Observations in Science
3.2 Chemical Formulas and Equations
Although the properties and uses of different explosives vary considerably, all involve chemical reactions. A realistic description of the chemistry of virtually any explosion would require us to consider many individual reactions. To discuss the science behind an explosion, we must be able to describe a chemical reaction concisely. In Chapter 2, we’ve already seen how chemical formulas provide a concise way to represent chemi- cal compounds, and we will now describe how chemical equations build on this to accomplish the same goal for chemical reactions.
Writing Chemical Equations
Chemical equations are designed to represent the transformation of one or more chemical species into new substances. To ensure that their meanings are clear, we follow a set of conventions when writing chemical equations. Each chemical equa- tion has two sides, and we usually envision the reaction as proceeding from left to right. The original materials are called the reactants and they appear on the left-hand side of the equation. The compounds that are formed from the reaction are called products and appear on the right-hand side of the equation. An arrow is used to rep- resent the changes that occur during the reaction. Thus we can write a completely generic chemical equation:
Reactants : Products
This would usually be read as “reactants go to products” or “reactants give products.”
We use chemical formulas to identify the specifi c reactants and products. The physical states of the compounds are often designated; (s) indicates a solid, (,) a liquid, (g) a gas, and (aq) a substance dissolved in water. We can write the reac- tion of hydrogen with oxygen, which is often done as a lecture demonstration, as a simple example.
2 H2(g) + O2(g) : 2 H2O(g)
All substances involved in this equation are gases, as indicated by the “(g)” nota- tion. One characteristic of this reaction is that it will only occur at high tempera- tures. As seen in the photos in Figure 3.3, the mixture of H2 and O2 is stable unless ignited. If our written equations are going to describe fully the chemical processes they represent, then we need ways to indicate such conditions in addition to identify- ing the chemicals involved. Generally, this is done by placing a symbol over the arrow.
Here we will introduce two such symbols: one for heat and one for light. A reaction that requires heat (or high temperatures) is indicated with a Δ (the Greek letter delta)
When a substance dissolves in water, the result is called an aqueous solution, and the “aq” label is short for aqueous.
When a substance dissolves in water, the result is called an aqueous solution, and the “aq” label is short for aqueous.
2 H2(g) ⫹ O2(g)
⫹
Reactants Products
2 H2O(g)
Figure 3.3 ❚ A balloon fi lled with a mixture of H2 and O2 explodes in a classroom demonstration. The gas mixture in the balloon is stable until ignited by the fl ame from a candle.
Once ignited, the balloon explodes in a ball of fl ame as the reaction takes place. The accompanying microscopic scale illustration shows the molecular species present before and after reaction.
Charles D. WintersPhotos: © Cengage Learning/
above the arrow, and one that is initiated by light energy is designated with an hn. We show two examples of these symbols below.
2 H2(g) + O2(g) 9:Δ 2 H2O(g) H2(g) + Cl2(g) 9:hn 2 HCl(g)
The fi rst equation is a more complete description of the reaction between hydro- gen and oxygen. In the second equation, light energy is used to initiate the reaction of hydrogen with chlorine, producing hydrogen chloride. Such a light-induced process is called a photochemical reaction.
Balancing Chemical Equations
If you look carefully at the reactions we’ve been writing, you’ll notice more than just the formulas and the states of the substances in the equation. Numerical information about the relative amounts of the substances involved is also given. For example, we included the number “2” in front of both the H2(g) and H2O(g) in the equation for the reaction between hydrogen and oxygen. Why are these numbers there, and what do they mean? If you’ve studied chemistry before in high school, you probably remember that these coeffi cients are often needed to balance a chemical equation.
The underlying premise of the chemical equation is that it is a written representa- tion of a chemical reaction. So any reasonable representation must be consistent with all of our observations of the actual reaction. One of the most fundamental laws of nature is the law of conservation of matter: matter is neither created nor destroyed.
If we specifi cally exclude nuclear reactions from our consideration, this law can be phrased more specifi cally. Atoms of one element are neither created nor destroyed in a chemical reaction. A chemical reaction simply rearranges the atoms present into new compounds. In its written representation of nature, therefore, the chemical equa- tion must not “create or destroy” atoms. To uphold this condition, we must have the same number of atoms of each element on both sides of the chemical equation (see Figure 3.4). An equation that does not meet this condition cannot accurately repre- sent the observed chemical reaction and is said to be unbalanced.
In many cases, the most effi cient way to balance the equations is “by inspection,”
which really means by trial and error. Even when this exploratory method is used, however, some systematic strategies can help make the balancing process easier. In all cases, the way we obtain balance is to introduce numbers in front of appropriate The symbol hn is used because it
represents the energy of the light, as we will learn in Chapter 6.
