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AlcoholAlcohol
Alcoholic beverages
• BeerBeer 10° = 4 vol. % 10° = 4 vol. % (3.16 g v 0.1 l = 15.8 g v 0,5 l)(3.16 g v 0.1 l = 15.8 g v 0,5 l)12° = 5 vol. % 12° = 5 vol. % (3.95 g v 0.1 l = 19.7 g v 0,5 l)(3.95 g v 0.1 l = 19.7 g v 0,5 l)
• WineWine – cca 11% – cca 11% (7.9 g v 0.1 l = 15.8 g v 0.2 l)(7.9 g v 0.1 l = 15.8 g v 0.2 l)
• SpiritsSpirits – 40 % – 40 % (31.6 g v 0.1 l = 15.8 g v 0,05 l)(31.6 g v 0.1 l = 15.8 g v 0,05 l)
t
‰=millesimal
Blood-alcohol concentration (BAC)
I. II.
Absorption Elimination
Peak
Absorption• Fullness and content of the stomach
• CO2 content
• Concentration
• Temperature
Metabolism of ethanol• ADH (alcoholdehydrogenase)• MEOS (microsomal ethanol
oxidation system)• Catalase
1. Alcohol
2. Acetaldehyde
3. CO2 and water
Velocity of eliminationVelocity of elimination
• 0.12 – 0.20 g/kg per hour
Analysis of alcohol1. Specific gas chromatography – physical
method
2. Non-specific Widmark´s proof – chemical method
Calculation
1. Backward calculation
2. Balance calculation
Backward calculation
peak
t
‰
Criminal act Sample
0,12 – 0,20 ‰ /hour
Alcohol (g) – absorption deficit
BAC ‰= Weight (kg) x reduction factor
Balance calculation (formula)
absorption deficit = 10%
reduction factor:♀= 0.6
♂= 0.7
Example
• Man (weight 70 kg) drank from 7.00 p.m. till 9.00 p.m. five snifters „shots“ (0.05 l) of vodka (40 vol. %).
• ? BAC at 11.00 p.m. ?
Example - result
79 – 7.9
Ct = —————— – 0.20 x 4 = 0.65 70 x 0.7
79 – 7.9
Ct = —————— – 0.12 x 4 = 0.97 70 x 0.7
Ct = 0.65 – 0,97 g/kg
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