[argorb]

Hello, I am new to chemistry. I am trying to understand (or rather get at least a good sense of the mathematical process with) moles, amus and Avogardo's constant.

What I understood so far is that: 1 mole (in "chemistry math") is equal to 1 douzen (in "regular math"). In order for a substance to have a mass of 1 mole, it has to contain 6.02*10^23 atomic mass units. Similar to: in order for eggs to be one douzen they have to be 12. So this way was made to make life easier on scientists and express somehow how much substance is there in one particular "group". Sorry if I explain bad, I am just getting it and somehow its still quite fuzzy for me. Anyways, using the periodic table, what came to realize is that the smallest amount of what a certain element from the table can have is AT LEAST 1 gram (being H with 1 atomic mass number). Moving on, I did this exercise and... hopefully you will correct me, as well as you might give some advices of how to clarify what is an amu and how it can be compared to like Hidrogen's atomic mass number (1.008).

Though, I did find out that 1 amu = 1.6605 * 10^(-27). I am using Khan's Academy videos to study. Below is the exercise with some `personal' noted explanation:

[BEGINNING OF PERSONAL NOTES]
What science from today is telling us, is that 1 mole (in chemistry) IS SIMILAR (not equal) to 1 douzen (in math from todays)
1 mole = 6.02 * 10^23 substance -- meaning that one substance in order to be a mole it has to contain AT LEAST 6.02*10^23 amus.

Example:
In one douzine there are 12 eggs.
One egg weights 60 grams.
Find out how much a douzen weights.
1 egg = 60g
1 douzine = 12 eggs
Mass of one douzen = 12 eggs * 60grams (60 grams each egg)
One douzen mass = 720g
1 dm (we particularly -- for the explanation sake -- invent ourselves the word: douzen mass) = 720g

In chemistry case, 1 douzen is the "similar equivalent" to one mole! Only that the value of one mole is not 12 but rather 6.02 * 10^23, meaning that, in order for a substance to have a mass of 1 mole it has to have at least 6.02*10^23 particles (amus).

Find out how much mass 5 moles of K2Cl2O4 has.
1 mole = 6.02*10^23
5 moles = 5 * 6.02*10^23
K2:
   Atomic mass number of one potassium = 1 * 39 amus
   Atomic mass number of two potassium atoms = 2 * 39 = 78 amus
Cl2:
    Atomic mass number of one atom of Chlorine = 35 amus
    Atomic mass number of two atoms of Chlorine = 2 * 35 = 70 amus
O4:
   Atomic mass number of one oxigen = 16 amus
   Atomic mass number of four atoms of oxigen = 4 * 16 = 64 amus

We write the exercise sentence with what information we got so far: Find out how much mass there is in 5 * 6.02*10^23 (5 moles) * (of) 78 + 70 + 64 amus (substance)           ||
                                             \/
5 * 6.02*10^23 * (K2:78 + Cl2:70 + O4:64 * (of) 1.6605 * 10^(-27) (the value of one amu)).
=> 5 * 6.02*10^23 * 212 * 1.6605 * 10^(-27) = 1.05959826kg * 1000g (1kg) = 1059g
[END OF PERSONAL NOTES]

Is all of what I wrote correct? And curious thing: as I was writing this thread, I realized that `amu' is a unit measure (that has value of 1.6605 * 10^(-27)) just like 1 km is measured in meters and it has the value of 1000m. Is this true or am I mistaking?
So 1 km = 1 mole and 1 amu  = 1 meter: 1 mole = 6.02*10^23 amus just like 1 km = 1000 meters. Is this true?

If I got it wrong somewhere, please clarify for me. Thank you very much and I apologize this long long thread.

[fledarmus]

The one thing that you seem to be getting wrong is that an amu is actually a unit of mass - it is not 1.6605 * 10^(-27), it is 1.6605 * 10^(-27) kg! That unit is important! This is the approximate mass of a single proton or neutron, and if you want to know how much a single molecule or atom weighs, these are the units you use.

But we rarely work with single molecules or atoms, because they are just way too small. We work with moles instead.

So your final comparison isn't really true:

So 1 km = 1 mole and 1 amu  = 1 meter: 1 mole = 6.02*10^23 amus just like 1 km = 1000 meters. Is this true?

