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  1. Define the mole in terms of a specific number of particles called Avogadro’s constant.

6 x 1023  is known as the Avogardo constant. The amount of substance with the Avogrado number of particles is called the mole. So a mole of any substance will contain 6 x 1023  particles (atoms/ions/molecules).

  1. Use the molar gas volume, taken as 24dm3 at room temperature and pressure.

At room temperature and pressure, the volume of a mole of any gas is 24dm3 or 24000cm3. This is called the molar gas volume.


  1. Calculate stoichiometric reacting masses and reacting volumes of solutions; solution concentrations will be expressed in mol / dm3 .  (Calculations involving the idea of limiting reactants may be set.)

You can find the number of moles of atoms, molecules or ions using the following formula:

No. of moles =  mass of substance taken  ÷ mass of one mole of the substance

You can use this triangle diagram on the left to help you calculate each of the variable. Just cover the variable you want to calculate with your hand, to get the formula



Example problems:

  • How many moles of water are there in 4.5 grams of water? The Mr of water is 18.

1 mole of water = 18g
No. of moles = mass of substance ÷ Mr
So 4.5 grams of water is 4.5 ÷ 18 = 0.25 mol (the abbbreviation for moles is mol)

  • Calculate the mass of oxygen needed to react with 12g of magnesium.
    Ar values: Mg=24, O=16.

Write down the balanced equation:  2Mg +O2 —————-> 2MgO

Calculate their masses by multiplying the no.of atoms with their respective Ar vlaues
2 x 24= 48g of Mg   &   2 x 16= 32g of O2

So now we know that 48g of magnesium reacts with 32g of oxygen.

Moles in 12g of magnesium= Mass ÷ Mr
12 ÷ 48 = 0.25 mol

Mass of oxygen needed = Moles of magnesium x Mr
0.25 x 32 = 8g of Oxygen
To find the volume of gas produced in a reaction using the formula:

Volume of gas (dm3) = number of moles of gas x 24.

You can use the triangle diagram on the left to help you calculate the different variables.


Example problems:

  • Calculate the mass of carbon dioxide present in 60 cm3 of carbon dioxide. Mr of CO2 = 44.

convert 60 cm3 to 0.06 dm3
moles of CO2 = volume of gas ÷ 24
0.06 dm3 ÷ 24 = 0.0025 moles

Mass of CO2 = moles x Mr
0.0025 x 44 = 0.11g of carbon dioxide

  • Calculate the volume of carbon dioxide which is produced when 2.8g of butene burns in excess air. The formula given is C4H8(g) + 6O2(g) ———-> 4CO2(g) + 4H2O(l)
    Mr of butene = 56

moles of butene = 2.8 ÷ 56 = 0.05 moles

Only butene’s Mr is given, so you cannot calculate the masses of the butene and oxygen that produces the carbon-dioxide. But looking at the equation you can see that 1 mole of butene produces 4 moles of CO2.    C4H8(g)  ——> 4CO2(g)

So, 0.05 moles of butene will produce 0.05 x 4 = 0.2 moles of CO2.

Volume of gas (dm3) = number of moles of gas x 24.

0.2 moles of CO2 will ocuppy 0.2 x 24dm3 = 4.8dm3


When a reaction is carried out, sometimes an excess of one reactant may be used. The reactant that is not in excess is called a limiting reagent. The reaction will stop when the limiting reagent is used up. The reactant with the lower no. of moles is the limiting reactant in any reaction.

Example problem:

  • 1.2g of magnesium is reacted with a solution containing 2.74g of hydrochloric acid. Which is the limiting reactant?
    Mg + 2HCl ————-> MgCl2 + H2
    Ar values:   Mg= 24,  Cl= 35.5,   H= 1

Mr of HCl = 35.5 + 1 = 36.1

No. of moles = Mass ÷ Mr
number of moles of magnesium = 1.2 ÷ 24 = 0.05 mol
No. of moles of hydrochloric acid = 2.7 ÷ 36.5 = 0.075 mol

From the equation, we understand that 1 mol of magnesium reacts with 2 mol of HCl. So to react completely, 0.05 mol of magnesium will need to react with 2 x 0.05 = 0.1 mol of HCl.

But we only have 0.75 mol of HCl. So HCl is the limiting reactant.

(Try doing the question on your own. This is a very confusing topic). Using this type of calculation you can also find out by how much one reactant is in excess.

In this question, magnesium is the excess reactant. From the equation we understand that 2 mol of HCl reacts with one mol of magnesium. So the 0.075 mol of HCl acid will have reacted with 0.075 ÷ 2 = 0.0375 mol of magnesium. After this the reaction stops because all of the HCl (limiting reagent) has been used up. The amount of magnesium that is in excess (not used up) is 0.05 (moles at the start of the reaction) – 0.0375 (moles used up) = 0.0125 mol.



Notes submitted by Lintha

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7 thoughts on “C4.1 – The Mole Concept

  1. on the first example problem, finding the mass of CO2 present in 60cm3 of CO2, i got stuck in the last part. This might be a stupid question but why is it 0.025 x 44 and not x 22 as the MR of CO2 is 22?


      1. Hi Madalena,
        Thank you so much for your question!!!
        It turns out that due to a typo, our team member wrote 22 instead of 44. The Mr of CO2 is actually 44 (Because Ar of C is 12 and Ar of O is 16, and 12 + 16 + 16 = 44). I’ll edit that right now.


  2. 24 dm cube is equal to 24000 cm cube

    “At room temperature and pressure, the volume of a mole of any gas is 24dm3 or 2400cm3.”


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