Chapter 3, Problem 105A

(a)

Interpretation Introduction

Interpretation: The mass percent of carbon and oxygen in different compounds needs to be compared.

Concept introduction: The mass % of an element in any compound is calculated by comparing the element’s mass with the total mass of the compound as shown below:

  $\text{mass \% of}\text{C=}\frac{\text{Mass}\text{of}\text{C}\text{in that}\text{compound}}{\text{Total}\text{mass}\text{of}\text{the}\text{compound}}$

Answer to Problem 105A

The $\text{mass \% of}\text{C}$ in charcoal, Indigo, Verdigris are $\text{100}\text{\%},\text{73}\text{\%}\text{a}\text{n}\text{d}\text{14}\text{\%}$ respectively.

Explanation of Solution

To calculate Mass percent:

  $\text{mass \% of}\text{C=}\frac{\text{Mass}\text{of}\text{C}\text{in that}\text{compound}}{\text{Total}\text{mass}\text{of}\text{the}\text{compound}}$

Mass Percent of Carbon in Charcoal, Indigo and Verdigris are:

For Charcoal:

  $\text{mass \% of}\text{C in Charcoal is 100\% as charcoal is made of carbon only}\text{.}$

For Indigo:

  $\begin{array}{c}\text{I}\text{n}\text{d}\text{i}\text{g}\text{o}={\text{C}}_{\text{1}\text{6}}{\text{H}}_{\text{1}\text{0}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\end{array}$

     $\begin{array}{c}=\text{16}(\text{12})+\text{10}(\text{1})+\text{2}(\text{14})+\text{2}(\text{16})\end{array}$

     $\begin{array}{c}=\text{262}\end{array}$

  $\text{mass}\text{\%}\text{o}\text{f}\text{C}=\frac{\text{1}\text{9}\text{2}.\text{1}\text{6}\text{g}\text{m}/\text{m}\text{o}\text{l}}{\text{2}\text{6}\text{2}\text{g}\text{m}/\text{m}\text{o}\text{l}}\times \text{1}\text{0}\text{0}=\text{7}\text{3}\text{\%}$

For Verdigris:

$\begin{array}{}$ \text{V}\text{e}\text{r}\text{d}\text{i}\text{g}\text{r}\text{i}\text{s}=\text{C}\text{u}{({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}})}_{\text{2}}.\text{2}\text{C}\text{u}{(\text{O}\text{H})}_{\text{2}}$\end{array}$

     $\begin{array}{c}=\text{63.5}+\text{2}(\text{2}\times \text{12}+\text{3}\times \text{1}+\text{2}\times \text{16})+\text{2}\left\{\text{63.5}+\text{2}(\text{16}+\text{1})\right\}\end{array}$

     $\begin{array}{c}=\text{63.5}+\text{2}(\text{59})+\text{2}\left\{\text{63.5}+\text{34}\right\}\end{array}$

     $\begin{array}{c}=\text{181.5}+\text{2}(\text{97.5})\end{array}$

     $\begin{array}{c}=\text{181.5}+\text{195}\end{array}$

     $\begin{array}{c}=\text{376.5}\end{array}$

$\text{mass \% of}\text{C}=\frac{\text{4}\text{8}.\text{0}\text{4}\text{g}\text{m}/\text{m}\text{o}\text{l}}{\text{3}\text{4}\text{3}.\text{1}\text{4}\text{g}\text{m}/\text{m}\text{o}\text{l}}\times \text{1}\text{0}\text{0}=\text{1}\text{4}\text{\%}$

Mass percent of oxygen in iron oxide and Egyptian blue are:

For Iron Oxide:

$\begin{array}{c}\text{Iron Oxide Red}={\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\end{array}$

       $\begin{array}{c}=\text{2(56)+3(16)}\end{array}$

       $\begin{array}{c}=\text{160}\end{array}$

Or,

  $\text{mass \%}\text{of}\text{O}=\frac{\text{48 gm/mol}}{\text{160gm/mol}}\text{×100}=\text{30\%}$

For Egyptian blue:

$\begin{array}{c}\text{E}\text{g}\text{y}\text{p}\text{t}\text{i}\text{a}\text{n}\text{b}\text{l}\text{u}\text{e}=\text{C}\text{a}\text{C}\text{u}\text{S}{\text{i}}_{\text{4}}{\text{O}}_{\text{1}\text{0}}\end{array}$

       $\begin{array}{c}=\text{40}+\text{63.5}+\text{4}(\text{28})+\text{10}(\text{16})\end{array}$

       $\begin{array}{c}=\text{375.5}\end{array}$

$\text{mass \% of}\text{O}=\frac{\text{160 gm/mol}}{\text{375}\text{.5 gm/mol}}\text{×100}=\text{43\%}$

(b)

Interpretation Introduction

Interpretation: The mass percent of carbon and oxygen in different compounds needs to be compared.

Concept introduction: The mass% of an element in any compound is calculated by comparing element’s mass with the total mass of the compound as shown below:

  $\text{mass \% of}\text{C}=\frac{\text{Mass}\text{of}\text{C}\text{in that}\text{compound}}{\text{Total}\text{mass}\text{of}\text{the}\text{compound}}$

Answer to Problem 105A

The $\text{mass}\text{\%}\text{o}\text{f}\text{O}$ in iron oxide and Egyptian blue are $\text{30}\text{\%}$ and $\text{43}\text{\%}$ respectively.

Explanation of Solution

Mass percent of oxygen in iron oxide and Egyptian blue are:

For Iron Oxide:

$\begin{array}{c}\text{Iron Oxide Red}={\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\end{array}$

       $\begin{array}{c}=\text{2(56)+3(16)}\end{array}$

       $\begin{array}{c}=\text{160}\end{array}$

$\text{mass \%}\text{of}\text{O}=\frac{\text{48 gm/mol}}{\text{160 gm/mol}}\text{×100=30\%}$

For Egyptian blue:

$\begin{array}{c}\text{Egyptian blue}=\text{C}\text{a}\text{C}\text{u}\text{S}{\text{i}}_{\text{4}}{\text{O}}_{\text{1}\text{0}}\end{array}$

       $\begin{array}{c}=\text{40}+\text{63.5}+\text{4}(\text{28})+\text{10}(\text{16})\end{array}$

       $\begin{array}{c}=\text{375.5}\end{array}$

$\text{mass \%}\text{of}\text{O}=\frac{\text{160 gm/mol}}{\text{375}\text{.5 gm/mol}}\text{×100}=\text{43\%}$