This video is on writing and balancing chemical equations. This first law here is very important when talking about balancing equations. It's a law you have heard of before, the law of conservation of mass. And what this is saying is that in all chemical reactions, atoms are neither created nor destroyed. That atoms may switch positions and may switch to what atoms they are bonded to, but you still need to have the same amount.
in your reactants and your products. So before we go into talking about writing chemical equations, we need to know that the left side of any chemical equation are the reactants. And then you have your arrow, which means produces or yields to form. And your right side, we have our products. So reactants on your left, products are on your right.
There are seven really important elements we need to memorize and those elements are H, N, O, F, C, L, B, R, and I. These seven elements are called diatomic elements. Diatomic elements, when they are by themselves in a chemical reaction, appear with a little number 2 next to them.
I'm not going to get into why they are diatomic or what that means. All you need to know, that is if hydrogen or any of those elements are by themselves in an equation, there needs to be a 2 next to them. They cannot exist by themselves. Only for those 7. Any other element, let's take iron for example. If we're just talking about iron, that is just Fe.
That can be alone. But the 7 diatomics cannot be. So that is very important to remember.
And then when we are writing equations, if we're given words to write equations from, we know how to use our nomenclature and our charges chart to form compounds using the crossover method or the least common denominator method. So here is an example. We have nitrogen.
Notice it's by itself. That's why there's a 2 there. We have...
oxygen by itself. Again, we have a two there, and that forms, see if you can figure out how to name that. That would be a binary molecular compound, two non-metals.
That would be nitrogen monoxide, mono for one oxygen. Something you might notice here are these G's in parentheses after each atom. That G stands for a gas.
So we have nitrogen gas plus oxygen gas, and we get nitrogen monoxide. Other letters you will see in parentheses, you could see an S for a solid, an L for a liquid. Or if you see an AQ, that means aqueous.
That just means something is dissolved in water. The number in front here, this red 2, that is called a coefficient. Coefficient tells you... the relative proportions you have of that molecule. If there is no number in front, such as the case of N2 and O2, there is understood to be a 1 there.
So an example here, we have propane. That's C3H8 plus oxygen, and that yields carbon dioxide and water. You will see later on in the PowerPoint that the type of reaction that this is is called a combustion reaction. But for now, it says identify the reactants and the products.
Reactants are on the left side of the arrow. Products are on the right side of the arrow. The next question says, identify the coefficients and explain what they signify in terms of molar ratios. So the coefficients we have here in front of propane, there is a 1. In front of oxygen is a 5. In front of carbon dioxide is a 3. And in front of water is a 4. And what does that mean? That means we have one molecule of propane.
And when that reacts with five molecules of... oxygen or moles of oxygen, it produces three moles of carbon dioxide and four moles of water. Anytime propane and oxygen is in a one to five ratio, we will always get three and four for carbon dioxide and water.
If you would multiply this one in five by two, so let's pretend for example, we had two moles of propane and 10 moles of oxygen, that three and four would also be multiplied by two. So that would be six and eight. So that is a direct relationship there.
The next question says identify the physical state of each reactant and product. So we are looking at the parentheses and it looks like all of them have G's after them. So they are all gases. So water in the form of a gas would just be your water vapor.
And lastly, and this will get into our balancing then, it says how many C, H, and O atoms are on each side of the arrow? So if we're looking at C, on the left side, we have three C's here. On the right side, there's a little one after the here. This coefficient tells you to multiply your subscripts by that number. So we are actually doing 3 times 1 there, which would be 3. So on the left and the right, we have 3 on both sides.
4 H's. On the left side, we have 8 here. On the right side, again we're going to take this 4 and multiply it by 2, and we have 8. Is that a coincidence that C's and H's are the same on the left and the right? I think not. And for O's here, on the left side, we have a 5 times 2, which is our 10. And on the right side, don't forget to count everything up on the right side.
So here we have 3 times 2, so we have 6 O's here. And then we also have 4 times a 1, which is 4. So 6 plus 4, we have a 10. This obeys the law of conservation of mass from the beginning of the video. The number of elements on the left and right side have to be the same.
And that is why we balance equations. A rather important part of balancing is that the coefficients need to be in the smallest whole number ratio. So for example, if we were balancing an equation and the coefficients came out to be a 2, 4, 6, and 8. Let's pretend those are the numbers.
You must reduce those. So in this case, they're all divisible by a 2. So this would have to become a 1, 2, 3, and 4. It has to be in the smallest whole number. So let's practice doing this.
People do this in different ways. Some people can just do it in their heads and kind of play with it. It's kind of like a puzzle you have to put together. And then other people like writing it down in longhand form. It's really up to you how you do it.
Always do H's and O's last. So balance H's and O's last because they appear in a lot of different molecules. So here I would start with my C's first. So if you want to kind of like draw a line. down the middle to separate the left and the right side, you can do that.
