Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 10, Problem 101QRT

(a)

Interpretation Introduction

Interpretation:

The empirical formula of the given hydrocarbon has to be determined.

Concept Introduction:

The hydrocarbon compounds are compounds which contains only carbon and hydrogen atoms.  The combustion of hydrocarbon compounds produces large amount of heat.  Combustion reaction of a hydrocarbon results in the formation of carbon dioxide and water.

(a)

Expert Solution
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Answer to Problem 101QRT

The empirical formula of given hydrocarbon is CH2.

Explanation of Solution

The mass of carbon dioxide produced is 5.287g.

The mass of water produced is 2.164g.

The molar mass of carbon dioxide is 44.01gmol1.

The molar mass of water is 18.02gmol1.

Use the expression to calculate number of moles.

  Numberofmoles=GivenmassMolarmass

Substitute 5.287g for mass of carbon dioxide and 44.01gmol1 for molar mass.

  Numberofmolesof carbondioxide=5.287g44.01gmol1=0.12mol

Therefore, the number of moles of carbon dioxide is 0.12mol.

Consider the combustion reaction of hydrocarbon as follows.

  CxHy+(x+y/4)O2xCO2+(y/2)H2O

Here, x is number of moles of carbon atoms and y is number of moles of hydrogen atoms.

When combustion of hydrocarbon takes place, the carbon atoms present in hydrocarbon gets converted to carbon dioxide molecule.  One molecule of carbon dioxide contains one carbon atom.  Therefore, the number of moles of carbon atoms present in hydrocarbon is equal to the moles of carbon dioxide.

Therefore, moles of carbon atom in hydrocarbon is 0.12mol.

Substitute 2.164g for mass of water and 18.02gmol1 for molar mass.

  Numberofmoleswater=2.164g18.02gmol1=0.12mol

Therefore, number of moles of water is 0.12mol.

When the combustion of hydrocarbon takes place the hydrogen atoms present in hydrocarbon gets converted into hydrogen atoms of water molecule.  One molecule of water contains two hydrogen atoms.  Therefore, the number of moles of hydrogen atom present in hydrocarbon is twice the number of moles of water.

Therefore, the number of moles of hydrogen atoms is 0.24mol.

The smallest number of moles is 0.12mol.

Use the expression to calculate mole ratio.

  Moleratio=NumberofmolesSmallestmole

Substitute 0.12mol for number of moles of carbon atom and 0.12mol for smallest mole.

  MoleratioofC=0.12mol0.12mol=1

Substitute 0.24mol for number of moles of hydrogen atom and 0.12mol for smallest mole.

  MoleratioofH=0.24mol0.12mol=2

Therefore, the empirical formula of given hydrocarbon is CH2.

(b)

Interpretation Introduction

Interpretation:

The given hydrocarbon is an alkane or alkene has to be identified.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 101QRT

The given hydrocarbon compound is an alkene.

Explanation of Solution

The general formula of alkane is CnH2n+2 and for alkene is CnH2n.  The empirical formula of given hydrocarbon is CH2.  Therefore, the given hydrocarbon is an alkene.

(c)

Interpretation Introduction

Interpretation:

The Lewis structure for given hydrocarbon has to be drawn.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 101QRT

The Lewis structure for given hydrocarbon is as follows.

Chemistry: The Molecular Science, Chapter 10, Problem 101QRT , additional homework tip  1

Another possible structure is as follows.

Chemistry: The Molecular Science, Chapter 10, Problem 101QRT , additional homework tip  2

Explanation of Solution

The given hydrocarbon has empirical formula CH2.  The molecular formula of the alkene can be C2H4 or C3H6.  The Lewis structure of these two possible hydrocarbons is as follows.

Chemistry: The Molecular Science, Chapter 10, Problem 101QRT , additional homework tip  3

Figure 1

Chemistry: The Molecular Science, Chapter 10, Problem 101QRT , additional homework tip  4

