August 13, 2020, 10:54:58 AM
Forum Rules: Read This Before Posting

### Topic: Chapter 1 Problem  (Read 6789 times)

0 Members and 1 Guest are viewing this topic.

#### dshipp17

• Regular Member
• Posts: 17
• Mole Snacks: +1/-0
##### Chapter 1 Problem
« on: March 02, 2017, 02:03:53 PM »
I came across this problem and my answer seems to either be slightly off or the textbook might be wrong: A sample of an ethanol–water solution has a volume of 54.2 cm3 and a mass of 49.6 g. What is the percentage of ethanol (by mass) in the solution? (Assume that there is no change in volume when the pure compounds are mixed.) The density of ethanol is 0.789 g/cm3 and that of water is 0.998 g/cm3.

Would you help me with this problem by working it out step by step? This problem shouldn't involve molar masses, although it could require using molar masses (e.g. no examination of the periodic table should be necessary); it should just be a conservation of mass problem.

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7452
• Mole Snacks: +523/-87
• Gender:
##### Re: Chapter 1 Problem
« Reply #1 on: March 02, 2017, 02:14:53 PM »
This is quite simple problem. two different volumes gives you  a known volume. Each volume has its own mass. Sum of masses is also known.
Compare your result with table of densities for mixture ethanol-water. You can find that this assumpion (Assume that there is no change in volume when the pure compounds are mixed.) is rather not valid.
AWK

#### dshipp17

• Regular Member
• Posts: 17
• Mole Snacks: +1/-0
##### Re: Chapter 1 Problem
« Reply #2 on: March 02, 2017, 02:56:58 PM »
That's precisely what I did; would you work out the calculation and give the answer? You'll either get my answer or the textbook answer; perhaps you would get the density for this mixture from a table of densities?

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7452
• Mole Snacks: +523/-87
• Gender:
##### Re: Chapter 1 Problem
« Reply #3 on: March 02, 2017, 03:17:05 PM »
According to Forum Rules (http://www.chemicalforums.com/index.php?topic=65859.0) you should show us your calculations.
AWK

#### dshipp17

• Regular Member
• Posts: 17
• Mole Snacks: +1/-0
##### Re: Chapter 1 Problem
« Reply #4 on: March 02, 2017, 05:46:53 PM »
Sorry about that; my mistake; I took V (total) = 54.2 and m (total) = 49.6; m (ethanol) = (0.789)(54.2) = 42.76 g; m (water) = (0.998)(54.2) = 54.09 g; m (solution) = 42.76 + 54.09 =96.85 g; %m (ethanol) = (42.76/96.85)*100%; %m (ethanol) = 44.15%; if I went wrong somewhere, please explain how I went wrong with my calculation; it seems slightly off from the textbook answer; not talking about something like possible significant figures causing a minor difference.

#### Borek

• Mr. pH
• Deity Member
• Posts: 25889
• Mole Snacks: +1693/-401
• Gender:
• I am known to be occasionally wrong.
##### Re: Chapter 1 Problem
« Reply #5 on: March 02, 2017, 06:02:14 PM »
If the V total is 54.2 mL, why do you use it for calculations as if it was just a volume of pure water?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### dshipp17

• Regular Member
• Posts: 17
• Mole Snacks: +1/-0
##### Re: Chapter 1 Problem
« Reply #6 on: March 02, 2017, 07:48:42 PM »
I'm not sure that I fully understand your question; could you elaborate? Please work the problem out and see where we may differ. I thought I accounted for both ethanol and water.

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7452
• Mole Snacks: +523/-87
• Gender:
##### Re: Chapter 1 Problem
« Reply #7 on: March 02, 2017, 08:10:52 PM »
Borek pointed out that this is wrong
Quote
(ethanol) = (0.789)(54.2)
And all further calculations are also wrong.
You should set a system of 2 equation with 2 variables as I told previously.
The real concentration of your alcohol solution is 49.4 % of ethanol (on the basis of density of solution).
Your assumption (approximation) should give you a result about ± 5% different (my guess).
« Last Edit: March 02, 2017, 08:35:55 PM by AWK »
AWK

#### dshipp17

• Regular Member
• Posts: 17
• Mole Snacks: +1/-0
##### Re: Chapter 1 Problem
« Reply #8 on: March 03, 2017, 07:53:45 AM »
Sorry, no, that's not the textbook answer and neither one of you solved the equation, as requested; how did you perform your calculation, was the question? The answer in the back of the textbook is 34%. And volume times density gives the mass; the ethanol density is 0.789g/mL and the total volume of the solution is 54.2 mL.