Instead, only the "k" part of your "km" is like the mole. k (kilo-) means "1000 of" and mole means "6.023x10²³ of" - so you could have km, kg, kL, kPa, etc., and you can have moles O₂, moles H₂, moles H₂O, etc. The comparison would be "1000 meters", "1000 grams", "1000 liters", "1000 pascals", compared to "6.023x10²³ molecules of oxygen", "6.023x10²³ molecules of hydrogen", "6.023x10²³ molecules of water". The conversion is that one mole of atoms in grams is the same number as 1 atom in amus - eg, 1 mole of O₂ would weigh about 32 gm (16+16) and one atom of O₂ would weigh about 32 amus.

You don't see amus much outside of high school and college. In biology, this unit is called the Dalton (Da) and is commonly used to describe the size of proteins (a single molecule of a 100 kDa protein would weigh 100,000 amus, while a mole of it would weigh 100,000 g)

I hope that helps instead of just muddying the water

[argorb]

The one thing that you seem to be getting wrong is that an amu is actually a unit of mass - it is not 1.6605 * 10^(-27), it is 1.6605 * 10^(-27) kg! That unit is important! This is the approximate mass of a single proton or neutron, and if you want to know how much a single molecule or atom weighs, these are the units you use.

But we rarely work with single molecules or atoms, because they are just way too small. We work with moles instead.

So your final comparison isn't really true:

So 1 km = 1 mole and 1 amu  = 1 meter: 1 mole = 6.02*10^23 amus just like 1 km = 1000 meters. Is this true?

Instead, only the "k" part of your "km" is like the mole. k (kilo-) means "1000 of" and mole means "6.023x10²³ of" - so you could have km, kg, kL, kPa, etc., and you can have moles O₂, moles H₂, moles H₂O, etc. The comparison would be "1000 meters", "1000 grams", "1000 liters", "1000 pascals", compared to "6.023x10²³ molecules of oxygen", "6.023x10²³ molecules of hydrogen", "6.023x10²³ molecules of water". The conversion is that one mole of atoms in grams is the same number as 1 atom in amus - eg, 1 mole of O₂ would weigh about 32 gm (16+16) and one atom of O₂ would weigh about 32 amus.

You don't see amus much outside of high school and college. In biology, this unit is called the Dalton (Da) and is commonly used to describe the size of proteins (a single molecule of a 100 kDa protein would weigh 100,000 amus, while a mole of it would weigh 100,000 g)

I hope that helps instead of just muddying the water

Thank you very much! With your reply I got a little bit more confused because I have to still `digest' it, but yes, I did forgot about the 1.6605*10^-27 kg = 1 amu. I will read your post more and more while checking my notes; it should get clear by grasping at it for a while. By the way, it did feel to me before that 1 k from the km example kind of applies (but not literally -- just for comaprision example or to get an idea of how is represented) to the mole.

Later edit
In the following example:

eg, 1 mole of O2 would weigh about 32 gm (16+16) and one atom of O2 would weigh about 32 amus."

You meant one diatomic oxygen or one molecule of O2, right? Thanks!

[fledarmus]

Later edit
In the following example:

eg, 1 mole of O2 would weigh about 32 gm (16+16) and one atom of O2 would weigh about 32 amus."

You meant one diatomic oxygen or one molecule of O2, right? Thanks!
[/quote]

Ooops - yes I did! Thanks - good catch!

[argorb]

Well, it seems after I thought I understood how all this "stuff" works, apparentely I lost it again, probably due to my bad background in math (I hate it because I do not understand it for the reason that I did not give it any chance at all or time, but I love it as the intelectual "logic" challenge -- dunno if this is interesting), when I stumbled upon some Moles exercises like the following:

Code: [Select]

1)      How many moles of sodium atoms correspond to 1.56x1021 atoms of sodium?

 

2)      How many moles of Al atoms are needed to combine with 1.58 mol of O atoms to make aluminum oxide, Al2O3?

 

3)      How many moles of Al are in 2.16 mol of Al2O3?

 

4)      Aluminum sulfate, Al2(SO4)3, is a compound used in sewage treatment plants.

a.     Construct a pair of conversion factors that relate moles of aluminum to moles of sulfur for this compound

b.     Construct a pair of conversion factors that relate moles of sulfur to moles of Al2(SO4)3

c.       How many moles of Al are in a sample of this compound if the sample also contains 0.900 mol S?

d.     How many moles of S are in 1.16 mol Al2(SO4)3?

