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Using Moles to Balance Equations (AQA GCSE Chemistry)

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Stewart Hird

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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 Tips and Tricks

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 Tips and Tricks

  • 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.

Ar(H) = 1   Ar(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 Ar)

begin mathsize 14px style open parentheses 10 over 1 close parentheses end style = 10

begin mathsize 14px style open parentheses 80 over 16 close parentheses end style = 5

Find the simplest whole number molar ratio

(Divide by the smallest number)

begin mathsize 14px style open parentheses 10 over 5 close parentheses end style = 2 

begin mathsize 14px style open parentheses 5 over 5 close parentheses end style = 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 Tips and Tricks

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

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    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.