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 Daltons 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 ¹ CuCl2
H2O ¹ H2O2
3C2H2 ¹ 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 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
|1 C||1 C|
|4 O||4 O|
|4 H||4 H|
Example 2 - Combustion of Octane
Lets 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 ½ 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 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