Chapter 14, Problem 117A

(a)

Interpretation Introduction

Interpretation: To calculate the molecular formula of each of the given compounds.

Concept Introduction: Molecular formula represents the number of atoms present in the given compound. Molecular formula is linked to gram molecular masses which are simple whole number multiples of the empirical formula. The molecular formula is given as

  $\frac{\text{molarmassofthecompound}}{\text{empericalformulamassofthecompound}}$

When the above formula is used it give rise to a whole number. The empirical formula will be multiplied with this whole number.

Answer to Problem 117A

  ${\text{C}}_4{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ is the molecular formula.

Explanation of Solution

The given molar mass of the compound is 88 grams and the empirical formula is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$

First calculate the empirical formula mass

The molar mass of one carbon atom is 12 g/mol and there are two carbons atoms. Therefore, the molar mass of carbon atom is

  $2\times \text{12}\frac{\text{g}}{\text{mol}}=24\frac{\text{g}}{\text{mol}}$

The molar mass of one hydrogen atom is 1 g/mol and there are four hydrogen atoms. Therefore, the molar mass of the four hydrogen atoms is

  $4\times \text{1}\frac{\text{g}}{\text{mol}}=4\frac{\text{g}}{\text{mol}}$

The molar mass of one oxygen atom is 16 g/mol and there is only one oxygen atom. Therefore, the molar mass of one oxygen atom is

  $1\times \text{16}\frac{\text{g}}{\text{mol}}=16\frac{\text{g}}{\text{mol}}$

The total empirical formula mass will be 44 g/mol

Now the molecular formula can be determined by

  $\frac{\text{88}\frac{\text{g}}{\text{mol}}}{\text{44}\frac{\text{g}}{\text{mol}}}=2$

Now multiply the empirical formula with the whole number therefore the molecular formula will be

  ${\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O)}\times {\text{2=C}}_4{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation: To calculate the molecular formula of each of the given compounds.

Concept Introduction: Molecular formula represents the number of atoms present in the given compound. Molecular formula is linked to gram molecular masses which are simple whole number multiples of the empirical formula. The molecular formula is given as

  $\frac{\text{molarmassofthecompound}}{\text{empericalformulamassofthecompound}}$

When the above formula is used it give rise to a whole number. The empirical formula will be multiplied with this whole number.

Answer to Problem 117A

  ${\text{C}}_8{\text{H}}_{\text{8}}$

Explanation of Solution

The given molar mass of the compound is 104 grams and the empirical formula is $\text{CH}$

First calculate the empirical formula mass

The molar mass of one carbon atom is 12 g/mol and there is only one carbons atoms. Therefore, the molar mass of carbon atom is

  $1\times \text{12}\frac{\text{g}}{\text{mol}}=12\frac{\text{g}}{\text{mol}}$

The molar mass of one hydrogen atom is 1 g/mol and there is only one hydrogen atoms. Therefore, the molar mass of the hydrogen atoms is

  $1\times \text{1}\frac{\text{g}}{\text{mol}}=1\frac{\text{g}}{\text{mol}}$

The total empirical formula mass will be 13 g/mol

Now the molecular formula can be determined by

  $\frac{104\frac{\text{g}}{\text{mol}}}{13\frac{\text{g}}{\text{mol}}}=8$

Now multiply the empirical formula with the whole number therefore the molecular formula will be

  $\text{(CH)}\times 8{\text{=C}}_8{\text{H}}_{\text{8}}$

(c)

Interpretation Introduction

Interpretation: To calculate the molecular formula of each of the given compounds.

Concept Introduction: Molecular formula represents the number of atoms present in the given compound. Molecular formula is linked to gram molecular masses which are simple whole number multiples of the empirical formula. The molecular formula is given as

  $\frac{\text{molarmassofthecompound}}{\text{empericalformulamassofthecompound}}$

When the above formula is used it give rise to a whole number. The empirical formula will be multiplied with this whole number.

Answer to Problem 117A

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}$

Explanation of Solution

The given molar mass of the compound is 90 grams

First calculate the empirical formula from the percent composition

Step 1: divide the precent composition mass with the molar mass of each atom.

The molar mass of a single carbon atom is 12 g/mol. Therefore,

  $\text{Carbon=}\frac{\text{26}\text{.7}}{\text{12}}\text{=2}\text{.225}$

The molar mass of hydrogen atom is 1 g/mol. Therefore,

  $\text{Hydrogen=}\frac{2.2}{1}\text{=2}\text{.2}$

The Molar mass of one oxygen atom is 16 g/mol. Therefore,

  $\text{Hydrogen=}\frac{71.1}{16}\text{=4}\text{.44}$

Step 2: Divide the obtained numbers with the smallest whole number which is 2.22 then it will be

  $\text{Carbon=}\frac{\text{2}\text{.22}}{2.22}\text{=1}$

  $\text{Hydrogen=}\frac{\text{2}\text{.22}}{2.22}\text{=1}$

  $\text{Oxygen=}\frac{\text{4}\text{.44}}{2.22}\text{=2}$

So, the atoms are in the ratio 1:1:2 therefore, the empirical formula will be

  ${\text{CHO}}_2$

Now calculate the empirical formula mass

The molar mass of one carbon atom is 12 g/mol and there is only one carbons atoms. Therefore, the molar mass of carbon atom is

  $1\times \text{12}\frac{\text{g}}{\text{mol}}=12\frac{\text{g}}{\text{mol}}$

The molar mass of one hydrogen atom is 1 g/mol and there is only one hydrogen atoms. Therefore, the molar mass of the hydrogen atoms is

  $1\times \text{1}\frac{\text{g}}{\text{mol}}=1\frac{\text{g}}{\text{mol}}$

The molar mass of one oxygen atom is 16 g/mol and there is only one oxygen atom. Therefore, the molar mass of two oxygen atom is

  $2\times \text{16}\frac{\text{g}}{\text{mol}}=32\frac{\text{g}}{\text{mol}}$

The total empirical formula mass will be 45 g/mol

Now the molecular formula can be determined by

  $\frac{90\frac{\text{g}}{\text{mol}}}{45\frac{\text{g}}{\text{mol}}}=2$

Now multiply the empirical formula with the whole number therefore the molecular formula will be

  ${\text{(CHO}}_{\text{2}}{\text{)×2=C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}$