Figure 2

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Chapter 10 Solutions

Chemistry: The Molecular Science

Ch. 10.4 - Prob. 10.8CECh. 10.4 - Prob. 10.9CECh. 10.4 - Prob. 10.10CECh. 10.4 - Prob. 10.11ECh. 10.5 - Prob. 10.12ECh. 10.5 - Prob. 10.4PSPCh. 10.5 - Prob. 10.13ECh. 10.6 - Prob. 10.14CECh. 10.6 - Prob. 10.5PSPCh. 10.6 - Prob. 10.6PSPCh. 10.6 - Prob. 10.7PSPCh. 10.6 - Prob. 10.8PSPCh. 10.6 - Prob. 10.9PSPCh. 10.6 - Prob. 10.15CECh. 10.6 - Prob. 10.16ECh. 10.7 - Prob. 10.17CECh. 10.7 - Prob. 10.18CECh. 10.7 - Prob. 10.19CECh. 10.7 - Prob. 10.20CECh. 10.7 - Prob. 10.10PSPCh. 10.7 - Prob. 10.21ECh. 10 - Prob. ISPCh. 10 - Prob. IISPCh. 10 - Prob. IIISPCh. 10 - Prob. 1QRTCh. 10 - Prob. 2QRTCh. 10 - Prob. 3QRTCh. 10 - Prob. 4QRTCh. 10 - Prob. 5QRTCh. 10 - Prob. 6QRTCh. 10 - Prob. 7QRTCh. 10 - Give two reasons why ethylene glycol has a higher...Ch. 10 - Prob. 9QRTCh. 10 - Prob. 10QRTCh. 10 - Prob. 11QRTCh. 10 - Prob. 12QRTCh. 10 - Prob. 13QRTCh. 10 - Prob. 14QRTCh. 10 - Prob. 15QRTCh. 10 - Prob. 16QRTCh. 10 - Prob. 17QRTCh. 10 - Prob. 18QRTCh. 10 - Prob. 19QRTCh. 10 - Prob. 20QRTCh. 10 - Prob. 21QRTCh. 10 - Prob. 22QRTCh. 10 - Prob. 23QRTCh. 10 - Prob. 24QRTCh. 10 - Prob. 25QRTCh. 10 - Prob. 26QRTCh. 10 - Prob. 27QRTCh. 10 - Prob. 28QRTCh. 10 - Prob. 29QRTCh. 10 - Prob. 30QRTCh. 10 - Prob. 31QRTCh. 10 - Prob. 32QRTCh. 10 - Prob. 33QRTCh. 10 - Prob. 34QRTCh. 10 - Prob. 35QRTCh. 10 - Prob. 36QRTCh. 10 - Prob. 37QRTCh. 10 - Prob. 38QRTCh. 10 - Prob. 39QRTCh. 10 - Prob. 40QRTCh. 10 - Prob. 41QRTCh. 10 - Prob. 42QRTCh. 10 - Prob. 43QRTCh. 10 - Prob. 44QRTCh. 10 - Prob. 45QRTCh. 10 - Prob. 46QRTCh. 10 - Prob. 47QRTCh. 10 - Beeswax contains this compound: Identify what...Ch. 10 - Prob. 49QRTCh. 10 - Prob. 50QRTCh. 10 - Prob. 51QRTCh. 10 - Prob. 52QRTCh. 10 - Prob. 53QRTCh. 10 - Prob. 54QRTCh. 10 - Prob. 55QRTCh. 10 - Prob. 56QRTCh. 10 - Prob. 57QRTCh. 10 - Prob. 58QRTCh. 10 - Prob. 59QRTCh. 10 - Prob. 60QRTCh. 10 - Prob. 61QRTCh. 10 - Prob. 62QRTCh. 10 - Prob. 63QRTCh. 10 - Prob. 64QRTCh. 10 - Prob. 65QRTCh. 10 - Prob. 66QRTCh. 10 - Prob. 67QRTCh. 10 - Prob. 68QRTCh. 10 - Prob. 69QRTCh. 10 - Prob. 70QRTCh. 10 - Prob. 71QRTCh. 10 - Prob. 72QRTCh. 10 - Prob. 73QRTCh. 10 - Prob. 74QRTCh. 10 - Prob. 75QRTCh. 10 - Prob. 76QRTCh. 10 - Prob. 77QRTCh. 10 - Prob. 78QRTCh. 10 - Prob. 79QRTCh. 10 - Identify and name all the functional groups in...Ch. 10 - Prob. 81QRTCh. 10 - Prob. 82QRTCh. 10 - Prob. 83QRTCh. 10 - Prob. 84QRTCh. 10 - Prob. 85QRTCh. 10 - Prob. 86QRTCh. 10 - Prob. 87QRTCh. 10 - Prob. 88QRTCh. 10 - Prob. 89QRTCh. 10 - Prob. 90QRTCh. 10 - Prob. 91QRTCh. 10 - Prob. 92QRTCh. 10 - Prob. 93QRTCh. 10 - Prob. 94QRTCh. 10 - Prob. 95QRTCh. 10 - Prob. 96QRTCh. 10 - Assume that a car burns pure octane. C8H18 (d =...Ch. 10 - Prob. 98QRTCh. 10 - Prob. 99QRTCh. 10 - Prob. 100QRTCh. 10 - Prob. 101QRTCh. 10 - Prob. 102QRTCh. 10 - Prob. 103QRTCh. 10 - Prob. 104QRTCh. 10 - Prob. 105QRTCh. 10 - Prob. 106QRTCh. 10 - Prob. 107QRTCh. 10 - Prob. 108QRTCh. 10 - Prob. 109QRTCh. 10 - Prob. 110QRTCh. 10 - Prob. 111QRTCh. 10 - Prob. 112QRTCh. 10 - Prob. 113QRTCh. 10 - Prob. 114QRTCh. 10 - Prob. 115QRTCh. 10 - Prob. 116QRTCh. 10 - Prob. 118QRTCh. 10 - Prob. 119QRTCh. 10 - Prob. 120QRTCh. 10 - Prob. 121QRTCh. 10 - Prob. 122QRTCh. 10 - Prob. 123QRTCh. 10 - Prob. 124QRTCh. 10 - Prob. 125QRTCh. 10 - Prob. 126QRTCh. 10 - Prob. 127QRTCh. 10 - Prob. 10.ACPCh. 10 - Prob. 10.BCPCh. 10 - Prob. 10.CCP
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