Chapter 11, Problem 76A

(a)

Interpretation Introduction

Interpretation: The empirical formula of each compound needs to be determined using the data on the graphs.

Concept Introduction: Empirical formula gives the simplest ratio of the number of atoms of different elements present in the compound.

Explanation of Solution

According to the data from the graph, the percent composition of the two compounds is as follows:

Compound 1: Iron-36.7%, sulphur-21.1% and, Oxygen-42.1%

Compound 2: Iron-41.1%, sulphur-23.6% and, Oxygen-35.3%

Considering 100 g of each compound, thus, the mass of iron, sulfur, and oxygen in both compounds will be:

Compound 1: Iron-36.7 g, sulphur-21.1 g and, Oxygen-42.1 g

Compound 2: Iron-41.1 g, sulphur-23.6 g and, Oxygen-35.3 g

Now, the number of moles can be calculated from mass and molar mass as follows:

  $n=\frac{m}{M}$

The molar mass of iron, sulfur, and oxygen is 55.85 g/mol, 32.065 g/mol, and 16 g/mol respectively.

The number of moles in each compound can be calculated as follows:

For compound 1:

  $\begin{array}{l}{n}_{\text{Fe}}=\frac{36.7\text{ g}}{55.85\text{ g/mol}}=0.6571\text{ mol}\\ n_{\text{S}}=\frac{21.1\text{ g}}{32.065\text{ g/mol}}=0.6580\text{ mol}\\ n_{\text{O}}=\frac{42.1\text{ g}}{16\text{ g/mol}}=2.63\text{ mol}\end{array}$

The simplest ratio of the number of moles can be determined as follows:

  $\begin{array}{c}\text{Fe}:\text{S}:\text{O}=0.6571:0.6580:2.63\\ =\frac{0.6571}{0.6571}:\frac{0.6580}{0.6571}:\frac{2.63}{0.6571}\\ =1:1:4\end{array}$

Thus, the empirical formula of compound 1 is ${\text{FeSO}}_{\text{4}}$ .

For compound 2:

  $\begin{array}{l}{n}_{\text{Fe}}=\frac{41.1\text{ g}}{55.85\text{ g/mol}}=0.736\text{ mol}\\ n_{\text{S}}=\frac{23.6\text{ g}}{32.065\text{ g/mol}}=0.736\text{ mol}\\ n_{\text{O}}=\frac{35.3\text{ g}}{16\text{ g/mol}}=2.20\text{ mol}\end{array}$

The simplest ratio of the number of moles can be determined as follows:

  $\begin{array}{c}\text{Fe}:\text{S}:\text{O}=0.736:0.736:2.20\\ =\frac{0.736}{0.736}:\frac{0.736}{0.736}:\frac{2.20}{0.736}\\ =1:1:3\end{array}$

Thus, the empirical formula of compound 2 is ${\text{FeSO}}_3$ .

Therefore, the empirical formula of compounds 1 and 2 is ${\text{FeSO}}_4$ and ${\text{FeSO}}_3$ respectively.

(b)

Interpretation Introduction

Interpretation: The name of each compound needs to be identified.

Concept Introduction: Molecular formula is used to represent the actual number of/moles of atoms of different elements present in the compound. If the molecular formula of a compound is known, its name can be identified.

Explanation of Solution

The empirical formula for compounds 1 and 2 is ${\text{FeSO}}_4$ and ${\text{FeSO}}_3$ respectively. Here, the empirical formula is equal to the molecular formula.

Here, in ${\text{FeSO}}_4$ , the oxidation state of Fe is II and ${\text{SO}}_4^{2-}$ is a sulfate ion; thus, name of the compound 1 is iron (II) sulfate or ferrous sulfate.

In ${\text{FeSO}}_3$ , the oxidation state of Fe is II and ${\text{SO}}_3^{2-}$ is a sulfite ion; thus, name of the compound 2 is iron (II) sulfite or ferrous sulfite.