Chapter 4, Problem 4.42P

4-42 Calculate the formula weight of:

(a) KCl

(b) Na₃PO₄

(c) Fe(OH)₂

(d) NaAl(SO₃)₂

(e) AL₂(SO₄)₃

(f) (NH₄)₂CO₃

Interpretation Introduction

(a)

Interpretation:

To calculate the formula weight for $K\text{C}\text{l}$

Concept Introduction:

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For, $\text{N}\text{a}\text{C}\text{l}$

The formula mass is calculated as sum of 1 Sodium atom and 1 mole of Chlorine atoms.

Answer to Problem 4.42P

The formula weight of $K\text{C}\text{l}$

= 74.5 amu.

Explanation of Solution

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

$K\text{C}\text{l}$

The formula mass is calculated as sum of 1 Potassium atom and 1 Chlorine atom.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massofK\text{C}\text{l}=Mass\text{ 1 }of\text{ Potass}\text{i}um\text{i}\text{O}n+1\text{C}\text{H}\text{l}\text{O}r\text{i}d\text{e}\text{i}\text{O}n\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{ K}\text{C}\text{l}=39+35.5\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{ K}\text{C}\text{l}=\text{ 74}.5\text{ amu}\hfill \end{array}$.

Interpretation Introduction

(b)

Interpretation:

To calculate the formula weights for $\text{N}{\text{a}}_{3}P{\text{O}}_4$

Concept Introduction:

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For e.g. - $\text{N}\text{a}\text{C}\text{l}$

The formula mass is calculated as sum of 1 Sodium atom and 1 mole of Chlorine atoms.

Answer to Problem 4.42P

The formula weight of $\text{N}{\text{a}}_{3}P{\text{O}}_4$

= 163 amu.

Explanation of Solution

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

$\text{N}{\text{a}}_{3}P{\text{O}}_4$

The formula mass is calculated as sum of 3 Sodium atom and 1 Phosphorous atom and 4 oxygen atom.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massof\text{N}{\text{a}}_{3}P{\text{O}}_4=3S\text{O}d\text{i}um\text{a}t\text{O}m +1 P\text{H}\text{O}sp\text{H}\text{O}r\text{O}us\text{a}t\text{O}m\text{ + }4oxyg\text{e}n\text{a}t\text{O}m.\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{N}{\text{a}}_{3}P{\text{O}}_4=\left(3\times 23\right)+\left(1\times 30\right)+\left(4\times 16\right)\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{N}{\text{a}}_{3}P{\text{O}}_4=\text{ 163 amu}\hfill \end{array}$.

Interpretation Introduction

(c)

Interpretation:

To calculate the formula weights for $\text{F}\text{e}{\left(\text{O}\text{H}\right)}_2$

Concept Introduction:

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For e.g. - $\text{F}\text{e}{\left(\text{O}\text{H}\right)}_2$

The formula mass is calculated as sum of 1 Iron atom, 2 oxygen atoms and 2 Hydrogen atoms.

Answer to Problem 4.42P

The formula weight of $\text{F}\text{e}{\left(\text{O}\text{H}\right)}_2$

= 90amu.

Explanation of Solution

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

$\text{F}\text{e}{\left(\text{O}\text{H}\right)}_2$

The formula mass is calculated as sum of 1 Iron atom, 2 oxygen atoms and 2 Hydrogen atoms.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massof\text{F}\text{e}{\left(\text{O}\text{H}\right)}_2=\text{ 1 }\text{i}r\text{O}n\text{a}t\text{O}m +\text{ 2 } oxyg\text{e}n\text{a}t\text{O}m\text{ + 2 }\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}m.\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{F}\text{e}{\left(\text{O}\text{H}\right)}_2=\left(1\times 56\right)+\left(2\times 16\right)+\left(2\times 1\right)\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{F}\text{e}{\left(\text{O}\text{H}\right)}_2=\text{ 90 amu}\hfill \end{array}$.

Interpretation Introduction

(d)

Interpretation:

To calculate the formula weights for $\text{N}\text{a}\text{a}\text{l}{\left(S{\text{O}}_3\right)}_2$

Concept Introduction:

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For e.g. - $\text{N}\text{a}\text{C}\text{l}$

The formula mass is calculated as sum of 1 Sodium atom and 1 mole of Chlorine atoms.

Answer to Problem 4.42P

The formula weight of $\text{N}\text{a}\text{a}\text{l}{\left(S{\text{O}}_3\right)}_2$

= 210 amu.

