Chapter 10.3, Problem 39SP

a.

Interpretation Introduction

Interpretation: The empirical formula of a compound that contains 94.1% O and 5.9% H should be calculated.

Concept Introduction: The percent composition represents the relative amount of the elements in a compound. It is represented as the percent by mass of each element in the compound.

Answer to Problem 39SP

The empirical formula of a compound that contains 94.1% O and 5.9% H is $\text{OH}$ .

Explanation of Solution

The percent by mass of each element in the compound represents the percent composition as:

  $\text{percent by mass of an element = }\frac{\text{mass of an element}}{\text{mass of the compound}}\times 100\%$

So, 100.0 g of the compound contains 94.1 g O and 5.9 g H.

By using the molar mass of an element, the ratio of masses can be changed to the ratio of moles by using the following conversion factor:

  $\text{mass of element (g) }\times \text{ molar mass of element (g/mol) = mol of element}$ …. (1)

The molar mass of $\text{O}$ and $\text{H}$ is 16.0 g/mol and 1.0 g/mol respectively.

The percent by mass of O and H can be converted to moles of O and H using the relation (1) as:

  $\begin{array}{l}\text{94}\text{.1 g O }\times \frac{1\text{ mol O}}{16.0\text{ g O}}\text{ = 5}\text{.88 mol O}\\ \text{5}\text{.9 g H }\times \frac{1\text{ mol H}}{1.0\text{ g H}}\text{ = 5}\text{.9 mol H}\end{array}$

The mole ratio of O to H is ${\text{O}}_{\text{5}\text{.88}}{\text{H}}_{\text{5}\text{.9}}$ .

Now, each molar quantity is divided by the smaller number of moles:

  $\begin{array}{l}\frac{\text{5}\text{.88 mol O}}{5.88}\text{ = 1 mol O}\\ \frac{\text{5}\text{.9 mol H}}{5.88}\text{ = 1}\text{.0 mol H}\end{array}$

Hence, the empirical formula of a compound that contains 94.1% O and 5.9% H is $\text{OH}$ .

b.

Interpretation Introduction

Interpretation: The empirical formula of a compound that contains 67.6% Hg, 10.8% S, and 21.6% O should be calculated.

Concept Introduction: The percent composition represents the relative amount of the elements in a compound. It is represented as the percent by mass of each element in the compound.

Answer to Problem 39SP

The empirical formula of a compound that contains 67.6% Hg, 10.8% S, and 21.6% O is ${\text{HgSO}}_4$ .

Explanation of Solution

The percent by mass of each element in the compound represents the percent composition as:

  $\text{percent by mass of an element = }\frac{\text{mass of an element}}{\text{mass of the compound}}\times 100\%$

So, 100.0 g of the compound contains 67.6 g Hg, 10.8 g S, and 21.6 g O.

By using the molar mass of an element, the ratio of masses can be changed to the ratio of moles by using the following conversion factor:

  $\text{mass of element (g) }\times \text{ molar mass of element (g/mol) = mol of element}$ …. (1)

The molar mass of $\text{Hg, S}$ and $\text{O}$ is 200.5 g/mol, 32.0, and 16.0 g/mol respectively.

The percent by mass of $\text{Hg, S}$ and $\text{O}$ can be converted to moles of $\text{Hg, S}$ and $\text{O}$ using relation (1) as:

  $\begin{array}{l}\text{67}\text{.6 g Hg }\times \frac{1\text{ mol Hg}}{\text{200}\text{.5 g Hg}}\text{ = 0}\text{.34 mol Hg}\\ \text{10}\text{.8 g S }\times \frac{1\text{ mol S}}{32.0\text{ g S}}\text{ = 0}\text{.34 mol S}\\ \text{21}\text{.6 g O }\times \frac{1\text{ mol O}}{16.0\text{ g O}}\text{ = 1}\text{.4 mol O}\end{array}$

The mole ratio of the given compound is ${\text{Hg}}_{\text{0}\text{.34}}{\text{S}}_{\text{0}\text{.34}}{\text{O}}_{\text{1}\text{.4}}$ .

Now, each molar quantity is divided by the smaller number of moles:

  $\begin{array}{l}\frac{\text{0}\text{.34 mol Hg}}{0.34}\text{ = 1 mol Hg}\\ \frac{\text{0}\text{.34 mol S}}{0.34}\text{ = 1}\text{.0 mol S}\\ \frac{\text{1}\text{.4 mol O}}{0.34}\simeq \text{ 4}\text{.0 mol O}\end{array}$

Hence, the empirical formula of a compound that contains 67.6% Hg, 10.8% S, and 21.6% O is ${\text{HgSO}}_4$ .