Chapter 10, Problem 88A

(a)

Interpretation Introduction

Interpretation: The empirical formula of compounds of given percent compositions needs to be determined.

Concept Introduction: An empirical formula of a compound has the simplest whole number ratio of atoms in the compound.

Explanation of Solution

The given percent composition is 42.9% C and 57.1% O. Considering 100 g of a compound, the mass of C and O in 100 g of the compound will be 42.9% and 57.1% respectively.

The number of moles of C and O can be calculated as follow:

  $n=\frac{m}{M}$

Since the molar mass of C and O is 12 g/mol and 16 g/mol respectively, thus,

  $\begin{array}{c}{n}_{\text{C}}=\frac{42.9\text{ g}}{12\text{ g/mol}}\\ =3.575\text{ mol}\end{array}$

Similarly,

  $\begin{array}{c}{n}_{\text{O}}=\frac{57.1\text{ g}}{16\text{ g/mol}}\\ =3.57\text{ mol}\end{array}$

Calculate the ratio of the number of moles of C and O.

  $\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{3.575}{3.57}=\frac{1}{1}$

Now, the empirical formula of the compound will be:

  ${\text{C}}_{\text{1}}{\text{O}}_1$

Or,

  $\text{CO}$ that is carbon monoxide.

(b)

Interpretation Introduction

Interpretation: The empirical formula of compounds of given percent compositions needs to be determined.

Concept Introduction: An empirical formula of a compound has the simplest whole number ratio of atoms in the compound.

Explanation of Solution

The given composition is 32.00% C, 42.66% O, 18.67% N, and 6.67% H.

Consider 100 g of the compound. Thus, the mass of C, O, N, and H will be 32.00 g, 42.66 g, 18.67 g, and 6.67 g respectively,

Now, the number of moles of C, O, N, and H can be calculated as follows:

  $n=\frac{m}{M}$

Here, molar mass of C, O, N and H is 12 g/mol, 16 g/mol, 14 g/mol and 1 g/mol.

Thus,

  $\begin{array}{l}{n}_{\text{C}}=\frac{32\text{ g}}{12\text{ g/mol}}\\ =2.67\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{O}}=\frac{\text{42}\text{.66 g}}{18\text{ g/mol}}\\ =2.67\text{ mol}\end{array}$

Similarly,

  $\begin{array}{c}{n}_{\text{N}}=\frac{\text{18}\text{.67 g}}{\text{14 g/mol}}\\ =1.34\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{H}}=\frac{\text{6}\text{.67 g}}{1\text{ g/mol}}\\ \text{=6}\text{.67 mol}\end{array}$

Now, the simplest ratio of atoms can be calculated as follows:

  $\begin{array}{c}{n}_{\text{C}}:n_{\text{O}}:n_{\text{N}}:n_H=2.67:2.67:1.34:6.637\\ =1:1:2:2.58\\ =1:1:2:\frac{5}{2}\\ =2:2:4:5\end{array}$

Thus, the empirical formula of the compound is ${\text{C}}_{\text{2}}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{5}}$

(c)

Interpretation Introduction

Interpretation: The empirical formula of compounds of given percent compositions needs to be determined.

Concept Introduction: An empirical formula of a compound has the simplest whole number ratio of atoms in the compound.

Explanation of Solution

The given composition is 71.72% Cl, 16.16% O, and 12.12 % C.

Considering 100 g of sample. Thus, the mass of Cl, O, and C will be 71.72 g, 16.16 g, and 12.12 g.

Number of moles of Cl, O, and C can be calculated as follows:

  $\begin{array}{c}{n}_{Cl}=\frac{m}{M}\\ =\frac{71.72\text{ g}}{35.5\text{ g/mol}}\\ =2.0\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{O}}=\frac{16.16\text{ g}}{16\text{ g/mol}}\\ =1.01\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{C}}=\frac{12.12\text{ g}}{12\text{ g/mol}}\\ =1.01\text{ mol}\end{array}$

Now, the simplest ratio of the compound will be:

  $\begin{array}{c}\mathrm{Cl}:\text{O}:\text{C}=2:1.01:1.01\\ =2:1:1\end{array}$

Thus, the empirical formula of a compound is ${\text{COCl}}_2$ .