A chemical “breathalyzer” test works because ethanol in the breath is oxidized by the dichromate ion (orange) to form acetic acid and chromium(III) ion (green). The balanced reaction is You analyze a breathalyzer test in which 4.2 $\mathrm{mg} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ was reduced. Assuming the volume of the breath was 0.500 \(\mathrm{L}\) at \(30 .^{\circ} \mathrm{C}\) and \(750 . \mathrm{mm}\) Hg, what was the mole percent alcohol of the breath?

To convert the mass of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) to moles, we need its molar mass. The molar mass of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) is 394.20 g/mol. Now we can convert the given mass to moles: \(\text{moles of }\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} = \frac{4.2\,\text{mg}}{394.20\,\text{g/mol}} = \frac{4.2\times 10^{-3}\,\text{g}}{394.20\,\text{g/mol}}\)

From the balanced chemical equation, we know that 3 moles of ethanol react with 2 moles of dichromate ion. Using this ratio, we can calculate the moles of ethanol consumed in the reaction: \(\text{moles of ethanol} = \frac{3\,\text{moles of ethanol}}{2\,\text{moles of }\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}} \times \text{moles of }\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\)

First, calculate the total moles of gas in the breath sample. We will use the ideal gas law for this: \(PV = nRT\) Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the moles, \(R\) is the universal gas constant, and \(T\) is the temperature. We are given: - \(P = 750\,\text{mm}\,\text{Hg}\) (converted to atm): \(P = \frac{750}{760}\,\text{atm}\) - \(V = 0.500\,\text{L}\) - \(T = 30^{\circ}\,\text{C}\) (converted to Kelvin): \(T = 273 + 30 = 303\,\text{K}\) - \(R = 0.0821\,\text{atm}\,\text{L}/\text{mol}\,\text{K}\) Now we can solve for \(n\): \[\text{moles of gases in breath} = n = \frac{PV}{RT} = \frac{\left(\frac{750}{760}\,\text{atm}\right)(0.500\,\text{L})}{(0.0821\,\text{atm}\,\text{L}/\text{mol}\,\text{K})(303\,\text{K})}\] Finally, we can determine the mole percent of ethanol: \(\text{mole percent of ethanol} = \frac{\text{moles of ethanol}}{\text{total moles of gases in breath}}\times 100\%\)