Chapter 3, Problem 3.65QE

(a)

Interpretation Introduction

Interpretation:

The number of moles in $\text{2}{\text{.2g K}}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be calculated.

Explanation of Solution

The given compound is ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The molecular mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is the sum of the atomic mass of each of its elements multiplied by the numbers of each type of atoms present:

    $\begin{array}{l}\text{2(K) 2× 39}\text{.098 u = 78}\text{.196u}\\ \text{4(O) 4× 15}\text{.999u = 63}\text{.996u}\\ \text{1 (S) 1×32}\text{.065u = 32}\text{.065u}\\ \text{----------------------------------}\\ {\text{Molecular mass K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{ = 174}\text{.257u}\end{array}$.

Given that the molar mass of a molecular substance is numerically equal to its molecular mass, the molar mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{174}\text{.257 g/mol}$.

  ${\text{Amount K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{ = 2}{\text{.2g K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\frac{{\text{1mol K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{174}\text{.257g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{=0}\text{.0126mol}$.

(b)

Interpretation Introduction

Interpretation:

The number of moles in $\text{6}{\text{.4g C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}$ has to be calculated.

Explanation of Solution

The given compound is ${\text{C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}$.

The molecular mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}$ is the sum of the atomic mass of each of its elements multiplied by the numbers of each type of atoms present:

    $\begin{array}{l}\text{8(C) 8× 12}\text{.07 u = 96}\text{.56u}\\ \text{12(H) 12× 1}\text{.007u = 12}\text{.084u}\\ \text{4 (N) 4×14}\text{.0067u = 56}\text{.0268u}\\ \text{----------------------------------}\\ {\text{Molecular mass C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}\text{ = 164}\text{.6708u}\end{array}$.

Given that the molar mass of a molecular substance is numerically equal to its molecular mass, the molar mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}$ is $\text{164}\text{.6708 g/mol}$.

  ${\text{Amount C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}\text{ = 6}{\text{.4g C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}\text{×}\frac{{\text{1mol C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}}{\text{164}{\text{.6708g C}}_{\text{8}}{\text{H}}_{\text{12}}{\text{N}}_{\text{4}}}\text{=0}\text{.0388mol}$.

(c)

Interpretation Introduction

Interpretation:

The number of moles in $\text{7}{\text{.13g Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}$ has to be calculated.

Explanation of Solution

The given compound is ${\text{Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}$.

The molecular mass of ${\text{Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}$ is the sum of the atomic mass of each of its elements multiplied by the numbers of each type of atoms present:

    $\begin{array}{l}\text{1(Fe) 1× 55}\text{.84 u = 55}\text{.84u}\\ \text{10(C) 10×12}\text{.07u = 120}\text{.7u}\\ \text{10 (H) 10×1}\text{.007u = 10}\text{.07u}\\ \text{----------------------------------}\\ {\text{Molecular mass Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}\text{ = 186}\text{.61u}\end{array}$.

Given that the molar mass of a molecular substance is numerically equal to its molecular mass, the molar mass of ${\text{Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}$ is $\text{186}\text{.61 g/mol}$.

  ${\text{Amount Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}\text{ = 7}{\text{.13g Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}\text{×}\frac{{\text{1mol Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}}{\text{186}{\text{.61g Fe(C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{)}}_{\text{2}}}\text{=0}\text{.0382mol}$.