Explanation of Solution

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

$\text{N}\text{a}\text{a}\text{l}{\left(S{\text{O}}_3\right)}_2$

The formula mass is calculated as sum of 1 Sodium atom, 1 Aluminium atom ,6 oxygen atoms and 2 Sulphur atoms.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massof\text{N}\text{a}\text{a}\text{l}{\left(S{\text{O}}_3\right)}_2=\text{ 1 }S\text{O}d\text{i}um\text{a}t\text{O}m +\text{1 }\text{a}\text{l}u\min \text{i}um\text{a}t\text{O}m +\text{ 6 } oxyg\text{e}n\text{a}t\text{O}m\text{ + 2 }Su\text{l}\text{F}ur\text{a}t\text{O}m.\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{N}\text{a}\text{a}\text{l}{\left(S{\text{O}}_3\right)}_2=\left(1\times 23\right)+\left(1\times 27\right)+\left(6\times 16\right)+\left(2\times 32\right)\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{N}\text{a}\text{a}\text{l}{\left(S{\text{O}}_3\right)}_2=\text{ 210 amu}\hfill \end{array}$.

Interpretation Introduction

(e)

Interpretation:

To calculate the formula weights for $\text{a}{\text{l}}_2{\left(S{\text{O}}_4\right)}_3$

Concept Introduction:

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For e.g. - $\text{N}\text{a}\text{C}\text{l}$

The formula mass is calculated as sum of 1 Sodium atom and 1 mole of Chlorine atoms.

Answer to Problem 4.42P

The formula weight of $\text{a}{\text{l}}_2{\left(S{\text{O}}_4\right)}_3$

=342amu.

Explanation of Solution

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

$\text{a}{\text{l}}_2{\left(S{\text{O}}_4\right)}_3$

The formula mass is calculated as sum of 2 Aluminium atom, 12 oxygen atoms and 3Sulfur atoms.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massof\text{a}{\text{l}}_2{\left(S{\text{O}}_4\right)}_3=\text{ 2 }\text{a}\text{l}u\min \text{i}um\text{a}t\text{O}m +\text{ 12 } oxyg\text{e}n\text{a}t\text{O}m\text{ + 3 }Su\text{l}\text{F}ur\text{a}t\text{O}m.\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{a}{\text{l}}_2{\left(S{\text{O}}_4\right)}_3=\left(2\times 27\right)+\left(12\times 16\right)+\left(3\times 32\right)\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof\text{a}{\text{l}}_2{\left(S{\text{O}}_4\right)}_3=\text{ 342 amu}\hfill \end{array}$.

Interpretation Introduction

(f)

Interpretation:

To calculate the formula weights for ${\left(\text{N}{\text{H}}_4\right)}_2\text{C}{\text{O}}_3$

Concept Introduction:

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For e.g. - $\text{N}\text{a}\text{C}\text{l}$

The formula mass is calculated as sum of 1 Sodium atom and 1 mole of Chlorine atoms.

Answer to Problem 4.42P

The formula weight of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$

= 96amu.

Explanation of Solution

Formula mass is the sum of masses of the atoms present in a molecule. These compounds are those that do not generally occur as a single molecule. For e.g. $\text{N}\text{a}\text{C}\text{l}$

${\left(\text{N}{\text{H}}_4\right)}_2\text{C}{\text{O}}_3$

The formula mass is calculated as sum of 2Nitrogen atom, 8Hydrogen atom, 3 oxygen atoms and 1 Carbon atom.

$\begin{array}{l}\text{F}\text{O}rmu\text{l}\text{a}massof{\left(\text{N}{\text{H}}_4\right)}_2\text{C}{\text{O}}_3=\text{ 2 }\text{N}\text{i}tr\text{O}g\text{e}n\text{a}t\text{O}m +8\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}m +\text{ 3 } oxyg\text{e}n\text{a}t\text{O}m\text{ + 1 }\text{C}\text{a}rb\text{O}n\text{a}t\text{O}m.\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof{\left(\text{N}{\text{H}}_4\right)}_2\text{C}{\text{O}}_3=\left(2\times 14\right)+\left(8\times 1\right)+\left(3\times 16\right)+\left(1\times 12\right)\hfill \\ \text{F}\text{O}rmu\text{l}\text{a}massof{\left(\text{N}{\text{H}}_4\right)}_2\text{C}{\text{O}}_3=\text{ 96 amu}\hfill \end{array}$.