in this video we're going to talk about how to write the chemical formula of ionic compounds so let's say if we have lithium chloride how can we write the formula of this ionic compound the first thing you want to do is write the charges of each ion lithium has a plus one charge all of the alkali metals in group one have a positive one charge as an ion chloride has a minus one charge now whenever the charges are the same you could simply write the elements together so therefore lithium chloride is simply li cl you don't need any subscripts if the charges are the same so let me give you another example of that calcium sulfide calcium is an alkaline earth metal found in group 2a of the periodic table so therefore it has a positive 2 charge sulfide has a -2 charge because the charges are the same you can simply write them together with no subscripts so it's simply called cas calcium sulfide now let's look at another similar example aluminum nitride aluminum has a positive three charge nitride has a negative three charge once again the charges are the same so you could simply write it as a l n aluminum nitride let's work on this one magnesium bromide magnesium like calcium is an alkaline earth metal found in group two and it has a plus two charge bromide is a halogen that has a minus one charge now notice that this time the charges are not the same they're different so in this situation you want to exchange the charges with subscripts and you got to reverse them so the plus two which is the charge on magnesium will become the subscript for bromine and the minus one which is the charge on bromine will become the subscript of mg so it's going to be mg1 br2 the negative sign does not carry over now if you have a one you don't need to write it so you could ignore the one so you could simplify and say this is mg br2 so that's what you need to do whenever the charges are not the same so let's look at another example sodium sulfide sodium has a positive one charge sulfide has a minus two charge so using the same method it's going to be na2 s1 but we don't have to write them once so it's going to be simply na2s try these two problems aluminum fluoride and also calcium phosphide aluminum is a metal found in group 3a so it has a positive 3 charge fluoride is the halogen for minus one charge so using the same method it's going to be al1 f3 or simply al f3 so once you work on a few examples and get the hang of it this process becomes a piece of cake calcium has a plus two charge phosphorus as an ion has a minus three charge so let's switch the charges with subscripts so it's going to be ca3p2 here's another two examples that you can work on with polyatomic ions so write the chemical formula of aluminum sulfate and also strontium phosphate aluminum as we mentioned before has a positive three charge sulfate is so4 with a negative two charge so let's use the same crisscross methods it's going to be al2 so4 3. now if you have multiple polyatomic ions you need to enclose it within a parenthesis so that's how you write the chemical formula of aluminum sulfate now strontium is strontium's sr it's an alkali earth metal with a positive two charge phosphate is po4 with a negative three charge so this is going to be called sr3 po4 times two so that's the chemical formula of strontium phosphate now let's incorporate some examples with transition metals along with roman numerals so write the chemical formula of iron ii chloride and also iron iii nitrate the chemical symbol of iron is fe the roman numerals tells you the charge on fe so the charge is positive two and chloride has a minus one charge and then use the same method so this is going to be fe1cl2 or simply fecl2 as you can see this is not difficult the next one rn3 nitrate this time fe has a positive three charge based on a roman numeral nitrate is a polyatomic ion with the formula no3 minus one so this is going to be fe1 no3 times 3 due to the positive 3 charge so we could simply write it as fe no3 3. so what do you think is this doable or is it like too difficult here's some more examples that you can work on copper one phosphate and also copper two phosphide so in the first example copper has a plus one charge phosphate is po4 with a negative three charge so this is going to be cu3 po4 times 1. if it's just a 1 as a subscript we really don't need to write the 1 nor do we need to write the parentheses so the answer is simply cu3 po4 so that's copper one phosphate the next one copper has a plus two charge phosphide has a minus three charge so this is going to be cu3 p2 and we can't modify the answer so this is it let's work on some harder examples 10 4 selenide and lead for per bromate so go ahead and work on those examples so now first example 10 has a plus 4 charge selenide which is associated with selenium has a minus two charge so initially this is going to be sn2 se4 now if you can reduce the numbers you should because they're both even you could divide both numbers by two so therefore the correct formula is going to be s n one which we don't have to write the one s e two so this is the answer the next one lead has a plot excuse me a positive 4 charge and per bromate is bro4 with a negative 1 charge so this is going to be pb1 bro4 times 4 or simply just pb bro4 4. go ahead and work on this one vanadium five oxide so in this example vanadium has a positive 5 charge oxygen has a negative 2 charge so using this method is going to be v2o5 that's vanadium 5 oxide you