Chapter 3, Problem 38E

(a)

Interpretation Introduction

Interpretation: The molar mass of aspartame is to be determined.

Concept introduction: Mole concept conveniently associates a compound’s amount (in moles) with the given mass and that substance’s molar mass.

It also helps in revealing the count of molecules that acquire certain substance’s moles. This is achieved by relating moles with Avogadro’s number as shown below.

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

Hence, mole concept aids the interconversion between various ways of representing substance’s amount like mass, moles, volume and so on.

Answer to Problem 38E

The molar mass of aspartame is $\text{294}\text{.304}\text{g/mol}$ .

Explanation of Solution

The given molecular formula of aspartame is ${\text{C}}_{\text{14}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ .

The molar mass of ${\text{C}}_{\text{14}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ can be calculated by the formula shown below.

  $M^{\prime }=n_{\text{C}}\times M_{\text{C}}+n_{\text{H}}\times M_{\text{H}}+n_{\text{N}}\times M_{\text{N}}+n_{\text{O}}\times M_{\text{O}}$  (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.
• $n_{\text{N}}$ is the number of atoms of nitrogen.
• $M_{\text{N}}$ is the molar mass of nitrogen.
• $n_{\text{O}}$ is the number of atoms of oxygen.
• $M_{\text{O}}$ is the molar mass of oxygen.
• $M^{\prime }$ is the molar mass of the ${\text{C}}_{\text{14}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ .

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 nitrogen is $\text{14}\text{.01}\text{g/mol}$ .

Substitute the known values in equation (1).

  $\begin{array}{c}{M}^{\prime }=\left(\left(14\times 12.01\right)+\left(18\times 1.008\right)+\left(2\times 14.01\right)+\left(5\times 16.00\right)\right)\text{g/mol}\\ =\left(168.14+18.144+28.02+80.0\right)\text{g/mol}\\ =\text{294}\text{.304}\text{g/mol}\end{array}$

Hence, the molar mass of aspartame is $\text{294}\text{.304}\text{g/mol}$ .

(b)

Interpretation Introduction

Interpretation: The moles of molecules present in $\text{10}\text{.0}\text{g}$ of aspartame are to be determined.

Concept introduction: Mole concept conveniently associates a compound’s amount (in moles) with the given mass and that substance’s molar mass as shown below.

  $\text{Moles}\text{of}\text{substance}=\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molar}\text{mass}\text{of}\text{substamce}}$

It also helps in revealing the count of molecules that acquire certain substance’s moles. This is achieved by relating moles with Avogadro’s number.

Hence, mole concept aids the interconversion between various ways of representing substance’s amount like mass, moles, volume and so on.

Answer to Problem 38E

The moles of molecules present in $\text{10}\text{.0}\text{g}$ of aspartame are $3.397\times {10}^{-2}\text{mol}$ .

Explanation of Solution

The moles of molecules of aspartame 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).

  $\begin{array}{c}n=\frac{10.0\text{g}}{\text{294}\text{.304}\text{g/mol}}\\ =3.397\times {10}^{-2}\text{mol}\end{array}$

Hence, the moles of molecules present in $\text{10}\text{.0}\text{g}$ of aspartame are $3.397\times {10}^{-2}\text{mol}$ .

  $$

(c)

Interpretation Introduction

Interpretation: The mass in grams of $1.56$ moles of aspartame is to be determined.

Concept introduction: Mole concept conveniently associates a compound’s amount (in moles) with the given mass and that substance’s molar mass as shown below.

  $\text{Moles}\text{of}\text{substance}=\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molar}\text{mass}\text{of}\text{substamce}}$

It also helps in revealing the count of molecules that acquire certain substance’s moles. This is achieved by relating moles with Avogadro’s number.Hence, mole concept aids the interconversion between various ways of representing substance’s amount like mass, moles, volume and so on.

Answer to Problem 38E

The mass in grams of $1.56$ moles of aspartame is $459.11\text{g}$ .

Explanation of Solution

The given moles of aspartame are $1.56$ mol.

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

  $\begin{array}{c}1.56=\frac{m}{\text{294}\text{.304}\text{g/mol}}\\ m=1.56\times \text{294}\text{.304}\text{g/mol}\\ =459.11\text{g}\end{array}$

Hence, the mass in grams of $1.56$ moles of aspartame is $459.11\text{g}$ .

(d)

Interpretation Introduction

Interpretation: The number of molecules present in $\text{5}\text{.0}\text{mg}$ of aspartame is to be determined.

Concept introduction: Mole concept conveniently associates a compound’s amount (in moles) with the given mass and that substance’s molar mass.

It also helps in revealing the count of molecules that acquire certain substance’s moles. This is achieved by relating moles with Avogadro’s number as shown below.

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

Hence, mole concept aids the interconversion between various ways of representing substance’s amount like mass, moles, volume and so on.

Answer to Problem 38E

The number of molecules present in $\text{5}\text{.0}\text{mg}$ of aspartame is $1.023\times {10}^{19}$ .

Explanation of Solution

The given mass of aspartame is $\text{5}\text{.0}\text{mg}$ .

The conversion of mg into g is shown below.

  $\begin{array}{c}1\text{mg}={10}^{-3}\text{g}\\ \text{5}\text{.0}\text{mg}=\text{5}\text{.0}\times {10}^{-3}\text{g}\end{array}$

In order to calculate the number of molecules, first calculate the moles of aspartame by substituting the mass and molar mass in equation (2).

