A. The meaning of a balanced chemical equation
A chemical equation is said to be balanced when there are the same numbers of each atom on the left side (the reactants) as on the right side (products). This is a consequence of the Law of Conservation of Matter: matter can neither be created nor destroyed. The equation is a "recipe" that tells you exactly how much of each ingredient you have to use in order to get the product.
B. Two important types of reactions
There are many ways to classify chemical reactions. For simplicity, let's just consider two of them now. These are combination reactions and combustion reactions.
C. Balancing Equations
Note that in each of the equations in part B, the number and type of atoms on the left-hand side of the equation is equal to the number and type on the right hand side. This is what it means to have a balanced equation.
For example, in the combustion of ethanol:
CH_{3}CH_{2}OH + 3 O_{2} ---> 2 CO_{2} + 3 H_{2}O
On the left we have one molecule of ethanol (CH_{3}CH_{2}OH) and 3 molecules of oxygen (O_{2}). On the right we have two molecules of carbon dioxide (CO_{2}) and three molecules of water (H_{2}O). So let's count atoms:
Left hand side: 2 C + 6 H + 1 O from ethanol, 6 O from 3 oxygen molecules
= 2 C + 6 H + 7 O
Right-hand side: 2 C + 4 O from two carbon dioxide molecules, 6 H and 3 O from 3 water molecules
= 2 C + 6 H + 7 O
The coefficients in a balanced equation are called the stoichiometric coefficients.
Many combination reactions can be balanced "by inspection." For example, white phosphorus (P_{4}) reacts with fluorine (F_{2}) to give PF_{5}. We can express this reaction by means of the "skeleton equation" (unbalanced) that just shows the reactants and the products without the correct numbers of each.
P_{4} + F_{2} ---> PF_{5} (not balanced)
It is clear that one P_{4} molecule can produce 4 molecules of PF_{5}. But if we have 4 molecules of PF_{5}, we need 20 F atoms, or 10 F_{2} molecules. Thus the balanced equation is
P_{4} + 10 F_{2} ---> 4 PF_{5} (balanced)
The rules for balancing combustion reactions are given in Kotz and Treichel, pp. 151-153 (4th ed.)
D. Use of the Balanced Equation: A Simple Example
Let's suppose I want to make sandwiches, each consisting of 2 slices of bread (Bd) and 3 slices of turkey (Tk). My "balanced equation" tells me how to make one sandwich:
2 Bd + 3 Tk ---> Bd_{2}Tk_{3}
Now, if I want to make 200 of these Bd_{2}Tk_{3 }sandwiches, how much Bd and Tk do I need?
200 Bd_{2}Tk_{3} x 2 Bd / 1 Bd_{2}Tk_{3} = 400 Bd
200 Bd_{2}Tk_{3} x 3 Tk / 1 Bd_{2}Tk_{3 }= 600 Tk
However, Bd is sold in loaves that have 20 Bd/loaf, and Tk comes in packages that have 8 slices/pkg. So I need to buy these quantities:
400 Bd x 1 loaf / 20 Bd = 20 loaves of Bd
600 Tk x 1 pkg / 8 Tk = 75 pkg Tk
E. Use of the Balanced Equation: A Chemical Example
The anticancer drug cis-platin has the composition Pt(NH_{3})_{2}Cl_{2}. It is made by the following reaction:
(NH_{4})_{2}PtCl_{4} + 2 NH_{3} ---> Pt(NH_{3})_{2}Cl_{2} + 2 NH_{4}Cl
This equation tells me that if I want to make a single Pt(NH_{3})_{2}Cl_{2} molecule, I need to take one formula unit of (NH_{4})_{2}PtCl_{4} and 2 molecules of NH_{3}. (This is like making a single sandwich). But I want to make enough to be useful for a cancer patient, so I'm likely to need not one molecule, but 6.022 x 10^{23} molecules (1 mole!) Thus I would need 6.022 x 10^{23} units of (NH_{4})_{2}PtCl_{4} (1 mole) and 2 x 6.022 x 10^{23 }molecules (2 moles) of NH_{3}.
Just as in our sandwich analogy in part D, we have to consider the "packaging" of our reactants. We don't count individual molecules, we count by weighing. The formula weights (molar masses) of our reactants and product are:
(NH_{4})_{2}PtCl_{4} 372.97 g/mol
NH_{3} 17.03 g/mol
Pt(NH_{3})_{2}Cl_{2} 300.05 g/molSo if I want to make one mole (300.05 g) of cis-platin, I will need one mole (372.97 g) of (NH_{4})_{2}PtCl_{4} and two moles of NH_{3} (34.06 g).
F. Stoichiometric Factors
Look at the equation for preparing cis-platin again:
(NH_{4})_{2}PtCl_{4} + 2 NH_{3} ---> Pt(NH_{3})_{2}Cl_{2} + 2 NH_{4}Cl
The ratio of the stoichiometric coefficients, 2 mol NH_{3} / mol (NH_{4})_{2}PtCl_{4}, is called the stoichiometric ratio or stoichiometric factor for this particular reaction. Other stoichiometric factors for this reaction are:
2 mol NH_{3} / mol Pt(NH_{3})_{2}Cl_{2 }2 mol NH_{3} / 2 mol NH_{4}Cl
and so on...Example: Suppose we want to make 100.0 g Pt(NH_{3})_{2}Cl_{2}. What mass of NH_{3} will be required?
Notice that we used the stoichiometric factor 2 mol NH_{3} / mol Pt(NH_{3})_{2}Cl_{2 }to solve this problem, after first converting the mass of Pt(NH_{3})_{2}Cl_{2} to moles.
This page is
http://chemiris.labs.brocku.ca/~chemweb/courses/chem180/CHEM1P80_Lecture_10.html
Created September 11, 2000 by M. F. Richardson
© Brock University, 2000