0:00:00.710,0:00:04.200 Like the concept of the mole,[br]balancing equations is one of 0:00:04.200,0:00:06.700 those ideas that you learn in[br]first-year chemistry class. 0:00:06.700,0:00:09.250 It tends to give a lot of[br]students a hard time, even 0:00:09.250,0:00:12.430 though it is a fairly[br]straightforward concept. 0:00:12.430,0:00:13.970 I think what makes it difficult[br]is that there's a 0:00:13.970,0:00:15.350 bit of an art to it. 0:00:15.350,0:00:17.790 So before we talk about[br]balancing chemical equations, 0:00:17.790,0:00:18.920 what is a chemical equation? 0:00:18.920,0:00:20.660 Well here's some examples right[br]here, and I have some 0:00:20.660,0:00:22.180 more in the rest[br]of this video. 0:00:22.180,0:00:24.750 But it essentially just[br]describes a chemical reaction. 0:00:24.750,0:00:25.720 You've got some aluminum. 0:00:25.720,0:00:30.600 You have some oxygen gas or a[br]diatomic oxygen molecule. 0:00:30.600,0:00:32.860 And then you end up with[br]aluminum oxide. 0:00:32.860,0:00:34.850 And you would say, ok, fine,[br]that's an equation. 0:00:34.850,0:00:35.830 It looks nice. 0:00:35.830,0:00:38.700 I have my reactants, or[br]the things that react. 0:00:38.700,0:00:40.240 These are the reactants. 0:00:40.240,0:00:42.510 And then I have the products[br]of this reaction. 0:00:42.510,0:00:44.090 What's left there to do? 0:00:44.090,0:00:45.400 Well you have a problem here. 0:00:45.400,0:00:47.770 The way I've written it right[br]now, I have one atom of 0:00:47.770,0:00:51.550 aluminum plus two atoms[br]of oxygen, right? 0:00:51.550,0:00:53.700 They're bonded to each other,[br]but there's two atoms of 0:00:53.700,0:00:54.250 oxygen here. 0:00:54.250,0:00:59.270 One molecule of diatomic oxygen,[br]or one molecule of 0:00:59.270,0:01:01.540 oxygen, but they have two[br]oxygen atoms here. 0:01:01.540,0:01:05.030 And when you add them together,[br]I have two atoms of 0:01:05.030,0:01:07.020 aluminum and three[br]atoms of oxygen. 0:01:07.020,0:01:09.550 So I have a different number of[br]aluminums on both sides of 0:01:09.550,0:01:10.510 this equation. 0:01:10.510,0:01:12.470 On this side, I have[br]one aluminum. 0:01:12.470,0:01:14.810 On this side, I have two[br]aluminums. And then I have a 0:01:14.810,0:01:15.850 different number of oxygens. 0:01:15.850,0:01:17.580 On this side, I have[br]two oxygens. 0:01:17.580,0:01:19.360 And on that side, I have[br]three oxygens. 0:01:19.360,0:01:22.700 So balancing equations is all[br]about fixing that problem, so 0:01:22.700,0:01:24.990 that I have the same number of[br]aluminum on both sides of the 0:01:24.990,0:01:28.245 equation, and the same number[br]of oxygens on both sides. 0:01:28.245,0:01:30.200 So let's try to do that. 0:01:30.200,0:01:31.960 I'll do it in orange. 0:01:31.960,0:01:35.100 So I said I have one aluminum[br]here and I have 0:01:35.100,0:01:36.600 two aluminums there. 0:01:36.600,0:01:39.130 So maybe a simple thing is just[br]to put a two out here. 0:01:39.130,0:01:42.990 So now I have two aluminums on[br]this side and I have two 0:01:42.990,0:01:44.290 aluminums on this side. 0:01:44.290,0:01:45.400 The aluminums look happy. 0:01:45.400,0:01:47.060 Now let's look at the oxygen. 0:01:47.060,0:01:49.880 Here I have two oxygens on the 0:01:49.880,0:01:51.080 left-hand side of the equation. 0:01:51.080,0:01:52.900 And on the right-hand[br]side of the equation 0:01:52.900,0:01:54.280 I have three oxygens. 