Method 3 involves the number of ions from molarity. Now we're going to say here problems involving number of ions and molarity can use a given amount and converting factors to isolate an end amount. Now for those of you who haven't watched my videos on dimensional analysis and conversion factors, these terms basically form the foundation for a lot of the calculations you'll see in chemistry. Your conversion factor is basically just when you have 2 different units connected together. That is seen in molarity because molarity itself represents moles per liter, two units connected together. Our given amount is just a value that's given to us that has only 1 unit; that would be represented by these liters here. And then our end amount is just simply the value or the amount that we're trying to find at the end.
Now if we look at this question, remember when we have the word 'of' in between numbers, it means that we have to multiply them together. So, we're going to say here our given amount where we're starting is 0.120 liters and we're gonna have to utilize conversion factors to get to our end amount. Our end amount we're looking for is moles of calcium ions. Now we need to cancel out liters here. We're able to do that by utilizing the molarity given to us. The molarity is 0.450 molar, which means 0.450 moles of calcium phosphate per 1 liter of solution. So that's our first conversion factor. Then liters cancel out.
Now we have to just change moles of calcium phosphate into moles of just calcium ion. So here for every one mole of our entire compound, we can see that from the formula there's a little 3 there, which means that I have 3 moles of calcium ions. So then if we multiply everything out that's on top, I'll be left with 0.162 moles of calcium ion as my final answer.