How to Write Ionic Compound Formulas: Step-by-Step Guide

Now let's try to write some formulas of ionic compounds. Now we know when you see ionic compounds that means we have

to consider a metal and non-metal. And also you should know that metals lose electrons and become cations. Nonmetals

gain electrons and become annions. Let's try to see magnesium and oxygen. Magnesium is a metal. Oxygen is a

nonmetal. It should be ionic. What type of ions they make? Magnesium always make two positive because it is in 2A. Oxygen

always make two negative because it's in 6 A. It will take two electrons and become negative two charge ion. There

are certain steps. Let's go over the steps. How to write a formula of ionic compound. First step is write the symbol

of the metal and its charge. We just found out symbol of the metal is mg and it charge is 2 plus followed by the

symbol of the nonmetal and its charge. Nonmetal is oxygen. It charge is -2. Let's go to the second step. Make the

magnitude of the charge of each ion. So magnitude of charge of each ion mean the magnitude of magnesium is two two

positive the magnitude is two. Similarly oxygen also in oxide ion magnitude of charge is two. Let's see what they say.

Make the magnitude of the charge of each ion without the sign. Yeah magnitude is not the sign become the subscript of the

other ion. What does that mean? is asking us to make the magnitude of the charge to become the subscript of the

other ion. You know subscript is the this is a subscript that is a subscript. Let's try to do what it has to do. Make

the magnitude of the charge the subscript of the other. Basically mg2 this two please understand this two is

coming from here. That's why it says make the magnity of the charge of each ion become the subscript of the other

ion. So magnesium mg2 because it's a substrate of the other ion. Similarly oxygen I will that two here the two in

oxygen means actually this one to be interchanged. Let's go to the next step. If possible if possible reduce the

subscript we have subset now two and two reduce the substitute give the ratio with the smallest whole number. So it's

asking look at this subscript and see if we can give smallest whole numbers. We have two and two ratios. We can make it

even smaller. We can make it 1 and one. Still small and whole ratio go because I divide both sides by two. The next one

is to check to make sure the sum of charges of the cations exactly cancels the sum of charges of the annions. I

just need to emphasize that when you are writing a formula unit in a Cl or any ionic compounds, this is the smallest

and neutral. There cannot be a charge. It's asking check to make sure sum of the charges of kions exactly cancels the

sum of the charges of annions. Let's see magnesium is positive2 oxygen is -2. There's one of each. Therefore, it

cancels out each other. Therefore, this MGO should be correct. These are the steps and that's how you write a

formula. Let's try to do one more. Write the formulas with ionic compounds that form from aluminum and oxygen. Magnesium

oxygen. We just did aluminum and oxygen. We have the steps. Write the symbol for metal and it charge. Metal is aluminum

L3+ followed by the nonmetal and it charge. Nonmetal is oxygen and it charge is -2

O2 minus. Make the magnitude of the charge of each ion becomes subscript. That means I'm going to exchange this

magnitude and then I will get Al23. This two is coming from oxygen. This three is coming from aluminum. The third

step is if possible reduce the subscript to a smallest whole numbers. We this is the smallest we can have two and three.

There's no way we can simplify this without having fractions. But we cannot have fractions. We have to have whole

numbers. Finally check to make sure the sum of the charges of cations exactly cancels the sum of the charges of

annions. Here we have two aluminum 2 * each aluminum is positive3 + 3 and other one is 3 * because there are three

oxygen that's that's what this subscript shows three times each oxygen is -2 this one is pos6 this one is -6 that cancels

out therefore Al23 is correct what are the formulas for compounds made out of following ions

potassium with nitride potassium ium is K plus K plus nitride is N minus these are the steps I will go very fast we

exchange them first then K3 is from this nitrogen and N1 but in chemistry we don't write one here similarly we did

not write K + one similarly we don't write here this is the chemical formula K3N and if you see we cannot simplify 3

and 1 and you can also check if canceling out because we have three potassium atoms each one is positive one

is going to be three positive and And this one is -3. Positive and negative cancels out. That is correct. K3 N.

Third one is calcium ion with bromide ion. Calcium is this one. Ca2 plus and bro mean bromide ion is this one. Br

minus. Let's interchange that then you will get C A this Br has negative one. We do not write one here. We do not

write one. And I will write B R 2 two from this calcium ch. Then we cannot simplify this one and two ratios. These

1 and two ratios we cannot simplify. And finally let's try to see the charges are zero. One calcium that is positive2 from

here we can see and there are two broines. Each one is -1 from here -1 positive2 and -2 they cancels out each

other. That is correct. I'm going to erase everything and C A B are 2. Finally aluminum and sulfide. Aluminum

is 3 plus a L3+ sulfide S2 minus we'll exchange L2 S3

and if you just see if it is going to cancel out. All right.