#### Borek

• Mr. pH
• Deity Member
• Posts: 25889
• Mole Snacks: +1693/-401
• Gender:
• I am known to be occasionally wrong.
##### Re: Chapter 1 Problem
« Reply #9 on: March 03, 2017, 08:39:32 AM »
m (ethanol) = (0.789)(54.2) = 42.76 g;
m (water) = (0.998)(54.2) = 54.09 g;
m (solution) = 42.76 + 54.09 =96.85 g;

Do you see that you have calculated the same solution to have three different masses at the same time? Don't you think it is impossible?

The answer in the back of the textbook is 34%.

With density tables I got 49.8% of ethanol. A bit different from what AWK got (chances are we used density tables for different temperature), but looks like the book answer is way off from the correct one.

Without density tables (assuming volumes are additive) I got 43% - not even close to the book answer either (unless it has digits switched).

how did you perform your calculation, was the question?

We try to guide you so that you will find how to solve the problem on your own, that's how the forum works. AWK gave you several hints about how to approach it.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### dshipp17

• Regular Member
• Posts: 17
• Mole Snacks: +1/-0
##### Re: Chapter 1 Problem
« Reply #10 on: March 03, 2017, 09:41:11 AM »
Well, the numbers are used that way, because the problem gave does conditions (e.g. the densities, volumes, and masses); I assumed that these conditions were given, because they're needed in the calculation; isn't this how problems are usually solved (e.g. the question states the conditions and you take those to solve the problem)? If I looked for something other than what's given in the problem, than either I'm going on a tangent or the person creating the problem is creating a bate and switch; but, yes, as I suspected, the answer in the back of the book must be wrong.

If you used a density table, how did you know which temperature to assume for a specific density? You could pick every temperature there. Why do we have to work out the problem but you not? Although not required, it would aid greatly if you extended the courtesy to show how you're deriving the answer. Also, if my calculations are off, I'll need you to be more specific and telling me. But, my answer took each of the masses, got the total mass, than calculated the mass percent of one of the variables. But, each of our answers are close to each others while the textbook answer is way off; if you used a density at a different temperature, than the density would be different from the one stated in the textbook and would explain the differences in our answers (e.g. we all used different densities).

#### Borek

• Mr. pH
• Deity Member
• Posts: 25889
• Mole Snacks: +1693/-401
• Gender:
• I am known to be occasionally wrong.
##### Re: Chapter 1 Problem
« Reply #11 on: March 03, 2017, 09:51:16 AM »
Instead of defending what you did try to think why it is nonsensical - can the same solution have three different masses at the same time? No, so your approach is clearly wrong. What can I add to that?

Solution has one mass, and it is given. Solution is a mixture of two substances - so the sum of their masses must be the mass of the solution. To solve the problem without density tables you have to assume volumes are additive - that is, the total volume is a sum of their volumes (that's not true, which is why the solution without density tables is only approximate). Express these thing with math and two variables, solve and you are done. You are told that for the second time.

When solving typical problems usually we assume temperatures around 20°C. Yes, it means sometimes we are a bit off, but it is still much better assumption that the one about volume additivity.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7452
• Mole Snacks: +523/-87
• Gender:
##### Re: Chapter 1 Problem
« Reply #12 on: March 03, 2017, 09:54:35 AM »
I got 34.2 mass percent of ethanol.
AWK

#### dshipp17

• Regular Member
• Posts: 17
• Mole Snacks: +1/-0
##### Re: Chapter 1 Problem
« Reply #13 on: March 03, 2017, 12:34:53 PM »
Well, I didn't calculate three separate masses; the masses were calculated based on the total volume of the solution; from there, the volumes were added; the calculation was based on the partial masses; without knowing how you would do you're calculations, you could basically say anything that gives an answer other than mine; so, I'm still not sure I can follow. Additionally, the problem says that the volume did not change but it didn't say anything about that mass. And, I'm more stating what the problem conditions were, not defending my answer while asking you to point out where I could be wrong; but, the answer has to be derivative of the conditions specified in the problem. But, I think I'm sensing what you're saying, but, there would be two unknown variables; this problem wasn't meant to require matrix algebra to solve, but, maybe it did; it's only in chapter 1 of a general chemistry book.

How did you you 34.2%? Please give some idea of your calculations.

#### Borek

• Mr. pH
• Deity Member
• Posts: 25889
• Mole Snacks: +1693/-401
• Gender:
• I am known to be occasionally wrong.
##### Re: Chapter 1 Problem
« Reply #14 on: March 03, 2017, 01:03:04 PM »
there would be two unknown variables; this problem wasn't meant to require matrix algebra to solve

Come on, solving system of equations with two unknowns doesn't require matrix algebra. Solve first equation for one unknown, put it into the second equation, solve for the remaining variable.

Well, I didn't calculate three separate masses

Sure you did - you got 42.76 g, 54.09 g and 96.85 g.

Quote
How did you you 34.2%? Please give some idea of your calculations.

Actually I am interested in seeing where this number comes from as well, I got 43 following the logic described in this thread.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info