Question

The chemical method of analysis in determination of the blood alcohol content (%BAC) is:

2K2Cr2O7 + 8H2SO4 + 3C2H5OH → 2Cr2(SO4)3 + 2K2SO4 + 3CH3COOH + 11H2O

1. During a Breathalyzer Test it was determined that 1.75mg of K2Cr2O7 was consumed by the above reaction. Calculate the number of milligrams of ethanol in the test sample.

2.Calculate the Blood Alcohol Content in this sample (%BAC).

Answer 1

Q1.

mass of K2Cr2O7 in g = 1.75*10^-3

MW K2Cr2O7 = 294.185 g/mol

mol = mass/MW = (1.75*10^-3)/294.185 = 0.000005948 mol of K2Cr2O7

ratio is:

2 mol of K2Cr2O7 : 3 mol of ethanol

so..

3/2*K2Cr2O7 -> 0.000008922 mol of ethanol

MW ethanol = 46 g/mol

mass = mol*MW = 0.000008922*46 = 0.000410412 grams = 0.000410412*10^3 mg = 0.410412 mg

Q2

BAC...

a typical lung is about 6 liters

so

M = mol/V = 0.000008922/6 = 0.000001487 Mol per liter

ins % --> mass/V = 0.000410412/6*100 = 0.068402 %in which legal limiti is 0.08 ... its fine to drive