In this lecture, I shall discuss the manipulation of quantities, in chemical reactions. You will want to have a periodic table and a calculator handy, in order to complete the end video questions. In this case, we're looking at a chemical reaction, converting the grams of one substance into moles of that substance, then using the mole ratios to convert between substances, and finally converting back to mass. [NOISE] Here's an example, let's start with a neutralization. How many grams of sodium hydroxide, are needed to completely react with exactly 100 grams of sulfuric acid? The first thing you need to do to answer this question, is write the balanced chemical equation. Go ahead and try to select the correct chemical equation from this list. [SOUND] Thank you for making your selection. In the reaction of sulfuric acid with sodium hydroxide, 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide, to create 1 mole of sodium sulfate, and 2 moles of water. [NOISE] The question asks us, how many grams of sodium hydroxide we need to completely react with 100 grams of sulfuric acid. Therefore, we need to convert the grams that we're given into moles, because we know the mole ratio. We know that 1 moles of sulfuric acid reacts with 2 moles, of sodium hydroxide. Let's start by converting the grams of sulfuric acid into moles. In order to do that, we need the formula mass for sulfuric acid. We can calculate this using a technique we learned in an earlier lecture. Hopefully you've become skilled enough to realize, that 1 mole of sulfuric acid has a mass of 98.1 grams. We can look at the balanced chemical reaction equation, and determine the ratio between the sodium hydroxide and sulfuric acid. So, we also need to know the molar mass, of the sodium hydroxide. 1 Mole of sodium hydroxide weighs exactly 40.0 grams. With this information in mind, are you able to calculate how many grams of sodium hydroxide are necessary, to react with exactly 100 grams of sulfuric acid? Go ahead and try now, and then we'll do the example together. We're starting with 100 grams of sulfuric acid. I need to first convert that into moles of sulfuric acid. So, I'll do that by dividing by the formula mass. This allows my grams of sulfuric acid to cancel out, and I'm left with just moles of sulfuric acid. The next step, involves using the mole ratios given in the chemical reaction equation. 1 Mole of sulfuric acid reacts with 2 moles of sodium hydroxide. I set this up in the calculation so that, my moles of sulfuric acid cancel out. Finally I need to convert the moles of sodium hydroxide, into grams of sodium hydroxide. And I do that using the molar mass of sodium hydroxide, which is 40 grams per mole here. That causes my moles of sodium hydroxide to cancel out, and by using simple unit analysis, I'm able to determine that I need 81.5 grams of sodium hydroxide, to completely react with my 100 grams of sulfuric acid. Remember the majority of chemical story problems, the contained reactions are simply a matter of recognizing the ratio, and using those ratios in your calculation. Let's do some more review and practice of this concept. Let's look at an important reaction that is used to make the ammonia that is in fertilizer. That fixed nitrogen allows crops to grow, and provides a large amount of food for people on the planet. Ammonia which has a chemical formula NH3 is sixth on the list of chemical compounds produced annually in the United States. It was first synthesized by Fritz Haber in his lab using a process where he was able to combine nitrogen gas, and hydrogen gas to make ammonia. This is now known as the Haber process. First, let's write a balanced chemical reaction equation for this process. This shows the reaction equipment that Fritz Haber used in his lab, and I've done a lot of high pressure chemistry, so I'm really fond of this type of equipment. Thank you for attempting to write the balanced chemical reaction on your own. First, you had to remember that hydrogen and nitrogen are both diatomic molecules. Next you had to go ahead and balance the equation, by putting the coefficient of two, in front of the ammonia. And then you need to go back and put a coefficient of 3 in front of the hydrogen gas. [SOUND] Next let's answer a stoichiometric calculation question. How much nitrogen is required to react with 5 grams of hydrogen gas? Well, here is the balanced chemical reaction equation. Can you do this calculation on your own? Go ahead and try now. Thank you for your submission. The five grams of hydrogen gas can be converted into mols of hydrogen gas, by dividing by the formula mass, for hydrogen, which is 2 grams per mol. Therefore, in this reaction, we're starting with 2.5 mols of hydrogen gas. We then need to convert that, to the number of mols of nitrogen that we need to react. I see up here that every 3 mols of hydrogen gas, reacts with 1 mol of nitrogen, remember there's that implied 1. So, the number of moles of nitrogen gas that reacts is, 1/3 the number of moles of hydrogen gas that reacts. So, we can take the 2.5 moles of hydrogen gas, multiply it times 1/3 and see that there's 0.83 moles of nitrogen gas. In the final step, we need the formula mass of nitrogen gas, which is 28 grams per mole. If we multiply the number of moles of nitrogen we have, times the formula mass, we see that this reaction uses 23 grams of nitrogen gas. Here's another story keyometric/ calculation question. How much ammonia is produced in this reaction? Again, we need the stoichiometric ratio. Here I see that 1 mole of nitrogen, reacts to produce 2 moles of ammonia. Since I know the number of moles of nitrogen that I have in this reaction, I can multiply that times 2, to see that this reaction produces 1.7 moles of ammonia, twice as many moles of ammonia as I had of nitrogen gas. In the final step, I need to convert that into grams of ammonia. I guess I could have answered in moles, because the question doesn't specify that I need to answer in grams. But if I wanted to convert to grams, I would simply multiply the 1.7 moles times the formula mass of NH3, which is 17 grams per mole, and I find out that this reaction produces 28 grams of ammonia. I can do a quick double check, to make sure that my calculations make sense. I know that there has to be conservation of matter, which means that the mass that went into the matter has to equal the mass that comes out of the reaction. In this case, I had 5.0 grams of hydrogen and 23 grams of nitrogen, or 28 grams of reactant, and, I also produced 28 grams of product. So I do have conservation of matter here. Here is an interesting reaction of three solids, and let's practice working it backwards, because as a chemist you are often asked to produce an amount of a chemical. Just like somebody might ask you to bring 28 cupcakes to a bake sale, in this case somebody might say to you, I would like you to synthesize 1.5 grams of magnesium silicate. Well, then you'd have to work the problem backwards, and figure out what quantities of each of the reagents were necessary. So in order to do that, I'm going to just work it backwards. The formula mass of magnesium silicate is 100.4 grams per mole. So if I divide 1.5 grams by 100.4 grams per mole, I see that I have 0.015 moles of magnesium silicate. Let's suppose I'm trying to determine how much silicon dioxide I need to begin with. Well, 6 moles of silicon dioxide are needed to produce 6 moles of magnesium silicate. That's handy, I have a 1 to 1 mole ratio there, 6 to 6 in this case, which simplifies to 1 to 1. So, if I have 0.015 moles of magnesium silicate produced, I must have needed to start with 0.015 moles of silicon dioxide. The silicon dioxide has a formula mass of 60.1 grams per mole. That's just adding up the mass of 1 silicon with the masses of 2 oxygens. So, if I multiply the number of moles of silicon dioxide I have that number, I see that I needed to start with 0.90 grams of silicone dioxide. Remember that the chemical equation, this recipe, is given up here, only works by number of moles. I can't plug a number of grams in here, and predict a number of grams directly. I always have to convert to moles using those formula masses. So, to summarize, we've learned how to do several different types of stoichiometric operations. In this particular video, we talked about converting from mass to moles, using mole ratios to go from moles to moles, and then going back to mass. So, really this whole process was using a combination of ratios. One of those types of ratios was the formula masses, which was in grams per mole. And the second type of ratio is a stoichiometric ratio given by the coefficients, in the chemical reaction equation.