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.

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, CH₃CH₂CH₂OH, burns in air, it reacts with oxygen, O₂, to form carbon dioxide, CO₂, and water, H₂O.

Write the balanced chemical reaction by the two methods:

1. by inspection
2. algebraically

Example - balance by inspection

Step 1: Identify the reactants and products.

Reactants: propyl alcohol, CH3CH2CH2OH, oxygen, O2
Products:  carbon dioxide, CO2, water, H2O

Step 2: Write the unbalanced chemical equation.

 CH3CH2CH2OH(l) + O2 (g)  CO2 (g) + H2O (l)

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

	No. of C	No. of H	No. of O
left-hand-side	3	8	3
right-hand-side	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, CH3CH2CH2OH.	

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, O2

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

1. carbon
2. hydrogen
3. oxygen

Step 5: Balance each type of atoms.

Balance carbon: CH3CH2CH2OH(l) + O2 (g)  3 CO2 (g) + H2O (l)
Balance hydrogen: CH3CH2CH2OH(l) + O2 (g)  3 CO2 (g) +  4 H2O (l)
Balance oxygen: CH3CH2CH2OH(l) + 9/2 O2 (g)  3 CO2 (g) +  4 H2O (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:
2 ( CH3CH2CH2OH(l) + 9/2 O2 (g)  3 CO2 (g) +  4 H2O (l) )

The balanced chemical equation is:

2 CH₃CH₂CH₂OH(l) + 9 O₂ (g) 6 CO₂ (g) + 8 H₂O (l)

Example - balance algebraically

Steps 1 and 2: Same as above.

Step 3: Assign variables as coefficients for each substance in the chemical equation.

Assign:
"a" for the coefficient of CH3CH2CH2OH
"b" for the coefficeient of O2 
"c" for the coefficeient of CO2
"d" for the coefficeient of H2O

a CH3CH2CH2OH(l) + b O2 (g)  c CO2 (g) +  d H2O (l)

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

	left-hand-side	right-hand-side
number of carbon atoms	3a	c
number of hydrogen atoms	8a	2d
number of oxygen atoms	a + 2b	2c + d

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

equation (1)	3a =  c
equation (2)	8a = 2 d
equation (3)	a + 2 b = 2 c + d

Let a = 1, and solve:
equation (1) becomes 3 = c
equation (2) becomes 8 = 2d, OR 4 = d
equation (3) becomes 1 + 2b = 6 + 4, OR b = 9/2

Substitute back into the chemical equation, we get CH₃CH₂CH₂OH(l) + 9/2 O₂ (g) 3 CO₂ (g) + 4 H₂O (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:
2 ( CH3CH2CH2OH(l) + 9/2 O2 (g)  3 CO2 (g) +  4 H2O (l) )

The balanced chemical equation is:

2 CH₃CH₂CH₂OH(l) + 9 O₂ (g) 6 CO₂ (g) + 8 H₂O (l)

Content suitability

BCIT courses: CHEM 0011