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Unformatted text preview: Solomon's Study Notes General Chemistry I
Fall 2011 Solomon Weiskop PhD [Balancing Chemical Equations]
General Chemistry 1 Study Notes & Practice Problems are available to print out by registering at www.solomonlinetutor.com Solomon Weiskop PhD Copyright 2011 1 1. Balancing Chemical Equations
A chemical reaction is described by a chemical equation. In any chemical reaction the number of each kind of atom is conserved during the reaction. This is reflected in the requirement that the corresponding chemical equation be "balanced". Most chemical equations can be balanced "by inspection". For example, consider the following (unbalanced) reaction: To balance it, we need to arrange that there are equal numbers of N atoms on both sides of the equation. Likewise, we must arrange that there are equal numbers of H atoms on both sides. A little playing around should show you that this equation can be balanced as follows: We now have 2 N's on both sides. We also have H's on both sides. Thus the equation is balanced (since we have equal numbers of N's and equal numbers of H's on both sides.) The numbers out in front are called stoichiometric coefficients Balancing a chemical equation means: Determining the stoichiometric coefficients that will make the numbers of each kind of atom the same on both sides of the equation After you are done, you should always check that indeed the numbers of each kind of atoms are equal on both sides. That way you know for sure that you have correctly balanced the equation. If the numbers of any kind of atom comes out differently on both sides then the equation is NOT balanced and you need to go back and find your error. 2 Often balancing chemical equations can be done "by inspection" (as was done above) and whenever that is possible, that is actually the preferable method, because it is quick. (But, of course, always check afterwards!) Sometimes however, if the equation is sufficiently complex, it might not be so obvious how to balance it "by inspection". You play around for a while and you don't get it to balance. In such cases, it is good to have a systematic method for balancing chemical equations. I will now present such a systematic method for balancing chemical equations: Algebraic Method of Balancing Chemical Equations I call it the "algebraic method" because this method approaches the problem of balancing chemical equations as an algebra problem (which is, in fact, what it really is!). In algebra, you have algebraic unknowns, represented as letters usually from the back of the alphabet , to solve for. When balancing a chemical equation, the algebraic unknowns are the stoichiometric coefficients I will here refer to them by letters from the front of the alphabet ( ) [Notice that I skipped . I do this to avoid a possible confusion with the element carbon C.] Our job will be to determine these algebraic unknowns (i.e. the stoichiometric coefficients: ). We'll do this by ensuring that the number of each kind of atom is equal on both sides of the chemical equation. 3 Let me illustrate with the simple reaction we began with: Ex.1: Here are the unknown stoichiometric coefficients (i.e. the algebraic `unknowns') that we are trying to determine. So we begin here with three algebraic unknowns. Now let's start to balance. We can start with N: On the left side we have N's. On the right side we have N's. We must have equal numbers of N's on both sides, so we must have Even though I don't yet know what is or what whatever they are, they must satisfy . Otherwise, I won't balance the N's! So we can now replace in our equation with equal!). is, I do know that (after all, they are Doing so reduces the number of unknowns from three ( ) down to two ( ) so we are getting closer to a solution! Next, we can balance H: On the left side we have H's. On the right side we have H's. We must have equal numbers of H's on both sides, so we must have which can be simplified (by cancelling out a factor of 2): We can now replace in our equation with ) to just Doing so reduces the number of unknowns further, from two ( one ( ). So that's it. We are done! Once you have it down to just one unknown, you are finished! 4 I can pick any number I like for and the equation will be balanced, but one normally wants low integers in a balanced chemical equation. So let's pick and we get: which we already know is correct because we checked the number of N's and H's on both sides earlier. (But remember always to check!) The algebraic method described above is not really needed for simple equations like the one we just balanced. For those you can (and should) just balance by inspection. However, for more complex equations the algebraic method can be very helpful! So let's do some more complex examples. 5 Ex. 2: You should always begin with an atom that only shows up once on each side. That helps keep the algebra simple. So in this example, we could begin with B or with H but not with O. O shows up twice on the right side, and that will make balancing O not simple. So you should postpone balancing O until after you've balanced the simple ones: B and H. Let's begin with B: Now let's balance H: Finally, we balance O: We're now down to one unknown so we are done. We can pick anything we like for . But if we want to have stoichiometric coefficients that are integers we will need to get rid of the fraction in front of O2. We can do this if we pick : Now check and make sure that the numbers of B, and of H, and of O are the same on both sides! When checking, it is a good idea to check in the same order that you balanced (i.e. in this case, check B, then H, then O). 6 Ex.3: The point of this example is to draw your attention to the need to be careful when counting H's here. In there are, of course, 7+1=8 H's! Let's begin with C because it is simple: Next, let's balance H, it is also simple: Finally balancing the not simple O: Pick : Check (in the same order that you balanced) that each different type of atom is now balanced: C: H: O: 7 Let's now get even more challenging! Ex. 4: The point of this example is to draw your attention to the need to be careful when counting N's and O's here. In there are 3 N's and O's! In this example, everything is simple except for O. So we'll leave O for last. We can start with C: We can balance N next: We can balance H next: Finally, we balance O: To get rid of the fractions pick : You should confirm that this is correct by checking the balance for each kind of atom. 8 Ex. 5: In this example only N and H are simple. Both C and O are not simple, so we'll postpone them till later. Let's start with N: Now we can balance H: Who should we balance next, C or O? Does it matter? Yes it matters! Can you see that if we next balance C we'll get an equation with three unknowns in it ( ) whereas if we next balance O we'll get an equation with just two unknowns in it ( ). That makes O simpler than C so we should balance O next (and leave C for dead last). Balancing O: Finally balancing C: Picking (to get rid of the fractions) yields: You should check this result on your own by making sure that the number of each kind of atoms is the same on both sides. 9 Ex.6: Consider the following (unbalanced) chemical equation: What is the coefficient of HF? To answer this question is it necessary to follow the whole systematic procedure? No! Since one of the coefficients has already been given in the problem, we can reason more quickly as follows: Since the coefficient of is 5, then the coefficient of also be 5 (otherwise there's no way to balance the S). must Since the coefficient of is 5, then the coefficient of must be 1 (otherwise there's no way to balance the Ca). Since the coefficient of is 1, then the coefficient of HF must be 1 (otherwise there's no way to balance the F). So we have answered the question. ...
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This note was uploaded on 03/12/2012 for the course CHEMISTRY 131 taught by Professor Lacey during the Fall '11 term at SUNY Stony Brook.
 Fall '11
 Lacey
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