Moles for chem:( (1 Viewer)

[kazemagic]

Sup,
So i got a question on moles for chem

For question b)
I did 0.037 times (40+32) and got 2.664, but the answer says its wrong - should be 5...
So I guess my logic of approach to this question is wrong. Here's my logic:
1 mole of calcium = 40 grams
1 mole of sulfur = 32 grams

If I add them up, it will tell me the mass of a mole of calcium sulfate which is 72 grams

Since the question asks you to find out 0.037 of a calcium sulfate mole: I just times 0.037 by 72.

And I got it wrong

Any help would be greatly appreciated!! Thanks

 

[kazemagic]

woops wrong category, soz modss

 

[nightweaver066]

Calcium Sulfate = , not .

 

[kazemagic]

Calcium Sulfate = , not .

oh wow im a tard arent i LOL
thanks brah

 

[kazemagic]

k got another question lol

for b, I don't really know how to do it, so heres my approach:
0.374 / 4 = 0.935 (amount of aluminium in that compound)
0.935 / 27 = 0.0346 (number of moles for aluminium)

The answer should be 2.8 x 10^-3 and I've no idea how to get that lol

Thanks

 

[deswa1]

It should contain the same number of moles of alumium as it does aluminium chloride as it does chlorine

 

[kazemagic]

It should contain the same number of moles of alumium as it does aluminium chloride as it does chlorine

thanks des
why though?
i thought moles were just a number of atoms. so shouldnt the number of aluminium atoms be lower than the atoms of chlorine and aluminium combined?

 

[deswa1]

thanks des
why though?
i thought moles were just a number of atoms. so shouldnt the number of aluminium atoms be lower than the atoms of chlorine and aluminium combined?

1 mole is Avogadro's number of molecules yes? So in 1 mole of alumium chloride, there are Avogadro's number of alumium chloride molecules. Likewise, in 1 mole of alumium chloride molecules, there must also be 1 mole of aluminium atoms in there and also 1 mole of chlorine atoms.

Your problem is that you are thinking of a mole as a pure number of atoms. That isn't correct. Its a number of molecules or atoms or whatever. For example, in one mole of molecular oxygen, there are actually 2x6.022x10^X (forgot what Avogadro's number is) oxygen atoms because molecular oxygen is O2.

That made no sense at all to me- hopefully you get something out of it haha

 

[kazemagic]

1 mole is Avogadro's number of molecules yes? So in 1 mole of alumium chloride, there are Avogadro's number of alumium chloride molecules. Likewise, in 1 mole of alumium chloride molecules, there must also be 1 mole of aluminium atoms in there and also 1 mole of chlorine atoms.

Your problem is that you are thinking of a mole as a pure number of atoms. That isn't correct. Its a number of molecules or atoms or whatever. For example, in one mole of molecular oxygen, there are actually 2x6.022x10^X (forgot what Avogadro's number is) oxygen atoms because molecular oxygen is O2.

That made no sense at all to me- hopefully you get something out of it haha

ohh yea it does make sense
thanks brah

 

[HSC2014]

And in ionic compounds, the empirical formula is referred to as a formula unit (instead of molecules/atoms). Can't think right now (bad explanation) but you'll run into it eventually.. ><

 

[Eyempro]

1 mole is Avogadro's number of molecules yes? So in 1 mole of alumium chloride, there are Avogadro's number of alumium chloride molecules. Likewise, in 1 mole of alumium chloride molecules, there must also be 1 mole of aluminium atoms in there and also 1 mole of chlorine atoms.

Your problem is that you are thinking of a mole as a pure number of atoms. That isn't correct. Its a number of molecules or atoms or whatever. For example, in one mole of molecular oxygen, there are actually 2x6.022x10^X (forgot what Avogadro's number is) oxygen atoms because molecular oxygen is O2.

That made no sense at all to me- hopefully you get something out of it haha

Haha! Although this may be confusing, it is fairly right Keep up the practice!

 

[Eyempro]

Did you get the answer? Use the number of moles from the whole compound (AlCl3) which you should have found from part a. Then find the mass of Al present (using mole the formula). Then you should have mass of Al. Then divide that by the molecular weight of Al. Take care with rounding. Pretty sure this is right (there may be other similar methods).
Btw I finished the HSC in 2012

 

[deswa1]

Did you get the answer? Use the number of moles from the whole compound (AlCl3) which you should have found from part a. Then find the mass of Al present (using mole the formula). Then you should have mass of Al. Then divide that by the molecular weight of Al. Take care with rounding. Pretty sure this is right (there may be other similar methods).
Btw I finished the HSC in 2012

Alternatively, use the number of moles from the whole compound and that by definition is the same as the number of moles of aluminium present. What you are doing is multiplying by molecular mass and then dividing by it. And I finished HSC in 2012 too