 Okay let's try a slightly harder one. What's the molar mass of aluminium hydroxide? As with the water example the first thing you need to do is write down the formula for the compound. Now aluminium has a valence of 3 plus, it's in group 3. The hydroxide ion has a charge of minus 1. So when we put those together that gives us the formula of AL brackets OH3. Now to work out its molar mass you can see I'm going to need the molar masses of aluminium, oxygen and hydrogen. So I'm going to write those down first. Now I have oxygen and hydrogen already, 15.999 and 1.008. Let me add aluminium to that, 26.981. Alright now I can put them together. Now one unit of aluminium hydroxide, remember I can't say a molecule because this is an ionic substance, contains 1 aluminium ion, 3 oxygen atoms and 3 hydrogen atoms. So to work out the molar mass of aluminium hydroxide I need to put together the mass of the aluminium plus 3 times the mass of the oxygen plus 3 times the mass of the hydrogen. And that gives me a final molar mass of 78.002 grams per mole. Alright now let's try a problem that involves us finding out how many moles of aluminium hydroxide are in a certain mass. I'm going to set this out in two different ways. One is just to use the formula I gave you before. The other will use the kind of format that I've also used in the videos on unit conversions. Now here's the problem. If I have 23.4 grams of aluminium hydroxide, how many moles of aluminium hydroxide is this? So first of all write down what you know. You know that the mass is 23.4 grams. You also know the molar mass of aluminium hydroxide because we've just calculated it. You might come at a problem like this cold and not know the molar mass straight away but it's always something that you can calculate as long as you've got a periodic table handy. So I have the mass, I have the molar mass and what I want to know is the moles. Now we know from before that the molar mass equals the mass over the moles. I can rearrange that to make moles the subject. I multiply both sides by moles and divide both sides by molar mass. And that gives me that the moles equals the mass over the molar mass. Then all I have to do is substitute in the values that I know and get the answer. And because my mass had three significant figures I'm going to round this off to three sig figs. Now I'm going to do the same problem but I'm going to set it out in a slightly different way. Again I need to write down what I know. So I have the mass, I have the molar mass and I want to know the moles. But I'm going to set it out a bit like a unit conversion. So first of all I will put down what I know as a fraction. 23.4 grams over one. And the conversion factor is my molar mass in grams per mole. But I have a mass that's in grams and I want to cancel out those grams and turn it into moles. So I'm going to flip my molar mass upside down. The 78.002 grams will go on the bottom and the one mole will go on the top. Just take a minute to look at this if this is confusing you. All I've done is take a conversion factor that looks like that and flipped it on its head. I've inverted it like that in exactly the same way as you invert your conversion factors when you're doing unit conversions. I can now cancel out the grams and I'll be left with moles when I do this conversion. So I multiply out the top and bottom of the fraction. I do the calculation and I come out with exactly the same answer as before. Now which of these two formats you use is up to you. My advice is if you have a tendency to mix up algebraic conversions and are a bit sloppy with your units use the second format, the one that's like unit conversions. It will keep you on the straight and narrow. Ideally though you would practice rearranging equations until you felt confident with it because we will use bigger and more complicated formulae later on in the course that will definitely need rearranging. However as you see both formats work perfectly.