In this module, we're going to look at stoichiometric relationships, this time involving the masses of substances. By the end of this module, you should be able to calculate the mass of an unknown quantity given the mass of a known quantity of substance. Now, we're going to go back to our diagram showing us the relationship between the unknown and the known substances. Only now, instead of stopping and starting with moles of our substances, we're going to go one step further out and look at the masses of our known and our unknown. Note, that we're still going to go to moles of our known. Use the coefficients of our balanced chemical equation. Find the moles of B, and then convert to the mass of B. This step here in the middle, converting from moles of A to moles of B using our coefficients for balanced chemical equation, is going to be done in every single problem. So let's look at an example of how we would do this. First, we have an equation. We do want to check to make sure it's balanced. And what I see, is that it's not balanced. Because immediately, I notice that there are two nitrogens on the right side of the equation but only one on the left side. So the first thing I need to do, is go through and balance the equation. I'm going to make a short list to tally up what I have on either side of the reaction. So I have nitrogen, oxygen, and hydrogen. And I have 1 nitrogen on the left, I have 3 oxygens on the left and 2 hydrogens. On the right side of my equation, I have 2 nitrogens, I have 4 oxygens and I have just 1 hydrogen. So, because I see oxygen in multiple substances, I'm actually going to balance that last. I'm going to start with my nitrogen, and I'm going to put a 2 in front of the NO2, so that changes my nitrogens to two, and it also changes my oxygens to now a total of five oxygens, because I have two times two is four, plus one is five. And I see that it doesn't have any effect on the hydrogen. Now, I'm going to again, skip over the oxygens, because it's present in every substance, and I'm going to balance my hydrogens. If I put a 2 in front of HNO3 to balance my hydrogens, I get the 2 hydrogens on the right side. I also see that it changes my nitrogens to 3 and it changes my oxygens to 7. So I have, 2 hydrogens on the right, two nitrogens plus one nitrogen for three nitrogens, six plus one, seven oxygens. So, now I see that again my nitrogens are out of balance. I change that to a three for the nitrogen. Now I have six, seven high oxygens, I still have 2 hydrogens on the left and now it appears that my equation is balanced, but I'm always going to go back through and double check to make sure it's correct. So, we have three nitrogens on the left. We have 3 nitrogens on the right. We have six plus one, seven oxygens on the left. We have six plus one equals seven oxygens on the right, we have two hydrogens on the left, and we have two hydrogens on the right. So, now that I know I have a balanced chemical equation, I can proceed with the stoichiometry calculation. In this case it's asking how many grams of nitric acid are produced from 100 grams of nitrogen dioxide with excess water. So the first thing I want to do, is pull out the information that seems to be useful in this problem. I see that I have HNO3 grams equals question mark. That's just a reminder that that's the answer I'm trying to find. It tells me I'm starting with 100 grams of NO2 is my starting material. Plus I have excess water. And the only reason we have to worry about it being excess water, is to know that we're not going to run out of the water. The NO2 will be able to react completely to form the maximum amount of HNO3. Now, I see that I'm going to have to convert between grams and moles of substances, so I want to find the molar mass. And I can do this with the information on the periodic table. Looking at the molar mass of nitrogen and oxygen, considering that I have two oxygens, I find the molar mass of NO2 is 46.01 grams per mole. And that, for HNO3, the molar mass is 63.01 grams per mole. Note, that I find the molar mass for the substance as written, excluding any coefficients. I'll take the coefficients into account when I do my calculation. But the molar mass of a substance is going to be the same regardless of what its coefficient is or isn't in a chemical equation. Now, I look and see which of these but, substances will allow me to find the moles of that substance. If I look at HNO3, I don't know the mass, I only know the molar mass. So, I'm not going to be able to determine the moles of HNO3. So, that's not going to be my starting point. For water, I know we have excess water, but I don't know a specific amount of water. So, that's not going to help me get to moles of some known quantity. The only thing that remains is my NO2. I have 100 grams of NO2. Now, given the molar mass of NO2, I can actually go from grams of NO2 to moles of NO2. And so I use the molar mass, putting the 46.01 grams on the bottom, so that my units will cancel out. So now, I can cancel out grams of NO2 with grams of NO2. If I stop my calculation at this point, I would have mols of NO2. Which tells me how much I started with, but doesn't tell me anything about the mass of HNO3, which is what is being asked for in the question. So now, I need to use my mol to mol ratio. Just as we did in the previous calculation, where we were going from moles of one substance to moles of another. I still have used by mass to get to moles. Now I can put in my mole ratio. Now I'm going to use the coefficients from my balanced equation, so I have 3 moles of NO2, from the coefficient of 3 in my equation, on the bottom. And I put that on the bottom, so that moles of NO2 will cancel with moles of NO2. On the top, I'm going to have 2 moles of HNO3, because that's what I'm trying to find. So now, moles of NO2 cancels with moles of NO2. And if I stopped here, I'd have moles of HNO3 as my answer. But I don't want moles of HNO3, I want grams of nitric acid or grams of HNO3. So, I need to do one more step to get from moles of HNO3 to grams of HNO3. And again, I'm going to use my molar mass, this time for the HNO3. And I know 63.01 grams per mole of HNO3. Now, my moles of HNO3 cancels with my moles of HNO3. I check back over my units and I see that the only units I have left remaining are grams of HNO3. Now, I can do the calculation by taking 100 times two times 63.01 divided by 46.01 divided by three, and what I find is that I get 91.30 grams of HNO3. I notice that my answers are on the same order of magnitude, 100 grams of NO2 to 91.3 grams of HNO3, so that seems to be a reasonable answer. Now let's look at an example problem involving the decomposition of sodium azide, which is used in airbags, and it produces as large volume of gas in a very short time frame. Let's look at our plan to figure out how we can get the mass of N2 produced from a given mass of sodium azide, or NAN3. Remember, that the only relationship we know between amounts of two different substances in an equation are the molar amounts. So, if I want to look at a relationship between two substances, I first have to convert to moles. Because if I know the moles of one substance, I can use my mole ratio for my balanced chemical equation to find the moles of the other substance, and then I can convert to grams. Now, we need to think about what we're going to put for the molar mass of nitrogen. We want to remember how nitrogen exists in nature. Remember, that nitrogen exists as a diatomic or N2. Therefore, when I'm looking for my molar mass of N2, I'm going to want to put it in there as 28.02 grams per mole. If we were referring to atomic nitrogen, we would need to specify that. Remember, for diatomic substances such as nitrogen or hydrogen, oxygen, when we say the name of the element we assume that it is in the form of the diatomic element. Now, let's look at the calculation to figure out how many grams of nitrogen can be produced from 10 grams of sodium azide. We find that we can produce 6.47 grams of nitrogen. Remember, that we're given our 10 grams of sodium azide, and we're asked to find the grams of nitrogen. Here we can set up our 10.0 grams of NaN3. We need the molar mass of sodium azide, so we use the values on the periodic table, and that gets us 65.01 grams per mole of NaN3. So now, our grams cancels with grams. Now we have moles of sodium azide. And now we need to use our mole to mole ratio from our balanced equation. Notice, that our coefficient tells us, we have 2 moles of sodium azide and we have 3 moles of Nitrogen. So, our moles of sodium azide will cancel. Now we have moles of nitrogen, and we're going to use our molar mass of 28.02 grams per mole of nitrogen to find the grams of nitrogen. And so what we find, is that we get 6.47 grams. Note, that I had to convert to moles first, before I could use my mole ratio. These numbers come from the balanced chemical equation. In the next module, we're going to look at limiting and excess reagents.
