Chapter 3, Problem 39E

(a)

Interpretation Introduction

Interpretation: The molar mass of chloral hydrate is to be calculated.

Concept introduction: Mole concept makes it quite possible to deal with subatomic particles. It helps in relating the macroscopic quantities involved in reaction with these small particles. This is achieved by using mole as the bridge between them. Mole is related to both by the expression shown below.

  $\begin{array}{c}n\times 6.022\times {10}^{23}\text{particles}=\text{Moles}\left(n\right)\\ =\frac{\text{Mass}\left(m\right)}{\text{Molar}\text{mass}\left(M\right)}\end{array}$

Mole is referred as the substance’s amount possessing mass equal to that in $\text{12}\text{g}$ of $\text{C}$ .

Answer to Problem 39E

The molar mass of chloral hydrate is $\text{165}\text{.4}\text{g/mol}$

Explanation of Solution

The given molecular formula of chloral hydrate is ${\text{C}}_2{\text{H}}_3{\text{Cl}}_3{\text{O}}_2$ .

The molar mass of ${\text{C}}_2{\text{H}}_3{\text{Cl}}_3{\text{O}}_2$ can be calculated by the formula shown below.

  $M^{\prime }=n_{\text{C}}\times M_{\text{C}}+n_{\text{H}}\times M_{\text{H}}+n_{\text{Cl}}\times M_{\text{Cl}}+n_{\text{O}}\times M_{\text{O}}$  (1)

Where,

• $n_{\text{C}}$ is the number of atoms of carbon.
• $M_{\text{C}}$ is the molar mass of carbon.
• $n_{\text{H}}$ is the number of atoms of hydrogen.
• $M_{\text{C}}$ is the molar mass of hydrogen.
• $n_{\text{Cl}}$ is the number of atoms of chlorine.
• $M_{\text{Cl}}$ is the molar mass of chlorine.
• $n_{\text{O}}$ is the number of atoms of oxygen.
• $M_{\text{O}}$ is the molar mass of oxygen.
• $M^{\prime }$ is the molar mass of the ${\text{C}}_2{\text{H}}_3{\text{Cl}}_3{\text{O}}_2$ .

The molar mass of carbon is $\text{12}\text{.01}\text{g/mol}$ .

The molar mass of hydrogen is $1.008\text{g/mol}$ .

The molar mass of oxygen is $16.00\text{g/mol}$ .

The molar mass of chlorine is $\text{35}\text{.45}\text{g/mol}$ .

Substitute the known values in equation (1).

  $\begin{array}{c}{M}^{\prime }=\left(2\times 12.01+3\times 1.008+3\times 35.45+2\times 16.00\right)\text{g/mol}\\ =\left(24.02+3.024+106.35+32.0\right)\text{g/mol}\\ =165.39\text{g/mol}\\ \approx \text{165}\text{.4}\text{g/mol}\end{array}$

Hence, the molar mass of chloral hydrate is $\text{165}\text{.4}\text{g/mol}$ .

(b)

Interpretation Introduction

Interpretation: The moles of chloral hydrate molecules present in $\text{500}\text{.0}\text{g}$ of chloral hydrate are to be calculated.

Concept introduction: Mole concept makes it quite possible to deal with subatomic particles. It helps in relating the macroscopic quantities involved in reaction with these small particles. This is achieved by using mole as the bridge between them. Mole is related to both by the expression shown below.

  $\begin{array}{c}n\times 6.022\times {10}^{23}\text{particles}=\text{Moles}\left(n\right)\\ =\frac{\text{Mass}\left(m\right)}{\text{Molar}\text{mass}\left(M\right)}\end{array}$

Mole is referred as the substance’s amount possessing mass equal to that in $\text{12}\text{g}$ of $\text{C}$ .

Answer to Problem 39E

The moles of chloral hydrate molecules present in $\text{500}\text{.0}\text{g}$ of chloral hydrate is $3.0229\text{mol}$ .

Explanation of Solution

The given mass of chloral hydrate is $\text{500}\text{.0}\text{g}$ .

The moles of molecules of chloral hydrate can be calculated by the formula shown below.

