Lecture 10

Chemical Equations and Stoichiometry

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:

CH3CH2OH + 3 O2 ---> 2 CO2 + 3 H2O

On the left we have one molecule of ethanol (CH3CH2OH) and 3 molecules of oxygen (O2). On the right we have two molecules of carbon dioxide (CO2) and three molecules of water (H2O). 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 (P4) reacts with fluorine (F2) to give PF5. 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.

P4 + F2 ---> PF5 (not balanced)

It is clear that one P4 molecule can produce 4 molecules of PF5. But if we have 4 molecules of PF5, we need 20 F atoms, or 10 F2 molecules. Thus the balanced equation is

P4 + 10 F2 ---> 4 PF5 (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 ---> Bd2Tk3

Now, if I want to make 200 of these Bd2Tk3 sandwiches, how much Bd and Tk do I need?

200 Bd2Tk3 x 2 Bd / 1 Bd2Tk3 = 400 Bd

200 Bd2Tk3 x 3 Tk / 1 Bd2Tk3 = 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(NH3)2Cl2. It is made by the following reaction:

(NH4)2PtCl4 + 2 NH3 ---> Pt(NH3)2Cl2 + 2 NH4Cl

This equation tells me that if I want to make a single Pt(NH3)2Cl2 molecule, I need to take one formula unit of (NH4)2PtCl4 and 2 molecules of NH3. (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 1023 molecules (1 mole!) Thus I would need 6.022 x 1023 units of (NH4)2PtCl4 (1 mole) and 2 x 6.022 x 1023 molecules (2 moles) of NH3.

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:

(NH4)2PtCl4 372.97 g/mol
NH3 17.03 g/mol
Pt(NH3)2Cl2 300.05 g/mol

So if I want to make one mole (300.05 g) of cis-platin, I will need one mole (372.97 g) of (NH4)2PtCl4 and two moles of NH3 (34.06 g).


F. Stoichiometric Factors

Look at the equation for preparing cis-platin again:

(NH4)2PtCl4 + 2 NH3 ---> Pt(NH3)2Cl2 + 2 NH4Cl

The ratio of the stoichiometric coefficients, 2 mol NH3 / mol (NH4)2PtCl4, is called the stoichiometric ratio or stoichiometric factor for this particular reaction. Other stoichiometric factors for this reaction are:

2 mol NH3 / mol Pt(NH3)2Cl2
2 mol NH3 / 2 mol NH4Cl
and so on...

Example: Suppose we want to make 100.0 g Pt(NH3)2Cl2. What mass of NH3 will be required?

Notice that we used the stoichiometric factor 2 mol NH3 / mol Pt(NH3)2Cl2 to solve this problem, after first converting the mass of Pt(NH3)2Cl2 to moles.


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Created September 11, 2000 by M. F. Richardson
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