Chapter 3, Problem 3.63QP

Ethylene glycol is used as an automobile antifreeze and in the manufacture of polyester fibers. The name glycol stems from the sweet taste of this poisonous compound. Combustion of 6.38 mg of ethylene glycol gives 9.06 mg CO₂ and 5.58 mg H₂O. The compound contains only C, H, and O. What are the mass percentages of the elements in ethylene glycol?

Interpretation Introduction

Interpretation:

Mass percentages of each elements present in $\text{6}\text{.38 mg}$ of Ethylene glycol should be determined.

Concept introduction:

Mass percentage:

The percent ration of mass of analyte that is present in a given sample with total mass of sample to give a mass percent of analyte present in a given sample.

$\text{Mass}\text{precent}\text{\%}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{of}\text{sample}}\times 100$

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ration between taken mass of compound and molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Empirical formula

The simplest ratio of the elements that are present in a molecule are representing as empirical formula.

Answer to Problem 3.63QP

The mass percentage of $\text{C}$ in $\text{ethylene glycol}$ is $\text{38}\text{.7}\text{\%}$

The mass percentage of $\text{H}$ in $\text{ethylene glycol}$ is $\text{9}\text{.79}\text{\%}$

The mass percentage of $\text{O}$ in $\text{ethylene glycol}$ is $51.5\text{\%}$

Explanation of Solution

To calculate the masses of Carbon, Hydrogen and Oxygen.

Molar mass of Carbon is $\text{12}\text{.01 g}$

Molar mass of Hydrogen is $\text{1}\text{.008 g}$

Molar mass of Oxygen is $\text{16}\text{.00 g}$

One ${\text{CO}}_{\text{2}}$ molecule has 1 Carbon atom.

Mass of Carbon present in $\text{9}{\text{.06 mg CO}}_{\text{2}}$ is,

$\begin{array}{l}\text{Mass}\text{of}\text{C}\text{=9}{\text{.06 mg CO}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{44}\text{.0}\text{g}{\text{CO}}_{\text{2}}}\text{×}\frac{\text{12}\text{.0}}{\text{1}\text{mol}\text{C}}\\ \text{=}\text{2}\text{.472}\text{mg}\text{C}\end{array}$

One ${\text{H}}_{\text{2}}\text{O}$ molecule has 2 Hydrogen atoms.

Mass of Hydrogen present in $\text{5}{\text{.58 mg H}}_{\text{2}}\text{O}$ is,

$\begin{array}{l}\text{Mass}\text{of}\text{H}\text{=5}{\text{.58 mg H}}_{\text{2}}\text{O×}\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{1}\text{.008}\text{H}}{\text{1}\text{mol}\text{H}}\\ \text{=}\text{0}\text{.6243}\text{mg}\text{H}\end{array}$

The given masses and molar masses of Carbon, Hydrogen are plugged in above equation to give masses of Carbon, Hydrogen are present in $\text{6}\text{.38 mg of ethylene glycol}$.

Mass of oxygen present in $\text{6}\text{.38 mg of ethylene glycol}$ is,

$\begin{array}{l}\text{=}\text{6}\text{.38}\text{mg-}\text{(2}\text{.472+0}\text{.6243)}\\ \text{=}\text{3}\text{.284}\text{mg}\text{O}\end{array}$

Subtract the calculated masses of Carbon, Hydrogen from $\text{6}\text{.38 mg of ethylene glycol}$ to give a mass of oxygen present in $\text{6}\text{.38 mg of ethylene glycol}$.

To calculate the mass percentage of Carbon, Hydrogen and Oxygen

To calculate the mass percentage of $\text{C}$ in $\text{ethylene glycol}$

$\begin{array}{l}\text{\%}\text{C}\text{in}\text{ethylene glycol}\text{=}\frac{\text{Mass}\text{of}\text{C}}{\text{Mass}\text{of}\text{Ethylene glycol}}\text{×100}\\ \text{=}\frac{\text{2}\text{.472}\text{mg}}{\text{6}\text{.38}\text{mg}}\text{×100}\\ \text{=}\text{38}\text{.7\%}\end{array}$

Molar masses of $\text{C}$ and $\text{ethylene glycol}$ are plugged in above equation to give percentage composition of $\text{C}$ present in $\text{ethylene glycol}$.

To calculate the mass percentage of $\text{H}$ in $\text{ethylene glycol}$

$\begin{array}{l}\text{\%}\text{H}\text{in}\text{ethylene glycol}\text{=}\frac{\text{Mass}\text{of}\text{H}}{\text{Mass}\text{of}\text{Ethylene glycol}}\text{×100}\\ \text{=}\frac{\text{0}\text{.6243}\text{mg}}{\text{6}\text{.38}\text{mg}}\text{×100}\\ \text{=}\text{9}\text{.79\%}\end{array}$

Molar masses of $\text{H}$ and $\text{ethylene glycol}$ are plugged in above equation to give percentage composition of $\text{H}$ present in $\text{ethylene glycol}$.

To calculate the mass percentage of $\text{H}$ in $\text{ethylene glycol}$

$\begin{array}{l}\text{\%}\text{O}\text{in}\text{ethylene glycol}\text{=}\frac{\text{Mass}\text{of}\text{O}}{\text{Mass}\text{of}\text{Ethylene glycol}}\text{×100}\\ \text{=}\frac{\text{3}\text{.284}\text{mg}}{\text{6}\text{.38}\text{mg}}\text{×100}\\ \text{=}\text{51}\text{.5\%}\end{array}$

Molar masses of $\text{O}$ and $\text{ethylene glycol}$ are plugged in above equation to give percentage composition of $\text{O}$ present in $\text{ethylene glycol}$.

Hence,

The mass percentage of $\text{C}$ in $\text{ethylene glycol}$ is $\text{38}\text{.7}\text{\%}$

The mass percentage of $\text{H}$ in $\text{ethylene glycol}$ is $\text{9}\text{.79}\text{\%}$

The mass percentage of $\text{O}$ in $\text{ethylene glycol}$ is $51.5\text{\%}$

Conclusion

Mass percentages of each elements present in $\text{6}\text{.38 mg}$ of Ethylene glycol was determined.