Chemical Reactions

Chemical Reaction – Involves a rearrangement of atoms, one or more substance is destroyed (reactant) and one or more substance is created (product).

Conservation of Mass – Atoms are neither created nor destroyed during a chemical reaction.

Balanced Chemical Equation – Tells us which substances are the products and which substances are the reactants, and the relative amounts of each.

Balancing Chemical Equations

In class demo Electrolysis of Water

The reaction that is taking place right now is the decomposition of water into hydrogen and oxygen.

Water Hydrogen + Oxygen

In this example water is the reactant (it will be completely destroyed if the reaction goes to completion), while hydrogen and oxygen are the products (they are created by this chemical reaction).

However, substances are described in chemical equations by chemical formulas, not by names. So we might write:

H2O H + O

Is this correct?

No. We have already learned that both hydrogen and oxygen exist as diatomic gases

H2O H2 + O2

Is this correct?

No. The equation as written above implies that one atom of oxygen is created by the reaction, this is a violation of Dalton’s conservation of mass. Therefore, we must balance the equation to satisfy the conservation of mass principle.

2H2O 2H2 + O2

It is very important to realize that the subscripts describe the atomic ratio within a given substance, these numbers are never changed when balancing a chemical equation. For example

CuCl CuCl2

H2O H2O2

3C2H2 C6H6

On the other hand changing the coefficients simply means that we are changing the relative amount of each substance. In the above example the equation tells us that for every water molecule that is destroyed, two molecules of hydrogen and one molecule of oxygen is created.

The coefficients in a balanced chemical equation tell us the stoichiometry of the reaction.

Combustion Reactions

Combustion reactions are classified as rapid reactions that produce a flame. Usually, a combustion reaction refers to the reaction of a substance with O2.

A very important class of reactions is the combustion of hydrocarbons (compounds containing hydrogen and oxygen) to form CO2 and H2O.

Example 1 - Combustion of Methane

In class demo – Combustion of Methane

Lets try to write a balanced equation for this reaction

CH4 + O2 CO2 + H2O

We will balance equations by trial and error. Usually, the best thing to do is to start with elements that only appear in one of the reactants and one of the products, if possible. In this case that would be C and H.

CH4 + O2 CO2 + 2H2O

Now we can look to balance the numbers of oxygen atoms.

CH4 + 2O2 CO2 + 2H2O

Lets check to see if the equation is balanced

Products Reactants
1 C 1 C
4 O 4 O
4 H 4 H


Example 2 - Combustion of Octane

Let’s look at the combustion of octane, which is the primary reaction in an automobile engine.

C8H18 + O2 CO2 + H2O

Balancing for carbon and hydrogen

C8H18 + O2 8CO2 + 9H2O

And then for oxygen

C8H18 + 12.5O2 8CO2 + 9H2O

It is not possible to have of a molecule so we multiply by a common factor to make all of the coefficients whole numbers

2C8H18 + 25O2 16CO2 + 18H2O


Example 3 - Combustion of Ethanol

We can also balance combustion reactions when the hydrocarbon contains oxygen. Consider the combustion of ethanol that I performed on the first day of class.

C2H6O + O2 CO2 + H2O

C2H6O + O2 2CO2 + 3H2O

C2H6O + 3O2 2CO2 + 3H2O

This combustion reaction also has an important commercial application because in addition to its use in alchoholic beverages, ethanol is also produced from agricultural crops and added to gasoline. This decreases the amount of crude oil which must be imported and reduces the exhaust emissions.

Combustion Analysis

Combustion reactions are also used to determine empirical formulas of compounds containing only carbon, hydrogen and oxygen. This analysis works on the following principles:


A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the empirical formula? If the compound has a molar mass of 156 g/mol what is the molecular formula?

0.2829 g CO2 (1 mol CO2/44.01 g CO2) (1 mol C/1 mol CO2) (12.01 g C/1 mol C)

= 0.07714 g C

0.1159 g H2O (1 mol H2O/18.02 g H2O) (2 mol H/1 mol H2O) (1.008 g H/1 mol H)

= 0.01297 g H

Mass Oxygen = Total Mass – Mass Carbon – Mass Hydrogen

Mass Oxygen = 0.1005 g – 0.07714 g – 0.01297 g

Mass Oxygen = 0.0104 g

Now we have the mass % information, the final step is to convert from mass % to an empirical formula.

0.07714 g C (1 mol C/12.01 g C) = 6.423 10-3 mol C ~ 10

0.01297 g H (1 mol H/1.008 g H) = 0.01287 mol H ~ 20

0.0104 g O (1 mol O/16.00 g O) = 6.50 10-4 mol O ~ 1

The empirical formula is C10H20O.

The MW corresponding to the empirical formula is

MW = 10 (12.01) + 20 (1.008) + 1 (16.00) = 156.3

Therefore C10H20O is also the molecular formula.


Combination and Decomposition Reactions

Earlier in class I demonstrated the electrolysis of water. This is an example of a decomposition reaction, where one substance reacts to produce two or more products.

Another example might be the decomposition of sodium azide NaN3, which is the mechanism responsible for inflating an air bag

NaN3(s) Na(s) + N2(g)


2NaN3(s) 2Na(s) + 3N2(g)

Note the symbols in parentheses refer to the state of each product/reactant

The opposite of a decomposition reaction is called a combination reaction, where two or more reactants combine to form a single product. For example:

2H2 + O2 2H2O