Chapter 2, Problem 2.76P

a)

Interpretation Introduction

Interpretation:Formula and molecular mass of ammonium sulfate should be determined.

Concept introduction:In the molecular formula, the exact number and type of constituent atoms are shown. Type of constituent atoms is represented by their atomic symbols and number of each constituent atom is represented in the right subscript of its atomic symbol.In a hypothetical molecule ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ , x number of $\text{A}$ atom and y number of $\text{B}$ atom combines to form the compound.

Formula to calculate molecular mass of ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ is as follows:

  $\text{Molecular}\text{mass}\text{of}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}=\left(\begin{array}{l}\left(\text{x}\right)\left(\text{Atomic}\text{mass}\text{of}\text{A}\right)+\\ \left(\text{y}\right)\left(\text{Atomic}\text{mass}\text{of}\text{B}\right)\end{array}\right)$

Explanation of Solution

In ammonium sulfate, one sulfate ion $\left({\text{SO}}_{\text{4}}^{\text{2-}}\right)$ .and two ammonium ion $\left({\text{NH}}_{\text{4}}^{\text{+}}\right)$ is attached by electrostatic attraction to maintain electrical neutrality. So the formula of ammonium sulfate is ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\left({\text{SO}}_{\text{4}}\right)$ .

Molecular mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\left({\text{SO}}_{\text{4}}\right)$ is calculated as follows:

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_2\left({\text{SO}}_{\text{4}}\right)=\left(\begin{array}{l}\left(2\right)\left(\text{Atomic}\text{mass}\text{of}\text{N}\right)+\\ \left(8\right)\left(\text{Atomic}\text{mass}\text{of}\text{H}\right)+\\ \left(1\right)\left(\text{Atomic}\text{mass}\text{of}\text{S}\right)+\\ \left(4\right)\left(\text{Atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right)\\ =\left[\begin{array}{l}\left(2\right)\left(14.01\text{ amu}\right)+\left(8\right)\left(1.008\text{ amu}\right)\\ +\left(1\right)\left(32.06\text{ amu}\right)+\left(4\right)\left(16\text{ amu}\right)\end{array}\right]\\ =28.02\text{ amu}+8.064\text{ amu}+32.06\text{ amu}+64\text{ amu}\\ =132.14\text{ amu}\end{array}$

b)

Interpretation Introduction

Interpretation: Formula and molecular mass of sodium dihydrogen phosphate should be determined.

Concept introduction:In the molecular formula, the exact number and type of constituent atoms are shown. Type of constituent atoms is represented by their atomic symbols and number of each constituent atom is represented in the right subscript of its atomic symbol.In a hypothetical molecule ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ , x number of $\text{A}$ atom and y number of $\text{B}$ atom combines to form the compound.

Formula to calculate molecular mass of ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ is as follows:

  $\text{Molecular}\text{mass}\text{of}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}=\left(\begin{array}{l}\left(\text{x}\right)\left(\text{Atomic}\text{mass}\text{of}\text{A}\right)+\\ \left(\text{y}\right)\left(\text{Atomic}\text{mass}\text{of}\text{B}\right)\end{array}\right)$

Explanation of Solution

In sodium dihydrogen phosphate, one dihydrogen phosphate ion $\left({\text{H}}_2{\text{PO}}_{\text{4}}^{-}\right)$ .and one sodium ion $\left({\text{Na}}^{+}\right)$ is attached by electrostatic attraction to maintain electrical neutrality. So the formula of sodiumdihydrogen phosphateis ${\text{NaH}}_{\text{2}}{\text{PO}}_{\text{4}}$ .

Molecular mass of ${\text{NaH}}_{\text{2}}{\text{PO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{4}}=\left(\begin{array}{l}\left(1\right)\left(\text{Atomic}\text{mass}\text{of}\text{Na}\right)\\ +\left(2\right)\left(\text{Atomic}\text{mass}\text{of}\text{H}\right)+\\ \left(1\right)\left(\text{Atomic}\text{mass}\text{of}\text{P}\right)\\ +\left(4\right)\left(\text{Atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right)\\ =\left[\begin{array}{l}\left(1\right)\left(22.99\text{ amu}\right)+\left(2\right)\left(1.008\text{ amu}\right)+\\ \left(1\right)\left(30.97\text{ amu}\right)+\left(4\right)\left(16\text{ amu}\right)\end{array}\right]\\ =22.99\text{ amu}+2.016\text{ amu}+30.97\text{ amu}+64\text{ amu}\\ =119.98\text{ amu}\end{array}$

c)

Interpretation Introduction

Interpretation: Formula and molecular mass of potassium bicarbonate should be determined.

Concept introduction:In the molecular formula, the exact number and type of constituent atoms are shown. Type of constituent atoms is represented by their atomic symbols and number of each constituent atom is represented in the right subscript of its atomic symbol.In a hypothetical molecule ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ , x number of $\text{A}$ atom and y number of $\text{B}$ atom combines to form the compound.

Formula to calculate molecular mass of ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ is as follows:

  $\text{Molecular}\text{mass}\text{of}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}=\left(\begin{array}{l}\left(\text{x}\right)\left(\text{Atomic}\text{mass}\text{of}\text{A}\right)+\\ \left(\text{y}\right)\left(\text{Atomic}\text{mass}\text{of}\text{B}\right)\end{array}\right)$

Explanation of Solution

In potassium bicarbonate,one bicarbonate $\left({\text{HCO}}_3^{-}\right)$ and one potassium ion $\left({\text{K}}^{+}\right)$ is attached by electrostatic attraction to maintain electrical neutrality. So the formula of potassium carbonateis ${\text{KHCO}}_3$ .

Molecular mass of ${\text{KHCO}}_3$ is calculated as follows:

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{KHCO}}_3=\left(\begin{array}{l}\left(1\right)\left(\text{Atomic}\text{mass}\text{of}\text{K}\right)+\\ \left(1\right)\left(\text{Atomic}\text{mass}\text{of}\text{H}\right)+\\ \left(1\right)\left(\text{Atomic}\text{mass}\text{of}\text{C}\right)+\\ \left(3\right)\left(\text{Atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right)\\ =\left[\begin{array}{l}\left(1\right)\left(39.10\text{ amu}\right)+\left(1\right)\left(1.008\text{ amu}\right)+\\ \left(1\right)\left(12.01\text{ amu}\right)+\left(3\right)\left(16\text{ amu}\right)\end{array}\right]\\ =39.10\text{ amu}+1.008\text{ amu}+12.01\text{ amu}+48\text{ amu}\\ =100.12\text{ amu}\end{array}$