What is mole in stoichiometry?

Mole in stoichiometry is defined as the amount that contains as many elementary parties as there are atoms in 12 grams of carbon 12.

Most calculations and analysis in chemistry are done with the concept of mole in chemistry.

I will love to analyse mole using different basic illustrations.

A mole in stoichiometry can be likened to a bunch of keys (in which you find so many keys), a school of birds (in which you find so many birds) etc

In the same light, a mole of atoms (you find so many atoms inside)

From the mole concept we can relate to any other parameter

Mole and Mass relationship

1 mole of a substance = molar mass of the substance


What is the mass of 3.5 moles of calcium oxide?


I mole of substance = molar mass of substance

CaO= 40 + 16= 56g

1 mole of CaO= 56g

3.5 moles of CaO= ×

X= 3.5 X 56

X= 196 grams of calcium oxide

Mole and volume relationship

1 mole of a substance =molar volume (22.4 dm3)


What volume is equivalent to 4.5 moles of carbon iv oxide gas?


1 mole of CO2= 22.4 dm3

4.5 mole of CO2= X

X= 4.5 X 22.4 = 100.8 dm3 of Carbon iv oxide

Mole and Number of particles (Avogadro’s number) relationship

I mole of substance = Avogadro’s number (6.02 X 1023 particles)


What’s the number of particles in 0.5 moles of sodium?


1 mole of sodium = Avogadro’s number (6.02 × 10 23)

0.5 mole of sodium = X

X= 3.01 x 1023 particles

What is stoichiometry?

Stoichiometry is the relationships and calculations based on balanced equations involving comparing reactants and products.

Mole -Mole Stoichiometry calculations

I’ll talk a little about the whole concept of the mole- mole calculations which is simply comparing the moles of the substances involved in an equation.

It is important to know that you can easily calculate mole per mole calculations using simple analysis.

Example 1

What is the mole of Zinc produced by reacting 3.3 moles of magnesium?

Mg + ZnCl2===MgCl2 + Zn


1 mole of Mg === 1 mole of zinc

3.3 moles of Mg=== x

X = 3.3 x 1 = 3.3 moles of zinc

Example 2

How many moles of ammonia can be produced from 3.3 moles of Nitrogen?

N2 + 3H2====2NH3


1 mole of Nitrogen (N2) ====== 2 moles of Ammonia (NH3)

3.3 moles of Nitrogen (N2) ===== x

X = 3.3 x 2 = 6.6 moles of ammonia is evolved

Example 3

What’s the mole of methane that will produce 1.2 moles of carbon iv oxide according to this equation CH4 + 2O2 ====CO2 + 2H2O?


1 mole of Methane (CH4) ==== 1 mole of carbon iv oxide (CO2)

X moles of methane ==== 1.2 mole of carbon iv oxide (CO2)

X= 1 mole x 1.2 mole /1mole =1.2 mole of methane

Mole to Mass Stoichiometry calculations

Mole to mass calculation is another area in the stoichiometry calculations that involve comparing and converting the mole gotten to mass.

Example 4

What is the mass of magnesium chloride produced when 2.5 moles of magnesium reacts with iron II chloride according to this equation Mg + ZnCl2===MgCl2 + Zn?


Mg + ZnCl2===MgCl2 + Zn

So we will start first by calculating the number of moles

So, I mole of Magnesium (Mg) ===== 1 mole of Magnesium Chloride (MgCl2)

      2.5 moles of Magnesium (Mg) ===== x

      X = 2.5 x 1/1 = 2.5 moles of Magnesium Chloride (MgCl2)

Then we convert the mole to mass

Since I mole of MgCl2 ===== 95g of MgCl2

                (24 +71=95g)

Therefore 2.5 mole of MgCl2==== x

X= 2.5 x 95=237.5g of Magnesium Chloride

Example 5

What mass of carbon monoxide is produced by the incomplete combustion of 2.2 moles of carbon according to this equation 2C + O2====2CO?


2C + O2====2CO

So we will start first by calculating the number of moles

So, 2 moles of Carbon (C) ===== 2 moles of Carbon monoxide (CO)

      2.2 moles of Carbon(C)===== x

      X = 2.2 x 2/2 = 2.2 moles of Carbon monoxide (CO)

Then we convert the mole to mass

Since I mole of CO===== 28g of CO

                (12 +16=28g)

Therefore 2.2 mole of CO==== x

X= 2.2 x 28=61.6g Carbon monoxide

In conclusion, mole in stoichiometry calculation is embedded in the principle of mole concept analysis. To fully solve the stoichiometry calculations easily then we need to fully understand the comparisons and analysis.

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