Chemistry Atoms First, Second Edition
Chemistry Atoms First, Second Edition
2nd Edition
ISBN: 9781308211657
Author: Burdge
Publisher: McGraw Hill
Question
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Chapter 5, Problem 5.68QP

(a)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

(a)

Expert Solution
Check Mark

Answer to Problem 5.68QP

Molecular mass of Li2CO3 = 66.949 amu

Explanation of Solution

To calculate: Molecular mass of Li2CO3

Molecular formula is given as Li2CO3

Molecular mass of Li2CO3 = 2(mass of Li  in amu) + 1(mass of C  in amu) + 3 (mass of O in amu)                                                    = 1(6.941 amu) + 1(12.0107 amu) + 3(15.9994)                                                    = 66.949amu The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of Li2CO3 was found to be 66.949amu.

(b)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

(b)

Expert Solution
Check Mark

Answer to Problem 5.68QP

Molecular mass of C2H6 = 54.0902 amu.

Explanation of Solution

To calculate: Molecular mass of C2H6

Molecular formula is given as C2H6

Molecular mass of C2H6 = 2(mass of C  in amu) + 6(mass of H  in amu)                                                      = 4(12.0107 amu) + 6(1.0079 amu)                                                      = 54.0902 amu

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of C2H6 was found to be 54.0902 amu.

(C)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

(C)

Expert Solution
Check Mark

Answer to Problem 5.68QP

Molecular mass of NF2 = 51.987 amu

Explanation of Solution

To calculate: Molecular mass of NF2

Molecular formula is given as NF2

Molecular mass of NF2 = 1(mass of N  in amu) + 2(mass of F  in amu)                                                = 1(14.007 amu) + 2(18.99 amu)                                                = 51.987 amu

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of NF2 was found to be 51.987 amu.

(d)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

(d)

Expert Solution
Check Mark

Answer to Problem 5.68QP

Formula mass of Al2O3 = 101.961 amu

Explanation of Solution

To calculate: Formula mass of Al2O3

Molecular formula is given as Al2O3

Formula mass of Al2S3 = 2(mass of Al  in amu) + 3(mass of O  in amu)                                                 = 2(26.9815 amu) + 3(15.9994 amu)                                                 = 101.961 amu

The formula mass of the given substance was calculated using atomic weights of each elements present in its formula unit.  The molecular mass of Al2O3 was found to be 101.961 amu.

(e)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

(e)

Expert Solution
Check Mark

Answer to Problem 5.68QP

Formula mass of Fe(NO3)3 = 241.8597 amu

Explanation of Solution

To calculate: Formula mass of Fe(NO3)3

Molecular formula is given as Fe(NO3)3

Formula mass of Fe(NO3)3 = 1(mass of Fe) + 3(1(mass of N  in amu) + 3(mass of O  in amu))                                                           = 55.845 amu +3(14.0067 + 3(15.9994))amu                                                           = 55.845 amu + 3(14.0067 + 47.9982)amu                                                           = 241.8597 amu

The formula mass of the given substance was calculated using atomic weights of each elements present in its formula unit.  The molecular mass of Fe(NO3)3 was found to be 241.8597 amu.

(f)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

(f)

Expert Solution
Check Mark

Answer to Problem 5.68QP

Molecular mass of PCl5 = 208.2388 amu

To calculate: Molecular mass of PCl5

Molecular formula is given as PCl5

Molecular mass of PCl5 = 1(mass of P  in amu) + 5(mass of Cl  in amu)                                                   = 1(30.9738 amu) + 5(35.453 amu)                                                   = 208.2388 amu 

Explanation of Solution

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of PCl5 was found to be 208.2388 amu.

(g)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

(g)

Expert Solution
Check Mark

Answer to Problem 5.68QP

Molecular mass of Mg3N2 = 100.9284 amu.

Explanation of Solution

To calculate: Molecular mass of Mg3N2

Molecular formula is given as Mg3N2

Molecular mass of Mg3N2 = 3(mass of Mg  in amu) + 2(mass of N in amu)                                                       = 3(24.305) + 2 (14.0067)                                                       = 100.9284 amu                                                    

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of Mg3N2 was found to be 100.9284 amu.

