EBK CHEMISTRY: AN ATOMS FIRST APPROACH
EBK CHEMISTRY: AN ATOMS FIRST APPROACH
2nd Edition
ISBN: 8220100552236
Author: ZUMDAHL
Publisher: CENGAGE L
Question
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Chapter 21, Problem 160IP

(a)

Interpretation Introduction

Interpretation: The empirical formula and the molecular formula of the given helicene are to be calculated. The balanced chemical reaction for the combustion of helicene is to be stated.

Concept introduction: Helicines are defined as the polycyclic aromatic compounds in which the aromatic ring is annulated to provide helically shaped molecule. The molecular formula of any organic compound is determined by using empirical formula when the percent of each element is given in the compound.

To determine: The empirical formula of the given helicene.

(a)

Expert Solution
Check Mark

Answer to Problem 160IP

Answer

The empirical formula of the given helicene is C5H3 .

Explanation of Solution

Explanation

The empirical formula of the given helicene is C5H3 .

Helicene is an aromatic compound so it must contain carbon and hydrogen atoms.

Given

Weight of CO2 is 0.5063 g .

Weight of compound is 0.1450 g .

The weight of carbon is calculated by using the formula,

Weight of carbon=Weight of CO2×Molecular weight of carbonMolecular weightof CO2

The molar mass of carbon is 12g/mol .

The molar mass of carbon dioxide is 44g/mol .

Substitute the value of weight of CO2 , molecular weight of carbon and CO2 to calculate the weight of carbon in the above formula.

Weight of carbon=Weight of CO2×Molecular weight of carbonMolecular weightof CO2=0.5063 g×12 g/mol44 g/mol=0.1380 g

The weight of carbon is 0.1380 g .

Therefore the weight of hydrogen is calculated as,

Weight of hygrogen=Weight of compoundWeight of carbon=0.1450 g0.1380 g=0.007 g

Therefore the number of moles of carbon and hydrogen is calculated by the formula,

Numberofmoles=GivenmassMolarmass

Substitute the value of the given mass and the molar mass, to calculate the number of moles of carbon and hydrogen in the above equation.

For carbon,

Numberofmolesofcarbon=GivenmassMolarmass=0.1380g12 g/mol=0.0115 mol

For hydrogen,

Numberofmolesofhydrogen=GivenmassMolarmass=0.007g1g/mol=0.007mol

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent.

For carbon (C) ,

C=0.0115mol0.007mol=1.642

For hydrogen (H) ,

H=0.007mol0.007mol=1

By multiplying each with 3 we get the whole number as,

For carbon

1.642×3=4.9265 .

For hydrogen

1×3=3 .

Hence, the empirical formula of the compound is C5H3 .

(b)

Interpretation Introduction

Interpretation: The empirical formula and the molecular formula of the given helicene are to be calculated. The balanced chemical reaction for the combustion of helicene is to be stated.

Concept introduction: Helicines are defined as the polycyclic aromatic compounds in which the aromatic ring is annulated to provide helically shaped molecule. The molecular formula of any organic compound is determined by using empirical formula when the percent of each element is given in the compound.

To determine: The molecular formula of the given helicene.

(b)

Expert Solution
Check Mark

Answer to Problem 160IP

Answer

The molecular formula of the given helicene is C20H12 .

Explanation of Solution

Explanation

The molecular formula of the given helicene is C20H12 .

Given

Molality is 0.0175 m .

Weight of solvent is 12.5 g .

Weight of solute is 0.0938 g .

Therefore the molecular weight of solute is calculated by the given expression.

Molality(m)=Weight of solute ×1000Molecular Weight of solute ×Weight of solvent

Substitute the values of weight of solute, molality and weight of solvent in the above expression.

