## Moles and Molarity HW Problem E.29b

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### Moles and Molarity HW Problem E.29b

The homework question for E29 b says "How many moles of Cl- ions are present in the sample?"
The sample is 8.61g of CuCl2 x 4H20 , the moles of this sample is 0.0417.
The answer to this problem is .0834 mol Cl-
Why do we have to multiply 0.0417 mols twice? Doesn't 0.0417 represent the entire entity of the molecule and not just Cl-?

Posts: 36
Joined: Mon Jun 17, 2019 7:23 am

### Re: Moles and Molarity HW Problem E.29b

We have to multiply 0.0417 twice to get mol Cl, because there are two mol Cl per mole of the compound (CuCl2(2H2O)):

mol Cl = 0.0417 mol CuCl2(2H2O) x (2 mol Cl)/1 mol CuCl2(H2O)) = 0.0834

Using molar ratio, we see that mole CuCl2(2H2O) cancels and we are left with mole Cl

Chris Charton 1B
Posts: 69
Joined: Mon Jun 17, 2019 7:23 am

### Re: Moles and Molarity HW Problem E.29b

In each molecule of CuCl2 there will be two ions of Cl, hence you need to double the amount of moles found of CuCl2 to find the number of moles of Cl- ions.

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