Metabolism of ethanol is primarily a two-step process. Ethanol is converted to acetaldehyde by alcohol dehydrogenase (ADH) and the resultant acetaldehyde is converted to acetate ion by aldehyde dehydrogenase (ALDH). Acetate is further metabolized to carbon dioxide and water or is eliminated directly, since it is highly water soluble. Other pathways can occur, but this is the primary route of elimination. The rate of this process depends primarily on genetics and nutrition. Body weight (or more correctly, body mass, as you say, but this terms is used mostly in the context of body mass index, which is ancillary to my point) is important because the degree of effect realized from any physiologically-active substance is dependent upon dose. Dose has a mass component of the physiologically-active substance (usually expressed in mg), a mass component from the receptor (i.e., the person taking the physiologically-active substance, usually expressed as kgbody weight), and a time component (usually expressed in days).
Thus, dose is expressed in mg of substance per kilogram of body weight per day. Mathematically, this is expressed as mg/kg-d. Note that mg/kg/d is a mathematically-ambiguous, but it is used by some because that's how the term is spoken. I digress.
So, if you drink a pint of beer that's 6% alcohol, that's 473.2 mL (0.06) = 28.4 mL of ethanol. Since ethanol has a density of 0.789 g/mL, this is equivalent to 22.4 g or 22,400 mg of ethanol. If you chug the beer and weigh 90 kg (198 pounds), your instantaneous dose is 22,400 mg / (90 kg) (1 day) = 249 mg/kg-d. If you weigh 45 kg (99 pounds), your instantaneous dose is 22,400 mg / (45 kg) (1 day) = 498 mg/kg-d. So, the relationship between dose and body weight is directly linear. If you weigh twice as much as someone else and if you ingest the same amount of a substance, your instantaneous dose will be half of the person that weighs half your weight.
Which is why midgets shouldn't drink gin. Thus endeth the lesson.
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