6 Chemistry questions: Stoichiometry. (+10pts!)?

I will try to help, but I've forgotten a lot of my chem stuff so I may not be able to do all of it. Luckily, I just found a periodic table with my bio stuff. Ok, question 1. First, to find the weight in grams that a mole of an element is, it's the same as that elements atomic weight. So for NH3 it would be 14+3x1=17g/mole. So take 62.1/17=3.653, and that's how many moles you've got. For question 2, start out the same way. One mole of F2 is 38g (because the atomic weight of fluorine is 19, then times 2). 13.7/38=.361 moles. One mole is equal to 6.02x10^23 of anything - molecules in this case. So times .361 to 6.02x10^23 and you get 2.173x10^23, the number of molecules of F2 that you have. For question three, we shall again consult our lovely periodic table. % composition is based on atomic weight I believe, so we'll find the weight of these two elements, N and H. Nitrogen has a weight of 14 amu (atomic m units,,,that's the unit it's in on the periodic table right? regardless, these are the numbers). and hydrogen's is just 1. Now times by how many atoms of each element are present to get 28 for N and 4 for H, making a total of 32. Next get percents for each, by dividing the individual weight of the element by the total. 28/32 is 87.5% for nitrogen and 4/32 is 12.5% for hydrogen. Question 4 is basically question 3 in reverse - start with out percents and work backwards. (I'm rounding numbers a little as I go, so if your teacher requires you to be more precise you may have to crunch some numbers again yourself). The weight of silver is 107.87 amu, phosphorus is 30.97 and oxygen is 16. This would be easier with a given molar m...I've got an answer, but it was mainly guess and check, so I won't bother walking you through all the work. I just experimented a little with various numbers of moles of each element till I found ones that fit the equation. Not too hard to do if you're handy with a calculator. Here's the answer I got Ag3PO4 (make all the numbers subscripts). I'm fairly certain that works. Question 5, you just need to know that empirical formulas are the simplest possible, so it's a general formula that can fit many different compounds. I won't go into further detail there, if you don't understand that already ask your teacher or a clmate. The weight of C2H3O2 is 59. The weight your compound should be is 119 (these numbers don't fit exactly, but it's close enough you can tell. Something must have been rounded a little, though C,H, and O all have weights quite close to the simple numbers I used). Do 119/59 and you get approximately 2. So that means the molecular formula for your compound is twice the empirical formula given - just multiply everything by two, giving you C4H6O4. I just checked and you've already got an answer that went over all your questions, so I won't go over question 6. I'm going to post the rest of this anyways though, just since I took the time to type it out; maybe it'll help, and if not I can't see how it'll hurt