 

5)      How many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3?

 

6)      What is the total number of atoms in 0.260 mol of glucose, C6H12O6?

8)      Determine the mass in grams of each of the following:

a.     1.35 mol Fe

b.     24.5 mol O

c.       0.876 mol Ca

d.     1.25 mol Ca3(PO4)2

e.      0.625 mol Fe(NO3)3

f.       0.600 mol C4H10

g.     1.45 mol (NH4)2CO3

 

9)      Calculate the number of moles of each compound:

a.     21.5 g CaCO3

b.     1.56 g NH3

c.       16.8 g Sr(NO3)2

d.     6.98 mg Na2CrO4

Answers (not mine):

Code: [Select]

1)      2.59x10-3mol Na atoms 

2)      1.05mol Al               

3)      4.32mol Al

4)      a. 2mol Al/3mol S            b. 3mol S/1mol Al2(SO4)3            c. 0.600mol Al     d. 3.48mol S

5)      0.0725mol N2 and 0.218mol H2

6)      3.76x1024 atoms

8)      a. 75.4g Fe            b. 392g O            c. 35.1g Ca            d. 388g Ca3(PO4)2       

e. 151g Fe(NO3)2      f. 34.9g C4H10               g. 139g (NH4)2CO3

9)      a. 0.215mol            b. 0.0916mol                c. 0.0794mol                d. 4.31x10-8mol

I am having a hard time to resolve especially the exercises which involes the nr^power. If someone could elude a little bit this mistery for me, it would be awesome. Right now I am in a rush, but I promise I will try again to resolve them and will come back with my different attempts at this.

Thanks again and I hope I am not annoying somehow, with my elementary noob questions about this domain.

By the way, exercises where taken from the following website:

Code: [Select]

http://academic.evergreen.edu/curricular/matterandmotion/chem_phys\practice_problems.htm

[fledarmus]

Not sure what you mean by nr^power, but once you start posting your solutions, I am sure you will have lots of people weighing in on what you get right!

To start you off, on question 1, "how many moles of sodium atoms correspond to 1.56 x 10²¹ moles of sodium", think about how you would answer the question, "how many dozens of eggs correspond to 7 eggs?" They are exactly the same question, only the numbers are different.

[argorb]

To start you off, on question 1, "how many moles of sodium atoms correspond to 1.56 x 10²¹ moles of sodium", think about how you would answer the question, "how many dozens of eggs correspond to 7 eggs?" They are exactly the same question, only the numbers are different.

Sorry for the bad paste that is in the [ code ] function, I forgot to edit and put the powers to the coresponding numbers, glad I put the link though. Anyways!

Of course it will be with fraction, 7 eggs is not a dozen; like we stated above 12 eggs makes it a dozen, so this makes it .58 from a dozen.
Thank you! That clarifies it for me, my answer is the following:

1 atom of sodium has the mass = 32 amus or 32 grams
6.02 x 10²³ x 32 (a whole mole of sodium) = 19264000000000000000000000
1.56 x 10²¹ x 32 (this is like a fraction from a mole) = 49920000000000000000000

So we divide like the following: 49920000000000000000000 / 19264000000000000000000000 and we get: 0.002591362 amus (1.6605 x 10^-27 kg), so we multiply this by 1000 (one kg) and we get: 2.5913

The exercise is telling us: 2.59x10 which means 25.9, and it appears that my answer is wrong? I don't understand the multiplication from 2.59x10

I will come back with a full post on the other problems, being careful to solve each exercise with patience. Thanks and I am very humble to you since you answer to my elementary questions about chemistry, I wish I knew more and post something interesting, exciting and challenging.

[fledarmus]

You don't need to go through grams to get your answer - did you have to use the weight of the eggs to find out how many dozen you had?

[argorb]

1. How many moles of sodium atoms correspond to 1.56x10²¹ atoms of sodium?

You don't need to go through grams to get your answer - did you have to use the weight of the eggs to find out how many dozen you had?

Good point. I looked over the problem again and I said to be a little cheater. It give us 1.56x10²¹ atoms of sodium, like you stated, we can treat this as the 7 eggs from a dozen. After that it told us "How many moles of sodium atoms", the number of the dozen we know it is correct, which is 6.02x10²³. We divide 1.56 by 6.02 and we get 0.259136213 which we multiply by 10; 0.259136213 x 10 = 2.59 . I do not know why I still have the feeling that I did this like a challenge instead of understanding what is really happening, since I did not know what to do with the ^power numbers like the ²³ from the Avogardo's constant and ²¹ from what the problem gave us.