0:01:54.280,0:01:55.560 What can I do here? 0:01:55.560,0:01:58.960 Well if I could kind of have[br]half atoms, I could just 0:01:58.960,0:02:01.840 multiply this by[br]one and a half. 0:02:01.840,0:02:03.910 1.5. 0:02:03.910,0:02:06.190 1.5 times 2 is 3. 0:02:06.190,0:02:09.600 So now I have three oxygens on[br]both sides of this equation 0:02:09.600,0:02:12.180 and I have two aluminums on both[br]sides of the equation. 0:02:12.180,0:02:13.210 Am I done? 0:02:13.210,0:02:15.940 Well, no, you can't have half[br]an atom, or one and 0:02:15.940,0:02:17.130 a half of an atom. 0:02:17.130,0:02:18.270 That's not cool. 0:02:18.270,0:02:20.480 So what you do is you just[br]multiply this so that you end 0:02:20.480,0:02:21.145 up with whole numbers. 0:02:21.145,0:02:23.230 So let's just multiply[br]both sides of this 0:02:23.230,0:02:26.070 whole equation by two. 0:02:26.070,0:02:37.700 And we have four aluminums plus[br]three oxygens yields-- we 0:02:37.700,0:02:40.210 multiplied everything[br]by two-- two 0:02:40.210,0:02:45.090 molecules of aluminum oxide. 0:02:45.090,0:02:47.780 Now, you might have been tempted[br]at some point in this 0:02:47.780,0:02:50.950 exercise to say, oh, well why[br]don't I just tweak part of the 0:02:50.950,0:02:51.860 aluminum oxide? 0:02:51.860,0:02:54.940 Why don't I put a 2/3 in[br]front of this oxygen. 0:02:54.940,0:02:56.640 You cannot do that. 0:02:56.640,0:02:58.380 The equation is as it is. 0:02:58.380,0:03:00.720 The molecule aluminum oxide[br]is aluminum oxide. 0:03:00.720,0:03:04.630 You can't change the relative[br]ratios of the aluminum and the 0:03:04.630,0:03:07.310 oxygen within the aluminum[br]oxide molecule. 0:03:07.310,0:03:11.320 You can just change the number[br]of aluminum oxide molecules as 0:03:11.320,0:03:13.460 a whole that you have,[br]in this case, two. 0:03:13.460,0:03:14.000 So what did we do? 0:03:14.000,0:03:14.800 We looked at the aluminum. 0:03:14.800,0:03:17.060 We said, OK, we need two[br]aluminums to have both sides 0:03:17.060,0:03:18.660 of that be two. 0:03:18.660,0:03:20.550 And then when we looked at[br]oxygen, we said, well, if I 0:03:20.550,0:03:23.310 multiply this by one and a half,[br]then that becomes three 0:03:23.310,0:03:26.340 oxygens here, because one and[br]a half times two oxygens. 0:03:26.340,0:03:27.880 And three oxygens there. 0:03:27.880,0:03:30.400 And then all we said is, oh, I[br]can't have a one half there, 0:03:30.400,0:03:32.540 so let me multiply both[br]sides by two. 0:03:32.540,0:03:37.360 And I ended up with four[br]aluminums plus three oxygen 0:03:37.360,0:03:41.980 molecules, or six oxygen atoms,[br]yields two molecules of 0:03:41.980,0:03:43.660 aluminum oxide. 0:03:43.660,0:03:46.070 Let's see if we can do[br]some more of these. 0:03:46.070,0:03:48.270 So here I have methane. 0:03:48.270,0:03:50.200 And that g in parentheses,[br]I just wanted to 0:03:50.200,0:03:50.930 expose you to that. 0:03:50.930,0:03:53.390 That just means it's a gas. 0:03:53.390,0:03:58.340 So I have methane gas plus[br]oxygen gas yields carbon 0:03:58.340,0:04:00.090 dioxide gas plus water. 0:04:00.090,0:04:02.410 That's an l there,[br]so liquid water. 0:04:02.410,0:04:04.640 So what can we do here? 0:04:04.640,0:04:07.440 So the general thing is, do[br]the complicated molecules 0:04:07.440,0:04:11.140 first, and then at the end you[br]can worry about the single 0:04:11.140,0:04:13.000 atom molecules. 