Chapter 4, Problem 4.102P

4-102 Aspartame, an artificial sweetener used as a sugar substitute in some foods and beverages, has the molecular formula C₁₄H₁₈N₂O₅.

(a) How many mg of aspartame are present in 3.72 × 10²⁶ molecules of aspartame?

(b) Imagine you obtain 25.0 mL of aspartame, which is known to have a density of 1.35 g/mL. How many molecules of aspartame are present in this volume?

(c) How many hydrogen atoms are present in 1.00 mg of aspartame?

(d) Complete the skeletal structure of aspartame, where all the bonded atoms are shown but double bonds, triple bonds, and/or lone pairs are missing.

(e) Identify the various types of geometries present in each central atom of aspartame using VSEPR theory.

(f) Determine the various relative bond angles associated with each central atom of aspartame using VSEPR theory.

(g) What is the most polar bond in aspartame?

(h) Would you predict aspartame to be polar or nonpolar?

(i) Is aspartame expected to possess resonance? Explain why or why not.

(j) Consider the combustion of aspartame, which results in formation of NO₂(g) as well as other expected products. Write a balanced chemical equation for this reaction.

(k) Calculate the weight of C02(g) that can be prepared from 1.62 g of aspartame mixed with 2.11 g of oxygen gas.

Interpretation Introduction

Interpretation:

The amount of aspartame in mg present in $3.72\times 10^{26}$ molecules of aspartame should be determined.

Concept Introduction:

1 mole of a molecule contains $6.02\times 10^{23}$ molecules ( $N_A$ ). Also, molar mass of a molecule is the mass of one mole of that molecule in grams.

Answer to Problem 4.102P

  $1.82\times 10^{8}mg$

Explanation of Solution

Molar mass of aspartame

  $\begin{array}{l}C14\times 12.0=168\\ H18\times 1.0=18\\ N2\times 14=28\\ \underset{\_}{O5\times 16=80}\\ C_{14}{H}_{18}{N}_2{O}_5=294g\end{array}$

Since $6.02\times {10}^{23}$ molecules of aspartame weighs = 294 g

1 molecule of aspartame weighs = $\frac{294}{6.02\times 10^{23}}g$

Therefore, $3.72\times 10^{26}$ molecules weight = $\frac{294}{6.02\times {10}^{23}}\times 3.72\times {10}^{26}g$

  $=181.67\times 10^{3}g$

We know that,

1 g = $10^{3}mg$

Therefore,

  $\begin{array}{l}181.67\times 10^{3}g=181.67\times 10^3\times 10^{3}mg\\ =181.67\times 10^{6}mg\end{array}$

Hence, weight of $3.72\times 10^{26}$ molecules is $1.82\times 10^{8}mg$ .

Interpretation Introduction

Interpretation:

The number of molecules of aspartame present in 25.0 mL of aspartame should be determined.

Concept Introduction:

The density of a substance is equal to the given mass per volume of that substance.

1 mole of a molecule contains Avogadro number of molecules ( $6.02\times 10^{23}$ ). Also, the molar mass of a molecule is the mass of one mole of that molecule in grams.

Answer to Problem 4.102P

  $6.9\times 10^{22}molecules$

Explanation of Solution

Molar mass of aspartame

  $\begin{array}{l}C14\times 12.0=168\\ H18\times 1.0=18\\ N2\times 14=28\\ \underset{\_}{O5\times 16=80}\\ C_{14}{H}_{18}{N}_2{O}_5=294g\end{array}$

Density of aspartame (d) = $\frac{Givenmass(m)}{Volume(V)}$

  $\begin{array}{l}d=1.35g/mL(Given)\\ V=25mL(Given)\\ d=\frac{m}{V}\\ m=d\times V\\ =1.35g/mL\times 25mL\\ =33.75g\end{array}$

Since 294g of aspartame contains $6.02\times 10^{23}$ molecules,

1g of aspartame contains= $\frac{6.02\times 10^{23}}{294}$ molecules

  $33.75$ g of aspartame contains= $\frac{6.02\times 10^{23}}{294}\times 33.75$

= $6.9\times 10^{22}$ molecules

Interpretation Introduction

Interpretation:

The number of hydrogen atoms present in $1.00$ mg of aspartame should be determined.

Concept Introduction:

1 molecule of aspartame contains 18 atoms of hydrogen. 1 mole of a molecule contains Avogadro number of molecules ( $6.02\times 10^{23}$ ). Also, molar mass of a molecule is the mass of one mole of that molecule in grams.

