Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 15, Problem 15.110P

(a)

Interpretation Introduction

Interpretation:

Compound A containing C, H and O in branched structure is optically active.  A 0.500 g of sample burns in excess O2 to yield 1.25 g of CO2 and 0.613 g of H2O.  The empirical formula of the compound has to be determined.

(a)

Expert Solution
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Explanation of Solution

Given,

  • 1.25 g of CO2
  • 0.613 g of H2O
  • Total sample = 0.500 g.

From the given mass, moles can be calculated.

For Carbon:

  Moles of C = (1.25 g CO2)(1 mol CO2440.1 g CO2)(1 mol C1 mol CO2)                   = 0.0284026 mol C

For Hydrogen:

  moles of H = (0.613 g H2O)(1 mol H2O18.02 g H2O)(2 mol H1 mol H2O)                  = 0.0680355 mol H

The mass of carbon and hydrogen can be calculated from the moles obtained.

For carbon:

  Mass of C (g) = (0.0284026 mol C)(12.01 g C1 mol C)                       = 0.341115 g C

For hydrogen:

  Mass of H (g) = (0.0680355 mol H)(1.008 g H1 mol H)                       = 0.06857978 g H

The mass of oxygen can be determined by subtracting the mass of carbon and hydrogen from the total mass of C, H and O.

  mass of O (g) = 0.500 g (C,H,O)-(0.341115 g C + 0.06857978 g H)                      = 0.0903052 g O

The mass of the oxygen is 0.0903052 g.

By using the mass of oxygen, moles can be determined.

  moles of O = (0.0903052 g O)(1 mol O16.00 g O)                   = 0.005644075 mol O

The moles of C, H and O are divided by smallest number of moles.

Carbon:

  0.0284026 mol C0.005644075 mol O = 5

Hydrogen:

  0.0680355 mol H0.005644075 mol O = 12

Oxygen:

  0.005644075 mol O0.005644075 mol O = 1

The empirical formula can be given as C5H12O.

(b)

Interpretation Introduction

Interpretation:

When 0.225 g of compound A vaporizes at 755 torr and 970C, the vapour occupies 78.0 mL.  The molecular formula of compound A has to be determined.

Concept Introduction:

The equation used to find the molecular weight of the compound is

  PV       = mRTmassmass     = mRTPV

Where, P = pressureV = volumeR = gas constantT = temperaturem = mass of compound

(b)

Expert Solution
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Explanation of Solution

The molecular weight of the compound can be calculated as

Given,

  • pressure = 755 torr
  • temperature = 970C
  • volume = 78.0 mL
  • total mass = 0.225 g

  mass = mRTPV         = (0.225g)(0.0821L.atmmol.K)((273+97)K)(755 torr)(78.0 mL)(1 mL10-3 L)(760 torr1 atm)         = 88.206 = 88.2 g/mol

The empirical formula weight can be calculated.

Empirical formula: C5H12O

Empirical mass: 88.15 g/mol.

As the molecular weight and empirical mass of nearly same, the molecular formula is C5H12O.

(c)

Interpretation Introduction

Interpretation:

On careful oxidation, Compound A yields a ketone.  The compound A has to be drawn and named.  The chiral centre has to be circled.

Concept Introduction:

Carbonyl compounds are formed by the oxidation of alcohols.  Ketone is formed on oxidation of secondary alcohol.

(c)

Expert Solution
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Explanation of Solution

The compound A is a secondary alcohol as it gives a ketone as product on oxidation.  The structure of the formula C5H12O can be

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 15, Problem 15.110P , additional homework tip  1

The chiral centre in the structure is the carbon bonded to four different groups, -CH3, -CH2, -OH and H.

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 15, Problem 15.110P , additional homework tip  2

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Chapter 15 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

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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