Chapter 8, Problem 48QAP

Calculate the percent by mass of the element listed first in the formulas for each of the following compounds.

l type="a">

copper(II) bromide, ${\text{CuBr}}_2$

i>copper(I) bromide, $\text{CuBr}$

i>iron(II) chloride, ${\text{FeCl}}_2$

i>iron(III) chloride, ${\text{FeCl}}_3$

i>cobalt(II) iodide, ${\text{Col}}_2$

i>cobalt(III) iodide, ${\text{CoI}}_3$

i>tin(II) oxide, $\text{SnO}$

i>tin(IV)oxide, ${\text{SnO}}_2$

Interpretation Introduction

(a)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 48QAP

$\text{Copper (II) bromide},\text{C}\text{u}\text{B}\text{r}2 \text{C}{\text{u}}^{2+ }\text{\%} =28.45\text{\%}$.

Explanation of Solution

A chemical compound is a collection of several atoms. Molar masses of each and every atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$

$\text{C}\text{opper (II) bromide},\text{C}\text{u}\text{B}\text{r}2 \text{C}{\text{u}}^{2+}\text{\%} =$

$\text{Mass of 1 mol of }\text{C}{\text{u}}^{2+} =1\times \text{ 63}\text{.55 }\text{g}=\text{ 63}\text{.55 }\text{g}$

$\text{Mass of 2 mol of }\text{B}{\text{r}}^{-}\text{ = 2 }\times \text{ 79}\text{.91 g = 159}\text{.82 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{C}\text{u}\text{B}\text{r}2\text{= 63}\text{.55+ 159}\text{.82 = 223}\text{.37 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{C}\text{u}\text{B}\text{r}2\text{= 223}{\text{.37 g mol}}^{\text{-1}}$

Mass of the copper present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{63}\text{.55 g}}{\text{1 mol}}=63.55\text{ g}$

Mass percent of $\text{C}{\text{u}}^{2+}$ = $\frac{\text{63}\text{.55 }\text{g}\text{C}{\text{u}}^{2+} }{\text{223}\text{.37 }\text{g} \text{C}\text{u}\text{B}\text{r}2}\times 100\text{\%}=\underset{\_}{\underset{\_}{28.45\text{\%}}}$.

Interpretation Introduction

(b)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 48QAP

$\text{C}\text{opper (I) bromide},\text{C}\text{u}\text{B}\text{r} \text{C}{\text{u}}^{+ }\text{\%} =44.29\text{\%}$.

Explanation of Solution

$\text{Copper (I) bromide},\text{C}\text{u}\text{B}\text{r} \text{C}{\text{u}}^{+}\text{\%} =$

$\text{Mass of 1 mol of }\text{C}{\text{u}}^{+}=1\times \text{ 63}\text{.55 }\text{g}=\text{ 63}\text{.55 }\text{g}$

$\text{Mass of 1 mol of }\text{B}{\text{r}}^{-}\text{ = 1 }\times \text{ 79}\text{.91 g = 79}\text{.91 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{C}\text{u}\text{B}\text{r}\text{= 63}\text{.55+79}\text{.91= 143}\text{.46 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{C}\text{u}\text{B}\text{r}\text{= 143}{\text{.46 g mol}}^{\text{-1}}$

Mass of the copper present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{63}\text{.55 g}}{\text{1 mol}}=63.55\text{ g}$

Mass percent of $\text{C}{\text{u}}^{+}$ = $\frac{\text{63}\text{.55 g }\text{C}{\text{u}}^{+}}{\text{143}\text{.46 g }\text{C}\text{u}\text{B}\text{r}}\times 100\text{\%}=\underset{\_}{\underset{\_}{44.29\text{\%}}}$.

Interpretation Introduction

(c)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 48QAP

$\text{I}\text{r}\text{o}\text{n}(\text{I}\text{I})\text{ chloride}{,}_{,}\text{F}\text{e}\text{C}\text{l}2 \text{F}{\text{e}}^{2+}\text{\%} = 40.08\text{\%}$.

