Chapter 21, Problem 155AE

Interpretation Introduction

Interpretation: The mole percent of alcohol analyzed by the ‘breath analyzer’ wherein 4.2 mg of potassium dichromate is oxidized from breath of 0.500 L at 30(C and 750 mm Hg needs to be determined.

Concept introduction:

Number of moles of a substance is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Answer to Problem 155AE

The mole percent of alcohol analyzed by breath analyzer is 0.11%

Explanation of Solution

The reaction of ethyl alcohol oxidized by dichromate ion giving acetic acid and chromium (III) is as follows:

  ${\text{3C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH(aq) + 2Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}{\text{(aq)+2H}}^{\text{+}}{}_{}\to {\text{3HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{(aq)+4Cr}}^{\text{3+}}{\text{(aq)+11H}}_{\text{2}}\text{O(l)}$ ….. (I)

But it is known that,

  $\text{Molar mass = }\frac{\text{Mass }}{\text{mole}}=grams/mole$

Given Mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is 4.2 mg = 0.0042 g

  $\begin{array}{l}{\text{Molar mass of K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{ = 294}\text{.18 g/mol}\\ \text{294}\text{.18 g/mol}=\frac{0.0042g}{moles}\\ \text{No}{\text{. of Moles of K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\frac{\text{0}\text{.0042 g}}{\text{294}\text{.18 g/mol}}\\ \text{ = 1}{\text{.43 x 10}}^{-5}\text{ mol}\end{array}$

Observing equation (i), it can be seen that 3 moles of ethyl alcohol is oxidized by 2 moles of potassium dichromate. This will help to calculate the number of moles of alcohol oxidized as given below

  $\begin{array}{l}\text{3 moles of ethyl alcohol }=\text{ 2 moles of potassium dichromate}\\ ‘\text{x}’\text{ moles of ethyl alcohol }=\text{ 1}{\text{.43 x 10}}^{-5}\text{moles of potassium dichromate}\\ \text{ 2}x\text{ =}3\text{x 1}{\text{.43 x 10}}^{-5}\text{moles }\\ x\text{ = }\frac{3}{2}\text{x 1}{\text{.43 x 10}}^{-5}\\ \text{ =2}{\text{.145 x 10}}^{-5}mol\end{array}$

From Ideal gas law $PV=nRT$

Where V = 0.50 L

P = 750 mm Hg = 0.98 atm

R = 0.082 L.atm /mol K

T = 30 + 273 K= 303 K

  $n=\frac{PV}{RT}=\frac{\text{0}\text{.98 x 0}\text{.50 }}{0.082\text{ x 303}}=\frac{0.49}{24.85}=0.019$ Mole

  $\begin{array}{l}\text{Total moles = Moles of breath + Moles of ethanol}\\ \text{ = 2}{\text{.145 x 10}}^{-5}mol\text{ + 0}\text{.019 mol}\\ \text{ = 0}\text{.00002145 + 0}\text{.019}\\ \text{ = 0}\text{.0190 mol}\end{array}$

  $\begin{array}{l}\text{Mole percent of ethanol = }\frac{\text{Moles of ethanol }}{\text{Total moles}}\text{x 100}\\ \text{ = }\frac{\text{2}{\text{.145 x 10}}^{-5}mol}{\text{0}\text{.0190 mol}}\text{ x 100}\\ \text{ = 0}\text{.0011 x100=0}\text{.11\%}\end{array}$

Conclusion

The mole percent of alcohol analyzed by the ‘breath analyzer’ is 0.11% when 4.2mg of Potassium dichromate is oxidized from breath of 0.500L at 30(C and 750mm Hg.