 Okay, the next thing you need to be able to do is to write the formula from the name. This is slightly trickier but as long as you learn the rules and practice a bit you won't find it too difficult. The steps are these. First you need to identify the ions that are involved. Then you need to write out the ion formulae including the charges on the ions. So let's start with an example, we'll start with magnesium bromide. Again, you might find it handy to have a periodic table on hand so that you can work out the charges on your ions. So the two ions that are involved here are magnesium and bromide. So first we work out the magnesium ion, we'll have the symbol MG, you'll find it's in group number two and it's a metal so it will be two plus. Bromide that comes from bromine has the symbol BR, you'll find it's in group number seven which means it gains one electron and becomes BR minus. Okay, so that's step number two. Step number three is we need to work out how many of each ion is needed for us to have a neutral compound. Remember overall that the ionic compound must have a charge of zero. Working out how many of each ion is needed can be done by trial and error. The idea is to find a common multiple, you're looking to multiply the positive and negative charges by a factor so that the total positive and the total negative charges are the same. So for instance here we will go we have a total positive charge of two plus and we have a total negative charge of one minus, clearly we need more negatives. So we're going to multiply the bromides by two, we need two of those, which means that our final formula must be one magnesium for two bromides. However, there is a neat shortcut that I'm going to show you. For simple ones like this you can probably do it just by inspection but the crossover method does get useful for slightly larger charges. So it works like this, you write out your two ions and you take the charge that's on the metal and you transfer that number down as the subscript for the anion and you take the charge of the anion and you make that the subscript of the cation. So what that means is that for us the magnesium, the charge on the bromide was one minus so we just have one magnesium and if there's just one of them you don't bother writing a one and for the bromide the magnesium had a charge of two plus so we take the two down to the bromide and that means we need two bromides. So you can see it gives us the same answer, this crossover method is good as a shortcut but it doesn't really explain to you why it works. Keep in mind that what you're trying to do is balance up the charges so that the total negative is equal to the total positive. Let's try another example, let's try and write the formula for calcium nitrate. First we work out the ion formulae, calcium is the element calcium it's in group two so it's going to have a charge of two plus. Nitrate is what now don't get this confused with nitride which is the nitrogen monatomic ion. Nitrate with the Ate that indicates that it's one of your polyatomic ions you should remember that its formula is NO3 with a charge of one minus. Okay if we do the crossover method for this the two is going to come down near the nitrate the one is going to come down near the calcium which is going to give us a formula of one calcium and two nitrates. Now we can't just write NO32 like this, this makes no sense at all. What we need to indicate is that we need two whole nitrate ions and the way we do that is to put them in brackets and put a little two outside which means two lots of everything that's inside the bracket. So that's our formula for calcium nitrate. Checking that it works by our sort of total charges method we've got one calcium that's a total positive charge of plus two, we've got two nitrates each of those is minus one so two times minus one equals minus two and so we've got plus two and minus two and it cancels out nicely. One final example, aluminium oxide. So aluminium is one of our ions it's in group three so it has a charge of plus three and oxygen is in group six that means it has to gain two electrons to get a full outer shell which means that the oxide ion has a charge of minus two. If we do the crossover method we simply take the three down to the oxygen and the two down to the aluminium and that gives us a formula of Al2O3. To check that that works in terms of charges we've got two aluminium's and each of them is three plus which equals a total positive charge of plus six and we have three oxygens and each of those is minus two. Three times minus two is minus six so plus six and minus six cancel out and we have our neutral compound.