The Mole Concept/Moles/Mass/M_r

This section is newly developed to provide you with a quick last minute pre-boards revision on the  Stoichiometry  leccon-The mole concelt.It  will reinforce the key concepts, so that you do not overlook even the smallest of concept in your board preparation:It also guides you into using the correct vocabulary required for answering your board exam papers.

1. What is a mole?         [5 different ways of defing it as per the CAIE}

One mole is the number of particles which is equal to the number of atoms in 12g of carbon 12

[or]

One mole of a substance is that substance that has a mass equal to its relative formula mass/relative atomic mass/ relative formula mass in grams

[or]

One mole is equal to 6.023 x 10²³ atoms /ions/molecules/particles or electrons.

[or]

A mole is the amount of the substance that has a volume equal to 24dm³ of a gas at r.t.p

[or]

One mole is tha mass in grams that has the avogadros number of particles.

 

Example

• 1 Mole of H will contain  6.023 x 10²³ atoms of hydrogen
• 1 mole of H₂ will contains( 6.023 x 10²³) x 2 atoms of H { because H₂ is diatomic) 
• 1 mole of  H₂ will conatin 6.023 x 10²³ moleules of hydrogen.

So if you have oberved carfully the fine difference that exists in the answers of the second and the thirs bullet point above. The second bullet point talks of  H  which is a single atom. While the third bullet point talks of H₂ which  consists of 2 H atoms. So be careful while attempting such questions.

How do you know which formula needs to be use  for calculating the mole?

The  answer is rather simple! It all depends on what information has been provided to you. the following examples will make the meaning more clearer.

• Question No:Q4c -May-June 2019-paper-41

Hydrochloric acid produces salts called chlorides. Magnesium carbonate reacts with hydrochloric acid to produce magnesium chloride.

MgCO₃ + 2HCl -->MgCl ₂ + H₂O + CO₂

A student used 50.00cm³ of 2.00mol/dm³ hydrochloric acid in an experiment to produce magnesium chloride.

Calculate the mass, in g, of magnesium carbonate needed to react exactly with 50.00cm³ of 2.00mol/dm³ hydrochloric acid using the following steps.

Calculate the number of moles of HCl present in 50.00cm3 of 2.00mol/dm³ HCl.

Determine the number of moles of MgCO₃ which would react with 50.00cm³ of 2.00mol/dm³ HCl.

Solution:

From the stocihiometry of the equation:

Every 1 mole of MgCO₃ reacts with every 2 moles of HCl

• We have the volume of hydrochloric acid (v)= 50.0cm³

• We also have the concentration of hydrochloric acid (C)= 2.00 mol/dm³

• Hence we will use the formula: Number of moles of HCl= concentration x volume to find the moles of HCl.
• hence number of moles of HCl= 2 x (50 /1000)=0.1 moles. Now why have we divided by 1000? Well you must always remember to keep the concentration in mol/dm³, volume in dm³ when trying to use  this formula.To convert cm³ to dm ³, divide by 1000 and vice verse.

 

Moles of MgCO₃ = half the number of moles of HCl= 0.5 x 0.1=0.05mol

Ans:. 0.05   mol

So here as you saw, the moles could be calculated because we took into account the original equation[ Reaction equation]

Calculate the relative formula mass, Mr , of MgCO₃.

Solution:

• Relative formula mass=M_r= Sum of all the relative atomic massess
• M_r MgCO₃= A_r (Mg) + A_r(C) + 3A_r(O) = 24+ 12+ 3(16)= 24+12+48=84

Mr of MgCO₃ = 84

 Calculate the mass of MgCO₃ needed to react exactly with 50.00cm^₃ of 2.00mol/dm³ HCl.

Solution:

• Mass of MgCO₃= Moles  of MgCO₃x M_r of MgCO₃= 0.05 x 84= 4.2g

mass = 4.2 g    [4]

 

So in a nutshell,

• you need to rememebr the formula at your finger tips.
• Analyse the given data
• Apply the correct fomula based on what information is given to you
• Use the appropriate units

Thats all in this mudule on  "  The Mole Concept/Moles/Mass/M_r"  Comming up soon, with yet another Revion Summary Topic. 

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