Chapter 18, Problem 18.CE

(a)

Interpretation Introduction

Interpretation:

For the given blue product, the value of molar absorptivity has to be calculated.

Concept introduction:

Beer-lambert law:

$\begin{array}{l}\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}\\ \text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

Answer to Problem 18.CE

The molar absorptivity $\left(\epsilon \right)$ value is $\text{4}\text{.493}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Mass}\text{of}{\text{NH}}_{\text{4}}\text{Cl}\text{is}\text{1}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{g}\text{.}\\ SampleAbsorbance\left(at625nm\right)\\ \text{Blank}\text{0}\text{.140}\\ \text{Reference}\text{0}\text{.308}\\ \text{Unknown}\text{0}\text{.592}\end{array}$

Calculate the concentration of ${\text{NH}}_{\text{4}}\text{Cl}$

$\frac{\frac{\text{1}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{g}}{\text{53}\text{.491}\text{g/mol}}}{\text{1}\text{.00}\text{L}}\text{=}\text{1}\text{.869}\text{×}{\text{10}}^{\text{-4}}\text{M}\text{.}$

Calculate the concentration of ${\text{NH}}_{\text{4}}\text{Cl}$ in the colored solution

$\left(\frac{\text{10}\text{mL}}{\text{50}\text{mL}}\right)\left(\text{1}\text{.869}\text{×}{\text{10}}^{\text{-4}}\text{M}\right)\text{=}\text{3}\text{.739}\text{×}{\text{10}}^{\text{-5}}\text{M}\text{.}$

Apply the Beer’s law to calculate the value of molar absorptivity $\left(\text{ε}\right)$

$\begin{array}{l}\text{ε}\text{=}\frac{\text{A}}{\text{bc}}\text{=}\frac{\left(\text{0}\text{.308}\text{-}\text{0}\text{.140}\right)}{\left(\text{1}\text{.000}\right)\left(\text{3}\text{.739}\text{×}{\text{10}}^{\text{-5}}\text{M}\right)}\\ \text{=}\text{4}\text{.493}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}\end{array}$

The value of molar absorptivity $\left(\epsilon \right)$ is $\text{4}\text{.493}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The weight percent of nitrogen in the protein sample has to be calculated.

Answer to Problem 18.CE

The weight percent of nitrogen in the protein sample is $16.1\text{\%}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Mass}\text{of}{\text{NH}}_{\text{4}}\text{Cl}\text{is}\text{1}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{g}\text{.}\\ \text{Concentration}\text{of}\text{reference}\text{solution}\text{is}\text{1}\text{.869}\text{×}{\text{10}}^{\text{-4}}\text{M}\text{.}\\ \text{mass}\text{of}\text{protein}\text{sample}\text{is}\text{4}\text{.37}\text{mg}\text{.}\\ SampleAbsorbance\left(at625nm\right)\\ \text{Blank}\text{0}\text{.140}\\ \text{Reference}\text{0}\text{.308}\\ \text{Unknown}\text{0}\text{.592}\end{array}$

Calculate the concentration of ${\text{NH}}_{\text{3}}$ in the unknown solution

$\begin{array}{l}\frac{\text{Absorbance}\text{of}\text{unknown}}{\text{Absorbance}\text{of}\text{reference}}\text{=}\frac{\text{0}\text{.592}\text{-}\text{0}\text{.140}}{\text{0}\text{.308}\text{-}\text{0}\text{.140}}\\ \text{=}\frac{\text{Concentration}\text{of}\text{unknown}}{\text{Concentration}\text{of}\text{reference}}\\ \Rightarrow \left(\frac{\text{0}\text{.452}}{\text{0}\text{.168}}\right)\left(\text{1}\text{.869}\text{×}{\text{10}}^{\text{-4}}\right)\text{=}\text{5}\text{.028}\text{×}{\text{10}}^{\text{-4}}\text{M}{\text{NH}}_{\text{3}}\text{.}\end{array}$

Calculate the mass of $\text{N}$ in $\text{100}\text{ml}$ unknown solution

$\text{5}\text{.028}\text{×}{\text{10}}^{\text{-5}}\text{mol}\text{N}\text{×}\text{14}\text{.0067}\text{=}\text{0}\text{.7043}\text{mg}\text{of}\text{N}\text{.}$

Calculate the weight percent of nitrogen in the unknown solution

$\begin{array}{l}\text{wt\%}\text{of}\text{N}\text{=}\frac{\text{100}\text{×}\left(\text{0}\text{.7043}\text{mg}\right)}{\text{4}\text{.37}\text{mg}}\\ \text{=}\text{16}\text{.1}\text{\%}\text{.}\end{array}$

The weight percent of nitrogen in the protein sample is $16.1\text{\%}\text{.}$