Chapter 2, Problem 111QRT

(a)

Interpretation Introduction

Interpretation:

The mass percent of nitrogen in caffeine has to be calculated.

Concept Introduction:

It is a way of representing the concentration of an element in a compound or a component in a mixture.  Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by $\text{100}\text{\%}$.

Explanation of Solution

The molecular formula of caffeine is ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$.

Mass of nitrogen per mole of ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$:

  $\text{2}\text{×}\text{14}\text{.0067}\text{g}\text{=}\text{56}\text{.0268}\text{N}$

Mass of one mole ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$:

  $\begin{array}{l}\text{(8}\text{×}\text{12}\text{.0107}\text{g/mol}\text{C)}\text{+}\text{(10}\text{×}\text{1}\text{.0079}\text{g/mol}\text{H)}\text{+}\text{(}\text{4}\text{×}\text{14}\text{.0067}\text{g/mol}\text{N)}\\ \text{+}\text{(2}\text{×}\text{15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{194}\text{.1902}\text{g/mol}\end{array}$

Mass percent of Nitrogen:

  $\frac{\text{56}\text{.0268}\text{g}\text{N}}{\text{194}\text{.1902}\text{g}\text{caffeine}}\text{×}\text{100\%}\text{=}\text{28}\text{.8515}\text{\%}\text{N}$

(b)

Interpretation Introduction

Interpretation:

The number of caffeine molecule that the student ingested has to be calculated.

Explanation of Solution

The molecular formula of caffeine is ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$.

Mass of one mole ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$:

  $\begin{array}{l}\text{(8}\text{×}\text{12}\text{.0107}\text{g/mol}\text{C)}\text{+}\text{(10}\text{×}\text{1}\text{.0079}\text{g/mol}\text{H)}\text{+}\text{(}\text{4}\text{×}\text{14}\text{.0067}\text{g/mol}\text{N)}\\ \text{+}\text{(2}\text{×}\text{15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{194}\text{.1902}\text{g/mol}\end{array}$

Amount of caffeine is given as $\text{212}\text{mg}$.

The number of molecules can be calculated as shown:

    $\begin{array}{l}\text{212}\text{mg}\text{caffeine}\text{×}\left(\frac{\text{1}\text{g}}{\text{1000}\text{mg}}\right)\text{×}\left(\frac{\text{1}\text{mol}\text{caffeine}}{\text{194}\text{.1902}\text{g}\text{caffeine}}\right)\\ \text{×}\left(\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{caffeine}\text{molecules}}{\text{1}\text{mol}\text{caffeine}}\right)\text{=}\text{6}\text{.57}\text{×}{\text{10}}^{\text{20}}\text{caffeine}\text{molecules}\end{array}$.

(c)

Interpretation Introduction

Interpretation:

The number of carbon atoms in the given mass of caffeine has to be calculated.

Explanation of Solution

The molecular formula of caffeine is ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$.

Mass of one mole ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$:

  $\begin{array}{l}\text{(8}\text{×}\text{12}\text{.0107}\text{g/mol}\text{C)}\text{+}\text{(10}\text{×}\text{1}\text{.0079}\text{g/mol}\text{H)}\text{+}\text{(}\text{4}\text{×}\text{14}\text{.0067}\text{g/mol}\text{N)}\\ \text{+}\text{(2}\text{×}\text{15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{194}\text{.1902}\text{g/mol}\end{array}$

Amount of caffeine is $\text{212}\text{mg}$.

The number of carbon atoms can be calculated as shown,

  $\text{6}\text{.57}\text{×}{\text{10}}^{\text{20}}\text{caffeine}\text{molecules}\text{×}\left(\frac{\text{8}\text{C}\text{atoms}}{\text{1}\text{caffeine}\text{molecule}}\right)\text{=}\text{5}\text{.26}\text{×}{\text{10}}^{\text{21}}\text{C}\text{atoms}$.

(d)

Interpretation Introduction

Interpretation:

An $\text{8}$-oz cup of regular coffee contains approximately $\text{100}\text{mg}$ caffeine.  Then how many times greater the caffeine concentration is in the $\text{5-}$ Hour Energy drink than in the regular coffee has to be calculated.

Explanation of Solution

The molecular formula of caffeine is ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$.

Mass of one mole ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$:

  $\begin{array}{l}\text{(8}\text{×}\text{12}\text{.0107}\text{g/mol}\text{C)}\text{+}\text{(10}\text{×}\text{1}\text{.0079}\text{g/mol}\text{H)}\text{+}\text{(}\text{4}\text{×}\text{14}\text{.0067}\text{g/mol}\text{N)}\\ \text{+}\text{(2}\text{×}\text{15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{194}\text{.1902}\text{g/mol}\end{array}$

Amount of caffeine is $\text{212}\text{mg}$.

Concentration in $\text{5-}$ Hour Energy drink:

  $\frac{\text{212}\text{mg}}{\text{1}\text{.93}\text{oz}}\text{=}\text{12}\text{.5}\frac{\text{mg}}{\text{oz}}$

Compare the concentrations, using a ratio:

  $\frac{\text{109}\text{.8}\frac{\text{mg}}{\text{oz}}}{\text{12}\text{.5}\frac{\text{mg}}{\text{oz}}}\text{=}\text{8}\text{.78}\text{=}\text{9}$.