Chapter 10, Problem 89A

(a)

Interpretation Introduction

Interpretation: The molecular formula for a given compound needs to be calculated.

Concept Introduction: For any compound, the empirical formula shows the simplest ratio of atoms of different elements present in the compound. On the other hand, the molecular formula shows the actual number of atoms of elements present in the compound.

Explanation of Solution

The given compound has 94.1% O and 5.9% H. Also, the molar mass of the compound is 34 g/mol.

Considering 100 g of the compound, the mass of oxygen and hydrogen atoms will be 94.1 g and 5.9 g respectively.

Calculate the number of moles of O and H as follow:

  $n_{\text{O}}=\frac{m}{M}$

Here, m is mass and M is the molar mass of oxygen.

Or,

  $\begin{array}{c}{n}_{\text{O}}=\frac{94.1\text{ g}}{16\text{ g/mol}}\\ =5.88\text{ mol}\end{array}$

Similarly, the number of moles of hydrogen can be calculated as follows:

  $\begin{array}{c}{n}_{\text{H}}=\frac{5.9\text{ g}}{1\text{ g/mol}}\\ =5.9\text{ mol}\end{array}$

Now, calculate the ratio of the number of moles of oxygen and hydrogen in the compound as follow:

  $\begin{array}{c}{n}_{\text{O}}:n_{\text{H}}=5.88:5.99\\ =1:1\end{array}$

Thus, oxygen and hydrogen are present in a 1:1 ratio. The empirical formula of the compound will be:

  $\text{HO}$

The empirical mass can be calculated as follows:

  $\begin{array}{c}{\text{M}}_{\text{H}}+M_{\text{O}}=1\text{ g/mol+16 g/mol}\\ =\text{17 g/mol}\end{array}$

Since the molar mass of the compound is 34 g/mol thus, the simplest ratio can be calculated as follows:

  $\begin{array}{c}{\text{O}}_x{\text{H}}_x\\ x=\frac{34}{17}\\ =2\end{array}$

Thus, the molecular formula of the compound is ${\text{H}}_2{\text{O}}_2$ .

(b)

Interpretation Introduction

Interpretation: The molecular formula for a given compound needs to be calculated.

Concept Introduction: For any compound, the empirical formula shows the simplest ratio of atoms of different elements present in the compound. On the other hand, the molecular formula shows the actual number of atoms of elements present in the compound.

Explanation of Solution

The given compound has 50.7% C, 4.2% H, and 45.1% O. Also, the molar mass of the compound is 142 g/mol. Considering 100 g of the compound, the mass of C, H, and O will be 50.7 g, 4.2 g, and 45.1 g respectively.

The number of moles of C, H, and O can be calculated as follows:

  $\begin{array}{l}{n}_{\text{C}}=\frac{50.7\text{ g}}{12\text{ g/mol}}=4.225\text{ mol}\\ n_{\text{H}}=\frac{4.2\text{ g}}{1.0\text{ g/mol}}=4.2\text{ mol}\\ n_{\text{O}}=\frac{45.1\text{ g}}{16\text{ g/mol}}=2.82\text{ mol}\end{array}$

Determine the simples ratio of the number of moles of atoms:

  $\begin{array}{c}{n}_{\text{C}}:n_{\text{H}}:n_{\text{O}}=4.225:4.2:2.82\\ =\frac{3}{2}:\frac{3}{2}:1\\ =3:3:2\end{array}$

Therefore, the empirical formula of the compound is ${\text{C}}_3{\text{H}}_3{\text{O}}_2$ .

Now, empirical mass is calculated as follows:

  $\begin{array}{c}{\text{C}}_3{\text{H}}_3{\text{O}}_2=3\left(12\text{ g/mol}\right)+3\left(1\text{ g/mol}\right)+2\left(16\text{ g/mol}\right)\\ =36+1+32\\ =69\text{ g/mol}\end{array}$

Since the molar mass of the compound is 142 g/mol thus, the molecular formula is calculated as follows:

  $\begin{array}{c}x=\frac{142}{69}\\ =2\end{array}$

Thus, the molecular formula of the compound will be ${\text{C}}_6{\text{H}}_6{\text{O}}_4$ .

(c)

Interpretation Introduction

Interpretation: The molecular formula for a given compound needs to be calculated.

Concept Introduction: For any compound, the empirical formula shows the simplest ratio of atoms of different elements present in the compound. On the other hand, the molecular formula shows the actual number of atoms of elements present in the compound.

Explanation of Solution

The given compound has 56.6% K, 8.7% C, and 34.7% O. Also, the molar mass of the compound is 138.2 g/mol.

Considering 100 g of the compound, the mass of K, C, and O will be 56.6 g, 8.7 g, and 34.7 g respectively.

The number of moles of K, C, and O can be calculated as follows:

  $\begin{array}{c}{n}_{\text{K}}=\frac{56.6\text{ g}}{39\text{ g/mol}}\\ =1.45\text{ mol}\end{array}$

Also,

  $\begin{array}{c}{n}_{\text{C}}=\frac{8.7\text{ g}}{12\text{ g/mol}}\\ =0.725\text{ mol}\end{array}$

And,

  $\begin{array}{c}{n}_{\text{O}}=\frac{34.7\text{ g}}{\text{16 g/mol}}\\ =2.17\text{ mol}\end{array}$

The ratio of the number of moles of K, C, and O will be:

  $\begin{array}{c}{n}_{\text{K}}:n_{\text{C}}:n_{\text{O}}=1.45:0.725:2.17\\ =2:1:3\end{array}$

Therefore, the empirical formula of the compound is ${\text{K}}_2{\text{CO}}_3$ .

The empirical mass can be calculated as follows:

  $\begin{array}{c}{\text{K}}_2{\text{CO}}_3=2\left(39\right)+12+3\left(16\right)\\ =78+12+48\\ =138\text{ g/mol}\end{array}$

The given molar mass is 138.2 g/mol which is equal to the empirical mass; thus, the molecular formula will be ${\text{K}}_2{\text{CO}}_3$ .