Explanation of Solution

$\text{I}\text{r}\text{o}\text{n}(\text{I}\text{I})\text{ chloride}{,}_{,}\text{F}\text{e}\text{C}\text{l}2 \text{F}{\text{e}}^{2+}\text{\%} =$

$\text{Mass of 1 mol of }\text{F}{\text{e}}^{2+}=1\times \text{ 55}\text{.85 }\text{g}=\text{ 55}\text{.85 }\text{g}$

$\text{Mass of 2 mol of }\text{C}{\text{l}}^{-}\text{ = 2 }\times \text{ 35}\text{.45 g = 70}\text{.9 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{F}\text{e}\text{C}\text{l}2 \text{ = 55}\text{.85+70}\text{.90= 126}\text{.7 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{F}\text{e}\text{C}\text{l}2\text{ = 126}{\text{.7 g mol}}^{\text{-1}}$

Mass of the carbon present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{55}\text{.85 g}}{\text{1 mol}}=55.85\text{ g}$

Mass percent of $\text{F}{\text{e}}^{2+}$ = $\frac{\text{55}\text{.85 g }\text{F}{\text{e}}^{2+}}{\text{126}\text{.7 g }\text{F}\text{e}\text{C}\text{l}2 }\times 100\text{\%}=\underset{\_}{\underset{\_}{40.08\text{\%}}}$.

Interpretation Introduction

(d)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 48QAP

$\text{I}\text{r}\text{o}\text{n}(\text{I}\text{I}\text{I})\text{ chloride}{,}_{,}\text{F}\text{e}\text{C}\text{l}3 \text{F}{\text{e}}^{3+}\text{\%} = 34.43\text{\%}$.

Explanation of Solution

$\text{I}\text{r}\text{o}\text{n}(\text{I}\text{I}\text{I})\text{ chloride}{,}_{,}\text{F}\text{e}\text{C}\text{l}3 \text{F}{\text{e}}^{3+}\text{\%} =$

$\text{Mass of 1 mol of }\text{F}{\text{e}}^{3+}=1\times \text{ 55}\text{.85 }\text{g}=\text{ 55}\text{.85 }\text{g}$

$\text{Mass of 3 mol of }\text{C}{\text{l}}^{-}\text{ = 3 }\times \text{ 35}\text{.45 g = 106}\text{.35 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{F}\text{e}\text{C}\text{l}3 \text{ = 55}\text{.85+106}\text{.35= 162}\text{.2 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{F}\text{e}\text{C}\text{l}3\text{ = 162}{\text{.2 g mol}}^{\text{-1}}$

Mass of the iron present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{55}\text{.85 g}}{\text{1 mol}}=55.85\text{ g}$

Mass percent of $\text{F}{\text{e}}^{3+}$ = $\frac{\text{55}\text{.85 g }\text{F}{\text{e}}^{3+}}{\text{162}\text{.2 g }\text{F}\text{e}\text{C}\text{l}3 }\times 100\text{\%}=\underset{\_}{\underset{\_}{34.43\text{\%}}}$.

Interpretation Introduction

(e)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 48QAP

$\text{C}\text{o}\text{b}\text{a}\text{l}\text{t}(\text{I}\text{I})\text{ iodide, }\text{C}\text{o}\text{I}2\text{C}{\text{o}}^{2+}\text{\%} =18.84\text{\%}$.

Explanation of Solution

$\text{C}\text{o}\text{b}\text{a}\text{l}\text{t}(\text{I}\text{I})\text{ iodide, }\text{C}\text{o}\text{I}2\text{C}{\text{o}}^{2+}\text{\%} =$

$\text{Mass of 1 mol of }\text{C}{\text{o}}^{2+}=1\times \text{ 58}\text{.93 }\text{g}=\text{ 58}\text{.93 }\text{g}$

$\text{Mass of 2 mol of }{\text{I}}^{-}\text{ = 2 }\times \text{ 126}\text{.9 g = 253}\text{.8 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{C}\text{o}\text{I}2\text{= 58}\text{.93+253}\text{.8= 312}\text{.73 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{C}\text{o}\text{I}2\text{= 312}{\text{.73 g mol}}^{\text{-1}}$

Mass of the cobalt present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{58}\text{.93 g}}{\text{1 mol}}=58.93\text{ g}$

Mass percent of $\text{C}{\text{o}}^{2+}$ = $\frac{\text{58}\text{.93 g }\text{C}{\text{o}}^{2+}}{\text{312}\text{.73 g }\text{C}\text{o}\text{I}2}\times 100\text{\%}=\underset{\_}{\underset{\_}{18.84\text{\%}}}$.

Interpretation Introduction

(f)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 48QAP

$\text{C}\text{o}\text{b}\text{a}\text{l}\text{t}(\text{I}\text{I}\text{I})\text{ iodide, }\text{C}\text{o}\text{I}3\text{C}{\text{o}}^{3+}\text{\%} =13.40\text{\%}$.

