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 1.56x10
21 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 6.02x10
23. 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 compoundSo 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)3We 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 6.02x10
23The 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!