Q100CP - Chemistry: Molecular Nature Of Matter And Change

Question

The oxidation of $\mathbf{1}\mathit{m}\mathit{o}\mathit{l}$  of glucose supplies enough metabolic energy to form $\mathbf{36}\mathit{m}\mathit{o}\mathit{l}$  of ATP. Oxidation of $\mathbf{1}\mathit{m}\mathit{o}\mathit{l}$  of a typical dietary fat like tristearin  $\mathbf{\left(}{\mathit{C}}_{\mathbf{57}}{\mathit{H}}_{\mathbf{116}}{\mathit{O}}_{\mathbf{6}}\mathbf{\right)}$ yields enough energy to form  $\mathbf{458}\mathit{m}\mathit{o}\mathit{l}$ of ATP. 

(a) How many molecules of ATP can form per gram of glucose? 

(b) Per gram of tristearin?

Step-by-Step Solution

Answer

(a) The number of molecules that can form is $1.20\times {10}^{23}$  per gram of glucose.

(b) The number of molecules that can form is  $3.07\times {10}^{23}$ per gram of tristearin

Step 1: Concept Introduction

The addition of oxygen or any other electronegative element, or the removal of hydrogen or any other electropositive element, is referred to as oxidation in classical or earlier concepts. According to the electronic idea, oxidation is defined as the loss of one or more electrons by an atom or ion.

Step 2: Per gram of Glucose

(a)

This problem is simple stoichiometry. First, write the given and possible values that will be used.

Molecular Mass of Glucose  ${\text{ = 180.2 g/mol C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$

  $1mol$of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ = 36 mol}$  of $ATP$ 

Avogadro’s Number  $=6.022\times {10}^{23}$

The number of molecules is calculated as –

 $$\begin{gathered} 1 g {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{=}\frac{1mol\times {\displaystyle {{\displaystyle \text{C}}}_{\text{6}}}{\displaystyle {{\displaystyle \text{H}}}_{\text{12}}}{\displaystyle {{\displaystyle \text{O}}}_{\text{6}}}}{180.2mol\times {\displaystyle {{\displaystyle \text{C}}}_{\text{6}}}{\displaystyle {{\displaystyle \text{H}}}_{\text{12}}}{\displaystyle {{\displaystyle \text{O}}}_{\text{6}}}}\text{×}\frac{36molATP}{1mol\times {\displaystyle {{\displaystyle \text{C}}}_{\text{6}}}{\displaystyle {{\displaystyle \text{H}}}_{\text{12}}}{\displaystyle {{\displaystyle \text{O}}}_{\text{6}}}}\text{×}\frac{6.022\times {10}^{23}molecules}{1mol ATP} \\ {\text{=1.20×10}}^{23}\text{molecules of ATP per gram glucose} \end{gathered}$$

 

Therefore, the value is obtained as$1.20\times {10}^{23}$  .

Step 3: Per gram of Tristearin

(b)

This problem is simple stoichiometry. First, write the given and possible values that will be used.

Molecular Mass of Tristearin  ${\text{ = 897.5 g/mol C}}_{57}{\text{H}}_{\text{116}}{\text{O}}_{\text{6}}$

 $1mol$ of${\text{C}}_{57}{\text{H}}_{\text{116}}{\text{O}}_{\text{6}}\text{ = 458 mol}$   of  $ATP$

Avogadro’s Number   $=6.022\times {10}^{23}$

The number of molecules is calculated as –

 $$\begin{gathered} 1 g {\text{C}}_{57}{\text{H}}_{\text{116}}{\text{O}}_{\text{6}}\text{=}\frac{1mol \times {\displaystyle {{\displaystyle \text{C}}}_{57}}{\displaystyle {{\displaystyle \text{H}}}_{\text{116}}}{\displaystyle {{\displaystyle \text{O}}}_{\text{6}}}}{897.5g\times {\displaystyle {{\displaystyle \text{C}}}_{57}}{\displaystyle {{\displaystyle \text{H}}}_{\text{116}}}{\displaystyle {{\displaystyle \text{O}}}_{\text{6}}}}\text{×}\frac{558mol ATP}{1mol {\displaystyle {{\displaystyle \text{C}}}_{57}}{\displaystyle {{\displaystyle \text{H}}}_{\text{116}}}{\displaystyle {{\displaystyle \text{O}}}_{\text{6}}}}\text{×}\frac{6.022\times {10}^{23}molecules}{1mol ATP} \\ {\text{=3.07×10}}^{23}\text{ molecules of ATP per gram tristearin} \end{gathered}$$

 

Therefore, the value is obtained as $3.07\times {10}^{23}$ .