2. How many moles of Al atoms are needed to combine with 1.58 mol of O atoms to make aluminum oxide, Al2O3?

I was already going through that crap again of calculating the molecular mass of the whole molecule of aluminium oxide. When I noticed that, for every 3 atoms of O there are 2 atoms of Al. I divided 2 by 3 and multiplied this by 1.58 mol and I got 1.053333333. I am still not sure yet, why I did this, but this seemed worthy enough, finding out the ratio of Al to O and then multiplying it with the moles the problem gave us.



3. How many moles of Al are in 2.16 mol of Al2O3?

1 mole of Al2O3 -> 102amu
number of Al2 from 1 mole of Al2O3 -> 2 x 27 -> 54 amu
1 mole of Al = 27 amu

After many tries, getting fuzzy and looking again over my notes, I already had the information to solve this.
We know that the number of Al2 from 1 mole of aluminium oxide is 54 -> this number represents the amu that is currently in 1 mole of Al2O3. We also know that 1 mole of Al is 27 amu. We divide 54 by 27 and multiplying it with the number of moles it gave us 2.16 resulting in 54 / 27 x 2.16 = 4.32   -- this type of substitution (words for actually numbers) and working with moles and so on are always getting me into blur.



4. Aluminum sulfate, Al2(SO4)3, is a compound used in sewage treatment plants.

a. Construct a pair of conversion factors that relate moles of aluminum to moles of sulfur for this compound
So this is telling us, how many moles of aluminium is there per moles of sulfur in this compound (Al2(SO4)3).
Well, we know that there is a diatomic Aluminium which is 54 grams per mole. We need to know how many grams of (SO4)3 we have in this compound.
3 x 32 + 4 x 3 x 16 = 288g since SO group cannot be splitted.
1 mole of Al = 27
1 mole of S = 32

Actually, I realised from the answer and what I first said in the beginning of this point, "So this is telling us, how many moles of aluminium is there per moles of sulfur in this compound (Al2(SO4)3)." -- it tells us directly that there are 2 Al per 3 SO4 so the ratio is 2/3 moles like the answer. Is this a correct way to think about it? Or I could have approached it differently?



b. Construct a pair of conversion factors that relate moles of sulfur to moles of Al2(SO4)3
We know from point `a.' that 1 mole of S = 32 and 1 mole of Al2(SO4)3 = 288, actually no, again I looked up at the answer and it is the same like point `a'.
We have 3 SO4 per 1 mole of Al2(SO4)3, so the relate moles of sulfur per moles of Al2(SO4)3 is: 3/1.



c. How many moles of Al are in a sample of this compound (Al2(SO4)3) if the sample also contains 0.900 mol S?
This is literally asking, how many moles of Al in this compound are, if the sample also contains 0.900 mol S. We know that there are 2 moles of Al in this compound (Al2). We also know that 1 mole of S in this compound = 96 (3 x 32) .. well I am stuck, I cannot figure it out.



d. How many moles of S are in 1.16 mol Al2(SO4)3?
1 mole of S in this compound = 3 x 32 is equal to 3 moles of normal S.
3 x 1.16 = 3.48





5. How many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3?
No clue. What "decomposition" even means? I imagined that if we take apart 0.145 mol of NH3 into its forming diatomic H2 and N2, how many moles of those particular diatoms will be formed. It seems this is very easy but I can't get it, god damn it!




6. What is the total number of atoms in 0.260 mol of glucose, C6H12O6?
C6 = 6 x 12 = 72
H12 = 12
O6 = 6 x 16 = 96
72 + 12 + 96 = 180
1 mole of C6H12O6 = 180g

Later edit: Funny, that I solve another problem with a member from here and I figured out that number of atoms is Avogardo's number, I was mistaking with with number of substance, but it is actually number of atoms in a substance: 0.260 x 180 = 46.8g x 6.02x10²³

The answer it gave us is 3.76x1024 atoms, how the heck it came up with this figure? 3.76x1024?




