calculate the number of shots of Captain Morgan’s rum

Question

I am facing a challenge, and this is my first question in this stack exchange. I have worked out the following problem but the answer I got (number of shots) is ridiculous. Could any of the chemist here please advise me where I went wrong?

Question: Assume that you are on vacation in Florida and enjoying the company of friends at a bar. Using the information in Table 80-1 in your text, calculate the number of shots of Captain Morgan’s rum required to take your blood ethanol concentration above twice the legal limit to drive. [Note: The legal limit in Florida is 80 mg/dL]. Assume that the shots are consumed in rapid succession (the wisdom of which you were later unable to explain). Show your complete calculation, including all inputs and assumptions.

My calculation:

Table reference

Let us assume that my binge-drinking has led to a blood alcohol concentration (BAC) of 162 mg/dL.

The formula for BAC is as follows:

`BAC = A/(R x W) – (0.15 x H)` where

• A = weight of pure ethanol consumed (g)
• R = Widmark’s rho factor (0.68 and 0.55 L/kg for men and women, respectively)
• W = body weight (kg)
• 0.15 = average ethanol elimination rate in humans
• H = drinking period (hr)

Inserting the values of 162 mg/dL for BAC, R = 0.55 L/kg, W = 50 kg, and H = 1 hr, the equation becomes:

162 = A/(0.55 x 50) – (0.15 x 1)
162 = A/27.5 – 0.15

Therefore,

A = 27.5(162 + 0.15)
    = 4459.125 g

According to table 80-1 in the book, one shot is 30 mL and contains 40% ethanol. So one shot contains 12 mL ethanol. Density of ethanol = 0.789 g/cm3 The volume of pure ethanol which I consumed = 4459.125/0.789 = 5651.62 mL So, no. of shots consumed = 5651.62/12 = 471, over the course of 1 hour

Conculsion: 471 shots of Captain Morgan Rum is absurd for any person. So I think I am missing something in my calculation. I'm not able to figure out what it is.

Answer 1

As Zhe mentioned, you have inconsistent units in your calculation, so your numerical results are not physically meaningful.

Take for instance this expression:

162 = A/(0.55 x 50) – (0.15 x 1)

Here, you have different units for each term; from the values you supplied, the units are

162 mg/dL = A/(0.55 L/kg x 50 kg) - 0.15 g/(L•h) x 1 h

[Note that the units for the elimination rate weren't stated in the question, but a quick search of the literature gave studies measuring it at about 0.01-0.02 g/(dL•h), so I infer the 0.15 value refers to g/L] If we use uniform units throughout, we get

1.62 g/L = A/27.5 L - 0.15 g/L

from where you get a more reasonable value for A:

A = (1.62 + 0.15) * 27.5 = 48.675 g

Each shot contains

V = 30 mL * 0.40 / 0.789 g/mL = 15.2 g

of ethanol, so you need to drink 48.675 / 15.2 = 3.2 shots of rhum over the course of an hour.

I'd suggest, however, that drinking shots in "rapid succession" implies a negligible drinking time (∆t ~ 0 h). In that case, A = 44.5 g and you need about 3 shots (2.93).