0:04:13.000,0:04:15.180 Because those are very[br]easy to play with. 0:04:15.180,0:04:17.139 And the reason why you do that,[br]whenever you change a 0:04:17.139,0:04:19.680 number here-- let's say we're[br]trying to engineer how many 0:04:19.680,0:04:22.680 carbons we have on both sides of[br]this equation-- if I set a 0:04:22.680,0:04:24.680 number here, I'm also changing[br]the number of hydrogens. 0:04:24.680,0:04:26.140 Then I'll have to play with[br]the hydrogens there. 0:04:26.140,0:04:28.720 And at the end, I'll have[br]some number of oxygens. 0:04:28.720,0:04:30.630 And then I have to tweak this[br]number right there. 0:04:30.630,0:04:33.770 So let's just start[br]with the carbons. 0:04:33.770,0:04:36.050 It seems complicated, but when[br]you go step by step and you 0:04:36.050,0:04:38.100 kind of play with things a[br]little bit, it should proceed 0:04:38.100,0:04:39.550 fairly smoothly. 0:04:39.550,0:04:41.070 So here I have two carbons. 0:04:43.680,0:04:45.790 And here I have, on[br]the right-hand 0:04:45.790,0:04:47.810 side, only one carbon. 0:04:47.810,0:04:49.790 So ideally I'd want two[br]carbons on both 0:04:49.790,0:04:50.780 sides of this equation. 0:04:50.780,0:04:53.930 So let me put a two out here. 0:04:53.930,0:04:55.700 So there you go, my[br]carbons are happy. 0:04:55.700,0:04:58.460 I now have two carbons[br]and two carbons. 0:04:58.460,0:04:59.850 Now let me move to[br]the hydrogens. 0:04:59.850,0:05:02.010 Remember, I wanted to do the[br]oxygens last, because I can 0:05:02.010,0:05:05.000 just set this to whatever I want[br]it to be without messing 0:05:05.000,0:05:08.070 up any of the other atoms. So on[br]this side of the equation I 0:05:08.070,0:05:11.370 have four hydrogen atoms. How[br]many hydrogen atoms do I have 0:05:11.370,0:05:12.690 on this side of the equation? 0:05:12.690,0:05:16.320 So I have four hydrogen[br]atoms here. 0:05:16.320,0:05:18.150 How many do I have[br]on this side? 0:05:18.150,0:05:21.070 Well, I have two hydrogen[br]atoms right there. 0:05:21.070,0:05:24.100 So I want to have four on both[br]sides, so let me put a two in 0:05:24.100,0:05:25.510 front of the water. 0:05:25.510,0:05:29.310 So now I have four hydrogen[br]atoms. Cool. 0:05:29.310,0:05:30.850 Finally, oxygen. 0:05:30.850,0:05:35.800 On the left-hand side I have two[br]oxygen atoms. And on the 0:05:35.800,0:05:37.780 right-hand side,[br]what do I have? 0:05:37.780,0:05:41.970 I have two times this one[br]oxygen, right here, so right 0:05:41.970,0:05:44.290 now that's two, right? 0:05:44.290,0:05:44.980 Two waters. 0:05:44.980,0:05:49.450 And in each water I have one[br]oxygen atom, but I have two 0:05:49.450,0:05:52.170 water molecules, so I[br]have two oxygens. 0:05:52.170,0:05:56.270 And then in the carbon dioxide[br]I have two oxygens in each 0:05:56.270,0:05:56.845 carbon dioxide. 0:05:56.845,0:05:58.950 And I have two carbon[br]dioxides, right? 0:05:58.950,0:06:01.290 When I put that magenta[br]two out front. 0:06:01.290,0:06:02.560 So how many oxygens do I have? 0:06:02.560,0:06:03.420 Two times two. 0:06:03.420,0:06:05.330 I have four oxygens. 0:06:05.330,0:06:07.820 So on the left-hand side[br]I have two oxygens. 0:06:07.820,0:06:11.020 On the right-hand side I have[br]six oxygens, two in the two 0:06:11.020,0:06:13.430 molecules of water and[br]four in the two 0:06:13.430,0:06:15.770 molecules of carbon dioxide. 