Explanation of Solution

$\text{C}\text{o}\text{b}\text{a}\text{l}\text{t}(\text{I}\text{I}\text{I})\text{ iodide, }\text{C}\text{o}\text{I}3\text{C}{\text{o}}^{3+}\text{\%} =$

$\text{Mass of 1 mol of }\text{C}{\text{o}}^{3+}=1\times \text{ 58}\text{.93 }\text{g}=\text{ 58}\text{.93 }\text{g}$

$\text{Mass of 3 mol of }{\text{I}}^{-}\text{ = 3 }\times \text{ 126}\text{.9 g = 380}\text{.7 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{C}\text{o}\text{I}3\text{= 58}\text{.93+380}\text{.7 = 439}\text{.63 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{C}\text{o}\text{I}3\text{= 439}{\text{.63 g mol}}^{\text{-1}}$

Mass of the cobalt present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{58}\text{.93 g}}{\text{1 mol}}=58.93\text{ g}$

Mass percent of $\text{C}{\text{o}}^{3+}$ = $\frac{\text{58}\text{.93 g }\text{C}{\text{o}}^{3+}}{\text{439}\text{.63 g }\text{C}\text{o}\text{I}3}\times 100\text{\%}=\underset{\_}{\underset{\_}{13.40\text{\%}}}$.

Interpretation Introduction

(g)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 48QAP

$\text{T}\text{in(II) oxide},\text{S}\text{n}\text{O} \text{S}{\text{n}}^{2+ }\text{\%} =88.12\text{\%}$.

Explanation of Solution

$\text{T}\text{in(II) oxide},\text{S}\text{n}\text{O} \text{S}{\text{n}}^{2+}\text{\%} =$

$\text{Mass of 1 mol of }\text{S}{\text{n}}^{2+}=1\times \text{ 118}\text{.7 }\text{g}=\text{ 118}\text{.7 }\text{g}$

$\text{Mass of 1 mol of }{\text{O}}^{2-}\text{ = 1 }\times \text{ 16}\text{.00 g = 16}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{S}\text{n}\text{O} \text{= 118}\text{.7+16}\text{.00= 134}\text{.7 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{S}\text{n}\text{O} \text{= 134}{\text{.7 g mol}}^{\text{-1}}$

Mass of the tin present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{118}\text{.7 g}}{\text{1 mol}}=118.7\text{ g}$

Mass percent of $\text{S}{\text{n}}^{2+}$ = $\frac{\text{118}\text{.7 g }\text{S}{\text{n}}^{2+}}{\text{134}\text{.7 g SnO }}\times 100\text{\%}=\underset{\_}{\underset{\_}{88.12\text{\%}}}$.

Interpretation Introduction

(h)

Interpretation:

The percent by mass of element listed first in the formula of a compound should be calculated.

Concept Introduction:

A chemical compound is a collection of several atoms. Molar masses of each atom collectively provide the molar mass of that compound.

$\text{mass fraction for a given element =}\frac{\text{mass of the element present in 1 mole of compound}}{\text{mass of 1 mole of compound}}$

Mass fraction for a given element can be converted into mass percent by multiplying $100\text{\%}.$.

Answer to Problem 48QAP

$\text{T}\text{in(IV) oxide},\text{S}\text{n}\text{O}2 \text{S}{\text{n}}^{4+}\text{\%} =76.73\text{\%}$.

Explanation of Solution

$\text{T}\text{in(IV) oxide},\text{S}\text{n}\text{O}2 \text{S}{\text{n}}^{4+}\text{\%} =$

$\text{Mass of 1 mol of }\text{S}{\text{n}}^{4+}=1\times \text{ 118}\text{.7}\text{g}=\text{ 118}\text{.7 }\text{g}$

$\text{Mass of 2 mol of }{\text{O}}^{2-}\text{= 2 }\times \text{ 16}\text{.00 g = 32}\text{.00 g}$

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}1\text{m}\text{o}\text{l}\text{o}\text{f}\text{S}\text{n}\text{O}2\text{= 118}\text{.7+36}\text{.00 = 154}\text{.70 }\text{g}$

$\text{M}\text{o}\text{l}\text{a}\text{r}\text{ mass of }\text{S}\text{n}\text{O}2\text{= 154}{\text{.70 g mol}}^{\text{-1}}$

Mass of the tin present in $1\text{m}\text{o}\text{l}$ of compound = $\text{1 mol × }\frac{\text{118}\text{.7 g}}{\text{1 mol}}=118.7\text{ g}$

Mass percent of $\text{S}{\text{n}}^{4+}$ = $\frac{\text{118}\text{.7 g }\text{S}{\text{n}}^{4+}}{\text{154}\text{.70 g }\text{S}\text{n}\text{O}2}\times 100\text{\%}=\underset{\_}{\underset{\_}{76.73\text{\%}}}$.