  $n=\frac{m}{M}$  (2)

Where,

• $n$ is the number of moles of molecules.
• $m$ is the given mass.
• $M$ is the molar mass.

Substitute the known values in equation (2).

  $\begin{array}{c}n=\frac{500.0\text{g}}{\text{165}\text{.4}\text{g/mol}}\\ =3.0229\text{mol}\end{array}$

Hence, the moles of molecules present in $\text{500}\text{.0}\text{g}$ of chloral hydrate are $3.0229\text{mol}$ .

(c)

Interpretation Introduction

Interpretation: The mass in grams of $2\times {10}^{-2}$ moles of chloral hydrate is to be calculated.

Concept introduction: Mole concept makes it quite possible to deal with subatomic particles. It helps in relating the macroscopic quantities involved in reaction with these small particles. This is achieved by using mole as the bridge between them. Mole is related to both by the expression shown below.

  $\begin{array}{c}n\times 6.022\times {10}^{23}\text{particles}=\text{Moles}\left(n\right)\\ =\frac{\text{Mass}\left(m\right)}{\text{Molar}\text{mass}\left(M\right)}\end{array}$

Mole is referred as the substance’s amount possessing mass equal to that in $\text{12}\text{g}$ of $\text{C}$ .

Answer to Problem 39E

The mass in grams of $2.0\times {10}^{-2}$ moles of chloral hydrate is $3.308\text{g}$ .

Explanation of Solution

The given moles of chloral hydrate are $2.0\times {10}^{-2}$ mol.

The mass of chloral hydrate can be calculated by substituting its moles and molar mass in equation (2).

  $\begin{array}{c}2.0\times {10}^{-2}=\frac{m}{165.4\text{g/mol}}\\ m=2.0\times {10}^{-2}\times 165.4\text{g/mol}\\ =3.308\text{g}\end{array}$

Hence, the mass in grams of $2.0\times {10}^{-2}$ moles of chloral hydrate is $3.308\text{g}$ .

(d)

Interpretation Introduction

Interpretation: The number of chlorine atoms present in $5.0\text{g}$ of chloral hydrate is to be calculated.

Concept introduction: Mole concept makes it quite possible to deal with subatomic particles. It helps in relating the macroscopic quantities involved in reaction with these small particles. This is achieved by using mole as the bridge between them. Mole is related to both by the expression shown below.

  $\begin{array}{c}n\times 6.022\times {10}^{23}\text{particles}=\text{Moles}\left(n\right)\\ =\frac{\text{Mass}\left(m\right)}{\text{Molar}\text{mass}\left(M\right)}\end{array}$

Mole is referred as the substance’s amount possessing mass equal to that in $\text{12}\text{g}$ of $\text{C}$ .

Answer to Problem 39E

The number of chlorine atoms present in $5.0\text{g}$ of chloral hydrate is $5.46\times {10}^{22}$ .

Explanation of Solution

The given mass of chloral hydrate is $5.0\text{g}$ .

In order to calculate the number of atoms of chlorine, first calculate the moles of chloral hydrateby substituting the mass and molar mass in equation (2).

  $\begin{array}{c}n=\frac{5.0\text{g}}{165.4\text{g/mol}}\\ =3.0229\times {10}^{-2}\text{mol}\end{array}$

As per mole concept, $1$ mole of chloral hydrate contains $3.0229\times {10}^{-2}$ molecules.

Therefore, $3.0229\times {10}^{-2}$ moles of chloral hydrate contain $3.0229\times {10}^{-2}\times 6.022\times {10}^{23}$ molecules.

Hence, the number of molecules present in $\text{5}\text{.0}\text{mg}$ of chloral hydrate is $1.82\times {10}^{22}$ .

As seen from the molecular formula, $1$ molecule of chloral hydrate contains $3$ chlorine atoms.

Therefore, $1.82\times {10}^{22}$ molecules of chloral hydrate contain $3\times 1.82\times {10}^{22}$ chlorine atoms.

Hence, the atoms of chlorine are $5.46\times {10}^{22}$ atoms.

(e)

Interpretation Introduction

Interpretation: The mass of chloral hydrate that contains $1.0$ g of $\text{Cl}$ is to be calculated.

