Chapter 9, Problem 89A

(a)

Interpretation Introduction

Interpretation: The ratio of lead contained in the two compounds for every 1 g of iodine is to be determined.

Concept Introduction: The law of multiple proportions states that, when two elements combine to form more than one compound, then the masses of the second element which combine with a fixed mass of the first element will always bear a simple whole number ratio.

Answer to Problem 89A

The ratio of lead present in two compounds for every 1 g of iodine is 2:1.

Explanation of Solution

The amount of lead present in compound I for every 1 g of iodine is 0.816 g. and that of lead present in compound II is 0.408 g. so the ratio between these two is 1.7:1 $\approx 2:1$ .

(b)

Interpretation Introduction

Interpretation: The formulas and names of the given compounds are to be determined.

Concept Introduction: The law of multiple proportions states that, when two elements combine to form more than one compound, then the masses of the second element which combine with a fixed mass of the first element will always bear a simple whole number ratio.

Answer to Problem 89A

The compound I is lead (II) iodide, having the formula ${\text{PbI}}_{\text{2}}$ . Compound II is lead (IV) iodide with the formula ${\text{PbI}}_{\text{4}}$ .

Explanation of Solution

Lead usually exists as ${\text{Pb}}^{\text{2+}}\text{and}{\text{Pb}}^{\text{4+}}$ ions while iodine is a monovalent anion. Therefore compound I have the name lead (II) iodide with formula ${\text{PbI}}_{\text{2}}$ and compound (II) has the name lead (IV) iodide with formula ${\text{PbI}}_{\text{4}}$ .