  $\begin{array}{c}n=\frac{\text{5}\text{.0}\times {10}^{-3}\text{g}}{\text{294}\text{.304}\text{g/mol}}\\ =1.6989\times {10}^{-5}\text{mol}\end{array}$

As per mole concept, $1$ mole of aspartame contains $6.022\times {10}^{23}$ molecules.

Therefore, $1.6989\times {10}^{-5}$ moles of aspartame contain $1.6989\times {10}^{-5}\times 6.022\times {10}^{23}$ molecules.

Hence, the number of molecules present in $\text{5}\text{.0}\text{mg}$ of aspartame is $1.023\times {10}^{19}$ .

(e)

Interpretation Introduction

Interpretation: The number of atoms of nitrogen present in $\text{1}\text{.2}\text{g}$ of aspartame is to be determined.

Concept introduction: Mole concept conveniently associates a compound’s amount (in moles) with the given mass and that substance’s molar mass.

It also helps in revealing the count of molecules that acquire certain substance’s moles. This is achieved by relating moles with Avogadro’s number as shown below.

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

Hence, mole concept aids the interconversion between various ways of representing substance’s amount like mass, moles, volume and so on.

Answer to Problem 38E

The number of atoms of nitrogen present in $\text{1}\text{.2}\text{g}$ of aspartame is $4.91\times {10}^{21}$ .

Explanation of Solution

The given mass of aspartame is $\text{1}\text{.2}\text{g}$ .

In order to calculate the number of atoms of nitrogen, first calculate the moles of aspartame by substituting the mass and molar mass in equation (2).

  $\begin{array}{c}n=\frac{\text{1}\text{.2}\text{g}}{\text{294}\text{.304}\text{g/mol}}\\ =4.0774\times {10}^{-3}\text{mol}\end{array}$

As per mole concept, $1$ mole of aspartame contains $6.022\times {10}^{23}$ molecules.

Therefore, $4.0774\times {10}^{-3}$ moles of aspartame contain $4.0774\times {10}^{-3}\times 6.022\times {10}^{23}$ molecules.

Hence, the number of molecules present in $\text{1}\text{.2}\text{g}$ of aspartame is $2.455\times {10}^{21}$ .

As seen from the molecular formula, $1$ molecule of aspartame contains $2$ nitrogen atoms.

Therefore, $2.455\times {10}^{21}$ molecules of aspartame contain $2\times 2.455\times {10}^{21}$ nitrogen atoms.

Hence, the atoms of nitrogen are $4.91\times {10}^{21}$ atoms.

(f)

Interpretation Introduction

Interpretation: The mass in grams of $1.0\times {10}^9$ molecules of aspartame is to be determined.

Concept introduction: Mole concept conveniently associates a compound’s amount (in moles) with the given mass and that substance’s molar mass.

It also helps in revealing the count of molecules that acquire certain substance’s moles. This is achieved by relating moles with Avogadro’s number as shown below.

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

Hence, mole concept aids the interconversion between various ways of representing substance’s amount like mass, moles, volume and so on.

Answer to Problem 38E

The mass in grams of $1.0\times {10}^9$ molecules of aspartame is $4.8854\times {10}^{-13}\text{g}$ .

Explanation of Solution

The given molecules of aspartame is $1.0\times {10}^9$ .

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

  $\text{Moles}\left(n\right)=\frac{\text{Number}\text{of}\text{molecules}}{\text{6}{\text{.022×10}}^{\text{23}}}$  (3)

Substitute the number of molecules in equation (3).

  $\begin{array}{c}\text{Moles}\left(n\right)=\frac{1.0\times {10}^9}{\text{6}{\text{.022×10}}^{\text{23}}}\\ =1.66\times {10}^{-15}\text{mol}\end{array}$

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

  $\begin{array}{c}1.66\times {10}^{-15}\text{mol}=\frac{m}{\text{294}\text{.304}\text{g/mol}}\\ m=1.66\times {10}^{-15}\text{mol}\times \text{294}\text{.304}\text{g/mol}\\ =4.8854\times {10}^{-13}\text{g}\end{array}$

Hence, the mass in grams of $1.0\times {10}^9$ molecules of aspartame is $4.8854\times {10}^{-13}\text{g}$ .

(g)

Interpretation Introduction

Interpretation: The mass in grams of $1$ molecule of aspartame is to be determined.

Concept introduction: Mole concept conveniently associates a compound’s amount (in moles) with the given mass and that substance’s molar mass.

It also helps in revealing the count of molecules that acquire certain substance’s moles. This is achieved by relating moles with Avogadro’s number as shown below.

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

Hence, mole concept aids the interconversion between various ways of representing substance’s amount like mass, moles, volume and so on.

Answer to Problem 38E

The mass in grams of $1$ molecule of aspartame is $4.8854\times {10}^{-22}\text{g}$ .

Explanation of Solution

The given molecules of aspartame is $1$ .

Substitute the number of molecules in equation (3) to find the moles of aspartame.

  $\begin{array}{c}\text{Moles}\left(n\right)=\frac{1.0}{\text{6}{\text{.022×10}}^{\text{23}}}\\ =1.66\times {10}^{-24}\text{mol}\end{array}$

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

  $\begin{array}{c}1.66\times {10}^{-24}\text{mol}=\frac{m}{\text{294}\text{.304}\text{g/mol}}\\ m=1.66\times {10}^{-24}\text{mol}\times \text{294}\text{.304}\text{g/mol}\\ =4.8854\times {10}^{-22}\text{g}\end{array}$

Hence, the mass in grams of $1$ molecule of aspartame is $4.8854\times {10}^{-22}\text{g}$ .