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 15.103P

(a)

Interpretation Introduction

Interpretation:

MTBE (methyl tertiary-butyl ether) is synthesized by the catalyzed reaction of 2-methylpropene with methanol.  The balanced equation for the synthesis of MTBE has to be written.

Concept Introduction:

Alkene reacts with alcohol to form ethers.  The alcohol forms alkoxide anion and it reacts with the alkene to form ether.

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 15, Problem 15.103P , additional homework tip  1

(a)

Expert Solution
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Explanation of Solution

MTBE is synthesized from 2-methylpropene and methanol.  The reaction can be given as follows,

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 15, Problem 15.103P , additional homework tip  2

The reaction mechanism is as follows

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 15, Problem 15.103P , additional homework tip  3

(b)

Interpretation Introduction

Interpretation:

If the government required auto fuel mixtures contain 2.7% oxygen by mass to reduce CO emissions, the number of grams of MTBE would have to be added to each 100 g of gasoline has to be given.

Concept Introduction:

Mass percentage:

It is the concentration of an element or component in the total compound.  It can be calculated as

  mass percentage =weight of the componenttotal weight of compound × 100

Mass:

  mass (g) = weight of the component × weight of the compoundmass percentage of component

(b)

Expert Solution
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Explanation of Solution

Given,

  Percentage of Oxygen = 2.7 %Total mass of MTBE   = 100 g

The mass percentage of oxygen in MTBE has to be calculated.

  mass percentage =weight of the componenttotal weight of compound × 100

The weight of oxygen is 16.0 g and total weight of MTBE is 88.15 g.  Substituting the values in the above equation, we get,

  mass percentage =weight of the oxygentotal weight of MTBE × 100 %mass percentage of oxygen in MTBE = 16.0g88.15g × 100 %                                                           = 18.150879 %

Mass of MTBE can be calculated using the equation

  mass (g) = weight of the component × weight of the compoundmass percentage of component

The weight of the oxygen is given as 2.7 g, total weight of MTBE is 100 g and the mass percentage of oxygen is calculated as 18.150879 g.

  mass of MTBE (g) = weight of the Oxygen × weight of the MTBEmass percentage of Oxygenmass of MTBE (g) = 2.7g(100 g of MTBE18.150879 g)                               = 14.8753 = 15g MTBE/100.g gasoline    

The mass of MTBE for 100 g of gasoline is 15 g.

(c)

Interpretation Introduction

Interpretation:

The number of litres of MTBE would be in each litre of fuel mixture has to be calculated (density of both gasoline and MTBE is 0.740 g/mL).

Concept Introduction:

The volume of the compound can be calculated by using the density of that compound.  Density can be calculated by using the equation

  Density = massvolumevolume = massDensity

(c)

Expert Solution
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Explanation of Solution

The given density of gasoline and MTBE is 0.740 g/mL.

Mass of MTBE for 100 g of gasoline is 14.8753 g.

The volume of MTBE can be calculated as

Volume of MTBE (L) = (14.8753g MTBE100g gasoline)(0.740 g MTBE1mL gasoline)(1mL10-3L)(1mL MTBE0.740 g MTBE)(10-3L1mL)                                    = 0.148753 L MTBE/L gasoline

The volume of MTBE in 1 L of gasoline is 0.148753L.

(d)

Interpretation Introduction

Interpretation:

The number of litres of air (21% O2 by volume ) needed at 240C and 1.00 atm to fully combust 1.00 L of MTBE 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 = nRTV    =nRTP

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

(d)

Expert Solution
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Explanation of Solution

The balanced equation of MTBE on combustion with O2 can be given as

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 15, Problem 15.103P , additional homework tip  4

The number of moles of oxygen required can be calculated as

Moles of O2 = (1.00 L MTBE)(1 mL10-3 L)(0.740 g MTBE1 mL)(1 mol MTBE88.15g  MTBE)(15 mol O22 mol MTBE)                     = 62.96086 mol O2

Given,

  Pressure       = 1.00 atmTemperature = 240CMoles of O2 = 62.96086 mol

The volume can be calculated as

  V = nRTPV = (62.96086 mol O2)(0.0821L.atmmol.K)((273+24)K)1.00 atm(100%21%)V = 7.310565×103 L air

The volume air required is 7.310565×103 L.

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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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