Chapter 3, Problem 111QRT

Interpretation Introduction

Interpretation:

The composition of the liquid has to be calculated.

Concept introduction:

Balancing reaction:

Balanced reaction is a chemical reaction in which number of atoms for each element in the reaction and the total charge are same on both reactant side and the product side.

Steps in balancing the information

• Step 1: Write the unbalanced equation
• Step 2: Find the coefficient to balance the equation.
• The coefficient should be reduced to the smallest whole number.

Combustion reaction:

Combustion is an exothermic reaction between a fuel (the reductant) and an oxidant. Oxidant is usually atmospheric oxygen.

Generally organic molecule undergoes combustion reaction which produces carbon dioxide and water molecule.

  ${\text{Hydrocarbons + O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

Number of moles can be calculated by using following formula,

  $\text{No}\text{.of Moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Explanation of Solution

Generally organic molecule undergoes combustion reaction which produces carbon dioxide and water molecule. Therefore,

1. (a) if the ethyl alcohol burns the balancing equation is given below,

  ${\text{C}}_2{\text{H}}_5{\text{OH + O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

When balancing the equation, altering the subscripts is restricted and changing the coefficients can be done. There are two carbon atoms in the left side of the reaction and one carbon in the right side of the reaction, therefore two molecule of carbon dioxide is added to the right side of the reaction Now, the balanced equation is given below.

    ${\text{C}}_2{\text{H}}_5{\text{OH + O}}_{\text{2}}\text{(g) }\to {\text{ 2 CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

There are six hydrogen atoms in the left side of the reaction and two hydrogen atoms in the right side of the reaction, therefore three molecule of water is added to the right side of the reaction. Now, the balanced equation is given below.

  ${\text{C}}_2{\text{H}}_5{\text{OH + O}}_{\text{2}}\text{(g) }\to {\text{ 2 CO}}_{\text{2}}{\text{(g) + 3 H}}_{\text{2}}\text{O(g)}$

There are three oxygen atoms in the left side of the reaction and seven oxygen atoms in the right side of the reaction, therefore three molecule of oxygen is added to the left side of the reaction. Now, the balanced equation is given below.

    ${\text{C}}_2{\text{H}}_5{\text{OH + 3 O}}_{\text{2}}\text{(g) }\to {\text{ 2 CO}}_{\text{2}}{\text{(g) + 3 H}}_{\text{2}}\text{O(g)}$

1. (b) if the ethyl alcohol burns the balancing equation is given below,

  ${\text{CH}}_3{\text{OH + O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

When balancing the equation, altering the subscripts is restricted and changing the coefficients can be done. There are four hydrogen atoms in the left side of the reaction and two hydrogen atoms in the right side of the reaction. Therefore, two molecule of water is added to the right side of the reaction. Now, the balanced equation is given below.

    ${\text{CH}}_3{\text{OH + O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(g)}$

There are three oxygen atoms in the left side of the reaction and four oxygen atoms in the right side of the reaction. Therefore, $\text{3/2}$ molecule of oxygen is added to the left side of the reaction. Now, the balanced equation is given below.

  ${\text{CH}}_3\text{OH + }\frac{3}{2}{\text{O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(g)}$

Multiply the reaction by 2 to get the whole numbers, now, the balanced equation is given below.

    ${\text{2CH}}_3{\text{OH + 3O}}_{\text{2}}\text{(g) }\to {\text{ 2 CO}}_{\text{2}}{\text{(g) + 4 H}}_{\text{2}}\text{O(g)}$

The statement is shown below,

The weight of ethyl alcohol and methyl alcohol is $\text{0}\text{.280 g}$, it burned and forms $\text{0}{\text{.385 g CO}}_{\text{2}}\left(\text{g}\right)$

Let consider,

X is the mass of the ethyl alcohol.

Y is the mass of methyl alcohol.

Total mass of sample =$\text{X + Y = 0}\text{.280 g}$.

