Determining Moles and Particles from Mass of a Compound

In summary: No, that’s the formula weight. You notice there is a 1:1 relationship between AlCl3 and Al+3. Each mole of AlCl3 produces a mole of Al+3. Half a mole of AlCl3 would be 133.3 g.
  • #36
chemisttree said:
No. You would multiply the stoichiometric ratio by the number of moles given (or calculated in your case).
0.279 * 1:3 ratio?
 
Physics news on Phys.org
  • #37
rachelmaddiee said:
0.279 * 1:3 ratio?
What textbook are you using for class?
 
  • #38
chemisttree said:
What textbook are you using for class?
656079E0-7DE3-4A12-B22D-5C721A5EDB29.png
 
  • #39
Go to chapter 12 and review stiochiometric ratios used in calculations. In your example the ratio 1:3 would be the ratio of AlCl3 to Cl- and the ratio 3:1 would be the ratio of Cl- to AlCl3. You have to lnow when to use either one of them. Units inform that choice.
From this point you must show your units in calculations so you don’t use the wrong ratio or conversion factor.
 
  • #40
rachelmaddiee said:
1 mole = 6.02 x 10^23 atoms

So how many Al atoms or ions does the 0.279 mol (which you calculated) contain?

Pretty easy question, one of us must have misunderstood what the question is.
 
  • #41
epenguin said:
So how many Al atoms or ions does the 0.279 mol (which you calculated) contain?

Pretty easy question, one of us must have misunderstood what the question is.
I thought the answer for part a and part b was 1 mole of aluminum ions and 3 moles of Chloride ions
 
  • #42
chemisttree said:
Go to chapter 12 and review stiochiometric ratios used in calculations. In your example the ratio 1:3 would be the ratio of AlCl3 to Cl- and the ratio 3:1 would be the ratio of Cl- to AlCl3. You have to lnow when to use either one of them. Units inform that choice.
From this point you must show your units in calculations so you don’t use the wrong ratio or conversion factor.
I can’t find the chapter
 
  • #43
Would it be 1.67 x 10^23?
 
  • #44
rachelmaddiee said:
1 mole = 6.02 x 10^23 atoms
OK let's consider AlCl3 to be a molecule.
From the above quote, how many molecules of AlCl3 are there in 0.279 mol?

How many molecules of anything are there in 0.279 moles of it?
 
  • #45
epenguin said:
OK let's consider AlCl3 to be a molecule.
From the above quote, how many molecules of AlCl3 are there in 0.279 mol?

How many molecules of anything are there in 0.279 moles of it?
Am I suppose to multiply 6.02 x 10^23 by the number of moles?
 
  • #46
rachelmaddiee said:
Am I suppose to multiply 6.02 x 10^23 by the number of moles?

Do you have no way of working out whether that is so or not except someone tells you? if you do that multiplication what will it tell you? OK a number, but what is it the number of and why?

I just have to guess why such an easy problem is holding you up so. One possibility is you have not understood what question they are asking. More probably or additionally it is a very common one found in this section again and again for questions of stoichiometry: that students have done many exercises in elementary school on something called "simple proportions". Only the lessons then were called "arithmetic" and they were self-contained excercises to get right but maybe not much idea given they would ever be any use for anything. When the same sort of calculation is called "chemistry" they can't do the calculations they did 10 years earlier any more or don't know what they mean.
 
  • #47
epenguin said:
Do you have no way of working out whether that is so or not except someone tells you? if you do that multiplication what will it tell you? OK a number, but what is it the number of and why?

I just have to guess why such an easy problem is holding you up so. One possibility is you have not understood what question they are asking. More probably or additionally it is a very common one found in this section again and again for questions of stoichiometry: that students have done many exercises in elementary school on something called "simple proportions". Only the lessons then were called "arithmetic" and they were self-contained excercises to get right but maybe not much idea given they would ever be any use for anything. When the same sort of calculation is called "chemistry" they can't do the calculations they did 10 years earlier any more or don't know what they mean.
I’ve found an example and attempted it. Hopefully everything is correct.
0.279 x 6.02 x 10^23 = 1.67 x 10^23 formula units AICI3

1.67 x 10^23 x 1 AI3+ ion = 1.67 x 10^23 AI3+ ions
1.67 x 10^23 x 3 CI- ion = 5.01 x 10^23 CI- ions
 
  • #48
18E39A97-5C0A-4E7C-A0D4-76216DD82C74.png
 
  • #49
That's right, it's not a guess or that way arbitrarily .

What does the new printed extract correspond to - worked solution in your textbook?
 
  • #50
epenguin said:
That's right, it's not a guess or that way arbitrarily .

What does the new printed extract correspond to - worked solution in your textbook?
What does that mean?
 
  • #51
rachelmaddiee said:
What does that mean?
I just wondered what it meant after working through this problem for two pages you produce a solution that looked from print style like textbook extract. What is #48?
 
  • #52
epenguin said:
I just wondered what it meant after working through this problem for two pages you produce a solution that looked from print style like textbook extract. What is #48?
Oh, yes I found an example in my book and followed it!
 

Similar threads

  • Biology and Chemistry Homework Help
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
2
Views
1K
  • Biology and Chemistry Homework Help
Replies
2
Views
3K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
14
Views
9K
  • Biology and Chemistry Homework Help
Replies
7
Views
4K
  • Biology and Chemistry Homework Help
Replies
6
Views
2K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
  • Biology and Chemistry Homework Help
Replies
16
Views
2K
Back
Top