Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
11th Edition
ISBN: 9781285869759
Author: Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher: Cengage Learning
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Textbook Question
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.

Chapter 4, Problem 4.102P, 4-102 Aspartame, an artificial sweetener used as a sugar substitute in some foods and beverages, has

(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.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The amount of aspartame in mg present in 3.72 ×1026 molecules of aspartame should be determined.

Concept Introduction:

1 mole of a molecule contains 6.02×1023 molecules ( NA ). Also, molar mass of a molecule is the mass of one mole of that molecule in grams.

Answer to Problem 4.102P

  1.82 ×108mg

Explanation of Solution

Molar mass of aspartame

  C 14 × 12.0 = 168H 18 ×1.0=18N 2×14=28O5×16=80_C14H18N2O5=294g

Since 6.02×1023 molecules of aspartame weighs = 294 g

1 molecule of aspartame weighs = 2946.02×1023g

Therefore, 3.72 ×1026 molecules weight = 2946.02×1023×3.72×1026g

  =181.67×103g

We know that,

1 g = 103mg

Therefore,

  181.67×103g=181.67×103×103mg=181.67×106mg

Hence, weight of 3.72×1026 molecules is 1.82×108mg .

(b)

Expert Solution
Check Mark
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 ( 6.02×1023 ). Also, the molar mass of a molecule is the mass of one mole of that molecule in grams.

Answer to Problem 4.102P

  6.9×1022molecules

Explanation of Solution

Molar mass of aspartame

  C 14 × 12.0 = 168H 18 ×1.0=18N 2×14=28O5×16=80_C14H18N2O5=294g

Density of aspartame (d) = Givenmass(m)Volume(V)

  d=1.35g/mL(Given)V=25mL(Given)d=mVm=d×V=1.35g/mL×25mL=33.75g

Since 294g of aspartame contains 6.02×1023 molecules,

1g of aspartame contains= 6.02×1023294 molecules

  33.75 g of aspartame contains= 6.02×1023294×33.75

= 6.9×1022 molecules

(c)

Expert Solution
Check Mark
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 ( 6.02×1023 ). Also, molar mass of a molecule is the mass of one mole of that molecule in grams.

Answer to Problem 4.102P

  3.68×1019atoms

Explanation of Solution

1 mole of aspartame contains 6.02×1023 molecules of aspartame.

1 molecule of aspartame contains 18 atoms of hydrogen.

Therefore, 1 mole of aspartame contains 6.02×1023×18 atoms of hydrogen.

The molar mass of aspartame

  C 14 × 12.0 = 168H 18 ×1.0=18N 2×14=28O5×16=80_C14H18N2O5=294g

294 g of aspartame contains 6.02×1023×18 atoms of hydrogen.

1g of aspartame contains= 6.02×1023×18294 atoms

= 3.68×1022 atoms of H

We know that,

1g = 1000mg

Therefore,

1 mg contains = 3.68×10221000 atoms of H

Hence, 1 mg of aspartame contains 3.68×1019 atoms of hydrogen.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

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

  Introduction to General, Organic and Biochemistry, Chapter 4, Problem 4.102P , additional homework tip  1

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

  Introduction to General, Organic and Biochemistry, Chapter 4, Problem 4.102P , additional homework tip  2

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:

  Introduction to General, Organic and Biochemistry, Chapter 4, Problem 4.102P , additional homework tip  3

(e)

Expert Solution
Check Mark
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

  Introduction to General, Organic and Biochemistry, Chapter 4, Problem 4.102P , additional homework tip  4

Explanation of Solution

  1. Each double-bonded carbon atom has three sigma bonds, therefore, it is sp2 hybridized. The central atom that undergoes sp2 hybridization has trigonal planar geometry.
  2. Each nitrogen atom has a three-sigma bond and one lone pair of electrons, therefore, it is sp3 hybridized. The central atom that undergoes sp3
  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 sp2 hybridized. The central atom that undergoes sp2hybridization has bent geometry.
  5. All the carbon atoms with all single bonds have four sigma bonds, therefore, it is sp3 hybridized. The central atom that undergoes sp3 hybridization has tetrahedral geometry.

(f)

Expert Solution
Check Mark
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

  Introduction to General, Organic and Biochemistry, Chapter 4, Problem 4.102P , additional homework tip  5

Explanation of Solution

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

(g)

Expert Solution
Check Mark
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.

(h)

Expert Solution
Check Mark
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.

(i)

Expert Solution
Check Mark
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:

  Introduction to General, Organic and Biochemistry, Chapter 4, Problem 4.102P , additional homework tip  6

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

(j)

Expert Solution
Check Mark
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

  C14H18N2O5(s)+18O2(g)14CO2(g)+9H2O(g)+2NO2(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:

  C14H18N2O5(s)+18O2(g)14CO2(g)+9H2O(g)+2NO2(g)

(k)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The weight of CO2(g) which is prepared from 1.62 g of aspartame mixed with 2.11 g 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

  C14H18N2O5(s)+18O2(g)14CO2(g)+9H2O(g)+2NO2(g)294g576g

From the equation, we get that

1 mole of aspartame requires 18 moles of O2 to give 14 moles of CO2 gas.

Given,

Mass of aspartame = 1.62g

Number of moles of aspartame = 1.62294

= 0.0055 moles

Mass of oxygen = 2.11 g

Number of moles of oxygen = 2.1132=0.065 mol

  0.0055 mol of aspartame will require oxygen= 18×0.0055 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 CO2 gas.

Therefore,

  0.0036 mol aspartame will give 0.0504 mol of CO2 gas.

