hi everyone my name is Dr Michelle banogan and I'm going to be doing the videos for this section we are talking about chemical reactions and it's quite exciting because so far in this semester we've learned about atoms and molecules and now we're ready to put them all together and have some reactions so in this particular video we are learning about how to write and balance chemical equations our learning goals are to derive chemical equations from narrative descriptions of chemical reactions and then also to learn to write and balance chemical equations in the molecular format so when we write chemical equations we're going to find that they need to be balanced and we'll talk about what that means later but these reactions will represent the identities and the relative quantities of the substances involved right the substances undergoing the chemical or physical change so we have a chemical reaction written here we can see that the symbols for our molecules and our atoms are written so we've got methane CH4 reacting with oxygen O2 and these two are our reactants they occur on the left hand side of the equation on the right hand side we have our products those substances that are formed in our reaction so these include carbon dioxide CO2 and also water H2O as our products we can also see that there are what we call coefficients and coefficients are the numbers in front of the different chemical species so we see on the left hand side for our reactant O2 we have a 2 in front of it which means that we need to have two oxygen molecules in order to react with one methane and so we can see this shown by these space filling models for the molecules right here the black sphere is carbon and the four white spheres are hydrogen so this is a methane molecule and it would need to react with two oxygen molecules right so here the red spheres represent oxygen atoms so together we have 202 molecules and we see that we also have a coefficient on our product side the product water will be generated to give us two water molecules for every one molecule of carbon dioxide that's formed in our reaction so we learned a few new important vocabulary words here for our chemical reactions we have substances that are on the left hand side of our chemical equation and we call these our reactants these are the substances undergoing a reaction on the right hand side of our chemical equation we discussed that these are our products these are the substances that are generated by the reaction and on both sides the substances are separated by plus signs so we see this on the reactant and the product side and then between them we have an arrow right the arrow separates the reactant side from the product side and then we also discussed that there are coefficients we see that there are twos right in front of the oxygen and Waters there's no coefficient in front of methane or CO2 and that's because it's implied that there would be a coefficient of one so typically we omit the one and we do not write it out and it's implied if there is no coefficient we're going to talk more about balancing chemical equations and that will involve trying to figure out the correct coefficients in our chemical equations but we need to know how to interpret what these coefficients mean so a few things about coefficients we need to make sure that we use the smallest possible whole number coefficients in our chemical equation and that simply means that if all of our coefficients are say a multiple of two then I could divide them all by two and that would give me the smallest possible coefficients if all of my coefficients are factors of 10 right then they're going to be very large coefficients so I'll need to divide all of them by 10 so that I have the lowest the lowest possible coefficients also the coefficients are there to represent the relative numbers of reactants and products and so we can interpret them really like ratios so there's different ways for us to interpret this particular chemical reaction right we could think about this reaction happening on the molecular scale so we could have one molecule of methane reacting with two molecules of oxygen right and this would give us one molecule of CO2 and two molecules of water as product we could also think of a larger scale right we could think about moles so this would represent one mole of methane reacting with two moles of oxygen to generate one mole of carbon dioxide and two moles of water in each case whether we're talking about molecules or moles right it's really about having the correct ratio of reactants and products we could also think about this being in dozens of molecules right so it's really just about making sure we have the right ratio to drive this home here we have just a snapshot of our reactant and product side mixtures at a given point in our chemical reaction and if these ratios need to be true for our reaction then what we should see in our reaction mixture should also represent that ratio so here we said that the black and white molecules are methane so you can see that we have three methane molecules in our reactant mixture and the red molecules are the oxygen so one two three four five six of those so even though we don't have the exact numbers that are represented in our chemical reaction the three methane molecules to six oxygen molecules is still a one to two ratio which is the ratio that we see in our chemical reaction and if we look at our product mixture we see the same thing that the ratio of molecules is the same as what we see in our balanced chemical equation so in our reactant mixture the carbon dioxide is represented by the black sphere in the Middle with the two oxygens on the side the two red spheres so we have one two three carbon dioxide molecules and the water molecules are the red sphere with the two white right so this is oxygen and then two hydrogens so one two three four five six water molecules and again you can see that three to six is the same as a one to two ratio which is the ratio that we see in our balanced chemical equation so how now do we balance chemical equations um when we have a balanced chemical equation we have equal numbers of atoms for each element on both the reactant and the product sides to achieve this we need to make sure that the