The structures of A, B and C has to be given. Concept Introduction: The ideal gas equation can be used to calculate the volume using pressure and temperature. The ideal gas equation is, PV = nRT n = PV RT Where, P = pressure V = volume R = gas constant T = temperature n = number of moles .

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Answer 1

Question

Definition Definition Class of organic compounds that contain a carboxyl group ( - COOH ) and has a general formula R - COOH or R - CO 2 H , where R refers to the alkyl, alkenyl, aryl, or other groups. They can undergo different chemical reactions, such as acid-base reactions, esterification, and oxidation. These are essential components of living organisms, playing important roles in metabolic processes, signaling, and as pharmaceuticals.

Chapter 15, Problem 15.107P

(a)

Interpretation Introduction

Interpretation:

The structures of A, B and C has to be given.

Concept Introduction:

The ideal gas equation can be used to calculate the volume using pressure and temperature. The ideal gas equation is,

$PV = nRTn =PVRT$

Where, $P = pressureV = volumeR = gas constantT = temperaturen = number of moles.$

(a)

Expert Solution

Explanation of Solution

The structures of compounds A, B and C can be determined by calculating the moles of the compounds from the given mass of the compounds.

The moles of A can be calculated by using the ideal gas formula.

Given,

$pressure = 1.00atm volume = 1.00Ltemperature = 1600C$

$moles of A (n) = PVRT (n) = (1.00 atm)(1.00L)(0.0821L.atmmol.K)(273+160)K (n) = 0.02812995 mol$

Molar mass of A can be calculated using the number of moles obtained.

Given: $2.48 g$ of A.

$molar mass of A = 2.48gA0.02812995mol = 88.16226 g/mol$

The molecular formula of A can be determined by the combustion analysis.

Given,

$0.409 g$ of $H2O$
$1.00 g of CO2$

Moles of $H$ can be calculated from the mass of $H2O$ .

$moles of H = (0.409 g H2O)(2 mol H18.02 g H2O) = 0.0454 mol H$

By using the moles of $H$ in $H2O$ , the mass of $H$ is calculated.

$mass of H = (0.0454 mol H)(1.008 g H1 mol H) = 0.04576 g H$

Moles of $C$ can be calculated from the mass of $CO2$ .

$moles of C = (1.00 g CO2)(1 mol C44.01 g CO2) = 0.0227 mol C$

By using the moles of $C$ in $CO2$ , the mass of $C$ is calculated.

$mass of C = (0.0227 mol C)(12.01 g C1 mol C) = 0.2726 g C$

Moles of $O$ can be calculated by subtracting the mass of $C$ and $H$ from the total mass.

$moles of O = (0.500 g A - (0.04576+0.2726) g O)(1 mol O16.00 g O) = 0.01135 mol O$

To get the moles of the $C$ , $H$ and $O$ , the moles calculated is divided by the smallest number.

$C: 0.0227 mol C0.01135 mol = 2H: 0.0454 mol H0.01135 mol = 4O: 0.01135 mol O0.01135 mol = 1$

The empirical formula of compound A is obtained as $C2H4O$ , with a molar mass $44.05 g/mol$ .

As the molar mass of compound A is calculated as $88.16 g/mol$ , the molecular formula of compound A is $C4H8O2$ .

As compound B is acidic, it will be a carboxylic acid. Moles of $COOH$ can be calculated using the given information.

Given,

$1.000 g$ Of B is neutralized with $33.9 mL of 0.5 M$ sodium hydroxide.

$moles of COOH = (0.5 mol NaOHL)(33.9 mL)(10-3 L1 mL)(1 mol COOH1 mol NaOH) = 0.01695 mol COOH$

The 1.00 g sample of compound B contains 0.01695 moles of $COOH$ .

$mass of COOH = 1.00 g0.01695 mol COOH = 59 g/mol COOH$

As the molar mass of compound A is $88.16 g/mol$ , the molar mass of compound B will be $118 g/mol$ , it will be a dicarboxylic acid. The molecular formula can be given as $C4H6O4$ .

Compound B on loss of water molecule by heating forms compound C. Hence, it will be an anhydride. The molecular formula of compound C can be given by the loss of water molecule from molecular formula of B.