Or what I like to do under the arrow, write down all of the elements you have, and then look on the left side of the equation. How many C's do we have? We have six C's, so I put that on the left side of C. Look on the right side, we only have one, so I put that on the right. Do the same thing with the H's.
I have 12 on the left. I have two on the right and then the O's make sure you count up all of the O's left I have four and then right I have three it looks like so we are not balanced we need these numbers to be the same on the left and right for each letter so I'm going to do C's first I need to add something to the coefficient of the C to get this to be a six so if I put a six here and I do 6 times 1. Notice I now have a 6 here. So my C's are now balanced.
What else did that 6 do? That 6 also changed my O's. So I have 12 O's here and then I still have my 1 O here. So my O's on the right now change to 13. Now my C's are good.
Now I'm going to look at H's or O's. Which one do you do first? It really depends on which one looks easier. So if I'm looking at this, my H's only appear at one spot on the left and the right. And my O's appear at literally every spot, four different spots.
So H's look easier to me. So I need more H's on the right, obviously, because I only have two. So how do I get this two to be a 12? Well, I would need to put a six.
front of there the coefficient have to be a 6 because now that 2 becomes a 12 but recall that this 6 also changes my O's as well so now if I'm recounting up my O's I have 12 here and I have 6 here so O's wise I now have 18 on the right so I'm referring back down to my little puzzle down here I have six C's, so I'm good there. I have 12 H's, so I'm good there. And O's, I need 14 more on the left. Two different spots that I could put O's. If I added it here, that would change my C's and H's as well.
So I don't really want to mess with that. I think I'm going to put something here. So if I already have two O's here, and I need a total of 18 there. What coefficient would have to go in front of the O2? I believe it would have to be an 8. Because 8 times 2 would be 16. And then 16 plus 2 would be an 18. So that would turn to an 18. And notice I am balanced.
There's no coefficient in front of the C6H12O2, so it's understood to be a 1 there. So you can put a 1 or you don't have to put anything. Either way is okay.
I cannot reduce any of these coefficients because of that 1. They are not all divisible by anything other than 1. So that is reduced all of the way and that's completely balanced. Here's one I'd recommend trying on your own before I give you the answer. So pause the video now. So if I'm doing this, I'm looking at my ends first, and I notice I have two ends on the right and one on the left.
So I'd want to put a 2 here. That would make that a 2. Now I'm looking at my H's and O's here. I'm going to choose to do my O's first.
So I need more O's on the right, so I would put a 2 here. Which would change that to a 2. But that would also change my H's to a 4. Now, my N's and my O's are good. But the tricky part here are my H's.
So I have 3 H's on the left and 4 on the right. If I add, if I make that 2 a 3, that would just give me a, actually my math is, I didn't change that. H's now I have a 6 over here. Right, the 2 times the 3 is a 6. So I need more H's on the right hand side. Now, if I need 6, I need to change this to a 3, correct?
So now I have 6 H's here. And now my O's I have 3 there. Now my O's here are an issue.
The left side, anytime I add a coefficient, it's changing by a factor of 2. So for example, if I just put a 2 here, I would now have 4 0s here, and that would be more than 3. And then I'd be going back and forth. So there's a way you can kind of speed this up. If I'm looking at 2 and 3, I'm thinking of the least common multiple of 2 and 3. Where will they meet first? They will meet at 6 first. So I want to make both these a 6. So if I want to make both of these a 6, I would put a 3 here.
to make that a 6. And then for this O over here, if I want to make that a 6, there's a 1 here right now, that would have to be a 6, because 6 times 1 is 6. That 6 makes my H's now be 12, and now my H's are messed up on the left, so I would need to make this 2 a 4 to make this H a 12. Now of course that changes my ends as well. So I now have four here. And now to make this a four, I would just put a two here.
And that is now balanced. So that may be a lot of back and forth, but it is just a puzzle that always works itself out in the end. So if you're going back and forth for a while, don't panic. Sometimes it just takes a little bit of time.
So balancing equations, that's a very important part of chemistry. Another important part here is being able to identify what type of reaction you're working with. So the main classifications, we have a combination reaction, a decomposition, a single replacement, also called a single displacement, and a double replacement, also called a double displacement.
I would argue another main type is combustion, but there are... these other classifications as well. We're going to be able to identify one from the other and they're not too hard to do.
So first a combination reaction. If you just think what combination means it's just combining two things. So your example here if we have element A plus element B it just becomes element AB.
So you're starting with two things and you have a single product that's a combination. Here are real life examples. So we have sodium chloride, we have our table salt on the bottom, and then the top equation we have what we saw before. Notice they are balanced.
A decomposition is the opposite of a combination reaction. So now we're starting with one compound, and we're breaking it up, we're decomposing it into its elements separated by a plus sign. So the big difference between combo and decomp is where the plus sign is at.
An example of a decomposition reaction, we'll see we have carbonic acid and that breaks apart to forming water and carbon dioxide. A single replacement reaction. This is when you have a single element by itself, and then you have a compound with two elements.