numbers of atoms for a given element is multiplied by the coefficient of any formula containing that element by the element subscript in the formula so here we can see if we look at this chemical reaction the oxygen has a coefficient of 2 and it has a subscript of 2 in the molecular formula so in order to calculate the number of oxygens that are present as reactant here we would need to multiply two by two and then finally if an element appears in more than one formula on a given side of the equation the number of atoms represented in each must be computed and then added together so here we have our formula that we've been working with right the reaction of methane with oxygen so let's make sure that this is a balanced chemical equation first we're going to look at Carbon we'll consider the number of carbon atoms on the reactant side and then the product side and if they are equal then we will know that the carbons are balanced so on the reactant side we have the carbon in methane there's only one carbon atom in each methane and we only have one methane molecule so not multiplying the coefficient by the number of methane atoms is one times one meaning we only have one carbon atom as a reactant but if we look on the product side we have carbon present in the carbon dioxide the coefficient is one and there's only one carbon in each carbon dioxide so this is also one times one and we can see that yes the carbons are balanced when we consider the hydrogens on the reactant side we have four hydrogens in each one carbon each one methane molecule the coefficient is one the subscript for hydrogen is four so we're Computing one times four we have four hydrogens as reactants and on the right hand side our product side we have hydrogen present in water so the coefficient is two we multiply this by the subscript for hydrogen in the molecule which is also two so 2 times 2 gives me 4 hydrogen atoms since both give us four we can see that this is balanced as well and then finally for the oxygen we see that oxygen is present as O2 at the reactant side the coefficient is two we multiply that by its subscript which is also 2 to give us four oxygen atoms present on the reactant side on the product side we see that oxygen is present both in the CO2 and also in the water so we're going to figure out how much oxygen is in each and then we will add them together so in the carbon dioxide the coefficient is 1 the subscript for the oxygen in this molecular formula is 2 which gives us two atoms of oxygen on the product side in the carbon dioxide and then for the water molecule the coefficient is 2 and the subscript for oxygen in this molecular formula is one so this is also two when I add these two together this is going to give me 4. since I had four oxygens on the reactant side and four in the product side for balance with respect to oxygen so let's get a little bit more practice balancing chemical equations we have an unbalanced chemical equation involving water being broken up into its elements hydrogen H2 and oxygen O2 we're told that this is unbalanced so we're going to first check and see how many of each species we have on the reactants and product side and then we'll inspect the reaction to figure out how we can change it to make it balanced so starting with the hydrogens on the reactant side we see that we've got one molecule of water and there are two hydrogens in each molecule giving me a total of two on the reactant side the hydrogen is present as H2 the coefficient is one the subscript is two which means that I have two hydrogen atoms on the product side so the hydrogens are balanced when we look at the oxygens we see that oxygen is present on the reactant side as water our coefficient is one for the oxygen the subscript would be one so I only have one oxygen atom on the product side I have oxygen presidents O2 my coefficient is 1 and I'm multiplying that by my subscript which is 2 to give me two oxygen atoms on the right hand side so we can see that no the oxygens are not balanced in order to balance this we can look at our reaction and think how can we change it so that the oxygens would become balanced we need more oxygen on the reactant side right because we're comparing one to two so we need more oxygen reactant so how can we change this so it says to achieve balance the coefficients of the equation must be changed as needed we know that the subscripts cannot be changed right that would change the actual identity of the species involved right it would change the molecules so what we can change is are the coefficients of these species so we can try to increase the amount of oxygen on the reactant Side by adding a 2 in front of the water so we're going to try two water molecules coming apart to give us hydrogen gas and oxygen gas now let's inspect this to see if it has solved our problems and balanced the oxygen so I'm going to use a little bit of a shortcut to write how many reactant and product atoms we have for each so I'm going to first inspect for hydrogen and I'm going to draw two dashes on the left hand side I'll write how many hydrogens I have as reactant and then on the right I'll write how many I have as product so again I'm doing the 2 write the coefficients for the water molecule multiplied by the subscript for the hydrogen so I can see that I have four hydrogens on the reactant side for the product side I can see I have one as my coefficient for the hydrogen gas my subscript is 2. so 1 times 2 gives me only two hydrogen atoms doing the same for oxygen we see our coefficient of 2 multiplied by 1 gives me only two oxygen atoms and on the right hand side I have a coefficient of 1 for the oxygen O2 gas and a subscript of two multiplying those together gives me two oxygen atoms so we've solved our problem for the oxygen and balanced it but our hydrogens are not balanced so looking at our reaction we can see that we need more hydrogen gas on the product side the hydrogen on the product side is in the form H2 so if we add a coefficient in front of that then that will allow us to hopefully balance our equation so I'm going to write in a 2 here and then we can see how this changed the number of hydrogen atoms right now we'll have 2 times our subscript 2 to give us four hydrogens nothing else should have