hehe, it was
me. This just took a long time to write, so I disappeared from the online list :P
ooooooo, fun stuff.
First of all, what do you mean by saying having things in units "u"? Is that just "units" or something?
EDIT: oh, I think it's AMUs, atomic mass units.
If I use dashes (--) in molecular formulae, it's just for clarity because I can't use subscript. (HTML is off :/)
Molecules - For starters, it's a good know that all the masses on the periodic table are in grams per mole. For instance, one mole of Carbon weighs about 12 g/mol. Also, 6.02 x 10ēģ is avagadro's constant, the number of molecules in a mole.
Look at your periodic table. You should see that Carbon has a mass of about 12 g/mol, and Oxygen has a mass of about 16 g/mol. One carbon, and two oxygen = 12 + 16 + 16 = 44 grams/mole. If you have one mole, then you have 44 grams. So 6.02 x 10ēģ molecules of carbon has a mass of 44 grams.
I don't know what the problem means by saying "a mass of 44 u". I guess it means that one mole of that substance would be or something.
I'll explain empirical formulas here; if you already know what they are, skip this paragraph.
The Empirical formula is like the smallest ingredient in the molecule. Suppose for instance, that we had an element Sock. Sock is a dimolecular element, so we only find it naturally as Sock2. After all, whoever heard of a single sock? (Besides a dryer :rofl: ) Even though the molecule is Sock2, the empirical formula is just Sock, because Sock2 is a multiple of Sock. The empirical formula is the simplest form of the molecule.
For example, glucose, C6-H12-06 has an empirical formula of CH20. That's the basic formula. Put in 6 of those, (CH20)6, and you get glucose.
First of all, you want to assume you have 100g of the compound. That is, you have 25.9 grams of Nitrogen, and 74.1 grams of Oxygen. If you want to compare the two, you have to have them in terms of moles because they weigh different amounts. Lets go through some stoichiometry.
25.9 grams of Nitrogen / (14.01 grams of Nitrogen/mol) = 1.85 mol nitrogen
74.1 grams of Oxygen / (16.00 grams of Oxygen/mol) = 4.63 mol oxygen
You get about 1.85 mol N, and 4.63 mol O. You can now compare the number of mols of the elements together, starting with the smallest.
There are only two here, so this will be easy. You have 4.63 mol O / 1.85 mol N which is about 2.5. So for every 1 Oxygen in the compound, there are about 2.5 Nitrogens. Multiply by two to get an even number, and you get 2 Oxygens for every 5 Nitrogens. Thus the empirical formula is O2N5. The ratio is 2 to 5.
This leads up to the next problem because, while the empirical formula can tell you what the compound is made up of, it isn't the actual formula. If you know the molecular mass, you can get the formula :)
This confused the hell out of me for a minute. I looked it up and found that the empirical formula is not CCINO, it is CClNO, that is Cl or clorine, not CI, or carbon iodide ;)
Ok, for starters, we have CClNO, which has a mass of 12.01 + 35.45 + 14.01 + 16.00 = 77.47. The only thing you have to do, is take the mass of the whole molecule and divide by the mass of the empirical formula. 232.42 / 77.47 is very close to 3. This means that you can put in 3 of the empirical formula. Put in three of each Carbon, Clorine, Nitrogen, and Oxygen to make the formula for Symclosene C3-Cl3-N3-O3, which is what it says on this page
. If you check the mass of that, it should be 232.42.
I hope that helped, and arrived in time for you to use it. Good luck with Chem. :arr: