E 23 Part C sixth edition

Moderators: Chem_Mod, Chem_Admin

Chloe Orsini 1K
Posts: 32
Joined: Fri Sep 28, 2018 12:18 am

E 23 Part C sixth edition

Postby Chloe Orsini 1K » Mon Oct 01, 2018 9:16 pm

In this problem it states:
F- ions in 25.2 kg of UF6
I started by converting 25.2 kgs into grams and then calculating the molar mass as 352.017 g*mol^-1.
I then divided 25200 by 352.017 and got 71.587
The answer multiplies this answer by 6 moles of F-1 to get 430 moles
Is this because it is F6 in the equation? Had we been solving for the Uranium ion instead would we have not multiplied it (because of the implied 1)?

jonathanjchang2E
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

Re: E 23 Part C sixth edition

Postby jonathanjchang2E » Mon Oct 01, 2018 9:31 pm

Yeah, since the question asks for the amount of F^- ions and there are 6 moles of F^- ions for every one mole of UF6 we would have to multiply the amount (in moles) of UF6 by 6. Also, yes for your second question.

hazelyang2E
Posts: 57
Joined: Fri Sep 28, 2018 12:23 am

Re: E 23 Part C sixth edition

Postby hazelyang2E » Mon Oct 01, 2018 9:40 pm

Yes, your inference is correct because when you are calculating the amount (in moles) of an element within a compound, you need to know the ratio of that specific element within your compound. So the calculation they used was:

71.587 mol UF6 x (6 mol F)/(1 mol UF6) = 429.48 mol F = 430. mol F

So, if we were asked to calculate the number of Uranium moles, you would technically multiply the number of moles (71.587) by one because the ratio of Uranium moles in UF6 is 1:1.


Return to “Accuracy, Precision, Mole, Other Definitions”

Who is online

Users browsing this forum: No registered users and 1 guest