General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 11, Problem 11.51QP

(a)

Interpretation Introduction

Interpretation:

Empirical formula of the given substance has to be predicted.

Concept introduction:

Steps to calculate empirical formula:

  • Convert the mass of elements into moles.
  • Divide each mole value by the smallest number of moles calculated.
  • Round to the nearest whole number.

Number of moles = Molarity × volume

Number of moles=MassMolarmass

(a)

Expert Solution
Check Mark

Explanation of Solution

Calculate moles of each given elements:

C:(37.5gC)1molC12.01gC=3.12molC

H:(3.2gH)1molH1.008 gH=3.17molH

F:(59.3gF)1molF19.0 gF=3.12molF

This gives the formula C3.12H3.17F3.12. Dividing the smallest number of moles gives the empirical formula, HCF

b)

Interpretation Introduction

Interpretation:

Does the substance behave as an ideal gas has to be predicted.

Concept introduction:

Ideal gas equation:PV=nRT An equation used to calculate either pressure, volume, temperature or number of moles of gas.

Boyle’s law: The pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

b)

Expert Solution
Check Mark

Explanation of Solution

When temperature and amount of gas are constant, the product of pressure times volume is constant (Boyle’s law).

For given pressure and volume values,

PV=nRT

If the number of moles and temperature are remains constant, then the product of pressure and volume should be same. If not then substances does not behave as an ideal gas.

(2.00atm)(0.322L)=0.664atm.L(1.50atm)(0.409L)=0.614atm.L(1.00atm)(0.564L)=0.564atm.L(0.50atm)(1.028L)=0.514atm.L

As shown above none of the values are same. Hence, the substances do not behave as an ideal gas.

c)

Interpretation Introduction

Interpretation:

The molecular formula has to be predicted.

Concept introduction:

Steps to calculate empirical formula:

  • Convert the mass of elements into moles.
  • Divide each mole value by the smallest number of moles calculated.
  • Round to the nearest whole number.

Number of moles = Molarity × volume

Number of moles=MassMolarmass

c)

Expert Solution
Check Mark

Explanation of Solution

Calculate moles of each given elements:

C:(37.5gC)1molC12.01gC=3.12molC

H:(3.2gH)1molH1.008 gH=3.17molH

F:(59.3gF)1molF19.0 gF=3.12molF

This gives the formula C3.12H3.17F3.12. Dividing the smallest number of moles gives the empirical formula, HCF

Now, let’s calculate moles using the ideal gas equation, and then calculate the molar mass.

n=PVRT=(0.50atm)(1.028L)(0.0821L.atm/K.mol)(363K)=0.0172mol

Molar mass = gofsubstancemolofsubstance=1.00g0.0172mol=58.1g/mol

The formula mass of C2H2F2 is 64g/mol, so is the molecular formula. Hence, the predicted molecular formula is reasonable.

d)

Interpretation Introduction

Interpretation:

Lewis structure of the molecule and its geometry has to be drawn and described.

Concept introduction:

Structural Isomerism: Structural Isomers are the structure of a molecule with same molecular formula but have different arrangements of bonds and atoms and position of double bond also changes from more substituted to less substituted or vice-versa.

Lewis structure: The bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.

Geometric isomers of Alkenes:

Cis-isomer: When two particular atoms (group of atoms) are adjacent to each other, the alkene is known as cis-isomer.

General Chemistry, Chapter 11, Problem 11.51QP , additional homework tip  1

Trans-isomer: When two particular atoms (group of atoms) are across from each other, the alkene is known as trans-isomer.

d)

Expert Solution
Check Mark

Explanation of Solution

Compound C2H2F2 formula is that of difluoroethylene. Three isomers are possible. Lewis structures are shown below as,

General Chemistry, Chapter 11, Problem 11.51QP , additional homework tip  2

The geometry of each carbon is trigonal planar. Arrangement of two identical fluorine atoms on the same side adjacent to each other known as cis-isomer. And represnted opposite side to each other known as trans-isomer.

e)

Interpretation Introduction

Interpretation:

The systematic name of the structure has to be written.

Concept introduction:

IUPAC Nomenclature of alkenes:

  • The longest continuous chain of carbon atoms is identified.
  • The substituent groups attached to the parent chain is identified. A substituent group contains group of atoms attached to the carbon atom of the chain.
  • While numbering the longest chain, the substituent should get least possible number.
  • Write the name of the compound; the parent name written as last part of the name. The name of the substituents is written as prefix and a hyphen separates the number that the substituents attached with carbon. More than one substituent should be written in alphabetical order.

Geometric isomers of Alkenes:

Cis-isomer: When two particular atoms (group of atoms) are adjacent to each other, the alkene is known as cis-isomer.

