Chapter 2, Problem 85A

(a)

Interpretation Introduction

Interpretation: Whether all the four given samples are the same compound or not needs to be explained.

Concept Introduction: A chemical substance made up of two or more two elements combined in a fixed ratio by mass is known as a compound. If the same atoms of elements are combined in different ratios, the compound will be different.

Explanation of Solution

The mass of iron and oxygen is given in four samples of a rust-colored substance. The approximate data can be obtained from the graph as follow:

Sample	Mass of iron (g)	Mass of oxygen (g)
1	5 g	2.1 g
2	8 g	3.8 g
3	15 g	6.2 g
4	18 g	7.9 g

In a compound, elements are present in a fixed ratio by mass. Thus, for the above data to belong to the same compound, the ratio of the mass of iron and oxygen must be the same.

The molar mass of iron is 55.85 g/mol and the molar mass of oxygen is 16 g/mol. The number of moles of iron and oxygen in sample 1 can be calculated as follows:

Sample 1:

  $\begin{array}{c}{n}_{\text{Fe}}=\frac{m_{\text{Fe}}}{M_{\text{Fe}}}\\ =\frac{\text{5 g}}{55.85\text{ g/mol}}\\ =0.08952\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{O}}=\frac{m_{\text{O}}}{M_{\text{O}}}\\ =\frac{\text{2}\text{.1 g}}{\text{16 g/mol}}\\ =0.1312\text{ mol}\end{array}$

The ratio will be:

  $\begin{array}{c}\frac{n_{\text{Fe}}}{n_{\text{O}}}=\frac{0.08952}{0.1312}\\ =\frac{1}{1.5}\\ =\frac{2}{3}\end{array}$

Thus, the formula of compound will be ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ .

Sample 2:

  $\begin{array}{c}{n}_{\text{Fe}}=\frac{m_{\text{Fe}}}{M_{\text{Fe}}}\\ =\frac{\text{8 g}}{55.85\text{ g/mol}}\\ =0.143\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{O}}=\frac{m_{\text{O}}}{M_{\text{O}}}\\ =\frac{\text{3}\text{.8 g}}{\text{16 g/mol}}\\ =0.24\text{ mol}\end{array}$

The ratio will be:

  $\begin{array}{c}\frac{n_{\text{Fe}}}{n_{\text{O}}}=\frac{0.143}{0.24}\\ =\frac{1}{1.66}\approx \frac{2}{3}\end{array}$

Thus, the formula of the compound will be ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ .

Sample 3:

  $\begin{array}{c}{n}_{\text{Fe}}=\frac{m_{\text{Fe}}}{M_{\text{Fe}}}\\ =\frac{\text{15 g}}{55.85\text{ g/mol}}\\ =0.268\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{O}}=\frac{m_{\text{O}}}{M_{\text{O}}}\\ =\frac{\text{6}\text{.2 g}}{\text{16 g/mol}}\\ =0.388\text{ mol}\end{array}$

The ratio will be:

  $\begin{array}{c}\frac{n_{\text{Fe}}}{n_{\text{O}}}=\frac{0.268}{0.388}\\ =\frac{1}{1.45}\approx \frac{2}{3}\end{array}$

Thus, the formula of the compound will be ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ .

Sample 4:

  $\begin{array}{c}{n}_{\text{Fe}}=\frac{m_{\text{Fe}}}{M_{\text{Fe}}}\\ =\frac{\text{18 g}}{55.85\text{ g/mol}}\\ =0.32\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{O}}=\frac{m_{\text{O}}}{M_{\text{O}}}\\ =\frac{\text{7}\text{.9 g}}{\text{16 g/mol}}\\ =0.494\text{ mol}\end{array}$

The ratio will be:

  $\begin{array}{c}\frac{n_{\text{Fe}}}{n_{\text{O}}}=\frac{0.32}{0.494}\\ =\frac{1}{1.54}\approx \frac{2}{3}\end{array}$

Thus, the formula of the compound will be ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ .

Since the formula of compounds of all four samples is the same thus, all four samples are the same compound.

(b)

Interpretation Introduction

Interpretation: A sample of similar material contains 9.9 g of iron and 3.4 g of oxygen. Whether the sample is the same as the other four samples or not needs to be explained.

Concept Introduction: A chemical substance made up of two or more two elements combined in a fixed ratio by mass is known as a compound. If the same atoms of elements are combined in different ratios, the compound will be different.

Explanation of Solution

In the question, the mass of iron is given 9.9 g and that of oxygen is 3.4 g. Calculate a number of moles of Fe and O as follows:

  $\begin{array}{c}{n}_{\text{Fe}}=\frac{m_{\text{Fe}}}{M_{\text{Fe}}}\\ =\frac{\text{9}\text{.9 g}}{55.85\text{ g/mol}}\\ =0.177\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{O}}=\frac{m_{\text{O}}}{M_{\text{O}}}\\ =\frac{\text{3}\text{.4 g}}{\text{16 g/mol}}\\ =0.2125\text{ mol}\end{array}$

The ratio will be:

  $\begin{array}{c}\frac{n_{\text{Fe}}}{n_{\text{O}}}=\frac{0.177}{0.2125}\\ =\frac{1}{1.2}\ne \frac{2}{3}\end{array}$

Thus, the formula of the compound is not ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and it is not the same compound.