View source for Balancing chemical equations

The purpose of balancing a chemical equation is to make sure that

the number of atoms of each type on the LEFT-HAND-SIDE 	= the number of atoms of each type on the RIGHT-HAND-SIDE

This is important because of the [[Law of conservation of mass|Law of Conservation of Mass]]. 

In the balancing of chemical equations, we are only allowed to insert coefficients to make the number of each type of atoms on both sides of the equation equal. You are not allowed to change the subscripts in the chemical formulas of the compounds.

Follow the two examples to balance the following chemical reaction:

:::When propyl alcohol,  CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH, burns in air, it reacts with oxygen, O<sub>2</sub>, to form carbon dioxide, CO<sub>2</sub>, and water, H<sub>2</sub>O. 
Write the balanced chemical reaction by the two methods:
# [[Balancing chemical equations#Example - balance by inspection|by inspection]]
# [[Balancing chemical equations#Example - balance algebraically|algebraically]]



==Example - balance by inspection==
Step 1: Identify the reactants and products. 
 Reactants: propyl alcohol, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH, oxygen, O<sub>2</sub>
 Products:  carbon dioxide, CO<sub>2</sub>, water, H<sub>2</sub>O

Step 2: Write the unbalanced chemical equation. 
  CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + O<sub>2</sub> (g) [[File:reactionarrow.gif]] CO<sub>2</sub> (g) + H<sub>2</sub>O (l)

Step 3: Count the number of each type of atoms on both sides of the arrow. 

{|width="400"
|
|<b>No. of C</b>
|<b>No. of H</b>	
|<b>No. of O</b>
|-
|<b>left-hand-side</b>	
|3
|8
|3
|-
|<b>right-hand-side</b>	
|1
|2
|3
|}

Step 4: Decide the order of atom(s) to be balanced.

a. Start with the compound that is made up of the most different kinds of elements. 
  propyl alcohol, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH.	

b. Pick the element(s) within the compound that appear only in one substance on each side of the arrow and balance it (them). 	
  carbon, C
  hydrogen, H
  (Oxygen appears in two substances on the right-hand-side, we leave that until later.)

c. Pick the element(s) within the compound that appear more than once on each side of the arrow and balance it (them).
  oxygen, O 	

d. Balance uncombined elements last.
   oxygen, O<sub>2</sub>

Step 4 summary: Balance the atoms on each side of the equation, by inserting coefficients in front of the substances in the  following order:

# carbon
# hydrogen
# oxygen

Step 5: Balance each type of atoms. 
 Balance carbon: CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red">3</font> CO<sub>2</sub> (g) + H<sub>2</sub>O (l)
 Balance hydrogen: CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red">3</font> CO<sub>2</sub> (g) +  <font color="red">4</font> H<sub>2</sub>O (l)
 Balance oxygen: CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + <font color="red">9/2</font> O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red">3</font> CO<sub>2</sub> (g) +  <font color="red">4</font> H<sub>2</sub>O (l)

Step 6:	Make sure that all the coefficients are whole numbers. Multiply all the coefficients by an appropriate factor to convert any fractions to whole numbers. 

 Remove the fractions by multiplying through by 2:
 <font color="red">2 (</font> CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + <font color="red">9/2</font> O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red">3</font> CO<sub>2</sub> (g) +  <font color="red">4</font> H<sub>2</sub>O (l) <font color="red">)</font>

The balanced chemical equation is: 

<b> <font color="red">2 </font>CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + <font color="red">9</font> O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red">6</font> CO<sub>2</sub> (g) +  <font color="red">8</font> H<sub>2</sub>O (l)</b>

==Example - balance algebraically==

Steps 1 and 2: Same as above.

Step 3: Assign variables as coefficients for each substance in the chemical equation.
 Assign:
 "<font color="red"><i>a</i></font>" for the coefficient of CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH
 "<font color="red"><i>b</i></font>" for the coefficeient of O<sub>2</sub> 
 "<font color="red"><i>c</i></font>" for the coefficeient of CO<sub>2</sub>
 "<font color="red"><i>d</i></font>" for the coefficeient of H<sub>2</sub>O

 <font color="red"><i>a</i></font> CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + <font color="red"><i>b</i></font> O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red"><i>c</i></font> CO<sub>2</sub> (g) +  <font color="red"><i>d</i></font> H<sub>2</sub>O (l)

Step 4: Determine the number of each type of atoms on each side of the arrow.

{|width="500"
|
|<b>left-hand-side</b>
|<b>right-hand-side</b>
|-
|<b>number of carbon atoms</b>	
|3<font color="red"><i>a</i></font>
|<font color="red"><i>c</i></font>
|-
|<b>number of hydrogen atoms</b>		
|8<font color="red"><i>a</i></font>
|2<font color="red"><i>d</i></font>
|-
|<b>number of oxygen atoms</b>	
|<font color="red"><i>a</i></font> + 2<font color="red"><i>b</i></font>
|2<font color="red"><i>c</i></font> + <font color="red"><i>d</i></font>
|}

Step 5: Write out the algebraic equations and solve for the coefficients.

 equation (1)	3<font color="red"><i>a</i></font> =  <font color="red"><i>c</i></font>
 equation (2)	8<font color="red"><i>a</i></font> = 2 <font color="red"><i>d</i></font>
 equation (3)	<font color="red"><i>a</i></font> + 2 <font color="red"><i>b</i></font> = 2 <font color="red"><i>c</i></font> + <font color="red"><i>d</i></font>

 Let <font color="red"><i>a</i></font> = 1, and solve:
 equation (1) becomes 3 = <font color="red"><i>c</i></font>
 equation (2) becomes 8 = 2<font color="red"><i>d</i></font>, OR 4 = <font color="red"><i>d</i></font>
 equation (3) becomes 1 + 2<font color="red"><i>b</i></font> = 6 + 4, OR <font color="red"><i>b</i></font> = 9/2

Substitute back into the chemical equation, we get
CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + <font color="red">9/2</font> O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red">3</font> CO<sub>2</sub> (g) +  <font color="red">4</font> H<sub>2</sub>O (l)

Step 6:	Make sure that all the coefficients are whole numbers. Multiply all the coefficients by an appropriate factor to convert any fractions to whole numbers. 

 Remove the fractions by multiplying through by 2:
 <font color="red">2 (</font> CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + <font color="red">9/2</font> O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red">3</font> CO<sub>2</sub> (g) +  <font color="red">4</font> H<sub>2</sub>O (l) <font color="red">)</font>

The balanced chemical equation is: 

<b> <font color="red">2 </font>CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH(l) + <font color="red">9</font> O<sub>2</sub> (g) [[File:reactionarrow.gif]] <font color="red">6</font> CO<sub>2</sub> (g) +  <font color="red">8</font> H<sub>2</sub>O (l)</b>


==Content suitability==
BCIT courses:
<span class="pops">[http://www.bcit.ca/study/outlines/chem0011  CHEM 0011]</span>