Answer to Problem 4.102P

  $3.68\times 10^{19}atoms$

Explanation of Solution

1 mole of aspartame contains $6.02\times 10^{23}$ molecules of aspartame.

1 molecule of aspartame contains 18 atoms of hydrogen.

Therefore, 1 mole of aspartame contains $6.02\times 10^{23}\times 18$ atoms of hydrogen.

The molar mass of aspartame

  $\begin{array}{l}C14\times 12.0=168\\ H18\times 1.0=18\\ N2\times 14=28\\ \underset{\_}{O5\times 16=80}\\ C_{14}{H}_{18}{N}_2{O}_5=294g\end{array}$

294 g of aspartame contains $6.02\times 10^{23}\times 18$ atoms of hydrogen.

1g of aspartame contains= $\frac{6.02\times 10^{23}\times 18}{294}$ atoms

= $3.68\times {10}^{22}$ atoms of H

We know that,

1g = 1000mg

Therefore,

1 mg contains = $\frac{3.68\times {10}^{22}}{1000}$ atoms of H

Hence, 1 mg of aspartame contains $3.68\times {10}^{19}$ atoms of hydrogen.

Interpretation Introduction

Interpretation:

The complete skeletal structure of aspartame with double bonds, triple bonds, and/or lone pairs should be drawn.

  

Concept Introduction:

Each element has a tendency to lose, gain or share sufficient electrons to achieve an electronic configuration having eight valence electrons. This rule is also known as the octet rule.

Answer to Problem 4.102P

  

Explanation of Solution

Octet of all the atoms in aspartame should be completed. According to an octet rule, the valency of carbon is four. Thus, it can form four single bonds or two double bonds or one double bond and two single bonds or one triple bond and one single bond with other atoms to complete its octet. Nitrogen has five electrons in its valence shell so it can form three bonds with other atoms to complete its octet. Oxygen has six electrons in its valence shell so it can form two bonds with other atoms and hydrogen can only form one single bond with other atoms. So, the structure of aspartame is as follows:

  

Interpretation Introduction

Interpretation:

To identify the different types of geometries present in each central atom of aspartame using VSEPR theory.

Concept Introduction:

The geometry of a molecule can be predicted by using the valence shell electron pair repulsion (VSEPR) model. According to this model, the valence electrons of an atom participate in the formation of bonds (double, single and triple) or they can be unshared. Every combination generates a negatively charged area of electron density around the nucleus. Since like charges repel each other, the different areas of electron density around a nucleus stretch out so that each one is present at some distance from the others.

Answer to Problem 4.102P

  

Explanation of Solution

1. Each double-bonded carbon atom has three sigma bonds, therefore, it is $s{p}^2$ hybridized. The central atom that undergoes $s{p}^2$ hybridization has trigonal planar geometry.
2. Each nitrogen atom has a three-sigma bond and one lone pair of electrons, therefore, it is $s{p}^3$ hybridized. The central atom that undergoes $s{p}^3$
3. hybridization has trigonal pyramidal geometry.
4. All the oxygen atoms bonded to hydrogen or methyl group have two sigma bonds and one lone pair of electrons, therefore, it is $s{p}^2$ hybridized. The central atom that undergoes $s{p}^2$hybridization has bent geometry.
5. All the carbon atoms with all single bonds have four sigma bonds, therefore, it is $s{p}^3$ hybridized. The central atom that undergoes $s{p}^3$ hybridization has tetrahedral geometry.

Interpretation Introduction

Interpretation:

The various relative bond angles associated with each central atom of aspartame using VSEPR theory should be determined.

Concept Introduction:

The bond angle of the molecules can be predicted by using the valence shell electron pair repulsion (VSEPR) model. According to this model, the valence electrons of an atom participate in the formation of bonds (double, single and triple) or they can be unshared. Every combination generates a negatively charged area of electron density around the nucleus. Since, like charges repel each other, the different areas of electron density around a nucleus stretch out so that each one is present at some distance from the others.

Answer to Problem 4.102P

  

Explanation of Solution

1. Each double bond carbon atom having trigonal planar geometry has bond angle $120^{{\rm O}}$ .
2. Each nitrogen atom having trigonal pyramidal geometry has a bond angle $109.5^{{\rm O}}$ .
3. All the oxygen atoms bonded with hydrogen atom having bent geometry has a bond angle $109.5^{{\rm O}}$ .
4. All the carbon atoms will all single bonds having tetrahedral geometry has bond angle $109.5^{{\rm O}}$ .