8. Determine the mass in grams of each of the following:

a.     1.35 mol Fe = 1 mole of Fe = 55.845g x 1.35 = 75.39g rounding it to 75.4g

b.     24.5 mol O = 1 mole of O = 16g x 24.5 = 392g

c.     0.876 mol Ca = 1 mole of Ca = 40 x 0.876 = 35.04g

d.     1.25 mol Ca3(PO4)2 = 1 mole of Ca3(PO4)2 = 3 x 40 + 2 x 31 + 8 x 16 = 310 x 1.25 = 387.5g

e.     0.625 mol Fe(NO3)3 = 1 mole of Fe(NO3)3 = 56 + 3 x 14 + 9 x 16 = 242 x 0.625 = 151.25g

f.     0.600 mol C4H10 = 1 mole of C4H10 = 4 x 12 + 10 = 58 x 0.600 = 34.8g

g.     1.45 mol (NH4)2CO3 = 1 mole of (NH4)2CO3 = 2 x 14 + 8 + 12 + 3 x 16 = 96 x 1.45 = 139.2g

 

9.  Calculate the number of moles of each compound:

a.     21.5 g CaCO3 = 1 mole of CaCO3 = 40 + 12 + 3 x 16 = 100g = 21.5/100 = 0.215 mole

b.     1.56 g NH3 = 1 mole of NH3 = 14 + 3 = 17g = 1.56/17 = 0.0917 mole

c.     16.8 g Sr(NO3)2 = 1 mole of Sr(NO3)2 = 88 + 2 x 14 + 6 x 16 = 212g = 16.8g/212g = 0.0792 mole

d.     6.98 mg Na2CrO4 = 1 mole of Na2CrO4 = 2 x 23 + 52 + 4 x 16 = 162g = 162000mg = 6.98 / 162000 = 0.000043 but the answer that is given to us is 4.31x10 which is 43.1?


As you can see, I am still having difficulties into understanding the concept required to solve those types of problems, probably because I cannot do this basic math using the words like 'moles' and getting confused by the requirement of the exercise, as I understood (probably not that much) the concept of mole, amu and Avogardo's number when I posted the opening thread post. Though I am feeling like this barrier and finding it hard to really understand the problems and solve them easily and it was supposed to be, right? Sorry for being such a `hard head' in understanding this, I hope I am not making you pull out the hair. Don't know why, but one time when I write and read posts like first one ( http://www.chemicalforums.com/index.php?topic=52071.msg192988#msg192988 ) I feel like I understood it, but when I put it in practice with problems and exercises, I'm getting confused and not being able to solve them.

Thanks again!

[fledarmus]

Much better - you're getting closer to understanding it.

I didn't realize you were having problems with the scientific notation (the "x 10^power part). Most calculators these days can handle that sort of notation. If your calculator can't, you will have to learn to do some exponent math. Two helpful things to remember:

1) Multiplying - when you multiply terms with exponents, if the base term is the same you can add the exponents. For example:
a^wxa^y = a^w+y
Dealing with scientific notation, this means
10^w + 10^y = 10^w+y.
So if you had to multiply
(1.56 x 10²¹)x(6.02x10²³)
 you would get
(1.56x6.02)x(10²¹x10²³)
 which would be
9.39x10²¹⁺²³, or 9.39x10⁴⁴

2) Dividing - when you divide terms with exponents, if the base term is the same you can subtract the exponents. For example:
a^w/a^y = a^w-y
Again, dealing with scientific notation, this means
10^w/10^y=10^w+y
So if you had to divide
1.56x10²¹/6.02x10²³
(which in fact is exactly what you want with your first problem)
That would give you
(1.56/6.02)x(10²¹/10²³)
which would be
(0.259)x10^21-23, or (0.259x10^-2)

3)Entering and leaving scientific notation - by definition, scientific notation is of the form a.aa...x10^b - that is, one digit in front of the decimal point, as many digits behind the decimal points as your precision allows, times 10 to some exponent. Since we use numbers in base ten, you can move the decimal point to the right or the left in your number by raising or lowering the exponent. For each place you move the decimal point to the left, you raise the exponent by 1. For each point you move the decimal to the right, you lower the exponent by 1.
So the answer from the previous division problem,
0.259x10^-2 would be the same as 2.59x10^-3 (which would be in proper scientific notation), or 0.00259

And that is actually the answer to your first question - 1.56x10²¹ atoms of sodium is 0.00259 moles.