Concept introduction: Mole concept makes it quite possible to deal with subatomic particles. It helps in relating the macroscopic quantities involved in reaction with these small particles. This is achieved by using mole as the bridge between them. Mole is related to both by the expression shown below.

  $\begin{array}{c}n\times 6.022\times {10}^{23}\text{particles}=\text{Moles}\left(n\right)\\ =\frac{\text{Mass}\left(m\right)}{\text{Molar}\text{mass}\left(M\right)}\end{array}$

Mole is referred as the substance’s amount possessing mass equal to that in $\text{12}\text{g}$ of $\text{C}$ .

Answer to Problem 39E

The mass of chloral hydrate that contains $1.0$ g of $\text{Cl}$ is $1.538\text{g}$ .

Explanation of Solution

The given mass of $\text{Cl}$ is $\text{1}\text{.0}\text{g}$ .

The number of moles of $\text{Cl}$ can be calculated by substituting its mass and molar mass in equation (2).

  $\begin{array}{c}\text{Moles}\left(n\right)=\frac{\text{1}\text{.0}}{\text{35}\text{.45}\text{g/mol}}\\ =0.028\text{mol}\end{array}$

Since, each molecule of chloral hydrate contains $\text{3}\text{Cl}$ atoms. Therefore, moles of chloral hydrate are calculated by the formula shown below.

  $n_{{\text{C}}_2{\text{H}}_3{\text{Cl}}_3{\text{O}}_2}=\frac{n_{\text{Cl}}}{3}$  (3)

Substitute the moles of chlorine in equation (3).

  $\begin{array}{c}{n}_{{\text{C}}_2{\text{H}}_3{\text{Cl}}_3{\text{O}}_2}=\frac{0.028}{3}\\ =0.0093\end{array}$

The mass of chloral hydrate can be calculated by substituting its moles and molar mass in equation (2).

  $\begin{array}{c}0.0093\text{mol}=\frac{m}{165.4\text{g/mol}}\\ m=0.0093\text{mol}\times 165.4\text{g/mol}\\ =1.538\text{g}\end{array}$

Hence, the mass of chloral hydrate that contains $1.0$ g of $\text{Cl}$ is $1.538\text{g}$ .

(f)

Interpretation Introduction

Interpretation: The mass in grams of $500$ molecules of aspartame is to be calculated.

Concept introduction: Mole concept makes it quite possible to deal with subatomic particles. It helps in relating the macroscopic quantities involved in reaction with these small particles. This is achieved by using mole as the bridge between them. Mole is related to both by the expression shown below.

  $\begin{array}{c}n\times 6.022\times {10}^{23}\text{particles}=\text{Moles}\left(n\right)\\ =\frac{\text{Mass}\left(m\right)}{\text{Molar}\text{mass}\left(M\right)}\end{array}$

Mole is referred as the substance’s amount possessing mass equal to that in $\text{12}\text{g}$ of $\text{C}$ .

Answer to Problem 39E

The mass in grams of $500$ molecules of chloral hydrate is $1.37\times {10}^{-19}\text{g}$ .

Explanation of Solution

The given molecules of chloral hydrate are $500$ .

The moles of chloral hydrate can be calculated by using the formula shown below.

  $\text{Moles}\left(n\right)=\frac{\text{Number}\text{of}\text{molecules}}{\text{6}{\text{.022×10}}^{\text{23}}}$  (4)

Substitute the number of molecules in equation (4).

  $\begin{array}{c}\text{Moles}\left(n\right)=\frac{500}{\text{6}{\text{.022×10}}^{\text{23}}}\\ =8.3\times {10}^{-22}\text{mol}\end{array}$

The mass of chloral hydrate can be calculated by substituting its moles and molar mass in equation (2).

  $\begin{array}{c}8.3\times {10}^{-22}\text{mol}\text{mol}=\frac{m}{165.4\text{g/mol}}\\ m=8.3\times {10}^{-22}\text{mol}\text{mol}\times 165.4\text{g/mol}\\ =1.37\times {10}^{-19}\text{g}\end{array}$

Hence, the mass in grams of $500$ molecules of chloral hydrate is $1.37\times {10}^{-19}\text{g}$ .