Interpretation Introduction

Interpretation:

The most polar bond in aspartame should be identified.

Concept Introduction:

A molecule is polar if it has polar bonds and the centers of its partial positive and partial negative charge do not coincide.

Answer to Problem 4.102P

O−H is the most polar bond in aspartame.

Explanation of Solution

More the electronegativity difference, more is polarity of the bond. In the structure of aspartame molecule, electronegativity difference between H and O atom is the most in O−H bond therefore, O−H bond is the most polar bond in aspartame.

Interpretation Introduction

Interpretation:

Whether aspartame is polar or non-polar should be predicted.

Concept Introduction:

A molecule is polar if it has polar bonds and if the centers of and partial negative charge and partial positive charge lie at different places within the molecule.

Answer to Problem 4.102P

Aspartame is a polar molecule.

Explanation of Solution

Aspartame has polar bonds such as N−C, N−H and O−H. Hence, it has dipole moment and thus, it is a polar molecule.

Interpretation Introduction

Interpretation:

Whether aspartame possess resonance should be determined. Explain why or why not.

Concept Introduction:

According to the theory of resonance, molecules and ions are explained by writing two or more Lewis structures and by considering the real molecule as a hybrid of these structures. An individual Lewis structure is called a contributing structure. They are also sometimes referred to as resonance structures or resonance contributors.

Answer to Problem 4.102P

Yes, aspartame molecule possesses resonance.

Explanation of Solution

Aspartame molecule contains a benzene ring as a part of it and benzene ring show resonance as follows:

  

Therefore, the aspartame molecule possesses resonance in the benzene part.

Interpretation Introduction

Interpretation:

The balanced chemical equation for combustion of aspartame should be written.

Concept Introduction:

Combustion reaction is a chemical reaction in which a compound reacts with oxygen to form mainly carbon dioxide and water.

Answer to Problem 4.102P

  $C_{14}{H}_{18}{N}_2{O}_5(s)+18O_2(g)\underset{}{\overset{}{\to }}14C{O}_2(g)+9H_{2}O(g)+2N{O}_2(g)$

Explanation of Solution

Combustion of aspartame releases carbon dioxide, water and nitrogen dioxide gases.

It contains 14 carbon atoms, 18 hydrogen atoms and 2 nitrogen atoms in one aspartame molecule, so balancing the same on right hand side of the equation we get:

  $C_{14}{H}_{18}{N}_2{O}_5(s)+18O_2(g)\underset{}{\overset{}{\to }}14C{O}_2(g)+9H_{2}O(g)+2N{O}_2(g)$

Interpretation Introduction

Interpretation:

The weight of $C{O}_2(g)$ which is prepared from $1.62g$ of aspartame mixed with $2.11g$ of oxygen gas should be determined.

Concept Introduction:

Combustion reaction is a chemical reaction in which a compound reacts with oxygen to form mainly carbon dioxide and water.

Answer to Problem 4.102P

  $2.217g$

Explanation of Solution

  $\begin{array}{l}{C}_{14}{H}_{18}{N}_2{O}_5(s)+18O_2(g)\underset{}{\overset{}{\to }}14C{O}_2(g)+9H_{2}O(g)+2N{O}_2(g)\\ 294g576g\end{array}$

From the equation, we get that

1 mole of aspartame requires 18 moles of $O_2$ to give 14 moles of $C{O}_2$ gas.

Given,

Mass of aspartame = $1.62g$

Number of moles of aspartame = $\frac{1.62}{294}$

= $0.0055$ moles

Mass of oxygen = $2.11$ g

Number of moles of oxygen = $\frac{2.11}{32}=0.065$ mol

  $0.0055$ mol of aspartame will require oxygen= $18\times 0.0055$ mol

= $0.099$ mol

However, given moles of oxygen are $0.065$ mol.

Hence, oxygen is the limiting reagent here.

For $0.065$ mol of oxygen available, the number of moles of aspartame required is $0.0036$ mol.

Since 1 mol of aspartame gives 14 mol of $C{O}_2$ gas.

Therefore,

  $0.0036$ mol aspartame will give $0.0504$ mol of $C{O}_2$ gas.

Number of moles= $\frac{mass}{molarmass}$

So, the weight of $C{O}_2$ formed = number of moles of $C{O}_2$ formed × molar mass of $C{O}_2$

= $0.0504\times 44$

= $2.217$ g