posted by Anonymous on .
I posted yesterday. Re: Chemistry (Sunday, October 4, 2009 at 12:49pm). Here's the information I left out.
Number 3 simply states: nickel forms a large number of ammonia salts Ni^2+(NH3)m(Br)n . p(H2O) where m, n, and p represent the coefficients in the formula
How do you know how many molecules ammonia has? (why did you say 2? –are you doing a trial and error?) How do you know to subtract Ni and 2NH3 and divide by the molar mass of H2O? Is there some kind of formula? (I tried searching online, but haven’t had any luck) How would I mention Bromine in the molecular formula? Like this: Ni^2+(NH3)2(Br)0 (H2O)4 ? Sorry for asking so much questions. Thank you so much for the help!
No apologies necessary. I tried to address what I knew was coming in my original response. First, about the bromine. I have no idea what kind of a hint that is since your post had NO bromine anywhere attached to anything. The only place I saw it mentioned was in the hint and it has absolutely no connection with what you posted.
With regard to the 2 in the NH3. My assumption, since Br was mentioned AND it was a multipart question, that you had failed to type in the earlier parts of the question and critical information was not listed. So I went with the last part of the question that said, draw (or design or determine or something like that) a structure that is compatible with the data. So I did that. I know that most of the coordination compounds of Ni complexes is 6, I stuck in 2 for NH3 and that made me multiply the equivalent weight by 2 to get 2 x 82.44 = 164.88. The rest of it is just reasoning.
If the molecule is 164.88, we subtract 1 Ni = 58.69 to leave 106.19, then we subtract 2 NH3 @ 17.03 each to leave 72.13, then we divide that by the molar mass of H2O (18.015) to make a stab at the number of water molecules which gives us 4.004. Good, it came out a whole number which can't be by chance. I round to 4.00 (pretty close to four places, eh?) and the formula becomes [Ni(NH3)(H2O)4]^2+. Then I look at the agreement with the molar mass and I say it is compatible with the data. If you have any clue as to the Br comment I would appreciate hearing about it. And if there was ANY other information in the problem, perhaps I can make something of it. (As an addendum, you should understand that when one works with equivalent weights, there is NO WAY to KNOW what the molecular weight is. The equivalent weight is done easily with mL x N x mew = grams and multiply the mew by 1000. But additional information must be presented (or in the lab it must be determined) to know the difference between the equivalent weight and the molecular weight. This problem didn't have that; only a statement to make the formula compatible with the data.)