# Blood Alcohol Level?

1. Feb 25, 2009

### Planarian

1. The problem statement, all variables and given/known data
A person will receive a DWI if his/her blood alcohol level (BAL) is 100mg per dL of blood or higher. Suppose a person is found to have a BAL of 0.033 mol of ethanol per liter of blood. Will the person recieve a DWI? Assuming the cop isn't cool and just follows you home.

2. Relevant equations
molarity and basic metric conversions.

3. The attempt at a solution
100mg/dL = 1000mg/dL = 1g/dL?

0.033 mol of C2H5OH (ethanol) x 46.07g/1 mol = 1.52g/L of C2H5OH

Therefore the drunk driver would get a ticket.

I think this is correct. I did the problem backwards. What's the best way to get mg/dL from the 0.033 mols of C2H5OH?

Thanks for the help :D

Last edited: Feb 25, 2009
2. Feb 26, 2009

### Staff: Mentor

Sure, and 2 = 7 :tongue2:

Conversion moles -> grams is OK, but I can't see L on the left side of the equation. Well, you probably meant mol/L, but omitting units is a sure way of making mistake one day.

3. Feb 26, 2009

### Planarian

I meant 100g/dL not 100mg/dL

You're right. 0.033 mol/L. it is in the original question, yes I for got to add that in.

-now that you have successfully shown me all my typos, can you please answer my original question? What is the correct way of answering this problem to come up with a solution of mg/dL. I can ask my professor if its too much trouble.

I'm trying to learn how to be more anal with my calculations. I know it is a big part of chemistry. I have a social science degree so math is my weak point.

4. Feb 26, 2009

### redargon

100g/dL still doesn't make sense.

1L = 10dL
therefore 100mg/dL = 1000mg/L = 1g/L(this is your alcohol limit in terms of Litres)

I'll assume your mol calculation is correct (it seems good), then there are 1.52g of Ethanol in 0.033 mol of Ethanol.

So the person has 1.52g of Ethanol per litre of blood. This is higher than the 1g/L limit, so he should be arrested for DUI.

5. Feb 26, 2009

### Staff: Mentor

You were on the right track, just the view was obscured by small errors. Sorry for not stating that clearly.

6. Feb 26, 2009

### Planarian

It's cool, I think I figured it out from your help. Please let me know if this is right.

So to get from 1.52 g/L of C2H5OH to mg/dL

1.52g/L (1000mg/1g per liter) = 1520 mg/L

(1520 mg) / (10 dL/ 1L) = 152 mg/dL of C2H5OH

The answer in the back of the book is 150 mg/dL so my molar mass calc.s are different from the books, but close enough.

I was doing this:
1520 mg/L = (1520mg) (10dL/1L) and getting 15200 mg/dL (I forgot that it is mg PER dL, ha. makes sense now)

__I do need to pay closer attention to units/states/detail. Thanks for pointing this out Borek. Criticism helps even though I don't enjoy it. I've made SEVERAL errors on my exams this same way costing me letter grades in some cases :grumpy:. Any good suggestions to avoid this, besides the obvious double checking? I'm slightly A.D.D.

7. Feb 26, 2009

### Staff: Mentor

Significant figures. You were told ethanol is 0.033M, so your answer should have 2 SF - that means 150 (or more correctly 1.5x102 mg/dL), not 152.

8. Feb 26, 2009

### Planarian

Sig. fig.s! ahhh yes. I forgot 150 is only 2 significant figures. tricky.

Thanks again.

9. Feb 27, 2009

### redargon

I'm not sure of the best way to stop silly mistakes, I still often make them. I try to set out every problem by starting with a template that I'm comfortable with and use very often. I make a sketch and write down all the information I know. I then jot down some formulae that I think may be handy. I stop here and do a quick double check to make sure I've written down the information correctly. Then I do my conversions from strange units, to units that make more sense to me and that I know will be easier to use (kg, N, l, m, seconds, Pa, etc) that kind of depends on what formulae you'll be using and what you're comfortable with. The thing that helps the most is just the following the routine and keeping the same pattern. Methodically working through the problem will stop your mind from trying to jump to the answer too quickly allowing silly mistakes to creep into the part that you're actually busy with.

10. Feb 27, 2009

### Planarian

11. Feb 27, 2009

### Planarian

BAL 100 mg/dL
Person 0.033 mol/L C2H5OH

0.033 mol/L * 46.07g C2H5OH/1 mol (cancel out mols) * 1000 mg/1 g (cancel out g) * 1 L /10 dL (cancel out L) = 152.031....sig. figs....[1.5 x10^2 mg/dL]

:D easy.