General Chemistry, Chapter 11, Problem 11.51QP , additional homework tip  3

Trans-isomer: When two particular atoms (group of atoms) are across from each other, the alkene is known as trans-isomer.

e)

Expert Solution
Check Mark

Explanation of Solution

Given name: cis-2-butene

Predict the longest continuous chain of carbon atoms:

General Chemistry, Chapter 11, Problem 11.51QP , additional homework tip  4

The parent name is ETHENE represent the longest chain of carbon atoms contains two carbons. The Suffix ‘ene’ represents presence of double bond at C-1.

Predict substituents and its location:

General Chemistry, Chapter 11, Problem 11.51QP , additional homework tip  5

The first compound structure has two fluorine atoms located at carbon-1. Hence the name can be written as substituent followed by parent name; 2,2-difluoroethene.

The second compound structure has two fluorine atoms located at carbon-1and 2. The term ‘cis-’ indicates two fluorine atoms are located adjacent to each other on same side. Hence the name can be written as substituent followed by parent name; cis-1,2-difluoroethene.

The third compound structure has two fluorine atoms located at carbon-1and 2. The term ‘trans-’ indicates two fluorine atoms are located opposite to each other. Hence the name can be written as substituent followed by parent name; trans-1,2-difluoroethene.

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Chapter 11 Solutions

General Chemistry

Ch. 11 - Prob. 11.2QPCh. 11 - Prob. 11.3QPCh. 11 - 11.4 What are structural isomers? Ch. 11 - Prob. 11.5QPCh. 11 - 11.6 Draw skeletal structures of the boat and...Ch. 11 - 11.7 Alkenes exhibit geometric isomerism because...Ch. 11 - 11.8 Why is it that alkanes and alkynes, unlike...Ch. 11 - Prob. 11.9QPCh. 11 - 11.10 Describe reactions that are characteristic...Ch. 11 - Prob. 11.11QPCh. 11 - Prob. 11.12QPCh. 11 - Prob. 11.13QPCh. 11 - Prob. 11.14QPCh. 11 - Prob. 11.15QPCh. 11 - Prob. 11.16QPCh. 11 - Prob. 11.17QPCh. 11 - 11.18 Draw Newman projections of four different...Ch. 11 - 11.19 Draw the structures of cis-2-butene and...Ch. 11 - 11.20 Would you expect cyclobutadiene to be a...Ch. 11 - Prob. 11.21QPCh. 11 - Prob. 11.22QPCh. 11 - 11.23 Sulfuric acid (H2SO4) adds to the double...Ch. 11 - Prob. 11.24QPCh. 11 - Prob. 11.25QPCh. 11 - Prob. 11.26QPCh. 11 - Prob. 11.27QPCh. 11 - Prob. 11.28QPCh. 11 - Prob. 11.29QPCh. 11 - 11.30 Benzene and cyclohexane both contain...Ch. 11 - Prob. 11.31QPCh. 11 - Prob. 11.32QPCh. 11 - Prob. 11.33QPCh. 11 - Prob. 11.34QPCh. 11 - Prob. 11.35QPCh. 11 - Prob. 11.36QPCh. 11 - Prob. 11.37QPCh. 11 - Prob. 11.38QPCh. 11 - Prob. 11.39QPCh. 11 - Prob. 11.40QPCh. 11 - Prob. 11.41QPCh. 11 - Prob. 11.42QPCh. 11 - Prob. 11.43QPCh. 11 - Prob. 11.44QPCh. 11 - Prob. 11.45QPCh. 11 - Prob. 11.46QPCh. 11 - Prob. 11.47QPCh. 11 - Prob. 11.48QPCh. 11 - Prob. 11.49QPCh. 11 - Prob. 11.50QPCh. 11 - Prob. 11.51QPCh. 11 - Prob. 11.52QPCh. 11 - Prob. 11.53QPCh. 11 - Prob. 11.54QPCh. 11 - Prob. 11.55QPCh. 11 - Prob. 11.56QPCh. 11 - Prob. 11.57QPCh. 11 - Prob. 11.58QPCh. 11 - Prob. 11.59QPCh. 11 - Prob. 11.60QPCh. 11 - Prob. 11.61QPCh. 11 - Prob. 11.62QPCh. 11 - Prob. 11.63QPCh. 11 - Prob. 11.64QPCh. 11 - Prob. 11.65QPCh. 11 - Prob. 11.66QPCh. 11 - Prob. 11.67QPCh. 11 - Prob. 11.68QPCh. 11 - Prob. 11.69QPCh. 11 - Prob. 11.70QPCh. 11 - Prob. 11.71QPCh. 11 - Prob. 11.72QPCh. 11 - 11.73 Octane number is assigned to gasoline to...Ch. 11 - Prob. 11.74SPCh. 11 - Prob. 11.75SPCh. 11 - Prob. 11.76SPCh. 11 - Prob. 11.77SPCh. 11 - Prob. 11.78SP
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