Chapter 8, Problem 119AP

Interpretation Introduction

(a)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 119AP

The percent by mass of sodium that is mentioned first in the formula of sodium azide, ${\text{NaN}}_{\text{3}}$ is $35.37\text{\%}$.

Explanation of Solution

Given compound is ${\text{NaN}}_{\text{3}}$

The mass of $1\text{mole}$ of $\text{Na}$ present in sodium azide is $22.99\text{g/mol}\times \text{1}\text{mol}=\text{22}\text{.99}\text{g}$.

The mass of $3\text{mole}$ of $\text{N}$ present in sodium azide is $14\text{g/mol}\times 3\text{mol}=4\text{2}\text{g}$.

Thus, the total mass of $1\text{mole}$ sodium azide is,

$\text{Total}\text{mass}\text{of}{\text{NaN}}_{\text{3}}=42+22.99=64.99\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of sodium and mass of sodium azide in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of Na}=\frac{\text{22}\text{.99}\text{g}}{\text{64}\text{.99}\text{g}}\times 100\text{\%}\\ =35.37\text{\%}\end{array}$

Thus, the percent by mass of sodium that is mentioned first in the formula of sodium azide, ${\text{NaN}}_{\text{3}}$ is $35.37\text{\%}$.

Interpretation Introduction

(b)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 119AP

The percent by mass of copper that is mentioned first in the formula of copper (II) sulfate, ${\text{CuSO}}_4$ is $39.82\text{\%}$.

Explanation of Solution

Given compound is ${\text{CuSO}}_4$

The mass of $1\text{mole}$ of $\text{Cu}$ present in copper (II) sulfate is $63.5\text{g/mol}\times \text{1}\text{mol}=63.5\text{g}$.

The mass of $1\text{mole}$ of $\text{S}$ present in copper (II) sulfate is $32\text{g/mol}\times \text{1}\text{mol}=32\text{g}$.

The mass of $4\text{mole}$ of $\text{O}$ present in copper (II) sulfate is $15.99\text{g/mol}\times 4\text{mol}=63.96\text{g}$.

Thus, the total mass of $1\text{mole}$ copper (II) sulfate is,

$\text{Total}\text{mass}\text{of}{\text{CuSO}}_4=63.5+32+63.96\text{g}=159.46\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of copper and mass of copper (II) sulfate in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of Cu}=\frac{\text{63}\text{.5}\text{g}}{159.46\text{g}}\times 100\text{\%}\\ =39.82\text{\%}\end{array}$

Thus, the percent by mass of copper that is mentioned first in the formula of copper (II) sulfate, ${\text{CuSO}}_4$ is $39.82\text{\%}$.

Interpretation Introduction

(c)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 119AP

The percent by mass of gold that is mentioned first in the formula of gold (III) chloride, ${\text{AuCl}}_{\text{3}}$ is $64.94\text{\%}$.

Explanation of Solution

Given compound is ${\text{AuCl}}_{\text{3}}$

The mass of $1\text{mole}$ of $\text{Au}$ present in gold (III) chloride is $196.97\text{g/mol}\times \text{1}\text{mol}=196.97\text{g}$.

The mass of $3\text{mole}$ of $\text{Cl}$ present in gold (III) chloride is $35.45\text{g/mol}\times 3\text{mol}=106.35\text{g}$.

Thus, the total mass of $1\text{mole}$ gold (III) chloride is,

$\text{Total}\text{mass}\text{of}{\text{AuCl}}_{\text{3}}=196.97+106.35=303.32\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of gold and mass of gold (III) chloride in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of Au}=\frac{\text{196}\text{.97}\text{g}}{\text{303}\text{.32}\text{g}}\times 100\text{\%}\\ =64.94\text{\%}\end{array}$

Thus, the percent by mass of gold that is mentioned first in the formula of gold (III) chloride, ${\text{AuCl}}_{\text{3}}$ is $64.94\text{\%}$.

Interpretation Introduction

(d)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 119AP

The percent by mass of silver that is mentioned first in the formula of silver nitrate, ${\text{AgNO}}_{\text{3}}$ is $63.51\text{\%}$.

Explanation of Solution

Given compound is ${\text{AgNO}}_{\text{3}}$

The mass of $1\text{mole}$ of $\text{Ag}$ present in silver nitrate is $107.87\text{g/mol}\times \text{1}\text{mol}=107.87\text{g}$.

The mass of $1\text{mole}$ of $\text{N}$ present in silver nitrate is $14\text{g/mol}\times \text{1}\text{mol}=14\text{g}$.

The mass of $3\text{mole}$ of $\text{O}$ present in silver nitrate is $15.99\text{g/mol}\times 3\text{mol}=47.97\text{g}$.

Thus, the total mass of $1\text{mole}$ silver nitrate is,

$\text{Total}\text{mass}\text{of}{\text{AgNO}}_3=107.87+14+47.97=169.84\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of gold and mass of gold (III) chloride in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of Ag}=\frac{\text{107}\text{.87}\text{g}}{169.84\text{g}}\times 100\text{\%}\\ =63.51\text{\%}\end{array}$

Thus, the percent by mass of silver that is mentioned first in the formula of silver nitrate, ${\text{AgNO}}_{\text{3}}$ is $63.51\text{\%}$.

Interpretation Introduction

(e)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 119AP

The percent by mass of rubidium that is mentioned first in the formula of rubidium sulfate, ${\text{Rb}}_2{\text{SO}}_4$ is $64.05\text{\%}$.