And look on the product side what happens here. What happens here is that A kicks B out of its place. So A started single, and then once the reaction took place, A is now with C. and now B is single.
So that is a single replacement. Something got, one thing got replaced. The reason A has the authority to do that is A wants to react with C more than B wants to react with C.
And we know that's from something called the activity series table. It's a table that tells you what is more reactive than the other thing. Here's a very common example used in high school laboratories. If you have a piece of magnesium, solid magnesium ribbon here, and you put it in hydrochloric acid, what happens is, let's label this, so we have, this would be our A, H would be our B, and C would be our C.
Remember, A kicks B out. So now you have A and C here, and B is by itself. Magnesium wants to react with chlorine more than hydrogen does. And we have a double replacement. Same idea as single, except now two things are switching spots.
So we have A and B together, and we have C and D together, and what happens is A and C switch spots. So we're left with A and D and C and B. Important to note that A and C are cations.
These come from our metals. That's why they go first in the products. And then B and D are the anions. They will always go second.
Now the order of this doesn't matter because it's multiplication, the commutative property. You could have C, B, and A, D, but C has to come before B and A has to come before D. Here's another common example, the top one used in high school labs.
So you have potassium iodide. This is a clear colorless solution mixed with lead nitrate, which again is a clear colorless solution. And what happens is pretty cool. When you mix them, you get a yellow precipitate. a solid forms from two clear liquids.
It's a pretty neat lab. So if we're labeling, we would have A and B, C and D, and then we have A, D, C, B. Okay, we have oxidation reduction reactions. We have a different chapter in the book that focuses on these. These are energy producing reactions in the cell. Cellular respiration is an example here.
Internal combustion engines are an example. Batteries, hydrogen fuel cells. This is used a lot in our everyday lives. And here is a little acronym to help you remember what oxidation reduction means.
Oil rig. So the O in oil stands for oxidation. and the oxidation is loss of electron of electrons reduction the rn rig is a gain of electrons you cannot have oxidation without reduction because as you can imagine when this loses the electrons where do they go they go over to here this gains the electron that it lost another acronym i like is Leo goes grr.
Leo the lion goes grr. Loss of electrons is oxidation. Gain of electrons is reduction. If you can remember that, you will be fine with oxidation reduction reactions.
Here's an example of one. You are looking at oxidation numbers in order to help you here. So an important concept here is if any element is by itself it has a zero, sorry, it has a zero what's called oxidation number.
And then whenever an element is in a compound, it takes on to charge whatever charge it has in that compound. So let's look at this here. Let's focus on Cu.
So Cu over here is by itself, right? So Cu is starting as a zero charge. What happens to Cu?
It's now over here with Cl2. And if you remember from nomenclature, In order to figure out what charge Cu has, we look at Cl, we look at the anion, that has a negative 1 charge, that's in group 7A, and there are 2 of them. So it has 2 negative 1 charges, so it has a total of negative 2 charge.
So Cu, in order to make this a neutral compound, Cu must be a plus 2 charge. So Cu went from a 0 to a plus 2. So would that have gained electrons or lost electrons? If it went from zero to positive, that loss, it has a loss of two electrons, two negatives loss.
And we know that oxidation is a loss of electrons. So that is why the Cu here is oxidized. Now if we look at silver.
So silver here, if Cl is a minus one, silver must be a plus one. So silver is starting as a plus one. And notice it's by itself over here now. So that is a zero. So silver went from a plus one to a zero.
So it became more negative. So it gained an electron. Now, why is there a 2 electrons here? That's because of the coefficient of a 2 up here.
We know gain of electrons is called reduction. Again, you will never have one without the other. So combustion reactions are a type of oxidation reduction.
reduction reactions. But let me show you the outline for combustion. Again, I think this is one, this should be classified as one of the main five types.
So we have a combination, a decomposition, a single and double replacement. I think combustion should be up here. Combustion requires some kind of fuel source, some kind of hydrocarbon, some sort of C and H.
The X and Y just means it can be any number. There has to be oxygen. involved with it and then most of the time you will have carbon dioxide and water and that is combustion classify the following reactions as Combo decomp single or double this would be a good spot to pause the video and do these and then check your answers Letter B here. I notice Two different pairs here and they're swept swapping This would be a double displacement.
Letter C, I'm starting with two things and I'm ending with one thing. There's no plus on the right-hand side. That is a combination reaction.
E, I'm starting with one thing and I'm ending with two things. No plus sign on the left. That is a decomposition. And G, a single person with a couple. And what happens?
They swap. Br is now single and you have a new couple that would be a single displacement reaction. And lastly, how do we know that this is a redox reaction?
Now look at G. This is the same thing. So are you wrong in saying single?
Not necessarily, but more specifically, why is this redox? Look what happens to Cl. It's starting by itself as a zero and it's ending as a negative one. So that gained electrons. That is reduction.
Look at your Br. That started as a minus one and it went to a zero. So that lost a negative.
That lost an electron. That is oxidation.