Number of moles= massmolarmass

So, the weight of CO2 formed = number of moles of CO2 formed × molar mass of CO2

= 0.0504×44

= 2.217 g

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Chapter 4 Solutions

Introduction to General, Organic and Biochemistry

Ch. 4.6 - Prob. 4.11PCh. 4.7 - Prob. 4.12PCh. 4.7 - Prob. 4.13PCh. 4.7 - Problem 4-14 Ethanol is produced industrially by...Ch. 4.7 - Prob. 4.15PCh. 4.7 - Prob. 4.16PCh. 4 - 4-17 Balance each equation.Ch. 4 - 4-18 Balance each equation.Ch. 4 - Prob. 4.19PCh. 4 - 4-20 Calcium oxide is prepared by heating...Ch. 4 - 4-21 The brilliant white light in some firework...Ch. 4 - Prob. 4.22PCh. 4 - 4-23 When solid carbon burns in a limited supply...Ch. 4 - Prob. 4.24PCh. 4 - 4-25 In the chemical test for arsenic, the gas...Ch. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - 4-28 Answer true or false. (a) A net ionic...Ch. 4 - 4-29 Balance these net ionic equations. (a)...Ch. 4 - 4-30 In the equation (a) Identify the spectator...Ch. 4 - 4-31 Predict whether a precipitate will form when...Ch. 4 - 4-32 When a solution of ammonium chloride is added...Ch. 4 - 4-33 When a solution of hydrochloric acid, HCl, is...Ch. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - 4-36 Using the solubility generalizations given in...Ch. 4 - 4-37 Answer true or false. (a) When a substance is...Ch. 4 - Prob. 4.38PCh. 4 - Prob. 4.39PCh. 4 - Prob. 4.40PCh. 4 - Prob. 4.41PCh. 4 - 4-42 Calculate the formula weight of: (a) KCl (b)...Ch. 4 - 4-43 Calculate the molecular weight of: (a)...Ch. 4 - 4-44 Answer true or false. (a) The mole is a...Ch. 4 - 4-45 Calculate the number of moles in: (a) 32 g of...Ch. 4 - 4-46 Calculate the number of grams in: (a) 1.77...Ch. 4 - 4-47 Calculate the number of moles of: (a) O atoms...Ch. 4 - 4-48 Calculate the number of moles of: (a) S2-...Ch. 4 - 4-49 Calculate the number of: (a) nitrogen atoms...Ch. 4 - 4-50 How many molecules are in each of the...Ch. 4 - 4-51 What is the mass in grams of each number of...Ch. 4 - 4-52 The molecular weight of hemoglobin is about...Ch. 4 - 4-53 A typical deposit of cholesterol, C27H46O, in...Ch. 4 - 4-54 Answer true or false. (a) Stoichiometry is...Ch. 4 - 4-55 For the reaction: (a) How many moles of N2...Ch. 4 - 4-56 Magnesium reacts with sulfuric acid according...Ch. 4 - 4-57 Chloroform, CHCl3, is prepared industrially...Ch. 4 - 4-58 At one time, acetaldehyde was prepared...Ch. 4 - 4-59 Chlorine dioxide, ClO2, is used for bleaching...Ch. 4 - 4-60 Ethanol, C2H6O, is added to gasoline to...Ch. 4 - 4-61 In photosynthesis, green plants convert CO2...Ch. 4 - 4-62 Iron ore is converted to iron by heating it...Ch. 4 - Prob. 4.63PCh. 4 - 4-64 Aspirin is made by the reaction of salicylic...Ch. 4 - 4-65 Suppose the preparation of aspirin from...Ch. 4 - 4-66 Benzene reacts with bromine to produce...Ch. 4 - 4-67 Ethyl chloride is prepared by the reaction of...Ch. 4 - 4-68 Diethyl ether is made from ethanol according...Ch. 4 - Prob. 4.69PCh. 4 - Prob. 4.70PCh. 4 - 4-71 Which of these reactions are exothermic, and...Ch. 4 - Prob. 4.72PCh. 4 - Prob. 4.73PCh. 4 - Prob. 4.74PCh. 4 - Prob. 4.75PCh. 4 - Prob. 4.76PCh. 4 - 4-77 To convert 1 mol of iron(III) oxide to its...Ch. 4 - 4-78 (Chemical Connections 4A) How does fluoride...Ch. 4 - Prob. 4.79PCh. 4 - Prob. 4.80PCh. 4 - 4-81 (Chemical Connections 4C) Balance the lithium...Ch. 4 - 4-82 When gaseous dinitrogen pentoxide, N2O5, is...Ch. 4 - Prob. 4.83PCh. 4 - Prob. 4.84PCh. 4 - Prob. 4.85PCh. 4 - 4-86 When an aqueous solution of Na3PO4 is added...Ch. 4 - Prob. 4.87PCh. 4 - 4-88 Chlorophyll, the compound responsible for the...Ch. 4 - 4-89 If 7.0 kg of is added to 11.0 kg of to form...Ch. 4 - 4-90 Lead(lI) nitrate and aluminum chloride react...Ch. 4 - 4-91 Assume that the average red blood cell has a...Ch. 4 - 4-92 Reaction of pentane, C5H12, with oxygen, O2,...Ch. 4 - 4-93 Ammonia is prepared industrially by the...Ch. 4 - 4-94 2,3,7,8-Tetrachlorodibenzo-p-dioxin (TCDD) is...Ch. 4 - Prob. 4.95PCh. 4 - Prob. 4.96PCh. 4 - Prob. 4.97PCh. 4 - Prob. 4.98PCh. 4 - Prob. 4.99PCh. 4 - Prob. 4.100PCh. 4 - Prob. 4.101PCh. 4 - 4-102 Aspartame, an artificial sweetener used as a...Ch. 4 - 4-103 Caffeine, a central nervous system...
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