0:06:15.770,0:06:18.460 So how do I make this[br]two into six? 0:06:18.460,0:06:20.410 I want to have six oxygens[br]on this side. 0:06:20.410,0:06:21.690 I want to them to have[br]six oxygens on the 0:06:21.690,0:06:22.855 left-hand side as well. 0:06:22.855,0:06:30.110 Well, if I put a three out here,[br]now I have six oxygens. 0:06:30.110,0:06:31.810 And our equation has[br]been balanced. 0:06:31.810,0:06:35.210 I have two carbons on this[br]side, two carbons on that 0:06:35.210,0:06:38.780 side, four hydrogens on this[br]side, four hydrogens on this 0:06:38.780,0:06:42.040 side, six oxygens on this side,[br]and then-- four plus 0:06:42.040,0:06:44.810 two-- six oxygens[br]on that side. 0:06:44.810,0:06:46.840 Next equation to balance. 0:06:46.840,0:06:49.270 This actually becomes quite[br]fun once you get 0:06:49.270,0:06:51.320 the knack of it. 0:06:51.320,0:06:58.510 So I have ethane plus[br]oxygen gas yielding 0:06:58.510,0:07:00.575 carbon dioxide and water. 0:07:00.575,0:07:03.230 So this is a combustion[br]process. 0:07:03.230,0:07:05.280 Let's look at the[br]carbons first. I 0:07:05.280,0:07:06.810 have two carbons here. 0:07:06.810,0:07:09.010 I have one carbon there. 0:07:09.010,0:07:11.050 So let me put a two here. 0:07:11.050,0:07:13.240 So now I have two carbons. 0:07:13.240,0:07:14.090 Fair enough. 0:07:14.090,0:07:15.900 Remember, I'm going to worry[br]about the oxygen last. This 0:07:15.900,0:07:16.910 one's actually not[br]too different 0:07:16.910,0:07:18.420 than the last problem. 0:07:18.420,0:07:21.130 I have six hydrogens here. 0:07:21.130,0:07:23.400 I only have two hydrogens[br]in the water. 0:07:23.400,0:07:28.090 So then we have three[br]water molecules. 0:07:28.090,0:07:30.470 And now I've balanced[br]out the hydrogens. 0:07:30.470,0:07:33.520 I have six hydrogens on both[br]sides of the equation. 0:07:33.520,0:07:36.060 And now let's deal[br]with the water. 0:07:36.060,0:07:40.330 Here on the right-hand side--[br]I'll do it in this orange 0:07:40.330,0:07:45.590 color-- I have two oxygens in[br]each carbon dioxide molecule. 0:07:45.590,0:07:48.780 And I have two molecules,[br]so I have four oxygens. 0:07:48.780,0:07:52.230 And I have one oxygen in[br]each water molecule. 0:07:52.230,0:07:56.240 And I have three molecules, so[br]I have three oxygens here. 0:07:56.240,0:07:57.580 Is that right? 0:07:57.580,0:08:00.190 Three oxygens on the[br]right-hand side. 0:08:00.190,0:08:02.060 Yep, three water molecules. 0:08:02.060,0:08:07.200 And then I have four oxygens[br]in the carbon dioxide. 0:08:07.200,0:08:07.860 Right. 0:08:07.860,0:08:09.790 So I have seven oxygens. 0:08:09.790,0:08:11.590 I have seven oxygens on[br]this side and I only 0:08:11.590,0:08:13.130 have two on this side. 0:08:13.130,0:08:15.575 So how can I make this[br]two into seven. 0:08:15.575,0:08:19.360 Well I could multiply it[br]by three and a half. 0:08:19.360,0:08:19.690 Right? 0:08:19.690,0:08:21.620 Remember, I just want to have[br]seven on both sides. 0:08:21.620,0:08:25.290 If I have three and a half[br]of these diatomic oxygen 0:08:25.290,0:08:27.640 molecules-- three and a half[br]times two is seven-- so now I 0:08:27.640,0:08:30.830 have seven oxygens on both[br]sides of the equation. 0:08:30.830,0:08:32.870 Four plus three and[br]seven here. 0:08:32.870,0:08:34.059 Two carbons. 0:08:34.059,0:08:35.820 And then six hydrogens. 0:08:35.820,0:08:38.429 And I'm almost done, except for[br]the fact that you can't 0:08:38.429,0:08:40.559 have fractions of molecules. 