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

Step-by-step solution

Determine the oxidation states of lead in the compound

In the compound \(\mathrm{Pb}_{3} \mathrm{O}_{4}\), there are three lead (Pb) atoms and four oxygen (O) atoms. Oxygen usually has an oxidation state of -2. So, the total negative charge contributed by oxygen atoms is: Total negative charge = (number of oxygen atoms) × (oxidation state of oxygen) Total negative charge = 4 × (-2) Total negative charge = -8 For the compound to be neutral, the total positive charge contributed by the lead atoms must balance this negative charge. Therefore, the total positive charge from the lead atoms must be +8.

Set up a system of linear equations to represent the lead(II) and lead(IV) quantities

Let x be the number of moles of lead(II) (with an oxidation state of +2) and y be the number of moles of lead(IV) (with an oxidation state of +4) in the compound. Now, we can use the mole ratios to set up two linear equations representing the total amount of lead atoms and the total charge from lead atoms: 1. x + y = 3 (Total lead atoms) 2. 2x + 4y = 8 (Total charge from lead atoms)

Solve the system of linear equations to determine the quantities of lead(II) and lead(IV)

We can use the substitution method to solve these linear equations. First, rewrite Equation 1 to make y the subject: y = 3 - x Now substitute this expression for y into Equation 2: 2x + 4(3 - x) = 8 Expand and simplify: 2x + 12 - 4x = 8 -2x = -4 Divide both sides by -2: x = 2 Now that we have found the value of x, substitute it back into the expression for y: y = 3 - 2 y = 1

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

Now that we have the values for x (moles of lead(II)) and y (moles of lead(IV)), we can determine the mole ratio of lead(II) to lead(IV) in the compound. Mole ratio (lead(II) to lead(IV)) = x : y Mole ratio (lead(II) to lead(IV)) = 2 : 1 So, the mole ratio of lead(II) to lead(IV) in \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) is 2:1.

Key concepts

Oxidation States

Understanding the oxidation states of elements within compounds is crucial for deciphering their chemical behaviors. The oxidation state, often referred to as the oxidation number, is an indicator of the degree of oxidation (loss of electrons) that an atom has undergone in a compound.

An element can have different oxidation states, reflecting its ability to donate or accept electrons during chemical reactions. For example, in the exercise problem involving Pb₃O₄, lead is present in two different oxidation states, +2 and +4. Determining these states is fundamental for analyzing the compound's composition and predicting its reactivity.

Chemical Compound Analysis

The chemical compound analysis is a vital tool in understanding the composition and properties of compounds like Pb₃O₄. By analyzing the proportions and types of atoms, chemists can unravel the structural makeup of a compound. This often involves calculations to determine the mole ratios of particular atoms or ions.

Through analysis, one can deduce the stoichiometry and also the oxidation states of the various components, as in the provided exercise where calculating the oxidation states and mole ratios leads to a deeper understanding of the chemical constitution of red lead.

Stoichiometry

The concept of stoichiometry is used to describe the quantitative relationships between reactants and products in chemical reactions. It's a cornerstone of chemical calculations and critical for understanding the mole ratios in compounds.

Using stoichiometry, students can calculate how much of each substance is required or produced in a reaction. The exercise with red lead is an example of stoichiometry in action; using mole ratios to find the relationship between lead(II) and lead(IV) in the compound showcases the practical application of this concept.

Balance of Charge in Compounds

The concept of the balance of charge in compounds is essential in chemistry. For a compound to be stable, it must be electrically neutral, which means the sum of the oxidation states (or ionic charges) of all atoms must equal zero.

In the exercise, the balance of charge is attained by the three lead atoms combined at two different oxidation states, giving a total positive charge which counteracts the total negative charge from the oxygen atoms. This concept helps explain why atoms combine in certain proportions and not others, leading to stable compounds with fixed compositions.