changed so now both my hydrogens and my oxygens are balanced let's try another example so here we have nitrogen and oxygen reacting to give me dinitrogen pentoxide n2o5 we're going to do the same thing so first I'll look at my nitrogens to see if they're balanced and my oxygens on the left hand side I have two nitrogens as reactants on the right hand side for products I also have two for the oxygens as reactant I have two and on the right hand side I have 5. so we're balanced with respect to nitrogen but not with respect to oxygen but I can see that I can balance these if I'm able to make the number of oxygens equal on both the left and right sides what I can do is notice that if I multiply the number of oxygens on the left hand side by 5 and I multiply the oxygens on the right hand side by 2 then I should have 10 on both sides right I'm using the lowest common multiple of the numbers we had initially so I can do this by changing my reaction so then I would have added a 5 in front of the O2 and A2 in front of the n2o5 so let's see how that has changed my coefficients so now in terms of the nitrogen I have 2 on the left hand side and I have 2 times 2 or 4 nitrogens on the left on the right hand side for the oxygens on the left hand side I have 10 right 5 times 2 and on the right hand side I have 2 times 5 also 10. so now we've succeeded in balancing our oxygens but the nitrogens are not balanced so we can look at our equation and see that we need more nitrogen on the left hand side nitrogen is present as N2 on the left hand side so if I'm able to add a 2 in front of it then that should balance our equation all right so going back and looking at my number of nitrogen atoms it should be 2 times 2 or 4 on my left hand side and the right hand side should now not have been changed so we now are balanced with respect to both the nitrogen and the oxygen let's do one more example we're going to see in this example that is sometimes convenient to use fractions while we're balancing but then in the end we want our coefficients to be the lowest whole numbers and so we'll need to multiply by something in order to get rid of our fractions and I'll show you what I mean so let's look at our reaction here we have a reaction involving carbon hydrogen and oxygen so we're reacting c2h6 with oxygen O2 to generate water and carbon dioxide initially we can go through and see how many of each species we have we've got carbon hydrogen and oxygen in terms of carbon on the left hand side I see I have two atoms of carbon on the right hand side I have only one in the carbon dioxide for hydrogen I have six hydrogens on the left hand side on the right hand side I have two hydrogens and for the oxygen on the left hand side I have two and on the right hand side I have oxygen present in both products so I have one oxygen present in water and I also have two oxygens in the carbon dioxide to give me a total of three so none of our atoms are none of our elements are balanced so we need to set about how to do this um for reactions evolving carbon hydrogen and oxygen I would recommend that you start by balancing the carbons so we're going to start by balancing the carbons so let's look at the carbons we said we have two on the left and one on the right so it would seem that we could add a coefficient of 2 for the CO2 and that should help us balance our carbons indeed now I've got two carbons on the right hand side but because there is oxygen also in the carbon dioxide this has changed the number of oxygens I have for my product side as well so now I have one oxygen in the water and a coefficient of 2 multiplied by the subscript of 2 for the oxygen gives me four so now I have a total of five oxygens on the right hand side after balancing our carbons I want to now look at the hydrogens so next we're going to balance the hydrogens I see that I've got six hydrogens on the left and two on the right if I add a 3 in front of my water molecule then that should indicate that we've got six hydrogens right so we didn't change anything on the Loft so we still have six hydrogens on the left and on the right we have three times two to give us a total of six hydrogens for the product side our last step is to balance the oxygen we see that we've got two oxygens on the left hand side and since I changed the number of water molecules and that also contains oxygen I need to reevaluate how many oxygens I have on the product side so in the water I have three right three times a subscript of one for the oxygen and in the carbon dioxide I have 2 times a subscript of two so that would be four so I have a total of seven oxygens on the right hand side now in order to balance this I need to add more oxygens on the left right on the reactant side I can see that if I multiply the 2 by 7 halves then I would achieve a number of 7 on the left and that's the number that I have on the right so if I can achieve this then the oxygens will also be balanced so I can use 7 halves as my coefficient for the oxygens and when I do this I have seven halves times two or seven oxygens on the left which equals the number of oxygens I said I have on the right so now we are balanced with respect to all of our elements but I have this fractional coefficient which is not preferred so when a balance is achieved with fractions you can then multiply all of your coefficients by some whole number in order to convert the fraction to an integer um and multiplying through all of your coefficients by the same number will keep the same ratios right which achieved the balanced chemical equation but will just help us to have them in integer format so to get rid of my seven halves I can see I just need to multiply by the number that I have in my denominator so because 2 is in my denominator If I multiply all of my coefficients by 2 then this should get rid of my fraction so I'm going to multiply by 2 to achieve integer coefficients so I'm going to write my balanced chemical equation down here now so if I multiply each coefficient by 2 I will have 2 for the C2 H6 If I multiply 7 halves by 2 I'm going to get 7 as my coefficient for the O2 If I multiply 2 by 3 for my coefficient for the waters I will get 6 and then finally 2 multiplied by 2 for the carbon dioxide will give me four carbon dioxide molecules