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Balancing Chemical Equations
P r eview
The lectures in this unit cover balancing & classifying chemical reactions: including redox, acid- base, and precipitation reactions. This lecture covers how to express and balance chemical equations and introduces the three general classes of chemical reactions.
Chemical Reactions
I. E xp r essi n g Chemical R e a ct ion s
We express chemical changes using chemical equations. In doing so we use our chemical formulas to express how atoms rearrange from reactants to products.
Equations are read from left to right: with the reactant(s) on the left, an arrow which tells what direction the reaction occurs in the middle, and product(s) on the right. For example, propane reacts with oxygen to form carbon dioxide and water. We write the reaction as follows:
C 3 H 8 + O 2 → CO 2 + H 2 O
Reactants Products
Before we continue, please note that elemental oxygen exists as a dimer: O 2. There are seven elements that exist naturally as diatomic species, meaning they exist as dimers of themselves. They are elemental hy d roge n, n it roge n , oxygen, fl uor in e, ch lorine, bro min e, and iod in e. You must remember these in order to use the correct form of these elements in chemical reactions!
II. Balancing Chemical E qu a t io n s
You might have been concerned with the chemical equation shown above. Let’s look at it again:
C 3 H 8 + O 2 → CO 2 + H 2 O
We know that the correct formula for propane is C 3 H 8 ; and we know that when things burn they consume oxygen and produce carbon dioxide and water. Each of the formulas given are correct. But if you look closely to the equation as written it appears there are more carbon atoms on one side than on the other. This is true of the oxygen atoms and hydrogen atoms as well! In other words, the equation is not ba la nced.
Balancing is a tool that helps us determine how many reactant molecules react to form how many product molecules. We balance an equation by systematically adding molecules or atoms on each side by inserting a coefficient in front of the atomic symbol or molecular formula. We continue to add or change these coefficients as needed until our reaction has the lowest ratio of reactants and products possible while still being completely balanced.
Balancing Chemical Equations
In the case of the previous reaction, we note that one molecule of C 3 H 8 provides three carbon atoms on the left hand side; So we must have three carbon atoms on the right hand side to balance the number of carbon atoms. We put a 3 in front of the formula on the right that contains carbon, as follows:
1 C 3H 8 + O 2 → 3 CO 2 + H 2 O
I have included the 1 in front of the C 3 H 8 molecule so you can see the relationship between the two molecules clearly. However, I need not include the 1 in front of the C 3 H 8 molecule. If there is no coefficient shown it is assumed the value is 1. Please note that we do NOT change t h e formula of the CO 2 compound. We cannot arbitrarily change that formula to C 3 O 2 to satisfy the number of carbon atoms required. Nor do we change the formula to C 3 O 6 , even though the coefficient of 3 does indicate there are now 3 carbon atoms and 6 oxygen atoms represented. That would change the identity of the product molecule. Instead we are merely stating that one molecule of C 3 H 8 must be capable of forming three molecules of CO 2 according to the reaction given, because there are three carbon atoms in each of these expressions.
Next we can balance the hydrogen atoms. The C 3 H 8 molecule on the left hand side includes 8 hydrogen atoms. So we must have 8 hydrogen atoms in the products to balance the reaction. Because there are two hydrogen atoms in every molecule of H 2 O, it needs a coefficient of 4 to represent a total balanced number of 8 hydrogen atoms:
C 3H 8 + O 2 → 3 CO 2 + 4 H 2 O
Finally we can balance our oxygen atoms. There are a total of 10 oxygen atoms on the right hand side of the reaction, given by 3 CO 2 molecules and 4 H 2 O molecules. In order to get 10 oxygen atoms on the left, I need a coefficient of 5 in front of the oxygen molecule:
C 3H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O
Now our equation is balanced. It shows 3 carbon atoms, 8 hydrogen atoms, and 10 oxygen atoms on each side.
