Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
Question
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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
Check Mark

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

Chemistry: The Molecular Nature of Matter and Change - Standalone book, 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

Chemistry: The Molecular Nature of Matter and Change - Standalone book, 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.

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 15, Problem 15.107P , additional homework tip  3

(b)

Expert Solution
Check Mark

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

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 15, Problem 15.107P , additional homework tip  4

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Chapter 15 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 15.4 - Prob. 15.4BFPCh. 15.4 - Prob. 15.5AFPCh. 15.4 - Prob. 15.5BFPCh. 15.4 - Prob. 15.6AFPCh. 15.4 - Prob. 15.6BFPCh. 15.4 - Prob. 15.7AFPCh. 15.4 - Prob. 15.7BFPCh. 15.6 - Prob. B15.4PCh. 15.6 - Prob. B15.5PCh. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Silicon lies just below carbon in Group 4A(14) and...Ch. 15 - What is the range of oxidation states for carbon?...Ch. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Define each type of isomer: (a) constitutional;...Ch. 15 - Prob. 15.9PCh. 15 - Prob. 15.10PCh. 15 - Prob. 15.11PCh. 15 - How does an aromatic hydrocarbon differ from a...Ch. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Draw structures from the following names, and...Ch. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Determine the type of each of the following...Ch. 15 - Prob. 15.42PCh. 15 - Prob. 15.43PCh. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Prob. 15.46PCh. 15 - Prob. 15.47PCh. 15 - Prob. 15.48PCh. 15 - Prob. 15.49PCh. 15 - Prob. 15.50PCh. 15 - Prob. 15.51PCh. 15 - Prob. 15.52PCh. 15 - Prob. 15.53PCh. 15 - Prob. 15.54PCh. 15 - Prob. 15.55PCh. 15 - Prob. 15.56PCh. 15 - Prob. 15.57PCh. 15 - Prob. 15.58PCh. 15 - Prob. 15.59PCh. 15 - Prob. 15.60PCh. 15 - Prob. 15.61PCh. 15 - Prob. 15.62PCh. 15 - Prob. 15.63PCh. 15 - Prob. 15.64PCh. 15 - Prob. 15.65PCh. 15 - Prob. 15.66PCh. 15 - Prob. 15.67PCh. 15 - Prob. 15.68PCh. 15 - Prob. 15.69PCh. 15 - Prob. 15.70PCh. 15 - Prob. 15.71PCh. 15 - Prob. 15.72PCh. 15 - Prob. 15.73PCh. 15 - Prob. 15.74PCh. 15 - Prob. 15.75PCh. 15 - Prob. 15.76PCh. 15 - Prob. 15.77PCh. 15 - Prob. 15.78PCh. 15 - Prob. 15.79PCh. 15 - Prob. 15.80PCh. 15 - Prob. 15.81PCh. 15 - Prob. 15.82PCh. 15 - Prob. 15.83PCh. 15 - Prob. 15.84PCh. 15 - Prob. 15.85PCh. 15 - Prob. 15.86PCh. 15 - Prob. 15.87PCh. 15 - What is the key structural difference between...Ch. 15 - Protein shape, function, and amino acid sequence...Ch. 15 - What linkage joins the monomers in each strand of...Ch. 15 - What is base pairing? How does it pertain to DNA...Ch. 15 - RNA base sequence, protein amino acid sequence,...Ch. 15 - Prob. 15.93PCh. 15 - Prob. 15.94PCh. 15 - Draw the structure of each of the following...Ch. 15 - Prob. 15.96PCh. 15 - Write the sequence of the complementary DNA strand...Ch. 15 - Prob. 15.98PCh. 15 - Prob. 15.99PCh. 15 - Prob. 15.100PCh. 15 - Prob. 15.101PCh. 15 - Amino acids have an average molar mass of 100...Ch. 15 - Prob. 15.103PCh. 15 - Prob. 15.104PCh. 15 - Some of the most useful compounds for organic...Ch. 15 - Prob. 15.106PCh. 15 - Prob. 15.107PCh. 15 - Prob. 15.108PCh. 15 - Prob. 15.109PCh. 15 - Prob. 15.110PCh. 15 - Prob. 15.111PCh. 15 - Prob. 15.112PCh. 15 - The polypeptide chain in proteins does not exhibit...Ch. 15 - Prob. 15.114PCh. 15 - Prob. 15.115PCh. 15 - Prob. 15.116PCh. 15 - Prob. 15.117PCh. 15 - Wastewater from a cheese factory has the following...Ch. 15 - Prob. 15.119P
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