Molecular formula of compound C = $(C4H6O4-H2O) = C4H4O3$ .

As the NMR spectrum of compound C gives only one peak, the compound must be symmetrical with only one type of hydrogens. The structures of compound B and compound C according to the NMR spectrum are given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 1

Compound A is not acidic. It will have an alcohol and aldehyde groups. The structure of compound A can be given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

Compound A is a controlled substance because it is metabolized to the weakly acidic date rape drug GHB, $C4H8O3$ . The structure and name of GHB has to be given.

Concept Introduction:

Oxidation:

Addition of oxygen atoms to the molecule or removal hydrogens from the molecule is said to be oxidation.

Aldehyde on oxidation forms carboxylic acid.

(b)

Expert Solution

Explanation of Solution

Compound A is 4-hydroxybutanal with an alcohol and aldehyde groups.

GHB is an acidic compound, hence compound A is oxidized and the aldehyde group converts to carboxylic acid group. The structure and name of GHB is

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 4

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Answer 2

Question

Definition Definition Class of organic compounds that contain a carboxyl group ( - COOH ) and has a general formula R - COOH or R - CO 2 H , where R refers to the alkyl, alkenyl, aryl, or other groups. They can undergo different chemical reactions, such as acid-base reactions, esterification, and oxidation. These are essential components of living organisms, playing important roles in metabolic processes, signaling, and as pharmaceuticals.

Chapter 15, Problem 15.107P

(a)

Interpretation Introduction

Interpretation:

The structures of A, B and C has to be given.

Concept Introduction:

The ideal gas equation can be used to calculate the volume using pressure and temperature. The ideal gas equation is,

$PV = nRTn =PVRT$

Where, $P = pressureV = volumeR = gas constantT = temperaturen = number of moles.$

(a)

Expert Solution

Explanation of Solution

The structures of compounds A, B and C can be determined by calculating the moles of the compounds from the given mass of the compounds.

The moles of A can be calculated by using the ideal gas formula.

Given,

$pressure = 1.00atm volume = 1.00Ltemperature = 1600C$

$moles of A (n) = PVRT (n) = (1.00 atm)(1.00L)(0.0821L.atmmol.K)(273+160)K (n) = 0.02812995 mol$

Molar mass of A can be calculated using the number of moles obtained.

Given: $2.48 g$ of A.

$molar mass of A = 2.48gA0.02812995mol = 88.16226 g/mol$

The molecular formula of A can be determined by the combustion analysis.

Given,

$0.409 g$ of $H2O$
$1.00 g of CO2$

Moles of $H$ can be calculated from the mass of $H2O$ .

$moles of H = (0.409 g H2O)(2 mol H18.02 g H2O) = 0.0454 mol H$

By using the moles of $H$ in $H2O$ , the mass of $H$ is calculated.

$mass of H = (0.0454 mol H)(1.008 g H1 mol H) = 0.04576 g H$

Moles of $C$ can be calculated from the mass of $CO2$ .

$moles of C = (1.00 g CO2)(1 mol C44.01 g CO2) = 0.0227 mol C$

By using the moles of $C$ in $CO2$ , the mass of $C$ is calculated.

$mass of C = (0.0227 mol C)(12.01 g C1 mol C) = 0.2726 g C$

Moles of $O$ can be calculated by subtracting the mass of $C$ and $H$ from the total mass.

$moles of O = (0.500 g A - (0.04576+0.2726) g O)(1 mol O16.00 g O) = 0.01135 mol O$

To get the moles of the $C$ , $H$ and $O$ , the moles calculated is divided by the smallest number.

$C: 0.0227 mol C0.01135 mol = 2H: 0.0454 mol H0.01135 mol = 4O: 0.01135 mol O0.01135 mol = 1$

The empirical formula of compound A is obtained as $C2H4O$ , with a molar mass $44.05 g/mol$ .

As the molar mass of compound A is calculated as $88.16 g/mol$ , the molecular formula of compound A is $C4H8O2$ .

As compound B is acidic, it will be a carboxylic acid. Moles of $COOH$ can be calculated using the given information.

Given,

$1.000 g$ Of B is neutralized with $33.9 mL of 0.5 M$ sodium hydroxide.

