Simple guide (IMO) to working out ions:
Main group (s, p block), look at the group and you should be able to work out how many electrons to remove or add to form a stable octet. For instance, magnesium is 1s2 2s2 2p6 3s2 (group 2), so it tries to lose two electrons. Similarly, fluorine is 1s2 2s2 2p5 (group 17), so it tries to gain an electron. Use the periodic table!
For polyatomic ions, it may be helpful to find other ways of knowing these ions. Ethanoate ion, for instance, is the conjugate base of ethanoic acid, which is CH3COOH (you'll need to know this for 1/2 anyway), so the ethanoate ion is CH3COO-. Phosphate has an 'ate' ending, indicating that the number of oxygens bonded to the phosphorus is a maximum of 4 (similar to sulfate, although permanganate and perhalates like perchlorate follow a slightly different custom as halogens can bond with from zero to four oxygens), so we have PO4 with some charge. Phosphorus has 5 valence electrons. If you think of phosphorus as losing all these electrons and bonding to four oxide ions (not quite true but it's a way of remembering it), you'll see that phosphate has a charge of -3. Similarly, sulfate has 4 oxygens as well and sulfur has 6 valence electrons, so thinking of sulfur losing all these electrons to form oxide ions and we have a charge of -2 (this is really an introduction to oxidation numbers hahaha). Thus, sulfate is SO42-.
Nitrogen can only bond to three oxygens, so nitrate consists of a nitrogen bonded to three oxygens. Five valence electrons, three oxides => -1 charge, so nitrate is NO3-.
Nitrite is a funny thing because it has the same charge as nitrate but one less oxygen (just remember that). Same with sulfite (SO32-).
If you have something like hydrogen sulfate, think adding a sulfate anion with a positive proton. Add a hydrogen and increase the charge by 1 to get HSO4-.
Transition metals (d block) can have differing charges, so the charge is always indicated by a letter. Cu(II) would then be a copper cation after losing two electrons.
I hope this made some sense.