E.23

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Isabel Bellon 4F
Posts: 30
Joined: Fri Sep 28, 2018 12:27 am

E.23

Postby Isabel Bellon 4F » Tue Oct 09, 2018 10:09 pm

I am not sure how to go about any parts of this problem, specifically how to relate the overall amount to the specific ion/atoms in moles. Any help would be appreciated!

705152867
Posts: 28
Joined: Fri Sep 28, 2018 12:18 am

Re: E.23

Postby 705152867 » Tue Oct 09, 2018 10:40 pm

You want to convert the grams given of each compound into the moles of each element/compound it asks you to calculate for. So, for example, in part a you're given 3.00g CuBr2; find its molar mass. Then to convert to moles of of Cu^2+ you have to know the mole-to-mole ratio of Cu^2+ and CuBr2. For every one mole of CuBr2, there is only one mole of Cu^2+. Multiply (3.00g CuBr2/molar mass CuBr2)*(1mol Cu^2+/1mol CuBr2). The answer is the product :)

Alicia Gibbons 1B
Posts: 31
Joined: Fri Sep 28, 2018 12:17 am

Re: E.23

Postby Alicia Gibbons 1B » Tue Oct 09, 2018 10:41 pm

I'll help you through part a!

Calculate the amount in moles of Cu2+ ions in 3.00g CuBr2.

First, we must go from grams of the molecule to moles of the molecule by dividing by the molar mass.
3.00 g CuBr2 x (1 mol CuBr2/223.35g/mol CuBr2)= 1.34 x 10^2 mol CuBr2
Now you have the moles of the molecule, but you have to find the moles of Cu2+ ions in the molecule.
Cu2+ has a subscript of 1, and because of that, there is 1 Cu2+ ion for every one molecule of CUBr2.
Thus, compare the ratio:
1.34 x 10^2 mol CuBr2 x (1 mol Cu2+/1 mol CuBr2)= 1.34 x 10^2 mol Cu2+

Alternatively, if you were solving for moles of Br-1 ion, you would have a different amount since Br2 is in a 2:1 ratio with the CuBr2 molecule.
mol Br-: 1.34 x 10^2 mol CuBr2 x (2 mol Br-/1 mol CuBr2) = 2.68 x 10^2 mol Br-


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