The symbol hn is used because it represents the energy of the light, as we will learn in Chapter 6.
If we also consider nuclear reactions, we must acknowledge that matter and energy can be interconverted. Nuclear chemistry is discussed in Chapter 14.
If we also consider nuclear reactions, we must acknowledge that matter and energy can be interconverted. Nuclear chemistry is discussed in Chapter 14.
Equation
Reactants Products
Molecules
Atoms
Four hydrogen
atoms
Four hydrogen
atoms Four
oxygen atoms
Four oxygen
atoms One
carbon atom
One carbon
atom
2 O2 CO2
+
2 H2O CH4+
Figure 3.4 ❚ The chemical reaction for the burning of methane (CH4) in oxygen illustrates the concept of atom balance for chemical equations. The equation is shown fi rst in symbolic form, with each compound represented by its chemical formula. The individual molecules are pictured in the center row and then are broken down into their constituent atoms. The number of atoms of each element is the same on the left- and right-hand sides.
formulas in the equation. Chemists use the term stoichiometry to refer to the vari- ous quantitative relationships between the amounts of reactants and products in a chemical reaction. So the numbers used to balance a chemical equation are called stoichiometric coeffi cients. The stoichiometric coeffi cient multiplies the number of atoms of each element in the formula unit of the compound that it precedes. Example Problem 3.1 shows one strategy for balancing a common class of reactions, the com- bustion of hydrocarbon fuels. Although these combustion reactions obviously release energy, in most instances they are carried out under conditions at which the reaction is not fast enough to produce a full-fl edged explosion.
E X A M P L E P RO B L E M 3 .1
Propane, C3H8, is used as a fuel for gas barbecue grills, where it burns in a controlled fashion. But if a mixture of propane and air is ignited in a closed space, like a gas pipeline, an explosion can easily result. In either of these cases, the propane combines with oxygen, O2, to form carbon dioxide and water. Write a balanced chemical equation describing this reaction.
Strategy This problem requires two steps. First we must read the problem and deter- mine which substances are reactants and which are products. This will allow us to write an unbalanced “skeleton” equation. Then we must proceed to balance the equation, making sure that the same number of atoms of each element appears on both sides.
Solution
Step 1: The problem notes that propane burns or explodes by combining with oxygen.
The reactants are therefore C3H8 and O2. We also are told that the products are CO2 and H2O. So we have enough information to write the unbalanced equation. (Here we will write “blanks” in front of each formula to emphasize that we still need to determine the coeffi cients.)
––––C3H8 + ––––O2 ––––CO2 + ––––H2O
Step 2: Balancing an equation like this is aided by making some observations. In this case, both carbon and hydrogen appear in only one place on each side of the equation.
Our fi rst steps will be to balance these two elements, because there will be no other way to adjust them. Let’s begin with carbon. (This choice is arbitrary.) To achieve the required balance, we can use stoichiometric coeffi cients of one for C3H8 and three for CO2, giving three carbon atoms on each side of the equation. At this point, we have
1 C3H8 + ––––O2 3 CO2 + ––––H2O
(The coeffi cient of one in front of the propane would normally be omitted, but is written explicitly here for emphasis.) Next we will balance hydrogen. The propane already has its coeffi cient set from our work on carbon, so it dictates that there are
eight H atoms to be accounted for by water on the product side of the equation. We need to insert a coeffi cient of four to achieve this balance.
1 C3H8 + ––––O2 3 CO2 + 4 H2O
This leaves us with only oxygen to balance. We have saved this for last because it appears in three compounds rather than two and also because it appears as an element rather than a compound on the left side of the equation, which makes it easy to adjust the number of oxygen atoms without upsetting the balance of the other elements.
Because there are ten O atoms on the product side of the equation, we need ten on the reactant side. We can easily achieve this by using 10/2 = 5 for the stoichiometric coeffi cient of the O2 reactant. This gives us the balanced chemical equation,
1 C3H8 + 5 O2 3 CO2 + 4 H2O
Normally, this would be written without showing the coeffi cient of one in front of the propane:
C3H8+ 5 O2: 3 CO2+ 4 H2O
Analyze Your Answer Our fi nal equation has three carbon atoms, eight hydrogen atoms, and ten oxygen atoms on each side, so it is balanced.
Check Your Understanding The reaction we wrote above describes the com- plete combustion of propane. Under many conditions, though, hydrocarbons such as propane may not burn cleanly or completely, and other products may be formed. One such possible reaction is the formation of formaldehyde (CH2O) and water. Write a balanced chemical equation for a reaction in which propane and oxygen gas form CH2O and H2O as products.