Number 2 in your problem set is right - all you need is the ratio between the number of atoms of O in your product and the number of atoms of Al in your product to find out how many moles of Al you need to combine with 1.58 mols O.

Number 3 is wrong - it is essentially the same problem as number 2, which you solved correctly. You don't need to go through the mass of the molecules, all you are concerned about is the number. There are 2 moles of Al atoms in 1 mole of Al₂O₃ molecules, so how many are in 2.16 moles?

All of the rest of the questions are similar until you get to question 8. You do not need the molecular weight or the atomic weight to find out molar ratios, just like you don't need the weight of the eggs to count out the dozens.

I haven't actually worked through all your numbers for 8 and 9 but your method looks right. I think for 9d they just left out the exponent - 6.98 mg couldn't possibly be 43 moles of any molecule, when 1 mole of the very lightest atom weighs 1 gram!

[argorb]

Much better - you're getting closer to understanding it.

I didn't realize you were having problems with the scientific notation (the "x 10^power part). Most calculators these days can handle that sort of notation. If your calculator can't, you will have to learn to do some exponent math. Two helpful things to remember:

1) Multiplying - when you multiply terms with exponents, if the base term is the same you can add the exponents. For example:
a^wxa^y = a^w+y
Dealing with scientific notation, this means
10^w + 10^y = 10^w+y.
So if you had to multiply
(1.56 x 10²¹)x(6.02x10²³)
 you would get
(1.56x6.02)x(10²¹x10²³)
 which would be
9.39x10²¹⁺²³, or 9.39x10⁴⁴

2) Dividing - when you divide terms with exponents, if the base term is the same you can subtract the exponents. For example:
a^w/a^y = a^w-y
Again, dealing with scientific notation, this means
10^w/10^y=10^w+y
So if you had to divide
1.56x10²¹/6.02x10²³
(which in fact is exactly what you want with your first problem)
That would give you
(1.56/6.02)x(10²¹/10²³)
which would be
(0.259)x10^21-23, or (0.259x10^-2)

3)Entering and leaving scientific notation - by definition, scientific notation is of the form a.aa...x10^b - that is, one digit in front of the decimal point, as many digits behind the decimal points as your precision allows, times 10 to some exponent. Since we use numbers in base ten, you can move the decimal point to the right or the left in your number by raising or lowering the exponent. For each place you move the decimal point to the left, you raise the exponent by 1. For each point you move the decimal to the right, you lower the exponent by 1.
So the answer from the previous division problem,
0.259x10^-2 would be the same as 2.59x10^-3 (which would be in proper scientific notation), or 0.00259

And that is actually the answer to your first question - 1.56x10²¹ atoms of sodium is 0.00259 moles.


Number 2 in your problem set is right - all you need is the ratio between the number of atoms of O in your product and the number of atoms of Al in your product to find out how many moles of Al you need to combine with 1.58 mols O.

Number 3 is wrong - it is essentially the same problem as number 2, which you solved correctly. You don't need to go through the mass of the molecules, all you are concerned about is the number. There are 2 moles of Al atoms in 1 mole of Al₂O₃ molecules, so how many are in 2.16 moles?

All of the rest of the questions are similar until you get to question 8. You do not need the molecular weight or the atomic weight to find out molar ratios, just like you don't need the weight of the eggs to count out the dozens.

I haven't actually worked through all your numbers for 8 and 9 but your method looks right. I think for 9d they just left out the exponent - 6.98 mg couldn't possibly be 43 moles of any molecule, when 1 mole of the very lightest atom weighs 1 gram!

Thank you for your time put in to evaluate my answers. At 9d there were 6.98 MIU(if i am not mistaken) - (µ) mg Na2CrO4, and since the copy paste didn't put miu (µ) here aswell, I got it wrong.

At 6, the answer was: 3.76x10²⁴ I thought its 3.76x1024, what an idiot I am!

For 4c, I still cannot figure it out, and same for exercise number 5.

And I understood what you said for question 3: 2 (2 Al atoms)x2.16(moles) = 4.32

I think I will make a break from moles,amus and Avogardo's and try a little Stiochiometry though they will involve this kind of math operations, so, I guess you cannot get rid of them and just have to live with them by doing exercises until you get it and it feels like it isn't a burden.

Thanks again, will review the ones I couldn't and try them again and again while checking what I did right and trying to understand why and what happend there, though I think as you said, I am getting closer to it.