$moles of COOH = (0.5 mol NaOHL)(33.9 mL)(10-3 L1 mL)(1 mol COOH1 mol NaOH) = 0.01695 mol COOH$

The 1.00 g sample of compound B contains 0.01695 moles of $COOH$ .

$mass of COOH = 1.00 g0.01695 mol COOH = 59 g/mol COOH$

As the molar mass of compound A is $88.16 g/mol$ , the molar mass of compound B will be $118 g/mol$ , it will be a dicarboxylic acid. The molecular formula can be given as $C4H6O4$ .

Compound B on loss of water molecule by heating forms compound C. Hence, it will be an anhydride. The molecular formula of compound C can be given by the loss of water molecule from molecular formula of B.

Molecular formula of compound C = $(C4H6O4-H2O) = C4H4O3$ .

As the NMR spectrum of compound C gives only one peak, the compound must be symmetrical with only one type of hydrogens. The structures of compound B and compound C according to the NMR spectrum are given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 1

Compound A is not acidic. It will have an alcohol and aldehyde groups. The structure of compound A can be given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

Compound A is a controlled substance because it is metabolized to the weakly acidic date rape drug GHB, $C4H8O3$ . The structure and name of GHB has to be given.

Concept Introduction:

Oxidation:

Addition of oxygen atoms to the molecule or removal hydrogens from the molecule is said to be oxidation.

Aldehyde on oxidation forms carboxylic acid.

(b)

Expert Solution

Explanation of Solution

Compound A is 4-hydroxybutanal with an alcohol and aldehyde groups.

GHB is an acidic compound, hence compound A is oxidized and the aldehyde group converts to carboxylic acid group. The structure and name of GHB is

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 4

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Answer 3

Question

Definition Definition Class of organic compounds that contain a carboxyl group ( - COOH ) and has a general formula R - COOH or R - CO 2 H , where R refers to the alkyl, alkenyl, aryl, or other groups. They can undergo different chemical reactions, such as acid-base reactions, esterification, and oxidation. These are essential components of living organisms, playing important roles in metabolic processes, signaling, and as pharmaceuticals.

Chapter 15, Problem 15.107P

(a)

Interpretation Introduction

Interpretation:

The structures of A, B and C has to be given.

Concept Introduction:

The ideal gas equation can be used to calculate the volume using pressure and temperature. The ideal gas equation is,

$PV = nRTn =PVRT$

Where, $P = pressureV = volumeR = gas constantT = temperaturen = number of moles.$

(a)

Expert Solution

Explanation of Solution

The structures of compounds A, B and C can be determined by calculating the moles of the compounds from the given mass of the compounds.

The moles of A can be calculated by using the ideal gas formula.

Given,

$pressure = 1.00atm volume = 1.00Ltemperature = 1600C$

$moles of A (n) = PVRT (n) = (1.00 atm)(1.00L)(0.0821L.atmmol.K)(273+160)K (n) = 0.02812995 mol$

Molar mass of A can be calculated using the number of moles obtained.

Given: $2.48 g$ of A.

$molar mass of A = 2.48gA0.02812995mol = 88.16226 g/mol$

The molecular formula of A can be determined by the combustion analysis.

Given,

$0.409 g$ of $H2O$
$1.00 g of CO2$

Moles of $H$ can be calculated from the mass of $H2O$ .

$moles of H = (0.409 g H2O)(2 mol H18.02 g H2O) = 0.0454 mol H$

By using the moles of $H$ in $H2O$ , the mass of $H$ is calculated.

$mass of H = (0.0454 mol H)(1.008 g H1 mol H) = 0.04576 g H$

Moles of $C$ can be calculated from the mass of $CO2$ .

$moles of C = (1.00 g CO2)(1 mol C44.01 g CO2) = 0.0227 mol C$

By using the moles of $C$ in $CO2$ , the mass of $C$ is calculated.

$mass of C = (0.0227 mol C)(12.01 g C1 mol C) = 0.2726 g C$

Moles of $O$ can be calculated by subtracting the mass of $C$ and $H$ from the total mass.

$moles of O = (0.500 g A - (0.04576+0.2726) g O)(1 mol O16.00 g O) = 0.01135 mol O$

To get the moles of the $C$ , $H$ and $O$ , the moles calculated is divided by the smallest number.

