Hi, all. I’m confused once again. What difference is there in balancing equations using coefficients, as opposed to not using them. How would that apply to Fe+O2 --> FeO ?
What is a balanced equation for Magnesium+Chlorine --> Magnesium Chloride ?
Are there any websites that have some useful tips for finding balanced equations?
Using coefficients shows in detail how the individual atoms are recombining into new molecules, not just appearing and disappearing as a coefficient-less equation might imply. It tells you how much of each reagent you need to put into a reaction to get a certain output, what the proportions must be to use all of each of them, how much of each will be left over if you don’t, and so forth.
Balancing these equations is not that difficult, really. Just count the number of atoms of one element appear on each side, multiply through to get the least common multiple, check another element to see if IT balances, multiply through again until it does, and you’ll get there eventually.
Can someone do the dirty work for me? I still don’t understand how I can multiply both sides of the reaction, without changing it.
::The former chemistry teacher enters the room::
Gosh. Where to begin. The toughest part here is to explain with out pictures.
K.
Lets go over some basic points.
Coefficients give the number of molecules (“groups” if you will).
Subscripts give the number of the specific type of atom chemically bonded to form the moleucle.
I assume by this that “to not using them” means “to use subscripts instead”.
You can’t do that. to change the subscript changes the compound. Example. CO[sub]2[/sub] is a molecule that has 2 Oxygen atoms chemically bonded to 1 Carbon atom. To change the subscript “creates” a completely different compound. Again, example. CO is a molecule that has 1 Oxygen atom chemically bonded to 1 Carbon atom. Though both molecules have the same components (carbon and oxygen) they are very different compounds because of the number of each type of atom. As you well know, CO[sub]2[/sub] is Carbon dioxide while CO is Carbon monoxide. This is why one can not change subscripts when balancing equations.
Ok, on to the next part. The coefficient gives the number of molecules of the stuff that is “used” in the chemical reaction. In your first example
Fe + O[sub]2[/sub] --> FeO
You have an equation that, when written out in words, says:
Iron reacts with Oxygen to form Iron Oxide (more correctly, it forms Iron II oxide, recall that Iron can have a +2 or +3 charge).
If you consider the number of atoms in the equation, you are currently showing that if you start with 1 atom of Iron and 2 atoms of Oxygen (bonded together in the diatomic molecule) you somehow end up with 1 atom of Iron and 1 atom of Oxygen. This can’t be correct, unless you’ve discovered some way to delete matter from existence. Where did the other atom of oxygen go?
Like the other dollar in the hotel puzzle, it didn’t “go” anywhere. The equation as written is quite qualitative; it lists formulas for the compounds/elements, but that’s it.
Since you start with 2 atoms of oxygen, you must finish with 2 atoms of oxygen. The only way this is possible, without changing subscripts (and therefore compounds) is if you actually end up with 2 molecules of FeO, as such:
Fe + O[sub]2[/sub] --> 2FeO
This shows that by having two molecules of Iron II oxide produced, all oxygen is accounted for. However, now it seems as if an atom of Iron was generated.
Recall that the the coefficient is distributed when calculating the number of atoms present. It stands to reason that if you have 2 molecules of FeO and each molecule has 1 Fe and 1 O you have a total of 2 Fe and 2 O.
The only way to end up with 2 Fe atoms is to start with 2 Fe atoms. Again, since you can’t change subscripts, you must change the coefficient on the reactant side to show 2 iron atoms.
2Fe + O[sub]2[/sub] --> 2FeO
now this shows that:
2 atoms of Iron combine with 1 molecule of Oxygen to form 2 molecules of Iron II oxide. All numbers of starting materials are accounted for after the reaction.
Things to watch out for:
Diatomic molecules (I[sub]2[/sub], Br[sub]2[/sub], Cl[sub]2[/sub], F[sub]2[/sub], O[sub]2[/sub], N[sub]2[/sub], and H[sub]2[/sub]
I’ve gone on far too long. (gosh and I didn’t even get to talk about Red-Ox reactions!!)
Feel free to print this out and show it to your Chem teacher for further explanation.
Also, feel free to e-mail me if you need more.
I’ve encountered this before…
The term “Chemical Equation” is a bit of a misnomer her. See, it isn’t the “equation” you’re used to in Algebra class. In Algebra, what you do to one side of an equation you must do to the other side.
Chem equations aren’t like that. You aren’t really “multiplying” the molecule by 2, you are stating that you have 2 of them. Subtle, yes, but a big difference. Try to look at chemical equations and balancing them as a chemistry thing and not having any association whatsoever with Algebra.
Er, what’s the formula for Magnesium Chloride?
Magnesium forms a +2 ion; Chlorine form a -1 ion.
Magnesium chloride, therefore is:
MgCl[sub]2[/sub]
I’m going home now; I’ll check back in with y’all tomorrow.
