Sciencemadness Discussion Board

Weight - Molarity of di/ter/quart-atomic elements

RU_KLO - 16-2-2024 at 09:05

Simple question, but I need to be certain.

In case of di/ter/quart..atomic elements, the weight needed for one mole of the element, is the (atomic) weight divided the di/ter/quart....

(sorry for poor presentation of the problem, maybe an example:

If I need a mole (in 1L) of Sulfur, and powdered sulfur is S8, atomic weight 32,
1) I need 32gr of powdered sulfur? (in 1 liter)
2) I need 4gr of powdered sulfur? (in 1 liter) (32/8)

Same as Iodine, (if I2 and atomic weight 126.9)
1) I need 126.9 gr of Iodine? (in 1 liter)
2) I need 63.45gr of Iodine? (in 1 liter)

thanks,



Texium - 16-2-2024 at 11:29

You're calculations are backwards. One mole of I weighs 126.9 g. One mole of I2 weighs 253.8 g. Ultimately how much you use depends on the stoichiometry of the reaction you're using it in.

For example, an electrophilic aromatic halogenation:

ArH + X2 ---> ArX + HX

You would need to use one mole of X2 per mole of ArH. If you only used half a mole, your reaction would not go to completion because half of the halogen becomes HX.

On the other hand, consider the reaction of sulfur and aluminum:
2 Al + 3 S ---> Al2S3

It would be equally valid to write it as:
16 Al + 3 S8 ---> 8 Al2S3

Either way you write it, the weight ratio of aluminum to sulfur is the same. Does that make sense to you?

RU_KLO - 17-2-2024 at 04:11

yes thanks.

My first mistake was not to write the formula stoichiometrically.

My idea was to make KI, from Iodine and KOH.
As I do not have much Iodine (less than a gram) was thinking how much KI could I get. But as I is diatomic, was not sure how many moles I have.

Then by looking at the stoi.... formula:
3I2 +6 KOH → KIO3 + 5KI + 3H2O

it seems that I need 6 M of I for making 5 M KI. (if 100% yield + if that is the only reaction - I know there are other possible reactions ; I2 + 2 KOH + H2O → 2 KI + 2 H2O + ½ O2; or other results: KIO;KIO2...)

So in the end, I will need a lot of Iodine to make KI....

Texium - 17-2-2024 at 06:23

You have iodine but you can’t buy KI? Normally, people make their iodine from KI or NaI, not the other way around.

RU_KLO - 19-2-2024 at 04:33

I extract Iodine from povidone-iodine tincture.

there are no other household material (here in Argentina) for getting iodine.

I asked for KI to chemical selling companies, they ask 150 usd (+ tax = usd 200 for 250gr KI) its to much for me. (they dont sell less than this...)

getting from Web (ebay, amazon, etc) is complicated because our customs is complicated, and importing chemicals by non chemical companies is rising red flags.

Texium - 19-2-2024 at 06:10

I have a lot of KI for sale at a better price: https://www.sciencemadness.org/whisper/viewthread.php?tid=16...

SplendidAcylation - 20-2-2024 at 06:39

RU_KLO raises an interesting point... I too would appreciate clarification on this topic.

Halogens, nitrogen, oxygen, and hydrogen exist as diatomic molecules, and they are always written as such, however, sulfur exists as S8, but we always write it simply as "S".

Of course, it doesn't matter in the slightest provided the stoichiometry adds up, but there must be a reason for this inconsistency?

For instance:

2Na + Cl2 --> 2NaCl

OR

Na + Cl --> NaCl

Both are balanced, however the latter would not be written because chlorine doesn't exist as a radical under normal conditions.


And yet, for sulfur, we would write:

Fe + S --> FeS

And not:

8Fe + S8 --> 8FeS


When you think about it, this is mighty strange.

Perhaps it is because sulfur can exist in many different allotropic forms, and even as a polymeric form, whereas the aforementioned diatomic molecules rarely exist in any other form?


Moreover, it does add some ambiguity when calculating stoichiometry when the formula is not explicitly written.

"Weigh out one mole of iodine" could mean one mole of iodine atoms, but usually it would refer to one mole of di-iodine.

On the other hand "weigh out one mole of sulfur" could mean one mole of S8, but more likely it would simply refer to one mole of S.


Am I just being dense or is this odd? :D

Texium - 20-2-2024 at 06:53

It is rather odd, but I think you pretty much summed it up here:
Quote: Originally posted by SplendidAcylation  
Perhaps it is because sulfur can exist in many different allotropic forms, and even as a polymeric form, whereas the aforementioned diatomic molecules rarely exist in any other form?
That, and because sometimes the diatomic nature of halogens is important to the way they react, such as in the substitution reaction example I showed above where one atom is oxidized and the other is reduced. For something like S8, it’s just not as relevant, and also makes the reaction look kind of ugly, having to multiply everything by 8.
Quote: Originally posted by SplendidAcylation  
Moreover, it does add some ambiguity when calculating stoichiometry when the formula is not explicitly written.

"Weigh out one mole of iodine" could mean one mole of iodine atoms, but usually it would refer to one mole of di-iodine.

On the other hand "weigh out one mole of sulfur" could mean one mole of S8, but more likely it would simply refer to one mole of S.
I’d just call that a poorly written procedure! :P