Chapter 8, Problem 30QAP

(a)

Interpretation Introduction

Interpretation : The formula and molar mass of barium perchlorate needs to be determined.

Concept Introduction : The molar mass of compound is calculated by taking sum of molar masses of all the atoms present in the compound.

Answer to Problem 30QAP

  ${\text{Ba(ClO}}_{\text{4}}{\text{)}}_{\text{2}}$ ; $MM=336.24\text{ g/mol}$

Explanation of Solution

Given Information:

  $\begin{array}{l}M{M}_{\text{Ba}}=137.33 \text{g/mol}\\ M{M}_{\text{Cl}}=35.9 \text{g/mol} \\ M{M}_{\text{O}}=16.00 \text{g/mol} \end{array}$

The formula of compound is ${\text{Ba(ClO}}_{\text{4}}{\text{)}}_{\text{2}}$ .

Now, calculate total molar mass

  $MM={\displaystyle \sum _{i=1}^{}M{M}_i}$

  $\begin{array}{l}MM=M{M}_{Ba}+2M{M}_{Cl}+8M{M}_O\\ MM=137.33+2\left(35.9\right)+8(16.0)\\ MM=336.24 \text{ g/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation : The formula and molar mass of magnesium sulfate needs to be determined.

Concept Introduction : The molar mass of compound is calculated by taking sum of molar masses of all the atoms present in the compound.

Answer to Problem 30QAP

  ${\text{MgSO}}_{\text{4}}$ ; $MM=120.37\text{ g/mol}$

Explanation of Solution

Given Information:

  $\begin{array}{l}M{M}_{\text{Mg}}=24.305\text{ g/mol}\\ M{M}_{\text{S}}=32.065 \text{ g/mol}\\ M{M}_{\text{O}}=16.00 \text{ g/mol}\end{array}$

The formula of the compound is ${\text{MgSO}}_{\text{4}}$

Now, calculate total molar mass

  $MM={\displaystyle \sum _{i=1}^{}M{M}_i}$

  $\begin{array}{l}MM=M{M}_{\text{Mg}}+M{M}_{\text{S}}+4M{M}_{\text{O}}\\ MM=24.305+32.065+4(16.0)\\ MM=120.37 \text{g/mol}\end{array}$

(c)

Interpretation Introduction

Interpretation : The formula and molar mass of lead (II) chloride needs to be determined.

Concept Introduction : The molar mass of compound is calculated by taking sum of molar masses of all the atoms present in the compound.

Answer to Problem 30QAP

  ${\mathrm{PbCl}}_2$ ; $MM=279\text{ g/mol}$

Explanation of Solution

Given Information:

  $\begin{array}{l}M{M}_{\text{Pb}}=207.20\text{ g/mol}\\ M{M}_{\text{Cl}}=35.9 \text{ g/mol} \end{array}$

First, the chemical formula of the compound must be stated. For Lead (II) Chloride, there must be the following: Lead and chlorine in the following amounts:

  ${\mathrm{PbCl}}_2$

Now, calculate total molar mass

  $MM={\displaystyle \sum _{i=1}^{}M{M}_i}$

  $\begin{array}{l}MM=M{M}_{\text{Pb}}+2M{M}_{\text{Cl}}\\ MM=207.20+2(35.90)\\ MM=279 \text{g/mol}\end{array}$

(d)

Interpretation Introduction

Interpretation : The formula and molar mass of copper (II) nitrate needs to be determined.

Concept Introduction : The molar mass of compound is calculated by taking sum of molar masses of all the atoms present in the compound.

Answer to Problem 30QAP

  ${\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ ; $MM=187.56 \text{g/mol}$

Explanation of Solution

Given Information:

  $\begin{array}{l}M{M}_{\text{Cu}}=63.546 \text{ g/mol}\\ M{M}_{\text{N}}=14.00 \text{ g/mol}\\ M{M}_{\text{O}}=16.00 \text{ g/mol} \end{array}$

The formula of the given compound is ${\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Now, calculate total molar mass

  $MM={\displaystyle \sum _{i=1}^{}M{M}_i}$

  $\begin{array}{l}MM=M{M}_{\text{Cu}}+2(M{M}_{\text{N}})+6M{M}_{\text{O}})\\ MM=63.546 +2(14.00)+6(16.00)\\ MM=187.56 \text{ g/mol}\end{array}$

(e)

Interpretation Introduction

Interpretation : The formula and molar mass of tin (IV) chloride needs to be determined.

Concept Introduction : Most of the compounds are ionic, since they are metal-nonmetal bonds.

Answer to Problem 30QAP

  ${\text{SnCl}}_{\text{4}}$ ; $MM=260.5 \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Explanation of Solution

Given Information:

  $\begin{array}{l}M{M}_{\text{Sn}}=118.71\text{ g/mol}\\ M{M}_{\text{Cl}}=35.9 \text{ g/mol} \end{array}$

The formula of the given compound is ${\text{SnCl}}_{\text{4}}$

Now, calculate total molar mass

  $MM={\displaystyle \sum _{i=1}^{}M{M}_i}$

  $\begin{array}{l}MM=M{M}_{Sn}+4M{M}_{Cl}\\ MM=118.71+4(35.90)\\ MM=260.5 \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\end{array}$