0:08:40.559,0:08:42.409 So what you do is just multiply[br]both sides of this 0:08:42.409,0:08:45.760 equation by two or three, or[br]whatever you need to multiply 0:08:45.760,0:08:46.960 it to get rid of[br]the fractions. 0:08:46.960,0:08:50.800 So if I multiply everything[br]by two, I end up with two 0:08:50.800,0:09:00.580 molecules of ethane plus seven[br]molecules of diatomic oxygen, 0:09:00.580,0:09:07.520 seven O2's, yielding two[br]molecules of carbon dioxide-- 0:09:07.520,0:09:10.200 oh, sorry, I'm multiplying[br]everything by two, so four 0:09:10.200,0:09:18.920 molecules of carbon dioxide,[br]plus six molecules of water. 0:09:18.920,0:09:21.600 And just to make sure that all[br]still works, if you want to, 0:09:21.600,0:09:22.630 you can check. 0:09:22.630,0:09:23.830 How much carbon do we have? 0:09:23.830,0:09:25.980 I have four carbons here. 0:09:25.980,0:09:28.180 I have four carbons here. 0:09:28.180,0:09:29.520 How much hydrogen do I have? 0:09:29.520,0:09:31.420 I have two times six. 0:09:31.420,0:09:33.630 I have 12 hydrogens here. 0:09:33.630,0:09:36.100 I have 12 hydrogens here. 0:09:36.100,0:09:37.640 How much oxygen do I have? 0:09:40.150,0:09:41.420 Let me do a different color. 0:09:41.420,0:09:44.190 I have 14 oxygens here. 0:09:44.190,0:09:48.040 And here I have eight oxygens. 0:09:48.040,0:09:50.860 And here I have-- six times[br]one-- six oxygens. 0:09:50.860,0:09:53.010 So six plus eight is 14. 0:09:53.010,0:09:56.000 So my equation has[br]been balanced. 0:09:56.000,0:09:58.230 So that one, a lot of people[br]might find that to be a hard 0:09:58.230,0:10:00.260 problem, because you have three[br]and a half, and just 0:10:00.260,0:10:02.130 going straight to this might[br]seem non-intuitive. 0:10:02.130,0:10:05.660 But if you just work from the[br]more complicated molecules and 0:10:05.660,0:10:07.980 you go atom by atom, and if you[br]end up with any fractions 0:10:07.980,0:10:10.775 you just multiply both sides by[br]some number to get rid of 0:10:10.775,0:10:13.030 the fraction, of and[br]then you're done. 0:10:13.030,0:10:13.590 All right. 0:10:13.590,0:10:15.850 Let's do another one. 0:10:15.850,0:10:17.950 So this one looks all hairy. 0:10:17.950,0:10:24.190 I have this iron oxide plus[br]sulfuric acid yields all this 0:10:24.190,0:10:25.260 hairy stuff. 0:10:25.260,0:10:30.000 But the key here to realize is[br]that this group right here, 0:10:30.000,0:10:35.170 this sulfate group,[br]stays together. 0:10:35.170,0:10:35.480 Right? 0:10:35.480,0:10:39.580 You have this SO4 there and[br]you have this SO4 there. 0:10:39.580,0:10:42.740 So to really simplify things for[br]our head, you can kind of 0:10:42.740,0:10:44.010 treat that like an atom. 0:10:44.010,0:10:45.410 So let's make a substitution. 0:10:45.410,0:10:48.230 We'll substitute that for x. 0:10:48.230,0:10:52.230 So if we rewrite this as--[br]I'll do it in a vibrant 0:10:52.230,0:11:03.390 color-- iron oxide plus H2--[br]there's only one sulfate 0:11:03.390,0:11:07.690 group here-- x. 0:11:07.690,0:11:15.070 And then that yields[br]two irons, this 0:11:15.070,0:11:15.940 molecule with two irons. 0:11:15.940,0:11:18.070 And it has three of these[br]sulfate groups. 0:11:21.140,0:11:22.390 And then plus water. 0:11:26.450,0:11:28.870 All I did is replace the[br]sulfate with an x. 0:11:28.870,0:11:32.040 And now we can treat that x like[br]an atom, and we can just 0:11:32.040,0:11:33.830 balance the equation. 