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


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.

  • 1
    $\begingroup$ Note that some of the masses are in grams, others are in milligrams, and still some more are in kilograms. You need to perform the correct conversion. The main issue I see is that you didn't bother to write out units in your computation, so there's no way to tell that you are making this mistake as you are making it. $\endgroup$ – Zhe Jul 30 '17 at 22:14
  • $\begingroup$ @Zhe I think I have mentioned all units and their conversions were already checked, yet I am pretty sure there some tiny mistake which I am not able to catch here. Please ask me which part of my question is not made clear, and I will correct it, as it will help the community as well. $\endgroup$ – Rene Duchamp Jul 30 '17 at 23:16
  • 2
    $\begingroup$ OK. Average alcohol elimination. What are the units? Also 162 is in mg/dL. Yet somehow you magically multiplied some numbers that were either in unknown units or in kg and without any conversion. $\endgroup$ – Zhe Jul 31 '17 at 0:28
  • 1
    $\begingroup$ @abhilash sukumari BAC is in mg/dl. You need to make each term on the right hand side match those units. Right away, the R factor being in kg/L will not lead to the right units. You also don't specify the units of 0.15 in the 2nd term, so its unclear if that term has the correct units. $\endgroup$ – Tyberius Jul 31 '17 at 0:31

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.