I'm in chemistry and I still have a hard time comprehending how to balance a chemical equation. What are the main steps and general things to know? thank you.
Basically, balanced chemical reactions must 1) have the same number of each element on each side, and 2) have the same net charge on each side. Consider the reaction $H_2 + O_2 → H_2O$. It is unbalanced because there are two oxygen atoms on the left side but only one on the right. If we assign variable coefficients to each reactant and product, $a H_2 + b O_2 → c H_2O$, we can set up a system of equations by multiplying the variables by the number of atoms in each species. For oxygen, $2b = c$ and for hydrogen, $2a = 2c$. While the system has infinitely many solutions, we take the lowest integer solution. Thus, the balanced equation is $2 H_2 + O_2 → 2 H_2O$.
Chemical reactions involving charges are a bit more complicated. A good explanation can be found this tutorial.
OK, here goes a semester of college chemistry in one post...
First off, remember the law of conservation of matter; in chemical reactions, matter is neither created nor destroyed. Therefore, every atom of every element on the left side of the arrow has to appear on the right side, with all their protons and neutrons intact. No strictly chemical reaction will cause a carbon to become a nitrogen by gaining a proton.
What can change in a reaction is the number of electrons each atom "owns". For that, remember the concept of valence shells; the number of electrons in the outermost "shell" of its orbitals. Most elements do not have a stable valence shell, and so will want to react in ways that increase their stability. For elements not in the "transition metal" groups, this is pretty straightforward; elements to the left of the transition metals generally want to lose electrons to empty their valence shell (acquiring the stable configuration of the next lightest noble gas), while elements to the right of the transition metals want to share or gain electrons to fill their valence shell, acquiring the configuration of the next heaviest noble gas. The transition metals are "stable" in various valence configurations based on their more complex d-orbital valence structure (the s and p orbitals exposed by the first two and last six columns of the table are more predictable).
When atoms outright gain or lose electrons, they become ions, and gain a charge based on the difference between the protons and electrons in their configuration. When they gain them, they become negative anions, and then they lose them, positive cations. Cations are then attracted to anions, and these charges must balance; a +2 cation such as Mg2+ or Ca2+ requires one -2 anion (carbonate, sulfate, hydrogen-phosphate) or two -1 anions (chloride, fluoride, bicarbonate).
The alkali metals, in the 1 and 2 groups, usually want to lose their one or two valence electrons; the heavier the alkali metal, the more the atom wants to lose the electrons, with cesium being the most reactive alkali metal that has a non-radioactive isotope. The elements in the halogen group (fluorine, chlorine, bromine, iodine) like to gain electrons, especially from the alkalis. The lighter the compound, the more it wants to gain an electron, with fluorine being the most aggressive electron receptor in the table (to the point where it will try to take electrons from other elements that don't want to give them up, like other halogens and noble gases).
Columns between the halogens and transition metals (from left to right, the boron group or triels, the carbon group or tetragens, nitrogen group or pnictogens, and oxygen group or chalcogens) are more of the "sharing" type; they'll take electrons outright from the alkalis, and if they really have to they'll give them up to a halogen like fluorine or chlorine, but they prefer a "give-and-take" relationship, and form shared-orbital "covalent bonds" with most other elements of these groups and with the transition metals. How strong these bonds are depend on what elements are being bonded and how many electrons each is sharing with its neighbor. A double or triple bond is very strong, while certain other combinations are strong for other reasons (bonds between carbon and the "Big 3" highly electronegative elements, namely oxygen, fluorine and chlorine, are commonly very strong). You can compute a balanced bond structure using valence electrons, much like calculating ionic charges; oxygen needs two valence electrons and so it will form no more than two bonds with other elements. When two atoms in these columns get together, expect to see a few multiple bonds; oxygen and nitrogen love forming double and triple bonds when they can, and carbon will oblige with these two and a few others, but typically prefers single bonds with other carbons and most other nonmetals (allowing it to form those long chains that create the backbone of life).
When determining what the reaction is, keep in mind that there has to be a driving force that causes an equilibrium shift. This can be a number of things, but basically, the reactants have to want to react and produce a product more than they want to stay where they are, and also more than any other component of their environment wants them not to react. Compounds generally want to react when the bonds of the newly-formed compounds are of lower energy and therefore more stable; such reactions are exothermic. Compounds can be forced to react when additional energy is added to the system (usually in the form of heat, sometimes with electricity to get the electrons jumping).
Compounds also generally want to react when they're attracted so strongly that nothing can keep them apart, so learn your solubility rules. Sodium chloride and aluminum bromide will both sit in solution together quite happily as dissociated ions; you can call some of it sodium bromide and some of it aluminum chloride, but because all four possible products are soluble and no electrons change hands, the solution is already in equilibrium and so there's no reaction. However, calcium chloride and sodium carbonate will react because one of the possible cation/anion combinations, calcium carbonate, is insoluble in water; these two will meet, bond, and precipitate out, leaving sodium chloride in the water.