$C: 0.0227 mol C0.01135 mol = 2H: 0.0454 mol H0.01135 mol = 4O: 0.01135 mol O0.01135 mol = 1$

The empirical formula of compound A is obtained as $C2H4O$ , with a molar mass $44.05 g/mol$ .

As the molar mass of compound A is calculated as $88.16 g/mol$ , the molecular formula of compound A is $C4H8O2$ .

As compound B is acidic, it will be a carboxylic acid. Moles of $COOH$ can be calculated using the given information.

Given,

$1.000 g$ Of B is neutralized with $33.9 mL of 0.5 M$ sodium hydroxide.

$moles of COOH = (0.5 mol NaOHL)(33.9 mL)(10-3 L1 mL)(1 mol COOH1 mol NaOH) = 0.01695 mol COOH$

The 1.00 g sample of compound B contains 0.01695 moles of $COOH$ .

$mass of COOH = 1.00 g0.01695 mol COOH = 59 g/mol COOH$

As the molar mass of compound A is $88.16 g/mol$ , the molar mass of compound B will be $118 g/mol$ , it will be a dicarboxylic acid. The molecular formula can be given as $C4H6O4$ .

Compound B on loss of water molecule by heating forms compound C. Hence, it will be an anhydride. The molecular formula of compound C can be given by the loss of water molecule from molecular formula of B.

Molecular formula of compound C = $(C4H6O4-H2O) = C4H4O3$ .

As the NMR spectrum of compound C gives only one peak, the compound must be symmetrical with only one type of hydrogens. The structures of compound B and compound C according to the NMR spectrum are given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 1

Compound A is not acidic. It will have an alcohol and aldehyde groups. The structure of compound A can be given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

Compound A is a controlled substance because it is metabolized to the weakly acidic date rape drug GHB, $C4H8O3$ . The structure and name of GHB has to be given.

Concept Introduction:

Oxidation:

Addition of oxygen atoms to the molecule or removal hydrogens from the molecule is said to be oxidation.

Aldehyde on oxidation forms carboxylic acid.

(b)

Expert Solution

Explanation of Solution

Compound A is 4-hydroxybutanal with an alcohol and aldehyde groups.

GHB is an acidic compound, hence compound A is oxidized and the aldehyde group converts to carboxylic acid group. The structure and name of GHB is

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 4

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Answer 4

Question

Definition Definition Class of organic compounds that contain a carboxyl group ( - COOH ) and has a general formula R - COOH or R - CO 2 H , where R refers to the alkyl, alkenyl, aryl, or other groups. They can undergo different chemical reactions, such as acid-base reactions, esterification, and oxidation. These are essential components of living organisms, playing important roles in metabolic processes, signaling, and as pharmaceuticals.

Chapter 15, Problem 15.107P

(a)

Interpretation Introduction

Interpretation:

The structures of A, B and C has to be given.

Concept Introduction:

The ideal gas equation can be used to calculate the volume using pressure and temperature. The ideal gas equation is,

$PV = nRTn =PVRT$

Where, $P = pressureV = volumeR = gas constantT = temperaturen = number of moles.$

(a)

Expert Solution

Explanation of Solution

The structures of compounds A, B and C can be determined by calculating the moles of the compounds from the given mass of the compounds.

The moles of A can be calculated by using the ideal gas formula.

Given,

$pressure = 1.00atm volume = 1.00Ltemperature = 1600C$

$moles of A (n) = PVRT (n) = (1.00 atm)(1.00L)(0.0821L.atmmol.K)(273+160)K (n) = 0.02812995 mol$

Molar mass of A can be calculated using the number of moles obtained.

Given: $2.48 g$ of A.

$molar mass of A = 2.48gA0.02812995mol = 88.16226 g/mol$

The molecular formula of A can be determined by the combustion analysis.

Given,

$0.409 g$ of $H2O$
$1.00 g of CO2$

Moles of $H$ can be calculated from the mass of $H2O$ .

$moles of H = (0.409 g H2O)(2 mol H18.02 g H2O) = 0.0454 mol H$

By using the moles of $H$ in $H2O$ , the mass of $H$ is calculated.

