Chapter 3, Problem 3.64QP

Interpretation Introduction

Interpretation:

Mass percentages of each elements present in $\text{5}\text{.23 mg}$ of $\text{Phenol}$ 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 3.64QP

The mass percentage of $\text{C}$ in $\text{Phenol}$ is $\text{76}\text{.54}\text{\%}$

The mass percentage of $\text{H}$ in $\text{Phenol}$ is $\text{6}\text{.44}\text{\%}$

The mass percentage of $\text{O}$ in $\text{Phenol}$ is $17.0\text{\%}$

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 $\text{14}{\text{.67 mg CO}}_{\text{2}}$ is,

$\begin{array}{l}\text{Mass}\text{of}\text{C}\text{=14}{\text{.67 mg CO}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{44}\text{.0}\text{g}{\text{CO}}_{\text{2}}}\text{×}\frac{\text{12}\text{.0}}{\text{1}\text{mol}\text{C}}\\ \text{=}4.0033\text{mg}\text{C}\end{array}$

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

Mass of Hydrogen present in $\text{3}{\text{.01 mg H}}_{\text{2}}\text{O}$ is,

$\begin{array}{l}\text{Mass}\text{of}\text{H}\text{=3}{\text{.01 mg H}}_{\text{2}}\text{O×}\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{1}\text{.008}\text{H}}{\text{1}\text{mol}\text{H}}\\ \text{=}0.3368\text{mg}\text{H}\end{array}$

The given masses and molar masses of Carbon, Hydrogen are plugged in above equation to give masses of Carbon, Hydrogen are present in $\text{5}\text{.23 mg}$ of $\text{Phenol}$.

Mass of oxygen present in $\text{Phenol}$ is,

$\begin{array}{l}\text{=}5.23\text{mg-}\text{(4}\text{.0033+0}\text{.3368)}\\ \text{=}0.8899\text{mg}\text{O}\end{array}$

Subtract the calculated masses of Carbon, Hydrogen from $\text{5}\text{.23 mg}$ of $\text{Phenol}$ to give a mass of Oxygen present in $\text{5}\text{.23 mg}$ of $\text{Phenol}$.

To calculate the mass percentage of Carbon, Hydrogen and Oxygen

To calculate the mass percentage of $\text{C}$ in $\text{5}\text{.23 mg}$ of $\text{Phenol}$

$\begin{array}{l}\text{\%}\text{C}\text{in}\text{Phenol}\text{=}\frac{\text{Mass}\text{of}\text{C}}{\text{Mass}\text{of}\text{Phenol}}\text{×100}\\ \text{=}\frac{\text{4}\text{.0033}\text{mg}}{\text{5}\text{.23}\text{mg}}\text{×100}\\ \text{=}\text{76}\text{.54\%}\end{array}$

Molar masses of $\text{C}$ and $\text{Phenol}$ are plugged in above equation to give percentage composition of $\text{C}$ present in $\text{5}\text{.23 mg}$ of $\text{Phenol}$.

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

$\begin{array}{l}\text{\%}\text{H}\text{in}\text{Phenol}\text{=}\frac{\text{Mass}\text{of}\text{H}}{\text{Mass}\text{of}\text{Phenol}}\text{×100}\\ \text{=}\frac{\text{0}\text{.3368}\text{mg}}{5.23\text{mg}}\text{×100}\\ \text{=}6.44\text{\%}\end{array}$

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

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

$\begin{array}{l}\text{\%}\text{O}\text{in}\text{Phenol}\text{=}\frac{\text{Mass}\text{of}\text{O}}{\text{Mass}\text{of}\text{Phenol}}\text{×100}\\ \text{=}\frac{0.8899\text{mg}}{5.23\text{mg}}\text{×100}\\ \text{=}17.0\text{\%}\end{array}$

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

Hence,

The mass percentage of $\text{C}$ in $\text{Phenol}$ is $\text{76}\text{.54}\text{\%}$

The mass percentage of $\text{H}$ in $\text{Phenol}$ is $\text{6}\text{.44}\text{\%}$

The mass percentage of $\text{O}$ in $\text{Phenol}$ is $17.0\text{\%}$

Conclusion

Mass percentages of each elements present in $\text{5}\text{.23 mg}$ of $\text{Phenol}$ was determined.