Calculate the amount (in moles) of (a) Cu^2+ ions in 3.00 g of CuBr2
The first time I solved this question, I moved on because my response matched. However, now that I am reviewing the questions I am rethinking why is the answer 0.0134 mol of Cu^-2 and not 0.0134/2 mol of Cu^-2.?
3.00 CuBr2/223.35 g CuBr2mol^-1 = 0.0134 mol CuBr2 and since the molar ratio is 1 Cu:2 Br shouldn't we divide the value by 2?
E23 Confused
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 58
- Joined: Wed Sep 30, 2020 9:44 pm
- Been upvoted: 1 time
-
- Posts: 110
- Joined: Wed Sep 30, 2020 9:54 pm
Re: E23 Confused
Hi Ruth!
The you are given the grams of CuBr2, which you can then convert to mols of CuBr2 using the molar mass. Because molar mass corresponds to the amount of grams in one mole of CuBr2, you can simply multiply the mols of CuBr2 by the molar ratio of 1 mol Cu : 1 mol CuBr2 to find mols of Cu^2+ ions.
I hope this helped. Let me know if you have any other questions.
The you are given the grams of CuBr2, which you can then convert to mols of CuBr2 using the molar mass. Because molar mass corresponds to the amount of grams in one mole of CuBr2, you can simply multiply the mols of CuBr2 by the molar ratio of 1 mol Cu : 1 mol CuBr2 to find mols of Cu^2+ ions.
I hope this helped. Let me know if you have any other questions.
-
- Posts: 116
- Joined: Wed Sep 30, 2020 9:39 pm
Re: E23 Confused
The ratio is actually 1:1 between moles of Cu ions and CuBR2 so you don’t have to divide the value by 2.
-
- Posts: 116
- Joined: Wed Sep 30, 2020 10:07 pm
Re: E23 Confused
We only look at the ratios between the same element when looking at the ratio of Cu and CuBr2. So there is 1 Cu : 1 Cu
-
- Posts: 95
- Joined: Wed Sep 30, 2020 9:49 pm
Re: E23 Confused
Because there is one atom of Cu in CuBr2, for every mole of CuBr2 you have one mole of Cu. I think you might be looking at the Br2 and thinkcing that's why you divide by however the fact that there are 2 Br atoms does not change the ratio between Cu and CuBr2.
Return to “Accuracy, Precision, Mole, Other Definitions”
Who is online
Users browsing this forum: No registered users and 6 guests