Molality(m)=Weight of solute ×1000Molecular Weight of solute ×Weight of solventMolecular Weight=Weight ofsolute×1000Molality×Weight ofsolvent=0.0938 g×10000.0175 m×12.5 g=248.8 g

Now, The empirical mass of C5H3 is 63 g . Then number of atoms ( n ) is calculated by using the formula,,

n=Molecular weightEmpirical weight

Substitute the value of molecular and empirical weight in the above expression.

n=Molecular weightEmpirical weight=248.863=4

Therefore the molecular formula is calculated as,

Molecular formula=(Empirical formula)n=(C5H3)4=C20H12

Hence, the molecular formula is C20H12

(c)

Interpretation Introduction

Interpretation: The empirical formula and the molecular formula of the given helicene are to be calculated. The balanced chemical reaction for the combustion of helicene is to be stated.

Concept introduction: Helicines are defined as the polycyclic aromatic compounds in which the aromatic ring is annulated to provide helically shaped molecule. The molecular formula of any organic compound is determined by using empirical formula when the percent of each element is given in the compound.

To determine: The balanced chemical reaction for the combustion of helicene.

(c)

Expert Solution
Check Mark

Answer to Problem 160IP

Answer

The balanced chemical reaction for the combustion of helicene is,

C20H12+ 23O220CO2+ 6H2O .

Explanation of Solution

Explanation

The balanced chemical reaction for the combustion of helicene is shown below.

