Chapter 2, Problem IISP

(a)

Interpretation Introduction

Interpretation:

The correct formula of ammonium dichromate, ammonium nitrate and trinitrotoluene has to be written.

Explanation of Solution

(i)Ammonium dichromate contains the ammonium cation, ${\text{NH}}_{\text{4}}^{\text{+}}$ and the dichromate anion ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}$.  The compound must have two $\text{+1}$ cations to balance the charge of the $\text{2-}$ anion, so its formula is ${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$.

(ii)Ammonium nitrate contains the ammonium cation, ${\text{NH}}_{\text{4}}^{\text{+}}$ and the nitrate anion, ${\text{NO}}_{\text{3}}^{\text{-}}$.  The compound must have one $\text{+1}$ cation to balance the charge of the $\text{1-}$ anion, so its formula is ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$.

(iii)Trinitrotoluene contains seven carbon atoms, five hydrogen atoms, three nitrogen atoms and six oxygen atoms, so its formula is ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$.

(b)

Interpretation Introduction

Interpretation:

Among the three compounds given, the compound which has the highest mass percent nitrogen has to be determined.

Explanation of Solution

(i)Ammonium dichromate:

Formula is. ${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$

One mole of ${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$ contains two moles of nitrogen.

Molar mass of ${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$ is:

$\text{=}\text{(2×14}\text{.0067}\text{g/mol}\text{N)}\text{}\text{+}\text{(8×1}\text{.0079}\text{g/mol}\text{H)}\text{+}\text{(2×51}\text{.9961}\text{g/mol}\text{Cr)}\text{}\text{+}\text{(7×15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{252}\text{.0646}\text{g/mol}\text{.}$

Mass of $\text{N}$ in one mole of ${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$

$\text{=}\text{2}\text{×}\text{14}\text{.0067}\text{g/mol}\text{N}\text{=}\text{28}\text{.0134}\text{g/mol}\text{N}$.

$\text{\%}\text{N}\text{=}\frac{\text{mass}\text{N}\text{per}\text{mol}}{\text{mass}\text{compound}\text{per}\text{mol}}\text{×}\text{100\%}\text{=}\frac{\text{28}\text{.0134}\text{g}\text{N}}{\text{252}\text{.0646}\text{compound}}\text{×}\text{100\%=}\text{11}\text{.1136}\text{\%}$

(ii)Ammonium nitrate:

Formula is ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$.

One mole of ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$ contains two moles of nitrogen.

Molar mass of ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$ is:

$\text{=}\text{(2×14}\text{.0067}\text{g/mol}\text{N)}\text{}\text{+}\text{(4×1}\text{.0079}\text{g/mol}\text{H)}\text{)}\text{}\text{+}\text{(3×15}\text{.9994}\text{g/mol}\text{O)}\text{=}80.0432\text{g/mol}\text{.}$

Mass of $\text{N}$ in one mole of ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$ $\text{=}\text{2}\text{×}\text{14}\text{.0067}\text{g/mol}\text{N}\text{=}\text{28}\text{.0134}\text{g/mol}\text{N}$.

$\text{\%}\text{N}\text{=}\frac{\text{mass}\text{N}\text{per}\text{mol}}{\text{mass}\text{compound}\text{per}\text{mol}}\text{×}\text{100\%}\text{=}\frac{\text{28}\text{.0134}\text{g}\text{N}}{\text{80}\text{.0432}\text{g}\text{compound}}\text{×}\text{100\%=}\text{34}\text{.9979}\text{\%}$

(iii)Trinitrotoluene:

Formula is ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$.

One mole of ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$ contains three moles of nitrogen.

Molar mass of ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$ is:

  $\text{=}\text{(7×}\text{12}\text{.0107}\text{g/mol}\text{C)}\text{+}\text{(3×14}\text{.0067}\text{g/mol}\text{N)}\text{}\text{+}\text{(5×1}\text{.0079}\text{g/mol}\text{H)}\text{)}\text{}\text{+}\text{(6×15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{227}\text{.1309}\text{g/mol}\text{.}$

Mass of $\text{N}$ in one mole of ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$ $\text{=}3\text{×}\text{14}\text{.0067}\text{g/mol}\text{N}\text{=}42.0201\text{g/mol}\text{N}$.

$\text{\%}\text{N}\text{=}\frac{\text{mass}\text{N}\text{per}\text{mol}}{\text{mass}\text{compound}\text{per}\text{mol}}\text{×}\text{100\%}\text{=}\frac{\text{42}\text{.0201}\text{g}\text{N}}{\text{227}\text{.1309}\text{g}\text{compound}}\text{×}\text{100\%=}\text{18}\text{.5004}\text{\%}\text{N}$

(c)

Interpretation Introduction

Interpretation:

Among the three compounds, the compound which has the lowest melting point has to be explained.

Explanation of Solution

Molecular compounds have lower melting points than ionic compounds.  ${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$ and ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$ are ionic compounds and ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$ is a molecular compound.  So trinitrotoluene will have the lowest melting point.

(d)

Interpretation Introduction

Interpretation:

Among the three compounds, the compound which conduct electricity at room temperature has to be explained.

Explanation of Solution

${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$, ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$ and ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$ are the given compounds and all the three compounds are solid at room temperature.  If the temperature were high enough to melt the ionic compounds ${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$, ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$ the molten ions would conduct electricity.  The molecular compound ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$ will not conduct electricity at any temperature.

(e)

Interpretation Introduction

Interpretation:

Among the three compounds, the compound which conducts electricity if they melted has to be given.

Explanation of Solution

If the temperature of the ionic compounds increased, then the ions will melt and in that molten state would electricity.  So ${{\text{(NH}}_{\text{4}}^{}\text{)}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{}$, ${\text{NH}}_{\text{4}}^{}{\text{NO}}_{\text{3}}^{}$ can conduct electricity in molten state.