$mass of H = (0.0454 mol H)(1.008 g H1 mol H) = 0.04576 g H$

Moles of $C$ can be calculated from the mass of $CO2$ .

$moles of C = (1.00 g CO2)(1 mol C44.01 g CO2) = 0.0227 mol C$

By using the moles of $C$ in $CO2$ , the mass of $C$ is calculated.

$mass of C = (0.0227 mol C)(12.01 g C1 mol C) = 0.2726 g C$

Moles of $O$ can be calculated by subtracting the mass of $C$ and $H$ from the total mass.

$moles of O = (0.500 g A - (0.04576+0.2726) g O)(1 mol O16.00 g O) = 0.01135 mol O$

To get the moles of the $C$ , $H$ and $O$ , the moles calculated is divided by the smallest number.

$C: 0.0227 mol C0.01135 mol = 2H: 0.0454 mol H0.01135 mol = 4O: 0.01135 mol O0.01135 mol = 1$

The empirical formula of compound A is obtained as $C2H4O$ , with a molar mass $44.05 g/mol$ .

As the molar mass of compound A is calculated as $88.16 g/mol$ , the molecular formula of compound A is $C4H8O2$ .

As compound B is acidic, it will be a carboxylic acid. Moles of $COOH$ can be calculated using the given information.

Given,

$1.000 g$ Of B is neutralized with $33.9 mL of 0.5 M$ sodium hydroxide.

$moles of COOH = (0.5 mol NaOHL)(33.9 mL)(10-3 L1 mL)(1 mol COOH1 mol NaOH) = 0.01695 mol COOH$

The 1.00 g sample of compound B contains 0.01695 moles of $COOH$ .

$mass of COOH = 1.00 g0.01695 mol COOH = 59 g/mol COOH$

As the molar mass of compound A is $88.16 g/mol$ , the molar mass of compound B will be $118 g/mol$ , it will be a dicarboxylic acid. The molecular formula can be given as $C4H6O4$ .

Compound B on loss of water molecule by heating forms compound C. Hence, it will be an anhydride. The molecular formula of compound C can be given by the loss of water molecule from molecular formula of B.

Molecular formula of compound C = $(C4H6O4-H2O) = C4H4O3$ .

As the NMR spectrum of compound C gives only one peak, the compound must be symmetrical with only one type of hydrogens. The structures of compound B and compound C according to the NMR spectrum are given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 1

Compound A is not acidic. It will have an alcohol and aldehyde groups. The structure of compound A can be given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

Compound A is a controlled substance because it is metabolized to the weakly acidic date rape drug GHB, $C4H8O3$ . The structure and name of GHB has to be given.

Concept Introduction:

Oxidation:

Addition of oxygen atoms to the molecule or removal hydrogens from the molecule is said to be oxidation.

Aldehyde on oxidation forms carboxylic acid.

(b)

Expert Solution

Explanation of Solution

Compound A is 4-hydroxybutanal with an alcohol and aldehyde groups.

GHB is an acidic compound, hence compound A is oxidized and the aldehyde group converts to carboxylic acid group. The structure and name of GHB is

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 4

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 15, Problem 15.107P

(a)

Interpretation Introduction

Interpretation:

The structures of A, B and C has to be given.

Concept Introduction:

The ideal gas equation can be used to calculate the volume using pressure and temperature. The ideal gas equation is,

$PV = nRTn =PVRT$

Where, $P = pressureV = volumeR = gas constantT = temperaturen = number of moles.$

(a)

Expert Solution

Explanation of Solution

The structures of compounds A, B and C can be determined by calculating the moles of the compounds from the given mass of the compounds.

The moles of A can be calculated by using the ideal gas formula.

Given,

$pressure = 1.00atm volume = 1.00Ltemperature = 1600C$

$moles of A (n) = PVRT (n) = (1.00 atm)(1.00L)(0.0821L.atmmol.K)(273+160)K (n) = 0.02812995 mol$

Molar mass of A can be calculated using the number of moles obtained.

Given: $2.48 g$ of A.

$molar mass of A = 2.48gA0.02812995mol = 88.16226 g/mol$

The molecular formula of A can be determined by the combustion analysis.

