: 2x + 4y - 8 = 0

Solve the equations for x and y

We now have two equations: 1) x + y = 3 2) 2x + 4y - 8 = 0 We solve these simultaneous equations to find the values of x and y. From equation 1, we can express x as x = 3 - y. Substituting this into equation 2, we get: 2(3 - y) + 4y - 8 = 0 Expanding and simplifying the equation, we obtain: 6 - 2y + 4y - 8 = 0 2y - 2 = 0 Now, divide by 2: y = 1 Now, substitute y back into equation 1 to find x: x + 1 = 3 x = 2

Calculate the mole ratio of lead(II) to lead(IV)

Now that we know the number of moles of lead(II) (x = 2) and lead(IV) (y = 1) in the compound, we can determine the mole ratio of lead(II) to lead(IV). The mole ratio is given by x:y, which is 2:1 in this case. Therefore, the mole ratio of lead(II) to lead(IV) in the compound Pb3O4 is 2:1.