 Why are moles so important for chemists? Well, imagine you're one yourself and you work in a chemical plant. One of your customers has ordered 100kg of copper sulfate crystals. To deliver this, you need to dissolve copper oxide in hot sulfuric acid. This will produce copper sulfate solution from which we can get the blue, hydrated copper sulfate crystals that you might be familiar with. Here's the problem though. How do you know what mass of copper oxide to place in your tub of acid? If you use too much, not all of the reactant will dissolve and you'll have to spend time and effort separating the undissolved solid from the solution. Too little and you simply won't produce enough product for your customer and fall behind schedule. Either way, you could be in trouble. But with the help of moles and the simple equation, problems like this can be easily solved. The first thing to do with any challenge like this is to look at the chemical equation and balance it. Fortunately that's not difficult here because as you can see, all we need to do is change the molar ratio of the water from 1 to 4. The molar ratios for all the other species are 1. You now know that every one mole of copper oxide reacts to form one mole of copper sulfate. 100 kilograms or 100,000 grams of the copper sulfate are required. We need to know how many moles of the product this represents and here's where the equation comes in. The number of moles needed equals the desired mass, 100,000 grams, divided by the relative molecular mass of the product we want, copper sulfate. Using your periodic table, you can work out that this is 63.5 for the copper atom, plus 32 for the sulfur, 9 times 16 for the oxygens, and 10 times 1 for the hydrogens, altogether making 249.5 grams per mole. 100,000 divided by 249.5 equals 400.8 moles. So this is the number of moles of copper oxide we need to add to the acid. Now to convert back to mass, and again this is straightforward, we just have to use the equation again. This time we know the number of moles, 400.8, and we can get the molecular mass of copper oxide, 63.5 plus 16 equals 79.5 grams per mole. To find the mass only a simple rearrangement is needed. 400.8 times 79.5 equals 31,860 grams of copper oxide, and there you have it, we've solved the problem. Let's look at another example, a famous one. Here's the harbour process for making ammonia from hydrogen and nitrogen. If you want to know more, we've a video all about this. Imagine one day at the chemical plant we produced 4,000 moles of ammonia with an excess of hydrogen gas. Whenever you see a reactant being described as in excess, that means there is more of it in the reaction vessel that is needed. So it isn't this chemical that's going to limit the amount of product formed, and you should focus on the other reactant, the nitrogen in this case. This is often given a special name, the yield limiting reagent. How many moles of nitrogen gas did we manage to make react? The first part of the challenge is to balance the equation. Pause the video and resume once you've had a go. Well, the answer is for every two moles of ammonia, we need one mole of nitrogen gas and three moles of hydrogen. If you worked it out as one mole of ammonia needs half a mole of hydrogen and three over two moles of nitrogen, that's quite right too. Can you work out how many moles of nitrogen gas we used to make the 4,000 moles of ammonia? Pause again and resume when you have the answer. Well, you simply have to divide the 4,000 by the molar ratio which is 2, so the answer is 2,000 moles of nitrogen. Did you get that one?