Given,

$0.409 g$ of $H2O$
$1.00 g of CO2$

Moles of $H$ can be calculated from the mass of $H2O$ .

$moles of H = (0.409 g H2O)(2 mol H18.02 g H2O) = 0.0454 mol H$

By using the moles of $H$ in $H2O$ , the mass of $H$ is calculated.

$mass of H = (0.0454 mol H)(1.008 g H1 mol H) = 0.04576 g H$

Moles of $C$ can be calculated from the mass of $CO2$ .

$moles of C = (1.00 g CO2)(1 mol C44.01 g CO2) = 0.0227 mol C$

By using the moles of $C$ in $CO2$ , the mass of $C$ is calculated.

$mass of C = (0.0227 mol C)(12.01 g C1 mol C) = 0.2726 g C$

Moles of $O$ can be calculated by subtracting the mass of $C$ and $H$ from the total mass.

$moles of O = (0.500 g A - (0.04576+0.2726) g O)(1 mol O16.00 g O) = 0.01135 mol O$

To get the moles of the $C$ , $H$ and $O$ , the moles calculated is divided by the smallest number.

$C: 0.0227 mol C0.01135 mol = 2H: 0.0454 mol H0.01135 mol = 4O: 0.01135 mol O0.01135 mol = 1$

The empirical formula of compound A is obtained as $C2H4O$ , with a molar mass $44.05 g/mol$ .

As the molar mass of compound A is calculated as $88.16 g/mol$ , the molecular formula of compound A is $C4H8O2$ .

As compound B is acidic, it will be a carboxylic acid. Moles of $COOH$ can be calculated using the given information.

Given,

$1.000 g$ Of B is neutralized with $33.9 mL of 0.5 M$ sodium hydroxide.

$moles of COOH = (0.5 mol NaOHL)(33.9 mL)(10-3 L1 mL)(1 mol COOH1 mol NaOH) = 0.01695 mol COOH$

The 1.00 g sample of compound B contains 0.01695 moles of $COOH$ .

$mass of COOH = 1.00 g0.01695 mol COOH = 59 g/mol COOH$

As the molar mass of compound A is $88.16 g/mol$ , the molar mass of compound B will be $118 g/mol$ , it will be a dicarboxylic acid. The molecular formula can be given as $C4H6O4$ .

Compound B on loss of water molecule by heating forms compound C. Hence, it will be an anhydride. The molecular formula of compound C can be given by the loss of water molecule from molecular formula of B.

Molecular formula of compound C = $(C4H6O4-H2O) = C4H4O3$ .

As the NMR spectrum of compound C gives only one peak, the compound must be symmetrical with only one type of hydrogens. The structures of compound B and compound C according to the NMR spectrum are given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 1

Compound A is not acidic. It will have an alcohol and aldehyde groups. The structure of compound A can be given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

Compound A is a controlled substance because it is metabolized to the weakly acidic date rape drug GHB, $C4H8O3$ . The structure and name of GHB has to be given.

Concept Introduction:

Oxidation:

Addition of oxygen atoms to the molecule or removal hydrogens from the molecule is said to be oxidation.

Aldehyde on oxidation forms carboxylic acid.

(b)

Expert Solution

Explanation of Solution

Compound A is 4-hydroxybutanal with an alcohol and aldehyde groups.

GHB is an acidic compound, hence compound A is oxidized and the aldehyde group converts to carboxylic acid group. The structure and name of GHB is

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 4

Answer 6

Chapter 15, Problem 15.107P

(a)

Interpretation Introduction

Interpretation:

The structures of A, B and C has to be given.

Concept Introduction:

The ideal gas equation can be used to calculate the volume using pressure and temperature. The ideal gas equation is,

$PV = nRTn =PVRT$

Where, $P = pressureV = volumeR = gas constantT = temperaturen = number of moles.$

(a)

Expert Solution

Explanation of Solution

The structures of compounds A, B and C can be determined by calculating the moles of the compounds from the given mass of the compounds.

The moles of A can be calculated by using the ideal gas formula.

Given,

$pressure = 1.00atm volume = 1.00Ltemperature = 1600C$

$moles of A (n) = PVRT (n) = (1.00 atm)(1.00L)(0.0821L.atmmol.K)(273+160)K (n) = 0.02812995 mol$

Molar mass of A can be calculated using the number of moles obtained.

