Using Moles to Balance Equations

Higher tier only

Stoichiometry refers to the numbers in front of the reactants and products in an equation, which must be adjusted to make sure that the equation is balanced
These numbers are called coefficients (or multipliers) and if we know the masses of reactants and products, the balanced chemical equation for a given reaction can be found by determining the coefficients
First, convert the masses of each reactant and product in to moles by dividing by the molar masses using the periodic table
If the result yields uneven numbers, then multiply all of the numbers by the same number, to find the smallest whole number for the coefficient of each species
- For example, if the resulting numbers initially were 1, 2 and 2.5, then you would multiply all of the numbers by 2, to give the whole numbers 2, 4 and 5
Then, use the molar ratio to write out the balanced equation

Worked example

64 g of methanol, CH₃OH, reacts with 96 g of oxygen gas to produce 88 g of carbon dioxide and 72 g of water.

Deduce the balanced equation for the reaction.

(C = 12, H = 1, O = 16)

Answer:

Step 1: Calculate the molar masses of the substances in the equation

CH₃OH = 32 g / mol O₂ = 32 g / mol

CO2 = 44 g / mol H₂O = 18 g / mol

Step 2: Divide the masses present by the molar mass to obtain the number of moles

CH₃OH = 64 g ÷ 32 g mol^-1 = 2 mol

O₂ = 96 g ÷ 32 g mol^-1 = 3 mol

CO₂ = 88 g ÷ 44 g mol^-1 = 2 mol

H₂O = 72 g ÷ 18 gmol^-1 = 4 mol

Step 3: Use the mole ratios to balance the equation

2CH₃OH + 3O₂⟶ 2CO₂ + 4H₂O

Examiner Tip

The molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction.

For the reaction of elements forming compounds, it can be necessary to determine the formula of one of the products in order to balance the equation
This involves calculating the empirical formula
1. Write each element involved
2. Determine the mass of each element
3. Write the atomic mass of each element
4. Calculate the number of moles of each element (divide by the atomic mass)
5. Calculate the ratio of elements (divide by the smallest answer)
6. Write the final empirical formula
With the formula of the product, it is then possible to write a balanced symbol equation

Examiner Tip

The empirical formula is the simplest whole number ratio of all elements in a substance
If the value that you calculate is very close to a whole number, then you can round it off
- e.g. 1.003 ≈ 1
If the value that you calculate is not close to a whole number, then you might need to double or triple all of the values you have
- 0.3324 and 1.003 would become 0.3324 x 3 ≈ 1 and 1.003 x 3 ≈ 3

Worked example

10 g of hydrogen and 80 g of oxygen react to form one product.

Deduce the balanced equation for the reaction.

A_r(H) = 1 A_r(O) = 16

Answer:

Step 1: Calculate the chemical formula of the product:

	hydrogen	oxygen
Write the mass of each element	10 g	80 g
Calculate the number of moles (Divide each mass by the A_r)	$(\frac{10}{1})$ = 10	$(\frac{80}{16})$ = 5
Find the simplest whole number molar ratio (Divide by the smallest number)	$(\frac{10}{5})$ = 2	$(\frac{5}{5})$ = 1

- So, the empirical formula of the product = H₂O
Step 2: Write the unbalanced chemical equation:
- _H₂ + _O₂ → _H₂O
Step 3: Using the calculations performed, deduce the balanced equation:
- Simplest moles of hydrogen = 2
- Simplest moles of oxygen = 1
Step 4: Apply the moles to the unbalanced equation:
- 2H₂ + 1O₂ → _H₂O
- 2H₂ + O₂ → _H₂O
Step 5: Balance the remainder of the equation:
- 2H₂ + O₂ → 2H₂O

Examiner Tip

At GCSE level, it is unlikely that you will be given the information in percentages but if you are just treat them as though the percentage value is the mass

Using Moles to Balance Equations (AQA GCSE Chemistry: Combined Science)

Revision Note