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Chapter 5 Solutions

Chemistry Atoms First, Second Edition

Ch. 5.2 - Write Lewis dot symbols for (a) fluoride ion (F),...Ch. 5.2 - Write Lewis dot symbols for (a) Ca2+, (b) N3 and...Ch. 5.2 - Indicate the charge on each of the ions...Ch. 5.2 - Prob. 1PPCCh. 5.2 - Prob. 5.2.1SRCh. 5.2 - Prob. 5.2.2SRCh. 5.2 - Prob. 5.2.3SRCh. 5.2 - Prob. 5.2.4SRCh. 5.3 - Arrange MgO, CaO, and SrO in order of increasing...Ch. 5.3 - Prob. 2PPA
Ch. 5.3 - Arrange the compounds NaF, MgO, and AlN in order...Ch. 5.3 - Prob. 2PPCCh. 5.3 - Prob. 5.3.1SRCh. 5.3 - Prob. 5.3.2SRCh. 5.4 - Prob. 5.3WECh. 5.4 - Prob. 3PPACh. 5.4 - Prob. 3PPBCh. 5.4 - Prob. 3PPCCh. 5.4 - Prob. 5.4WECh. 5.4 - Prob. 4PPACh. 5.4 - Prob. 4PPBCh. 5.4 - Prob. 4PPCCh. 5.4 - Prob. 5.4.1SRCh. 5.4 - Prob. 5.4.2SRCh. 5.4 - Prob. 5.4.3SRCh. 5.4 - Prob. 5.4.4SRCh. 5.5 - Prob. 5.5WECh. 5.5 - Prob. 5PPACh. 5.5 - Draw (a) a space-filling molecular model of carbon...Ch. 5.5 - Prob. 5PPCCh. 5.5 - Prob. 5.6WECh. 5.5 - Prob. 6PPACh. 5.5 - Prob. 6PPBCh. 5.5 - Prob. 6PPCCh. 5.5 - Prob. 5.5.1SRCh. 5.5 - Prob. 5.5.2SRCh. 5.6 - Name the following binary molecular compounds: (a)...Ch. 5.6 - Prob. 7PPACh. 5.6 - Prob. 7PPBCh. 5.6 - Prob. 7PPCCh. 5.6 - Prob. 5.8WECh. 5.6 - Give the molecular formula for each of the...Ch. 5.6 - Prob. 8PPBCh. 5.6 - Draw a molecular model of sulfur trioxide.Ch. 5.6 - Prob. 5.6.1SRCh. 5.6 - Prob. 5.6.2SRCh. 5.6 - Prob. 5.6.3SRCh. 5.6 - Prob. 5.6.4SRCh. 5.7 - Prob. 5.9WECh. 5.7 - Name the following ionic compounds: (a) Na2SO4,...Ch. 5.7 - Prob. 9PPBCh. 5.7 - Prob. 9PPCCh. 5.7 - Name the following species: (a) BrO4, (b) HCO3,...Ch. 5.7 - Prob. 10PPACh. 5.7 - Prob. 10PPBCh. 5.7 - Prob. 10PPCCh. 5.7 - Prob. 5.11WECh. 5.7 - Prob. 11PPACh. 5.7 - Prob. 11PPBCh. 5.7 - Prob. 11PPCCh. 5.7 - Prob. 5.7.1SRCh. 5.7 - Prob. 5.7.2SRCh. 5.7 - Prob. 5.7.3SRCh. 5.7 - Prob. 5.7.4SRCh. 5.8 - Calculate the molecular mass or the formula mass,...Ch. 5.8 - Calculate the molecular or formula mass of each of...Ch. 5.8 - Prob. 12PPBCh. 5.8 - Prob. 12PPCCh. 5.8 - Prob. 5.8.1SRCh. 5.8 - Prob. 5.8.2SRCh. 5.8 - Prob. 5.8.3SRCh. 5.8 - Prob. 5.8.4SRCh. 5.8 - Prob. 5.8.5SRCh. 5.8 - Prob. 5.8.6SRCh. 5.9 - Prob. 5.13WECh. 5.9 - Prob. 13PPACh. 5.9 - Prob. 13PPBCh. 5.9 - Prob. 13PPCCh. 5.9 - Prob. 5.9.1SRCh. 5.9 - Prob. 5.9.2SRCh. 5.10 - Prob. 5.14WECh. 5.10 - Prob. 14PPACh. 5.10 - Prob. 14PPBCh. 5.10 - Prob. 5.15WECh. 5.10 - Prob. 15PPACh. 5.10 - Prob. 15PPBCh. 5.10 - Prob. 15PPCCh. 5.10 - Prob. 5.16WECh. 5.10 - Prob. 16PPACh. 5.10 - Prob. 16PPBCh. 5.10 - Prob. 16PPCCh. 5.10 - Prob. 5.10.1SRCh. 5.10 - Prob. 5.10.2SRCh. 5.10 - Prob. 5.10.3SRCh. 5.10 - Prob. 5.10.4SRCh. 5 - Define the term compound and explain how a...Ch. 5 - Prob. 5.2QPCh. 5 - Prob. 5.3QPCh. 5 - Prob. 5.4QPCh. 5 - Use the second member of each group from Group 1A...Ch. 5 - Without referring to Figure 5.1, write Lewis dot...Ch. 5 - Prob. 5.7QPCh. 5 - Indicate the charge on each of the ions...Ch. 5 - To what group of the periodic table does element X...Ch. 5 - Explain what ionic bonding is.Ch. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Explain why the chemical formulas of ionic...Ch. 5 - Prob. 5.21QPCh. 5 - Give the formulas and names of the compounds...Ch. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Sulfur reacts with fluorine to produce three...Ch. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Give an example of a case in which two molecules...Ch. 5 - Prob. 5.40QPCh. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Identify the following as elements or compounds:...Ch. 5 - Prob. 5.44QPCh. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - Prob. 5.49QPCh. 5 - Describe how the naming of molecular binary...Ch. 5 - Define the term acid.Ch. 5 - Prob. 5.52QPCh. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - Prob. 5.58QPCh. 5 - Prob. 5.59QPCh. 5 - Prob. 5.60QPCh. 5 - Prob. 5.61QPCh. 5 - Prob. 5.62QPCh. 5 - Prob. 5.63QPCh. 5 - Prob. 5.64QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Prob. 5.76QPCh. 5 - Prob. 5.77QPCh. 5 - Prob. 5.78QPCh. 5 - All the substances listed here are fertilizers...Ch. 5 - Prob. 5.80QPCh. 5 - Molar mass is numerically equivalent to molecular...Ch. 5 - Prob. 5.82QPCh. 5 - Prob. 5.83QPCh. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - The density of water is 1.00 g/mL at 4C. 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