Given: $2.48 g$ of A.

$molar mass of A = 2.48gA0.02812995mol = 88.16226 g/mol$

The molecular formula of A can be determined by the combustion analysis.

Given,

$0.409 g$ of $H2O$
$1.00 g of CO2$

Moles of $H$ can be calculated from the mass of $H2O$ .

$moles of H = (0.409 g H2O)(2 mol H18.02 g H2O) = 0.0454 mol H$

By using the moles of $H$ in $H2O$ , the mass of $H$ is calculated.

$mass of H = (0.0454 mol H)(1.008 g H1 mol H) = 0.04576 g H$

Moles of $C$ can be calculated from the mass of $CO2$ .

$moles of C = (1.00 g CO2)(1 mol C44.01 g CO2) = 0.0227 mol C$

By using the moles of $C$ in $CO2$ , the mass of $C$ is calculated.

$mass of C = (0.0227 mol C)(12.01 g C1 mol C) = 0.2726 g C$

Moles of $O$ can be calculated by subtracting the mass of $C$ and $H$ from the total mass.

$moles of O = (0.500 g A - (0.04576+0.2726) g O)(1 mol O16.00 g O) = 0.01135 mol O$

To get the moles of the $C$ , $H$ and $O$ , the moles calculated is divided by the smallest number.

$C: 0.0227 mol C0.01135 mol = 2H: 0.0454 mol H0.01135 mol = 4O: 0.01135 mol O0.01135 mol = 1$

The empirical formula of compound A is obtained as $C2H4O$ , with a molar mass $44.05 g/mol$ .

As the molar mass of compound A is calculated as $88.16 g/mol$ , the molecular formula of compound A is $C4H8O2$ .

As compound B is acidic, it will be a carboxylic acid. Moles of $COOH$ can be calculated using the given information.

Given,

$1.000 g$ Of B is neutralized with $33.9 mL of 0.5 M$ sodium hydroxide.

$moles of COOH = (0.5 mol NaOHL)(33.9 mL)(10-3 L1 mL)(1 mol COOH1 mol NaOH) = 0.01695 mol COOH$

The 1.00 g sample of compound B contains 0.01695 moles of $COOH$ .

$mass of COOH = 1.00 g0.01695 mol COOH = 59 g/mol COOH$

As the molar mass of compound A is $88.16 g/mol$ , the molar mass of compound B will be $118 g/mol$ , it will be a dicarboxylic acid. The molecular formula can be given as $C4H6O4$ .

Compound B on loss of water molecule by heating forms compound C. Hence, it will be an anhydride. The molecular formula of compound C can be given by the loss of water molecule from molecular formula of B.

Molecular formula of compound C = $(C4H6O4-H2O) = C4H4O3$ .

As the NMR spectrum of compound C gives only one peak, the compound must be symmetrical with only one type of hydrogens. The structures of compound B and compound C according to the NMR spectrum are given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 1

Compound A is not acidic. It will have an alcohol and aldehyde groups. The structure of compound A can be given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 2

Answer 7

Chapter 15, Problem 15.107P

(a)

Interpretation Introduction

Interpretation:

The structures of A, B and C has to be given.

Concept Introduction:

The ideal gas equation can be used to calculate the volume using pressure and temperature. The ideal gas equation is,

$PV = nRTn =PVRT$

Where, $P = pressureV = volumeR = gas constantT = temperaturen = number of moles.$

Answer 8

(a)

Expert Solution

Explanation of Solution

The structures of compounds A, B and C can be determined by calculating the moles of the compounds from the given mass of the compounds.

The moles of A can be calculated by using the ideal gas formula.

Given,

$pressure = 1.00atm volume = 1.00Ltemperature = 1600C$

$moles of A (n) = PVRT (n) = (1.00 atm)(1.00L)(0.0821L.atmmol.K)(273+160)K (n) = 0.02812995 mol$

Molar mass of A can be calculated using the number of moles obtained.

Given: $2.48 g$ of A.

$molar mass of A = 2.48gA0.02812995mol = 88.16226 g/mol$

The molecular formula of A can be determined by the combustion analysis.

