Question

Astandard solution containing $\mathbf{6}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{M}$iodoacetone and $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{Mp}$-dichlorobenzene (an internal standard) gave peak areas of 395 and 787, respectively, in a gas chromatogram. A $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{-}\mathit{m}\mathit{l}$unknown solution of iodoacetone was treated with $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{mL}$of $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{Mp}$-dichlorobenzene and the mixture was diluted to. Gas chromatography gave peak areas of 633 and 520 for iodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mL}$of original unknown.

Short answer

$\mathrm{In}3.00\mathrm{ml}\mathrm{of}\mathrm{the}\mathrm{original}\mathrm{solution},\mathrm{the}\mathrm{concentration}\mathrm{of}\mathrm{lodoacetone}\mathrm{is}0.41\mathrm{\mu M}.$

Step-by-step solution

Definition of iodoacetone

$$\begin{gathered} •\mathrm{C}3\mathrm{H}5\mathrm{IO}\mathrm{is}\mathrm{the}\mathrm{chemical}\mathrm{formula}\mathrm{for}\mathrm{iodoacetone},\mathrm{an}\mathrm{organoiodine}\mathrm{compound}. \\ •\mathrm{Under}\mathrm{normal}\mathrm{circumstances},\mathrm{the}\mathrm{material}\mathrm{is}\mathrm{a}\mathrm{white}\mathrm{liquid}\mathrm{that}\mathrm{is}\mathrm{soluble}\mathrm{in}\mathrm{ethanol}. \end{gathered}$$

Step 2: Determine the concentration of iodoacetone in the 3.00 mL of originalunknown

It is necessary to calculate the concentration of lodoacetone in 3.00ml of the original solution.

The formula for quantitative analysis with internal standard can be found here,

$\frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{{\mathrm{A}}_{\mathrm{s}}}{\left[\mathrm{S}\right]}$

Here ${\mathrm{A}}_{\mathrm{x}}=$signal region of analyte

${\mathrm{A}}_{\mathrm{s}}$=area of internal standard

[X] = analyte's concentration

[S]= concentration of internal standard

F= response factor

To figure out how much lodoacetone is in 3.00ml of the original solution

Given,

Lodoacetone molarity $=6.3\times {10}^{-8}\mathrm{M}$

P-dichlorobenzene molarity $=2.0\times {10}^{-7}\mathrm{M}\mathrm{Peak}\mathrm{areas}=395\&787$

Unknown volume of lodoacetone solutionrole="math" localid="1654853858008" $=3.00ml$

p-dichlorobenzene Volume and Molarity$=0.100\mathrm{ml}\&1.6\times {10}^{-5}\mathrm{M}$

Lodoacetone has a peak area of 633 square metres.

p-dichlorobenzene has a peak area of =520.

The response factor is determined as follows,

$$\begin{gathered} \frac{395}{\left[63nM\right]}=\mathrm{F}\left(\frac{787}{\left[200\mathrm{nM}\right]}\right) \\ \mathrm{F}=1.59 \end{gathered}$$

When an internal standard is coupled with an unknown, the concentration is determined as,

$$\begin{gathered} \frac{0.100\mathrm{ml}}{10.00\mathrm{ml}}\left(1.6\times {10}^{-5}\mathrm{M}\right)=0.16\mathrm{\mu M} \\ \mathrm{Lodoacetone}\mathrm{concentration}\mathrm{is}\mathrm{determined}\mathrm{as}, \\ \frac{633}{\left[\mathrm{iodoacentone}\right]}=1.59\left(\frac{520}{0.16\mathrm{\mu M}}\right) \\ \left[\mathrm{iodoacentone}\right]=0.122\mathrm{\mu M} \\ \mathrm{In}\mathrm{the}\mathrm{original}\mathrm{unknown}\mathrm{solution},\mathrm{the}\mathrm{concentration}\mathrm{of}\mathrm{lodoacetone}\mathrm{is}\mathrm{determined}\mathrm{as}, \\ \left[\mathrm{iodoacentone}\right]=\frac{10.00}{3.00}\left(-0.122\mathrm{\mu M}\right) \\ \left[\mathrm{iodoacentone}\right]=0.41\mathrm{\mu M} \\ \mathrm{In}3.00\mathrm{ml}\mathrm{of}\mathrm{the}\mathrm{original}\mathrm{solution},\mathrm{the}\mathrm{concentration}\mathrm{of}\mathrm{lodoactone}=0.41\mathrm{\mu M} \\ \mathrm{The}3.00\mathrm{ml}\mathrm{of}\mathrm{the}\mathrm{original}\mathrm{solution},\mathrm{the}\mathrm{concentration}\mathrm{of}\mathrm{lodoacetone}\mathrm{was} \\ \mathrm{determined}\mathrm{and}\mathrm{found}\mathrm{to}\mathrm{be}0.41\mathrm{\mu M} \end{gathered}$$