Chemistry Determining Moles and Particles from Mass of a Compound

AI Thread Summary
The discussion focuses on calculating the number of ions from the mass of aluminum chloride (AlCl3). The molar mass of AlCl3 is determined to be 133.33 g/mol, leading to the calculation of 0.279 moles of AlCl3 from a mass of 37.2 g. Participants emphasize the importance of writing out the ionization equation to visualize the relationship between moles of AlCl3 and its ions, noting a 1:3 ratio of Al3+ to Cl-. The final calculations show that 0.279 moles of AlCl3 yield approximately 1.67 x 10^23 Al3+ ions and 5.01 x 10^23 Cl- ions. The discussion highlights the significance of understanding stoichiometric ratios in chemical calculations.
rachelmaddiee
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Homework Statement
I need help with a question
Relevant Equations
N/A
mass = 37.2 g AICI3 (given)
number of ions = AI3+ Ions (unknown)
number of ions = CI- Ions (unknown)
mass = g/formula unit AICI3 (unknown)
The ratio of AI3+ Ions to CI- Ions is 1:3
Molar Mass AICI3
(1 x 26.98 g/mol AI) + (3 x 35.45 g/mol CI) = 133.33 g/mol AICI3
molar mass = 133.33 g/mol AICI3
Divide the given mass of aluminum chloride by its molar mass.
37.2 AICI3/133.33 mol = 0.279 mol AICI3

Is this correct?
DB1914AA-E7F8-4E2D-BD90-0F60A2250649.jpeg
 
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So you have moles of AlCl3. Now you should write out the balanced equation showing the production of ions from the AlCl3.
 
Note: it is not AICI3, it is AlCl3 (small L, not capital I). Please revise what the chemical symbols of elements are.
 
You have got the number of moles of AlCl3. Which we will consider made of Al3+ and Cl- ions. The question (as I understand it) asks what is the number of these ions in the sample.

(Hint: a very large number).
 
epenguin said:
You have got the number of moles of AlCl3. Which we will consider made of Al3+ and Cl- ions. The question (as I understand it) asks what is the number of these ions in the sample.

(Hint: a very large number).
1 mole of aluminum ions and 3 moles of Chloride ions
 
rachelmaddiee said:
1 mole of aluminum ions and 3 moles of Chloride ions
Nope. You really should just write out the ionization equation. It will help you visualize how to proceed.
 
chemisttree said:
Nope. You really should just write out the ionization equation. It will help you visualize how to proceed.
Are you looking at question 36?
 
Yes.
When I was teaching AP Chemistry, I required my students to write out the relevant equation to help them visualize and organize their thoughts. I’m not being pedantic. I know it works. Try it yourself if you don’t believe me.
 
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chemisttree said:
Yes.
When I was teaching AP Chemistry, I required my students to write out the relevant equation to help them visualize and organize their thoughts. I’m not being pedantic. I know it works. Try it yourself if you don’t believe me.
I don’t understand what I’m looking for.
 
  • #10
AB2 —-> A+2 + 2B -

This would be an example. Put what you are given above the equation and what you need to calculate beneath the equation.

Like this:

3.2 g
AB2 —-> A+2 + 2B -
? mol ?mol ?mol
? # ? # ? #
 
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  • #11
rachelmaddiee said:
I don’t understand what I’m looking for.

If I ask how many atoms there are in a mole of an element does that help make sense of the Question?
 
  • #12
epenguin said:
If I ask how many atoms there are in a mole does that help make sense of the Question?
Nowadays they are asked for “how many particles” rather than how many atoms or molecules. It needlessly complicates things IMO but that’s the way it is.
 
  • #13
1 mole = 6.02 x 10^23 atoms
 
  • #14
rachelmaddiee said:
1 mole = 6.02 x 1023 atoms
1 mole = 6.02 x 1023 atoms
1 mole = 6.02 x 1023 particles
 
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  • #15
chemisttree said:
1 mole = 6.02 x 1023 atoms
Sorry, a mistake!
 
  • #16
chemisttree said:
AB2 —-> A+2 + 2B -

This would be an example. Put what you are given above the equation and what you need to calculate beneath the equation.

Like this:

3.2 g
AB2 —-> A+2 + 2B -
? mol ?mol ?mol
? # ? # ? #
This is confusing me. I can’t find any examples in my book.
 
  • #17
Which is why I’m going through it with you here.
 
  • #18
chemisttree said:
Which is why I’m going through it with you here.
37.2 g
AICI3 —> AI+3 + 3CI -
 
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  • #19
Can you please give me an example of How should it be written when I get my result?
 
  • #20
image.jpg

OK. You are asked to find the number of ions of aluminum in part a. To do that you must calculate the number of moles of Al first. You have already calculated the moles of AlCl3 so you are on your way!
 
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  • #21
From the equation above how many moles of Al+3 are produced from each mole of AlCl3?
 
  • #22
chemisttree said:
From the equation above how many moles of Al+3 are produced from each mole of AlCl3?
133.34g/mol?
 
  • #23
rachelmaddiee said:
133.34g/mol?
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 produces half a mole of Al+3. How about 0.279 moles of AlCl3?
 
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  • #24
chemisttree said:
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 produces half a mole of Al+3. How about 0.279 moles of AlCl3?
Yes, I’ve solved for. That
 
  • #25
Ok then. How many moles of Cl- do you get from 0.279 moles of AlCl3?
 
  • #26
chemisttree said:
Ok then. How many moles of Cl- do you get from 0.279 moles of AlCl3?
Wouldn’t it be 0.279 moles too?
 
  • #27
rachelmaddiee said:
Wouldn’t it be 0.279 moles too?
Each mole of AlCl3 produces how many moles of Cl-? Refer to the equation.
 
  • #28
chemisttree said:
Each mole of AlCl3 produces how many moles of Cl-? Refer to the equation.
3?
 
  • #29
Right! So how many moles from 0.279 moles of AlCl3?
 
  • #30
chemisttree said:
Right! So how many moles from 0.279 moles of AlCl3?
I don’t. Understand how to find that?
 
  • #31
Half a mole of AlCl3 would produce 3 X 1/2 or 1.5 mol of Cl-.
 
  • #32
chemisttree said:
Half a mole of AlCl3 would produce 3 X 1/2 or 1.5 mol of Cl-.
So it’s 1.5?
 
  • #33
Yeah, IF you had half a mole of AlCl3...
So, how many moles from 0.279 moles of AlCl3?
 
  • #34
chemisttree said:
Yeah, IF you had half a mole of AlCl3...
So, how many moles from 0.279 moles of AlCl3?
So 3 x 1?
 
  • #35
rachelmaddiee said:
So 3 x 1?
No. You would multiply the stoichiometric ratio by the number of moles given (or calculated in your case).
 
  • #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?
 
  • #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?
 

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