Chapter 22, Problem 171CP

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

${\text{3C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(aq\right)+2{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(aq\right)+{\text{2H}}^{\text{+}}\left(aq\right)\to 3{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+4{\text{Cr}}^{\text{3+}}\left(aq\right)+{\text{ 11H}}_2\text{O}\left(l\right)$You analyze a breathalyzer test in which 4.2 mg K₂Cr₂O₇ was reduced. Assuming the volume of the breath was 0.500 L at 30.°C and 750. mm Hg, what was the mole percent alcohol of the breath?

Interpretation Introduction

Interpretation: The mole percent of alcohol of the breath is to be calculated.

Concept introduction: Number of moles is defined as the mass divided by the molecular of the element. Mole percent is calculated by knowing the mass and mole fractions of any element.

To determine: The mole percent of alcohol of the breath.

Answer to Problem 171CP

Answer

The mole percent of alcohol of the breath is $\underset{\_}{0.12\%}$.

Explanation of Solution

Explanation

The number of moles is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Given,

$\begin{array}{c}{\text{Mass of K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{ = 4}\text{.2 mg}\\ \text{= 0}\text{.0042 g}\end{array}$

${\text{Molar mass of K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{ = 298}\text{.18 g/mol}$

Therefore, moles of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is calculated by using the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the value of the given mass and the molar mass, to calculate the number of moles of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$, in the above equation.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{\text{0}\text{.0042 g}}{\text{294}\text{.18 g/mol}}\\ =1.427\times {10}^{-5}\text{mol}\end{array}$

The given reaction is,

$3{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(aq\right)+2{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(aq\right)+2{\text{H}}^{\text{+}}\left(aq\right)\to 3{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\left(aq\right)+4{\text{Cr}}^{\text{3+}}\left(aq\right)+11{\text{H}}_{\text{2}}\text{O}\left(l\right)$

According to the stated reaction, 3 moles of alcohol react with 2 moles of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

Therefore,

$\begin{array}{c}\text{Moles of ethanol}\text{that}\text{react}\text{with}{\text{ K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{\text{= Moles of K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{×}\frac{\text{3}}{\text{2}}\\ =1.427\times {10}^{-5}\times \frac{3}{2}\\ \text{= 2}{\text{.415×10}}^{\text{-5}}\text{mol}\end{array}$

The idea gas equation is,

$PV=n\text{R}T$

Given,

The volume of the breath (V) is $0.50\text{L}$.

$\begin{array}{c}\text{Pressure}\left(\text{P}\right)=7\text{50 mmHg}\\ \text{= 0}\text{.98 atm}\end{array}$

$\begin{array}{c}\text{Gas constant}\left(\text{R}\right)\text{ = 0}\text{.0821 L}\text{.atm/mol}\text{.K}\\ \text{Temperature}\left(\text{T}\right)\text{ = 30+273}\\ \text{= 303K}\end{array}$

Rearrange the ideal gas equation to calculate the number of moles.

$\text{Mole}\left(n\right)\text{=}\frac{PV}{\text{R}T}$

Substitute the value of $P,V,\text{R}$ and $T$ in the above equation.

$\text{Moles of alcohol of breath = }\frac{\text{0}\text{.98}\text{atm}\times \text{0}\text{.50}\text{L}}{\begin{array}{l}\text{0}\text{.0821}\text{L}\text{.atm/mol}\text{.K}\times \text{303}\text{K}\\ \text{= 0}\text{.019}\text{mol}\end{array}}$

Therefore,

$\begin{array}{c}\text{Total moles =Moles of ethanol + Moles of breath}\\ \text{= 2}{\text{.415×10}}^{-\text{5}}\text{mol + 0}\text{.019}\text{mol}\\ \text{= 0}\text{.01902}\text{mol}\end{array}$

The mole percent of alcohol of the breath is calculated as,

$\begin{array}{c}\text{Mole percent = }\left(\frac{\text{Moles of ethanol}}{\text{Total moles }}\right)\text{×100}\\ \text{= }\left(\frac{\text{2}{\text{.415×10}}^{-\text{5}}\text{mol}}{\text{0}\text{.01902 mol}}\right)\text{×100}\\ \text{= }\underset{\_}{0.12\%}\end{array}$

Conclusion

Conclusion

The mole percent of alcohol of the breath is $\underset{\_}{0.12\%}$.