Buenos Tacos.
Thanx Spritle. That’s probably enough for the OP, but…
Then the balanced equation should be:
Mg + Cl[sub]2[/sub] —> MgCl[sub]2[/sub]
No big numbers needed.
I think I’m in love.
Just a slight addendum to Spritles answer .
Most chemical reactions either consume energy (usually provided in the form of heat) or produce energy (heat, light and several other interesting side effects). So to fully balance the equations one should add a KJ component to one side of the equation.
There are some further complications that can crop up due to
[List=A]
[li]Catalysts, which appear in the same quantities and in the same configuration on both sides of the equation[/li][li]‘Ignition energy’ needed to start a reaction.[/li][/List=A]
Unfortunately I do not have my chemistry texts handy so I can’t provide exact numbers but here is your MgCl[sub]2[/sub] equation.
Mg +Cl[sub]2[/sub] + aKJ —> MgCl[sub]2[/sub] +bKJ
I am however fairly certain that a < b is true in this case.
And of course, in real life processes, reactions don’t always go to nice, neat species. For example, while it is customary to think of hydrogen burning as being:
2H[sub]2[/sub] + O[sub]2[/sub] -> 2H[sub]2[/sub]O,
in reality, it looks more like:
100,000H[sub]2[/sub] + 50,000O[sub]2[/sub] -> aH[sub]2[/sub]O + bHO + cH[sub]2[/sub]O[sub]2[/sub] + dH[sub]2[/sub] + eO[sub]2[/sub] + fH[sub]3[/sub]O…
where “a,b,c,d,e,f” are non-zero numbers.
as many unstable and stable species are created. This will depend a tremendous amount on the Equilibrium constants of the various sub-reactions, the amount of material reacting, mixing, temperature of reaction, reaction time, and so forth.
This may not be the time to mention it, but there is a systematic method for balancing more complex equations. I will give you details if you want, but warn you that you need to memorize a fair number of valences before you can use it.
I realize you probably have to master enough to pass a class, or some such. But if you really can’t gain traction with the subject, don’t worry about it; they usually balance themselves without human intervention.
Does anyone else get the impression that we might have just done KV’s chemistry homework for him? Asking us to do the dirty work for specific examples of his choosing sounds mightily suspiscious to me. Might I just, as moderator, remind everyone that we really shouldn’t be doing other folks’ work for them?
I know this, but how are we to tell in some cases, really? This OP was worded such that it is impossible to tell, IMO, whether it is a “homework question” or not. I know two specific examples were called out - the Fe and Mg ones, but still… Perhaps the best answer is not to answer the specific examples, but to show by another example, so someone still has to learn something to apply the answer. I don’t know.
ABSOLUTELY!! That’s why (if you noticed) I went through one example, [his] first (the formula equation). I stayed away from the second (the word equation). The only reason I gave the formula for Magnesium chloride is because IJGrieve (not the OP) asked.
Also, because I figured KV was asking homework questions, I gave [him] a thorough explanation of balancing equations by inspection. I left out catalyist, spectator ions, energies and Red-ox reactions since the last thing [he] needed was more confusion.
We teachers is smart! 
BTS, Britt formula equations don’t show catalysts on both sides of the equation. If a catalyist is needed to start a reaction, it is written above the “yields” sign ( → ) in the equation. In fact, nothing is written “equally” on both sides of a reaction. Recall that a reaction expresses chemical change. If the stuff doesn’t change, it doesn’t get placed in the reaction. The perfect example of this would be spectator ions in an ionic equation. This kinda gets into Anthracite’s example of the complex organic reation. Not really, though, since Anth was trying to make a point about complex reactions more than trying to give some specific example. (In this case, H[sub]2[/sub] and O[sub]2[/sub] will ONLY combine to form H[sub]2[/sub]O.) For those of you following along at home, an example of an Organic Combustion Reaction would be the combustion of octane in your car. Octane reacts with Oxygen to form gobs of stuff, from carbon dioxide to carbon monoxide to water.
Oh, yeah, forgot…
Sue Duhnym, how YOU doin’?
and that “BTS” above should be “BTW”. The management apologizes for any inconvenience.
Possibly, but at least in this case, the OP would have to wade through the explanation to find the “answer”, probably learning more than if he had just done the homework himself. (Ha, ha! Made you learn! ;))
Nope, not a homework question, extra credit;)
Sorry about that. Adding to the confusion once again.
Given what Google turns up if you search on Chemistry,
It must be fun trying to set homework now that students can find the answers to virtually everything on-line (lets hope only the bright, motivated ones bother).
Oh and if anyone wants more information on balancing chemical equations can I recommend:
http://www.chemistry.vt.edu/chem-ed/courses/equil/toc.html