Given,

$0.409 g$ of $H2O$
$1.00 g of CO2$

Moles of $H$ can be calculated from the mass of $H2O$ .

$moles of H = (0.409 g H2O)(2 mol H18.02 g H2O) = 0.0454 mol H$

By using the moles of $H$ in $H2O$ , the mass of $H$ is calculated.

$mass of H = (0.0454 mol H)(1.008 g H1 mol H) = 0.04576 g H$

Moles of $C$ can be calculated from the mass of $CO2$ .

$moles of C = (1.00 g CO2)(1 mol C44.01 g CO2) = 0.0227 mol C$

By using the moles of $C$ in $CO2$ , the mass of $C$ is calculated.

$mass of C = (0.0227 mol C)(12.01 g C1 mol C) = 0.2726 g C$

Moles of $O$ can be calculated by subtracting the mass of $C$ and $H$ from the total mass.

$moles of O = (0.500 g A - (0.04576+0.2726) g O)(1 mol O16.00 g O) = 0.01135 mol O$

To get the moles of the $C$ , $H$ and $O$ , the moles calculated is divided by the smallest number.

$C: 0.0227 mol C0.01135 mol = 2H: 0.0454 mol H0.01135 mol = 4O: 0.01135 mol O0.01135 mol = 1$

The empirical formula of compound A is obtained as $C2H4O$ , with a molar mass $44.05 g/mol$ .

As the molar mass of compound A is calculated as $88.16 g/mol$ , the molecular formula of compound A is $C4H8O2$ .

As compound B is acidic, it will be a carboxylic acid. Moles of $COOH$ can be calculated using the given information.

Given,

$1.000 g$ Of B is neutralized with $33.9 mL of 0.5 M$ sodium hydroxide.

$moles of COOH = (0.5 mol NaOHL)(33.9 mL)(10-3 L1 mL)(1 mol COOH1 mol NaOH) = 0.01695 mol COOH$

The 1.00 g sample of compound B contains 0.01695 moles of $COOH$ .

$mass of COOH = 1.00 g0.01695 mol COOH = 59 g/mol COOH$

As the molar mass of compound A is $88.16 g/mol$ , the molar mass of compound B will be $118 g/mol$ , it will be a dicarboxylic acid. The molecular formula can be given as $C4H6O4$ .

Compound B on loss of water molecule by heating forms compound C. Hence, it will be an anhydride. The molecular formula of compound C can be given by the loss of water molecule from molecular formula of B.

Molecular formula of compound C = $(C4H6O4-H2O) = C4H4O3$ .

As the NMR spectrum of compound C gives only one peak, the compound must be symmetrical with only one type of hydrogens. The structures of compound B and compound C according to the NMR spectrum are given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 1

Compound A is not acidic. It will have an alcohol and aldehyde groups. The structure of compound A can be given as

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 2

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Compound A is a controlled substance because it is metabolized to the weakly acidic date rape drug GHB, $C4H8O3$ . The structure and name of GHB has to be given.

Concept Introduction:

Oxidation:

Addition of oxygen atoms to the molecule or removal hydrogens from the molecule is said to be oxidation.

Aldehyde on oxidation forms carboxylic acid.

(b)

Expert Solution

Explanation of Solution

Compound A is 4-hydroxybutanal with an alcohol and aldehyde groups.

GHB is an acidic compound, hence compound A is oxidized and the aldehyde group converts to carboxylic acid group. The structure and name of GHB is

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 4

Answer 13

(b)

Interpretation Introduction

Interpretation:

Compound A is a controlled substance because it is metabolized to the weakly acidic date rape drug GHB, $C4H8O3$ . The structure and name of GHB has to be given.

Concept Introduction:

Oxidation:

Addition of oxygen atoms to the molecule or removal hydrogens from the molecule is said to be oxidation.

Aldehyde on oxidation forms carboxylic acid.

Answer 14

(b)

Expert Solution

Explanation of Solution

Compound A is 4-hydroxybutanal with an alcohol and aldehyde groups.

GHB is an acidic compound, hence compound A is oxidized and the aldehyde group converts to carboxylic acid group. The structure and name of GHB is

Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics, Chapter 15, Problem 15.107P , additional homework tip 4