0:11:33.830,0:11:36.040 Let's see, on the left-hand[br]side, how 0:11:36.040,0:11:37.300 many irons do we have? 0:11:37.300,0:11:39.000 We have two irons here. 0:11:39.000,0:11:40.820 We have two irons there. 0:11:40.820,0:11:45.190 So the irons look balanced,[br]at first glance. 0:11:45.190,0:11:47.630 Let's deal with the oxygens. 0:11:47.630,0:11:52.280 So if we have three oxygens[br]here-- let me do it in a 0:11:52.280,0:11:59.660 different color-- and we only[br]have one oxygen here. 0:11:59.660,0:12:03.480 Remember, there were some[br]oxygens in this x group, but 0:12:03.480,0:12:05.590 the sulfate group stayed[br]together, so we can just treat 0:12:05.590,0:12:08.700 those separately. 0:12:08.700,0:12:10.820 We want to have three oxygens[br]on the right-hand side, as 0:12:10.820,0:12:13.250 well, so let's stick[br]a three here. 0:12:13.250,0:12:14.500 Oops. 0:12:19.710,0:12:21.040 So let me put that[br]three there. 0:12:21.040,0:12:23.960 So now we have three[br]oxygens, as well. 0:12:23.960,0:12:28.733 And then finally, let's[br]look at the hydrogens. 0:12:35.950,0:12:37.160 Let's see, how many hydrogens[br]do we have here. 0:12:37.160,0:12:41.270 We have six hydrogens now,[br]on the right-hand side. 0:12:41.270,0:12:42.840 Six hydrogens here. 0:12:42.840,0:12:43.725 Three times two. 0:12:43.725,0:12:46.770 And we want to have six[br]hydrogens here, so we have to 0:12:46.770,0:12:48.910 have three of these molecules. 0:12:48.910,0:12:52.200 And then finally, let's look at[br]the sulfate group, that x. 0:12:52.200,0:12:53.790 We have three x's here. 0:12:57.090,0:13:01.350 And lucky for us, we have[br]three x's over there. 0:13:01.350,0:13:03.420 So our equation has[br]been balanced. 0:13:03.420,0:13:05.760 And if we want to write it back[br]in terms of the sulfate 0:13:05.760,0:13:09.490 terms, we can just un-substitute[br]the x, and we're 0:13:09.490,0:13:16.450 left with one molecule of iron[br]oxide plus three molecules of 0:13:16.450,0:13:21.540 sulfuric acid, H2SO4 -- I just[br]un-substituted the x with SO4 0:13:21.540,0:13:28.570 -- yields one of these[br]molecules, SO4 3. 0:13:32.270,0:13:39.540 I just un-substituted the x plus[br]three molecules of water. 0:13:39.540,0:13:40.670 Let's do one more. 0:13:40.670,0:13:43.800 And then I think we'll[br]be all balanced out. 0:13:43.800,0:13:47.750 Carbon dioxide plus hydrogen gas[br]yields methane plus water. 0:13:47.750,0:13:51.700 So let's deal with the carbon[br]first. I have one carbon here, 0:13:51.700,0:13:52.540 one carbon there. 0:13:52.540,0:13:53.940 The carbons look happy. 0:13:53.940,0:13:54.890 Let's look at the oxygen. 0:13:54.890,0:13:56.920 I have two oxygens here. 0:13:56.920,0:13:58.990 I have one oxygen there. 0:13:58.990,0:14:02.350 So I want two oxygens here so[br]let me stick a two there. 0:14:02.350,0:14:04.300 And then let's deal with the[br]hydrogens last, because this 0:14:04.300,0:14:06.650 is the easiest one to play with[br]because it doesn't affect 0:14:06.650,0:14:11.030 any of the other atoms. So if[br]I have two here and I have 0:14:11.030,0:14:14.060 four, plus four here, right? 0:14:14.060,0:14:15.865 I have four hydrogens there. 0:14:15.865,0:14:18.160 And I have four hydrogens[br]there. 0:14:18.160,0:14:20.790 So I need eight hydrogens[br]on the left-hand side. 0:14:20.790,0:14:22.440 So I just put a four there. 0:14:22.440,0:14:24.350 We're all done balancing. 0:14:24.350,0:14:27.260 Anyway, hopefully you[br]found that useful.