The balanced reaction for the combustion of helicene is,

C20H12+ 23O220CO2+ 6H2O

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Chapter 21 Solutions

EBK CHEMISTRY: AN ATOMS FIRST APPROACH

Ch. 21 - Prob. 11RQCh. 21 - Prob. 12RQCh. 21 - Prob. 1QCh. 21 - Prob. 2QCh. 21 - What is wrong with the following names? Give the...Ch. 21 - Prob. 4QCh. 21 - Prob. 5QCh. 21 - Prob. 6QCh. 21 - Prob. 7QCh. 21 - Prob. 8QCh. 21 - Prob. 9QCh. 21 - Prob. 10QCh. 21 - Prob. 11QCh. 21 - Prob. 12QCh. 21 - Prob. 13ECh. 21 - Prob. 14ECh. 21 - Draw all the structural isomers for C8H18 that...Ch. 21 - Draw all the structural isomers for C8H18 that...Ch. 21 - Prob. 17ECh. 21 - Prob. 18ECh. 21 - Draw the structural formula for each of the...Ch. 21 - Prob. 20ECh. 21 - Prob. 21ECh. 21 - Prob. 22ECh. 21 - Prob. 23ECh. 21 - Prob. 24ECh. 21 - Name each of the following alkenes. a. CH2 = CH ...Ch. 21 - Name each of the following alkenes or alkynes. a....Ch. 21 - Prob. 27ECh. 21 - Prob. 28ECh. 21 - Prob. 29ECh. 21 - Prob. 30ECh. 21 - Name each of the following. a. b. CH3CH2CH2CCl3 c....Ch. 21 - Prob. 32ECh. 21 - There is only one compound that is named...Ch. 21 - Prob. 34ECh. 21 - Prob. 35ECh. 21 - Prob. 36ECh. 21 - Prob. 37ECh. 21 - Prob. 38ECh. 21 - Prob. 39ECh. 21 - Prob. 40ECh. 21 - Draw all structural and geometrical (cistrans)...Ch. 21 - Prob. 42ECh. 21 - Prob. 43ECh. 21 - Prob. 44ECh. 21 - If one hydrogen in a hydrocarbon is replaced by a...Ch. 21 - There are three isomers of dichlorobenzene, one of...Ch. 21 - Prob. 47ECh. 21 - Prob. 48ECh. 21 - Prob. 49ECh. 21 - Minoxidil (C9H15N5O) is a compound produced by...Ch. 21 - Prob. 51ECh. 21 - Prob. 52ECh. 21 - Name all the alcohols that have the formula...Ch. 21 - Prob. 54ECh. 21 - Prob. 55ECh. 21 - Prob. 56ECh. 21 - Prob. 57ECh. 21 - Prob. 58ECh. 21 - Prob. 59ECh. 21 - Prob. 60ECh. 21 - Prob. 61ECh. 21 - Prob. 62ECh. 21 - Prob. 63ECh. 21 - Prob. 64ECh. 21 - Prob. 65ECh. 21 - Prob. 66ECh. 21 - Prob. 67ECh. 21 - Prob. 68ECh. 21 - Prob. 69ECh. 21 - Complete the following reactions. a. CH3CO2H +...Ch. 21 - Prob. 71ECh. 21 - Prob. 72ECh. 21 - Prob. 73ECh. 21 - Prob. 74ECh. 21 - Prob. 75ECh. 21 - The polyester formed from lactic acid, is used for...Ch. 21 - Prob. 77ECh. 21 - Prob. 78ECh. 21 - Prob. 79ECh. 21 - Prob. 80ECh. 21 - Prob. 81ECh. 21 - Prob. 82ECh. 21 - Prob. 83ECh. 21 - Prob. 84ECh. 21 - Prob. 85ECh. 21 - Prob. 86ECh. 21 - Prob. 87ECh. 21 - Prob. 88ECh. 21 - Prob. 89ECh. 21 - Prob. 90ECh. 21 - Prob. 91ECh. 21 - Prob. 92ECh. 21 - Prob. 93ECh. 21 - Prob. 94ECh. 21 - Prob. 95ECh. 21 - Prob. 96ECh. 21 - Prob. 97ECh. 21 - Prob. 98ECh. 21 - Prob. 99ECh. 21 - Prob. 100ECh. 21 - Prob. 101ECh. 21 - Prob. 102ECh. 21 - Prob. 103ECh. 21 - Prob. 104ECh. 21 - Prob. 105ECh. 21 - Prob. 106ECh. 21 - Which base will hydrogen-bond with uracil within...Ch. 21 - Prob. 108ECh. 21 - The base sequences in mRNA that code for certain...Ch. 21 - Prob. 110ECh. 21 - Prob. 111AECh. 21 - Prob. 112AECh. 21 - Prob. 113AECh. 21 - Prob. 114AECh. 21 - Prob. 115AECh. 21 - Prob. 116AECh. 21 - Prob. 117AECh. 21 - Prob. 118AECh. 21 - Prob. 119AECh. 21 - Prob. 120AECh. 21 - Prob. 121AECh. 21 - Prob. 122AECh. 21 - Prob. 123AECh. 21 - Prob. 124AECh. 21 - Prob. 125AECh. 21 - Prob. 126AECh. 21 - Prob. 127AECh. 21 - Prob. 128AECh. 21 - Prob. 129AECh. 21 - Prob. 130AECh. 21 - Prob. 131AECh. 21 - Prob. 132AECh. 21 - Prob. 133AECh. 21 - Prob. 134AECh. 21 - When heat is added to proteins, the hydrogen...Ch. 21 - Prob. 136AECh. 21 - Prob. 137CWPCh. 21 - Prob. 138CWPCh. 21 - Prob. 139CWPCh. 21 - Name each of the following alkenes and alkynes. a....Ch. 21 - a. Name each of the following alcohols. b. Name...Ch. 21 - Prob. 142CWPCh. 21 - Prob. 143CWPCh. 21 - Prob. 144CWPCh. 21 - Prob. 145CPCh. 21 - Prob. 146CPCh. 21 - Prob. 147CPCh. 21 - Prob. 148CPCh. 21 - Prob. 149CPCh. 21 - Prob. 150CPCh. 21 - Prob. 151CPCh. 21 - Prob. 152CPCh. 21 - Prob. 153CPCh. 21 - Prob. 154CPCh. 21 - Stretch a rubber band while holding it gently to...Ch. 21 - Alcohols are very useful starting materials for...Ch. 21 - Prob. 157CPCh. 21 - Prob. 158CPCh. 21 - Prob. 159IPCh. 21 - Prob. 160IPCh. 21 - Prob. 161MPCh. 21 - Prob. 162MP
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