C4.1 – The Mole Concept

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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).

2. 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.

 

3. 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 M_r of water is 18.

1 mole of water = 18g
No. of moles = mass of substance ÷ M_r
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.
  A_r 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 A_r 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 ÷ M_r
12 ÷ 48 = 0.25 mol

Mass of oxygen needed = Moles of magnesium x M_r
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. M_r 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 M_r
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)
  M_r 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
  A_r values:   Mg= 24,  Cl= 35.5,   H= 1

Mr of HCl = 35.5 + 1 = 36.1

No. of moles = Mass ÷ M_r
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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