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.

Explanation of Solution

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

The reaction mechanism is as follows

(b)

Interpretation Introduction

Interpretation:

If the government required auto fuel mixtures contain $\text{2}\text{.7\%}$ oxygen by mass to reduce $\text{CO}$ emissions, the number of grams of MTBE would have to be added to each $\text{100 g}$ 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

  $\text{mass percentage =}\frac{\text{weight of the component}}{\text{total weight of compound}}\text{ × 100}$

Mass:

  $\text{mass (g) = weight of the component × }\frac{\text{weight of the compound}}{\text{mass percentage of component}}$

Explanation of Solution

Given,

  $\begin{array}{l}\text{Percentage of Oxygen = 2}\text{.7 \%}\\ \\ \text{Total mass of MTBE = 100 g}\end{array}$

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

  $\text{mass percentage =}\frac{\text{weight of the component}}{\text{total weight of compound}}\text{ × 100}$

The weight of oxygen is $\text{16}\text{.0 g}$ and total weight of MTBE is $\text{88}\text{.15 g}$.  Substituting the values in the above equation, we get,

  $\begin{array}{l}\text{mass percentage =}\frac{\text{weight of the oxygen}}{\text{total weight of MTBE}}\text{ × 100 \%}\\ \\ \text{mass percentage of oxygen in MTBE = }\frac{\text{16}\text{.0g}}{\text{88}\text{.15g}}\text{ × 100 \%}\\ \text{ = 18}\text{.150879 \%}\end{array}$

Mass of MTBE can be calculated using the equation

  $\text{mass (g) = weight of the component × }\frac{\text{weight of the compound}}{\text{mass percentage of component}}$

The weight of the oxygen is given as $\text{2}\text{.7 g}$, total weight of MTBE is $\text{100 g}$ and the mass percentage of oxygen is calculated as $\text{18}\text{.150879 g}$.

  $\begin{array}{l}\text{mass of MTBE (g) = weight of the Oxygen × }\frac{\text{weight of the MTBE}}{\text{mass percentage of Oxygen}}\\ \\ \text{mass of MTBE (g) = 2}\text{.7g}\left(\frac{\text{100 g of MTBE}}{\text{18}\text{.150879 g}}\right)\\ \\ \text{ = 14}\text{.8753 = 15g MTBE/100}\text{.g gasoline }\end{array}$

The mass of MTBE for $\text{100 g}$ of gasoline is $\text{15 g}$.

(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 $\text{0}\text{.740 g/mL}$).

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

  $\begin{array}{l}\text{Density = }\frac{\text{mass}}{\text{volume}}\\ \\ \text{volume = }\frac{\text{mass}}{\text{Density}}\end{array}$

Explanation of Solution

The given density of gasoline and MTBE is $\text{0}\text{.740 g/mL}$.

Mass of MTBE for $\text{100 g}$ of gasoline is $\text{14}\text{.8753 g}$.

The volume of MTBE can be calculated as

$\begin{array}{l}\text{Volume of MTBE (L) = }\left(\frac{\text{14}\text{.8753g MTBE}}{\text{100g gasoline}}\right)\left(\frac{\text{0}\text{.740 g MTBE}}{\text{1mL gasoline}}\right)\left(\frac{\text{1mL}}{{\text{10}}^{\text{-3}}\text{L}}\right)\left(\frac{\text{1mL MTBE}}{\text{0}\text{.740 g MTBE}}\right)\left(\frac{{\text{10}}^{\text{-3}}\text{L}}{\text{1mL}}\right)\\ \\ \text{ = 0}\text{.148753 L MTBE/L gasoline}\end{array}$

The volume of MTBE in $\text{1 L}$ of gasoline is $\text{0}\text{.148753L}$.

(d)

Interpretation Introduction

Interpretation:

The number of litres of air (${\text{21\% O}}_{\text{2}}\text{ by volume}$ ) needed at ${\text{24}}^{\text{0}}\text{C}$ and $\text{1}\text{.00 atm}$ to fully combust $\text{1}\text{.00 L}$ 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

  $\begin{array}{l}\text{PV = nRT}\\ \\ \text{V =}\frac{\text{nRT}}{\text{P}}\end{array}$

Where, $\begin{array}{l}\text{P = pressure}\\ \\ \text{V = volume}\\ \\ \text{R = gas constant}\\ \\ \text{T = temperature}\\ \\ \text{n = number of moles}\end{array}$

Explanation of Solution

The balanced equation of MTBE on combustion with ${\text{O}}_{\text{2}}$ can be given as

The number of moles of oxygen required can be calculated as

$\begin{array}{l}{\text{Moles of O}}_{\text{2}}\text{ = }\left(\text{1}\text{.00 L MTBE}\right)\left(\frac{\text{1 mL}}{{\text{10}}^{\text{-3}}\text{ L}}\right)\left(\frac{\text{0}\text{.740 g MTBE}}{\text{1 mL}}\right)\left(\frac{\text{1 mol MTBE}}{\text{88}\text{.15g MTBE}}\right)\left(\frac{{\text{15 mol O}}_{\text{2}}}{\text{2 mol MTBE}}\right)\\ \\ \text{ = 62}{\text{.96086 mol O}}_{\text{2}}\end{array}$

Given,

  $\begin{array}{l}\text{Pressure = 1}\text{.00 atm}\\ \\ {\text{Temperature = 24}}^{\text{0}}\text{C}\\ \\ {\text{Moles of O}}_{\text{2}}\text{ = 62}\text{.96086 mol}\end{array}$

The volume can be calculated as

  $\begin{array}{l}\text{V = }\frac{\text{nRT}}{\text{P}}\\ \\ \text{V = }\frac{\left(\text{62}{\text{.96086 mol O}}_{\text{2}}\right)\left(\text{0}\text{.0821}\frac{\text{L}\text{.atm}}{\text{mol}\text{.K}}\right)\left(\left(\text{273+24}\right)\text{K}\right)}{\text{1}\text{.00 atm}}\left(\frac{\text{100\%}}{\text{21\%}}\right)\\ \\ \text{V = 7}{\text{.310565×10}}^{\text{3}}\text{ L air}\end{array}$

The volume air required is $\text{7}{\text{.310565×10}}^{\text{3}}\text{ L}$.