  $\text{Y = 0}\text{.280 – X}$

Molar mass of carbon dioxide = $\text{44}\text{.0095 g/mol}$,

Moles of carbon dioxide is calculated as follows,

  $\begin{array}{l}\text{Moles of the carbon dioxide}\text{=}\text{0}\text{.385}\text{g×}\frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{44}\text{.0095}\text{g}{\text{CO}}_{\text{2}}}\\ \text{Moles of the carbon dioxide}\text{=}\text{8}{\text{.748×10}}^{\text{-3}}\text{mol}\end{array}$

Molar mass of ethanol = $\text{46}\text{.0682 g/mol}$, Molar mass of methanol  = $\text{32}\text{.0417 g/mol}$.

According to the balance equation, one mole of ethanol reacts with three moles of oxygen gives two moles of two moles of carbon dioxide and three moles of water.

Similarly, two moles of methanol reacts with three moles of oxygen gives two moles of two moles of carbon dioxide and four moles of water.

Moles of carbon is calculated as follows,

  $\begin{array}{l}\text{8}\text{.748}\text{×}{\text{10}}^{\text{-3}}\text{mol}{\text{CO}}_{\text{2}}\text{=X}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH×}\frac{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{\text{46}\text{.0682}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{×}\frac{\text{2}\text{mol}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\\ \text{+}\\ \text{Y}\text{g}{\text{CH}}_{\text{3}}\text{OH×}\frac{\text{1}\text{mol}{\text{CH}}_{\text{3}}\text{OH}}{\text{32}\text{.0417}\text{g}{\text{CH}}_{\text{3}}\text{OH}}\text{×}\frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}{\text{CH}}_{\text{3}}\text{OH}}\\ \\ \text{8}\text{.748}\text{×}{\text{10}}^{\text{-3}}\text{mol}{\text{CO}}_{\text{2}}\text{=}\text{0}\text{.0434}\text{X}\text{+}\text{0}\text{.0312}\text{Y}\end{array}$

Substitute the Y value in the equation,

  $\begin{array}{l}\text{8}\text{.748}\text{×}{\text{10}}^{\text{-3}}\text{=}\text{0}\text{.0434}\text{X}\text{+}\text{0}\text{.0312}\text{Y}\\ \text{8}\text{.748}\text{×}{\text{10}}^{\text{-3}}\text{=}\text{0}\text{.0434}\text{X}\text{+}\text{0}\text{.0312}\text{(0}\text{.280}\text{-}\text{X)}\\ \text{8}\text{.748}\text{×}{\text{10}}^{\text{-3}}\text{=}\text{0}\text{.0434}\text{X}\text{+}\text{8}\text{.738}\text{×}{\text{10}}^{\text{-3}}\text{-}\text{0}\text{.0312}\text{X}\\ \text{8}\text{.748}\text{×}{\text{10}}^{\text{-3}}\text{-}\text{8}\text{.738}\text{×}{\text{10}}^{\text{-3}}\text{=}\text{0}\text{.0122X}\\ \text{1}\text{.0×}{\text{10}}^{\text{-5}}\text{=}\text{0}\text{.0122X}\\ \text{X = 8}{\text{.197×10}}^{\text{-4}}\end{array}$

Total mass of sample = $\text{X + Y = 0}\text{.280 g}$

  $\begin{array}{l}\text{X + Y = 0}\text{.280 g}\\ \text{8}{\text{.197×10}}^{\text{-4}}\text{ + Y = 0}\text{.280 g}\\ \text{Y = 0}\text{.279 g}\end{array}$.

Percentage of methanol is calculated as follows,

    $\begin{array}{l}\text{Composition of the Methanol}\text{=}\frac{\text{0}\text{.279}\text{g}\text{methanol}}{\text{0}\text{.280}\text{g}\text{liquid}}\text{×}\text{100}\text{\%}\\ \text{Composition of the Methanol}\text{=}\text{99}\text{.64}\text{\%}\end{array}$

Percentage of ethanol is calculated as follows,

    $\begin{array}{l}\text{Composition of the ethanol}\text{=}\frac{\text{8}{\text{.197×10}}^{\text{-4}}\text{g}\text{ethanol}}{\text{0}\text{.280}\text{g}\text{liquid}}\text{×}\text{100}\text{\%}\\ \text{Composition of the ethanol}\text{=}\text{0}\text{.293}\text{\%}\end{array}$

The composition of the methanol is $\text{99}\text{.64}\text{\%}$ and ethanol is $\text{0}\text{.293}\text{\%}\text{.}$