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### E.23 Part A and C

Posted: Sun Oct 25, 2020 4:51 pm
The question asked to calculate the moles of:

a) Ca2+ ions in 3.00 g of CaBr2

c) F- ions in 25.2 kg of UF6

I seem to be getting the calculations involving ions wrong. Is there something I possible could be not considering and could someone walk through these questions?
I'm finding the molar mass of the molecule then the percentage of Ca or F, then multiplying the percentage by the given mass, then dividing the resulting mass by the molar mass of the individual atom to find the moles.

### Re: E.23 Part A and C

Posted: Sun Oct 25, 2020 5:10 pm
This isn't a mass percent problem, it only consists of dimensional analysis. So, for part A the process of solving the question would be:

3.00g CaBr2 x (1 mole CaBr2 / 199.197g) x (1 mole Ca2+ / 1 mole CaBr2)

You would repeat the same steps for part C, just make sure to convert kg to g first. Hopefully this helps :)

### Re: E.23 Part A and C

Posted: Sun Oct 25, 2020 5:17 pm
I'm pretty sure part A is CuBr2, but either way Megan's steps are correct. If you follow the same steps for part c, (make sure to convert kg of UF6 to g), you should get 430. mol F^- as your answer

### Re: E.23 Part A and C

Posted: Sun Oct 25, 2020 5:25 pm
You don't need to find the mass percentage. You're right that you find the molar mass of the molecules, but afterwards just use molar ratios. Divide the given masses (make sure they're in grams) by molar mass to get the moles of the molecule, then use the molar ratio between the atom and the molecule to find the atom's moles.

### Re: E.23 Part A and C

Posted: Sun Oct 25, 2020 8:12 pm
Megan Chan 2I wrote:This isn't a mass percent problem, it only consists of dimensional analysis. So, for part A the process of solving the question would be:

3.00g CaBr2 x (1 mole CaBr2 / 199.197g) x (1 mole Ca2+ / 1 mole CaBr2)

You would repeat the same steps for part C, just make sure to convert kg to g first. Hopefully this helps :)

Got it thanks! One more small thing is how did you get 199.197g? When I'm putting in numbers from the periodic table I don't end up with that.