Let’s try balancing another reaction:
FeCl 3 → Fe + Cl 2
We have 3 chlorine atoms on the left and two on the right. In this case, the easiest way to balance them is to take the lowest common multiple for the number of chlorine atoms on each side. The lowest common multiple is 6 since one side has 2 and the other has 3. So we put a 2 in front^ of the FeCl 3 and a 3 in front of the Cl 2. That balances our chlorine atoms like so:
2 FeCl 3 → Fe + 3 Cl 2
Balancing Chemical Equations
eliminate the fraction by multiplying the entire equation by 2. This retains the completely balanced nature of the equation while eliminating the fraction involved.
2 C 4H 10 + 13 O 2 → 8 CO 2 + 10 H 2 O
And we are done!
Let’s do one more equation using polyatomic ions:
Ca(OH) 2 + H 2 SO 4 → CaSO 4 + H 2 O
This equation has two polyatomic ions on the left side: a hydroxy ion in the Ca(OH) 2 and a sulfate ion in the H 2 SO 4. On the right side we have another compound with a sulfate ion: CaSO 4. Because the SO 4 is on both sides of the equation we can balance it as a whole entity. Because the hydroxide ion does not appear on the right hand side, we cannot do the same for it.
As written, the equation has one sulfate ion on the left and one on the right. These are already balanced. We can track our progress in analyzing the reaction like so:
Ca(OH) 2 + 1 H2SO 4 → 1 CaSO 4 + H 2 O
We may also notice that there is 1 calcium ion on the left and 1 on the right – so these are already balanced against each other, like so:
1 Ca(OH) 2 + 1 H2SO 4 → 1 CaSO 4 + H 2 O
All that remains is to determine how many water molecules we need to balance the hydrogen and oxygen atoms. The left hand side of the equations contains 4 hydrogen atoms and 2 oxygen atoms (we are not counting the oxygen atoms in the sulfate, they were already balanced!). So we need a coefficient of 2 in front of our water and we are done:
Ca(OH) 2 + H 2 SO 4 → CaSO 4 + 2 H 2 O
Balancing equations is mostly a matter of trial and error. There is no one correct way to do it: the most important thing is getting the right answer! Checking your work is as simple as counting atoms.
Take a minute to complete section I of your Chemical Equations worksheet. This has you balance many chemical reactions. Lots of practice makes this easier!
Balancing Chemical Equations
III. Classifying Chemical R e a ct ion s
There are many thousands of chemical reactions known, so it becomes useful to try and broadly categorize them into general classes of reactions. The following are three general classification types that help us understand chemical reactions.
1) Oxidation-Reduction reactions occur when elements either gain or lose electrons.
If the charge on an atom changes, it has undergone oxidation or reduction and
that process or reaction is an oxidation-reduction reaction. All reactions
involving covalent compounds are oxidation-reduction reactions.
Examples:
Mg + O 2 → Mg O
Al 2 O 3 → Al + O 2
C 4 H 10 O + O 2 → CO 2 + H 2 O
Reactions where neutral elements make ionic compounds (also known as synthesis or combination reactions).
Reactions where ionic compounds revert to the neutral elements (also known as decomposition reactions).
Reactions where atoms in covalent compounds are rearranged. (The example here is also known as a combustion reaction).
2) Precipitation reactions occur when soluble elements in an aqueous solution
combine to form an insoluble compound. Here we use phases to show what is
happening. The soluble elements are given a designation of (aq) to indicate they
are dissolved in aqueous, or water-based, solution. The insoluble compound is
given a designation of (s) to indicate it is a solid and thus not part of the solution.
Examples:
Pb(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) (^) → PbCO 3 (s) + Na NO 3 ( a q)
In this reaction solid PbCO 3 is the insoluble solid formed. It is the precipitate.
K 2 SO 4 (aq) + BaCl 2 (aq) (^) → KCl(aq) + B a S O 4 (s)
In this reaction solid BaSO 4 is the insoluble solid formed. It is the precipitate.