From a paper I wrote on the algebraic method for balancing chemical equations:
Instead of the inspection method, shown in most chemistry texts, a more general method of balancing chemical reaction equations is the algebraic method, proposed by the Russian academician, V. A. Kistiakovskij, in 1919. Let’s illustrate the method using the following chemical reaction, which is a complete oxidation (or burning) of pentane fuel in pure oxygen to give carbon dioxide and water as the only products:
C5H12 + O2 = CO2 + H2O
The idea is to write down the answer, only using algebraic letters instead of numbers, and then solve the system of equations resulting to obtain the numerical values of the coefficients:
aC5H12 + bO2 = cCO2 + dH2O
It turns out that most of the time, there is one more variable (the algebraic coefficients we set) than equations (different atom balances), and in these cases, we can set a = 1 and solve the equations using this starting value. This works because as all good chemists know, the compounds in a chemical equation react and form in ratio amounts, always (the mole concept).
The next step in the process is to set up the system of equations to solve, which are just balances of atoms of each element on the left and right sides of the original equation (required according to the law of conservation of mass).
In each case, the number of atoms of a particular element is equal to the coefficient in front of the compound containing that element multiplied by the subscript on the element in that compound, added to any other instances of the element in other compounds on either side of the equal sign in the original equation.
For the example, we have balances on carbon (C), hydrogen (H) and oxygen (O):
C: 5a = c
H: 12a = 2d
O: 2b = 2c + d
Substituting a = 1, we have:
5 = c (so c = 5)
12 = 2d (so d = 6)
2b = 2•5 + 6 = 16 (so b = 8)
Substituting the values obtained from the coefficients back into the original equation gives:
C5H12 + 8O2 = 5CO2 + 6H2O
The last step is to do a quick check to make sure our algebra was done correctly:
C: 5 = 5 (check)
H: 12 = 6•2 (check)
O: 8•2 = 5•2 + 6 (check)
This method can be employed for the vast majority of chemical reactions studied in high school and even beginning college chemistry, in both the inorganic and organic worlds. More importantly, it can be automated, and so it is used as the algorithm that does the calculating for on-line and off-line, computer-based chemical equation balancers. But it is simple enough that for most chemical reactions studied in high school, it can be used by students with some knowledge of algebra, given a pencil and paper, and for more difficult reactions, can be completed faster than the inspection method.
It should be noted that fractional coefficients are often obtained using the substitution a = 1 shown above. Since the convention is that coefficients of chemical equations are whole numbers, to get the final answer it is often required to multiply the set of coefficients through by the lowest common multiple (LCM or LCD) of all the denominators of the fractional values.
An example used in several papers in the Journal of Chemical Education from the 1930s considers the famous experiment of putting a copper penny in a flask of nitric acid, which dissolves the penny, producing copper nitrate, while giving off nitric oxide (which quickly oxidizes in air to the brown gas nitrogen dioxide):
Cu + HNO3 = NO + Cu(NO3)2 + H2O
Using the algebraic method, we write down the answer using algebraic letters as coefficients:
aCu + bHNO3 = cNO + dCu(NO3)2 + eH2O
Then do the atomic balances on each element (I always go from left to right):
Cu (copper): a = d
H (hydrogen): b = 2e
N (nitrogen): b = c + 2d (remember that the subscripts occurring after parentheses are distributed into the elements inside like exponents in algebra)
O (oxygen): 3b = c + 6d + e
Now set a = 1 and solve:
1 = d (so d = 1)
Substituting b = 2e and d = 1 into the nitrogen and oxygen balances gives:
5e = c + 6 (oxygen balance) – 2e = c + 2 (nitrogen balance) 3e = 0 + 4 (subtracting the bottom equation from the top one)
So e = 4/3, and therefore b = 8/3 (from the hydrogen balance) and c = b – 2d (from the nitrogen balance) or c = 8/3 – 2 = 2/3. The coefficients are therefore:
a = 1
b = 8/3
c = 2/3
d = 1
e = 4/3
The LCM of the denominators of the fractions is 3, so we multiply through by 3, giving:
a = 3
b = 8
c = 2
d = 3
e = 4
The balanced equation is:
3Cu + 8HNO3 = 2NO + 3Cu(NO3)2 + 4H2O
Our quick check shows:
Cu: 3 =3 (check)
H: 8 = 4•2 (check)
N: 8 = 2 + 2•3 (check)
O: 8•3 = 2 + 6•3 + 4 (check)