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# Tips for Writing Formulas of Ionic Compounds
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Here are some tips, and tricks for writing chemical formulas for ionic compounds. So the first thing you want to do is make sure you write out the formulas for the cations, which is the positive ions, and the anion which is the negative ion.

Then you want to make sure that the charges cancel out. That's because in ionic compounds, the total charge of your ions should equal 0. Should be neutral.

So say for example, if we're given that we have Sodium Phosphate. So you look at those ions. Sodium is Na+, Phosphate is PO4 3-. If you take a look, so this is +1 for Sodium, and this is -3 for Phosphate. So that would mean you would need 3 Sodiums to make it so it's Na3PO4.

Now here is the trick. If you take a look here, you can take the charge of the Phosphate, put that as a subscript on the Sodium. You can pick the charge of the Sodium, and put that as the subscript outside the Phosphate. Now there is only one Phosphate, so you don't need to write the 1 there, but there is a 3 there. So that's the way to get the charges to cancel out.

Now if you also have say Ammonium Sulfite. So you have NH4+, and then Sulfite is SO3 2-. So the 2 would become the subscript on the Ammonium. The 1 would become a subscript on the Sulfite. So you have (NH4)2 SO3. There's only one of that.

Now keep in mind if you have more than one element, you need parenthesis, except when you only have one. So like in the Sulfite we only have one of it, so we don't need the parenthesis. Then just like in the Phosphate, in the previous one, you don't need parenthesis. Single elements like Na, no parenthesis needed.

So just a couple more. So say if we have Strontium, and we have Arsenate. So Strontium is Sr2+, Arsenate is AsO4 3- charge. So we have Sr, then we take the charge of the Arsenate which is 3, and then we have Arsenate AsO4. Since the charge of this Strontium, that's 2, so we'll put that as a subscript. So in effect, you have 3 Strontium. So total, this would be +6, because it's 2 times 3 atoms.

Then here for the Arsenate, it would be -6, because basically it's -3 times 2. So we have +2 times 3. So +6 or -6. If you add those up, those equal 0, a neutral charge. So if you can do more practice, then you can actually take the charges, and then you don't have to worry about writing them out.

One more; Barium Oxide, Ba2+, O2-. So the charges are already equal, but if we use our method where we have Ba, and then we write a 2. Then you can have O and a 2. You would actually simplify, because the charges were already equal. So this is not the right answer. So it'd actually be just BaO.

Then the last one; Copper(II) nitride. So Copper(II), the Roman numeral, II, tells you the charge, so thus is Cu and then Nitride is -3. So we put a 3 there. Copper is +2 s we write N2. That would be your answer. So you'd have Cu3N2. So you have all your answers here.

So remember, using the short-cut; you take the charge of the other one and it swaps, and becomes a subscript of the other ion. So hopefully, if you remember that the total charge of these ions need to equals 0. Remember parenthesis, only around multiple elements, only if you have more than one of that particular ion. So hopefully these tips, and tricks help you in writing chemical formulas for ionic compounds. Have a good one.

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