Question

A chemical "breathalyzer" test works because ethyl alcohol in the breath is oxidized by the dichromate ion (orange) to form acetic acid and chromium(III) ion (green).The balanced reaction is

$\mathbf{3}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{O}\mathbf{H}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{C}{\mathbf{r}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\to }\mathbf{n}\mathbf{3}\mathbf{H}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{C}{\mathbf{r}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{11}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

You analyze a breath analyzer test in which${\mathbf{\text{4.2mg ofK}}}_{\mathbf{\text{2}}}{\mathbf{\text{Cr}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{7}}}$was reduced. Assuming the volume of the breath was$\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{L}\mathbf{}\mathbf{at}\mathbf{}\mathbf{30}\mathbf{°}\mathbf{C}\mathbf{}\mathbf{and}\mathbf{}\mathbf{750}\mathbf{}\mathbf{mmHg}$ what was the mole percent alcohol of the breath?

Short answer

The mole percent alcohol of the breath 0.11 %.

Step-by-step solution

The balanced reaction

Balanced reaction is shown below:

$\mathbf{3}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{C}{\mathbf{r}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$$\mathbf{3}\mathit{H}{\mathit{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{C}{\mathbf{r}}_{\mathbf{3}}\mathbf{+}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{11}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

Amount of dichromate ion

To findthe amount of dichromate ion,

$$\begin{gathered} \mathrm{m}\left({\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)=4.2\mathrm{mg} \\ =4.2\times {10}^{-3}\mathrm{g} \end{gathered}$$

We can calculate the amount of dichromate ion from the mass of reduced potassium dichromate, and according to that amount, we can calculate the amount of alcohol in the analyzed breath sample.

$$\begin{gathered} \mathrm{n}\left({\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)=\frac{\mathrm{m}\left({\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)}{\mathrm{M}\left({\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)}=\frac{4.2\times {10}^{-3}\mathrm{g}}{294.185{\mathrm{gmol}}^{-1}}=1.4\times {10}^{-5}\mathrm{mol} \\ \mathrm{n}\left({\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right)=\mathrm{n}\left({\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)=1.4\times {10}^{-5}\mathrm{mol} \\ \mathrm{n}\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)=\frac{3}{2}\times \mathrm{n}\left({\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right)=2.1\times {10}^{-5}\mathrm{mol} \end{gathered}$$

Amount of substance in the breath sample

To find the amount of substance in the breath sample,

$$\begin{gathered} \mathrm{V}\left(\mathrm{breath}\right)=0.500\mathrm{L} \\ \mathrm{V}=0.5\times {10}^{-3}{\mathrm{m}}^3 \\ \mathrm{T}=303.15\mathrm{K} \\ \mathrm{P}=99991.8\mathrm{Pa} \end{gathered}$$

From the given data we can calculate the amount of substance in the breath sample using the ideal gas law.

role="math" localid="1664186003987" $$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}} \\ \mathrm{n}=\frac{99991.8\mathrm{Pa}\times 0.5\times {10}^{-3}{\mathrm{m}}^3}{8.314.\mathrm{J}/\mathrm{Kmol}\times 303.15\mathrm{K}} \\ \mathrm{n}=0.0198\mathrm{mol} \end{gathered}$$

Alcohol in breath

To find percentage of alcohol in breath,

$$\begin{gathered} \%\left(\mathrm{alcohol}\mathrm{in}\mathrm{breath}\right)=\frac{\mathrm{n}\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)}{\mathrm{n}\left(\mathrm{breath}\right)}\times 100\% \\ =\frac{2.1\times {10}^{-3}\mathrm{mol}}{0.0198\mathrm{mol}}\times 100\% \\ \%\left(\mathrm{alcohol}\mathrm{in}\mathrm{breath}\right)=0.11\% \end{gathered}$$

Hence, the mole percent alcohol of the breath$\text{0.11\%}$.