Explanation of Solution

Given compound is ${\text{Rb}}_2{\text{SO}}_4$

The mass of $2\text{mole}$ of $\text{Rb}$ present in rubidium sulfate is $85.47\text{g/mol}\times 2\text{mol}=170.94\text{g}$.

The mass of $1\text{mole}$ of $\text{S}$ present in rubidium sulfate is $32\text{g/mol}\times \text{1}\text{mol}=32\text{g}$.

The mass of $4\text{mole}$ of $\text{O}$ present in rubidium sulfate is $15.99\text{g/mol}\times 4\text{mol}=63.96\text{g}$.

Thus, the total mass of $1\text{mole}$ rubidium sulfate is,

$\text{Total}\text{mass}\text{of}{\text{Rb}}_2{\text{SO}}_4=170.94+32+63.96=266.9\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of rubidium and mass of rubidium sulfate in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of Rb}=\frac{\text{170}\text{.94}\text{g}}{\text{266}\text{.9}\text{g}}\times 100\text{\%}\\ =64.05\text{\%}\end{array}$

Thus, the percent by mass of rubidium that is mentioned first in the formula of rubidium sulfate, ${\text{Rb}}_2{\text{SO}}_4$ is $64.05\text{\%}$.

Interpretation Introduction

(f)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 119AP

The percent by mass of sodium that is mentioned first in the formula of sodium chlorate, ${\text{NaClO}}_{\text{3}}$ is $21.61\text{\%}$.

Explanation of Solution

Given compound is ${\text{NaClO}}_{\text{3}}$

The mass of $1\text{mole}$ of $\text{Na}$ present in sodium chlorate is $22.99\text{g/mol}\times 1\text{mol}=22.99\text{g}$.

The mass of $1\text{mole}$ of $\text{Cl}$ present in sodium chlorate is $35.45\text{g/mol}\times \text{1}\text{mol}=35.45\text{g}$.

The mass of $3\text{mole}$ of $\text{O}$ present in sodium chlorate is $15.99\text{g/mol}\times 3\text{mol}=47.97\text{g}$.

Thus, the total mass of $1\text{mole}$ sodium chlorate is,

$\text{Total}\text{mass}\text{of}{\text{NaClO}}_{\text{3}}=22.99+35.45+47.97=106.41\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of sodium and mass of sodium chlorate in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of Na}=\frac{\text{22}\text{.99}\text{g}}{106.41\text{g}}\times 100\text{\%}\\ =21.61\text{\%}\end{array}$

Thus, the percent by mass of sodium that is mentioned first in the formula of sodium chlorate, ${\text{NaClO}}_{\text{3}}$ is $21.61\text{\%}$.

Interpretation Introduction

(g)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 119AP

The percent by mass of nitrogen that is mentioned first in the formula of nitrogen triiodide, ${\text{NI}}_{\text{3}}$ is $3.55\text{\%}$.

Explanation of Solution

Given compound is ${\text{NI}}_{\text{3}}$

The mass of $1\text{mole}$ of $\text{N}$ present in nitrogen triiodide is $14\text{g/mol}\times 1\text{mol}=14\text{g}$.

The mass of $3\text{mole}$ of $\text{I}$ present in nitrogen triiodide is $126.9\text{g/mol}\times 3\text{mol}=380.7\text{g}$.

Thus, the total mass of $1\text{mole}$ nitrogen triiodide is,

$\text{Total}\text{mass}\text{of}{\text{NI}}_{\text{3}}=14+380.7=394.7\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of nitrogen and mass of nitrogen triiodide in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of N}=\frac{\text{14}\text{g}}{394.7\text{g}}\times 100\text{\%}\\ =3.55\text{\%}\end{array}$

Thus, the percent by mass of nitrogen that is mentioned first in the formula of nitrogen triiodide, ${\text{NI}}_{\text{3}}$ is $3.55\text{\%}$.

Interpretation Introduction

(h)

Interpretation:

The percent by mass of the element that is mentioned first in the formula of the given compound is to be stated.

Concept Introduction:

The mass percent composition shows the relative amount of each and every element present in the compound.

The mass percent of any element in the given compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$.

Answer to Problem 119AP

The percent by mass of cesium that is mentioned first in the formula of cesium bromide, $\text{CsBr}$ is $62.45\text{\%}$.

Explanation of Solution

Given compound is $\text{CsBr}$

The mass of $1\text{mole}$ of $\text{Cs}$ present in cesium bromide is $132.9\text{g/mol}\times 1\text{mol}=132.9\text{g}$.

The mass of $1\text{mole}$ of $\text{Br}$ present in cesium bromide is $79.9\text{g/mol}\times \text{1}\text{mol}=79.9\text{g}$.

Thus, the total mass of $1\text{mole}$ cesium bromide is,

$\text{Total}\text{mass}\text{of}\text{CsBr}=132.9+79.9=212.8\text{g}$.

The mass percent of an element present in the compound is calculated by the formula,

$\text{\%}\text{}\text{Mass}=\frac{\text{Mass}\text{of}\text{element}\text{in}\text{one}\text{mole}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\times 100\text{\%}$

Substitute the respective values of mass of cesium and mass of cesium bromide in the above expression.

$\begin{array}{c}\text{\%}\text{}\text{Mass of Cs}=\frac{\text{132}\text{.9}\text{g}}{212.8\text{g}}\times 100\text{\%}\\ =62.45\text{\%}\end{array}$

Thus, the percent by mass of cesium that is mentioned first in the formula of cesium bromide, $\text{CsBr}$ is $62.45\text{\%}$.