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 102QRT

(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
Check Mark

Answer to Problem 102QRT

The empirical formula of given hydrocarbon is C7H16.

Explanation of Solution

The mass of carbon dioxide produced is 7.720g.

The mass of water produced is 3.612g.

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 7.720g for mass of carbon dioxide and 44.01gmol1 for molar mass.

  Numberofmolescarbondioxide=7.720g44.01gmol1=0.175mol

Therefore, moles of carbon dioxide is 0.175mol.

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 moles of carbon atoms present in hydrocarbon is equal to the moles of carbon dioxide.

Therefore, number of moles of carbon atoms in hydrocarbon is 0.175mol.

Substitute 3.612g for mass of water and 18.02gmol1 for molar mass.

  Numberofmoleswater=3.612g18.02gmol1=0.2mol

Therefore, number of moles of water is 0.2mol.

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 0.4mol.

The smallest moles is 0.175mol.

Use the expression to calculate mole ratio.

  Moleratio=NumberofmolesSmallestmole

Substitute 0.175mol for number of moles of carbon atom and 0.175mol for smallest mole.

  MoleratioofC=0.175mol0.175mol=1

Substitute 0.4mol for number of moles of hydrogen atom and 0.175mol for smallest mole.

  MoleratioofH=0.4mol0.175mol=2.286

The mole ratio of carbon to that of hydrogen is as follows.

  MoleratioofCMoleratioofH=12.286

To convert this ration to a whole number, multiply the ratio by 7.

  MoleratioofCMoleratioofH=716.008

Therefore, the empirical formula of given hydrocarbon is C7H16.

(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 102QRT

The given hydrocarbon compound is an alkane.

Explanation of Solution

The general formula of alkane is CnH2n+2 and for alkene is CnH2n.  The empirical formula of given hydrocarbon is C7H16. Therefore, the given hydrocarbon is an alkane.

(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 102QRT

The lewis structure for given hydrocarbon is as follows.

Chemistry: The Molecular Science, Chapter 10, Problem 102QRT , additional homework tip  1

Explanation of Solution

The given hydrocarbon has empirical formula C7H16.  The given hydrocarbon is an alkane compound. The alkane compound contains linear chain of carbon and hydrogen atoms.

Therefore, the Lewis structure for the given hydrocarbon is as follows.

Chemistry: The Molecular Science, Chapter 10, Problem 102QRT , additional homework tip  2

Figure 1

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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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